American Structural Design Calculation

STRUCTURAL DESIGN SOFTWARE

This web site provides structural design spreadsheets, which created using Microsoft Excel. Each spreadsheet contains both the formulas used and the reference code sections, as well as graphic drawings. The Example is intended for reuse and is loaded with floating comments as well as ActiveX pull-down menus for variable choices. All intermediate calculations are intended for submittal with the calculations to explain the results of the input. It is free to download, by click spreadsheet name, for non-business. For a package professional version of entire listed spreadsheets, Special Buy, the current price is less than $13 per spreadsheet. And single copy of each professional spreadsheet is available with $33. Our licenses are "on your honor". We trust our customers to decide if they need to purchase more licenses for multi-user.

Special Buy (User's Book) A Package Only $1260 (Total 104 software listed)

Concrete DesignSMRF-ACI Column

Circular Column

Column Supporting Discontinuous

Corbel

Composite Member

Development & Splice in Concrete

Friction

Prestressed Member

Mechanical Unit Anchorage (Concrete & Wood) PT Concrete Floor

Shear Wall - IBC Shear Wall - CBC Tilt-up Panel Wall Pier Slab Anchorage To Concrete Beam Deep Beam Punching Coupling Beam

Masonry DesignMasonry Shear Wall (ACI 530)

Bending Post at Top Wall

Anchorage To Masonry (ACI 530 & UBC) Horizontal Bending Wall

Development & Splice in Masonry

Masonry Shear Wall (UBC 97)

Masonry Bearing Wall (ACI 530 & UBC) Girder at Wall

Foundation Design Pad Footing Flagpole

Deep Footing Footing for DSA & OSHPD Boundary Spring Generator Plain Concrete Footing Conventional Slab on Grade PT Slab on Ground Concrete Pile Footing At Piping

Eccentric Footing Wall Footing Combined Footing Grade Beam Concrete Retaining Wall Retaining Wall for DSA & OSHPD Masonry Retaining /Fence Wall Restrained Retaining Wall

Page 1 of 3

11/24/2005http://www.engineering-international.com/

Technical Support You will receive, by email attachment, your purchased spreadsheets within 48 hours. For package purchaser, please provide an email box with 40 MB space available. The purchaser’s name will be put on left top of each spreadsheet to replace the existing logo. Macro is required when opening the spreadsheets to get full functions. We will constantly enhance the

Masonry Beam (ACI 530 & UBC)

Elevator Wall for DSA & OSHPD

Masonry Column (ACI 530 & UBC)

Wood DesignWood Joist Top Plate Connection

Wood Bolt Connection

Shear Wall with Openings

Toe Nail

Diaphragm-Ledger-CMU

Drag Forces

Double Joist Wood Beam Wood Column

Wood Shear Wall

Wood Diaphragm

Sub-Diaphragm

Steel DesignMetal Studs (Joist, Beam, Wall, Column) Web-Tapered Girder

Composite Floor Beam

Base Plate

Beam Connection

Brace Connection

WF-Opening

Bolts Connections

Weld Connections

Roof Deck

Floor Deck

Steel Joists

Joist Girder

Steel Stair

OCBF-IBC OCBF-UBC SCBF-IBC SCBF-UBC EBF-IBC EBF-UBC OMRF-IBC OMRF-UBC SMRF-IBC SMRF-UBC Beam Gravity Beam with Torsion Plate Girder

Lateral Analysis & Design LoadsWind ASCE 7-98 (for IBC 2000) Wall Lateral Force-UBC

Guardrail

Sign

Snow (ASCE 7-98, ASCE 7-02, & UBC 97)

Live Load Flexible Diaphragm Opening Lateral Frame Formulas

Wind ASCE 7-02 (for IBC 2003)

Wind-UBC

Seismic-IBC 2000

Seismic-IBC 2003

Seismic-UBC

Wall Lateral Force-IBC (IBC 2000 & IBC 2003)

Page 2 of 3


programs. The licensed owners will receive updated spreadsheets if requested, or new spreadsheets for the package owners only. When you email us your questions, please tell us your name, purchased license number, phone number, and give a decent problem description.

Disclaimer We do not provide unprotected spreadsheet, original software code. DO NOT UNPROTECT the spreadsheets, using "brute force" methods or VBA procedure in particular, otherwise some steps and database will be inadequate at random times. We intend that the analysis contained in the spreadsheets is accurate and reliable, but it is entirely the responsibility of the program user to verify the accuracy and applicability of any results obtained from the spreadsheets. Daniel T. Li Engineering International’s entire liability shall be limited to the purchase price of the spreadsheets.

Copyright © 2002-2005 Daniel T. Li Engineering International, All Rights Reserved.

Page 3 of 3


PROJECT : PAGE : CLIENT : DESIGN BY :

JOB NO. : DATE : REVIEW BY :

INPUT DATA DESIGN SUMMARYCOLUMN WIDTH c1 = 5 in FOOTING WIDTH B = 3.00 ftCOLUMN DEPTH c2 = 5 in FOOTING LENGTH L = 4.00 ftBASE PLATE WIDTH b1 = 16 in FOOTING THICKNESS T = 12 in

BASE PLATE DEPTH b2 = 16 in LONGITUDINAL REINF. 3 # 5 @ 15 in o.c.

FOOTING CONCRETE STRENGTH fc' = 2.5 ksi TRANSVERSE REINF. 4 # 5 @ 14 in o.c.

REBAR YIELD STRESS fy = 60 ksi

AXIAL DEAD LOAD PDL = 25 k

AXIAL LIVE LOAD PLL = 4.5 k

LATERAL LOAD (0=WIND, 1=SEISMIC) = 1 Seismic,SDSEISMIC AXIAL LOAD PLAT = -6 k, SD

SURCHARGE qs = 0 ksf

SOIL WEIGHT ws = 0.11 kcf

FOOTING EMBEDMENT DEPTH Df = 2 ft

FOOTING THICKNESS T = 12 in

ALLOW SOIL PRESSURE Qa = 2.5 ksf

FOOTING WIDTH B = 3 ftFOOTING LENGTH L = 4 ftBOTTOM REINFORCING # 5

THE PAD DESIGN IS ADEQUATE.

ANALYSISDESIGN LOADS (IBC SEC.1605.3.2 & ACI 318-02 SEC.9.2.1)CASE 1: DL + LL P = 30 kips 1.2 DL + 1.6 LL Pu = 37 kipsCASE 2: DL + LL + E / 1.4 P = 25 kips 1.2 DL + 1.0 LL + 1.0 E Pu = 29 kipsCASE 3: 0.9 DL + E / 1.4 P = 18 kips 0.9 DL + 1.0 E Pu = 17 kips

CHECK SOIL BEARING CAPACITY (ACI 318-02 SEC.15.2.2)CASE 1 CASE 2 CASE 3

2.50 ksf, 2.14 ksf, 1.56 ksf

q MAX < k Q a , [Satisfactory]

where k = 1 for grovity loads, 4/3 for lateral loads.

DESIGN FOR FLEXURE (ACI 318-02 SEC.15.4.2, 10.2, 10.3.5, 10.5.4, 7.12.2, 12.2, & 12.5)

2.5

LONGITUDINAL TRANSVERSEd 8.69 8.38b 36 48

q u,max 3.10 3.10Mu 11.35 7.00

ρ 0.001 0.000ρmin 0.001 0.001As 0.39 0.25

ReqD 2 # 5 1 # 5Max. Spacing 18 in o.c. 18 in o.c.

USE 3 # 5 @ 15 in o.c. 4 # 5 @ 14 in o.c.ρmax 0.013 0.013

Check ρprod < ρmax [Satisfactory] [Satisfactory]

Pad Footing Design Based on ACI 318-02

DanielTian Li

(0.15 )SMAX S

PTq q w

BL= + + − =

''2

0.85 1 10.383

M uf c b fd cf y

ρ

− − =

'10.85

MAX

f c uf u ty

β ερε ε

=+

40.0018 ,

3MIN

TMIN

dρρ =

(cont'd)CHECK FLEXURE SHEAR (ACI 318-02 SEC.9.3.2.3, 15.5.2, 11.1.3.1, & 11.3)

LONGITUDINAL TRANSVERSEVu 7.80 4.52

φ 0.75 0.75φVn 23.5 30.2

Check Vu < φVn [Satisfactory] [Satisfactory]

CHECK PUNCHING SHEAR (ACI 318-02 SEC.15.5.2, 11.12.1.2, 11.12.6, & 13.5.3.2)

= 97.42 kips

where φ = 0.75 (ACI 318-02, Section 9.3.2.3 )βc = ratio of long side to short side of concentrated load = 1.00

b0 = c1 + c2 + b1 + b2 + 4d = 76.1 in

Ap = b0 d = 649.4 in2

y = MIN(2 , 4 / βc , 40 d / b0) = 2.0

29.40 ft-kips < φ V n [Satisfactory]

( ) '2 y fV A pn cφ φ= +

1 1 2 211, max 2 2

b c b cd dV Puu BL + + = − + + =

'2 bd fV n cφ φ=




BASE PLATE DEPTH b2 = 16 in LONGITUDINAL REINF., TOP 1 # 5

FOOTING CONCRETE STRENGTH fc' = 2.5 ksi LONGITUDINAL REINF., BOT. 23 # 5 @ 8 in o.c.

REBAR YIELD STRESS fy = 60 ksi TRANSVERSE REINF., BOT. 6 # 5 @ 15 in o.c.



LATERAL LOAD (0=WIND, 1=SEISMIC) = 1 Seismic,SDSEISMIC AXIAL LOAD PLAT = 1 k, SD

SEISMIC MOMENT LOAD MLAT = 15 ft-k, SD

SEISMIC SHEAR LOAD VLAT = 10 k, SD

SURCHARGE qs = 0.1 ksf




ALLOW SOIL PRESSURE Qa = 3 ksf

FOOTING WIDTH B1 = 10 ft

B2 = 6 ft

FOOTING LENGTH L1 = 6 ft

L2 = 1 ft

REINFORCING SIZE # 5

THE FOOTING DESIGN IS ADEQUATE.

ANALYSISDESIGN LOADS (IBC SEC.1605.3.2 & ACI 318-02 SEC.9.2.1)CASE 1: DL + LL P = 55 kips 1.2 DL + 1.6 LL Pu = 67 kips

M = 136 ft-kips Mu = 168 ft-kipse = 2.5 ft, fr cl ftg eu = 2.5 ft, fr cl ftg

CASE 2: DL + LL + E / 1.4 P = 55 kips 1.2 DL + 1.0 LL + 1.0 E Pu = 66 kipsM = 149 ft-kips Mu = 179 ft-kipse = 2.7 ft, fr cl ftg eu = 2.7 ft, fr cl ftg

CASE 3: 0.9 DL + E / 1.4 P = 46 kips 0.9 DL + 1.0 E Pu = 46 kipsM = 125 ft-kips Mu = 130 ft-kipse = 2.7 ft, fr cl ftg eu = 2.8 ft, fr cl ftg

CHECK OVERTURNING FACTOR

MR / MO = 4.5 > F = 1 / (0.9x1.4) [Satisfactory]

Where MO = MLAT + VLAT Df - PLATL2 = 34 k-ft

Pftg = (0.15 kcf) T B L = 16.80 k, footing weight

Psoil = ws (Df - T) B L = 12.32 k, soil weight

MR = PDLL2 + 0.5 (Pftg + Psoil) L = 152 k-ft

F = 1 / (0.9x1.4) for seismic, IBC 1605.3.2

FOR REVERSED LATERAL LOADS,

MR / MO = 13.9 > F = 1 / (0.9x1.4) [Satisfactory]

Where MO = MLAT + VLAT Df - PLATL1 = 29 k-ft

MR = PDLL1 + 0.5 (Pftg + Psoil) L = 402 k-ft

Daniel

Eccentric Footing Design Based on ACI 318-02

Tian Li

(cont'd)CHECK SOIL BEARING CAPACITY (ACI 318-02 SEC.15.2.2)

Service Loads CASE 1 CASE 2 CASE 3P 54.5 55.2 45.7 ke 2.5 2.7 2.7 ft (from center of footing)

qs B L 11.2 11.2 0.0 k, (surcharge load)

(0.15-ws)T B L 4.5 4.5 4.0 k, (footing increased)Σ P 70.2 70.9 49.7 k

eL 1.9 > L/6 2.1 > L/6 2.5 > L/6 fteB 1.6 < B/6 1.6 < B/6 1.8 < B/6 ftqL 30.0 33.7 33.6 k / ft

qmax 3.0 3.3 3.5 ksf

qallow 3.0 4.0 4.0 ksf

Where

[Satisfactory]

DESIGN FLEXURE & CHECK FLEXURE SHEAR (ACI 318-02 SEC.15.4.2, 10.2, 10.3.5, 10.5.4, 7.12.2, 12.2, 12.5, 15.5.2, 11.1.3.1, & 11.3)

16

FACTORED SOIL PRESSURE Factored Loads CASE 1 CASE 2 CASE 3

Pu 67.2 65.5 46.0 k

eu 2.5 2.7 2.8 ft

γ qs B L 17.9 11.2 0.0 k, (factored surcharge load)

γ[0.15T + ws(Df - T)]BL 34.9 34.9 26.2 k, (factored footing & backfill loads)

Σ Pu 120.1 111.6 72.2 k

eu 1.4 > L/6 1.6 > L/6 1.8 > L/6 ft

qu, max 2.381 2.450 1.770 ksf

FOOTING MOMENT & SHEAR AT LONGITUDINAL SECTIONS FOR CASE 10 0.25 L1 0.50 L1 0.75 L1 ColL ColR 0.25 L2 0.50 L2 0.75 L2 L

0 1.50 3.00 4.50 5.56 6.44 6.25 6.50 6.75 7.00

0 0 0 0 0 -29.4 -16.8 -33.6 -50.4 -67.2

0 0.0 0.0 0.0 0.0 67.2 67.2 67.2 67.2 67.2

2.56 2.56 2.56 2.56 2.56 2.56 2.56 2.56 2.56 2.56

0 -2.9 -11.5 -25.9 -39.6 -53.0 -50.0 -54.1 -58.3 -62.7

0 3.8 7.7 11.5 14.2 16.5 16.0 16.6 17.3 17.9

4.99 4.99 4.99 4.99 4.99 4.99 4.99 4.99 4.99 4.99

0 -5.6 -22.5 -50.5 -77.2 -103.4 -97.5 -105.5 -113.7 -122.3

0 7.5 15.0 22.5 27.8 32.1 31.2 32.4 33.7 34.9

0.00 0.51 1.02 1.53 1.89 2.19 2.13 2.21 2.30 2.38

0 189.5 288.9 316.3 302.3 275.2 282.0 272.8 262.9 252.2

0 -48.2 -84.1 -107.8 -117.2 -120.3 -120.0 -120.3 -120.4 -120.1

0 181.1 254.9 239.8 185.5 89.3 117.7 79.7 40.5 0

0 -36.9 -61.5 -73.8 -75.2 -4.5 -5.6 -4.0 -2.2 0

Pu,ftg & fill (klf)

Vu,surch (k)

Section

Xu (ft, dist. from left of footing)

Mu,col (ft-k)

Vu,col (k)

Pu,surch (klf)

Mu,surch (ft-k)

Vu,soil (k)

Σ Σ Σ Σ Mu (ft-k)

Σ Σ Σ Σ Vu (kips)

Mu,ftg & fill (ft-k)

Vu,ftg & fill (k)

qu,soil (ksf)

Mu,soil (ft-k)

''2

0.85 1 10.383

M uf c b fd cf y

ρ

− − =

40.0018 ,

3MIN

TMIN

dρρ =

( )

( )

61

,6

2,

3(0.5 ) 6

L

L

L

LL

ePLL for e

LqP L

for eL e

Σ + ≤=

Σ>

−

61

,6

2,

3(0.5 ) 6

BL

BMAX

LB

B

eqBB for e

q Bq B

for eB e

+ ≤=

>

−

( )

( )

61

,6,

2,

3 (0.5 ) 6

MAX

euPu LLfor euBLqu

LPu for euB L eu

Σ + ≤=

Σ >

−

'10.85

MAX

f c uf u ty

β ερε ε

=+

(cont'd)FOOTING MOMENT & SHEAR AT LONGITUDINAL SECTIONS FOR CASE 2

0 0.25 L1 0.50 L1 0.75 L1 ColL ColR 0.25 L2 0.50 L2 0.75 L2 L

0 1.50 3.00 4.50 5.56 6.44 6.25 6.50 6.75 7.00

0 0 0 0 0 -13.7 -1.4 -17.8 -34.1 -50.5

0 0.0 0.0 0.0 0.0 65.5 65.5 65.5 65.5 65.5

1.60 1.60 1.60 1.60 1.60 1.60 1.60 1.60 1.60 1.60

0 -1.8 -7.2 -16.2 -24.8 -33.2 -31.3 -33.8 -36.5 -39.2

0 2.4 4.8 7.2 8.9 10.3 10.0 10.4 10.8 11.2

4.99 4.99 4.99 4.99 4.99 4.99 4.99 4.99 4.99 4.99

0 -5.6 -22.5 -50.5 -77.2 -103.4 -97.5 -105.5 -113.7 -122.3

0 7.5 15.0 22.5 27.8 32.1 31.2 32.4 33.7 34.9

0.00 0.52 1.05 1.57 1.95 2.25 2.19 2.27 2.36 2.45

0 175.5 263.2 282.2 264.2 235.4 242.4 232.9 222.8 212.0

0 -47.0 -81.4 -103.3 -111.1 -112.8 -112.8 -112.8 -112.4 -111.6

0 168.1 233.6 215.4 162.2 85.1 112.3 75.9 38.5 0

0 -37.1 -61.7 -73.6 -74.4 -4.9 -6.1 -4.4 -2.4 0

FOOTING MOMENT & SHEAR AT LONGITUDINAL SECTIONS FOR CASE 30 0.25 L1 0.50 L1 0.75 L1 ColL ColR 0.25 L2 0.50 L2 0.75 L2 L

0 1.50 3.00 4.50 5.56 6.44 6.25 6.50 6.75 7.00

0 0 0 0 0 -5.1 3.5 -8.0 -19.5 -31.0

0 0.0 0.0 0.0 0.0 46.0 46.0 46.0 46.0 46.0

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

3.74 3.74 3.74 3.74 3.74 3.74 3.74 3.74 3.74 3.74

0 -4.2 -16.8 -37.9 -57.9 -77.6 -73.1 -79.1 -85.3 -91.7

0 5.6 11.2 16.8 20.8 24.1 23.4 24.3 25.3 26.2

0.00 0.00 0.76 1.14 1.41 1.63 1.58 1.64 1.71 1.77

0 0.0 167.5 175.2 160.3 139.1 144.2 137.4 130.2 122.7

0 0.0 -55.2 -69.2 -73.6 -73.7 -74.0 -73.6 -73.0 -72.2

0 -4.2 150.6 137.3 102.3 56.4 74.6 50.3 25.4 0

0 5.6 -44.0 -52.3 -52.7 -3.6 -4.6 -3.3 -1.8 0

DESIGN FLEXURE

Location Mu,max d (in) ρmin ρreqD ρmax smax use ρprovD

Top Longitudinal -4.2 ft-k 9.69 0.0001 0.0001 0.0129 no limit 1 # 5 0.0002Bottom Longitudinal 254.9 ft-k 8.69 0.0025 0.0041 0.0129 18 23 # 5 @ 8 in o.c. 0.0043Bottom Transverse 1 ft-k / ft 8.38 0.0004 0.0003 0.0129 18 6 # 5 @ 15 in o.c. 0.0026

[Satisfactory]CHECK FLEXURE SHEAR

Direction Vu,max φVc = 2 φ b d (fc')0.5 check Vu < φ Vc

Longitudinal 75.2 k 125 k [Satisfactory]Transverse 4.3 k / ft 8 k / ft [Satisfactory]


Case Pu Mu b1 b2 b0 γv βc y Af Ap R J vu (psi) φ vc1 67.2 168.0 18.9 18.9 0.5 0.4 1.0 2.0 112.0 4.4 1.5 1.9 105.3 150.02 65.5 178.8 18.9 18.9 0.5 0.4 1.0 2.0 112.0 4.4 1.4 1.9 102.7 150.03 46.0 130.0 18.9 18.9 0.5 0.4 1.0 2.0 112.0 4.4 1.0 1.9 72.2 150.0

[Satisfactory]where φ = 0.75 (ACI 318-02, Section 9.3.2.3 )

Section


Mu,col (ft-k)

Vu,col (k)

Pu,surch (klf)

Mu,surch (ft-k)

Vu,surch (k)

Pu,ftg & fill (klf)




Vu,ftg & fill (k)

qu,soil (ksf)

Mu,soil (ft-k)

Vu,soil (k)

Section


Mu,col (ft-k)

Vu,col (k)

Pu,surch (klf)

Mu,surch (ft-k)

Vu,surch (k)

Pu,ftg & fill (klf)

Vu,soil (k)




Vu,ftg & fill (k)

qu,soil (ksf)

Mu,soil (ft-k)

0.5 1( )

231 21 3

6 1 1

1 2

R bMP uu vpsivu JAP

db d bJb b

b bPuRA f

γ−= +

= + +

=

( )2 1 21

12 113 2

db bAP

vbb

BLA f

γ

= +

= −+

=

( )

( ) ( )

'( ) 2

42, , 40

0

, 0.5 0.5 , 0.5 0.50 1 1 1 2 2 2

psi y fvc c

dy MIN

bc

AP d db b c b b c bd

φφ

β

= +

=

= = + + = + +



INPUT DATAWALL LENGTH Lw = 30.5 ft

WALL HIGHT h = 42 ft

WALL THICKNESS t = 12 in

FOOTING LENGTH L = 38.5 ft

L1 = 4 ft

FOOTING WIDTH B = 10 ft

FOOTING THICHNESS T = 24 in

FOOTING EMBEDMENT DEPTH D = 3 ft

ALLOWABLE SOIL PRESSURE qa = 2.5 ksf

DEAD LOAD AT TOP WALL Pr,DL = 54.115 kips

LIVE LOAD AT TOP WALL Pr,LL = 54.115 kips

TOP LOAD LOCATION a = 2 ft

WALL SELF WEIGHT Pw = 26.04 kips

LATERAL LOAD TYPE (0=wind,1=seismic) 1 seismic

SEISMIC LOAD (E/1.4 , ASD) F = 43.21 kips THE FOOTING DESIGN IS ADEQUATE.

CONCRETE STRENGTH fc' = 3 ksi


TOP BARS, LONGITUDINAL 3 # 5

BOTTOM BARS, LONGITUDINAL 11 # 10

BOTTOM BARS, TRANSVERSE # 6 @ 12 in o.c.

ANALYSISCHECK OVERTURNING FACTOR

F = MR / MO = 1.53 > 1.0 / 0.9 for seismic [Satisfactory]

Where Pf = 111.65 kips (footing self weight)

MO = F (h + D) = 1944 ft-kips (overturning moment)

MR = (Pr,DL) (L1 + a) + Pf (0.5 L) + Pw (L1 + 0.5Lw) = 2975 ft-kips (resisting moment without live load)

CHECK SOIL CAPACITY (ALLOWABLE STRESS DESIGN)

Ps = 77 kips (soil weight in footing size)

P = (Pr,DL + Pr,LL) + Pw + (Pf - Ps) = 168.92 kips (total vertical net load)

MR = (Pr,DL + Pr, LL) (L1 + a) + Pf (0.5 L) + Pw (L1 + 0.5Lw) = 3300 ft-kips (resisting moment with live load)

e = 0.5 L - (MR - MO) / P = 11.23 ft (eccentricity from middle of footing)

= 1.40 ksf < 4 / 3 qa

[Satisfactory]

Where e = 11.23 ft, > (L / 6)

CHECK FOOTING CAPACITY (STRENGTH DESIGN)

Mu,R = 1.2 [Pr,DL (L1 + a) + Pf (0.5 L) + Pw (L1 + 0.5Lw)] + 0.5 Pr, LL(L1 + a) = 3733 ft-kips

Mu,o = 1.4 F(h + D) = 2722 ft-kips

Pu = 1.2 (Pr,DL + Pf + Pw ) + 0.5 Pr, LL = 257 kips

eu = 0.5L - (Mu,R - Mu,O) / Pu = 15.32 ft

4.37 ksf 2.5

Footing Design of Shear Wall Based on ACI 318-02

DanielTian Li

61

,6

2,

3 (0.5 ) 6

MAX

eP

LL for eq BL

P Lfor e

B L e

+ ≤

=

>−

,

61

,6

2,

3 (0.5 ) 6

uu

uu MAX

uu

u

eP LL for e

q BLLP for e

B L e

+ ≤

= =

>−

(cont'd)

BENDING MOMENT & SHEAR AT EACH FOOTING SECTIONSection 0 1/10 L 2/10 L 3/10 L 4/10 L 5/10 L 6/10 L 7/10 L 8/10 L 9/10 L L

Xu (ft) 0 3.85 7.70 11.55 15.40 19.25 23.10 26.95 30.80 34.65 38.50

Pu,w (klf) 0.0 0.0 23.3 16.9 10.5 4.0 -2.4 -8.8 -15.2 0.0 0.0

Mu,w (ft-k) 0 0 -188 -720 -1503 -2440 -3438 -4400 -5232 -5839 -6314

Vu,w (kips) 0 0 -98 -175 -228 -255 -259 -237 -191 -123 -123

Pu,f (ksf) 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3

Mu,f (ft-k) 0 -26 -103 -232 -413 -645 -928 -1264 -1651 -2089 -2579

Vu,f (kips) 0 -13 -27 -40 -54 -67 -80 -94 -107 -121 -134

qu (ksf) -4.4 -2.9 -1.5 -0.1 0.0 0.0 0.0 0.0 0.0 0.0 0.0

Mu,q (ft-k) 0 288 1012 1961 2951 3941 4931 5922 6912 7902 8893

Vu,q (kips) 0 141 226 257 257 257 257 257 257 257 257

Σ Mu (ft-k) 0 263 722 1008 1035 856 565 258 29 -26 0

Σ Vu (kips) 0 127 102 42 -24 -65 -82 -74 -41 13 0

Location Mu,max d (in) ρreqD ρprovD Vu,max φVc = 2 φ b d (fc')0.5

Top Longitudinal -26 ft-k 20.69 0.0001 0.0004 127 kips 231 kips

Bottom Longitudinal 1035 ft-k 20.37 0.0049 0.0057 127 kips 228 kips

Bottom Transverse 7 ft-k / ft 19.36 0.0018 0.0019 3 kips / ft 22 kips / ft

Where ρ min = 0.0018

0.0155 [Satisfactory]

''2

0.85 1 10.383

uc

c

y

Mfb fd

fρ

− −

=

-500

0

500

1000

1500

M

-100

0

100

200

V

'10.85

MAX

f c uf u ty

β ερε ε

= =+



INPUT DATA COL#1 COL#2

COLUMN WIDTH c1 = 18 18 inCOLUMN DEPTH c2 = 18 18 inAXIAL DEAD LOAD PDL = 13 26 k

AXIAL LIVE LOAD PLL = 6.25 12.5 k

LATERAL LOAD (0=WIND, 1=SEISMIC) = 1 Seismic SDSEISMIC AXIAL LOAD, SD PLAT = -300 300 k

SEISMIC SHEAR LOAD, SD VLAT = 84.14 86.38 k

SEISMIC MOMENT, SD MLAT = 4.578 4.578 k-ft



ALLOWABLE SOIL PRESSURE Qa = 2 ksf

DISTANCE TO LEFT EDGE L1 = 36 ft

DISTANCE BETWEEN COLUMNS S = 30 ftDISTANCE TO RIGHT EDGE L2 = 36 ft

FOOTING WIDTH B = 7.5 ft

FTG EMBEDMENT DEPTH Df = 5 ft



SOIL WEIGHT ws = 0.11 kcf BAND WIDTH be = 7.5 ft

LONGITUDINAL REINFORCING BAR SIZE # 10 LONG. REINF AT TOP 10 # 10 @ 9 in o.c., cont.TRANSVERSE REINFORCING BAR SIZE # 5 LONG. REINF AT BOTTOM 13 # 10 @ 7 in o.c., cont.

TRANS. REINF. AT BAND WIDTH 7 # 5 @ 14 in o.c., bottomDESIGN SUMMARYFOOTING LENGTH L = 102.00 ftFOOTING WIDTH B = 7.50 ftFOOTING THICKNESS T = 48 in THE FOOTING DESIGN IS ADEQUATE.

ANALYSIS

DESIGN LOADS (IBC SEC.1605.3.2 & ACI 318 SEC.9.2.1) SERVICE LOADS COL # 1 COL # 2 TOTALCASE 1 : DL + LL P = 19 k 39 k 58 k

( e = 5.00 ft, fr CL ftg )CASE 2 : DL + LL + E / 1.4 P = -195 k 253 k 58 k

M = 3.3 ft-k 3.3 ft-k 6.5 ft-k( e = 116.43 ft, fr CL ftg )

V = 60 k 62 k 122 kCASE 3 : 0.9 DL + E / 1.4 P = -203 k 238 k 35 k


V = 60 k 62 k 122 kFACTORED LOADS CASE 1 : 1.2 DL + 1.6 LL Pu = 26 k 51 k 77 k

( eu = 5.00 ft, fr CL ftg )

CASE 2 : 1.2 DL + 1.0 LL + 1.0 E Pu = -278 k 344 k 66 kMu = 4.6 ft-k 4.6 ft-k 9.2 ft-k

( eu = 142.44 ft, fr CL ftg )

Vu = 84 k 86 k 171 kCASE 3 : 0.9 DL + 1.0 E Pu = -288 k 323 k 35 k

Mu = 4.6 ft-k 4.6 ft-k 9.2 ft-k( eu = 261.67 ft, fr CL ftg )

Vu = 84 k 86 k 171 k

CHECK OVERTURNING FACTORMR / MO = 2.99081 > F = 1 / (0.9x1.4) [Satisfactory]

Where MO = MLAT 1 + MLAT 2 + (VLAT 1 + VLAT 2) Df - PLAT 1(L - L1) - PLAT 2L2 = 9862 k-ftPftg = (0.15 kcf) T B L = 459.00 k, footing weightPsoil = ws (Df - T) B L = 84.15 k, soil weightMR = PDL 1(L - L1) + PDL 2L2 + 0.5 (Pftg + Psoil) L = 29495 k-ft


Combined Footing Design Based on ACI 318-02

DanielTian Li

(cont'd)CHECK SOIL BEARING CAPACITY (ACI 318 SEC.15.2.2)

Service Loads CASE 1 CASE 2 CASE 3P 57.8 57.8 35.1 ke 5.0 116.4 188.3 ft

qs B L 76.5 76.5 0.0 k, (surcharge load)(0.15-ws)T B L 122.4 122.4 110.2 k, (footing increased)

Σ P 256.7 256.7 145.3 ke 1.1 < L/6 26.2 > L/6 45.5 > L/6 ft

qmax 0.4 0.9 2.4 ksfqallow 2.0 2.7 2.7 ksf

Where [Satisfactory]

DESIGN FLEXURE & CHECK FLEXURE SHEAR (ACI 318 SEC.15.4.2, 10.2, 10.5.4, 7.12.2, 12.2, 12.5, 15.5.2, 11.1.3.1, & 11.3)

FACTORED SOIL PRESSURE Factored Loads CASE 1 CASE 2 CASE 3

Pu 76.8 65.6 35.1 keu 5.0 142.4 261.7 ft

γ qs B L 130.1 76.5 0.0 k, (factored surcharge load)

γ [0.15 T + ws (Df - T)] B L 760.4 651.8 488.8 k, (factored footing & backfill loads)Σ Pu 967.3 793.8 523.9 k

eu 0.4 < L/6 11.8 < L/6 17.5 > L/6 ftqu, max 1.294 1.756 1.391 ksf

FOOTING MOMENT & SHEAR AT LONGITUDINAL SECTIONS FOR CASE 10 0.5 L1 L1 left L1 right 0.2 S 0.4 S 0.6 S 0.8 S L2 left L2 right 0.5 L2 L0 18.00 36.00 36.00 42.00 48.00 54.00 60.00 66.00 66.00 84.00 102.000 0 0 0 -154 -307 -461 -614 -768 -768 -2,150 -3,5330 0.0 0.0 25.6 25.6 25.6 25.6 25.6 25.6 76.8 76.8 76.8

1.28 1.28 1.28 1.28 1.28 1.28 1.28 1.28 1.28 1.28 1.28 1.280 -207 -826 -826 -1125 -1469 -1859 -2295 -2777 -2777 -4498 -66330 23.0 45.9 45.9 53.6 61.2 68.9 76.5 84.2 84.2 107.1 130.1

7.455 7.455 7.455 7.455 7.455 7.455 7.455 7.455 7.455 7.455 7.455 7.4550 -1208 -4831 -4831 -6575 -8588 -10869 -13419 -16237 -16237 -26301 -387810 134.2 268.4 268.4 313.1 357.8 402.6 447.3 492.0 492.0 626.2 760.4

1.23 1.25 1.26 1.26 1.26 1.26 1.27 1.27 1.27 1.27 1.28 1.290 1505 6035 6035 8222 10749 13617 16827 20380 20380 33103 489460 -167.4 -336.2 -336.2 -392.8 -449.6 -506.5 -563.5 -620.7 -620.7 -793.3 -967.30 90 378 378 369 385 428 499 598 598 154 00 -10.3 -21.9 3.7 -0.6 -4.9 -9.4 -14.1 -18.9 32.3 16.8 0

Vu,soil (k)Σ Σ Σ Σ Mu (ft-k)Σ Σ Σ Σ Vu (kips)

Vu,col (k)

Vu,surch (k)

qu,soil (ksf)Mu,soil (ft-k)

SectionXu (ft)

Pu,surch (klf)Mu,surch (ft-k)

Pu,ftg & fill (klf)Mu,ftg & fill (ft-k)Vu,ftg & fill (k)

Mu,col (ft-k)

''2

0.85 1 10.383

M uf c b fd cf y

ρ

− − =

40.0018 ,

3MIN

TMIN

dρρ =

( )

( )

61

,6

2,

3 (0.5 ) 6

MAX

eP

LL for eBLq

P Lfor e

B L e

Σ + ≤=

Σ>

−

( )

( )

61

,6,

2,

3 (0.5 ) 6

MAX

euPu LLfor euBLqu

LPu for euB L eu

Σ + ≤=

Σ >

−

0100200300400500600700

M

-40

-20

0

20

40

V

'10.85

MAX

f c uf u ty

β ερε ε

=+


0 0.5 L1 L1 left L1 right 0.2 S 0.4 S 0.6 S 0.8 S L2 left L2 right 0.5 L2 L0 18.00 36.00 36.00 42.00 48.00 54.00 60.00 66.00 66.00 84.00 102.000 0 0 5 1,673 3,342 5,011 6,680 8,349 8,345 7,174 5,9940 0.0 0.0 -278.2 -278.2 -278.2 -278.2 -278.2 -278.2 65.6 65.6 65.6

0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.750 -122 -486 -486 -662 -864 -1094 -1350 -1634 -1634 -2646 -39020 13.5 27.0 27.0 31.5 36.0 40.5 45.0 49.5 49.5 63.0 76.5

6.39 6.39 6.39 6.39 6.39 6.39 6.39 6.39 6.39 6.39 6.39 6.390 -1035 -4141 -4141 -5636 -7361 -9317 -11502 -13917 -13917 -22544 -332410 115.0 230.0 230.0 268.4 306.7 345.1 383.4 421.7 421.7 536.8 651.8

0.32 0.57 0.83 0.83 0.91 1.00 1.08 1.16 1.25 1.25 1.50 1.760 491 2375 2375 3419 4709 6267 8117 10282 10282 18890 311480 -60.3 -154.7 -154.7 -193.8 -236.7 -283.4 -333.9 -388.2 -388.2 -573.9 -793.80 -666 -2252 -2247 -1205 -174 868 1946 3080 3076 874 00 68.3 102.3 -175.9 -172.1 -172.2 -176.0 -183.7 -195.1 148.6 91.4 0

FOOTING MOMENT & SHEAR AT LONGITUDINAL SECTIONS FOR CASE 30 0.5 L1 L1 left L1 right 0.2 S 0.4 S 0.6 S 0.8 S L2 left L2 right 0.5 L2 L0 18.00 36.00 36.00 42.00 48.00 54.00 60.00 66.00 66.00 84.00 102.000 0 0 5 1734 3464 5194 6924 8654 8,649 8,026 7,3950 0.0 0.0 -288.3 -288.3 -288.3 -288.3 -288.3 -288.3 35.1 35.1 35.1

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.000 0 0 0 0 0 0 0 0 0 0 00 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

4.79 4.79 4.79 4.79 4.79 4.79 4.79 4.79 4.79 4.79 4.79 4.790 -776 -3106 -3106 -4227 -5521 -6987 -8627 -10438 -10438 -16908 -249310 86.3 172.5 172.5 201.3 230.0 258.8 287.6 316.3 316.3 402.6 488.8

0.00 0.23 0.48 0.48 0.56 0.64 0.73 0.81 0.89 0.89 1.14 1.390 77 706 706 1143 1732 2494 3452 4629 4629 9695 175360 -14.0 -61.5 -61.5 -84.9 -111.9 -142.7 -177.3 -215.6 -215.6 -352.9 -523.90 -700 -2400 -2395 -1350 -325 700 1749 2844 2840 813 00 72.3 111.0 -177.3 -171.9 -170.2 -172.2 -178.0 -187.6 135.8 84.7 0

DESIGN FLEXURELocation Mu,max d (in) ρmin ρreqD ρmax smax(in) use ρprovD

Top Longitudinal -2400 ft-k 45.37 0.0019 0.0030 0.0155 no limit 10 # 10 @ 9 in o.c., cont. 0.0031Bottom Longitudinal 3080 ft-k 44.37 0.0019 0.0041 0.0155 18 13 # 10 @ 7 in o.c., cont. 0.0041

Bottom Transverse, be 1 ft-k / ft 43.42 0.0006 6.9E-06 0.0155 18 7 # 5 @ 14 in o.c. 0.0006

[Satisfactory]

Σ Σ Σ Σ Mu (ft-k)Σ Σ Σ Σ Vu (kips)

Vu,ftg & fill (k)qu,soil (ksf)Mu,soil (ft-k)Vu,soil (k)

Mu,surch (ft-k)Vu,surch (k)

Pu,ftg & fill (klf)Mu,ftg & fill (ft-k)

Xu (ft)Mu,col (ft-k)Vu,col (k)

Pu,surch (klf)

Section

Mu,col (ft-k)Vu,col (k)


Mu,ftg & fill (ft-k)Vu,ftg & fill (k)qu,soil (ksf)Mu,soil (ft-k)Vu,soil (k)


Pu,surch (klf)Mu,surch (ft-k)Vu,surch (k)

Pu,ftg & fill (klf)

SectionXu (ft)

-3000-2000-1000

01000200030004000

M

-300

-200

-100

0

100

200

V

-3000

-2000

-1000

0

1000

2000

3000

4000

M

-300

-200

-100

0

100

200

V

(cont'd)CHECK FLEXURE SHEAR


Longitudinal 195 k 328 k [Satisfactory]Transverse 0 k / ft 43 k / ft [Satisfactory]

CHECK PUNCHING SHEAR (ACI 318 SEC.15.5.2, 11.12.1.2, 11.12.6, & 13.5.3.2)

Column Case Pu Mu b1 b2 γv βc y Af Ap R J vu (psi) φ vc

1 25.6 0.0 61.4 61.4 0.4 1.0 2.0 56.3 74.1 11.9 363.8 1.3 164.3Col. 1 2 0.0 0.0 61.4 61.4 0.4 1.0 2.0 56.3 74.1 0.0 363.8 0.0 164.3

3 0.0 0.0 61.4 61.4 0.4 1.0 2.0 56.3 74.1 0.0 363.8 0.0 164.31 51.2 0.0 61.4 61.4 0.4 1.0 2.0 56.3 74.1 23.8 363.8 2.6 164.3

Col. 2 2 343.7 4.6 61.4 61.4 0.4 1.0 2.0 56.3 74.1 160.1 363.8 17.3 164.33 323.4 4.6 61.4 61.4 0.4 1.0 2.0 56.3 74.1 150.6 363.8 16.3 164.3

[Satisfactory]

0.5 1( )

231 21 3

6 1 1

1 2

R bMP uu vpsivu JAP

db d bJb b

b bPuRA f

γ−= +

= + +

=

( )2 1 21

12 113 2

db bAP

v bb

BbA f e

γ

= +

= −+

=

( )

( ) ( )

'( ) 2

42, , 40

0

, ,0 1 1 2 2

psi yv fc c

dy MIN

bcAP d db b c b cd

φφ

β

= +

=

= = + = +


JOB NO. : DATE : REVIEW BY : Seismic Design for Combined Footing, Based on ACI 318-02

DESIGN SUMMARY


REBAR YIELD STRESS fy = 60 ksiFOOTING WIDTH W = 90 inFOOTING THICKNESS D = 48 inDISTANCE BETWEEN COLUMNS L = 30 ft

COMBINED FOOTING LONGITUDINAL REINFORCINGTOP 12 # 10

( d = 43.74 in )( 1 Layer)

BOTTOM 13 # 10( d = 43.74 in )

( 1 Layer) 7.5

COMBINED FOOTING HOOPS (ACI 21.3.3)LOCATION AT END AT SPLICELENGTH 96 in 70 in

( 2h ) MAX0.075fyαβγdb/[(fc')0.5(c+Ktr)/db], 12

BAR 7 Legs # 5 7 Legs # 5SPACING @ 10 in o.c. @ 4 in o.c.

MIN(d/4, 8db, 24dt, 12) MIN(d/4, 4)

THE SEISMIC DESIGN IS ADEQUATE.

ANALYSISCHECK GB SECTION REQUIREMENTS (ACI 21.3.1)

Ln=L - c1 = 28.50 ft > 4 d = 14.58 ft [Satisfactory]

W / D = 1.88 > 0.3 [Satisfactory]W = 90 in > 10 in [Satisfactory]

< c1+1.5D = 90 in [Satisfactory]

CHECK SEISMIC FLEXURAL REQUIREMENTS(ACI 21.3.2.1) ρtop = 0.004 > ρmin=MIN[3(fc')

0.5/fy, 200/fy ]= 0.003 [Satisfactory]

< ρmax = 0.025 [Satisfactory]

ρbot = 0.004 > ρmin = 0.003 [Satisfactory]


(ACI 21.3.2.2) Mn,top > (1/2)Mn,bot [Satisfactory]

where Mn,bot = ρbot bd2fy (1 - 0.588ρbot fy/fc') = 3433 ft-kips

Mn,top = ρtop bd2fy (1 - 0.588ρtop fy/fc') = 3181 ft-kips

CHECK GB SHEAR STRENGTH (ACI 21.3.4)Ve = (Mpr, top + Mpr,bot) / Ln = 286.5 kips < 8φ(fc')

0.5bd = 1293.7 kips [Satisfactory]

< φ[2(fc')0.5bd + Avfyd/s ] = 750.5 kips [Satisfactory]

where Mpr,top = ρtop bd2fy (1.25 - 0.919ρtop fy/fc') = 3929 ft-kips

Mpr,bot = ρbot bd2fy (1.25 - 0.919ρbot fy/fc') = 4235 ft-kipsφ = 0.75 (ACI 9.3.2.3)

Av = 2.17 in2

DanielTian Li



Grade Beam Design Based on ACI 318-02

INPUT DATA & DESIGN SUMMARY


REBAR YIELD STRESS fy = 60 ksiSQUIRE PAD SIZE B = 8 ft

T = 16 inGRADE BEAM SIZE W = 36 in

D = 36 inCOLUMN DISTANCE L = 22 ft

GRADE BEAM EXTENSION Le = 5 ft

FRAME AXIAL LOADS, ASD PD,1 = 25 kips PD,2 = 25 kips (Dead Load)

PL,1 = 15 kips PL,2 = 15 kips (Live Load)

PE,1 = -30 kips PE,2 = 30 kips (Seismic Load)

SEISMIC SHEAR LOADS, ASD VE,1 = 50 kips VE,2 = 30 kips (Seismic Load)

SEISMIC MOMENTS, ASD ME,1 = 50 ft-kips ME,2 = 50 ft-kips (Seismic Load)

ALLOW SOIL PRESSURE Qa = 2.5 ksfPAD REINFORCING 8 # 6 @ 12 o.c., Each Way, Bottom.

GRADE BEAM LONGITUDINAL REINFORCINGTOP 7 # 7

( d = 31.94 in )( 1 Layer) -30

BOTTOM 7 # 7( d = 31.94 in )

( 1 Layer)GRADE BEAM HOOPS (ACI 21.3.3)

LOCATION AT END AT SPLICELENGTH 72 in 48 in


BAR 4 Legs # 5 4 Legs # 5(Legs to alternate long bars supported, ACI 7.10.5.3)

SPACING @ 7 in o.c. @ 4 in o.c.

MIN(d/4, 8db, 24dt, 12) MIN(d/4, 4)

THE GRADE BEAM DESIGN IS ADEQUATE.

ANALYSISCHECK OVERTURNING AT CENTER BOTTOM OF PAD 2

MO = ME,1 + ME,2 + (VE,1+VE,2)(D+T) - PE,1L = 1106.7 ft-kips

MR = (PD,1 + γconc B2T) L + 0.5γconc(L + 2Le) L D W = 1306.8 > MO / 0.9 = 1230 ft-kips [Satisfactory]

CHECK SOIL BEARING CAPACITY

1.55 ksf, (net pressure) < 4/3 Qa [Satisfactory]

where γconc = 0.15 kcf

γsoil = 0.11 kcf

CHECK PAD FLEXURAL REINFORCING

0.0020 <ρprovd = 0.0030

where d = 12.25 in

Qu,max = 1.5 Qmax = 2.33 ksf, factor 1.5 for SD

Mu = 0.125 (B-W)2 B Qu,max = 128 ft-kips

ρmax = 0.0155 (ACI 10.2.7.3 & 10.3.5)

ρmin = 0.0018 (ACI 7.12.2.1) [Satisfactory]

DanielTian Li

( ) ( )2,2 ,2

2 2

0.5D L CONC SOILOMAX

eT WD LP P B LMQLB B

γ γ+ + − + + = + =

''2

0.85 1 10.383

uc

c

y

MfB fd

fρ

− −

= =

(cont'd)CHECK PAD ONE WAY SHEAR CAPACITY

Vu < φVn [Satisfactory]

where Vu = 0.5 (B - W) B Qu,max - d B Qu,max = 27.6 kips

φVn = φ 2 d B (fc')0.5 = 96.6 kips

φ = 0.75

CHECK GB SECTION REQUIREMENTS (ACI 21.3.1)

Pu = 1.5(VE,1 - VE,2) = 30 kips < 0.1Agfc' = 388.8 kips [Satisfactory]

Ln=L - B = 14.00 ft > 4 d = 10.65 ft [Satisfactory]

W / D = 1.00 > 0.3 [Satisfactory]

W = 36 in > 10 in [Satisfactory]

< B+1.5D = 150 in [Satisfactory]

CHECK GB FLEXURAL REQUIREMENTS

(ACI 21.3.2.1) ρtop = 0.004 > ρmin=MIN[3(fc')0.5/fy, 200/fy ]= 0.003 [Satisfactory]





where Mn,bot = ρbot bd2fy (1 - 0.588ρbot fy/fc') = 642 ft-kips > Mu,bot / φ [Satisfactory]

Mn,top = ρtop bd2fy (1 - 0.588ρtop fy/fc') = 642 ft-kips > Mu,top / φ [Satisfactory]φ = 0.9

453 ft-kips

217 ft-kips

where 1.95 ksf, (full ASD pressure)

QMIN = 0.38 ksf, (full ASD pressure)

Factor 1.5 is for SD

CHECK GB SHEAR STRENGTH (ACI 21.3.4)

Ve = (Mpr, top + Mpr,bot) / Ln = 113.3 kips < 8φ(fc')0.5bd = 377.8 kips [Satisfactory]




Av = 1.24 in2

( ) ( ) ( )( )2, , ,1 ,1 ,1 ,1,1 ,1 ,1 ,21.5 0.5 / 0.5u top GB wt D L E EPAD E E EMIN MIN MAX MINL D D TQ Q Q QWt V V VM M P P P B M = + + + + − − − − + + + =

( )2,2 ,2

2 2

0.5D L CONCOMAX


γ+ + + + = + =

( ) ( ) ( ) ( )2, , ,2 ,2 ,2 ,2,1 ,2 ,1 ,21.5 0.5 / 0.5u bot GB wt D L E EPAD E E EMAX MAX MAX MINL D D TQ Q Q QWt V V VM M P P P B M = − − + + + − − − − + + + =



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 3 ksiREBAR YIELD STRESS fy = 60 ksiLATERAL SOIL PRESSURE Pa=ka γb = 30 pcf (equivalent fluid pressure)BACKFILL SPECIFIC WEIGHT γb = 100 pcfSATURATED SPECIFIC WEIGHT γsat = 118 pcfWATER TABLE h = 6 ftPASSIVE PRESSURE Pp = 450 psf / ftSURCHARGE WEIGHT ws = 200 psfFRICTION COEFFICIENT µ = 0.3ALLOW SOIL PRESSURE Qa = 3 ksfTHICKNESS OF TOP STEM tt = 8 inTHICKNESS OF KEY & STEM tb = 12 inTOE WIDTH LT = 3 ftHEEL WIDTH LH = 6 ftHEIGHT OF TOP STEM HT = 4 ftHEIGHT OF BOT. STEM HB = 4 ftFOOTING THICKNESS hf = 12 inKEY DEPTH hk = 12 inSOIL OVER TOE hp = 12 inTOP STEM REINF. (As,1) # 6 @ 20 in o.c., (Caution > 18"o.c. max. ACI 14.3.5)As,1 LOCATION (0=at soil face, 1=at middle) 0 at soil faceBOT. STEM REINF. (As,2) # 7 @ 8 in o.c.As,2 LOCATION (0=at soil face, 1=at middle) 0 at soil faceTOP REINF.OF FOOTING (As,3) # 6 @ 20 in o.c., (Caution > 18"o.c. max. ACI 7.12.2.2)BOT. REINF.OF FOOTING (As,4) # 5 @ 14 in [THE WALL DESIGN IS ADEQUATE.]

ANALYSISSERVICE LOADS

Hb = 0.5 Pa h2 + h Pa H + 0.5 [Pa (γsat - γw) / γb +γw] H2

= 1.44 kipsWhere h = 6 ft , H = 3 ft

Hs = ws Pa (HT + HB + hf) / γb = 0.54 kips 12Hp = 0.5 Pp (hp + hf + hk)2 = 2.03 kipsWs = ws (LH + tb - tt) = 1.27 kipsWb = Wb1 + Wb2 = 4.40 kips

Where Wb1 = 3.73 kips , Wb2 = 0.67 kipsWf = hf (LH + tb + LT) γc = 1.50 kipsWk = hk tb γc = 0.15 kipsWw,t = tt HT γc = 0.40 kipsWw,b = tb HB γc = 0.60 kips

FACTORED LOADSγHb = 1.6 Hb = 2.30 kipsγHs = 1.6 Hs = 0.86 kipsγWs = 1.6 Ws = 2.03 kips OVERTURNING MOMENT

γWb = 1.2 Wb = 5.28 kips H γH y H yγWf = 1.2 Wf = 1.80 kips Hb 1.44 2.30 2.69 3.87γWk = 1.2 Wk = 0.18 kips Hs 0.54 0.86 4.50 2.43γWw,t = 1.2 Ww,t = 0.48 kips ΣΣΣΣ 1.98 3.16 6.30γWw,b = 1.2 Ww,b = 0.72 kips

RESISTING MOMENT

W x W x γW x

Ws 1.27 6.83 8.66 13.85Wb 4.40 6.90 30.38 36.46 OVERTURNING FACTOR OF SAFETYWf 1.50 5.00 7.50 9.00Wk 0.15 3.50 0.53 0.63 8.02 > 1.5

Ww,t 0.40 3.33 1.33 1.60 [Satisfactory]Ww,b 0.60 3.50 2.10 2.52

ΣΣΣΣ 8.32 50.50 64.06

DanielTian Li

Retaining Wall Design Based on ACI 318-02

0.7210.49

5.281.80

0.48

γH y

6.193.8910.07

0.18

γW

2.03

WxSF

HyΣ= =Σ

CHECK SOIL BEARING CAPACITY (ACI 318-02 SEC.15.2.2)

= ft = -0.31 ft

= 0.67 ksf < Q a [Satisfactory]

CHECK FLEXURE CAPACITY, AS,1 & AS,2, FOR STEM (ACI 318-02 SEC.15.4.2, 10.2, 10.5.4, 7.12.2, 12.2, & 12.5)h = 6 ft , H' = 2 ft

A = ws Pa / γb = 60 plf B = h Pa = 180 plf

C = [Pa (γsat - γw) / γb + γw] H' =158 plf

At base of top stem Mu = 1.28 ft-kips Vu = 0.77 kips Pu = 0.48 kips

At base of bottom stem Mu = 7.27 ft-kips Vu = 2.46 kips Pu = 1.20 kips

At top stem At base of bottom stem

= 3.98 ft-kips , 24.25 ft-kips> M u > M u

[Satisfactory] [Satisfactory]where d = 4.56 in , 8.56 in

b = 12 in , 12 inφ = 0.7 , 0.7As = 0.264 in2 , 0.9 in2

ρ = 0.005 0.009

= 0.015 0.015> ρ > ρ

[Satisfactory] [Satisfactory]= 0.003 0.002

< ρ < ρ[Satisfactory] [Satisfactory]

CHECK SHEAR CAPACITY FOR STEM (ACI 318-02 SEC.15.5.2, 11.1.3.1, & 11.3)At top stem At base of bottom stem

= 4.50 kips , 8.44 kips

> V u > V u

[Satisfactory] [Satisfactory]where φ = 0.75 (ACI 318-02, Section 9.3.2.3 )

CHECK HEEL FLEXURE CAPACITY, AS,3, FOR FOOTING (ACI 318-02 SEC.15.4.2, 10.2, 10.3.5, 10.5.4, 7.12.2, 12.2, & 12.5)

= 0.015 = 0.001

10.002L Wx Hy

eW

Σ − Σ= −Σ

61

,6

,3 (0.5 ) 6

MAX

eW

LL for eq BL

W Lfor e

B L e

Σ + ≤

= Σ >

−

T HbL tL L= + +

'2allowable cbd fV φ=

0.00182

fMIN

hd

ρ =

'1.7S uy

n S yc

fA PdfM A

bfφ φ

− = −

0.0018MIN

td

ρ =

'10.85

MAX

f c uf u ty

β ερε ε

=+

'10.85

MAX

f c uf u ty

β ερε ε

=+

= 7.21 ft-kips

= 0.002

where d = 8.63 in qu, toe = 0.96 ksfeu = -0.15 ft qu, heel = 1.14 ksfS = n/a qu, 3 = 0.71 ksf

( A S, 3 ) required = in2 / ft < A S, 3 [Satisfactory]

CHECK TOE FLEXURE CAPACITY, AS,4, FOR FOOTING (ACI 318-02 SEC.15.4.2, 10.2, 10.3.5, 10.5.4, 7.12.2, 12.2, & 12.5)

= 0.015 = 0.001

= 3.21 ft-kips

where d = 8.69 inqu, 4 = 0.77 ksf

= 0.001


CHECK KEY CAPACITY FOR FOOTING

1.5 (Hb + Hs ) = 2.96 kips < Hp + µ ΣW = 4.52 kips[Satisfactory]

Techincal References: 1. Alan Williams: "Structuiral Engineering Reference Manual", Professional Publications, Inc, 2001. 2. Alan Williams: "Structuiral Engineering License Review Problems and Solutions", Oxford University Press, 2003.

0.19

0.11

' ,3'2

0.85 1 10.383

uc

c

y

Mfb fd

fρ

− −

=

( ) 2,3 ,

,32

,3

2,

2 6 6

,2 6 6

Hu u heelH Hs b f u

u

uH Hs b f u

bq q L LL L forw w w eL

Mbq S LL L forw w w e

L

γ γ γ

γ γ γ

+ + + − ≤ = + + − >

' ,4'2

0.85 1 10.383

uc

c

y

Mfb fd

fρ

− −

=

4 0.0018,

3 2f

MINhMINd

ρρ =

( ) 2 2,4 ,

,4

2

6 2Tu u toe T

u f

bq q L LwM

Lγ

+= −

'10.85

MAX

f c uf u ty

β ερε ε

=+



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 3 ksiREBAR YIELD STRESS fy = 60 ksiLATERAL SOIL PRESSURE Pa = 30 pcf (equivalent fluid pressure)PASSIVE PRESSURE Pp = 450 psf / ftSURCHARGE WEIGHT ws = 200 psfFRICTION COEFFICIENT µ = 0.3ALLOW SOIL PRESSURE Qa = 3 ksfTHICKNESS OF TOP STEM tt = 8 inTHICKNESS OF KEY & STEM tb = 12 inTOE WIDTH LT = 3 ftHEEL WIDTH LH = 6 ftHEIGHT OF TOP STEM HT = 6.5 ftHEIGHT OF BOT. STEM HB = 6.3 ftFOOTING THICKNESS hf = 12 inKEY DEPTH hk = 12 inSOIL OVER TOE hp = 12 inTOP STEM REINF. (As,1) 1 # 6 @ 16 in o.c., at middleBOT. STEM REINF. (As,2) 2 # 7 @ 8 in o.c., at each faceTOP REINF.OF FOOTING (As,3) # 6 @ 10 inBOT. REINF.OF FOOTING (As,4) # 5 @ 14 in [THE WALL DESIGN IS INADEQUATE.]


Hb = 0.5 Pa (HT + HB + hf)2 = 2.86 kipsHs = ws Pa (HT + HB + hf) / γb = 0.83 kipsHp = 0.5 Pp (hp + hf + hk)2 = 2.03 kipsWs = ws (LH + tb - tt) = 1.27 kipsWb = [HT (LH + tb - tt) + HB LH] γb = 7.90 kipsWf = hf (LH + tb + LT) γc = 1.50 kipsWk = hk tb γc = 0.15 kips 6Ww,t = tt HT γc = 0.65 kipsWw,b = tb HB γc = 0.95 kips

FACTORED LOADSγHb = 1.6 Hb = 4.57 kipsγHs = 1.6 Hs = 1.32 kipsγWs = 1.6 Ws = 2.03 kips OVERTURNING MOMENT

γWb = 1.2 Wb = 9.48 kips H γH y H yγWf = 1.2 Wf = 1.80 kips Hb 2.86 4.57 4.60 13.14γWk = 1.2 Wk = 0.18 kips Hs 0.83 1.32 6.90 5.71γWw,t = 1.2 Ww,t = 0.78 kips ΣΣΣΣ 3.68 5.90 18.85γWw,b = 1.2 Ww,b = 1.13 kips

RESISTING MOMENT

W x W x γW x



ΣΣΣΣ 12.41 76.75 95.56

Retaining Wall Design Based on ACI 318-02

γH y

21.029.1430.17

0.18

γW

15.40

2.039.481.80

0.781.13

DanielTian Li

WxSF

HyΣ= =Σ


= ft = 0.33 ft


CHECK FLEXURE CAPACITY, AS,1 & AS,2, FOR STEM (ACI 318-02 SEC.15.4.2, 10.2, 10.3.5, 10.5.4, 7.12.2, 12.2, & 12.5)

At top stem At base of bottom stem= 4.23 ft-kips , 24.64 ft-kips

= 0.78 kips , 1.91 kips

= 4.26 ft-kips , 24.29 ft-kips> M u < M u

[Satisfactory] [Unsatisfactory]where d = 4.00 in , 8.56 in

b = 12 in , 12 inφ = 0.7 , 0.7As = 0.33 in2 , 0.9 in2

ρ = 0.007 0.009

= 0.015 0.015> ρ > ρ



CHECK SHEAR CAPACITY FOR STEM (ACI 318-02 SEC.15.5.2, 11.1.3.1, & 11.3)

At top stem At base of bottom stem= 1.64 kips , 5.16 kips

= 3.94 kips , 8.44 kips

> V > V[Satisfactory] [Satisfactory]

where φ = 0.75 (ACI 318-02, Section 9.3.2.3 )


= 0.015 = 0.001

10.002L Wx Hy

eW

Σ − Σ= −Σ

61

,6

,3 (0.5 ) 6

MAX

eW

LL for eq BL

W Lfor e

B L e

Σ + ≤

= Σ >

−

T HbL tL L= + +

u wWP γ=


2

2a as

b

yy wP PV γγ

= +

0.00182

fMIN

hd

ρ =

3 2

6 2a a s

u

b

y y wP PM γ

γ

= +

'1.7S uy

n S yc

fA PdfM A

bfφ φ

− = −

0.0018MIN

td

ρ =

'10.85

MAX

f c uf u ty

β ερε ε

=+

'10.85

MAX

f c uf u ty

β ερε ε

=+

= 18.42 ft-kips

= 0.005

where d = 8.63 in qu, toe = 2.24 ksfeu = 0.75 ft qu, heel = 0.84 ksfS = n/a qu, 3 = 1.53 ksf


CHECK TOE FLEXURE CAPACITY, AS,4, FOR FOOTING (ACI 318-02 SEC.15.4.2, 10.2, 10.5.4, 7.12.2, 12.2, & 12.5)

= 0.015 = 0.001

= 8.46 ft-kips

where d = 8.69 inqu, 4 = 1.71 ksf

= 0.002




Techincal References: 1. Alan Williams: "Structuiral Engineering Reference Manual", Professional Publications, Inc, 2001. 2. Alan Williams: "Structuiral Engineering License Review Problems and Solutions", Oxford University Press, 2003.

0.50

0.22

' ,3'2

0.85 1 10.383

uc

c

y

Mfb fd

fρ

− −

=

( ) 2,3 ,

,32

,3

2,

2 6 6

,2 6 6

Hu u heelH Hs b f u

u

uH Hs b f u



L

γ γ γ

γ γ γ

+ + + − ≤ = + + − >

' ,4'2

0.85 1 10.383

uc

c

y

Mfb fd

fρ

− −

=

4 0.0018,

3 2f

MINhMINd

ρρ =

( ) 2 2,4 ,

,4

2

6 2Tu u toe T

u f

bq q L LwM

Lγ

+= −

'10.85

MAX

f c uf u ty

β ερε ε

=+



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 3 ksiREBAR YIELD STRESS fy = 60 ksiLATERAL SOIL PRESSURE Pa = 45 pcf (equivalent fluid pressure)PASSIVE PRESSURE Pp = 300 psf / ftSEISMIC GROUND SHAKING PE = 34.3 psf / ft, ASD

(PE only for H > 12 ft, CBC 1611A.6)SURCHARGE WEIGHT ws = 100 psfFRICTION COEFFICIENT µ = 0.35ALLOW SOIL PRESSURE Qa = 3 ksf, (w/o 4/3 increasing)THICKNESS OF TOP STEM tt = 20 inTHICKNESS OF KEY & STEM tb = 20 inTOE WIDTH LT = 4.67 ftHEEL WIDTH LH = 7.5 ftHEIGHT OF TOP STEM HT = 8 ftHEIGHT OF BOT. STEM HB = 8 ftFOOTING THICKNESS hf = 24 inKEY DEPTH hk = 38 inSOIL OVER TOE hp = 6 inTOP STEM REINF. (As,1) # 10 @ 5 in o.c.As,1 LOCATION ( 1 = at middle, 2 = at soil face) 2 at soil faceBOT. STEM REINF. (As,2) # 10 @ 5 in o.c.As,2 LOCATION ( 1 = at middle, 2 = at soil face) 2 at soil face [THE WALL DESIGN IS ADEQUATE.]TOP REINF.OF FOOTING (As,3) # 8 @ 10 inBOT. REINF.OF FOOTING (As,4) # 6 @ 18 in


Hb = 0.5 Pa (HT + HB + hf)2 = 7.29 kipsHs = ws Pa (HT + HB + hf) / γb = 0.81 kips

where γb = 100 pcf, soil weightHp = 0.5 Pp (hp + hf + hk)2 = 4.82 kipsHE = 0.5 PE (HT + HB)2 = 4.39 kipsWs = ws (LH + tb - tt) = 0.75 kipsWb = [HT (LH + tb - tt) + HB LH] γb = 12.00 kipsWf = hf (LH + tb + LT) γc = 4.15 kipsWk = hk tb γc = 0.79 kipsWw,t = tt HT γc = 2.00 kipsWw,b = tb HB γc = 2.00 kips

FACTORED LOADSγHb = 1.7 Hb = 12.39 kipsγHs = 1.7 Hs = 1.38 kipsγHE = 1.4 HE = 6.15 kips OVERTURNING MOMENT

γWs = 1.7 Ws = 1.28 kips H γH y H yγWb = 1.4 Wb = 16.80 kips Hb 7.29 12.39 6.00 43.74γWf = 1.4 Wf = 5.81 kips HE 4.39 6.15 12.67 55.61γWk = 1.4 Wk = 1.11 kips Hs 0.81 1.38 9.00 7.29γWw,t = 1.4 Ww,t = 2.80 kips ΣΣΣΣ 12.49 19.92 106.6γWw,b = 1.4 Ww,b = 2.80 kips

Retaining Wall Design Based on CBC 2001 Chapter A

74.36

12.39164.61

77.86

γH y

DanielTian Li

(cont'd)RESISTING MOMENT

W x W x γW x



ΣΣΣΣ 21.69 183.69 259.44

CHECK SOIL BEARING CAPACITY (CBC 2001 SEC.1915A.2.2)

= ft

= 3.37 ft < L / 4 = 3.46 ft, (CBC 1611A.6)[Satisfactory]

2.04 ksf < 4 / 3 Q a = 4.00 ksf [Satisfactory]

CHECK FLEXURE CAPACITY, AS,1 & AS,2, FOR STEM (CBC 2001 1915A.4.2, 1910A.2, 1910A.5.4, 1907A.12.2, 1912A.2, & 1912A.5)


= 2.80 kips , 5.60 kips

= 143.19 ft-kips , 143.68 ft-kips> M u > M u

[Satisfactory] [Satisfactory]where d = 16.37 in , 16.37 in

b = 12 in , 12 inφ = 0.7 , 0.7As = 3.048 in2 , 3.048 in2

ρ = 0.016 0.016

= 0.016 0.016> ρ > ρ



CHECK SHEAR CAPACITY FOR STEM (CBC SEC.1915A.5.2, 1911A.1.3.1, & 1911A.3)

At top stem At base of bottom stem= 7.67 kips , 17.16 kips

= 18.29 kips , 18.29 kips


where φ = 0.85 (CBC 2001, Section 1909A.3.2.3 )

1.11

γW

30.59

13.84

1.2816.805.81

2.802.80

WxSF

HyΣ= =Σ

2L Wx Hy

eW

Σ − Σ= −Σ

T HbL tL L= + +

u wWP γ=


( )2 22 2a as E

b

y H yPyy wP PVγ

γγ

−= + +

( )23 2 36 2 6a a s E

u

b

H yyPy y wP PM

γγ

γ −

= + +

'1.7S uy

n S yc

fA PdfM A

bfφ φ

− = −

'10.85 87

0.7587

cMAX

y y

ff fβρ

= +

0.0018MIN

td

ρ =

61

,6

,3 (0.5 ) 6

MAX

eW

LL for eq BL

W Lfor e

B L e

Σ + ≤

= = Σ >

−

(cont'd)CHECK HEEL FLEXURE CAPACITY, AS,3, FOR FOOTING (CBC 2001 1915A.4.2, 1910A.2, 1910A.5.4, 1907A.12.2, 1912A.2, & 1912A.5)

= 0.016 = 0.001

= 78.06 ft-kips

= 0.0047.5

where d = 20.50 in qu, toe = 3.29 ksfeu = 3.82 ft qu, heel = n/a ksfS = 2.96 ft qu, 3 = 1.05 ksf


CHECK TOE FLEXURE CAPACITY, AS,4, FOR FOOTING (CBC 2001 1915A.4.2, 1910A.2, 1910A.5.4, 1907A.12.2, 1912A.2, & 1912A.5)

= 0.016 = 0.001

= 25.29 ft-kips

where d = 20.63 inqu, 4 = 1.638 ksf

= 0.001




0.28

0.88

'10.85 87

0.7587

cMAX

y y

ff fβρ

= +

' ,3'2

0.85 1 10.383

uc

c

y

Mfb fd

fρ

− −

=

0.00182

fMIN

hd

ρ =

( ) 2,3 ,

,32

,3

2,

2 6 6

,2 6 6

Hu u heelH Hs b f u

u

uH Hs b f u



L

γ γ γ

γ γ γ

+ + + − ≤ = + + − >

'10.85 87

0.7587

cMAX

y y

ff fβρ

= +

' ,4'2

0.85 1 10.383

uc

c

y

Mfb fd

fρ

− −

=

4 0.0018,

3 2f

MINhMINd

ρρ =

( ) 2 2,4 ,

,4

2

6 2Tu u toe T

u f

bq q L LwM

Lγ

+= −



INPUT DATA & DESIGN SUMMARYSPECIAL INSPECTION ( 0=NO, 1=YES ) 1 YesTYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiCONCRETE STRENGTH fc' = 4.5 ksiREBAR YIELD STRESS fy = 60 ksiLATERAL SOIL PRESSURE Pa = 30 pcf (equivalent fluid pressure)PASSIVE PRESSURE Pp = 400 psf / ftSURCHARGE WEIGHT ws = 100 psfSERVICE LATERAL FORCE wLat = 20 psfFRICTION COEFFICIENT µ = 0.3ALLOW SOIL PRESSURE Qa = 3 ksfTHICKNESS OF TOP STEM tt = 8 inTHICKNESS OF KEY & STEM tb = 12 inTOE WIDTH LT = 3 ftHEEL WIDTH LH = 6 ftHEIGHT OF FENCE STEM HF = 4 ftHEIGHT OF TOP STEM HT = 3.3 ftHEIGHT OF BOT. STEM HB = 5.5 ftFOOTING THICKNESS hf = 12 inKEY DEPTH hk = 12 in [THE WALL DESIGN IS ADEQUATE.]SOIL OVER TOE hp = 12 inTOP STEM REINF. (As,1) # 6 @ 16 in o.c.As,1 LOCATION (0=at soil face, 1=at middle, 2=at each face) 2 at each faceBOT. STEM REINF. (As,2) # 7 @ 8 in o.c.As,2 LOCATION (0=at soil face, 1=at middle, 2=at each face) 2 at each faceTOP REINF.OF FOOTING (As,3) # 6 @ 20 in o.c., (Caution > 18" o.c. max. ACI 7.12.2.2)BOT. REINF.OF FOOTING (As,4) # 5 @ 14 in


Hb = 0.5 Pa (HT + HB + hf)2 = 1.44 kipsHs = ws Pa (HT + HB + hf) / γb = 0.29 kipsHp = 0.5 Pp (hp + hf + hk)2 = 1.80 kipsHLat = wLat (HF + HT + HB - hp) = 0.24 kipsWs = ws (LH + tb - tt) = 0.63 kipsWb = [HT (LH + tb - tt) + HB LH] γb = 5.39 kips 8Wf = hf (LH + tb + LT) γc = 1.50 kipsWk = hk tb γc = 0.15 kipsWw,t = tt (HT +HF) γm = 0.54 kipsWw,b = tb HB γm = 0.61 kips

FACTORED LOADSγHb = 1.6 Hb = 2.30 kipsγHs = 1.6 Hs = 0.47 kipsγHLat = 1.6 HLat = 0.38 kipsγWs = 1.6 Ws = 1.01 kips OVERTURNING MOMENT

γWb = 1.2 Wb = 6.47 kips H γH y H yγWf = 1.2 Wf = 1.80 kips Hb 1.44 2.30 3.27 4.71γWk = 1.2 Wk = 0.18 kips Hs 0.29 0.47 4.90 1.44γWw,t = 1.2 Ww,t = 0.64 kips HLat 0.24 0.38 7.90 1.86γWw,b = 1.2 Ww,b = 0.73 kips ΣΣΣΣ 1.97 3.15 8.01

RESISTING MOMENT

W x W x γW x



ΣΣΣΣ 8.81 53.64 66.090.73

Daniel

γH y

7.53

2.98

Retaining Wall Design Based on ACI 530-02 & ACI 318-02

γW

Tian Li

12.82

2.30

1.016.471.80

0.640.18

10.83

WxSF

HyΣ= =Σ


= ft = -0.18 ft


CHECK FLEXURE CAPACITY FOR MASONRY STEM (ACI 530 2.3.3)


= 0.54 kips , 1.14 kips

At top stem At base of bottom stemwhere te = 7.63 in , 11.63 in

d = 4.26 in , 8.19 inbw = 12 in , 12 inFb = 0.495 ksi , 0.495 ksiFs = 24 ksi , 24 ksiAs = 0.33 in2 , 0.9 in2

ρ = 0.006 , 0.009Em = 1350 ksi , 1350 ksiEs = 29000 ksi , 29000 ksin = 21.48 21.48k = 0.41 0.46

and M allowable = 1.55 ft-kips , 6.25 ft-kips

> M > M[Satisfactory] [Satisfactory]

CHECK SHEAR CAPACITY FOR MASONRY STEM (ACI 530 2.3.5)At top srem At base of bottom stem

= 0.41 kips , 1.66 kips

= 1.98 kips , 3.81 kips



= 0.023 = 0.001

10.002L Wx Hy

eW

Σ − Σ= −Σ

61

,6

,3 (0.5 ) 6

MAX

eW

LL for eq BL

W Lfor e

B L e

Σ + ≤

= Σ >

−

T HbL tL L= + +

3 2

6 2a a s

Lat

b

y y wP PM Mγ= + +

wP W=

1,

2 3 2 3 2 3e e

allowable b s swkd kd kdt tMIN kd d P d d PbM F A F

= − − − − + −

( )' , 50allowable w md MIN fV b=

2

2a as

Lat

b

yy wP PV Vγ= + +

0.00182

fMIN

hd

ρ ='

10.85MAX

f c uf u ty

β ερε ε

=+

= 8.35 ft-kips

= 0.002




= 0.023 = 0.001

= 3.95 ft-kips

where d = 8.69 inqu, 4 = 0.90 ksf

= 0.001



1.5 (Hb + Hs + HLat) = 2.96 kips < Hp + µ ΣW = 4.44 kips[Satisfactory]

Techincal References: 1. "Masonry Designers' Guide, Third Edition" (MDG-3), The Masonry Society, 2001. 2. Alan Williams: "Structural Engineering Reference Manual", Professional Publications, Inc, 2001. 3. Alan Williams: "Structural Engineering License Review Problems and Solutions", Oxford University Press, 2003.

0.22

0.13

' ,3'2

0.85 1 10.383

uc

c

y

Mfb fd

fρ

− −

=

( ) 2,3 ,

,32

,3

2,

2 6 6

,2 6 6

Hu u heelH Hs b f u

u

uH Hs b f u



L

γ γ γ

γ γ γ

+ + + − ≤ = + + − >

' ,4'2

0.85 1 10.383

uc

c

y

Mfb fd

fρ

− −

=

4 0.0018,

3 2f

MINhMINd

ρρ =

( ) 2 2,4 ,

,4

2

6 2Tu u toe T

u f

bq q L LwM

Lγ

+= −

'10.85

MAX

f c uf u ty

β ερε ε

=+



INPUT DATA & DESIGN SUMMARYSPECIAL INSPECTION ( 0=NO, 1=YES ) 1 YesTYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiCONCRETE STRENGTH fc' = 2.5 ksiREBAR YIELD STRESS fy = 60 ksiLATERAL SOIL PRESSURE Pa = 45 pcf (equivalent fluid pressure)PASSIVE PRESSURE Pp = 350 psf / ftSURCHARGE WEIGHT ws = 100 psfSERVICE LATERAL FORCE wLat = 30.1 psfSERVICE GRAVITY LOAD P = 0.386 kips / ftECCENTRICITY e = 6 inFRICTION COEFFICIENT µ = 0.35ALLOW SOIL PRESSURE Qa = 2.25 ksfTHICKNESS OF STEM t = 8 inTOE WIDTH LT = 0.917 ftHEEL WIDTH LH = 0.917 ftHEIGHT OF FENCE STEM HF = 5 ftHEIGHT OF STEM H = 7 ftRESTRAINED HEIGHT HR = 9 ftFOOTING THICKNESS hf = 12 inRESTRAINED BOTTOM ? (1=Yes, 0=No) 1 Yes

KEY DEPTH hk = 20 <=No ReqD [THE WALL DESIGN IS INADEQUATE.]SOIL OVER TOE hp = 6 inSTEM REINF. (As,1) # 8 @ 8 in o.c.As,1 LOCATION (0=at inside face, 1=at middle, 2=at each face) 0 at inside faceBOT. REINF.OF FOOTING (As,2) # 4 @ 24 inTOP REINF.OF FOOTING (As,3) # 4 @ 18 in


Hb = 0.5 Pa H2 = 1.10 kipsHs = ws Pa H / γb = 0.32 kipsHp = 0.5 Pp (hp + hf + hk)2 = 1.75 kipsHLat = wLat (HF + H - hp) = 0.35 kipsWs = ws LH = 0.09 kipsWb = H LH γb = 0.64 kipsWf = hf (LH + t + LT) γc = 0.38 kipsWk = hk t γc = 0.17 kipsWw = t ( HF + H ) γm = 0.88 kipsRT = 0.5HLat(HF/HR + hp/HR +H/HR) + Pe/HR

+ 0.5HsH/HR + HbH/3HR = 0.67 kipsRB = HLat + Hs + Hb - RT = 1.09 kipsVB = Ww + P = 1.27 kips

FACTORED LOADSγHb = 1.6 Hb = 1.76 kipsγHs = 1.6 Hs = 0.50 kipsγHLat = 1.6 HLat = 0.55 kips γRT = 1.6 RT = 1.07 kipsγWs = 1.6 Ws = 0.15 kips γRB = 1.6 RB = 1.75 kipsγWb = 1.2 Wb = 0.77 kipsγWf = 1.2 Wf = 0.45 kips OVERTURNING MOMENT

γWk = 1.2 Wk = 0.20 kips H γH y H yγWw = 1.2 Ww = 1.06 kips RB 1.09 1.75 1.00 1.09 1.75γP = 1.6 P = 0.62 kips ΣΣΣΣ 1.09 1.75 1.09

RESISTING MOMENT

W x W x γW x

Ws 0.09 2.04 0.19 0.30Wb 0.64 2.04 1.31 1.57 OVERTURNING FACTOR OF SAFETYWf 0.38 1.25 0.47 0.56

Wk 0.17 1.25 0.21 0.25 > 1.5

P 0.39 1.25 0.48 0.77 [Satisfactory]Ww 0.88 1.25 1.10 1.32

ΣΣΣΣ 2.54 3.76 4.78

Daniel

Restrained Retaining Wall Design Based on ACI 530-02 & ACI 318-02

0.45

Tian Li

0.621.06

γW

γH y

1.75

0.20

3.24

3.44

0.150.77

WxSF

HyΣ= =Σ


= ft = ft

= 1.51 ksf< Q a

[Satisfactory]

CHECK FLEXURE CAPACITY FOR MASONRY STEM (ACI 530 2.3.3) 5

S = Pa-1 ( PaH + wLat + Hs / H) + [( PaH + wLat + Hs / H)2-2Pa(RB +wLat hp)]0.5

= 3.58 ft

P = VB - Ww S / (H + HF) = 1.00 kips, @ Mmax section

MMax = S RB - 0.5 HS S2 / H - 0.5 Pa S3 - Pa (H-S) S2/6 - 0.5 wLat (S - hp)2 = 2.12 ft-kips

2.06 ft-kips

< Mwhere te = 7.63 in ρ = 0.024 [Unsatisfactory]

d = 4.13 in Em = 1350 ksibw = 12 in Es = 29000 ksiFb = 0.495 ksi n = 21.48Fs = 24 ksi k = 0.62As = 1.185 in2

CHECK SHEAR CAPACITY FOR MASONRY STEM (ACI 530 2.3.5)At restrained stem At bottom of stem

= 0.58 kips , 1.09 kips

= 1.97 kips , 1.92 kips



= 0.013 = 0.001

= 0.27 ft-kips

= 0.000



0.202.50

0.13

2L Wx Hy

eW

Σ − Σ= −Σ

61

,6

,3 (0.5 ) 6

MAX

eW

LL for eq BL

W Lfor e

B L e

Σ + ≤

= Σ >

−

T HbL tL L= + +

1,

2 3 2 3 2 3e e

allowable b s swkd kd kdt tMIN kd d P d d PbM F A F

= − − − − + − =

( )' , 50allowable w md MIN fV b=

. .V Max Horiz Shear=

' ,3'2

0.85 1 10.383

uc

c

y

Mfb fd

fρ

− −

=

0.00182

fMIN

hd

ρ =

( ) 2,3 ,

,32

,3

2,

2 6 6

,2 6 6

Hu u heelH Hs b f u

u

uH Hs b f u



L

γ γ γ

γ γ γ

+ + + − ≤ = + + − >

'10.85

MAX

f c uf u ty

β ερε ε

=+


= 0.013 = 0.000

= 0.78 ft-kips

where d = 8.75 inqu, 2 = 1.53 ksf

= 0.000



1.5 RB = 1.64 kips < Hp + µ ΣW = N/A (Restrained)[Satisfactory]

Techincal References: 1. "Masonry Designers' Guide, Third Edition" (MDG-3), The Masonry Society, 2001. 2. Alan Williams: "Structural Engineering Reference Manual", Professional Publications, Inc, 2001. 3. Alan Williams: "Structural Engineering License Review Problems and Solutions", Oxford University Press, 2003.

0.03

' ,2'2

0.85 1 10.383

uc

c

y

Mfb fd

fρ

− −

=

4 0.0018,

3 2f

MINhMINd

ρρ =

( ) 2 2,4 ,

,2

2

6 2Tu u toe T

u f

bq q L LwM

Lγ

+= −

'10.85

MAX

f c uf u ty

β ερε ε

=+




IS FOOTING RESTRAINED @ GRADE LEVEL ? (1=YES,0=NO) 0 no PLATERAL FORCE @ TOP OF POLE P = 29.843 kHEIGHT OF POLE ABOVE GRADE H = 62.11 ftDIAMETER OF POLE FOOTING B = 6 ft HLATERAL SOIL BEARING CAPACITY S = 0.2 ksf / ftISOLATED POLE FACTOR (IBC 1804.3.1 or UBC note 3 on Tab 18-I-A) F = 2FIRST TRIAL DEPTH ===> D = 8 ft

Use 6 ft dia x 19.75 ft deep footing unrestrained @ ground level D

ANALYSIS

LATERAL BEARING @ BOTTOM :LATERAL BEARING @ D/3 :

, FOR NONCONSTRAINEDREQUIRD DEPTH :

, FOR CONSTRAINED

NONCONSTRAINED CONSTRAINEDLATERAL FORCE @ TOP OF POLE P => 29.84 k 29.84 kHEIGHT OF POLE ABOVE GRADE H => 62.1 ft 62.1 ftDIAMETER OF POLE FOOTING B => 6.00 ft 6.00 ftLATERAL SOIL BEARING CAPACITY FS => 0.40 ksf / ft 0.40 ksf / ft

1ST TRIAL TRY D1 => 8.00 ft 8.00 ft

LAT SOIL BEARING @ 1/3 D S1 => 1.06 ksf 1.06 ksf

LAT SOIL BEARING @ 1.0 D S3 => 3.20 ksf 3.20 ksf CONSTANT 2.34P/(BS1) A => 11.02 - - REQD FOOTING DEPTH RQRD D => 33.38 ft 20.26 ft

2ND TRIAL : TRY D2=> 20.69 ft 14.13 ft LAT SOIL BEARING @ 1/3 D S1 => 2.73 ksf 1.86 ksf


3RD TRIAL : TRY D3=> 19.97 ft 14.69 ft LAT SOIL BEARING @ 1/3 D S1 => 2.64 ksf 1.94 ksf


4TH TRIAL : TRY D4=> 19.80 ft 14.82 ft LAT SOIL BEARING @ 1/3 D S1 => 2.61 ksf 1.96 ksf


5TH TRIAL : TRY D5=> 19.77 ft 14.85 ft LAT SOIL BEARING @ 1/3 D S1 => 2.61 ksf 1.96 ksf


T. LiFlagpole Footing Design Based on Chapter 18 both IBC & UBC

Daniel

3

1 3

1

3

0.332.34

4.361 1

2

4.25

FSDS

S SP

ABS

A HA

DPH

BS

==

=

+ +

=



INPUT DATAPEDESTAL DIAMETER c = 60 inSQUARE FOOTING LENGTH L = 18 ft

FOOTING EMBEDMENT DEPTH Df = 10.5 ft

FOOTING THICKNESS T = 18 inWATER TABLE h = 8 ft



AXIAL DEAD LOAD PDL = 73.1 k


LATERAL LOAD (0=WIND, 1=SEISMIC) = 0 Wind,ASDWIND AXIAL LOAD PLAT = -79.175 k, ASD, uplift

WIND MOMENT LOAD MLAT = 867.41 ft-k, ASD

WIND SHEAR LOAD VLAT = 11.95 k, ASD


BBACKFILL SOIL WEIGHT ws = 0.1 kcf


FOOTING REINFORCING SIZE # 10PEDESTAL VERT. REINF. SIZE 22 # 10 verticalPEDESTAL SHEAR. REINF. # 4 spiral @ 3 in o.c.

DESIGN SUMMARYTOP FOOTING REINF., E. WAY => 6 # 10BOT. FOOTING REINF., E. WAY => 29 # 10 @ 7 in o.c. THE FOOTING DESIGN IS ADEQUATE.

ANALYSISDESIGN LOADS (IBC SEC.1605.3.2 & ACI 318-02 SEC.9.2.1)CASE 1: DL + LL P = 439 kips 1.2 DL + 1.6 LL Pu = 672 kipsCASE 2: DL + LL + 1.3 W P = 336 kips 1.2 DL + LL + 1.6 W Pu = 326 kips

M = 1128 ft-kips Mu = 1388 ft-kipse = 3.4 ft, fr cl ftg eu = 4.3 ft, fr cl ftg

CASE 3: DL + LL + 0.65 W P = 387 kips 0.9 DL+ 1.6 W Pu = -61 kipsM = 434 ft-kips Mu = 1388 ft-kipse = 1.1 ft, fr cl ftg eu = -22.8 ft, fr cl ftg


MR / MO = 2.28444 > F = 1.5 [Satisfactory]

Where MO = MLAT + VLAT Df - 0.5 PLATL = 1705 k-ft

Pftg = (0.15 kcf) [T L2 + (π c2/ 4)(Df - T)] = 99.41 k, footing weight

wsat = ws + 0.018 kcf = 0.118 kcf, saturated soil weight

wwater = 0.0625 kcf, water specifc weight

Psoil = [ws MIN(h, Df -T)+ (wsat - wwater) MAX(Df - T - h, 0)] (L2 - π c2/ 4) = 260.38 k, soil wt

MR = 0.5 (PDL + Pftg + Psoil) L = 3896 k-ft

F = 1.5 for wind, IBC 1609.1.3

CHECK UPLIFT CAPACITY

FGravity / FUplift = 7.55168 > 1.0 [Satisfactory]

Where FUplift = - PLAT = 79.2 k

Pftg = 99.41 k, footing weight

Psoil = 425.40 k, soil weight with 30o pyramid

FGravity = PDL+ Pftg+ Psoil = 597.9 k 4

Tian LiDaniel

Deep Footing Design Based on ACI 318-02

(cont'd)CHECK SOIL BEARING CAPACITY (ACI 318-02 SEC.15.2.2)

Service Loads CASE 1 CASE 2 CASE 3P 438.5 335.6 387.0 ke 0.0 3.4 1.1 ft, (from center of footing)

qs L2 32.4 32.4 32.4 k, (surcharge load)

∆Pftg 33.1 33.1 33.1 k, (footing increased)Σ P 504.0 401.1 452.6 kΣ M 0.0 1290.8 496.4 k-ft, (VLat included)

e 0.0 < L/6 3.2 > L/6 1.1 < L/6 ft

qmax 1.6 2.6 1.9 ksf


Where

[Sati


0 0.09 L 0.18 L 0.27 L ColL ColR 0.73 L 0.82 L 0.91 L L

0 1.63 3.25 4.88 6.50 11.50 13.13 14.75 16.38 18

0 0 0 0 0 772.52 242.06 -288.41 -818.87 -1349.3

0 0.0 0.0 0.0 0.0 326.4 326.4 326.4 326.4 326.4

1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80

0 -2.4 -9.5 -21.4 -38.0 -119.0 -155.0 -195.8 -241.3 -291.6

0 2.9 5.9 8.8 11.7 20.7 23.6 26.6 29.5 32.4

23.99 23.99 23.99 23.99 23.99 23.99 23.99 23.99 23.99 23.99

0 -31.669 -126.68 -285.02 -506.71 -1586.1 -2066 -2609.2 -3215.8 -3885.7

0 39.0 78.0 116.9 155.9 275.8 314.8 353.8 392.8 431.7

0.81 1.10 1.40 1.69 1.99 2.89 3.19 3.48 3.78 4.07

0 21.5 95.3 235.5 456.0 1787.6 2480.9 3325.9 4336.5 5526.7

0 -27.9 -64.4 -109.5 -163.3 -382.9 -471.9 -569.5 -675.7 -790.6

0 -12.56 -40.888 -70.959 -88.746 854.97 501.96 232.45 60.449 0

0 14.0 19.4 16.2 4.3 240.1 193.0 137.3 73.0 0

FOOTING MOMENT & SHEAR AT LONGITUDINAL SECTIONS FOR CASE 30 0.09 L 0.18 L 0.27 L ColL ColR 0.73 L 0.82 L 0.91 L L

0 1.63 3.25 4.88 6.50 11.50 13.13 14.75 16.38 18.00

0 0 0 0 0 1740.8 1839.8 1938.7 2037.7 2136.6

0 0.0 0.0 0.0 0.0 -60.9 -60.9 -60.9 -60.9 -60.9

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

17.99 17.99 17.99 17.99 17.99 17.99 17.99 17.99 17.99 17.99

0 -23.752 -95.007 -213.77 -380.03 -1189.6 -1549.5 -1956.9 -2411.9 -2914.3

0 29.2 58.5 87.7 116.9 206.9 236.1 265.3 294.6 323.8

0.00 0.00 0.00 0.00 0.00 2.10 2.40 2.70 3.00 3.29

0 0 0 0 0 1491.4 1361.3 1189.6 990.32 777.68

0 0.0 0.0 0.0 0.0 -291.0 -297.0 -294.4 -283.0 -262.9

0 -23.8 -95.0 -213.8 -380.0 2042.6 1651.6 1171.4 616.1 0

0 29.2 58.5 87.7 116.9 -145.0 -121.8 -89.9 -49.3 0

DESIGN FLEXURE

Location Mu,max d (in) ρmin ρreqD ρmax smax use ρprovD

Top Longitudinal -380.0 ft-k 15.37 0.0021 0.0017 0.0155 no limit 6 # 10 0.0023Bottom Longitudinal 2042.6 ft-k 14.37 0.0023 0.0118 0.0155 18 29 # 10 @ 7 in o.c. 0.0119

[Satisfactory]CHECK FLEXURE SHEAR


Longitudinal 242.8 k 255 k [Satisfactory]

CHECK FOOTING PUNCHING SHEAR (ACI 318-02 SEC.15.5.2, 11.12.1.2, 11.12.6, & 13.5.3.2)

Case Pu Mu b0 γv βc y Af Ap R J vu (psi) φ vc

1 704.2 0.0 233.6 0.4 1.0 2.0 324.0 23.3 65.6 113.7 27.4 164.32 358.2 1559.9 233.6 0.4 1.0 2.0 324.0 23.3 33.4 113.7 31.0 164.33 -37.0 1559.9 233.6 0.4 1.0 2.0 324.0 23.3 -3.4 113.7 15.6 164.3


Vu,soil (k)




Vu,ftg & fill (k)

qu,soil (ksf)

Mu,soil (ft-k)

Pu,surch (klf)

Mu,surch (ft-k)

Vu,surch (k)

Pu,ftg & fill (klf)

Section


Mu,pedestal (ft-k)

Vu,pedestal (k)



Vu,ftg & fill (k)

qu,soil (ksf)

Mu,soil (ft-k)

Vu,soil (k)

Mu,surch (ft-k)

Vu,surch (k)

Pu,ftg & fill (klf)



Mu,pedestal (ft-k)

Vu,pedestal (k)

Pu,surch (klf)

Section

[ ]

( )

( )

0.5( )

2 30.5

2 3

2

4

d cR MP uu vpsivu JAP

d c dJ d c d

d cPuRA f

γ

π

π

+−= +

+ = + +

+=

00.4

2

dbAP

v

A Lf

γ=

=

=

( )

( )

'( ) 2

42, ,40

0

0

psi yv fc c

dy MIN

bcc db

φφ

β

π

= +

=

= +

(cont'd)CHECK PEDESTAL REINF. LIMITATIONS

ρmax = 0.08 (ACI 318-02, Section 10.9) ρprovd = 0.011ρmin = 0.01 (ACI 318-02, Section 10.9) [Satisfactory]

smax = 3 (ACI 318-02, Section 7.10.4.3) sprovd = 3 insmin = 1 (ACI 318-02, Section 7.10.4.3) [Satisfactory]


JOB NO. : DATE : REVIEW BY : Grade Beam Design Based on ACI 318 & CBC 1633A.2.12

DESIGN CRITERIA1. THE CBC 1633A.2.12, FACTOR Ω0, IS ONLY FOR FOOTING SHEAR AND BENDING CHECK, AND CONNECTION DESIGN.

USING (Ω0/1.4) E TO CHECK SOIL CAPACITY AND OVERTURNING IS ADEQUATE.2. WHEN LATERAL FRAME LOADS ARE NOT BIG, USING SQUIRE PADS WITH GRADE BEAM TIED THEM TOGETHER

WILL COST LESS THAN COMBINED FOOTING.THE SQUIRE PADS TAKE VERTICAL LOAD TO CENTRIC SOIL PRESSURE,WHILE THE BEAM BALANCED MOMENTS AND NEGLECTING SOIL PRESSURE ON THE BEAM. SEE THIS SHEET BELOW.

3. WHEN LATERAL FRAME LOADS ARE BIG, USING COMBINED FOOTING IS EFFICIENT. SEE SHEETS OF COMBINED FOOTING & COMBINED FOOTING OPTION FOR MORE INFORMATION.

INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 3 ksi

REBAR YIELD STRESS fy = 60 ksiSQUIRE PAD SIZE B = 7 ft

T = 12 inGRADE BEAM SIZE W = 36 in

D = 36 inCOLUMN DISTANCE L = 22 ftGRADE BEAM EXTENSION Le = 8 ft

FRAME AXIAL LOADS, ASD PD,1 = 25 kips PD,2 = 25 kips (Dead Load)PL,1 = 15 kips PL,2 = 15 kips (Live Load)

SEISMIC AXIAL LOADS, SD PE,1 = -15 kips PE,2 = 30 kips (Seismic Load)

SEISMIC SHEAR LOADS, SD VE,1 = 35 kips VE,2 = 35 kips (Seismic Load)

SEISMIC MOMENTS, SD ME,1 = 50 ft-kips ME,2 = 50 ft-kips (Seismic Load)

SEISMIC AMPLIFICATION Ω0 = 2.2 , (CBC 1633A.2.12)

ALLOW SOIL PRESSURE Qa = 2.5 ksfPAD REINFORCING 10 # 8 @ 8 o.c., Each Way, Bottom.

GRADE BEAM LONGITUDINAL REINFORCINGTOP 7 # 8

( d = 31.88 in )( 1 Layer)

BOTTOM 7 # 7( d = 31.94 in )

( 1 Layer)GRADE BEAM HOOPS (ACI 21.3.3)

LOCATION AT END AT SPLICELENGTH 72 in 55 in


BAR 4 Legs # 5 4 Legs # 5(Legs to alternate long bars supported, ACI 7.10.5.3)

SPACING @ 7 in o.c. @ 4 in o.c.MIN(d/4, 8db, 24dt, 12) MIN(d/4, 4)

THE GRADE BEAM DESIGN IS ADEQUATE.

ANALYSISCHECK OVERTURNING AT CENTER BOTTOM OF PAD 2

MO = (Ω0/1.4)[ME,1 + ME,2 + (VE,1+VE,2)(D+T) - PE,1L] = 1115.7 ft-kips

MR = (PD,1 + γconc B2T) L + 0.5γconc(L + 2Le) L D W = 1276.0 > MO / 0.9 = 1240 ft-kips [Satisfactory]

CHECK SOIL BEARING CAPACITY

2.03 ksf, (net pressure) < 4/3 Qa [Satisfactory]

where γconc = 0.15 kcf

γsoil = 0.11 kcf

DanielTian Li

( ) ( )2,2 ,2

2 2

0.5D L CONC SOILOMAX


γ γ+ + − + + = + =

(cont'd)

CHECK PAD FLEXURAL REINFORCING

0.0048 <ρprovd = 0.0118

where d = 8.00 inQu,max = 1.5 Qmax = 3.05 ksf, (SD, CBC 1915A. 2.1)

Mu = 0.125 (B-W)2 B Qu,max = 109 ft-kips

ρmax = 0.0160 (ACI 10.2.7.3 & 10.3.3)

ρmin = 0.0018 (ACI 7.12.2.1) [Satisfactory]

CHECK PAD ONE WAY SHEAR CAPACITYVu < φVn [Satisfactory]

where Vu = 0.5 (B - W) B Qu,max - d B Qu,max = 28.4 kips

φVn = φ 2 d B (fc')0.5 = 62.6 kips

φ = 0.85

CHECK GB SECTION REQUIREMENTS (ACI 21.3.1)Pu = Ω0(VE,1 - VE,2) = 0 kips < 0.1Agfc' = 388.8 kips [Satisfactory]

Ln=L - B = 15.00 ft > 4 d = 10.65 ft [Satisfactory]

W / D = 1.00 > 0.3 [Satisfactory]W = 36 in > 10 in [Satisfactory]

< B+1.5D = 138 in [Satisfactory]

CHECK GB FLEXURAL REQUIREMENTS






where Mn,bot = ρbot bd2fy (1 - 0.588ρbot fy/fc') = 642 ft-kips > Mu,bot / φ [Satisfactory]

Mn,top = ρtop bd2fy (1 - 0.588ρtop fy/fc') = 831 ft-kips > Mu,top / φ [Satisfactory]φ = 0.9

-171 ft-kips

-489 ft-kips

where 2.52 ksf, (full ASD pressure)

QMIN = 0.94 ksf, (full ASD pressure)

Factor 1.5 is for SD






Av = 1.24 in2

''2

0.85 1 10.383

uc

c

y

MfB fd

fρ

− −

= =

( )2,2 ,2

2 2

0.5D L CONCOMAX


γ+ + + + = + =

( ) ( )( )0 0 0 02, , ,1 ,1 ,1 ,1,1 ,1 ,1 ,21.5 0.5 / 0.5

1.4 1.4 1.4 1.4u top GB wt D L E EPAD E E EMIN MIN MAX MINL D D TQ Q Q QWt V V VM M P P P B M Ω Ω Ω Ω= + + + + − − − − + + + =

( ) ( )( )0 0 0 02, , ,2 ,2 ,2 ,2,1 ,2 ,1 ,21.5 0.5 / 0.5

1.4 1.4 1.4 1.4u bot GB wt D L E EPAD E E EMAX MAX MAX MINL D D TQ Q Q QWt V V VM M P P P B M

Ω Ω Ω Ω= − − + + + − − − − + + + =



DESIGN CRITERIA1. THE CBC 1633A.2.12, FACTOR Ω0, IS ONLY FOR FOOTING SHEAR AND BENDING CHECK, AND CONNECTION DESIGN.

USING (Ω0/1.4) E TO CHECK SOIL CAPACITY AND OVERTURNING IS ADEQUATE.

2. BASED ON CBC 1915A.2.1, USING ASD SOIL PRESSURE, SECTION FORCES AT (Ω0/1.4) E ABOVE, MULTPLY BY 1.5 TO DESIGNREINFORCEING CONCRETE.

3. IF THE FOOTING TO HAVE TO COMPLY WITH SEISMIC REQUIREMENTS OF CONCRETE FRAME BEAM, ACI 318 SECTION 21,SEE SHEETS OF COMBINED FOOTING OPTION FOR MORE INFORMATION.

INPUT DATA COL#1 COL#2COLUMN WIDTH c1 = 14 14 inCOLUMN DEPTH c2 = 14 14 inAXIAL DEAD LOAD PDL = 50.05 50.91 kAXIAL LIVE LOAD PLL = 19.21 21.92 kSEISMIC AXIAL LOAD, SD PLAT = -170 170 kSEISMIC SHEAR LOAD, SD VLAT = 117.6 117.6 k

SEISMIC MOMENT, SD MLAT = 3.27 3.27 k-ft

SEISMIC AMPLIFICATION FACTOR Ω0 = 2.2 ,(Tab. 16A-N)



ALLOWABLE SOIL PRESSURE Qa = 2 ksf

DISTANCE TO LEFT EDGE L1 = 11.75 ft

DISTANCE BETWEEN COLUMNS S = 18 ftDISTANCE TO RIGHT EDGE L2 = 29 ft

FOOTING WIDTH B = 10.17 ft

FTG EMBEDMENT DEPTH Df = 4.5 ft


SURCHARGE qs = 0 ksf

SOIL WEIGHT ws = 0.11 kcf BAND WIDTH be = 10.17 ft

LONGITUDINAL REINFORCING BAR SIZE # 10 LONG. REINF AT TOP 8 # 10 @ 16 in o.c., cont.TRANSVERSE REINFORCING BAR SIZE # 5 LONG. REINF AT BOTTOM 21 # 10 @ 5 in o.c., cont.

TRANS. REINF. AT BAND WIDTH 30 # 5 @ 4 in o.c., bottomDESIGN SUMMARYFOOTING LENGTH L = 58.75 ftFOOTING WIDTH B = 10.17 ftFOOTING THICKNESS T = 42 in THE FOOTING DESIGN IS ADEQUATE.

ANALYSISDESIGN LOADS (CBC SEC.1612A.3.2 & ACI 318 SEC.9.2.1)

SERVICE LOADS COL # 1 COL # 2 TOTALCASE 1 : DL + LL P = 69 k 73 k 142 k

( e = -8.40 ft, fr CL ftg )CASE 2 : DL + LL + (Ω0 / 1.4) E P = -198 k 340 k 142 k


V = 185 k 185 k 370 kCASE 3 : 0.9 DL + (Ω0 / 1.4) E P = -222 k 313 k 91 k


V = 185 k 185 k 370 k


MR / MO = 2.31003 > F = 1 / (0.9x1.4)

[Satisfactory]Where F = 1 / (0.9x1.4) for seismic, IBC 1605.3.2

Pftg = (0.15 kcf) T B L = 313.68 k, footing weightPsoil = ws (Df - T) B L = 65.72 k, soil weightMR = PDL 1(L - L1) + PDL 2L2 + 0.5 (Pftg + Psoil) L = 14974 k-ftMO = (Ω0 /1.4)[MLAT 1 + MLAT 2 + (VLAT 1 + VLAT 2) Df - PLAT 1(L - L1) - PLAT 2L2] = 6482 k-ft

CHECK SOIL BEARING CAPACITY (ACI 318 SEC.15.2.2)

Service Loads CASE 1 CASE 2 CASE 3P 142.1 142.1 90.9 ke -8.4 25.5 44.5 ft

qs B L 0.0 0.0 0.0 k, (surcharge load)(0.15-ws)T B L 83.6 83.6 75.3 k, (footing increased)

Σ P 225.7 225.7 166.1 ke -5.3 < L/6 16.1 > L/6 24.3 > L/6 ft

qmax 0.2 1.1 2.2 ksfqallow 2.0 2.7 2.7 ksf

[Satisfactory]

DanielTian Li

Combined Footing Design Based on CBC 2001 Chapter A

(cont'd)Where

DESIGN FLEXURE & CHECK FLEXURE SHEAR (ACI 318 SEC.15.4.2, 10.2, 10.5.4, 7.12.2, 12.2, 12.5, 15.5.2, 11.1.3.1, & 11.3)

SERVICE SOIL PRESSURE WITH FOOTING WEIGHT ASD Loads CASE 1 CASE 2 CASE 3

P 142.1 142.1 90.9 ke -8.4 25.5 44.5 ft

qs B L 0.0 0.0 0.0 k, (surcharge load)

[0.15 T + ws (Df - T)] B L 379.4 379.4 341.5 k, (footing & backfill loads)

Σ P 521.5 521.5 432.3 ke -2.3 < L/6 7.0 < L/6 9.3 < L/6 ft

q, max 0.669 1.493 1.414 ksf

FOOTING MOMENT & SHEAR AT LONGITUDINAL SECTIONS FOR CASE 1 , ASD0 0.5 L1 L1 left L1 right 0.2 S 0.4 S 0.6 S 0.8 S L2 left L2 right 0.5 L2 L

0 5.88 11.75 11.75 15.35 18.95 22.55 26.15 29.75 29.75 44.25 58.750 0 0 0 -249 -499 -748 -997 -1,247 -1,247 -3,307 -5,3670 0.0 0.0 69.3 69.3 69.3 69.3 69.3 69.3 142.1 142.1 142.1

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.000 0 0 0 0 0 0 0 0 0 0 00 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

6.46 6.46 6.46 6.46 6.46 6.46 6.46 6.46 6.46 6.46 6.46 6.460 -111 -446 -446 -761 -1160 -1642 -2208 -2858 -2858 -6323 -111450 37.9 75.9 75.9 99.1 122.4 145.6 168.9 192.1 192.1 285.8 379.4

1.08 1.04 1.00 1.00 0.97 0.95 0.92 0.90 0.87 0.87 0.77 0.670 187 737 737 1248 1886 2649 3534 4536 4536 9702 165120 -63.1 -123.8 -123.8 -159.8 -194.8 -229.0 -262.2 -294.5 -294.5 -415.4 -521.5

0 75 291 291 237 228 259 328 432 432 72 00 -25.2 -47.9 21.3 8.6 -3.2 -14.1 -24.1 -33.2 39.7 12.4 0


0 5.88 11.75 11.75 15.35 18.95 22.55 26.15 29.75 29.75 44.25 58.750 0 0 5 718 1,430 2,142 2,855 3,567 3,572 1,512 -5480 0.0 0.0 -197.9 -197.9 -197.9 -197.9 -197.9 -197.9 142.1 142.1 142.1

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.000 0 0 0 0 0 0 0 0 0 0 00 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

6.46 6.46 6.46 6.46 6.46 6.46 6.46 6.46 6.46 6.46 6.46 6.460 -111 -446 -446 -761 -1160 -1642 -2208 -2858 -2858 -6323 -111450 37.9 75.9 75.9 99.1 122.4 145.6 168.9 192.1 192.1 285.8 379.4

0.25 0.38 0.50 0.50 0.58 0.65 0.73 0.80 0.88 0.88 1.19 1.490 52 236 236 433 706 1065 1520 2081 2081 5618 116930 -18.8 -45.1 -45.1 -64.8 -87.3 -112.6 -140.7 -171.5 -171.5 -324.0 -521.5

0 -60 -210 -205 389 976 1565 2166 2790 2795 808 00 19.1 30.8 -167.1 -163.5 -162.8 -164.9 -169.7 -177.3 162.7 103.9 0

Section

X (ft)

P,surch (klf)M,surch (ft-k)

P,ftg & fill (klf)M,ftg & fill (ft-k)

V,soil (k)

q,soil (ksf)M,soil (ft-k)

M,col (ft-k)V,col (k)

V,surch (k)

V,ftg & fill (k)

Σ Σ Σ Σ M (ft-k)Σ Σ Σ Σ V (kips)

Section

X (ft)M,col (ft-k)

V,col (k)P,surch (klf)

M,surch (ft-k)V,surch (k)

P,ftg & fill (klf)M,ftg & fill (ft-k)

Σ Σ Σ Σ M (ft-k)Σ Σ Σ Σ V (kips)

V,ftg & fill (k)q,soil (ksf)M,soil (ft-k)

V,soil (k)

''2

0.85 1 10.383

M uf c b fd cf y

ρ

− − =

'10.85 87

0.7587MAX

f cf fy y

βρ

= +

0.0018MIN

Td

ρ =

( )

( )

61

,6

2,

3 (0.5 ) 6

MAX

eP

LL for eBLq

P Lfor e

B L e

Σ + ≤=

Σ>

−

( )

( )

61

,6,

2,

3 (0.5 ) 6

MAX

euPu LLfor euBLqu

LPu for euB L eu

Σ + ≤=

Σ >

−

0

100

200

300

400

500

M

-60-40-20

0204060

V

-1000

0

1000

2000

3000

M

(cont'd)


0 5.88 11.75 11.75 15.35 18.95 22.55 26.15 29.75 29.75 44.25 58.750 0 0 5 805 1604 2404 3203 4003 4,008 2,691 1,3730 0.0 0.0 -222.1 -222.1 -222.1 -222.1 -222.1 -222.1 90.9 90.9 90.9

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.000 0 0 0 0 0 0 0 0 0 0 00 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

5.81 5.81 5.81 5.81 5.81 5.81 5.81 5.81 5.81 5.81 5.81 5.810 -100 -401 -401 -685 -1044 -1478 -1987 -2572 -2572 -5690 -100310 34.1 68.3 68.3 89.2 110.1 131.1 152.0 172.9 172.9 257.2 341.5

0.03 0.17 0.31 0.31 0.39 0.48 0.56 0.65 0.73 0.73 1.07 1.410 14 88 88 183 331 542 826 1197 1197 3779 86580 -6.1 -20.4 -20.4 -33.3 -49.2 -68.3 -90.5 -115.7 -115.7 -248.9 -432.3

0 -86 -314 -308 303 892 1468 2043 2628 2633 780 00 28.1 47.9 -174.2 -166.2 -161.2 -159.3 -160.6 -164.9 148.0 99.2 0

DESIGN FLEXURE, Mu = 1.5 M, (CBC 1915A.2.1 )

Location Mu,max d (in) ρmin ρreqD ρmax smax(in) use ρprovD

Top Longitudinal -470 ft-k 39.37 0.0019 0.0006 0.0160 no limit 8 # 10 @ 16 in o.c., cont. 0.0021Bottom Longitudinal 4193 ft-k 38.37 0.0020 0.0055 0.0160 18 21 # 10 @ 5 in o.c., cont. 0.0057

Bottom Transverse, be 5 ft-k / ft 37.42 0.0020 6.1E-05 0.0160 18 30 # 5 @ 4 in o.c. 0.0020

[Satisfactory]CHECK FLEXURE SHEAR, Vu = 1.5 M, (CBC 1915A.2.1 )


Longitudinal 266 k 436 k [Satisfactory]Transverse 3 k / ft 42 k / ft [Satisfactory]

CHECK PUNCHING SHEAR (ACI 318 SEC.15.5.2, 11.12.1.2, 11.12.6, & 13.5.3.2)

Column Case Pu Mu b1 b2 γv βc y Af Ap R J vu (psi) φ vc

1 103.9 0.0 51.4 51.4 0.4 1.0 2.0 103.4 53.4 18.4 185.2 11.1 186.2Col. 1 2 0.0 7.7 51.4 51.4 0.4 1.0 2.0 103.4 53.4 0.0 185.2 0.2 186.2

3 0.0 7.7 51.4 51.4 0.4 1.0 2.0 103.4 53.4 0.0 185.2 0.2 186.21 109.2 0.0 51.4 51.4 0.4 1.0 2.0 103.4 53.4 19.4 185.2 11.7 186.2

Col. 2 2 510.0 7.7 51.4 51.4 0.4 1.0 2.0 103.4 53.4 90.5 185.2 54.8 186.23 469.4 7.7 51.4 51.4 0.4 1.0 2.0 103.4 53.4 83.3 185.2 50.4 186.2

[Satisfactory]

V,soil (k)

Σ Σ Σ Σ M (ft-k)

P,surch (klf)M,surch (ft-k)V,surch (k)

P,ftg & fill (klf)

Section

X (ft)

Σ Σ Σ Σ V (kips)

M,ftg & fill (ft-k)V,ftg & fill (k)q,soil (ksf)M,soil (ft-k)

M,col (ft-k)V,col (k)

0.5 1( )

231 21 3

6 1 1

1 2

R bMP uu vpsivu JAP

db d bJb b

b bPuRA f

γ−= +

= + +

=

( )2 1 21

12 113 2

db bAP

v bb

BbA f e

γ

= +

= −+

=

( )

( ) ( )

'( ) 2

42, , 40

0

, ,0 1 1 2 2

psi yv fc c

dy MIN

bc

AP d db b c b cd

φφ

β

= +

=

= = + = +

-200

-100

0

100

200

V

-500

0

500

1000

1500

2000

2500

3000

M

-200

-100

0

100

200

V


JOB NO. : DATE : REVIEW BY : Seismic Design for Combined Footing Based on ACI 318

DESIGN SUMMARY


REBAR YIELD STRESS fy = 60 ksiFOOTING WIDTH W = 122.04 inFOOTING THICKNESS D = 42 inDISTANCE BETWEEN COLUMNS L = 18 ft

COMBINED FOOTING LONGITUDINAL REINFORCINGTOP 14 # 10

( d = 37.74 in )( 1 Layer)

BOTTOM 21 # 10( d = 37.74 in )

( 1 Layer)

COMBINED FOOTING HOOPS (ACI 21.3.3)LOCATION AT END AT SPLICELENGTH 84 in 70 in


BAR 11 Legs # 5 11 Legs # 5SPACING @ 9 in o.c. @ 4 in o.c.

MIN(d/4, 8db, 24dt, 12) MIN(d/4, 4)

THE SEISMIC DESIGN IS ADEQUATE.

ANALYSISCHECK GB SECTION REQUIREMENTS (ACI 21.3.1)

Ln=L - c1 = 16.83 ft > 4 d = 12.58 ft [Satisfactory]

W / D = 2.91 > 0.3 [Satisfactory]W = 122.04 in > 10 in [Satisfactory]

CHECK SEISMIC FLEXURAL REQUIREMENTS(ACI 21.3.2.1) ρtop = 0.004 > ρmin=MIN[3(fc')

0.5/fy, 200/fy ]= 0.003 [Satisfactory]





where Mn,bot = ρbot bd2fy (1 - 0.588ρbot fy/fc') = 4690 ft-kips

Mn,top = ρtop bd2fy (1 - 0.588ρtop fy/fc') = 3203 ft-kips






Av = 3.41 in2

DanielTian Li



Footing Design Based on CBC 1633A.2.12

CBC 1633A.2.12 Foundations and superstructure-to-foundation connectionsThe foundation shall be capable of transmitting the design base shear and the overturning forces from the structure into the supporting soil.The In addition, the foundation and the connection of the superstructure elements to the foundation shall have the strength to resist,

in addition to gravity loads, the lesser of the following seismic loads:1. The strength of the superstructure elements.2. The maximum forces that would occur in the fully yielded structural system.3. o times the forces in the superstructure elements due to the seismic forces as prescribed in this chapter.

EXCEPTIONS:1. Where structures are designed using R 2.2 such as for inverted pendulum-type structures.2. When it can be demonstrated that inelastic deformation of the foundation and superstructure-tofoundation connection will not result

in a weak story or cause collapse of the structure.3. Where the basic structural system as described in Table 16A-N consists of light-framed walls with shear panels.

Where moment resistance is assumed at the base of the superstructure elements, the rotation and flexural deformation of the foundation aswell as deformation of the superstructure-to-foundation connection shall be considered in the drift and deformation compatibility analyses.

NOTETHERE IS A STATEWIDE CONSENSUS THAT THIS SECTION WILL ONLY BE APPLICABLE TO MEMENT FRAMES (SMRF AND OMRF),ECCENTRIC BRACED FRAMES (EBF), AND HEAVILY LOADED CONCENTRIC BRACED FRAMES (OCBF AND SCBF). THIS SECTIONWILL NOT BE APPLICABLE TO WOOD SHEAR WALL OR CONCRETE OR MASONRY SHEAR WALLS EXCEPT WHERE THESE WALLSARE VERY HEAVILY LOADED. FOR WOOD SHEAR WALLS, VERY HEAVILY LOADED IS DEFINED TO BE OVER 600 PLF OF SHEARIN A WALL.

ALTERNATE DESIGN METHODIF USING THE SOFTWARE STILL GET UNACCEPTABLE BIG FOOTING SIZE, JUST DESIGN THE CONNECTIONS BETWEEN DRAG ANDLATERAL FRAME WITHOUT OVER STRENGTH, OR DESIGN DIAPHRAGM SHEAR CAPACITY AT THE LATERAL FRAME SIDES ASSMALLER AS POSSIBLE. THEN DEVELOP THE SUPERSTRECTURE STRENGTH TO FOUNDATION LOADS. USE REGULAR SOFTWAREAT www.engineering-international.com TO DESIGN THE FOOTING. (CBC 1633A.2.12 Item 1 & 2)

30

2

DanielTian Li


JOB NO. : REVIEW BY :

INPUT DATA & DESIGN SUMMARYFOUNDATION LENGTH L = 64 ftFOUNDATION WIDTH B = 30 ftGRID TRIBUTARY AREA A = 1 ft2

MODULUS OF SUBGRADE K1 = 100 lb / in3

(Obtained from the soil report for 1' x 1' sf plate load test, in the absence of a soil report obtain from table below )

INSIDE SPRING VALUE 5.3 kips / jointEDGE SPRING VALUE 2.6 kips / jointCORNER SPRING VALUE 1.3 kips / joint

ANALYSIS

( for B = L )= 36.8 lb / in3

( for B < L )

= 26.7 lb / in3 = 46 k / ft3

= 36.8 lb / in3 = 64 k / ft3

TYPICAL VALUES OF MODULUS OF SUBGRADE REACTIONS, K1 (lb / in3 )MOISTURE CONTENT

TYPE OF MATERIAL 1 to 4% 5 to 8% 9 to 12% 13 to 16% 17 to 20% 21 to 24% 25 to 28% > 28%

Silts and clays (liquid limit >50) - 175 150 125 100 75 50 25 (OH, CH, MH )

Silts and clays (liquid limit <50) - 200 175 150 125 100 75 50 (OL, CL, ML )

Silty and clay sands 300 250 225 200 150 - - - (SM & SH )

Gravelly sands Over 300 300 250 - - - - - (SW & SP )

Silty and clayey gravels Over 300 Over 300 300 - - - - - (GM & GC )

Gravel and sand Over 300 Over 300 - - - - - -

Mat Boundary Spring GeneratorDATE :

DanielT. Li

2

11 1B+1

=k k2B

2

12 1

0.5L+1B+1 B=k k

2B 1.5

11s

12

k=kk




BASE PLATE DEPTH b2 = 4 in

FOOTING CONCRETE STRENGTH fc' = 2.5 ksiREBAR YIELD STRESS fy = 60 ksiAXIAL DEAD LOAD PDL = 2 k


LATERAL LOAD (0=WIND, 1=SEISMIC) = 1 Seismic,SDSEISMIC AXIAL LOAD PLAT = 6.5 k, SD

SURCHARGE qs = 0 ksfSOIL WEIGHT ws = 0.11 kcfFOOTING EMBEDMENT DEPTH Df = 0.50 ftFOOTING THICKNESS T = 5 inALLOW SOIL PRESSURE Qa = 1 ksfFOOTING WIDTH B = 3 ftFOOTING LENGTH L = 3 ft

THE FOOTING DESIGN IS ADEQUATE.

ANALYSISDESIGN LOADS (IBC 2000 SEC.1605.3.2 & ACI 318-02 SEC.9.2.1)CASE 1: DL + LL P = 7 k 1.2 DL + 1.6 LL Pu = 10 kCASE 2: DL + LL + E / 1.4 P = 11 k 1.2 DL + 1.0 LL + 1.0 E Pu = 13 kCASE 3: 0.9 DL + E / 1.4 P = 7 k 0.9 DL + 1.0 E Pu = 8 k

CHECK SOIL BEARING CAPACITY (ACI 318-02 SEC.15.2.2)CASE 1 CASE 2 CASE 3

= 0.74 ksf, 1.25 ksf, 0.73 ksf

q MAX < k Q a , [Satisfactory]

where k = 1 for grovity loads, 4/3 for lateral loads.

DESIGN FOR FLEXURE (ACI 318-02 SEC.22.5.1)

= 1.72 ft-kips

where φ = 0.55 (ACI 318-02, Section 9.3.5 )S = elastic section modulus of section = 150 in3

= 1.71 ft-kips < φ M n [Satisfactory]

CHECK FLEXURE SHEAR (ACI 318-02 SEC.22.5.4)

= 6.60 kips

where φ = 0.55 (ACI 318-02, Section 9.3.5 )

= 3.91 kips < φ V n [Satisfactory]

CHECK PUNCHING SHEAR (ACI 318-02 SEC.22.5.4)

= 13.53 kips

where φ = 0.55 (ACI 318-02, Section 9.3.5 )βc = ratio of long side to short side of concentrated load = 1.43

= 12.52 ft-kips < φ V n [Satisfactory]

Plain Concrete Footing Design Based on ACI 318-02

MAX

Pq

BL=

( )' '5 , 0.85n c cMIN S Sf fMφ φ φ=

( )2,max0.5 0.25 0.251 1

2u

u

L Tb c PM

L

− − −=

'43n c BTfVφ φ=

( ) ,max1 10.5 0.25 0.25 u

uPL TV b c

L= − − −

( )'1 2 1 2

4 8, 2.66 4

3 3n cc

MIN T TfV c c b bφ φβ

= + + + + +

1 1 2 2,ma x

11

2 2uub c b cT TV P

BL + + = − + +



1. DESIGN METHODS1.1 DIVIDE AN IRREGULAR FOUNDATION PLAN INTO OVERLAPPING RECTANGLES AND USING

THIS SPREADSHEET DESIGN EACH RECTANGULAR SECTION SEPARATELY.1.2 THE POST-TENSION INSTITUTE (PTI) METHOD IS ACCEPTABLE FOR THE DESIGN OF

NONPRESTRESSED SLAB ON GRADE (ACI 360, 2.4). THE DESIGNER MAY SELECT EITHERNONPRESTRESSED REINFORCENENT USING THIS SPREADSHEET, OR POST-TENSIONED REINFORCEMENT IF REQUIRED (ACI 360, 8.6).

2. INPUT DATA & DESIGN SUMMARY 2.1 SOILS PROPERITIES

ALLOWABLE SOIL-BEARING PRESSURE qallow = 2000 psf

EDGE MOISTURE VARIATION DISTANCE em = 4 ft, for center lift

= 4.5 ft, for edge lift

DIFFERENTIAL SOIL MOVEMENT ym = 2.68 in, for center lift

= 0.3 in, for edge lift 2.2 STRUCTURAL DATA AND MATERIALS PROPERITIES

SLAB LENGTH L = 164 ftSLAB WIDTH B = 125 ftSLAB THICHNESS t = 5 inPERIMETER LOADING P = 270 plf

MAX BEARING LOADING ON THE SLAB Pb = 270 plf

ADDED DEAD LOAD DL = 50 psfLIVE LOAD LL = 125 psf

AVERAGE STIFFENING BEAM SPACING, L DIRECTION SL = 30 ft

AVERAGE STIFFENING BEAM SPACING, B DIRECTION SB = 30 ft THE DESIGN IS ADEQUATE.

STIFFENING BEAM DEPTH h = 24 inSTIFFENING BEAM WIDTH b = 20 in

CONCRETE STRENGTH f'c = 3 ksi

SLAB REINFORCEMENT # 4 @ 18 in o.c., with 1.5 in clear from top of slab, each way.

3. ASSUME A TRIAL SECTION 3.1 ASSUME BEAM DEPTH AND SPACING

ALLOWABLE DIFFERENTIAL DEFLECTION, FOR CENTER LIFT, AT L DIRECTION ALLOWABLE DIFFERENTIAL DEFLECTION, FOR CENTER LIFT, AT B DIRECTION∆allow = 12 MIN(L, 6β) / C∆ = 1.60 in ∆allow = 12 MIN(B, 6β) / C∆ = 1.60 in

Where β = 8 ft Where β = 8 ftC∆ = 360 C∆ = 360

ALLOWABLE DIFFERENTIAL DEFLECTION, FOR EDGE LIFT, AT L DIRECTION ALLOWABLE DIFFERENTIAL DEFLECTION, FOR EDGE LIFT, AT B DIRECTION∆allow = 12 MIN(L, 6β) / C∆ = 0.80 in ∆allow = 12 MIN(B, 6β) / C∆ = 0.80 in


BEAM DEPTH, FOR CENTER LIFT, AT L DIRECTION BEAM DEPTH, FOR CENTER LIFT, AT B DIRECTION

h = [(ym L)0.205 SB1.059 P0.523 em

1.296 / 380 ∆allow ]0.824 = 13.56 in h = [(ym B)0.205 SL1.059 P0.523 em

1.296 / 380 ∆allow ]0.824 = 12.95 in

BEAM DEPTH, FOR EDGE LIFT, AT L DIRECTION BEAM DEPTH, FOR EDGE LIFT, AT B DIRECTION

h = [L0.35 SB0.88 em

0.74 ym0.76 / 15.9 ∆allow P0.01]1.176 = 8.47 in h = [B0.35 SL

0.88 em0.74 ym

0.76 / 15.9 ∆allow P0.01]1.176 = 7.58 in

GOVERNING h = 13.56 in < ACTUAL h = 24.00 in [Satisfactory]

3.2 DETERMINE SECTION PROPERTIESL DIRECTION B DIRECTION

As = 17 in2n = 6 beams As = 22 in2

n = 7 beams

Es / Ec = 9.29 yb = 18.75 in 2.68 Es / Ec = 9.29 yb = 19.00 in

CGS = 21.75 in St = 64268 in3CGS = 22.25 in St = 80834 in3

A = 9935 in2Sb = 17995 in3

A = 12703 in2Sb = 21276 in3

I = 337410 in4 I = 404232 in4

4. CALCULATE MAXIMUM APPLIED SERVICE MOMENTS 4.1 CENTER LIFT MOMENT AT L DIRECTION CENTER LIFT MOMENT AT B DIRECTION

ML = A0 (B em1.238 + C) = 4.96 ft-kips / ft MB = (58 + em) ML / 60, for L /B > 1.1

Where A0 = (L0.013 SB0.306 h0.688 P0.534 ym

0.193) / 727 = 0.891 MB = ML, for L /B < 1.1

B = 1, for em < 5

B = MIN[(ym - 1) / 3, 1], for em > 5

C = 0, for em < 5

C = MAX[8 - (P - 613) / 255] (4 - ym) / 3], 0, for em > 5

4.2 EDGE LIFT MOMENT AT L DIRECTION EDGE LIFT MOMENT AT B DIRECTION

ML = SB0.10 (h em)0.78 ym

0.66 / (7.2 L0.0065 P0.04) = 2.63 ft-kips / ft MB = h0.35 (19 + em) ML / 57.75, for L /B > 1.1

MB = ML, for L /B < 1.1

5. CHECK FLEXURAL CONCRETE STRESSES 5.1 ALLOWABLE CONCRETE STRESSES

FLEXURAL TENSILE STRESS ft,allow = - 6 (fc')0.5 = -0.329 ksi

FLEXURAL COMPRESSIVE STRESS fc,allow = - 0.45 fc' = 1.350 ksi

5.2 TOP STRESS, FOR CENTER LIFT MOMENT, AT L DIRECTION TOP STRESS, FOR CENTER LIFT MOMENT, AT B DIRECTION

f = - ML / St = -0.116 ksi f = - MB / St = -0.125 ksi

> ft,allow [Satisfactory] > ft,allow [Satisfactory]

< fc,allow [Satisfactory] < fc,allow [Satisfactory]

Tian LiDaniel

= 1.00

= 0.00

Design of Conventional Slabs on Expansive Soil Grade Based on ACI 360

Then f Then f

ft-kips / ft

ft-kips / ft

= 5.12

= 3.25

5.3 BOTTOM STRESS, FOR CENTER LIFT MOMENT, AT L DIRECTION BOTTOM STRESS, FOR CENTER LIFT MOMENT, AT B DIRECTION

f = ML / Sb = 0.413 ksi f = MB / Sb = 0.474 ksi



5.4 TOP STRESS, FOR EDGE LIFT MOMENT, AT L DIRECTION TOP STRESS, FOR EDGE LIFT MOMENT, AT B DIRECTION

f = - ML / Sb = -0.219 ksi f = - MB / Sb = -0.301 ksi



5.5 BOTTOM STRESS, FOR EDGE LIFT MOMENT, AT L DIRECTION BOTTOM STRESS, FOR EDGE LIFT MOMENT, AT B DIRECTION

f = ML / St = 0.061 ksi f = MB / St = 0.079 ksi



6. CHECK DIFFERENTIAL DEFLECTIONS 6.1 RELATIVE STIFFNESS LENGTH AT L DIRECTION RELATIVE STIFFNESS LENGTH AT B DIRECTION

β = (Ec I / Es)1/4 / 12 = 12.624 ft β = (Ec I / Es)1/4 / 12 = 13.208 ft

Where Ec = (0.5) 57000 (fc')0.5 = 1561009 psi Where Ec = (0.5) 57000 (fc')0.5 = 1561009 psi

Es = 1000 psi, soil Es = 1000 psi, soil

6.2 ALLOWABLE DIFFERENTIAL DEFLECTION AT L DIRECTION ALLOWABLE DIFFERENTIAL DEFLECTION AT B DIRECTIONFOR CENTER LIFT FOR CENTER LIFT∆allow = 12 MIN(L, 6β) / C∆ = 2.64 in ∆allow = 12 MIN(B, 6β) / C∆ = 2.52 in

Where C∆ = 360 Where C∆ = 360

FOR EDGE LIFT FOR EDGE LIFT∆allow = 12 MIN(L, 6β) / C∆ = 1.32 in ∆allow = 12 MIN(B, 6β) / C∆ = 1.26 in


6.3 EXPECTED DIFFERENTIAL DEFLECTION WITHOUT PRESTRESSING EXPECTED DIFFERENTIAL DEFLECTION WITHOUT PRESTRESSINGFOR CENTER LIFT, AT L DIRECTION FOR CENTER LIFT, AT B DIRECTION

∆0 = (ym L)0.205 SB1.059 P0.523 em

1.296 / (380 h1.214) = 0.80 in ∆0 = (ym B)0.205 SL1.059 P0.523 em

1.296 / (380 h1.214) = 0.76 in

< ∆allow < ∆allow

[Satisfactory] [Satisfactory]FOR EDGE LIFT, AT L DIRECTION FOR EDGE LIFT, AT B DIRECTION

∆0 = L0.35 ym0.76 SB

0.88 em0.74 / (15.9 h0.85 P0.01) = 0.58 in ∆0 = B0.35 ym

0.76 SL0.88 em

0.74 / (15.9 h0.85 P0.01) = 0.53 in


[Satisfactory] [Satisfactory]

7. CHECK SHEAR CAPACITY 7.1 APPLIED SERVICE LOAD SHEAR AT L DIRECTION APPLIED SERVICE LOAD SHEAR AT B DIRECTION

FOR CENTER LIFT FOR CENTER LIFT

VL = L0.09 SB0.71 h0.43 P0.44 ym

0.16 em0.93 / 1940 = 1.786 kips/ft VB = B0.19 SL

0.45 h0.20 P0.54 ym0.04 em

0.97 / 1350 = 1.327 kips/ft

FOR EDGE LIFT FOR EDGE LIFT

VL = L0.07 h0.4 P0.03 ym0.67 em

0.16 / (3.0 SB0.015) = 1.084 kips/ft VB = B0.07 h0.4 P0.03 ym

0.67 em0.16 / (3.0 SL

0.015) = 1.063 kips/ft

7.2 ALLOWABLE CONCRETE SHEAR STRESS, AT L DIRECTION ALLOWABLE CONCRETE SHEAR STRESS, AT B DIRECTION

vc = 2 (fc')0.5 = 0.110 ksi vc = 2 (fc')0.5 = 0.110 ksi

7.3 SHEAR STRESS OF RIBBED FOUNDATION, AT L DIRECTION SHEAR STRESS OF RIBBED FOUNDATION, AT B DIRECTIONFOR CENTER LIFT FOR CENTER LIFT

v = V B / (n h b) = 0.078 ksi < vc v = V L / (n h b) = 0.065 ksi < vc

[Satisfactory] [Satisfactory]FOR EDGE LIFT FOR EDGE LIFT



8. CHECK SOIL BEARING 8.1 APPLIED LOADING

SLAB WEIGHT 150 L B t = 1281250 lbs BEAM BEARING AREA (b)(Total Length) = 3028.333333 ft2

ADDED DL DL L B = 1025000 lbs SOIL PRESSURE q = Total Load / THE AREA = 1893 psf

LIVE LOAD LL L B = 2562500 lbs < qallow

BEAM WEIGHT 150 (h-t) b (Total Length) = 708146 lbs [Satisfactory]PERIMETER LOAD P (2L + 2B) = 156060 lbs

9. CHECK SLAB STRESS DUE TO LOAD-BEARING PARTITIONS 9.1 RELATIVE STIFFNESS LENGTH AT L DIRECTION RELATIVE STIFFNESS LENGTH AT B DIRECTION

Mmax = Pb β / 4 = 2.03 ft-kips / ft Mmax = Pb β / 4 = 2.03 ft-kips / ft

Where β = MIN[(Ec t3 / 3 ks)0.25, SB] = 30 ft Where β = MIN[(Ec t3 / 3 ks)0.25, SL] = 30 ft

ks = 4 lb / in3ks = 4 lb / in3

9.2 TENSILE STRESS AT L DIRECTION TENSILE STRESS AT B DIRECTION

f = - Mmax / 2 t2 = -0.041 ksi f = - Mmax / 2 t2 = -0.041 ksi


Techincal References: 1. "Design of Slabs on Grade, ACI Committee 360", American Concrete Institute, 1997. 2. "Design and Construction of Post-Tensioned Slab-on-Ground, Second Edition", The Post-Tensioning Institute, 2004. 3. "1997 Uniform Building Code, Volume 2", International Conference of Building Officials, 1997.

Then f Then f

Then f

Then f Then f

Then f





EXPECTED SETTLEMENT BY GEOTECHICAL ENR δ = 0.75 in 1.2 STRUCTURAL DATA AND MATERIALS PROPERITIES

SLAB LENGTH L = 164 ftSLAB WIDTH B = 125 ftSLAB THICHNESS t = 5 inPERIMETER LOADING P = 270 plf




AVERAGE STIFFENING BEAM SPACING, B DIRECTION SB = 30 ft



SLAB REINFORCEMENT # 4 @ 18 in o.c., with 1.5 in clear from top of slab, each way.

THE DESIGN IS ADEQUATE.2. DETERMINE SECTION PROPERTIES

L DIRECTION B DIRECTIONn = 6 I = 338782 in4 n = 7 I = 406345 in4

As = 17 in2yb = 18.80 in As = 22 in2

yb = 19.05 in

Es / Ec = 18.58 St = 65100 in3Es / Ec = 18.58 St = 82096 in3

CGS = 21.75 in Sb = 18024 in3CGS = 22.25 in Sb = 21330 in3

A = 10090 in2 A = 12906 in2

3. CALCULATE MAXIMUM APPLIED SERVICE MOMENTSL DIRECTION B DIRECTION

McsL = (δ / ∆nsL)0.5 MnsL = 0.83 ft-kips / ft McsS = McsL (970-h) / 880 = 0.90 ft-kips / ft

Where MnsL = h1.35 SB0.36 / 80 L0.12 P0.10 = 0.96 ft-kips / ft

∆nsL = L1.28 SB0.80 / 133 h0.28 P0.62 = 1.00 ∆nsL = L1.28 SB

0.80 / 133 h0.28 P0.62 = 0.70





f = ML / St = 0.019 ksi f = MB / St = 0.021 ksi




f = - ML / Sb = -0.069 ksi f = - MB / Sb = -0.083 ksi




β = (Ec I ∆nsL / Es δ )1/4 / 12 = 13.571 ft β = (Ec I ∆nsB / Es δ )

1/4 / 12 = 13.020 ft


Es = 1000 psi, soil Es = 1000 psi, soil

I = 338782 in4 I = 406345 in4

5.2 ALLOWABLE DIFFERENTIAL DEFLECTION AT L DIRECTION ALLOWABLE DIFFERENTIAL DEFLECTION AT B DIRECTION∆allow = 12 MIN(L, 6β) / C∆ = 0.51 in ∆allow = 12 MIN(B, 6β) / C∆ = 0.49 in

Where C∆ = 1920 Where C∆ = 1920

5.3 DIFFERENTIAL DEFLECTION WITHOUT PRESTRESSING DIFFERENTIAL DEFLECTION WITHOUT PRESTRESSING

∆cs = δ en01.78 - 0.103 h -1.65E-03 P + 3.95E-07 P P) = 0.25 in ∆cs = δ en

01.78 - 0.103 h -1.65E-03 P + 3.95E-07 P P) = 0.25 in




VcsL = (δ / ∆nsL)0.30 VnsL = 0.260 kips/ft VcsS = VcsL (116-h) / 94 = 0.255 kips/ft

Where VnsL = h0.90 (PSB)0.30 / 550 L0.10 = 0.284 kips/ft


vc = 2 (fc')0.5 = 0.110 ksi vc = 2 (fc')0.5 = 0.110 ksi

6.3 SHEAR STRESS OF RIBBED FOUNDATION, AT L DIRECTION SHEAR STRESS OF RIBBED FOUNDATION, AT B DIRECTION



Tian LiDaniel

Design of Conventional Slabs on Compressible Soil Grade Based on ACI 360

Then f Then f

Then f Then f

(cont'd)7. CHECK SOIL BEARING 7.1 APPLIED LOADING

SLAB WEIGHT 150 L B t = 1281250 lbs BEAM BEARING AREA (b)(Total Length) = 2982 ft2






Where β = MIN[(Ec h3 / 3 ks)0.25, SB] = 30.000 ft Where β = MIN[(Ec h3 / 3 ks)0.25, SB] = 30.000 ft



f = - Mmax / 2 t2 = -0.041 ksi f = - Mmax / 2 t2 = -0.041 ksi




1. DESIGN METHODS1.1 DIVIDE AN IRREGULAR FOUNDATION PLAN INTO OVERLAPPING RECTANGLES AND USING

THIS SPREADSHEET DESIGN EACH RECTANGULAR SECTION SEPARATELY.



EDGE MOISTURE VARIATION DISTANCE em = 5.5 ft, for center lift

= 2.5 ft, for edge lift

DIFFERENTIAL SOIL MOVEMENT ym = 3.608 in, for center lift

= 0.752 in, for edge liftSLAB-SUBGRADE FRICTION COEFFICIENT µ = 0.75

2.2 STRUCTURAL DATA AND MATERIALS PROPERITIESSLAB LENGTH L = 42 ftSLAB WIDTH B = 24 ftSLAB THICHNESS t = 4 inPERIMETER LOADING P = 1040 plf




AVERAGE STIFFENING BEAM SPACING, B DIRECTION SB = 12 ft

STIFFENING BEAM DEPTH h = 24 inSTIFFENING BEAM WIDTH b = 10 in THE DESIGN IS ADEQUATE.

CONCRETE STRENGTH f'c = 2.5 ksi SUGGESTED RATIO OF EXPECTED ELONGATION IS 0.00777

SLAB PRESTRESSING TENDONS, L DIRECTION 5 tendons w/ 0.153 in2 at each tendon. CONVERTED UNIFORM THICKNESS IS 6.48 inchSLAB PRESTRESSING TENDONS, B DIRECTION 7 tendons w/ 0.153 in2 at each tendon.

TENDON IN THE BOTTOM OF EACH BEAM 0 tendons w/ 0 in2 (only for edge lift governing required)EFFECTIVE PRESTRESS AFTER ALL LOSSES EXCEPT SG fe = 174 ksi

3. ASSUME A TRIAL SECTION 3.1 ASSUME BEAM DEPTH AND SPACING

ALLOWABLE DIFFERENTIAL DEFLECTION, FOR CENTER LIFT, AT L DIRECTION ALLOWABLE DIFFERENTIAL DEFLECTION, FOR CENTER LIFT, AT B DIRECTION∆allow = 12 MIN(L, 6β) / C∆ = 1.40 in ∆allow = 12 MIN(B, 6β) / C∆ = 0.80 in

Where β = 8 ft Where β = 8 ftC∆ = 360 2.5 C∆ = 360

ALLOWABLE DIFFERENTIAL DEFLECTION, FOR EDGE LIFT, AT L DIRECTION ALLOWABLE DIFFERENTIAL DEFLECTION, FOR EDGE LIFT, AT B DIRECTION∆allow = 12 MIN(L, 6β) / C∆ = 0.70 in ∆allow = 12 MIN(B, 6β) / C∆ = 0.40 in


BEAM DEPTH, FOR CENTER LIFT, AT L DIRECTION BEAM DEPTH, FOR CENTER LIFT, AT B DIRECTION

h = [(ym L)0.205 SB1.059 P0.523 em

1.296 / 380 ∆allow ]0.824 = 14.28 in h = [(ym B)0.205 SL1.059 P0.523 em

1.296 / 380 ∆allow ]0.824 = 23.57 in

BEAM DEPTH, FOR EDGE LIFT, AT L DIRECTION BEAM DEPTH, FOR EDGE LIFT, AT B DIRECTION

h = [L0.35 SB0.88 em

0.74 ym0.76 / 15.9 ∆allow P0.01]1.176 = 2.51 in h = [B0.35 SL

0.88 em0.74 ym

0.76 / 15.9 ∆allow P0.01]1.176 = 4.52 in

GOVERNING h = 23.57 in < ACTUAL h = 24.00 in [Satisfactory]

3.2 DETERMINE SECTION PROPERTIESL DIRECTION B DIRECTION

n = 3 yb = 17.89 in n = 4 yb = 18.59 in

A = 1752 in2St = 12824 in3

A = 2816 in2St = 20674 in3

I = 78347 in4Sb = 4379 in3

I = 111827 in4Sb = 6015 in3

CGS = 22.00 in e = 4.11 in CGS = 22.00 in e = 3.41 in

4. CALCULATE MAXIMUM APPLIED SERVICE MOMENTS 4.1 CENTER LIFT MOMENT AT L DIRECTION CENTER LIFT MOMENT AT B DIRECTION

ML = A0 (B em1.238 + C) = 11.51 ft-kips / ft MB = (58 + em) ML / 60, for L /B > 1.1

Where A0 = (L0.013 SB0.306 h0.688 P0.534 ym

0.193) / 727 = 1.439 MB = ML, for L /B < 1.1

B = 1, for em < 5

B = MIN[(ym - 1) / 3, 1], for em > 5

C = 0, for em < 5

C = MAX[8 - (P - 613) / 255] (4 - ym) / 3], 0, for em > 5

4.2 EDGE LIFT MOMENT AT L DIRECTION EDGE LIFT MOMENT AT B DIRECTION

ML = SB0.10 (h em)0.78 ym

0.66 / (7.2 L0.0065 P0.04) = 2.66 ft-kips / ft MB = h0.35 (19 + em) ML / 57.75, for L /B > 1.1

MB = ML, for L /B < 1.1





f= Pr / A - ML / St + Pr e / St = -0.172 ksi f = Pr / A - MB / St + Pr e / St = -0.219 ksi

Where Pr = Pe - SG = 97.65 kips Where Pr = Pe - SG = 150.89 kips

Pe = fe Aps = 133.11 kips Pe = fe Aps = 186.35 kips / ft

SG = Wslab µ / 2000 = 35.46 kips SG = Wslab µ / 2000 = 35.46 kips



Tian LiDaniel

Design of PT Slabs on Expansive Soil Ground Based on Specification of PTI

0.83

Then f Then f

ft-kips / ft

ft-kips / ft

= 12.18

= 3.01

= 0.87

=


f = Pr / A + ML / Sb - Pr e / Sb = 0.721 ksi f = Pr / A + MB / Sb - Pr e / Sb = 0.989 ksi



5.4 TOP STRESS, FOR EDGE LIFT MOMENT, AT L DIRECTION TOP STRESS, FOR EDGE LIFT MOMENT, AT B DIRECTION

f = Pr / A - ML / Sb - Pr e / Sb = -0.211 ksi f = Pr / A - MB / Sb - Pr e / Sb = -0.284 ksi



5.5 BOTTOM STRESS, FOR EDGE LIFT MOMENT, AT L DIRECTION BOTTOM STRESS, FOR EDGE LIFT MOMENT, AT B DIRECTION

f= Pr / A + ML / St + Pr e / St = 0.147 ksi f = Pr / A + MB / St + Pr e / St = 0.152 ksi




β = (Ec I / Es)1/4 / 12 = 8.566 ft β = (Ec I / Es)1/4 / 12 = 9.363 ft


Es = 1000 psi Es = 1000 psi

6.2 ALLOWABLE DIFFERENTIAL DEFLECTION AT L DIRECTION ALLOWABLE DIFFERENTIAL DEFLECTION AT B DIRECTIONFOR CENTER LIFT FOR CENTER LIFT∆allow = 12 MIN(L, 6β) / C∆ = 1.40 in ∆allow = 12 MIN(B, 6β) / C∆ = 0.80 in


FOR EDGE LIFT FOR EDGE LIFT∆allow = 12 MIN(L, 6β) / C∆ = 0.70 in ∆allow = 12 MIN(B, 6β) / C∆ = 0.40 in


6.3 EXPECTED DIFFERENTIAL DEFLECTION WITHOUT PRESTRESSING EXPECTED DIFFERENTIAL DEFLECTION WITHOUT PRESTRESSINGFOR CENTER LIFT, AT L DIRECTION FOR CENTER LIFT, AT B DIRECTION

∆0 = (ym L)0.205 SB1.059 P0.523 em

1.296 / (380 h1.214) = 0.74 in ∆0 = (ym B)0.205 SL1.059 P0.523 em

1.296 / (380 h1.214) = 0.78 in

FOR EDGE LIFT, AT L DIRECTION FOR EDGE LIFT, AT B DIRECTION

∆0 = L0.35 ym0.76 SB

0.88 em0.74 / (15.9 h0.85 P0.01) = 0.21 in ∆0 = B0.35 ym

0.76 SL0.88 em

0.74 / (15.9 h0.85 P0.01) = 0.19 in

6.4 DEFLECTION CAUSED BY PRESTRESSING AT L DIRECTION DEFLECTION CAUSED BY PRESTRESSING AT B DIRECTION

∆p = Pe e β2 / (2 Ec I) = 0.026 in ∆p = Pe e β2 / (2 Ec I) = 0.025 in

6.5 COMPARE EXPECTED TO ALLOWABLE DIFFERENTIAL DEFLECTION COMPARE EXPECTED TO ALLOWABLE DIFFERENTIAL DEFLECTIONFOR CENTER LIFT, AT L DIRECTION FOR CENTER LIFT, AT B DIRECTION

∆0 - ∆p = 0.72 in < ∆allow ∆0 - ∆p = 0.76 in < ∆allow

[Satisfactory] [Satisfactory]FOR EDGE LIFT, AT L DIRECTION FOR EDGE LIFT, AT B DIRECTION

∆0 + ∆p = 0.23 in < ∆allow ∆0 + ∆p = 0.22 in < ∆allow



FOR CENTER LIFT FOR CENTER LIFT

VL = L0.09 SB0.71 h0.43 P0.44 ym

0.16 em0.93 / 1940 = 2.105 kips/ft VB = B0.19 SL

0.45 h0.20 P0.54 ym0.04 em

0.97 / 1350 = 1.965 kips/ft

FOR EDGE LIFT FOR EDGE LIFT

VL = L0.07 h0.4 P0.03 ym0.67 em

0.16 / (3.0 SB0.015) = 1.752 kips/ft VB = B0.07 h0.4 P0.03 ym

0.67 em0.16 / (3.0 SL

0.015) = 1.681 kips/ft


vc = 1.7 (fc')0.5 + 0.2 fp = 0.096 ksi vc = 1.7 (fc')0.5 + 0.2 fp = 0.096 ksi

Where fp = 0.056 ksi > 50 psi Where fp = 0.054 ksi > 50 psi


7.3 SHEAR STRESS OF RIBBED FOUNDATION, AT L DIRECTION SHEAR STRESS OF RIBBED FOUNDATION, AT B DIRECTIONFOR CENTER LIFT FOR CENTER LIFT


[Satisfactory] [Satisfactory]FOR EDGE LIFT FOR EDGE LIFT










Where β = MIN[(Ec t3 / 3 ks)0.25, SB] = 12 ft Where β = MIN[(Ec t3 / 3 ks)0.25, SL] = 14 ft



f = Pr / A - Mmax / 2 t2 = -0.197 ksi f = Pr / A - Mmax / 2 t2 = -0.242 ksi


Then f Then f

Then f Then f

Then f Then f

10. CONVERT UNIFORM THICKNESS

H= MAX[ ( I / L)1/3, ( I / B)1/3 ] = 6.48 in

11. SUGGEST RATIO OF EXPECTED ELONGATION

r = fe / 0.8 Eps = 7.77E-03

Where Eps = 28000 ksi

Techincal References: 1. "Design and Construction of Post-Tensioned Slab-on-Ground, Second Edition", The Post-Tensioning Institute, 2004. 2. "1997 Uniform Building Code, Volume 2", International Conference of Building Officials, 1997.





EXPECTED SETTLEMENT BY GEOTECHICAL ENR δ = 0.75 inSLAB-SUBGRADE FRICTION COEFFICIENT µ = 0.75

1.2 STRUCTURAL DATA AND MATERIALS PROPERITIESSLAB LENGTH L = 40 ftSLAB WIDTH B = 38 ftSLAB THICHNESS t = 4 inPERIMETER LOADING P = 840 plf



AVERAGE STIFFENING BEAM SPACING, L DIRECTION SL = 13.333 ft

AVERAGE STIFFENING BEAM SPACING, B DIRECTION SB = 12.667 ft



SLAB PRESTRESSING TENDONS, L DIRECTION 8 tendons w/ 0.153 in2 at each tendon. THE DESIGN IS ADEQUATE.SLAB PRESTRESSING TENDONS, B DIRECTION 8 tendons w/ 0.153 in2 at each tendon. SUGGESTED RATIO OF EXPECTED ELONGATION IS 0.00777TENDON IN THE BOTTOM OF EACH BEAM 0 tendons w/ 0 in2 CONVERTED UNIFORM THICKNESS IS 6.21 inchEFFECTIVE PRESTRESS AFTER ALL LOSSES EXCEPT SG fe = 174 ksi

2. DETERMINE SECTION PROPERTIESL DIRECTION B DIRECTION

n = 4 yb = 18.34 in n = 4 yb = 18.47 in

A = 2624 in2St = 19294 in3

A = 2720 in2St = 19992 in3

I = 109177 in4Sb = 5952 in3

I = 110544 in4Sb = 5985 in3

CGS = 22.00 in e = 3.66 in CGS = 22.00 in e = 3.53 in

3. CALCULATE MAXIMUM APPLIED SERVICE MOMENTSL DIRECTION B DIRECTION

McsL = (δ / ∆nsL)0.5 MnsL = 3.20 ft-kips / ft McsS = McsL (970-h) / 880 = 3.44 ft-kips / ft

Where MnsL = h1.35 SB0.36 / 80 L0.12 P0.10 = 0.75 ft-kips / ft

∆nsL = L1.28 SB0.80 / 133 h0.28 P0.62 = 0.04 ∆nsL = L1.28 SB

0.80 / 133 h0.28 P0.62 = 0.04





f = Pr / A + ML / St + Pr e / St = 0.168 ksi f = Pr / A + MB / St + Pr e / St = 0.170 ksi

Where Pr = Pe - SG = 161.14 kips Where Pr = Pe - SG = 161.14 kips

Pe = fe Aps = 213 kips Pe = fe Aps = 213 kips

SG = Wslab µ / 2000 = 51.83 kips SG = Wslab µ / 2000 = 51.83 kips




f = Pr / A - ML / Sb - Pr e / Sb = -0.283 ksi f = Pr / A - MB / Sb - Pr e / Sb = -0.312 ksi




β = (Ec I ∆nsL / Es δ )1/4 / 12 = 4.595 ft β = (Ec I ∆nsB / Es δ )

1/4 / 12 = 4.581 ft


Es = 1000 psi Es = 1000 psi

I = 109177 in4 I = 110544 in4

5.2 ALLOWABLE DIFFERENTIAL DEFLECTION AT L DIRECTION ALLOWABLE DIFFERENTIAL DEFLECTION AT B DIRECTION∆allow = 12 MIN(L, 6β) / C∆ = 0.17 in ∆allow = 12 MIN(B, 6β) / C∆ = 0.17 in

Where C∆ = 1920 Where C∆ = 1920

5.3 DIFFERENTIAL DEFLECTION WITHOUT PRESTRESSING DIFFERENTIAL DEFLECTION WITHOUT PRESTRESSING

∆cs = δ en01.78 - 0.103 h -1.65E-03 P + 3.95E-07 P P) = 0.12 in ∆cs = δ en

01.78 - 0.103 h -1.65E-03 P + 3.95E-07 P P) = 0.12 in



Tian LiDaniel

Design of PT Slabs on Compressible Soil Ground Based on Specification of PTI

Then f Then f

Then f Then f


VcsL = (δ / ∆nsL)0.30 VnsL = 0.850 kips/ft VcsS = VcsL (116-h) / 94 = 0.832 kips/ft

Where VnsL = h0.90 (PSB)0.30 / 550 L0.10 = 0.355 kips/ft


vc = 1.7 (fc')0.5 + 0.2 fp = 0.105 ksi vc = 1.7 (fc')0.5 + 0.2 fp = 0.105 ksi

Where fp = 0.061 ksi > 50 psi Where fp = 0.059 ksi > 50 psi


6.3 SHEAR STRESS OF RIBBED FOUNDATION, AT L DIRECTION SHEAR STRESS OF RIBBED FOUNDATION, AT B DIRECTION










Where β = MIN[(Ec h3 / 3 ks)0.25, SB] = 12.667 ft Where β = MIN[(Ec h3 / 3 ks)0.25, SB] = 13.333 ft



f = Pr / A - Mmax / 2 t2 = -0.206 ksi f = Pr / A - Mmax / 2 t2 = -0.206 ksi


9. CONVERT UNIFORM THICKNESS

H= MAX[ ( I / L)1/3, ( I / B)1/3 ] = 6.21 in

10. SUGGEST RATIO OF EXPECTED ELONGATION

r = fe / 0.8 Eps = 7.77E-03

Where Eps = 28000 ksi



DESIGN CRITERIA1. SEND TO GEOTECHNICAL ENGINEER THE BASE FORCES OF COLUMNS AND THE PERMISSIBLE DEFLECTION AT TOP PILE.

(PERMISSIBLE DEFLECTION 0.25 in SUGGESTED)

2. ASSUME FIX HEAD CONDITION IF Ldh & Lhk COMPLY WITH THE TENSION DEVELOPMEMNT.

3. FROM SOIL REPORT, DETERMINE PILE PATTERN, LENGTH, AND MAX SECTION FORCES OF SINGLE PILE, Pu, Mu, & Vu.

3. PILE CAPS SHALL BE INTERCONNECTED BY TIES WITH SDS/10 TIMES AXIAL CAPACITY OF VERT COLUMN LOADING.

(IBC 1808.2.23.1)

INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 5 ksi

VERT. REBAR YIELD STRESS fy = 60 ksiPILE DIAMETER D = 20 inPILE LENGTH L = 35 ft

FACTORED AXIAL LOAD Pu = 480 k

FACTORED MOMENT LOAD Mu = 200 ft-k

FACTORED SHEAR LOAD Vu = 20 kPILE VERT. REINF. 8 # 7PILE SPIRAL REINF. # 3 @ 3 in o.c.

( 1.5 in o.c. at each end.)

( L dh = 8 in & L hk = 14 in )

THE PILE DESIGN IS ADEQUATE.

ANALYSIS

CHECK PILE LIMITATIONSfc' = 5 ksi > 2.5 ksi [Satisfactory] (IBC 1810.1.1)

D = 20 in > MAX( L / 30 , 12 in) [Satisfactory] (IBC 1810.3.2)

CHECK FLEXURAL & AXIAL CAPACITY

20

Drilled Cast-in-place Pile Design Based on ACI 318-02

DanielTian Li

ε

ε

( )'

'

2

'

'

2 0.85, 57 , 29000

0.85 2 , 0

0.85 ,

,

,

C

C

CC

C

S

fksifE Ec so

Ec

c c forf c of oo

forf c oforEss s y

fforf s yy

ε

ε ε ε εεε

ε εε ε ε

ε ε

= = =

− < < =

≥

≤= >

(cont'd)

φ φ φ φ Pn (kips) φφφφ Mn (ft-kips)

AT COMPRESSION ONLY 954 0AT MAXIMUM LOAD 954 75AT 0 % TENSION 699 172

φ Pn (k) AT 25 % TENSION 571 197AT 50 % TENSION 464 205AT ε t = 0.002 304 203

AT BALANCED CONDITION 297 205AT ε t = 0.005 82 189

AT FLEXURE ONLY 0 144

AT TENSION ONLY -259 0

φ Mn (ft-k)

φ Pmax =0.85 φ [ 0.85 fc' (Ag - Ast) + fy Ast] = 953.65 kips., (at max axial load, ACI 318-02, Sec. 10.3.6.1)

where φ = 0.70 (ACI 318-02, Sec.9.3.2.2) > Pu [Satisfactory]

Ag = 314 in2. Ast = 4.80 in2.

a = Cbβ1 = 8 in (at balanced strain condition, ACI 10.3.2)

φ = 0.57 + 67 εt = (ACI 318-02, Fig. R9.3.2)

where Cb = d εc / (εc + εs) = 10 in εt = 0.002069 εc =d = 16.2 in, (ACI 7.7.1) β1 = 0.8 ( ACI 318-02, Sec. 10.2.7.3 )

φ Mn = 0.9 Μn = 144 ft-kips @ Pn = 0, (ACI 318-02, Sec. 9.3.2) ,& εt,max = 0.004, (ACI 318-02, Sec. 10.3.5)

φ Mn = 204 ft-kips @ Pu = 480 kips > Mu [Satisfactory]

ρmax = 0.08 (ACI 318-02, Section 10.9) ρprovd = 0.015

ρmin = 0.005 (IBC, Section 1810.1.1.2) [Satisfactory]

CHECK SHEAR CAPACITY

φ Vn = φ (Vs + Vc) = 57 kips, (ACI 318-02 Sec. 11.1.1)

> Vu [Satisfactory]where φ = 0.75 (ACI 318-02 Sec. 9.3.2.3)

A0 = 206 in2. Av = 0.22 in2. fy = 40 ksi

Vc = 2 (fc')0.5A0 = 29.1 kips, (ACI 318-02 Sec. 11.3.1)

Vs = MIN (d fy Av / s , 4Vc) = 47.5 kips, (ACI 318-02 Sec. 11.5.6.2)

smax = 3 (ACI 318-02, Section 7.10.4.3) sprovd = 3 in

smin = 1 (ACI 318-02, Section 7.10.4.3) [Satisfactory]

DETERMINE FIX HEAD CONDITION

10 db = 8 in(ACI 318-02 12.5.2)

L hk = 14 in, (ACI 318-02, Fig. R12.5)

where db = 0.88 in

ρ required / ρ provided = 0.8 ( A s,reqd / A s,provd , ACI 318, 12.2.5)

α = 1.0 (ACI 318-02 12.2.4)

β = 1.0 (1.2 for epoxy-coated, ACI 318-02 12.2.4) γ = 1.0 (0.8 for # 6 or smaller, 1.0 for other) λ = 1.0 (normal weight)c = 3.4 in, min(d' , 0.5s), (ACI 318-02, 12.2.4)

Ktr = (Atr fyt / 1500 s n) = 0 (ACI 318-02, 12.2.4)

(c + Ktr ) / db = 2.5 < 2.5 , (ACI 318-02, 12.2.3) η = 0.7 (#11 or smaller, cover > 2.5" & side >2.0", ACI 318-02 12.5.3)

0.003

0.709

-400

-200

0

200

400

600

800

1000

1200

0 50 100 150 200 250

'

0.02, , 68

requird b ydh b

provided c

d fMAX indL

f

ρ βλη

ρ

= =



INPUT DATA & DESIGN SUMMARYCOLUMN WIDTH c1 = 5 in

COLUMN DEPTH c2 = 5 in

BASE PLATE WIDTH b1 = 16 in

BASE PLATE DEPTH b2 = 16 in

FOOTING CONCRETE STRENGTH fc' = 2.5 ksi



AXIAL LIVE LOAD PLL = 25 k

LATERAL LOAD (0=WIND, 1=SEISMIC) = 1 Seismic,SDSEISMIC AXIAL LOAD PLAT = 20 k, SD

SEISMIC MOMENT LOAD MLAT = 96 ft-k, SD

SEISMIC SHEAR LOAD VLAT = 2 k, SD




FOOTING MIDDLE THICKNESS T = 18 inSOIL COVER THICKNESS D = 12 in


SQUARE FOOTING LENGTH L = 7 ftREINFORCING SIZE # 5

MIDDLE BOTTOM EACH WAY : 9 # 5 @ 9 in o.c. THE FOOTING DESIGN IS ADEQUATE.

ANALYSISDESIGN LOADS (IBC SEC.1605.3.2 & ACI 318-02 SEC.9.2.1)CASE 1: DL + LL P = 65 kips

M = 0 ft-kipsCASE 2: DL + LL + E / 1.4 P = 79 kips

M = 69 ft-kipsCASE 3: 0.9 DL + E / 1.4 P = 50 kips

M = 69 ft-kips

CHECK OVERTURNING FACTORMR / MO = 6.4 > F = 1 / (0.9x1.4) [Satisfactory]

Where MO = MLAT + VLAT Df - 0.5 PLAT L = 32 k-ft

Pconc = (0.15 kcf) L2 [T + 2 (Df - D - T) /3] = 13.48 k, footing wt

Psoil = ws D L2 = 5.39 k, soil weight

MR = 0.5 PDLL + 0.5 (Pconc + Psoil) L = 206 k-ft



Service Loads CASE 1 CASE 2 CASE 3P 65.0 79.3 50.3 k

qs L2 4.9 4.9 0.0 k, (surcharge load)

P conc - soil 3.6 3.6 3.2 k, (footing increased)Σ P 73.5 87.8 53.5 kΣ M 0.0 68.6 68.6 ft - k

qmin 2.250 > 0 1.441 > 0 0.393 > 0 ksf, net pressure

q3 2.250 2.272 1.223 ksf, net pressure

q2 2.250 3.102 2.054 ksf, net pressure

qmax 2.250 3.933 2.884 ksf, net pressure


Design of Footing at Piping Based on ACI 318-02

Tian LiDaniel

(cont'd)Where

[Satisfactory]

DESIGN FLEXURE & CHECK FLEXURE SHEAR (ACI 318-02 SEC.15.4.2, 10.2, 10.3.5, 10.5.4, 7.12.2, 12.2, 12.5, 15.5.2, 11.1.3.1, & 11.3)

18

Service Loads CASE 1 CASE 2 CASE 3V 36.7 57.5 40.3 k, flexure shearM 69.7 111.6 79.1 ft - k, flexure moment

DESIGN FLEXURE

Location Mu,max = 1.5 M d (in) ρmin ρreqD ρmax smax use ρprovD

Middle Bottom Each Way 167.3 ft-k 14.69 0.0022 0.0021 0.0129 18 9 # 5 @ 9 in o.c. 0.0023[Satisfactory]

CHECK FLEXURE SHEAR

Direction Vu,max = 1.5 V φVc = 2 φ b d (fc')0.5 check Vu < φ Vc

Pipe Direction 86.2 k 93 k [Satisfactory]


Case Pu Mu b1 b2 b0 γv βc y Af Ap R J vu (psi) φ vc1 97.5 0.0 25.2 25.2 100.8 0.4 1.0 2.0 32.7 10.3 0.0 8.2 65.9 150.02 118.9 102.9 25.2 25.2 100.8 0.4 1.0 2.0 32.7 10.3 0.0 8.2 117.0 150.03 75.4 102.9 25.2 25.2 100.8 0.4 1.0 2.0 32.7 10.3 0.0 8.2 87.6 150.0


Pu = 1.5 PcolMu = 1.5 Mcol

''2

0.85 1 10.383

M uf c b fd cf y

ρ

− − =

40.0018 ,

3MIN

TMIN

dρρ =

max 32

min 32

3 1620.5

133 162

0.513

P Mq

LLP M

qLL

Σ Σ = +

Σ Σ = −

0.5 1( )

231 21 3

6 1 1

0

R bMP uu vpsivu JAP

db d bJb b

R

γ−= +

= + +

=

( )2 1 21

12 113 2

2 23

db bAP

vbb

A Lf

γ

= +

= −+

=

( )

( ) ( )

'( ) 2

42, , 40

0

, 0.5 0.5 , 0.5 0.50 1 1 1 2 2 2

psi y fvc c

dy MIN

bc

AP d db b c b b c bd

φφ

β

= +

=

= = + + = + +

2 max min

3 max min

2 13 31 23 3

q q q

q q q

= + = +

'10.85

MAX

f c uf u ty

β ερε ε

=+


JOB NO. : DATE : REVIEW BY : Seismic Design for Special Moment Resisting Frame Based on ACI 318-02


10.17

CONCRETE STRENGTH fc' = 3 ksi DISTRIBUTED UNFACTED LOADS D = 4.1 kips / ft

REBAR YIELD STRESS fy = 60 ksi L = 0.6 kips / ftBEAM LENGTH BET. COL. CENTERS L = 25 ft SECTION MOMENTS & SHEARS AT FACE OF COL. (ft-kips, kips)BEAM SIZE b = 24 in MA VA MB VB

h = 36 in D -130 45 -180 -45COLUMN SIZE c1 = 36 in L -25 7 -25 -7

c2 = 24 in QE 665 61 -665 -61

SEISMIC PARAMETER SDS = 0.44REDUNDANCY FACTOR ρ = 1.15LONGITUDINAL REINFORCING

SECTION A MID SPAN BTOP 9 # 9 3 # 9 9 # 9

( d = 33.31 in ) ( d = 33.31 in ) ( d = 33.31 in )( 1 Layer) ( 1 Layer) ( 1 Layer)

BOTTOM 5 # 9 5 # 9 5 # 9( d = 33.31 in ) ( d = 33.31 in ) ( d = 33.31 in )

( 1 Layer) ( 1 Layer) ( 1 Layer)

HOOP & STIRRUP LOCATIONS (ACI 21.3.3)LOCATION AT END, S1 AT MID, S2 AT SPLICE, S3

LENGTH 72 in 120 in 48 in( 2h ) (L-4h-c1) 1.3 MAX0.075fyαβγdb/[(fc')

0.5(c+Ktr)/db], 12TYPE Hoops Stirrups HoopsBAR 5 Legs # 5 3 Legs # 5 5 Legs # 5

(Legs to alternate long bars supported, ACI 7.10.5.3)SPACING @ 8 in o.c. @ 16 in o.c. @ 4 in o.c.

MIN(d/4, 8db, 24dt, 12) ( d/2 ) MIN(d/4, 4)

THE BEAM DESIGN IS ADEQUATE.

ANALYSISDESIGN LOADS U = (1.2+0.2SDS)D + ρQE + 1.0L (ACI 9-5)

AT SECTION A, FACE OF COLUMN AT SECTION B, FACE OF COLUMN

Vu = 135.1 kips or -5.2 kips Vu = -135.1 kips or 5.2 kips

Mu = 572.3 ft-kips or -957.2 ft-kips Mu = -1021.6 ft-kips or 507.9 ft-kipsAT MIDDLE OF THE SPAN

Vu = 0.0 kips or 0.0 kips

Mu = 131.1 ft-kips or 131.1 ft-kips

U = (0.9-0.2SDS)D + ρQE (ACI 9-7)AT SECTION A, FACE OF COLUMN AT SECTION B, FACE OF COLUMN

Vu = 106.7 kips or -33.6 kips Vu = -106.7 kips or 33.6 kips

Mu = 659.2 ft-kips or -870.3 ft-kips Mu = -910.9 ft-kips or 618.6 ft-kipsAT MIDDLE OF THE SPAN

Vu = 0.0 kips or 0.0 kips

Mu = 75.6 ft-kips or 75.6 ft-kips

U = 1.2D + 1.6L (ACI 9-2)AT SECTION A, FACE OF COLUMN AT MIDDLE OF THE SPAN AT SECTION B, FACE OF COLUMN

Vu = 65.2 kips Vu = 0.0 kips Vu = -65.2 kips

Mu = -196.0 ft-kips Mu = 129.8 ft-kips Mu = -256.0 ft-kips

DanielTian Li

cont'dCHECK SECTION REQUIREMENTS (ACI 21.3.1)

Pu < 0.1Agfc' [Satisfactory]

Lu=L-c1 = 22.00 ft > 4 d = 11.10 ft [Satisfactory]

b / h = 0.67 > 0.3 [Satisfactory]b = 24 in > 10 in [Satisfactory]

< c2+1.5h = 78 in [Satisfactory]

CHECK FLEXURAL REQUIREMENTSAT SECTION A, FACE OF COLUMN





(ACI 21.3.2.2) Mn,bot > (1/2)Mn,top [Satisfactory]

where Mn,bot = ρbot bd2fy (1 - 0.588ρbot fy/fc') = 772 ft-kips > Mu / φ [Satisfactory]

Mn,top = ρtop bd2fy (1 - 0.588ρtop fy/fc') = 1301 ft-kips > Mu / φ [Satisfactory]φ = 0.9

AT SECTION B, FACE OF COLUMN

(ACI 21.3.2.1) ρtop = 0.011 > ρmin = 0.003 [Satisfactory]




(ACI 21.3.2.2) Mn,bot > (1/2)Mn,top [Satisfactory]

where Mn,bot = ρbot bd2fy (1 - 0.588ρbot fy/fc') = 772 ft-kips > Mu / φ [Satisfactory]

Mn,top = ρtop bd2fy (1 - 0.588ρtop fy/fc') = 1301 ft-kips > Mu / φ [Satisfactory]AT MIDDLE OF THE SPAN

(ACI 21.3.2.1) ρtop = 0.004 > ρmin = 0.003 [Satisfactory]




(ACI 21.3.2.2) Mn,top > (1/4)Mn,max [Satisfactory]

Mn,bot > (1/4)Mn,max [Satisfactory]

where Mn,top = ρtop bd2fy (1 - 0.588ρtop fy/fc') = 478 ft-kips > Mu / φ [Satisfactory]

Mn,bot = ρbot bd2fy (1 - 0.588ρbot fy/fc') = 772 ft-kips > Mu / φ [Satisfactory]

Mn,max = 1301 ft-kips

CHECK SHEAR STRENGTH (ACI 21.3.4)FOR SEISMIC LOAD ACTING TO THE LEFTVe = (Mpr, A, top + Mpr, B,bot) / Ln + VgL = 178.7 kips < 8φ(fc')


Ve - dwu = 162.4 kips < φ[Vc + Avfyd/s1 ] = 290.4 kips [Satisfactory]

Ve - (2h + d)wu = 127.1 kips < φ[2(fc')0.5bd + Avfyd/s2 ] = 152.8 kips [Satisfactory]

where Vc = 2(fc')0.5bd = 0.0 kips, (Per ACI 21.3.4.2, Vc = 0, if (Ve -VgL) 50% Ve AND Pu < Ag fc' / 20 )

Mpr, A, top = ρtop bd2fy (1.25 - 0.919ρtop fy/fc') = 1564 ft-kips

Mpr, B ,bot = ρbot bd2fy (1.25 - 0.919ρbot fy/fc') = 945 ft-kips

Ln = L - c1 = 22.0 ft

wu = (1.2+0.2SSD)D+1.0L = 5.9 kips / ft, (for CBC, only D + L, without factor)

VgL = wuLn / 2 = 64.7 kipsφ = 0.75 (ACI 9.3.2.3)

Av = 1.55 in2 @ end , 0.93 in2 @ mid of beamFOR SEISMIC LOAD ACTING TO THE RIGHTVe = (Mpr, A, bot + Mpr, B,top) / Ln + VgL = 178.7 kips < 8φ(fc')


Ve - dwu = 162.4 kips < φ[Vc + Avfyd/s1 ] = 290.4 kips [Satisfactory]

Ve - (2h - d)wu = 110.8 kips < φ[2(fc')0.5bd + Avfyd/s2 ] = 152.8 kips [Satisfactory]

where Mpr, A ,bot = ρbot bd2fy (1.25 - 0.919ρbot fy/fc') = 945 ft-kips

Mpr, B, top = ρtop bd2fy (1.25 - 0.919ρtop fy/fc') = 1564 ft-kips

Techincal References: 1. Alan Williams: "Seismic and Wind Forces, Structural Design Examples", International Code Council, 2003. 2. SEAOC: "2000 IBC Structural/Seismic Design Manual - Volume 3", International Code Council, 2003. 3. S.K. Ghosh, Xuemei Liang: "Seismic and Wind Design of Concrete Buildings", Portland Cement Association, 2005.






COLUMN CLEAR HEIGHT Hn = 10.17 ft

COLUMN SIZE c1 = 36 in

c2 = 36 in

SEISMIC PARAMETER SDS = 0.44REDUNDANCY FACTOR ρ = 1.15

SECTION MOMENTS & SHEARS AT END OF COL. (ft-kips, kips)P Mtop V Mbot

D 692 -4 -20 -2L 87 -1 -1 -1 36

QE 2 900 93 500

LONGITUDINAL REINFORCINGSECTION TOP BOTTOM

LEFT 6 # 8 6 # 8( d = 33.38 in ) ( d = 33.38 in )

( 1 Layer) ( 1 Layer)RIGHT 6 # 8 6 # 8

( d = 33.38 in ) ( d = 33.38 in )( 1 Layer) ( 1 Layer)

TOTAL 20 # 8 20 # 8BARS ( 8 # 8 at sides ) ( 8 # 8 at sides )

TRANSVERSE REINFORCMENT FOR CONFINEMENT (ACI 21.4.4 & 21.4.3)LOCATION AT END, Lo AT MID AT SPLICE, SLENGTH 36 in 4 in + 4 in 43 in

MAX( c1, Hn/16, 18) 1.3Max(0.075)fydb/[(fc')0.5(c+Ktr)/db],12

TYPE Hoops Hoops HoopsBAR SIZE 4 Legs # 5 4 Legs # 5 4 Legs # 5

(Legs to alternate long bars supported, ACI 7.10.5.3)SPACING @ 4 in o.c. @ 6 in o.c. @ 4 in o.c.

MINc1/4, 6db, MAX[MIN(4+(14-hx)/3, 6), 4] MIN( 6db, 6 ) Same as END Lo

THE COLUMN DESIGN IS ADEQUATE.

ANALYSISDESIGN LOADS U = (1.2+0.2SDS)D + ρQE + 1.0L (ACI 9-5) U = (0.9-0.2SDS)D + ρQE (ACI 9-7)

Pu = 980.6 kips Mu,top = 1028.8 ft-kips Pu = 564.2 kips Mu,top = 1031.8 ft-kips

Vu = 80.2 kips Mu,bot = 571.4 ft-kips Vu = 90.7 kips Mu,bot = 573.4 ft-kips U = 1.2D + 1.6L (ACI 9-2)

Pu = 969.6 kips Mu,top = -6.4 ft-kips

Vu = -25.6 kips Mu,bot = -4.0 ft-kips

CHECK SECTION REQUIREMENTS (ACI 21.4.1)

Pu = 564.2 kips > 0.1Agfc' = 388.8 kips [Satisfactory]

cmin =MIN(c1, c2) = 36 in > 12 in [Satisfactory]

cmin / cmax = 1.00 > 0.4 [Satisfactory]

CHECK TRANSVERSE REINFORCING AT END OF COLUMN (ACI 21.4.4)

Ash = 1.24 in2 > MAX[ 0.09shcfc' / fyh , 0.3shc(Ag/Ach-1)fc' / fyh ] = 0.73 in2

[Satisfactory] where s = MAX[MIN(c1/4, 6db, 4+(14-hx)/3, 6), 4] = 5 in

hc = c1 - 2Cover - dt = 32.4 in

Ach = (c1-3)(c2-3) = 1089.0 in2

CHECK FLEXURAL REINFORCING (ACI 21.4.3.1)

AT TOP SECTION ρtotal = 0.013 > ρmin = 0.010 [Satisfactory]

DanielTian Li

(cont'd)

AT BOTTOM SECTION ρtotal = 0.013 > ρmin = 0.010 [Satisfactory]

AT SPLICE SECTION ρtotal = 0.026 < ρmax = 0.060 ( 0.030 should be used for lap splice existed.)

[Satisfactory]CHECK CAPACITY SUBJECTED TO BENDING AND AXIAL LOAD APPLIED LOADING (ACI 10.12.3)

LOADS 1 2 3 4 5 6

Pu (kips) 980.6 980.6 564.2 564.2 969.6 969.6

Mu (ft-kips) 1028.8 571.4 1031.8 573.4 6.4 4.0

δns = Cm/[1-Pu/(0.75Pc)] 1.018 1.018 1.011 1.011 1.018 1.018

δnsMu (ft-kips) 1047.8 581.9 1042.6 579.4 6.5 4.1

φMn (ft-kips) @ Pu 1325.3 1566.8 1293.9 1564.8 1104.9 1103.0

where EI = 0.4EcIg / (1+βd) = 0.25 EcIgPc = π2EI / (kLu)2

SUMMARY OF LOAD VERSUS MOMENT CAPACITIES (ACI 10.2 & 10.3)

CAPACITY φ φ φ φ Pn (kips) φφφφ Mn (ft-kips)AT AXIAL LOAD ONLY 2191 0AT MAXIMUM LOAD 2191 568AT 0 % TENSION 1959 805AT 25 % TENSION 1647 1028AT 50 % TENSION 1392 1162AT ε t = 0.002 1008 1307AT BALANCED CONDITION 987 1318AT ε t = 0.005 687 1651AT FLEXURE ONLY 0 1100

φ Pn (kips)

φ Mn (ft-kips)

All load points to be within capacity diagram. [Satisfactory]

CHECK SHEAR STRENGTH (ACI 21.4.5)

Ve = MAX[ (Mpr, left, top + Mpr, right,bot) / Hn , Vu,max] = 246.7 kips

< 8φ(fc')0.5c2d = 394.9 kips [Satisfactory]

< φ[2(fc')0.5c2d + Avfyd/smid ] = 409.1 kips [Satisfactory]

where ρtop,left = 0.004 > ρmin=MIN[3(fc')0.5/fy, 200/fy ]= 0.003 [Satisfactory]

ρbot,right = 0.004 > ρmin = 0.003 [Satisfactory]

Mpr, left, top = MIN [1.25Mn,col,max , 0.5 (Mpr,top beam, left + Mpr,top beam, right) ] = 1254 ft-kips

Mpr, right, bot = MIN [1.25Mn,col,max , 0.5 (Mpr,bot beam, left + Mpr,bot beam, right) ] = 1254 ft-kipsφ = 0.75 (ACI 9.3.2.3)

Av = 1.24 in2

Techincal References: 1. Alan Williams: "Seismic and Wind Forces, Structural Design Examples", International Code Council, 2003. 2. SEAOC: "2000 IBC Structural/Seismic Design Manual - Volume 3", International Code Council, 2003. 3. David A. Fanella: "Design of Concrete Buildings for Earthquake and Wind Forces", Portland Cement Association, 1998.

0

500

1000

1500

2000

2500

0 200 400 600 800 1000 1200 1400 1600 1800



CHECK STRONG COLUMN - WEAK BEAM (ACI 21.4.2.2)ΣMc = Mn,top @ Pu, top + Mn,bot @ Pu, bot = 4077.7 ft-kips Mn,top @ Pu, top

> 1.2ΣMg = 1.2(Mn,top + Mn,bot) = 2486.5 ft-kips [Satisfactory]

where Mn,top @ Pu, top = 2038.9 ft-kips

Mn,bot @ Pu, bot = 2038.9 ft-kips Mn,top Mn,bot

Mn,top = 1300.5 ft-kips, (slab bars included, ACI 318-02)

Mn,bot = 771.5 ft-kips

Note: For UBC 97, Mc & Mg shall be at the center of the joint with φ factors,

which means ΣMc > (0.9/0.7)1.2ΣMg. Mn,bot @ Pu, bot

CHECK JOINT CAPACITY (ACI 21.5)Ve = 1.25fy(As,top + As,bot) - (Mpr,top + Mpr,bot)/Hn = 803.3 kips

where As,top = 9.00 in2

As,bot = 5.00 in2As,top

Mpr,top = 1563.6 ft-kips

Mpr,bot = 945.2 ft-kips

Hn = 10.17 ft Mpr,top Mpr,bot

φVn = k Aj (fc')0.5 = 804.5 kips > Ve [Satisfactory]

where Aj = c1 MIN(b+c1 , c2) = 864 in2

k = 20 (20 for four faces, 15 for three faces, & 12 for others) As,botφ = 0.85 (ACI 9.3.4 c)

THEJOINT DESIGN IS ADEQUATE.

Techincal References: 1. Alan Williams: "Seismic and Wind Forces, Structural Design Examples", International Code Council, 2003. 2. SEAOC: "2000 IBC Structural/Seismic Design Manual - Volume 3", International Code Council, 2003. 3. David A. Fanella: "Design of Concrete Buildings for Earthquake and Wind Forces", Portland Cement Association, 1998.

DanielTian Li



INPUT DATA DESIGN SUMMARYCONCRETE STRENGTH fc' = 4 ksi SHEAR WALL LENGTH L = 25.00 ftREBAR YIELD STRESS fy = 60 ksi SHEAR WALL THICKNESS t = 16.00 in

BULB END WIDTH B = 36.00 inFACTORED AXIAL LOAD Pu = 1472 k BULB END DEPTH D = 36.00 inFACTORED MOMENT LOAD Mu = 52214 ft-k BULB REINFORCING 32 # 11FACTORED SHEAR LOAD Vu = 50 k WALL HORIZ. REINF 2 # 5 @ 12 in o.c.

WALL VERT. REINF 2 # 5 @ 12 in o.c.LENGTH OF SHEAR WALL L = 25 ft HOOP REINF - WIDTH, B, DIR. 5 # 5 @ 6 in o.c.THICKNESS OF WALL t = 16 in HOOP REINF - LENGTH DIR. 5 # 5 @ 6 in o.c.DEPTH AT FLANGE D = 36 inWIDTH AT FLANGE B = 36 in THE WALL DESIGN IS ADEQUATE.TOTAL WALL HEIGHT TO TOP hw = 148 ftREINF. BARS AT BULB 32 # 11WALL DIST. HORIZ. REINF. 2 # 5 @ 12 in. o.c.WALL DIST. VERT. REINF. 2 # 5 @ 12 in. o.c.HOOP REINF - WIDTH, B, DIR. 5 legs of # 5HOOP REINF - LENGTH DIR. 5 legs of # 5

ANALYSIS

φ Pn (k)

25

φ Mn (ft-k)

CHECK SHEAR CAPACITYTHE MINIMUM REINFORCEMENT RATIOS ARE GIVEN BY ACI 318-02 SECTION 21.7.2.1 AND SECTION 14.3 AS

(ρn )min. = 0.0020 [ for Acv (fc')0.5 = kips > Vu , and bar size #5 horizontal]

(ρV )min. = 0.0012 [ for Acv (fc')0.5 = kips > Vu , and bar size #5 horizontal]where Acv = 4800 in2 (gross area of concrete section bounded by web thickness and length in the shear direction)

THE PROVIDED REINFORCEMENT RATIOS AREρn = 0.0032 > (ρn )min. [Satisfactory]

ρV = 0.0032 > (ρV )min. [Satisfactory]

The proposed spacing is less than the maximum permissible value of 18 in and is satisfactory. Since wall Vu < 2 Acv (fc')0.5 ,one curtain reinforcement required. (ACI 318-02, Sec.21.7.2.2)THE DESIGN SHEAR FORCE IS GIVEN BY ACI 318-02 SECTION 21.7.4.1 & 21.7.4.4 AS

φVn =MIN [ φ Acv (αc (fc')0.5 + ρn fy), φ 8 Acv (fc')0.5 ] = 922.29 kips > Vu [Satisfactory]where φ = 0.6

αc = 2.0 ( for hw / L = > 2 )

ρV > ρn [Satisfactory] (only for hw / L > 2.0, ACI 318-02 Sec. 21.7.4.3)

Shear Wall Design Based on ACI 318-02

DanielTian Li

5.92

303.58

303.58

L

t

B

D

0

2000

4000

6000

8000

10000

12000

14000

16000

0 20000 40000 60000 80000 100000 120000 140000

CHECK FLEXURAL & AXIAL CAPACITYTHE MAXIMUN DESIGN AXIAL LOAD STRENGTH, ACI 318-02, Sec. 21.7.5.1 & Eq.(10-2), IS

φ Pmax =0.8 φ [ 0.85 fc' (Ag - Ast) + fy Ast] = 14336 kips. > Pu [Satisfactory]where φ = 0.65 (ACI 318-02, Sec.9.3.2.2)

Ag = 6240 in2.Ast = 112.24 in2.

THE DESIGN MOMENT CAPACITY AT MAXIMUM AXIAL LOAD STRENGTH ARE FROM 0 TO 30305 ft-kips.

FOR THE BALANCED STRAIN CONDITION UNDER COMBINED FLEXURE AND AXIAL LOAD, THE MAXIMUM STRAIN IN THE CONCRETEAND IN THE TENSION REINFORCEMENT MUST SIMULTANEOUSLY REACH THE VALUES SPECIFIED IN ACI 318-02 SEC. 10.3.2

AS εc = 0.003 AND εt = fy / Es = 0.002069 . THE DEPTH TO THE NEUTRAL AXIS AND EQUIVALENT RECTANGULAR CONCRETE

STRESS BLOCK ARE GIVEN BY

Cb = d εc / (εc + εs) = 167 in a = Cb β1 = 142 in φ = 0.48 + 83 εt = 0.652 (ACI 318-02, Fig. R9.3.2)

where d = (L-0.5D) = 282 in β1 = 0.85 ( ACI 318-02, Sec. 10.2.7.3 )THE DESIGN AXIAL AND MOMENT CAPACITIES AT THE BALANCED STRAIN CONDITION ARE 6580 kips AND 87164 ft-kips.

IN ACCORDANCE WITH ACI SEC. 9.3.2 THE DESIGN MOMENT CAPACITY WITHOUT AXIAL LOAD IS φ Mn = 0.9 Μn = 69914 kips.

SUMMARY OF LOAD VERSUS MOMENT CAPACITIES ARE SHOWN IN THE TABLE BELOW, AND THEY ARE PLOTTEDON THE INTERACTION DIAGRAM ABOVE.

φ φ φ φ Pn (kips) φφφφ Mn (ft-kips)AT AXIAL LOAD ONLY = 14336 0AT MAXIMUM LOAD = 14336 30305AT 0 % TENSION = 12262 46464AT 25 % TENSION = 10458 61850AT 50 % TENSION = 8972 72337

AT ε t = 0.002 = 6708 86103AT BALANCED CONDITION = 6580 87164

AT ε t = 0.005 = 6112 118161AT FLEXURE ONLY = 0 69914

THE DESIGN FORCES Pu & Mu ARE ALSO PLOTTED ON THE INTERACTION DIAGRAM. FROM THE INTERACTION DIAGRAM,THE ALLOWABLE MOMENT AT AN AXIAL LOAD Pu IS

φ Mn = 85496 kips. > Mu [Satisfactory]

where φ = Min[0.9, Max(0.48 + 83 εt , 0.65)] = 0.900 (ACI 318-02, Fig. R9.3.2)

CHECK BOUNDARY ZONE REQUIREMENTSAN EXEMPTION FROM THE PROVISION OF BOUNDARY ZONE CONFINEMENT REINFORCEMENT IS GIVEN BY ACI SECTION 21.7.6.2,21.7.6.3, and 21.7.6.5(a) PROVIDED THAT

c < (L hw) / (600 δu) and fc < 0.2 fc' [Unsatisfactory]

where c = 35 in. ( distance from the extreme compression fiber to neutral axis at Pu & Mn loads. )δu = 12.4 in. ( design displacement, assume 0.007hw as a conservative short cut, see ACI 318-02 Sec. 21.7.6.2a. )

fc = (Pu / A) + (Mu y / I ) = ksi. ( the maximun extreme fiber compressive stress at Pu & Mu loads. )y = 150 in. ( distance from the extreme compression fiber to neutral axis at Pu & Mu loads. )A = in2. ( area of transformed section. )

I = in4. ( moment of inertia of transformed section. )

And the longitudinal reinforcement ratio at the wall end = 0.039 > 400 / fy [Unsatisfactory]HENCE SPECIAL BOUNDARY ZONE DETAILING REQUIRED !

The boundary element length = MAX( c-0.1L, 0.5c ) = 17.31 in. ( ACI 318-02, Sec. 21.7.6.4 )

The maximum hoop spacing = MIN[ B/4 , 6db , 6 , 4+(14-hx)/3 ] = 6 in.o.c. ( ACI 318-02, Sec. 21.4.4.2 & 21.7.6.5a )

The required hoop reinforcement Ash, B DIR = (0.09 s hc fc' ) / fyh = 0.291 in2. < # 5 provided [Satisfactory]

( ACI 318-02, Eq.21-4 ) Ash, L DIR = (0.09 s hc fc' ) / fyh = 0.305 in2. < # 5 provided [Satisfactory]

714368017848

1.588



INPUT DATA DESIGN SUMMARYCONCRETE STRENGTH fc' = 4 ksi SHEAR WALL LENGTH L = 25.00 ftREBAR YIELD STRESS fy = 60 ksi SHEAR WALL THICKNESS t = 16.00 in

BULB END WIDTH B = 36.00 inFACTORED AXIAL LOAD Pu = 1472 k BULB END DEPTH D = 36.00 inFACTORED MOMENT LOAD Mu = 52214 ft-k BULB REINFORCING 32 # 11FACTORED SHEAR LOAD Vu = 50 k WALL HORIZ. REINF 2 # 5 @ 12 in o.c.

WALL VERT. REINF 2 # 5 @ 12 in o.c.LENGTH OF SHEAR WALL L = 25 ft HOOP REINF - WIDTH, B, DIR. 5 # 5 @ 6 in o.c.THICKNESS OF WALL t = 16 in HOOP REINF - LENGTH DIR. 5 # 5 @ 6 in o.c.DEPTH AT FLANGE D = 36 inWIDTH AT FLANGE B = 36 in THE WALL DESIGN IS ADEQUATE.WALL UNSUPPORT HEIGHT Lu = 22 ftTOTAL WALL HEIGHT TO TOP hw = 148 ftREINF. BARS AT BULB 32 # 11WALL DIST. HORIZ. REINF. 2 # 5 @ 12 in. o.c.WALL DIST. VERT. REINF. 2 # 5 @ 12 in. o.c.HOOP REINF - WIDTH, B, DIR. 5 legs of # 5HOOP REINF - LENGTH DIR. 5 legs of # 5

ANALYSIS

φ Pn (k)

22

φ Mn (ft-k)CHECK SHEAR CAPACITYTHE MINIMUM REINFORCEMENT RATIOS ARE GIVEN BY ACI 318-95 SECTION 21.6.2.1 AND SECTION 14.3 AS

(ρn )min. = 0.0020 [ for Acv (fc')0.5 = kips > Vu , and bar size #5 horizontal]

(ρV )min. = 0.0012 [ for Acv (fc')0.5 = kips > Vu , and bar size #5 horizontal]where Acv = 4800 in2 (gross area of concrete section bounded by web thickness and length in the shear direction)

THE PROVIDED REINFORCEMENT RATIOS AREρn = 0.0032 > (ρn )min. [Satisfactory]

ρV = 0.0032 > (ρV )min. [Satisfactory]

The proposed spacing is less than the maximum permissible value of 18 in and is satisfactory. Since wall Vu < 2 Acv (fc')0.5 ,one curtain reinforcement required. (ACI 318-95, Sec.21.7.2.2)THE DESIGN SHEAR FORCE IS GIVEN BY ACI 318-95 SECTION 21.7.4.1 & 21.7.4.4 AS

φVn =MIN [ φ Acv (αc (fc')0.5 + ρn fy), φ 8 Acv (fc')0.5 ] = 922.29 kips > Vu [Satisfactory]where φ = 0.6 , (conservatively, CBC 1909A.3.4.1)

αc = 2.0 ( for hw / L = > 2 )

ρV > ρn [Satisfactory] (only for hw / L > 2.0, ACI 318-95 Sec. 21.7.4.3)

5.92

303.58

303.58

Shear Wall Design Based on ACI 318-95 / CBC 2001

DanielTian Li

L

t

B

D

0

2000

4000

6000

8000

10000

12000

14000

16000

18000

0 20000 40000 60000 80000 100000

CHECK FLEXURAL & AXIAL CAPACITYTHE MAXIMUN DESIGN AXIAL LOAD STRENGTH, ACI 318-95, Sec. 21.7.5.1 & Eq.(10-2), IS

φ Pmax =0.8 φ [ 0.85 fc' (Ag - Ast) + fy Ast] = 15439 kips. > Pu [Satisfactory]where φ = 0.70 (ACI 318-02, Sec.9.3.2.2)

Ag = 6240 in2.Ast = 112.24 in2.

THE DESIGN MOMENT CAPACITY AT MAXIMUM AXIAL LOAD STRENGTH ARE FROM 0 TO 32636 ft-kips.

FOR THE BALANCED STRAIN CONDITION UNDER COMBINED FLEXURE AND AXIAL LOAD, THE MAXIMUM STRAIN IN THE CONCRETEAND IN THE TENSION REINFORCEMENT MUST SIMULTANEOUSLY REACH THE VALUES SPECIFIED IN ACI 318-95 SEC. 10.3.2

AS εc = 0.003 AND εt = fy / Es = 0.002069 . THE DEPTH TO THE NEUTRAL AXIS AND EQUIVALENT RECTANGULAR CONCRETE

STRESS BLOCK ARE GIVEN BY

Cb = d εc / (εc + εs) = 167 in a = Cb β1 = 142 in φ = 0.70 (ACI 318-02, Sec.9.3.2.2)

where d = (L-0.5D) = 282 in β1 = 0.85 ( ACI 318-95, Sec. 10.2.7.3 )THE DESIGN AXIAL AND MOMENT CAPACITIES AT THE BALANCED STRAIN CONDITION ARE 7068 kips AND 93621 ft-kips.IN ACCORDANCE WITH ACI SEC. 9.3.2 THE DESIGN MOMENT CAPACITY WITHOUT AXIAL LOAD IS φ Mn = 0.9 Μn = 69914 kips.


φ φ φ φ Pn (kips) φφφφ Mn (ft-kips)AT AXIAL LOAD ONLY = 15439 0AT MAXIMUM LOAD = 15439 32636AT 0 % TENSION = 13205 50038AT 25 % TENSION = 11262 66608AT 50 % TENSION = 9662 77901

AT ε t = 0.002 = 7223 92726AT BALANCED CONDITION = 7068 93621

AT SMALL AXIAL, φ 0.1 fc' Ag = 1747 51568AT FLEXURE ONLY = 0 69914

THE DESIGN FORCES Pu & Mu ARE ALSO PLOTTED ON THE INTERACTION DIAGRAM. FROM THE INTERACTION DIAGRAM,THE ALLOWABLE MOMENT AT AN AXIAL LOAD Pu IS

φ Mn = 76282 kips. > Mu [Satisfactory]

where φ = Max[0.9 - 2Min(Pu , 0.7Pb) / (fc' Ag) , 0.7] = 0.782 (ACI 318-95, 9.3.2.2)

DETERMINE WHETHER THE WALL CAN RESIST SEISMIC LOADS, (CBC 2001, 1921.6.6.3)Pu = 1472 k < 0.35 Ag fc' = 8736 k [Satisfactory]

CHECK BOUNDARY ZONE REQUIREMENTSAN EXEMPTION FROM THE PROVISION OF BOUNDARY ZONE CONFINEMENT REINFORCEMENT IS GIVEN BY ACI SECTION 21.6.6.6 PROVIDED THAT

Pu = 1472 k < 0.10 Ag fc' = 2496 k

and either

Mu / (Vu L ) = 41.8 > 1.0 [Unsatisfactory]

orVu = 50 k < 3 Acv (fc')2 = 910.7 k and Mu / (Vu L ) = 41.8 > 3.0

[Unsatisfactory]HENCE SPECIAL BOUNDARY ZONE DETAILING REQUIRED !

The boundary element width required = Lu / 16 = 17 in < B ( ACI 318-95, Sec. 21.6.6.6 ) [Satisfactory]The boundary element length required = 18.00 in. ( ACI 318-95, Sec. 21.6.6.6 )

The maximum hoop spacing = MIN[ 6db , 6 ] = 6 in.o.c. ( ACI 318-95, Sec. 21.6.6 )

The required hoop reinforcement Ash, B DIR = (0.09 s hc fc' ) / fyh = 0.291 in2. < # 5 provided [Satisfactory]

( ACI 318-95, Eq.21-10 ) Ash, L DIR = (0.09 s hc fc' ) / fyh = 0.305 in2. < # 5 provided [Satisfactory]



INPUT DATA & DESIGN SUMMARY bt diaphragm P1 P2CONCRETE STRENGTH f 'c = 4 ksi

REBAR STRENGTH fy = 60 ksi

OUT-OF-PLANE FORCE w1 = 15.429 psf, for wall e1OUT-OF-PARAPET w2 = 46.286 psf, for parapet

LEDGER DL P1DL = 0.8 k/ft

LEDGER LL P1LL = 0.8 k/ft design panel h wall loadDIST. FROM FACE e1 = 6 in

PARAPET DL P2DL = 0.18 k/ft

PARAPET LIVE LOAD P2LL = 0 k/ft beELEMENT WIDTH be = 1 ft

TRIBUTARY WIDTH bt = 1 ft

WALL THICKNESS t = 8 in

WALL VERT. SPAN h = 22 ft

PARAPET HEIGHT hp = 3 ft REVEAL THICKNESS = 0.75 in

VERT. REINF.- MIDDLE 1 LAYER # 5 @ 12 in, o.c.

HORIZ. REINF. - MIDDLE 1 LAYER # 5 @ 12 in, o.c.

ANALYSISCHECK VERTICAL FACTORED LOAD LESS THAN 0.06fc'Ag (ACI 318-02 Sec. 14.8.2.6)

= 3.78 k < 0.06fc'Ag = 20.88 k

[SATISFACTORY]

CHECK VERTICAL REINFORCEMENT LESS THAN 0.6ρb (ACI 318-02 Sec. 14.8.2.3)

= >[SATISFACTORY]

CHECK Mcr LESS THAN φMn

= 0.8823 (ACI 318-02 Sec. 9.3.2.2)

49.87 k-in < = 64.25 k-in

(ACI 318-02 Sec. 9.5.2.3) [SATISFACTORY]

CHECK WALL STRENGTH

= 27 in4 3 = 24.5 k-in

(ACI 318-02 Sec. 14.8.3)

= 35.23 k-in < = 64.25 k-in

[SATISFACTORY]

(ACI 318-02 Sec. 14.8.3, EQ14-6)

CHECK SERVICE LOAD OUT-OF-PLANE DEFLECTION (ACI 318-02 Sec. 14.8.3, EQ14-9)

= 19.2 k-in = 2.9 k

= 24.422 k-in = 512 in4

= 0.10 in < = 1.8 in [SATISFACTORY]

CHECK PARAPET STRENGTH

= 4 k-in < = 64 k-in [SATISFACTORY]

0.017 0.007

Daniel

Tilt-up Panel Design based on ACI 318 - 02

Tian Li

( )1.2 1.6 1.2 1.6 0.61 1 2 2u DL LL DL LL P thw bP P P P P= + + + +

'

max

0.85 870000.6 0.6

87000b c

by y

ff fβρ ρ

= =

+ actualρ =

( )'

0.250.9 , 0.65

0.1 ,u

g bc

PMAXMIN f A P

φφ

= −

'7.5 gccr

t

f IM

y= =

2sen y

adM fAφφ = −

( )2

2

3e

cr seb cn d cI A= − +

211.6 0.5

18 2

tua u

e tw b hM P

+= +

( )25

10.75 48

uau

u

c cr

MM

hPE I

=−

nMφ

251

48

sa

s

c cr

MMhP

E I

=−

2548s

c e

M hE I

=∆ 150h

( )2

1 0.51 1

8 2t

sa DL LLe tw b h

M P P+= + + ( )0.51 1 2 2s DL LL DL LL P thw bP P P P P= + + + +

3 3

1 ,e g cr gM Mcr crMINI I I IM M

= + −

0.92sen y

adM fAφ = −

22

,1.6

2t p

u parapetw b h

M =


JOB NO. : DATE : REVIEW BY : Wall Pier Design Based on CBC 2001 / UBC 1997


CONCRETE STRENGTHfc' = 3 ksi

REBAR YIELD STRESSfy = 60 ksi

WALL PIER LENGTHL = 5 ft

WALL PIER HIGHT 60H = 12 ft

WALL PIER THICKNESSt = 10 in

VERTICAL EDGE BARS, As2 # 9

TRANSVERSE REINFORCEMENT, Av THE DESIGN IS ADEQUATE.2 # 4 @ 6 in, o.c. (at each face.)

FACTORED AXIAL LOAD Pu = 60 kips

FACTORED SHEAR FORCE Vu = 60 kips, (in plane)

ANALYSIS

CHECK WALL PIER DEFINITION (CBC 2001 1902A)L / t = 6.00 within [2.5 , 6] & H / L = 2.40 > 2

[Satisfactory]

CHECK SHEAR STRENGTH (SEC 1921.6.13.2, 1921.4.51, & 1921.3.4.2)Ve = (Mpr, left, top + Mpr, right,bot) / H + Vu = 175.1 kips

< 8φ(fc')0.5bd = 216.7 kips [Satisfactory]

< φ[Vc + Avfyd/s ] = 197.8 kips [Satisfactory]where d = 58.19 in

ρleft = 0.003 > ρmin=MIN[3(fc')0.5/fy, 200/fy ]= 0.003 [Satisfactory]

ρright = 0.003 > ρmin = 0.003 [Satisfactory]

Mpr, left, top = ρleft bd2fy (1.25 - 0.919ρleft fy/fc') = 691 ft-kips

Mpr, right ,bot = ρright bd2fy (1.25 - 0.919ρright fy/fc') = 691 ft-kipsφ = 0.85 (Sec 1909.3.2.3)

Av = 0.4 in2

Vc = 2(fc')0.5bd = 0.0 kips, (Per Sec 1921.3.4.2, Vc = 0, if (Ve -Vu) 50% Ve AND Pu < Ag fc' / 20 )

DanielTian Li



f'c = ksi t = 7.25 infy = ksi cC = 1 in

REBAR As, in2/ft d, in a, in T, k/ft φφφφMn, ft-k/ft# 6 @ 24 " O.C. 0.220 5.88 0.288 13.20 5.67# 6 @ 22 " O.C. 0.240 5.88 0.314 14.40 6.18# 6 @ 20 " O.C. 0.264 5.88 0.345 15.84 6.77# 6 @ 18 " O.C. 0.293 5.88 0.383 17.60 7.50# 6 @ 16 " O.C. 0.330 5.88 0.431 19.80 8.40# 6 @ 14 " O.C. 0.377 5.88 0.493 22.63 9.55# 6 @ 12 " O.C. 0.440 5.88 0.575 26.40 11.06# 6 @ 10 " O.C. 0.528 5.88 0.690 31.68 13.14# 6 @ 8 " O.C. 0.660 5.88 0.863 39.60 16.17# 6 @ 6 " O.C. 0.880 5.88 1.150 52.80 20.99# 6 @ 4 " O.C. 1.320 5.88 1.725 79.20 29.77# 6 @ 2 " O.C. 2.640 5.88 3.451 158.40 49.30

# 5 @ 24 " O.C. 0.155 5.94 0.203 9.30 4.07# 5 @ 22 " O.C. 0.169 5.94 0.221 10.15 4.43# 5 @ 20 " O.C. 0.186 5.94 0.243 11.16 4.87# 5 @ 18 " O.C. 0.207 5.94 0.270 12.40 5.40# 5 @ 16 " O.C. 0.233 5.94 0.304 13.95 6.05# 5 @ 14 " O.C. 0.266 5.94 0.347 15.94 6.89# 5 @ 12 " O.C. 0.310 5.94 0.405 18.60 8.00# 5 @ 10 " O.C. 0.372 5.94 0.486 22.32 9.53# 5 @ 8 " O.C. 0.465 5.94 0.608 27.90 11.79# 5 @ 6 " O.C. 0.620 5.94 0.810 37.20 15.44# 5 @ 4 " O.C. 0.930 5.94 1.216 55.80 22.30# 5 @ 2 " O.C. 1.860 5.94 2.431 111.60 39.52

f'c = ksi t = 7.25 infy = ksi cC = 1.5 in

REBAR As, in2/ft d, in a, in T, k/ft φφφφMn, ft-k/ft# 5 @ 24 " O.C. 0.155 5.44 0.182 9.30 3.73# 5 @ 22 " O.C. 0.169 5.44 0.199 10.15 4.06# 5 @ 20 " O.C. 0.186 5.44 0.219 11.16 4.46# 5 @ 18 " O.C. 0.207 5.44 0.243 12.40 4.94# 5 @ 16 " O.C. 0.233 5.44 0.274 13.95 5.55# 5 @ 14 " O.C. 0.266 5.44 0.313 15.94 6.31# 5 @ 12 " O.C. 0.310 5.44 0.365 18.60 7.33# 5 @ 10 " O.C. 0.372 5.44 0.438 22.32 8.74# 5 @ 8 " O.C. 0.465 5.44 0.547 27.90 10.81# 5 @ 6 " O.C. 0.620 5.44 0.729 37.20 14.15# 5 @ 4 " O.C. 0.930 5.44 1.094 55.80 20.47# 5 @ 2 " O.C. 1.860 5.44 2.188 111.60 36.35

# 4 @ 24 " O.C. 0.100 5.50 0.118 6.00 2.45# 4 @ 22 " O.C. 0.109 5.50 0.128 6.55 2.67# 4 @ 20 " O.C. 0.120 5.50 0.141 7.20 2.93# 4 @ 18 " O.C. 0.133 5.50 0.157 8.00 3.25# 4 @ 16 " O.C. 0.150 5.50 0.176 9.00 3.65# 4 @ 14 " O.C. 0.171 5.50 0.202 10.29 4.17# 4 @ 12 " O.C. 0.200 5.50 0.235 12.00 4.84# 4 @ 10 " O.C. 0.240 5.50 0.282 14.40 5.79# 4 @ 8 " O.C. 0.300 5.50 0.353 18.00 7.19# 4 @ 6 " O.C. 0.400 5.50 0.471 24.00 9.48# 4 @ 4 " O.C. 0.600 5.50 0.706 36.00 13.90# 4 @ 2 " O.C. 1.200 5.50 1.412 72.00 25.89

604.5

Concrete Slab Capacity Based on ACI 318-02 / CBC 2001

5605

DanielTian Li



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 4 ksiSPECIFIED STRENGTH OF FASTENER fut = 60 ksi (The strength of most fastenings is likely to be controlled by the embedment strength rather than the steel strength, so it is usually economical to use ASTM A307 Grade A fastener.)FACTORED DESIGN LOAD Nu = 7.4 kEFFECTIVE EMBEDMENT DEPTH hef = 3.5 inFASTENER DIAMETER d = 0.625 inFASTENER HEAD TYPE 2 Heavy Square ( 1=Square, 2=Heavy Square, 3=Hex, 4=Heavy Hex, 5=Hardened Washers ) [THE FASTENER DESIGN IS ADEQUATE.]

ANALYSISEFFECTIVE AREA OF FASTENER Ase = 0.226 in2

BEARING AREA OF HEAD Ab = 0.822 in2

CHECK FASTENER TENSILE STRENGTH :

= 9.763 k > Nu [Satisfactory]

where : φ = 0.9CHECK CONCRETE BREAKOUT STRENGTH :

0.625 = 7.454 k > Nu [Satisfactory]

where : φ = 0.75 AN/Ano and ψ2 terms are 1.0 for single fasteners away form edges. ψ3 term is 1.0 for location where concrete cracking is likely to occur (i.e., bottom of the slab)

CHECK PULLOUT STRENGTH :


where : φ = 0.75 ψ4 term is 1.0 for location where concrete cracking is likely to occur.

EVALUATE SIDE-FACE BLOWOUT :Since this fastener is located far from a free edge of concrete (c>0.4hef ) this type of failure mode is not applicable.

REQUIRED EDGE DISTANCES AND SPACINGS TO PRECLUDE SPLITTING FAILURE :Since this fastener is located far from a free edge of concrete (c>0.4hef ) this type of failure mode is not applicable.

Summary of Dimensional Properties of FastenersEffective

Gross Area ofArea of Threaded

Fastener Fastener Heavy Heavy Hardened( in2 ) ( in2 ) Square Hex Washers

0.250 1/4 0.049 0.032 0.142 0.201 0.117 0.167 0.2580.375 3/8 0.110 0.078 0.280 0.362 0.164 0.299 0.408 Techincal Reference:0.500 1/2 0.196 0.142 0.464 0.569 0.291 0.467 0.690 1. Ronald Cook, "Strength Design0.625 5/8 0.307 0.226 0.693 0.822 0.454 0.671 1.046 of Anchorage to Concrete," PCA,0.750 3/4 0.442 0.334 0.824 1.121 0.654 0.911 1.252 1999.0.875 7/8 0.601 0.462 1.121 1.465 0.891 1.188 1.8041.000 1 0.785 0.606 1.465 1.855 1.163 1.501 2.3561.125 1 1/8 0.994 0.763 1.854 2.291 1.472 1.851 2.982 1.250 1 1/4 1.227 0.969 2.288 2.773 1.817 2.237 3.6821.375 1 3/8 1.485 1.160 2.769 3.300 2.199 2.659 4.4551.500 1 1/2 1.767 1.410 3.295 3.873 2.617 3.118 5.3011.750 1 3/4 2.405 1.900 - - - 4.144 6.5412.000 2 3.142 2.500 - - - 5.316 7.903

HexSquare( in )

FastenerDiameter

Bearing Area of Heads, Nuts, and Washers( Ab ) ( in

2 )

Single Tension Fastener Away from Edges Based on ACI 318-02

DanielTian Li

( )0.8ses utn fN Aφ φ=

( )' 1.52 3 2 3 24N N

cb b efcNo No

A A fN N hA A

φ φ φψ ψ ψ ψ= =

( )'4 8bpn cfN Aφ φψ=



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 4 ksiSPECIFIED STRENGTH OF FASTENER fut = 60 ksi (The strength of most fastenings is likely to be controlled by the embedment strength rather than the steel strength, so it is usually economical to use ASTM A307 Grade A fastener.)FACTORED DESIGN LOAD Nu = 11.7 kEFFECTIVE EMBEDMENT DEPTH hef = 4.5 inFASTENER DIAMETER d = 0.5 inFASTENER HEAD TYPE 2 Heavy Square ( 1=Square, 2=Heavy Square, 3=Hex, 4=Heavy Hex, 5=Hardened Washers )ECCENTRICITY e = 2 inFASTENER CENTER-TO-CENTER SPACING s = 6 inDIST. FR. THE OUTER FASTENERS TO EDGE c = 3 in

ANALYSIS [THE FASTENER DESIGN IS ADEQUATE.]TOTAL NUMBER OF FASTENERS n = 4EFFECTIVE AREA OF FASTENER Ase = 0.142 in2

BEARING AREA OF HEAD Ab = 0.569 in2, ( or determined from manufactures's catalogs.)CHECK HIGHTEST TENSILE STRENGTH :

= 6.134 k > = 4.875 [Satisfactory]


= 11.773 k > Nu [Satisfactory]where : φ = 0.75

ψ3 term is 1.0 for location where concrete cracking is likely to occur.CHECK PULLOUT STRENGTH OF SINGLE STUD :

= 13.656 k > Nu, max, 1 stud [Satisfactory]


EVALUATE SIDE-FACE BLOWOUT :c > 0.4hef [Satisfactory]

Since the fasteners are located far from a free edge of concrete c>0.4hef this type of failure mode is not applicable.REQUIRED EDGE DISTANCES AND SPACINGS TO PRECLUDE SPLITTING FAILURE :

Since a welded, headed fastener is not torqued, the minimun cover requirements of ACI 318 Sec. 7.7 apply.CoverProvd > CoverReqd [Satisfactory]




0.250 1/4 0.049 0.032 0.142 0.201 0.117 0.167 0.2580.375 3/8 0.110 0.078 0.280 0.362 0.164 0.299 0.4080.500 1/2 0.196 0.142 0.464 0.569 0.291 0.467 0.6900.625 5/8 0.307 0.226 0.693 0.822 0.454 0.671 1.0460.750 3/4 0.442 0.334 0.824 1.121 0.654 0.911 1.252 Techincal Reference:0.875 7/8 0.601 0.462 1.121 1.465 0.891 1.188 1.804 1. Ronald Cook, "Strength Design1.000 1 0.785 0.606 1.465 1.855 1.163 1.501 2.356 of Anchorage to Concrete," PCA,1.125 1 1/8 0.994 0.763 1.854 2.291 1.472 1.851 2.982 1999.1.250 1 1/4 1.227 0.969 2.288 2.773 1.817 2.237 3.6821.375 1 3/8 1.485 1.160 2.769 3.300 2.199 2.659 4.4551.500 1 1/2 1.767 1.410 3.295 3.873 2.617 3.118 5.3011.750 1 3/4 2.405 1.900 - - - 4.144 6.5412.000 2 3.142 2.500 - - - 5.316 7.903

Group of Tension Fasteners Near an Edge with Eccentricity Based on ACI 318-02

HexSquare( in )

FastenerDiameter


2 )

DanielTian Li

( ),1 0.8ses stud utfN Aφ φ=

( ) ( )' 1.51 2 3 32

1 0.30.7 24

2 1.59 13

N Ncbg b efc

No efef

ef

cA A fN N he hA hh

φ φ φψ ψ ψ ψ

= = + +


( ),max,1

2uu stud

s eNN

ns

+=



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 4 ksiSPECIFIED STRENGTH OF FASTENER fut = 60 ksi (The strength of most fastenings is likely to be controlled by the embedment strength rather than the steel strength, so it is usually economical to use ASTM A307 Grade A fastener.)FACTORED DESIGN TENSION LOAD Nu = 2.47 kFACTORED DESIGN SHEAR LOAD Vu = 0.65 kEFFECTIVE EMBEDMENT DEPTH hef = 7 inFASTENER DIAMETER d = 0.5 in [THE FASTENER DESIGN IS ADEQUATE.]FASTENER HEAD TYPE 3 Hex ( 1=Square, 2=Heavy Square, 3=Hex, 4=Heavy Hex, 5=Hardened Washers )DIST. BETWEEN THE FASTENER AND EDGE c = 1.75 in

ANALYSISEFFECTIVE AREA OF FASTENER Ase = 0.142 in2

BEARING AREA OF HEAD Ab = 0.291 in2, ( or determined from manufactures's catalogs.)

CHECK FASTENER TENSILE STRENGTH :




ψ3 term is 1.0 for location where concrete cracking is likely to occur.CHECK PULLOUT STRENGTH :



CHECK SIDE-FACE BLOWOUT STRENGTH : (Since c > 0.4hef , this type of failure mode is applicable.)


where : φ = 0.75DETERMINE DESIGN TENSILE STRENGTH :

= 6.134 K

CHECK FASTENER SHEAR STRENGTH :

= 4.601 k > Vu [Satisfactory]

where : φ = 0.9CHECK CONCRETE BREAKOUT STRENGTH FOR SHEAR LOAD :


where : φ = 0.75 ψ7 term is 1.0 for location where concrete cracking is likely to occur. AV/AVo and ψ6 terms are 1.0 for single shear fastener not influenced by more than one free edge. term is load bearing length of the anchor for shear, not to exceed 8d.

Single Fastener in Tension and Shear Near an Edge Based on ACI 318-02

DanielTian Li


( ) ( )' 1.52 3 32

0.30.7 24

1.59N N

cb b efcNo efef

cA A fN N hhA h

φ φ φψ ψ ψ

= = +


( )'160 bsb cc fN Aφ φ=

( )min , , ,n s cb pn sbN N N N Nφ φ φ φ φ=

0.6 ses utn fV Aφ φ=

0.2' 1.5

6 7 6 7 7V Vcb b c

Vo Vo

lA A d fV V cdA A

φ φ φψ ψ ψ ψ = =

(Cont'd)CHECK PRYOUT STRENGTH FOR SHEAR LOAD :

= 18.448 k > Vu [Satisfactory]where : φ = 0.75

ψ3 term is 1.0 for location where concrete cracking is likely to occur.kcp = 2.0 for hef > 2.5 in.

DETERMINE DESIGN TENSILE STRENGTH :

= 0.824 K

CHECK TENSION AND SHEAR INTERACTION :Since Nu > 0.2 φ Nn and

Vu > 0.2 φ Vn the full design strength is not permitted.

The interaction equation must be used

1.19 < 1.2 [Satisfactory]





HexSquare( in )

FastenerDiameter


2 )

( ) ( )' 1.52 3 32

0.30.7 24

1.59N N

cp cp b cp efcNo efef

cA A fV k N k hhA h

φ φ φψ ψ ψ

= = +

( )min , ,n s cb cpV V V Vφ φ φ φ=

u u

n n

N VN Vφ φ

+ =



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 4 ksiSPECIFIED STRENGTH OF FASTENER fut = 58 ksi (The strength of most fastenings is likely to be controlled by the embedment strength rather than the steel strength, so it is usually economical to use ASTM A307 Grade A fastener.)FACTORED DESIGN TENSION LOAD Nu = 21.3 kFACTORED DESIGN SHEAR LOAD Vu = 2 kEFFECTIVE EMBEDMENT DEPTH hef = 20 inFASTENER DIAMETER d = 1 inFASTENER HEAD TYPE 4 Heavy Hex ( 1=Square, 2=Heavy Square, 3=Hex, 4=Heavy Hex, 5=Hardened Washers )FASTENER CENTER-TO-CENTER SPACING s = 7 inDIST. BETWEEN THE FASTENER AND EDGE c1 = 14 inDIST. BETWEEN THE FASTENER AND EDGE c2 = 9 in

[THE FASTENER DESIGN IS ADEQUATE.]ANALYSISNUMBER OF FASTENERS n = 2EFFECTIVE AREA OF FASTENER Ase = 0.606 in2

BEARING AREA OF HEAD Ab = 1.501 in2, ( or determined from manufactures's catalogs.)

CHECK THE FASTENERS TENSILE STRENGTH :




ψ1 term is 1.0 for no eccentricity in the connection. ψ3 term is 1.0 for location where concrete cracking is likely to occur.

CHECK PULLOUT STRENGTH :



CHECK SIDE-FACE BLOWOUT STRENGTH :cmin > 0.4hef [Satisfactory]

Since the fasteners are located far from a free edge of concrete, c>0.4hef ,this type of failure mode is not applicable.DETERMINE DESIGN TENSILE STRENGTH :

= 45.225 K

CHECK FASTENERs SHEAR STRENGTH :


where : φ = 0.9

Group of Tension and Shear Fasteners Near Two Edges Based on ACI 318-02

DanielTian Li


( ) ( )'min 1.51 2 3 1 32

0.30.7 24

1.59N N

cbg b efcNo efef

cA A fN N hhA h

φ φ φψ ψ ψ ψ ψ

= = +

( )'4 8bpn cn fN Aφ φ ψ=

( )min , ,n s cb pnN N N Nφ φ φ φ=


CHECK CONCRETE BREAKOUT STRENGTH FOR SHEAR LOAD : (Cont'd)


ψ5 term is 1.0 for no eccentricity in the connection. ψ7 term is 1.0 for location where concrete cracking is likely to occur. term is load bearing length of the anchor for shear, not to exceed 8d.

CHECK PRYOUT STRENGTH FOR SHEAR LOAD :



DETERMINE DESIGN TENSILE STRENGTH :

= 19.243 K

REQUIRED EDGE DISTANCES AND SPACINGS TO PRECLUDE SPLITTING FAILURE :Since headed cast-in-place fasteners are not like to be highly torqued, the minimun cover requirements of ACI 318 Sec. 7.7 apply.CoverProvd > CoverReqd [Satisfactory]

CHECK TENSION AND SHEAR INTERACTION :Since Nu > 0.2 φ Nn and

Vu < 0.2 φ Vn the full tension design strength is permitted.

The interaction equation must be used






HexSquare( in )

FastenerDiameter


2 )

( )( ) 0.2' 1.5

5 6 7 5 72

1.5 1 1.5 1 2 20.7 0.3 7 1

4.5 1.5 11V

cbg b cVo

c c s c c lA d fV V cdccA

φ φ φψ ψ ψ ψ ψ + + = = +

( ) ( )'min 1.52 3 32

0.30.7 24

1.59N N


cA A fV k N k hhA h

φ φ φψ ψ ψ

= = +


u u

n n

N VN Vφ φ

+ =



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 4 ksiSPECIFIED STRENGTH OF FASTENER fut = 58 ksi (The strength of most fastenings is likely to be controlled by the embedment strength rather than the steel strength, so it is usually economical to use ASTM A307 Grade A fastener.)FACTORED COLUMN AXIAL LOAD Pu = 0 kFACTORED TOTAL BASE SHEAR LOAD Vu = 2 kFACTORED COLUMN MOMENT Mu = 14.58 ft-kEFFECTIVE EMBEDMENT DEPTH hef = 16 inFASTENER DIAMETER d = 0.875 inFASTENER HEAD TYPE 4 Heavy Hex ( 1=Square, 2=Heavy Square, 3=Hex, 4=Heavy Hex, 5=Hardened Washers )FASTENER CENTER-TO-CENTER SPACING s = 12 inDIST. BETWEEN FASTENER & EDGE c1 = 7 inDIST. BETWEEN FASTENER & EDGE c2 = 12 in

[THE FASTENER DESIGN IS ADEQUATE.]

ANALYSISNUMBER OF GOVERNING FASTENERS n = 2 16MAX TENSION ON GOV'G FASTENERS Nu = Mu / (s - 2.0") - Pu / 2 = 17.50 kSHEAR FORCE ON GOV'G FASTENERS Vu = Vu, total / 2 = 1 kEFFECTIVE AREA OF ONE FASTENER Ase = 0.462 in2

BEARING AREA OF ONE HEAD Ab = 1.188 in2, ( or determined from manufactures's catalogs.)

CHECK THE GOVERNING FASTENERS TENSILE STRENGTH :


where : φ = 0.9CHECK CONCRETE BREAKOUT STRENGTH OF GOVERNING FASTENERS :


ψ1 term is 1.0 for no eccentricity in the connection. ψ3 term is 1.0 for location where concrete cracking is likely to occur.

CHECK PULLOUT STRENGTH OF GOVERNING FASTENERS :



CHECK SIDE-FACE BLOWOUT STRENGTH :cmin > 0.4hef [Satisfactory]

Since the fasteners are located far from a free edge of concrete, c>0.4hef ,this type of failure mode is not applicable.DETERMINE DESIGN TENSILE STRENGTH OF GOVERNING FASTENERS :

= 37.056 K

CHECKGOVERNING FASTENERS SHEAR STRENGTH :


where : φ = 0.9 (for built-up grout pads, φ = 0.8 x 0.9, ACI 318-02 D6.1.3)

Column Base Fasteners Near Two Edges Based on ACI 318-02

DanielTian Li


( ) ( )'min 1.51 2 3 1 32

0.30.7 24

1.59N N

cbg b efcNo efef

cA A fN N hhA h

φ φ φψ ψ ψ ψ ψ

= = +

( )'4 8bpn cn fN Aφ φ ψ=

( )min , ,n s cb pnN N N Nφ φ φ φ=


CHECK CONCRETE BREAKOUT STRENGTH FOR SHEAR LOAD : (Cont'd)


ψ5 term is 1.0 for no eccentricity in the connection. ψ7 term is 1.0 for location where concrete cracking is likely to occur. term is load bearing length of the anchor for shear, not to exceed 8d.

CHECK PRYOUT STRENGTH FOR SHEAR LOAD ON GOVERNING FASTENERS :



DETERMINE DESIGN TENSILE STRENGTH OF GOVERNING FASTENERS :

= 14.938 K

REQUIRED EDGE DISTANCES AND SPACINGS TO PRECLUDE SPLITTING FAILURE :Since headed cast-in-place fasteners are not like to be highly torqued, the minimun cover requirements of ACI 318 Sec. 7.7 apply.CoverProvd > CoverReqd [Satisfactory]

CHECK TENSION AND SHEAR INTERACTION OF GOVERNING FASTENERS :Since Nu > 0.2 φ Nn and

Vu < 0.2 φ Vn the full tension design strength is permitted.The interaction equation must be used






HexSquare( in )

FastenerDiameter


2 )

( )( ) 0.2' 1.5

5 6 7 5 72

1.5 1 1.5 1 2 20.7 0.3 7 1

4.5 1.5 11V

cbg b cVo

c c s c c lA d fV V cdccA

φ φ φψ ψ ψ ψ ψ + + = = +

( ) ( )'min 1.52 3 32

0.30.7 24

1.59N N


cA A fV k N k hhA h

φ φ φψ ψ ψ

= = +


u u

n n

N VN Vφ φ

+ =


JOB NO. : DATE : REVIEW BY : Concrete Beam Design Based on ACI 318-02

INPUT DATAf 'c = 2.5 ksi Mu = 272.5 ft-k

Main Bar fy = 60 ksi Vu = 46.4 k

Stirrup fy = 60 ksi Tu = 110 ft-kb = 24 in, (ACI 8.10.2)

Top bars 3 # 7 h = 30 in

Bot bars 7 # 7 bw = 24 in

hf = 0 inStirrup size ==> # 4 @ 10 in o.c.No. of legs = 4 d (optional) = in The design is inadequateDESIGN SUMMARY 2Main bar top (Compressive Reinf.) : Use 3 # 7 Stirrups : Use # 4 @ 10 in. o.c. (4 legs)Main bar bottom (Tensile Reinf.) : Use 7 # 7 ( 1 layer )GOVERNING CASE ANALYSIS

= 0.003 = 0.85 = 0.0129

= 27.6 in = 0.90 = 29000 ksi

= 2.4 in = 0.0033 0.0063

24

Case 1 Case 2 Case 3 Case 4

CASE 2 APPLICABLE

FLEXURAL ANALYSISCase 1 Analysis :

in in2

Case 2 Analysis :

2.72 in 2.31 in2

Case 3 Analysis :

in in2

in2 ksi

Case 4 Analysis :

in in2

in2 ksi

DanielTian Li

uε 1β

d φ

minρ

SA'd

2 20.85 '

u

C

Ma d db fφ

= − − =0.85 'C

S

y

ab fA

f= =

( )2 20.85f ' 2

fuw f

w C

hMa d b dd b hfb

= − − − − − =

( )0.85 ' w f w fCS

y

a bf b h b hA

f

+ −= =

0.004u

u

da ε

ε= =

+ ( )

0.85 '2

'' '

u C

S

S

aba dfM

Ad df

φ

φ

− − = =

−

' 0.85 '' S C

S S

y y

baf fA A

f f= + =

( )

( )

0.85 ' 0.85 '2 2

'' '

fu w w fC C

S

S

a ha d b df fb b hMA

d df

φ

φ

− − + − − = =

−

( )0.85 ''' w f wCS

S S

y y

b af b h bfA A

f f

− + = + =

'1' S uS

a df Eaβ

ε−

= =

'1' S uS

a df Eaβ

ε−

= =

maxρ

SE

sw

w

Adb

ρ = =

0.004u

u

da ε

ε= =

+

SHEAR ANALYSIS (cont'd)Check section limitation (ACI 11.5.6.8) Determine concrete capacity (ACI 11.3.1.1 or 11.3.2.1)

66.15 k46.4 < 248.1 k [Satisfactory]

where φ = 0.7572.06 k ,<== applicable

Check shear reinforcement (ACI 11.5)

where 54.07

0.391

= 0.240 in2 / ft < 0.960 in2 / ft [Satisfactory]

Check spacing limits for shear reinforcement (ACI 11.5.4)

-10.20 k

= 13 > S = 10 in[Satisfactory]

TORSION ANALYSISCheck section limitation (ACI 11.6.3.1)

where φ = 0.75 (ACI 9.3.2.3)

Ph = 80 in, (perimeter of centerline of outermost closedtransverse torsional reinforcement.)

Aoh = 391 in2 (area enclosed by centerline of the outermost

0.412 > 0.375 [Unsatisfactory] closed transverse torsional reinforcement.)

Check if torsional reinforcement required (ACI 11.6.1)

where be = MIN(h-hf , 4hf) = 0 in, (one side, ACI 6.1.1)

Pcp = 108 in, (outside perimeter of the concrete crosssection.)

110.0 > 15.0 ft-k Acp = 720 in2 (area enclosed by outside perimeter of

Torsional reinforcement reqD. concrete cross section.)

Check the max factored torque causing cracking (ACI 11.6.2.2)

110.0 > 60.0 Torsional reinforcement reqD for full torque.

Determine the area of one leg of a closed stirrup (ACI 11.6.3.6)

0.00 in2 / ft

Determine the corresponding area of longitudinal reinforcement (derived from ACI 11.6.3.7 & 11.6.5.3)

0.00 in2

22

'821.7

C

cw

V T P Vu u h fd db bAw oh

φ + ≤ +

2' cp

u ccp

AfTP

φ

≤

0 02 1.7t u u

hyv yv

A T Ts f fA Aφ φ

= = =

'5 25, ,cpyv yvct t w

L h h

yL yL yL yv

f ffA bA AMAX MAXA P Ps sf f f f

= − =

'2C w cd fV b= ='10u w cd fV bφ≤

Re

0 ,2

50,

2

,

cu

w cu c

qD y

u cc u

y

Vfor V

A b Vv for V Vs f

V V for V Vdf

φ

φ φ

φ φφ

< = ≤ ≤

− ≤

( )1.9 2500C wwA B dV bρ= + =

( )' , 100cA MIN f= =

, 1.0u

u

dVB MINM

= =

ProvD

Avs

=

u Cs

V VV

φφ

−= =

'

max,'

( , 24) 42

( , 12) 44

s w c

shear

s w c

dMIN for d fV b

Sd

MIN for d fV b

≤= >

2'4 cp

u ccp

AfTP

φ

≤

(cont'd)Determine minimum combined area of longitudinal reinforcement

AL, top = As' +0.5AL = 0.00 in2< actual [Satisfactory]

AL, bot = As +0.5AL = 2.31 in2< actual [Satisfactory]

Determine minimum diameter for longitudinal reinforcement (ACI 11.6.6.2)

dbL = MAX(S/24, 0.375) = 0.42 in < 0.88 in [Satisfactory]

Determine minimum combined area of stirrups (ACI 11.6.5.2 & 11.6.6.1)

At / S = MAX[(Av+2At)S, 50bw/fyv] = 0.24 in2 / ft < actual [Satisfactory]

Smax, tor = MIN[(Ph/8, 12) = 10 in

SreqD = MIN(Smax,shear , Smax,tor) = 10 in > actual [Satisfactory]


JOB NO. : DATE : REVIEW BY : Concrete Beam Design Based on ACI 318-99/95

INPUT DATAf 'c = 2.5 ksi Mu = 272.5 ft-k

Main Bar fy = 60 ksi Vu = 46.4 k

Stirrup fy = 60 ksi Tu = 110 ft-kb = 24 in, (ACI 8.10.2)

Top bars 3 # 7 h = 30 in

Bot bars 7 # 7 bw = 24 in

hf = 0 inStirrup size ==> # 4 @ 10 in o.c.No. of legs = 4 d (optional) = in The design is adequate.DESIGN SUMMARY 2Main bar top (Compressive Reinf.) : Use 3 # 7 Stirrups : Use # 4 @ 10 in. o.c. (4 legs)Main bar bottom (Tensile Reinf.) : Use 7 # 7 ( 1 layer )GOVERNING CASE ANALYSIS

= 0.003 = 0.85 = 0.0134

= 27.6 in = 0.90 = 29000 ksi

= 2.4 in = 0.0033 0.0063

30

Case 1 Case 2 Case 3 Case 4

CASE 2 APPLICABLE

FLEXURAL ANALYSISCase 1 Analysis :

in in2

Case 2 Analysis :

2.72 in 2.31 in2

Case 3 Analysis :

in in2

in2 ksi

Case 4 Analysis :

in in2

in2 ksi

DanielTian Li

uε 1β

d φ

minρ

SA'd

2 20.85 '

u

C

Ma d db fφ

= − − =0.85 'C

S

y

ab fA

f= =

( )2 20.85f ' 2

fuw f

w C

hMa d b dd b hfb

= − − − − − =

( )0.85 ' w f w fCS

y

a bf b h b hA

f

+ −= =

10.75 u

yu

S

da

f

E

β ε

ε= =

+ ( )

0.85 '2

'' '

u C

S

S

aba dfM

Ad df

φ

φ

− − = =

−

' 0.85 '' S C

S S

y y

baf fA A

f f= + =

10.75 u

yu

S

da

f

E

β ε

ε= =

+

( )

( )

0.85 ' 0.85 '2 2

'' '

fu w w fC C

S

S

a ha d b df fb b hMA

d df

φ

φ

− − + − − = =

−

( )0.85 ''' w f wCS

S S

y y

b af b h bfA A

f f

− + = + =

'1' S uS

a df Eaβ

ε−

= =

'1' S uS

a df Eaβ

ε−

= =

maxρ

SE

sw

w

Adb

ρ = =

SHEAR ANALYSIS (cont'd)Check section limitation (ACI 11.5.6.8) Determine concrete capacity (ACI 11.3.1.1 or 11.3.2.1)

66.15 k46.4 < 281.1 k [Satisfactory]

where φ = 0.8572.06 k ,<== applicable

Check shear reinforcement (ACI 11.5)

where 54.07

0.391

= 0.240 in2 / ft < 0.960 in2 / ft [Satisfactory]

Check spacing limits for shear reinforcement (ACI 11.5.4)

-17.48 k

= 13 > S = 10 in[Satisfactory]

TORSION ANALYSISCheck section limitation (ACI 11.6.3.1)

where φ = 0.85 (ACI 9.3.2.3)

Ph = 80 in, (perimeter of centerline of outermost closedtransverse torsional reinforcement.)

Aoh = 391 in2 (area enclosed by centerline of the outermost

0.412 < 0.425 [Satisfactory] closed transverse torsional reinforcement.)

Check if torsional reinforcement required (ACI 11.6.1)

where be = MIN(h-hf , 4hf) = 0 in, (one side, ACI 6.1.1)

Pcp = 108 in, (outside perimeter of the concrete crosssection.)

110.0 > 17.0 ft-k Acp = 720 in2 (area enclosed by outside perimeter of

Torsional reinforcement reqD. concrete cross section.)

Check the max factored torque causing cracking (ACI 11.6.2.2)

110.0 > 68.0 Torsional reinforcement reqD for full torque.

Determine the area of one leg of a closed stirrup (ACI 11.6.3.6)

0.00 in2 / ft

Determine the corresponding area of longitudinal reinforcement (derived from ACI 11.6.3.7 & 11.6.5.3)

0.00 in2

22

'821.7

C

cw

V T P Vu u h fd db bAw oh

φ + ≤ +

2' cp

u ccp

AfTP

φ

≤

0 02 1.7t u u

hyv yv

A T Ts f fA Aφ φ

= = =

'5 25, ,cpyv yvct t w

L h h

yL yL yL yv

f ffA bA AMAX MAXA P Ps sf f f f

= − =

'2C w cd fV b= ='10u w cd fV bφ≤

Re

0 ,2

50,

2

,

cu

w cu c

qD y

u cc u

y

Vfor V

A b Vv for V Vs f

V V for V Vdf

φ

φ φ

φ φφ

< = ≤ ≤

− ≤

( )1.9 2500C wwA B dV bρ= + =

( )' , 100cA MIN f= =

, 1.0u

u

dVB MINM

= =

ProvD

Avs

=

u Cs

V VV

φφ

−= =

'

max,'

( , 24) 42

( , 12) 44

s w c

shear

s w c

dMIN for d fV b

Sd

MIN for d fV b

≤= >

2'4 cp

u ccp

AfTP

φ

≤

(cont'd)Determine minimum combined area of longitudinal reinforcement

AL, top = As' +0.5AL = 0.00 in2< actual [Satisfactory]

AL, bot = As +0.5AL = 2.31 in2< actual [Satisfactory]

Determine minimum diameter for longitudinal reinforcement (ACI 11.6.6.2)

dbL = MAX(S/24, 0.375) = 0.42 in < 0.88 in [Satisfactory]

Determine minimum combined area of stirrups (ACI 11.6.5.2 & 11.6.6.1)

At / S = MAX[(Av+2At)S, 50bw/fyv] = 0.24 in2 / ft < actual [Satisfactory]

Smax, tor = MIN[(Ph/8, 12) = 10 in

SreqD = MIN(Smax,shear , Smax,tor) = 10 in > actual [Satisfactory]

PROJECT : PAGE :

CLIENT : DESIGN BY :


INPUT DATACONCRETE STRENGTH fc' = 3 ksiREBAR YIELD STRESS fy = 60 ksiFACTORED SHEAR LOAD Vu = 120 kFACTORED MOMENT LOAD Mu = 200 ft-kWIDTH b = 12 inEFFECTIVE DEPTH d = 32 inCLEAR SPAN Ln = 12 ft

VERTICAL REINF. 1 # 3 @ 6 in o.c.

HORIZONTAL REINF. 2 # 4 @ 10 in o.c.

TENSION REINFORCEMENT 3 # 5

ANALYSIS

CHECK SECTION2.5 > = 1.94 [SATISFACTORY]

[ACI 11-29]CHECK CAPACITY

= k

(ACI Sec. 11.8.7)

= 57.73 k

(ACI Sec. 11.8.8)

k > 120.00 k

(ACI Sec. 11.1.1) [SATISFACTORY]

84.63

121.01

Deep Beam Design

DanielT. Li

2.53.5 u

u

MdV

−

' '2.5 25003.5 1.9 , 6u s uc c c

u u

A VMMIN bd bdf fVdV bM

= − +

( )c sV Vφ + =

≥

≥

1 11

12 12

n nv vh

s y

L LA A

d ddfVs h

+ − = +

uV =



INPUT DATACOLUMN WIDTH c1 = 16 in

COLUMN DEPTH c2 = 16 in

SLAB CONCRETE STRENGTH fc' = 4 ksi

SLAB THICKNESS T = 16 in

DROP CAP / PANEL THICKNESS Td = 8 in

BAR SIZE AT TOP SLAB # 10PUNCHING REINFORCING 2 # 5 each way

8

ANALYSISPUNCHING CAPACITY FOR COLUMN (ACI 318-02 SEC.11.5.6.4, 11.12.1.2, & 11.12.6)

683.089 kips

where φ = 0.75 (ACI 318-02, Section 9.3.2.3 )βc = ratio of long side to short side of concentrated load = 1.00

d = T + Tc - 2" cover - 2 (0.5 db) = 20.73 in

b0 = 2b1 + 2b2 = 2(c1 + d) + 2(c2 + d) = 146.9 in

Ap = b0 d = 3046 in2

y = MIN(2 , 4 / βc , 40 d / b0) = 2.0

Av = 2.48 in2

α = 45 0

fy = 60 ksi

PUNCHING CAPACITY FOR POINT LOAD (ACI 318-02 SEC.11.12.1.2, & 11.12.6)

96.6 kips


d = T - 2" cover - 2 (0.5 db) = 12.73 in

b0 = d π = 39.99 in

Ap = b0 d = 509 in2

Slab Punching Design Based on ACI 318-02

DanielTian Li

( ) ( )' '02 sin , 3, y MIN df f fbP A Au col p vc y cφ α≤ + + =

'4, int fP Au po pcφ≤ =



INPUT DATACOLUMN DIAMETER D = 22 in

SLAB CONCRETE STRENGTH fc' = 4 ksi

SLAB THICKNESS T = 16 in

DROP CAP / PANEL THICKNESS Td = 8 in

BAR SIZE AT TOP SLAB # 10PUNCHING REINFORCING 4 # 8 each way

ANALYSISPUNCHING CAPACITY FOR COLUMN (ACI 318-02 SEC.11.5.6.4, 11.12.1.2, & 11.12.6)

1056 kips


d = T + Tc - 2" cover - 2 (0.5 db) = 20.73 in

b0 = (D + d) π = 134.2 in

Ap = b0 d = 2783 in2

Av = 12.64 in2

α = 45 0

fy = 60 ksi

PUNCHING CAPACITY FOR POINT LOAD (ACI 318-02 SEC.11.12.1.2, & 11.12.6)

96.6 kips


d = T - 2" cover - 2 (0.5 db) = 12.73 in

b0 = d π = 39.99 in

Ap = b0 d = 509 in2

Slab Punching Design Based on ACI 318-02

DanielTian Li

( )' '04 sin , 3, MIN df f fbP A Au col p vc y cφ α≤ + =

'4, int fP Au po pcφ≤ =



Coupling Beam Design Based on ACI 318-02 / IBC 2003

INPUT DATACONCRETE STRENGTH fc' = 4 ksi


FACTORED SHEAR LOAD Vu = 254 k

WIDTH b = 16 in 72

OVERALL DEPTH h = 72 in

CLEAR SPAN L = 6 ft

DIAGONAL BARS, Ad 4 # 9

DIAGONAL TIES, Ash # 4 @ 4

HORIZONTAL BARS, Avh 16 # 4 ( 2 layers @ 10 in o.c. )

VERTICAL TIES, Av 2 legs # 3 @ 6 in o.c.

ANALYSISCHECK DIAGONAL BARS REQUIREMENT (ACI 318-02, 21.7.7.2 & 21.7.7.3)

< 4< 2

Vu < 291 k [Coupling Beam Permitted]

CHECK DIAGONAL BARS (ACI 318-02, 21.7.7.4)

Vu < 255 k [Satisfactory]

CHECK TIES AROUND DIAGONAL BARS (ACI 318-02, 21.4.4)

0.384 in2 <[Satisfactory]

CHECK VERTICAL TIES (ACI 318-02, 11.8.5)


CHECK HORIZONAL BARS (ACI 318-02, 11.8.4)


DETERMINE LONGITUDIANL BARS (ACI 318-02, 10.5.1)

3.07 in2 ==> ( 4 # 8 longitudinal bars both top & bottom)

(Note: These bars are not recommended by SEAOC to be used, and are not shown in figure above.)

L / h = 1.00

Tian LiDaniel

' '0.3 0.091 ,gc cc c

shchy y

f fAsh shMAXAf fA

= − =

,sh provdA

0.0015v bsA = = ,v provdA

0.0025vh bhA = = ,v provdA

'

,min

3200, c

s

y y

bd fbdMAXA

f f

= =

'4n cbh fVφ = =

'(2 sin , 10 )dn y cMIN bhf fV Aφ φ α= =



Coupling Beam Design Based on CBC 2001 / UBC 1997

INPUT DATACONCRETE STRENGTH fc' = 4 ksi


FACTORED SHEAR LOAD Vu = 254 k

WIDTH b = 16 in

OVERALL DEPTH h = 72 in 16

CLEAR SPAN L = 6 ft

DIAGONAL BARS, Ad 4 # 9

DIAGONAL TIES, Ash # 4 @ 4

HORIZONTAL BARS, Avh 16 # 4 ( 2 layers @ 10 in o.c. )

VERTICAL TIES, Av 2 legs # 3 @ 6 in o.c.

ANALYSISCHECK DIAGONAL BARS REQUIREMENT (SEC.1921.6.10.2)

L / d = 1.02 < 4

where d = 70.25 in

Vu < 284 k [Coupling Beam Permitted]

CHECK DIAGONAL BARS (SEC.1921.6.10.2)

Vu < 255 k [Satisfactory]

CHECK TIES AROUND DIAGONAL BARS (SEC. 1921.4.4)


CHECK VERTICAL TIES (SEC. 1911.8.9)


CHECK HORIZONAL BARS (SEC. 1911.8.10)


DETERMINE LONGITUDIANL BARS (SEC. 1910.5.1)

3.07 in2 ==> ( 4 # 8 longitudinal bars both top & bottom)

(Note: These bars are not recommended by SEAOC to be used, and are not shown in figure above.)

Tian LiDaniel

' '0.3 0.091 ,gc cc c

shchy y

f fAsh shMAXAf fA

= − =

'(2 sin , 10 )dn y cMIN bdf fV Aφ φ α= =

,sh provdA

0.0015v bsA = = ,v provdA

0.0025vh bhA = = ,v provdA

'

,min

3200, c

s

y y

bd fbdMAXA

f f

= =

'4n cbd fVφ = =



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 4.2 ksi


COLUMN DIMENSIONS cx = 36 in

cy = 40 in


FACTORED MAGNIFIED MOMENT Mu,x = 800 ft-k

Mu,y = 100 ft-k

FACTORED SHEAR LOAD Vu,x = 130 k

Vu,y = 150 kCOLUMN VERT. REINF. 8 # 10 at x dir.

9 # 10 at y dir.SHEAR REINF. 4 legs,# 4 @ 12 in o.c., x dir. THE COLUMN DESIGN IS ADEQUATE.

2 legs,# 4 @ 12 in o.c., y dir.

ANALYSIS

φ Pn (k)


AT AXIAL LOAD ONLY 3791 0AT MAXIMUM LOAD 3791 1058AT 0 % TENSION 3343 1492AT 25 % TENSION 2773 1958AT 50 % TENSION 2290 2251

AT ε t = 0.002 1564 2583

AT BALANCED CONDITION 1521 2609

φ Mn (ft-k) AT ε t = 0.005 667 3275

AT FLEXURE ONLY 0 2726CHECK FLEXURAL & AXIAL CAPACITY

Mu = ( Μu,x2 + Mu,y

2 )0.5 = ft-k, (combined bending load.) θ = 7.13o , (the direction of combined load.)



Ag = 1440 in2. Ast = 38.10 in2.


φ = 0.48 + 83 εt = (ACI 318-02, Fig. R9.3.2)

where Cb = d εc / (εc + εs) = 24 in εt = 0.002069 εc = 0.003

d = 41.2149 in, (ACI 7.7.1) β1 = 0.84 ( ACI 318-02, Sec. 10.2.7.3 )

φ Mn = 0.9 Μn = 2726 ft-kips @ Pn = 0, (ACI 318-02, Sec. 9.3.2) ,& εt,min = 0.004, (ACI 318-02, Sec. 10.3.5)

φ Mn = ft-kips @ Pu = 1700 kips > Mu [Satisfactory]


ρmin = 0.01 (ACI 318-02, Section 10.9) [Satisfactory]

CHECK SHEAR CAPACITY (ACI 318-02 Sec. 11.1.1, 11.3.1, & 11.5.6.2)

φ Vn = φ (Vs + Vc) (ACI 318-02 Sec. 11.1.1)

> Vu [Satisfactory]

where φ = 0.75 (ACI 318-02 Sec. 9.3.2.3) fy = 60 ksi

d A0 Av Vc = 2 (fc')0.5A0 Vs = MIN (d fy Av / s , 4Vc) φ Vnx 37.37 1345 0.80 174.4 149.5 243y 33.37 1335 0.40 173.0 66.7 180



Concrete Column Design Based on ACI 318-02

Daniel

(Total 30 # 10)

806.226

2521

Tian Li

0.652

0

500

1000

1500

2000

2500

3000

3500

4000

0 500 1000 1500 2000 2500 3000 3500

ε

εθ

( )''

2

'

'

2 0.85, 57 , 29000

0.85 2 , 0

0.85 ,

,

,

C

C

CC

C

S

fksifE Ec so

Ec

c c forf c of oo

forf c oforEss s y

fforf s yy

ε

ε ε ε εεε

ε εε ε ε

ε ε

= = =

− < < =

≥

≤= >



INPUT DATA & DESIGN SUMMARYEFFECTIVE LENGTH FACTOR k = 1.6 , (ACI 10.12.1 or 10.13.1)

COLUMN UNSUPPORTED LENGTH Lu = 12 ft

LARGER FACTORED MOMENT M2 = 200 ft-kSMALLER FACTORED END MOMENT M1 = 100 ft-k, (positive if single curvature.)


COLUMN DIMENSIONS h = 20 inb = 20 in


SUMMATION FOR ALL VERTICAL LOADS IN THE STORY Σ Pu = 1200 k

SUMMATION FOR ALL CRITICAL LOADS IN THE STORY Σ Pc = 13600 k, (ACI Eq. 10-10)

THE MAGNIFIED MOMENT: Mu = 236.7 ft-k , Sway

ANALYSISMAGNIFIED MOMENT - NONSWAY

r = 0.3 h = 6.0 in, ACI 10.11.2

k Lu / r = 38.4 > 34 - 12(M1 / M2) = 28 < = = Slenderness effect must be considered. (ACI 10.12.2)

Ec = 57000 (fc')0.5 = 3605.0 ksi, ACI 8.5.1

Ig = b h3 / 12 = 13333 in4

1E+07 k-in2 , ACI 10.12.3

2234.2 k, ACI Eq (10-10)

M2,min = MAX[ M2 , Pu (0.6+0.03 h) ] = 200 ft-k, ACI 10.12.3.2

Cm = MAX[ 0.6 + 0.4 (M1 / M2, min) , 0.4 ] = 0.8 , ACI 10.12.3.1

1.05 , ACI Eq (10-9)

Mu, ns = δns M2, min = 210.2 ft-k, ACI Eq (10-8) > 1.05 M2 = 210.0 ft-k [Unsatisfactory] ,(ACI 10.11.4.1)

The column is sway. See calculation as follows.

MAGNIFIED MOMENT - SWAY

k Lu / r = 38.4 > 22 < = = Slenderness effect must be considered. (ACI 10.13.1)

36 1.13 , ACI Eq (10-18) & 10.13.6 (c)

Ag = b h = 400 in2

Lu / r = 24.00 < 35 / [Pu / (fc' Ag)]0.5 = 70.00 [Satisfactory] , (ACI 10.13.5)

M2s = M2 = 200.0 ft-k, as given

M2ns = 5% M2s = 10.0 ft-k, assumed conservatively

Mu, s = M2ns + δs M2s = 236.7 ft-k, ACI Eq (10-16)

Note: For column subject to bending about both principal axis, the moment about each axis shall be magnified separately basedon the conditions corresponding to that axis. (ACI 10.11.6)

Magnified Moment Calculation for Concrete Column Based on ACI 318-02

DanielTian Li

0.4 0.40.25

1 1 0.6

E I E Ic g c gEI E Ic gdβ

= = = =+ +

( )2

2

EIPc

k Lu

π= =

, 1.01

0.75

CmMAXns PuPc

δ

= = −

1, 1.0 , 2.5

10.75

MIN MAXs PuPc

δ

= =

Σ − Σ


JOB NO. : DATE : REVIEW BY : Rectangular Concrete Column Design

INPUT DATAf 'c = 4 ksify = 60 ksix = 48 iny = 32 in

Bar Size ==> # 11No. of Asx = 8 8 # 11 No. of Asy = 9 9 # 11

Total Bars ==> # 30 # 11 = 3.0%

Pu = 1700 k

Mux = 2900 ft-k ex = 20.5 in

Muy = 1200 ft-k ey = 8.5 in

CHECK COLUMN CAPACITY BY THE BRESLER METHOD

1700.0 < 4407.9 ok 2428.6 < 3145.4 ok 1.199 > 1.0 NG

ANALYSIS

= 0.003 d = 45.295 in 29.295 ind' = 2.705 in 2.705 inAs = 12.48 in2

14.04 in2

= 0.70 b = 32 in 48 in

Mn = 4142.9 ft-k 1714.3 ft-k

= 0.85 Mno = 5485.1 ft-k 3864.7 ft-kPo = 3820.9 k 5457.1 k

MOMENT STRENGTH (Pn=0)

4.10078 in 3.70173 in

29612.2 psi 23425.7 psi

4034.01 ft-k 3052.89 ft-k

AXIAL LOAD STRENGTH (Mn=0)

7871.28 k = 7871.28 k

BALANCED CONDITION

22.7862 in 14.7372 in

60000 psi 60000 psi

2479.13 k 2405.11 k

5515.32 ft-k 3881.41 ft-k

DanielTian Li

x DIRECTION y DIRECTIONuε

1β

φ

( )2' '' '

1

'1

' 3.4

1.7

s s s s su uy c

c

bf f f dA A E A A E A Es s suyc

bf

βε ε ε

β

− + − += =

''

s us

c df Ec

ε− = =

( )' ' ''110.85

2n sc s

ccb d df f dM A

ββ = − + − =

'0.85 ( )o g st stc yf fP A A A= − + =

1sub

su y

dEa

fE

βεε

= =+

'' 1

su ysusb

su

fEdf Ed E

εε

ε +

= − =

' ''0.85b s sbc sb ybf f faP A A= + − =

' ' ''0.852 2 2 2

bb s sbc sb y

h h dab hf f fa dM A A = − + − − − =

11 1 1n

ox oy o

P

P P P

≤+ − 1.0nynx

nox noy

MMM M

+ ≤( ) ( )0.80 0.70u oP P≤



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = 5 ksiREBAR YIELD STRESS fy = 60 ksiCOLUMN DIAMETER D = 20 inFACTORED AXIAL LOAD Pu = 480 kFACTORED MAGNIFIED MOMENT Mu = 200 ft-kFACTORED SHEAR LOAD Vu = 20 kCOLUMN VERT. REINF. 8 # 7COLUMN SPIRAL REINF. # 3 @ 3 in o.c.


ANALYSIS

φ Pn (k)


AT AXIAL LOAD ONLY 954 0AT MAXIMUM LOAD 954 80AT 0 % TENSION 768 159AT 25 % TENSION 638 197AT 50 % TENSION 528 216

AT ε t = 0.002 366 227

φ Mn (ft-k) AT BALANCED CONDITION 360 230

AT ε t = 0.005 152 232

AT FLEXURE ONLY 0 165CHECK FLEXURAL & AXIAL CAPACITY



Ag = 314 in2. Ast = 4.80 in2.


φ = 0.57 + 67 εt = (ACI 318-02, Fig. R9.3.2)

where Cb = d εc / (εc + εs) = 10 in εt = 0.002069 εc =d = 17.7 in, (ACI 7.7.1) β1 = 0.8 ( ACI 318-02, Sec. 10.2.7.3 )

φ Mn = 0.9 Μn = 165 ft-kips @ Pn = 0, (ACI 318-02, Sec. 9.3.2) ,& εt,min = 0.004, (ACI 318-02, Sec. 10.3.5)

φ Mn = 219 ft-kips @ Pu = 480 kips > Mu [Satisfactory]


ρmin = 0.01 (ACI 318-02, Section 10.9) [Satisfactory]

CHECK SHEAR CAPACITYφ Vn = φ (Vs + Vc) = 65 kips, (ACI 318-02 Sec. 11.1.1)

> Vu [Satisfactory]where φ = 0.75 (ACI 318-02 Sec. 9.3.2.3)

A0 = 246 in2. Av = 0.22 in2. fy = 40 ksi

Vc = 2 (fc')0.5A0 = 34.7 kips, (ACI 318-02 Sec. 11.3.1)

Vs = MIN (d fy Av / s , 4Vc) = 51.9 kips, (ACI 318-02 Sec. 11.5.6.2)



Circular Column Design Based on ACI 318-02

DanielTian Li

0.003

0.709

0

200

400

600

800

1000

1200

0 50 100 150 200 250

ε

ε

( )'

'

2

'

'

2 0.85, 57 , 29000

0.85 2 , 0

0.85 ,

,

,

C

C

C

C

C

S

fksifE Ec so

Ec

c c forf c of oo

forf c oforEss s y

fforf s yy

ε

ε ε ε εεε

ε εε ε ε

ε ε

= = =

− < < =

≥

≤= >



INPUT DATA & DESIGN SUMMARYEFFECTIVE LENGTH FACTOR k = 1 , (ACI 10.12.1 or 10.13.1)

COLUMN UNSUPPORTED LENGTH Lu = 12 ft

LARGER FACTORED MOMENT M2 = 200 ft-kSMALLER FACTORED END MOMENT M1 = 12 ft-k, (positive if single curvature.)


COLUMN DIAMETER D = 20 in


SUMMATION FOR ALL VERTICAL LOADS IN THE STORY Σ Pu = 1200 k

SUMMATION FOR ALL CRITICAL LOADS IN THE STORY Σ Pc = 13600 k, (ACI Eq. 10-10)

THE MAGNIFIED MOMENT: Mu = 200.0 ft-k , Nonsway

ANALYSISMAGNIFIED MOMENT - NONSWAY

r = 0.25 D = 5.0 in, ACI 10.11.2

k Lu / r = 28.8 < 34 - 12(M1 / M2) = 33.28 < = = Slenderness effect may be ignored. (ACI 10.12.2)

Ec = 57000 (fc')0.5 = 3605.0 ksi, ACI 8.5.1

Ig = π D4 / 64 = 7854 in4

7E+06 k-in2 , ACI 10.12.3

3369.1 k, ACI Eq (10-10)

M2,min = MAX[ M2 , Pu (0.6+0.03 D) ] = 200 ft-k, ACI 10.12.3.2

Cm = MAX[ 0.6 + 0.4 (M1 / M2, min) , 0.4 ] = 0.624 , ACI 10.12.3.1

1.00 , ACI Eq (10-9)

Mu, ns = δns M2, min = 200.0 ft-k, ACI Eq (10-8) < 1.05 M2 = 210.0 ft-k [Satisfactory] ,(ACI 10.11.4.1)

The column is nonsway. Ignore following calculations.

MAGNIFIED MOMENT - SWAY < = = Not apply

k Lu / r = 28.8 > 22 < = = Slenderness effect must be considered. (ACI 10.13.1)

7 1.13 , ACI Eq (10-18) & 10.13.6 (c)

Ag = π D2 / 4 = 314 in2

Lu / r = 28.80 < 35 / [Pu / (fc' Ag)]0.5 = 62.04 [Satisfactory] , (ACI 10.13.5)

M2s = M2 = 200.0 ft-k, as given

M2ns = 5% M2s = 10.0 ft-k, assumed conservatively

Mu, s = M2ns + δs M2s = 236.7 ft-k, ACI Eq (10-16)

Magnified Moment Calculation for Circular Column Based on ACI 318-02

DanielTian Li

0.4 0.40.25

1 1 0.6

E I E Ic g c gEI E Ic gdβ

= = = =+ +

( )2

2

EIPc

k Lu

π= =

, 1.01

0.75

CmMAXns PuPc

δ

= = −

1, 1.0 , 2.5

10.75

MIN MAXs PuPc

δ

= =

Σ − Σ


JOB NO. : DATE : REVIEW BY : Design of Column Supporting Discontinuous System Based on IBC 2003 / CBC 2001



REBAR YIELD STRESS fy = 60 ksiCOLUMN CLEAR HEIGHT h = 16 ftCOLUMN SIZE c1 = 24 in

c2 = 24 inSEISMIC COEFFICIENT (Tab 16-N) R = 5.5AMPLIFICATION FACTOR (Tab 16-N) Ω0 = 2.8

DESIGN LEVEL STORY DISPLACEMENT ∆S = 0.2 inCOLUMN LOADS, ASD (ft-kips, kips)

P Mtop V Mbot

DL 40 0 0 0LL 20 0 0 0

E/1.4 100 80 20 80

LONGITUDINAL REINFORCINGSECTION TOP BOTTOM

LEFT 4 # 8 4 # 8( d = 21.50 in ) ( d = 21.50 in )

( 1 Layer) ( 1 Layer)RIGHT 4 # 8 4 # 8

( d = 21.50 in ) ( d = 21.50 in )( 1 Layer) ( 1 Layer)

THE COLUMN DESIGN IS ADEQUATE.TRANSVERSE REINFORCMENT FOR CONFINEMENT

3 Legs # 4 @ 5 in, o.c., full height (ACI 318-02 4.4.5 / CBC 1921.4.4.5)

ANALYSISDESIGN CRITERIA

1. since the column supported reaction from discontinued stiff member, CBC 1633.2.4 &1921.4.4.5 apply.2. since the column is not part of the lateral force resisting system, CBC 1921.7 apply.

3. since the column Ld required into top & 12" at least into footing per CBC 1921.4.4.5, a fixed-fixed condition should be used..

DESIGN LOADS

U1 = 1.2 D + f1 L +1.0 Ω0 Eh , (CBC 1630.8.2.1, 12-17 & 30-2) U2 = 0.9 D ± 1.0 Ω0 Eh , (CBC 1630.8.2.1, 12-18 & 30-2)

Pu = 450.0 kips Mu,top = 313.6 ft-kips Pu = 428.0 kips Mu,top = 313.6 ft-kips

Vu = 78.4 kips Mu,bot = 313.6 ft-kips Vu = 78.4 kips Mu,bot = 313.6 ft-kips

f1 = 0.5

U3 = 1.4 D + 1.7 L , (CBC 1909.2.1)

Pu = 90.0 kips Mu,top = 0.0 ft-kips

Vu = 0.0 kips Mu,bot = 0.0 ft-kips

CHECK CAPACITY SUBJECTED TO BENDING AND AXIAL LOAD

LOADING U1,top U1,bot U2,top U2,bot U3,top U3,bot

Pu (kips) 450.0 450.0 428.0 428.0 90.0 90.0

Mu (ft-kips) 313.6 313.6 313.6 313.6 0.0 0.0

δns = Cm/[1-Pu/(0.75Pc)] 1.116 1.116 1.110 1.110 1.021 1.021

δnsMu (ft-kips) 349.9 349.9 348.0 348.0 0.0 0.0

φMn (ft-kips) @ Pu 419.8 474.9 476.2 476.2 404.7 404.7

where EI = 0.4EcIg / (1+βd) = 0.25 EcIgPc = π2EI / (kLu)2

SUMMARY OF LOAD VERSUS MOMENT CAPACITIES (for ACI 318-02 10.2 & 10.3 only)

CAPACITY φ φ φ φ Pn (kips) φφφφ Mn (ft-kips)AT AXIAL LOAD ONLY 1047 0AT MAXIMUM LOAD 1047 175AT 0 % TENSION 893 272AT 25 % TENSION 745 336AT 50 % TENSION 622 376AT ε t = 0.002 431 425AT BALANCED CONDITION 420 428AT ε t = 0.005 256 537AT FLEXURE ONLY 0 405

DanielTian Li

φ Pn (kips) (cont'd)

24

φ Mn (ft-kips)

All load points to be within capacity diagram. [Satisfactory]

DETERMINE INDUCED MOMENT IN THE COLUMN

452.6664 ft-kips, at top & bot of column

where E c = 57000 (fc')0.5 = 3122 ksi, ACI 318-02 8.5.1 / CBC 1908.5.1

I g = c1 c23 / 12 = 27648 in4

I c = 0.5 I g = 13824 in4 , CBC 1633.2.4

∆ M = MAX( 0.7 R ∆s , 0.0025 h) = 0.77 in, CBC (30-17) & 1633.2.4

P = PDL + PLL = 60 kips, CBC 1633.2.4

∆ = 0.5 ∆M = 0.39 in, CBC (30-17)

CHECK REQUIREMENTS OF NOT PART OF THE LATERAL RESISTING SYSTEM

M u = 1.2 MDL + 1.0 MLL + Mcol = 452.6664 ft-kips, (CBC 1921.7.2, Mu = 1.4 MDL + 1.4 MLL + Mcol )

> φ Mn = 404.7 kips [Satisfactory]

Pu, max = 450.0 kips > 0.1Agfc' = 172.8 kips [Satisfactory]Per ACI 318-02 21.11.3.3 / CBC 1921.7.3.3, the column shall satisfy ACI 318-02 21.4.4, 21.4.5 and 21.5.2.1.

CHECK SECTION REQUIREMENTS (ACI 318-02 21.4.1 / CBC 1921.4.1)

cmin =MIN(c1, c2) = 24 in > 12 in [Satisfactory]

cmin / cmax = 1.00 > 0.4 [Satisfactory]

CHECK TRANSVERSE REINFORCING AT END OF COLUMN (ACI 318-02 21.4.4 / CBC 1921.4.4)

Ash = 0.60 in2 > MAX[ 0.09shcfc' / fyh , 0.3shc(Ag/Ach-1)fc' / fyh ] = 0.47 in2

[Satisfactory] where s = MAX[MIN(c1/4, 6db, 4+(14-hx)/3, 6), 4] = 5 in

hc = c1 - 2Cover - dt = 20.5 in

Ach = (c1-3)(c2-3) = 441.0 in2

CHECK FLEXURAL REINFORCING (ACI 318-02 21.4.3.1 / CBC 1921.4.3.1)

ρtotal = 0.018 > ρmin = 0.010 [Satisfactory]


CHECK SHEAR STRENGTH (ACI 318-02 21.4.5 / CBC 1921.4.5)

Ve = MAX[ (Mpr, left, top + Mpr, right,bot) / Hn , Vu,max] = 114.5 kips

< 8φ(fc')0.5c2d = 169.6 kips [Satisfactory]

< φ[2(fc')0.5c2d + Avfyd / s] = 158.5 kips [Satisfactory]

where ρtop,left = 0.006 > ρmin=MIN[3(fc')0.5/fy, 200/fy ]= 0.003 [Satisfactory]

ρbot,left = 0.006 > ρmin = 0.003 [Satisfactory]

Mpr, left, top = 1.25Mn,col,max = 916 ft-kips φ = 0.75 (0.85, if CBC 1909.3.2.3)

Mpr, right, bot = 1.25Mn,col,max = 916 ft-kips Av = 0.6 in2

DETERMINE SEISMIC TENSION DEVELOPMENT, Ld, INTO THE TOP PER ACI 318-02 21.4.4.5 / CBC 1921.4.4.5

17 db = 17 in, (ACI 318-02 21.5.4.1)

59 db = 59 in, (ACI 318-02 21.5.4.2)

where db = 1 in β = 1.0 , (1.2 for epoxy-coated, ACI 318-02 21.5.4.4 & 12.2.4)

0

200

400

600

800

1000

1200

0 100 200 300 400 500 600

26 ME Ic c PM col

h∆ + ∆ ==

( )3.5 , 12d dhMAX inL L β= =

', , 68

65

b ydh b

c

fdMAX indL

f

= =



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH f c ' = 5 ksi

REBAR YIELD STRESS f y = 60 ksi

FACTORED SHEAR LOAD V u = 80 k

FACTORED TENSILE LOAD N uc = 40 k

WIDTH b = 15 inEFFECTIVE DEPTH d = 20 inOVERALL DEPTH h = 22 inSHEAR SPAN a = 4 in

EDGE DEPTH h c = 12 in

PRIMARY REINFORCEMENT 3 # 7CLOSED STIRRUPS 3 # 3 ( spacing 4 in o.c. )

ANALYSISCHECK DIMENSIONAL REQUIREMENTS (ACI Sec. 11.9.1 & 11.9.2)

a / d = 0.20 < 1 [SATISFACTORY]Nuc / Vu = 0.50 < 1 [SATISFACTORY]

hc = 12.00 > 0.5 d [SATISFACTORY]

CHECK SECTION (ACI Sec. 11.9.3.2.1)

V u < = 180 k [SATISFACTORY]

where φ = 0.75

CHECK REINFORCEMENT

= 0.889 in2 = 1.270 in2

(ACI Sec. 11.9.3.4) (ACI Sec. 11.7.4.3)where φ = 0.75 µ = 1.4

= 0.373 in2= in-k

(ACI Sec. 10.2)

= 1.735 in2 < 1.800 in2

(ACI Sec. 11.9.3.5) [SATISFACTORY]

= 0.423 in2 < 0.660 in2

(ACI Sec. 11.9.4) [SATISFACTORY]

Tian Li

400

Corbel Design Based on IBC 03 / ACI 318-02

Daniel

''2

0.85 1 10.383

uc

cf

y

Mbd fb fd

Af

− −

=

0.2,uc u

ny y

N VMAXAf fφ φ

=

uvf

y

VA

fφµ=

'(0.2 , 0.8 )cMIN bd bdfφ φ

( )u u uca h dV NM = + −

'2 0.04, ,

3vf c

s f n n

y

fAMAX bdA A A Af

= + +

2s n

hA A

A−= ,h provdA =

≤

,s provdA =



INPUT DATA & DESIGN SUMMARYFLANGE STRENGTH fc,'f = 3 ksiGIRDER STRENGTH fc,'w = 6 ksiREBAR YIELD STRESS fy = 60 ksiEFFECTIVE SPAN L = 25 ftGIRDER SPACING S = 2.5 ftGIRDER WIDTH bw = 4 inFLANGE DEPTH hf = 2 inGIRDER DEPTH hw = 10 inFACTORED SECTION SHEAR Vu = 16 kPROPPED ? (1=YES, 2=NO) = 2 NoINTERFACE TIES 2 # 4 @ 16 in o.c.CONTACT SURFACES (1=CLEAR, 2=INTENTIONALLY ROUGHENED) 2 Intentionally RoughenedEFFECTIVE PRESTRESS Pe = 52 k REMOVAL OF PROPS MProp = 8.58 ft-kGIRDER SELF WEIGHT MG = 3.3 ft-k FORMWORK MSht = 2 ft-kFLANGE CONCRETE MF = 5 ft-k SUPERIMPOSED D+L MW = 15.5 ft-k

ANALYSISFLANGE MODULUS Ef = 3122 ksi GIRDER PROPERTIES : Ac = 40 in2

GIRDER MODULUS Ew = 4415 ksi Ic = 333 in2

MODULAR RATIO, Ew / Ef n = 1.41 Sb = Sb = 66.67 in3

30.0 in 21.2 in

8.09 in 1089 in4

278 in3 569 in3 135 in3

0.006 > 0.001

<= applicable Vu < 19.8 k

[SATISFACTORY]

ft = 1.854 ksifb = 0.746 ksifct = 0.412 ksifcb = -0.458 ksi >

fci(flan) = 0.201 ksifci(web) = 2.138 ksi <

[SATISFACTORY]

Daniel

Prestressed Composite Section DesignT. Li

( ) , 16 ,4f eff w fL

MIN Sb b h = + =

( )( )

f efff trans

bb

n= =

c

Ayy

A= =

2 2cc cI A Ay yI = + − =

ccct

c c

IS

yh= =

−cc

ciw c

IS

yh= =

−cc

cb

c

IS

y= =

vv

w

Asb

ρ = =min

50

yfρ = =

( )'

80

260 0.6 , 500

, 0.2 , 800

w

nh w wv y

vf c cy c

db

MIN d dfV b b

MIN f fA A A

φ

φ φ λρ

φ µ

= +

nhVφ =

'6 cf−'0.6 cf



NON-SEISMIC TENSION DEVELOPMENTS

15 db = 12 in, (ACI 318-02 12.5.2)

26 db = 20 in, (ACI 318-02 12.2.3)

L hk = 12 in, (ACI 318-02, Fig. R12.5)

where Bar size # 6db = 0.75 in

ρ required / ρ provided = 1 ( A s,reqd / A s,provd , ACI 318-02, 12.2.5)

fy = 60 ksi

f'c = 3 ksi

α = 1.0 (1.3 for bottom cover more than 12", ACI 318-02 12.2.4)

β = 1.0 (1.2 for epoxy-coated, ACI 318-02 12.2.4) γ = 0.8 (0.8 for # 6 or smaller, 1.0 for other) λ = 1.0 (1.3 for light weight)c = 3.4 in, min(d' , 0.5s), (ACI 318-02, 12.2.4)

Ktr = (Atr fyt / 1500 s n) = 0 (ACI 318-02, 12.2.4) , (50 bw / 1500 n, for CBC 2001)

(c + Ktr ) / db = 2.5 < 2.5 , (ACI 318-02, 12.2.3) η = 0.7 (#11 or smaller, cover > 2.5" & side >2.0", ACI 318-02 12.5.3)

SEISMIC TENSION DEVELOPMENTS

17 db = 13 in where Bar size # 6

(ACI 318-02 21.5.4.1) db = 0.75 in

fy = 60 ksi

59 db = 44 in, (ACI 318-02 21.5.4.2) f'c = 3 ksi

L hk = 12 in, (ACI 318-02, Fig. R12.5) β = 1.0(1.2 for epoxy-coated, ACI 318-02 21.5.4.4 & 12.2.4)

NON-SEISMIC COMPRESSION DEVELOPMENT

21 db = 16 in

(ACI 318-02 12.3)


ρ required / ρ provided = 0.95 (ACI 318-02 12.3.3 a)

fy = 60 ksi

f'c = 3 ksi

η = 1.0 (for spiral, 0.75, ACI 318-02 12.3.3 b)

Development of Reinforcement Based on ACI 318-02

DanielTian Li

'

0.075, 12requird b y

dtrprovided

cb

d fMAX inL

c Kfd

ρ αβγλρ

= =

+

'

0.02, , 68

requird b ydh b

provided c

d fMAX indL

f

ρ βλη

ρ

= =

( )3.5 , 12d dhMAX inL L β= =

', , 68

65

b ydh b

c

fdMAX indL

f

= =

'

0.02, 0.3 , 8brequird y requird

dc b yprovided providedc

fdMAX infdL

f

ρ ρη η

ρ ρ

= =



NON-SEISMIC TENSION SPLICE

ACI 318-02 Tab. R12.15.2As,provd / As,reqd 50% , Staggered 100% Lap

2 Class A Class B< 2 Class B Class B

L s = 1.3 L d = 30 db = 18 in

(for Class A, 1.0 L d , ACI 318-02, 12.15.1)

23 db = 14 in(ACI 318-02 12.15.1 & 12.2.3)


fy = 60 ksi

f'c = 4 ksi α = 1.0 (1.3 for bottom cover more than 12", ACI 318-02 12.2.4) β = 1.0 (1.2 for epoxy-coated, ACI 318-02 12.2.4) γ = 0.8 (0.8 for # 6 or smaller, 1.0 for other) λ = 1.0 (1.3 for light weight)c = 1.7 in, min(d' , 0.5s), s = see fig above, (ACI 318-02, 12.2.4)

Ktr = (Atr fyt / 1500 s n) = 0 (ACI 318-02, 12.2.4) , (50 bw / 1500 n, for CBC 2001)

(c + Ktr ) / db = 2.5 < 2.5 , (ACI 318-02, 12.2.3)

SEISMIC TENSION SPLICE

L s = 1.3 L d = 66 db = 42 in

(for Class A, 1.0 L d , ACI 318-02, 12.15.1)

15 db = 9 in(ACI 318-02 21.5.4.1)

51 db = 32 in, (ACI 318-02 21.5.4.2)


fy = 60 ksi

f'c = 4 ksi

β = 1.0(1.2 for epoxy-coated, ACI 318-02 21.5.4.4 & 12.2.4)

NON-SEISMIC COMPRESSION SPLICE

30 db = 23 in, (ACI 318-02 12.16.1)


fy = 60 ksi

f'c = 3 ksi η = 1.0 (for fc' < 3 ksi, 4/3, ACI 318-02 12.16.1)

Tian LiSplice of Reinforcement Based on ACI 318-02

Daniel

'

0.075, 12b y

dtr

cb

d fMAX inL

c Kfd

αβγλ = =

+

( )3.5 , 12d dhMAX inL L β= =

', , 68

65

b ydh b

c

fdMAX indL

f

= =

( )0.0005 , 12b ys MAX infdL η= =


JOB NO. : DATE : REVIEW BY : Shear Friction Reinforcing Design Based on ACI 318-02


FACTERED FRICTION FORCE Vu = 500 kips

CONCRTET STRENGTH fc' = 4 ksi

REINFORCEMENT STRENGTH fy = 60 ksi

FRICTION COEFFICIENT [See Table below] µ = 0.6

SHEAR PLANE THICKNESS t = 8 inSHEAR PLANE LENGTH L = 20 ftDOWEL SIZE # 5

Use # 5 Dowels @ 4.0 in o.c.

ANALYSISCHECK SHEAR STRENGTH LIMITATION (Sec. 11.7.5)

φφφφVn = φ φ φ φ MAX( 0.2fc' Ac , 800 Ac ) = 1152 kips > Vu

Where φ = 0.75 [Section 9.3.2.3] [SATISFACTORY]

Ac = 1920 in2

THE REQUIRED AREA OF SHEAR-TRANSFER REINFORCEMENT IS GIVEN BY SECTION R11.7.4.1 AS

Avf = Vu / (φφφφ fy µµµµ) = 18.5 in2

COEFFICIENT OF FRICTION FOR NORMALWEIGHT CONCRETE [Sec.11.7.4.3] µµµµConcrete place monolithically 1.40 Concrete placed against hardened concrete with surface intentionally roughened 1.00 Concrete placed against hardened concrete NOT intentionally roughened 0.60 <=Concrete anchored to as-rolled structural steel by headed stud or by reinforcing bars 0.70

DanielT. Li


JOB NO. : DATE : REVIEW BY : Shear Friction Reinforcing Design Based on ACI 318-02

INPUT DATA & DESIGN SUMMARY RuFACTERED FRICTION FORCE VERTICAL Ru = 86 kips

FACTERED FRICTION FORCE HORIZONTAL Tu = 34 kips

CONCRTET STRENGTH fc' = 4 ksi

REINFORCEMENT STRENGTH fy = 60 ksi

BEARING WIDTH a = 5 in 5.0 inBEARING THICKNESS b = 18 in TuREINFORCEMENT 2 - leg ties # 3ASSUME A POTENTIAL CRACK PLANE WITH AN ANGLE OF αf = 70 degrees

14 in

Use 6 # 3 with (2) - leg closed tiesDistribute over a vert. distance of 14 in

ANALYSIS

FRICTION COEFFICIENT [Sec.11.7.4.3] µ = 1.4

FORCES ON INCLINED PLANE:Vu = Ru sin αf + Tu cos αf = 92.4 kips < φVn = φ MAX( 0.2fc' Ac , 800 Ac ) = 160.6 kips

Where φ = 0.75 [Section 9.3.2.3] [SATISFACTORY] (Sec. 11.7.5)Ac = 268 in2

Nu = Tu sin αf - Ru cos αf = 2.5 kips ( net tension )

SHEAR FRICTION REINFORCEMENT IS GIVEN BY EQ(11-26) ASAvf =Vu / [ φ fy (µ sin αf + cos αf )] = 1.24 in2

REINFORCEMENT TO RESIST TENSION IS GIVEN BY SEC. 11.7.7 ASAn = Nu / (φ fy sin αf) = 0.06 in2

TOTAL REINFORCEMENT: As = Avf + An = = 1.30 in2

DanielT. Li



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH f 'c = 6 ksiREBAR STRENGTH fy = 60 ksiTENDON TENSILE STRENGTH fpu = 270 ksiTENDON YIELD STRENGTH fpy = 243 ksiCOMPRESSION REINF. 3 # 8TENSION REINF. 4 # 8SHEAR STIRRUP REINF. 2 legs, # 4 @ 12 in o. c.PRESTRESSING TENDONS 14 strends (each 0.5 in diameter & 0.153 in2 area )WIDTH OF COMPRESSION FACE b = 20 inDIST. TO CENTROID OF COMPRESSION d' = 3 inDIST. TO CENTROID OF PRESTRESSED dp = 22.5 inDISTANCE TO CENTROID OF TENSION d = 25.5 inHEIGHT OF SECTION h = 28 inSPAN LENGTH L = 82 ftMOMENT DUE TO SELF-WEIGHT MG = 45 ft-k TENDON FORCE IMMEDIATELY AFTER PRESTRESS TRANSFER, Pi = 375 kMOMENT DUE TO DEAD LOAD MD = 250 ft-kMOMENT DUE TO LIVE LOAD ML = 125 ft-k TENDON FORCE AT SERVICE LOAD AFTER ALLOWANCE LOSSES, Pe = 300 kSECTION LOCATION ( 0 or 1 ) 0 at midspanPRESTRESSING METHOD ( 0, 1 or 2 ) 2 post-tensioned & bondedFACTORED SHEAR FORCE Vu = 175 kFACTORED TORSIONAL MOMENT Tu = 240 k Additional 4 #8 Longitudinal Reinforcement Required for Torsion

TRANSFER DESIGN STAGEECCENTRICITY e = 8.50 inSECTION AREA Ac = 560 in2

SECTION MODULUS OF TOP St = 2613 in3

SECTION MODULUS OF BOT. Sb = 2613 in3

MIN. TOP FIBER STRESS - Fti = -0.232 ksiMAX. BOT. FIBER STRESS Fbi = 3.600 ksiMAX. PERMISSIBLE STRESS Fsi = 199.260 ksi

-0.343 ksi < - Fti

[MIN. As' REQUIRED]

1.683 ksi < Fbi

[SATISFACTORY]

f si = 175.070 ksi < Fsi (As')reqd = 0.543 in2< (As')provd

[SATISFACTORY] [SATISFACTORY]

SERVICEABILITY DESIGN STAGEMIN. TOP FIBER STRESS Fte = 0.6fc' = 3.600 ksi, for total loads

Fte, G+D = 0.45fc' = 2.700 ksi, for sustained loads only

MAX. BOT. FIBER STRESS -Fbe = -6(fc')0.5 = -0.465 ksi, for normal cover to ACI Sec. 7.7.3.1

(for 50% increased cover to ACI Sec. 7.7.3.2 & with deflection checked by cracked scetion properties.)

1.489 ksi < Fte

[SATISFACTORY]

0.915 ksi < Fte, G+D

[SATISFACTORY]

-0.417 ksi > -Fbe

[SATISFACTORY]

STRENGTH DESIGN STAGECOMPRESSION ZONE FACTOR β1 = 0.750TENDON TYPE FACTOR γp = 0.280RATIO OF TENSION REINF. ρ = 0.006RATIO OF COMPR. REINF. ρ' = 0.005RATIO OF PRESTR. REINF. ρp = 0.005

INDEX OF TENSION REINF. ω = 0.062INDEX OF COMPR. REINF. ω' = 0.046INDEX OF PRESTR. REINF. ωp = 0.196

STRESS IN BONDED TENDONS :

246.639 ksi

STRESS IN UNBONDED TENDONS :

Not applicable Not applicable

5.64 in 0.21 < 0.36β1

[SATISFACTORY]

785 ft-k > 547 ft-k

[SATISFACTORY]

DanielT. Li

Flexural Design for Prestressed Member

1 Giti

c t t

e Mf PS SA

= − + =

1 Gibi

c b b

e Mf PS SA

= + − =

1 G D Lete

c t t

e M M Mf PS SA

+ += − + =

1 G D Lebe

c b b

e M M Mf PS SA

+ += + − =

,

1 G Dete G D

c t t

e M Mf PS SA

+ += − + =

'

'0.85ps s sps s s

c

f f fA A Aa

bf

+ −= =

εc = 0.003

A ps f ps

A s f y

a=β1

c

0.85 f c' ab

0.85 f c'

A s' f y

( )'0.85p

p p

a d

d dωω ω

= + − =

( )''

1

1 , 0.17p pups pu p

pc

dfMINf f

f d

ωγ ωρβ

− = − × + =

'

10 , , 60100

cps se y se

p

fMINf f f f

ρ

= + + + =

'

10 , , 30300

cps se y se

p

fMINf f f f

ρ

= + + + =

''

2 2 2n ps s spps y y

a a adf f fd dM A A Aφ φ = − + − + − =

'11.2 1.2 7.5cr eb c

c b

efSM P

SA

= + + =



INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH f 'c = 6 ksiREBAR STRENGTH fy = 60 ksiTENDON TENSILE STRENGTH fpu = 270 ksiTENDON YIELD STRENGTH fpy = 243 ksiCOMPRESSION REINF. 3 # 8TENSION REINF. 4 # 8SHEAR STIRRUP REINF. 2 legs, # 4 @ 12 in o. c.PRESTRESSING TENDONS 14 strends (each 0.5 in diameter & 0.153 in2 area )WIDTH OF COMPRESSION FACE b = 20 inDIST. TO CENTROID OF COMPRESSION d' = 3 inDIST. TO CENTROID OF PRESTRESSED dp = 22.5 inDISTANCE TO CENTROID OF TENSION d = 25.5 inHEIGHT OF SECTION h = 28 inSPAN LENGTH L = 82 ftFACTORED SHEAR FORCE Vu = 175 k TENDON FORCE IMMEDIATELY AFTER PRESTRESS TRANSFER, Pi = 375 kFACTORED TORSIONAL MOMENT Tu = 240 kSECTION LOCATION ( 0 or 1 ) 0 at midspan TENDON FORCE AT SERVICE LOAD AFTER ALLOWANCE LOSSES, Pe = 300 kPRESTRESSING METHOD ( 0, 1 or 2 ) 2 post-tensioned & bonded

SHEAR DESIGN

22.50 in 77.46 psi

= 174.28 k

= 45.00 k = 0.200 in2

= 16.88 in

0.247828254

= 0.248 , Case 3 applicable

TORSION DESIGN

Acp = 560 in2

Pcp = 96 in Tu > 215.27 k

fpc = 0.536 ksiAoh = 404 in2 Thus, Torsional Reinf. Reqd.Ph = 82 in

0.006 in 2.9315 in2

Additional 4 #8 Longitudinal Reinforcement Required for Torsion

0.033 in < 0.033 in

[SATISFACTORY]

DanielT. Li

Shear & Torsion Design for Prestressed Member

( )0.8 , pd MAX h d= =

' ' '

'

0.6 700 1 , , 5 , 2 , 0.4

2 , 0.4

u pc c c se se

uc

c se se

V dMAX MIN MIN bd bd bd forf f f f fMV

bd forf f f

+ ≥ =

<

', 8v y

s c

dfAMIN bd fV

S

=

( )' '100 ,cprovd

MINf f c = =

(min)

50, , 0.4

80

50, 0.4

ps pu

se sey y

v

se sey

dSfAbS bMAX for f f

f dfA

bSfor f f

f

≥ =

<

( ) ( )( )

(min),

, (min)

inf . , 1:2

, 2 :2

, , 3 :

, 4 :

cu

cv u cv requd

v cal v c u S c

S c u

Vno shear re requd for case V

Vfor case V VAA

MAX for case V V V VA A

unsatisfactory for case V V V

φ

φ φ

φ φφ

< ≤ ≤=

≤ ≤ + + ≤

( )( )

'

max'

0.75 , 24 , 4

0.375 , 12 , 4

s c

s c

MIN d for bd fVS

MIN d for bd fV

≤= >

2'

'1

4

cp pcc

cpc

fAfP f

φ

+ =

( )1.7 cos 37.5t u

oh yv

A TS fAφ

= =° ( )

'2

5 2537.5 , max ,cot

cpyv yvct tL h h

yL yL yL yv

f ff bAA AMAXA P PS Sf f f f

= ° − =

Re

2 50,v t

Total qD yv

bA A At MAXS S f

+ = = ProvD

AtS

=


JOB NO. : DATE : REVIEW BY : Mechanical Equipment Anchorage to Concrete Based on IBC 2003

INPUT DATA & DESIGN SUMMARYKWIK BOLT - 3 DIAMETER φ = 1/2 in. Per ICC ESR-1385

ANCHOR DEPTH hef = 2 1/4 in. Per ICC ESR-1385

EDGE DISTANCE c = 6 in

EQUIPMENT WEIGHT W = 25 kipsCENTER OF MASS H = 8.00 ft, 2/3 total height

ANCHORAGE LENGTH L = 37.5 ftANCHOR # ALONG L DIR NL = 13 per lineANCHOR SPACING SL = L / (NL - 1) = 38 in

ANCHORAGE WIDTH B = 20 ftANCHOR # ALONG B DIR NB = 7 per lineANCHOR SPACING SB = B / (NB - 1) = 40 in

[THE ANCHORAGE, KWIK BOLT - 3, DESIGN IS ADEQUATE.]

ANALYSISALLOWABLE TENSION & SHEAR VALUES (ICC ESR-1385, Table 3 or 5)

Pt = 1284 lbs

Vt = 1745 lbs

SPACING & EDGE REQUIREMENTS (ICC ESR-1385, Table 2)

Scr = 5 1/8 in

Smin = 2 1/4 in

Ccr = 6 3/4 in, shear, 4 in, tension

Cmin = 3 3/8 in, shear, 2 1/4 in, tension

DESIGN LOADS

FH = Fp = (KH) MAX 0.3SDSIpW , MIN[ 0.4apSDSIp(1+2z/h)/Rp W , 1.6SDSIpW ] , (ASCE 7-02, 9.6.1.3)

= 1.3 MAX 0.24W , MIN[ 0.65W , 1.30W ] where SDS = 0.54 (IBC 1615.1.3, page 322)

= 0.84 W , (SD) Ip = 1.5 (IBC Tab 1604.5 & 1621.1.3)

= 0.60 W , (ASD) = 15.04 kips ap = 1 (ASCE 7-02 Tab 9.6.2.2)

Rp = 1.5 (ASCE 7-02 Tab 9.6.2.2)

FV = KV Fp = 0.20 W , (ASD) = 5.01 kips, up & down z = h ft

h = 36 ftKH = 1.3 (IBC 1621.1, page 342)KV = 0.333 (vert. seismic factor)

MAXIMUM OVERTURNING MOMENT AT ANCHOR EDGEMOT = Fp H = 120.34 ft-kipsMRES = (0.9W - Fv ) (0.5B) = 174.91 ft-kips > MOT no tension anchors.

CHECK TENSION CAPACITY

Ps = [(FV -0.9 W) / A + MOT y / I ] = -59 lbs / bolt < Pt KDSA Kseismic Kedge-space [SATISFACTORY]

where A = 2(NL + NB) -4 = 36 (total bolts)

I = MIN(ΣXi2 , ΣYi

2 ) = 406400 in2-bolts (B direction governs)

y = 0.5 B = 120 inKDSA = 0.8 (Factor adapted ICBO / ICC value)

Kseismic = 4 / 3 (IBC 1605.3.2, page 274)

Kedge-space = [0.8 + 0.2(c - Cmin)(Ccr - Cmin)] [0.6 + 0.4(S - Smin)(Scr - Smin)] = 1.00 (ICC ESR-1385, Tab.2 footnotes)


Vs = FH / A = 418 lbs / bolt < Vt KDSA Kseismic Kedge-space [SATISFACTORY]

where Kedge-space = [0.5 + 0.5(c - Cmin)(Ccr - Cmin)] [0.9 + 0.1(S - Smin)(Scr - Smin)] = 1.00 (ICC ESR-1385, Tab.2 footnotes)

CHECK COMBINED LOADING CAPACITY (ICC ESR-1385 Sec. 4.1)

(Ps / Pt)5/3 + (Vs / Vt)

5/3 = 0.083 < 1.00 [SATISFACTORY]

DanielTian Li


JOB NO. : DATE : REVIEW BY : Mechanical Equipment Anchorage to Concrete Based on CBC 2001

INPUT DATA & DESIGN SUMMARYKWIK BOLT-II DIAMETER φ = 1/2 in. Per ICC ESR-1355

ANCHOR DEPTH hef = 2 2/3 in. Per ICC ESR-1355


EQUIPMENT WEIGHT W = 6.8 kipsCENTER OF MASS H = 4.00 ft, 2/3 total height

ANCHORAGE LENGTH L = 18.25 ftANCHOR # ALONG L DIR NL = 10 per lineANCHOR SPACING SL = L / (NL - 1) = 24 in

ANCHORAGE WIDTH B = 10 ftANCHOR # ALONG B DIR NB = 6 per lineANCHOR SPACING SB = B / (NB - 1) = 24 in

[THE ANCHORAGE, KWIK BOLT-II, DESIGN IS ADEQUATE.]

1/2ANALYSISALLOWABLE TENSION & SHEAR VALUES (ICC ESR-1355, Table 3 or 4)

Pt = 960 lbs

Vt = 1840 lbs

SPACING & EDGE REQUIREMENTS (ICC ESR-1355, Table 2)

Scr = 4 1/2 in

Smin = 2 1/2 in

Ccr = 6 3/4 in, shear, 3 3/8 in, tension

Cmin = 3 3/8 in, shear, 2 1/4 in, tension

DESIGN LOADS

FH = Fp = MAX 0.7CaIpW , MIN[ apCaIp(1+3hx/hr)/Rp W , 4CaIpW ] , (CBC 1632.2)

= MAX 0.43W , MIN[ 4.05W , 2.43W ] where Ca = 0.528

= 2.43 W , (SD) Ip = 1.15

= 1.73 W , (ASD) = 11.80 kips ap = 2.5 (CBC Tab 16A-O)

Rp = 1.5 (CBC Tab 16A-O & 1632A.2)

FV = Fp / 3 = 3.93 kips, up & down, CBC Tab. 16A-O footnote 20 hx = hr ft

hr = 36 ft

MAXIMUM OVERTURNING MOMENT AT ANCHOR EDGEMOT = Fp H = 47.19 ft-kipsMRES = (0.9W - Fv ) (0.5B) = 10.94 ft-kips < MOT ,therefore design tension anchors.


Ps = [(FV -0.9 W) / A + MOT y / I ] = 359 lbs / bolt < Pt KDSA Kseismic Kedge-space [SATISFACTORY]




y = 0.5 B = 60 inKDSA = 0.8 (DSA adapted ICBO / ICC value)

Kseismic = 4 / 3 (CBC 1612.3.2)

Kedge-space = [0.8 + 0.2(c - Cmin)(Ccr - Cmin)] [0.7 + 0.3(S - Smin)(Scr - Smin)] = 1.00 (ICC ESR-1355, Tab.2 footnotes)



where Kedge-space = [0.5 + 0.5(c - Cmin)(Ccr - Cmin)] [0.7 + 0.3(S - Smin)(Scr - Smin)] = 1.00 (ICC ESR-1355, Tab.2 footnotes)

CHECK COMBINED LOADING CAPACITY (ICC ESR-1355 Sec. 4.1)

(Ps / Pt)5/3 + (Vs / Vt)

5/3 = 0.251 < 1.00 [SATISFACTORY]

DanielTian Li


JOB NO. : DATE : REVIEW BY : Mechanical Equipment Anchorage to Concrete Based on CBC 2001

INPUT DATA & DESIGN SUMMARYRED HEAD DIAMETER φ = 1/2 in. Per ICBO ER-1372

ANCHOR DEPTH hef = 4 1/8 in. Per ICBO ER-1372


EQUIPMENT WEIGHT W = 13.35 kipsCENTER OF MASS H = 4.33 ft, 2/3 total height

ANCHORAGE LENGTH L = 30.5 ftANCHOR # ALONG L DIR NL = 6 per line 4.00ANCHOR SPACING SL = L / (NL - 1) = 73 in

ANCHORAGE WIDTH B = 7.29 ftANCHOR # ALONG B DIR NB = 2 per lineANCHOR SPACING SB = B / (NB - 1) = 87 in

[THE ANCHORAGE DESIGN IS NOT ADEQUATE.]

ANALYSISALLOWABLE TENSION & SHEAR VALUES (ICBO ER-1372, Table 3)

Pt = 1787.5 lbs

Vt = 1810 lbs

SPACING & EDGE REQUIREMENTS (ICBO ER-1372, Table 4 & 5)

Sfull = 6 3/16 in, spacing required to max value.

Smin = 3 1/8 in

Efull = 5 3/16 in, shear, 3 1/8 in, tension (edge to max value.)

Emin = 1 9/16 in, shear, 1 9/16 in, tension

DESIGN LOADS



= 2.02 W , (SD) Ip = 1.15

= 1.45 W , (ASD) = 19.30 kips ap = 2.5 (CBC Tab 16A-O)

Rp = 1.5 (CBC Tab 16A-O & 1632A.2)


hr = 16 ft

MAXIMUM OVERTURNING MOMENT AT ANCHOR EDGEMOT = Fp H = 83.63 ft-kipsMRES = (0.9W - Fv ) (0.5B) = 20.34 ft-kips < MOT ,therefore design tension anchors.


Ps = [(FV -0.9 W) / A + MOT y / I ] = 1447 lbs / bolt < Pt KDSA Kseismic Kedge-space [SATISFACTORY]




y = 0.5 B = 43.74 inKDSA = 0.8 (DSA adapted ICBO / ICC value)

Kseismic = 4 / 3 (CBC 1612.3.2)

Kedge-space = [0.65 + 0.35(c - Emin)(Efull - Emin)] [0.7 + 0.3(S - Smin)(Sfull - Smin)] = 1.00 (ICBO ER-1372, Tab.4 & 5)



where Kedge-space = [0.2 + 0.8(c - Emin)(Efull - Emin)] [0.4 + 0.6(S - Smin)(Sfull - Smin)] = 1.00

(ICBO ER-1372, Tab.4 & 5, assuming, in Fig 2, a line between E2 & E conservatively.)

CHECK COMBINED LOADING CAPACITY (ICBO ER-1372 Sec. 2.4)

(Ps / Pt)5/3 + (Vs / Vt)

5/3 = 1.369 > 1.00 [UNSATISFACTORY]

DanielTian Li


JOB NO. : DATE : REVIEW BY : Mechanical Equipment Anchorage to Wood Roof Based on NDS 1997

INPUT DATA & DESIGN SUMMARYLAG SCREW DIAMETER φ = 1/2 in, NDS APP. LLAG SCREW LENGTH Long = 4 in, NDS APP. L

EQUIPMENT WEIGHT W = 1.4 kipsCM HEIGHT H = 8.25 ft, 2/3 total height

ANCHORAGE LENGTH L = 8 ftLAG SCREW # ALONG L DIR NL = 3 per lineLAG SCREW SPACING SL = L / (NL - 1) = 48 in

ANCHORAGE WIDTH B = 8 ftLAG SCREW # ALONG B DIR NB = 2 per lineLAG SCREW SPACING SB = B / (NB - 1) = 96 in

[THE ANCHORAGE, USING 1/2" x 4" LAG SCREWS, IS ADEQUATE.]

ANALYSISALLOWABLE TENSION & SHEAR VALUES

ZII' = CD Cd Cg ZII = 219 lbs

where CD = 1.33 (1.6 for NDS, 1.33 for CBC) 1.5

Cd = p / (8D) = 0.814

p = S + (T-E) - (Diaphragn Thk) - ts = 3.256 in

S + (T-E) = 3.688 in (NDS Appendix L, page 169)Diaphragm Thk = 0.298 in (0.298 for 15/32, 0.319 for 19/32, UBC Table 23-2-H, page 3-420)

ts = 0.134 in (10 gage, NDS Tab 9.3B, page 82)

Cg = 0.92 (NDS Tab 7.3.6C, page 48)

ZII = 220 lbs (NDS Tab 9.3B, page 82)

Z' = CD Cd Cg Z = 159 lbs

Z = 160 lbs (NDS Tab 9.3B, page 82)

(W ' p) = CD W (T-E) = 1100 lbswhere W = 378 lbs / in (NDS Tab 9.2A, page 78)

(T-E) = 2.188 in (NDS Appendix L, page 169)

DESIGN LOADS



= 0.81 W , (SD) Ip = 1.15

= 0.58 W , (ASD) = 0.81 kips ap = 1 (CBC Tab 16A-O)

Rp = 3 (CBC Tab 16A-O)


hr = 36 ft

MAXIMUM OVERTURNING MOMENT AT ANCHOR EDGEMOT = Fp H = 6.68 ft-kipsMRES,L = (0.9W - Fv ) (0.5B) = 3.96 ft-kips < MOT ,therefore design tension anchors.MRES,B = (0.9W - Fv ) (0.5B) = 3.96 ft-kips < MOT ,therefore design tension anchors.

TENSION LOAD AT CORNER LAG SCREW

TL = (FV -0.9 W) / A + MOT y / I = 252 lbs / bolt TB = (FV -0.9 W) / A + MOT y / I = 113 lbs / bolt

where A = 2(NL + NB) -4 = 6 (total bolts) where I = ΣYi2 = 13824 in2-bolts

I = ΣXi2 = 9216 in2-bolts y = 0.5 B = 48 in

y = 0.5 L = 48 in

SHEAR LOAD AT EACH LAG SCREWV = FH / A = 135 lbs / bolt

CHECK CORNER SCREW CAPACITY AT COMBINED LATERAL AND WITHDRAWAL LOADS

Z' = ZII'(W' p) / (ZII'sin2α + W'p cos2α) (NDS 9.3-6)

= 581 lbs / bolt > [TL2 + V2 ]0.5 = 286 lbs / bolt [SATISFACTORY]

where α = tan-1(TL / V) = 61.87o

Z' = Z'(W' p) / (Z'sin2α + W'p cos2α) (NDS 9.3-6)

= 246 lbs / bolt > [TB2 + V2 ]0.5 = 176 lbs / bolt [SATISFACTORY]

where θ = tan-1(TB / V) = 40.01o

DanielTian Li



INPUT DATALATERAL DIAPHRAGM FORCE : Fp = 380 plf

DIMENSIONS: X1 = 72 ft , X2 = 24 ft , X3 = 16 ft

Y1 = 20 ft , Y2 = 20 ft , Y3 = 20 ft

DESIGN SUMMARYTHE MAXIMUM STRAP FORCE, T = C = 4.66 kips

THE MAXIMUM SHEAR STRESS, v x = 466 plf

v y = 480 plf

ANALYSIS

16

DanielTian Li

Flexible Diaphragm with an Opening

(cont'd)REACTIONS : VL = 21.280 kips , VR = 21.280 kips

HATCHED AREA : a = X1 + 0.5 X2 = 84.00 ft , b = 0.5 Y2 + Y3 = 30.00 ft

Y = Y1 + Y2 + Y3 = 60.00 ft , w = Fp b / Y = 190.0 plf

M = a VL - 0.5 Fpx a2 = 446.9 ft-k, total moment at middle opening , V = b VL / Y = 10.640 kips

F2 = M / Y = 7448 lbf , F9 = F2 = 7448 lbf

F5 = ( F2 b - 0.5 w a2 ) / a = -5320 kips , F15 = w a - V + F5 = 0 kips

INDIVIDUAL PANEL X (ft) Y (ft) vx (plf) vy (plf) NO. FORCE (lbf) NO. FORCE (lbf)

1 72.00 20.00 138 480 F1 9956 F15 0

2 12.00 20.00 209 266 F2 7448 F16 0

3 12.00 20.00 323 380 F3 3572 F17 -2512

4 16.00 20.00 223 299 F4 3040 F18 -1034

5 72.00 10.00 103 103 F5 -5320 F19 -4655

6 16.00 10.00 466 466 F6 7600 F20 1108

7 72.00 10.00 103 103 F7 2508 F21 -2508

8 16.00 10.00 466 466 F8 -3876 F22 3876

9 72.00 20.00 138 480 F9 7448 F23 3040

10 12.00 20.00 209 266 F10 7448 F24 5320

11 12.00 20.00 323 380 F11 2512 F25 7600

12 16.00 20.00 223 299 F12 1034 F26 9956

F13 4655 F27 3572

F14 -1108

Techincal References:

1. Kelly E. Cobeen: "Structuiral Engineering Review Workshop", BYA publications, 2005.



1. DESIGN METHODS1.1 BREAKDOWN OF FLOOR INTO DESIGN STRIPS IN TWO DIRECTIONS. USE THIS SOFEWARE TO GET REQUIRED EFFECTIVE

POST-TENSIONING FORCEES IN EACH DIRECTIONS.1.2 BANDED ALL TENDONS AT COLUMN IN ONE DIRECTION, AND DISTRIBTUTED IN THE OTHER DIRECTION. ON THE DESIGN

DRAWINGS, SPECIFY TOTAL REQUIRED EFFECTIVE POST-TENSIONING FORCEES AT BANDED TENDONS, AND UNIFORMFORCES IN DISTRIBTUTED TENDONS.

1.3 WHEN TENDON LESS THAN 140 FT, STRESS AT ONE END. OTHERWISE STRESS AT BOTH ENDS.1.4 THE VARIOUS LIVE LOADING CONDITIONS SHOULD BE CONSIDED BY INPUT LL ZERO AT SOME SPANS.

2. INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH f'c = 4.5 ksi


TENDON PROPERTIES fpu = 270 ksi

fpy = 243 ksi

fse = 175 ksi

dia = 1 / 2 in

Aps = 0.153 in2

SLAB THICKNESS t = 8 in THE DESIGN IS ADEQUATE.TRIBUTARY WIDTH, PERPENDICULAR W = 30 ftCOLUMN WIDTH c1 = 22 inCOLUMN DEEPTH c2 = 22 inTOP BAR SIZE AT COLUMN # 5BOTTOM CONTIUOUS BAR SIZE # 4

Location Left Mid of L1 Support Mid of L2 Support Mid of LTyp Support Mid of Ln-1 Support Mid Ln Right

Span (ft) 20 30 30 30 20DL (psf) 120 120 120 120 120LL (psf) 75 75 75 75 75

Balanced DL (60%-80% suggested) 75% 75% 80% 75% 75% dCGS (in, from bottom) 4 3.33 7.5 2.5 7.5 2.5 7.5 2.5 7.5 3.33 4REQD EFFECTIVE PT (k / ft) 22.31 24.30 25.92 24.30 22.31 Total if banded (kips) 669.4 729.0 777.6 729.0 669.4

Tendons 1 / 2 in Dia @ 14 in o.c. 1 / 2 in Dia @ 13 in o.c. 1 / 2 in Dia @ 12 in o.c. 1 / 2 in Dia @ 13 in o.c. 1 / 2 in Dia @ 14 in o.c. Total Number if banded 26 28 30 28 26Top Bars at Column 7 # 5 , L = 3.03 ft 10 # 5 , L = 9.39 ft 22 # 5 , L = 9.39 ft 22 # 5 , L = 9.39 ft 10 # 5 , L = 9.39 ft 7 # 5 , L = 3.03 ftBot Bars, Cont., E. Way Not ReqD # 4 @ 45in. o.c. Not ReqD # 4 @ 45in. o.c. Not ReqDRequired column cap thk, (in) Not ReqD Not ReqD Not ReqD Not ReqD Not ReqD Not ReqD

3. DESIGN LOADS & SECTION FORCES


MDL (ft-k / ft) 0.00 2.14 -7.72 5.01 -9.26 4.24 -9.26 5.01 -7.72 2.14 0.00 MLL (ft-k / ft) 0.00 1.34 -4.83 3.13 -5.78 2.65 -5.78 3.13 -4.83 1.34 0.00

Balanced Load (psf, uplift) -90 -90 -96 -90 -90

Balanced MBal (ft-k / ft) 0.00 -1.65 5.71 -3.66 7.23 -3.57 7.23 -3.66 5.71 -1.65 0.00

Required Effective PT (k / ft) 22.31 24.30 25.92 24.30 22.31 Tendon Spacing (in) 14 13 12 13 14

Primary MFe (ft-k / ft) 0.00 -1.25 6.51 7.56 -3.24 7.56 6.51 -1.25 0.00

7.09 -3.04 7.09 7.09 -3.04 7.09

Secondary MSec (ft-k / ft) 0.00 0.40 0.80 0.33 0.33 0.33 0.80 0.40 0.00

1.38 0.62 -0.14 -0.14 0.62 1.38

4. CHECK SERVICE LOAD STRESSES


A (in2 / ft) 96 96 96 96 96 96 96 96 96 96 96 S (in3 / ft) 128 128 128 128 128 128 128 128 128 128 128

F / A (ksi) 0.232 0.232 0.232 0.270 0.270 0.270 0.232 0.232 0.2320.253 0.253 0.253 0.253 0.253 0.253

Check 125 psi < F /A < 275 psi [Satisfactory] (ADAPT suggestion)

M / S + F / A , (ksi) 0.232 0.279 0.043 0.080 0.333 0.080 0.043 0.279 0.232for load combination (DL + PT) 0.064 0.380 0.063 0.063 0.380 0.064 - M / S + F / A , (ksi) 0.232 0.186 0.421 0.460 0.207 0.460 0.421 0.186 0.232for load combination (DL + PT) 0.442 0.126 0.443 0.443 0.126 0.442

Check ft < 7.5 (fc')0.5 and fc < 0.45 fc' [Satisfactory] (ACI 318-02, 18.3.3 & 18.4.2a), where 7.5 (fc')0.5 = 0.503 0.45 fc' = 2.025

M / S + F / A , (ksi) 0.232 0.404 -0.409 -0.462 0.582 -0.462 -0.409 0.404 0.232 for load (DL + LL + PT) -0.388 0.674 -0.479 -0.479 0.674 -0.388 - M / S + F / A , (ksi) 0.232 0.061 0.874 1.002 -0.042 1.002 0.874 0.061 0.232 for load (DL + LL + PT) 0.895 -0.167 0.985 0.985 -0.167 0.895

Check ft < 7.5 (fc')0.5 and fc < 0.6 fc' [Satisfactory] (ACI 318-02, 18.3.3 & 18.4.2b), where 7.5 (fc')0.5 = 0.503 0.6 fc' = 2.700

Tian LiDesign of Post-Tensioned Concrete Floor Based on ACI 318-02

Daniel

(cont'd)5. CALCULATE NON-PRESTRESSED REINFORCEMENT


Max. Nc (k / ft), (ACI 318, 18.0) 0.000 1.599 0.134 1.599 0.000

As (in2 / ft), (ACI 318, 18.9.3.2) 0.000 0.053 0.000 0.053 0.000

Bottom Bars, Each Way Not ReqD # 4 @ 45in. o.c. Not ReqD # 4 @ 45in. o.c. Not ReqD

Max. Acf (in2), (ACI 318, 18.0) 2880 2880 2880 2880 2880 2880

As' (in2), (ACI 318, 18.9.3.2) 2.160 2.160 2.160 2.160 2.160 2.160

Top Bars at Column 7 # 5 7 # 5 7 # 5 7 # 5 7 # 5 7 # 5 L (ft), (ACI 318, 18.9.4.1) 3.03 9.39 9.39 9.39 9.39 3.03

6. CHECK FLEXURAL CAPACITY BY STRENGTH DESIGN METHOD


Factored Mu (ft-k / ft) 0.00 5.11 -16.19 -20.03 9.67 -20.03 -16.19 5.11 0.00

1.2 MDL + 1.6 MLL + 1.0 MSec -15.61 11.64 -20.50 -20.50 11.64 -15.61

dp (in) 4.00 4.67 7.50 5.50 7.50 5.50 7.50 5.50 7.50 4.67 4.00

ρp 0.00273 0.00234 0.00146 0.00170 0.00232 0.00170 0.00146 0.00234 0.00273

0.00157 0.00214 0.00157 0.00157 0.00214 0.00157

L / dp 60.00 51.39 40.00 65.45 48.00 65.45 48.00 65.45 40.00 51.39 60.00

fps (ksi) 185.64 186.56 190.44 188.24 185.88 188.24 190.44 186.56 185.64

(ACI 318, 18.7.2, b & c) 191.62 203.09 191.62 191.62 203.09 191.62

Aps (in2 / ft) 0.131 0.131 0.131 0.153 0.153 0.153 0.131 0.131 0.131

Actual Area 0.141 0.14 0.14 0.141 0.14 0.14 d (in) 6.63 6.75 6.63 6.75 6.63 6.75 6.63 6.75 6.63 6.75 6.63

a (in) 0.21 0.40 0.67 0.85 0.55 0.85 0.67 0.40 0.210.64 0.65 0.88 0.88 0.65 0.64

Required As (in2 / ft) 0.000 0.018 0.000 0.018 0.000

Bottom Bars, Each Way Not ReqD # 4 @ 136in. o.c. Not ReqD # 4 @ 136in. o.c. Not ReqD

Actual As (in2 / ft) 0.000 0.053 0.000 0.053 0.000

Required As' (in2) 0.000 2.933 6.655 6.655 2.933 0.000

Top Bars at Column Not ReqD 10 # 5 22 # 5 22 # 5 10 # 5 Not ReqD

Actual As' (in2) 2.170 3.100 6.820 6.820 3.100 2.170

φ Mn (ft-k / ft) -9.23 8.21 -16.34 -21.62 11.14 -21.62 -16.34 8.21 -9.23

Actual Capacity -17.51 12.68 -20.66 -20.66 12.68 -17.51

Check φ Mn > Mu [Satisfactory]

pt, (ACI 318, 18.8.1) 0.0433 0.0262 0.0246 0.0188 0.0216 0.0188 0.0246 0.0262 0.0433

c(dp - a / β1) / (a / β1) 0.0261 0.0180 0.0181 0.0181 0.0180 0.0261

Check pt > 0.005 [Satisfactory] (ACI 318, 18.8.1)

7. CHECK PUNCHING SHEAR CAPACITY BY STRENGTH DESIGN METHOD


RDL (k) 35.99 90.01 108.00 108.00 90.01 35.99 RLL (k) 22.49 56.26 67.50 67.50 56.26 22.49

RSec (k) -26.99 -67.51 -83.70 -83.70 -67.51 -26.99

Vu =1.2 RDL + 1.6 RLL + 1.0 Rsec 52.18 130.52 153.90 153.90 130.52 52.18

Required b0d, (ACI 318, 11-36) 228.48 565.70 655.06 655.06 565.70 228.48Required d, (ACI 318, 11.0) 3.16 5.20 5.88 5.88 5.20 3.16

For φ Vn > Vu, the required 0.00 0.00 0.00 0.00 0.00 0.00

column cap thickness, tcap (in) Not ReqD Not ReqD Not ReqD Not ReqD Not ReqD Not ReqD

Note: 1. The column moments are negligible for gravity punching design. Lateral loads, seismic and wind, should be supported by shear walls. Using equivalent frames to support lateral loads is not suggested. 2. By inspection, the deflections of slab do not govern PT concrete floor design. Otherwise, using PT concrete floor is inadequate for larger live load. (ACI 318, 9.5.4.1)

Techincal References: 1. "Design of Post-Tensioned Slabs Using Unbonded Tendons, Third Edition", The Post-Tensioning Institute, 2004. 2. "Design, Construction and Maintenance of Cast-in-Place Post-Tensioned Concrete Parking Structures, First Edition", The Post-Tensioning Institute, 2001. 3. Bijan O. Aalami & Allan Bommer, "Design Fundamentals of Post-Tensioned Concrete Floors, First Edition", The Post-Tensioning Institute, 1999.







2. INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH f'c = 5 ksi



fpy = 243 ksi

fse = 175 ksi

dia = 1 / 2 in

Aps = 0.153 in2


Location Left Mid of L1 Support Mid of L2 Support Mid of L3 Support Mid L4 Right

Span (ft) 20 30 30 20DL (psf) 120 120 120 120LL (psf) 75 75 75 75

Balanced DL (60%-80% suggested) 75% 80% 80% 75% dCGS (in, from bottom) 4 3.33 7.5 2.5 7.5 2.5 7.5 3.33 4REQD EFFECTIVE PT (k / ft) 22.31 25.92 25.92 22.31 Total if banded (kips) 669.4 777.6 777.6 669.4

Tendons 1 / 2 in Dia @ 14 in o.c. 1 / 2 in Dia @ 12 in o.c. 1 / 2 in Dia @ 12 in o.c. 1 / 2 in Dia @ 14 in o.c. Total Number if banded 26 30 30 26Top Bars at Column 7 # 5 , L = 3.03 ft 9 # 5 , L = 9.39 ft 22 # 5 , L = 9.39 ft 9 # 5 , L = 9.39 ft 7 # 5 , L = 3.03 ftBot Bars, Cont., E. Way Not ReqD Not ReqD # 4 @ 378in. o.c. Not ReqDRequired column cap thk, (in) Not ReqD Not ReqD Not ReqD Not ReqD Not ReqD



MDL (ft-k / ft) 0.00 2.21 -7.59 4.85 -9.71 4.85 -7.59 2.21 0.00 MLL (ft-k / ft) 0.00 1.38 -4.74 3.03 -6.07 3.03 -4.74 1.38 0.00

Balanced Load (psf, uplift) -90 -96 -96 -90

Balanced MBal (ft-k / ft) 0.00 -1.54 5.93 -3.92 7.84 -3.92 5.93 -1.54 0.00

Required Effective PT (k / ft) 22.31 25.92 25.92 22.31 Tendon Spacing (in) 14 12 12 14

Primary MFe (ft-k / ft) 0.00 -1.25 6.51 7.56 -3.24 7.56

7.56 -3.24 7.56 6.51 -1.25 0.00

Secondary MSec (ft-k / ft) 0.00 0.29 0.58 -0.28 0.68 1.63

1.63 0.68 -0.28 0.58 0.29 0.00



A (in2 / ft) 96 96 96 96 96 96 96 96 96 S (in3 / ft) 128 128 128 128 128 128 128 128 128

F / A (ksi) 0.232 0.232 0.232 0.270 0.270 0.2700.270 0.270 0.270 0.232 0.232 0.232


M / S + F / A , (ksi) 0.232 0.295 0.077 0.095 0.358 0.114for load combination (DL + PT) 0.114 0.358 0.095 0.077 0.295 0.232 - M / S + F / A , (ksi) 0.232 0.170 0.388 0.445 0.182 0.426for load combination (DL + PT) 0.426 0.182 0.445 0.388 0.170 0.232

Check ft < 7.5 (fc')0.5 and fc < 0.45 fc' [Satisfactory] (ACI 318-02, 18.3.3 & 18.4.2a), where 0.45 fc' = 0.530 0.45 fc' = 2.250

M / S + F / A , (ksi) 0.232 0.425 -0.368 -0.474 0.642 -0.330 for load (DL + LL + PT) -0.330 0.642 -0.474 -0.368 0.425 0.232 - M / S + F / A , (ksi) 0.232 0.040 0.833 1.014 -0.102 0.870 for load (DL + LL + PT) 0.870 -0.102 1.014 0.833 0.040 0.232


DanielTian Li

Design of Post-Tensioned Concrete Floor Based on ACI 318-02



Max. Nc (k / ft), (ACI 318, 18.0) 0.000 0.672 0.672 0.000

As (in2 / ft), (ACI 318, 18.9.3.2) 0.000 0.000 0.000 0.000

Bottom Bars, Each Way Not ReqD Not ReqD Not ReqD Not ReqD

Max. Acf (in2), (ACI 318, 18.0) 2880 2880 2880 2880 2880

As' (in2), (ACI 318, 18.9.3.2) 2.160 2.160 2.160 2.160 2.160

Top Bars at Column 7 # 5 7 # 5 7 # 5 7 # 5 7 # 5 L (ft), (ACI 318, 18.9.4.1) 3.03 9.39 9.39 9.39 3.03



Factored Mu (ft-k / ft) 0.00 5.14 -16.12 -21.63 11.35 -15.06 5.14 0.00

1.2 MDL + 1.6 MLL + 1.0 MSec -15.06 11.35 -21.63 -16.12

dp (in) 4.00 4.67 7.50 5.50 7.50 5.50 7.50 4.67 4.00

ρp 0.00273 0.00234 0.00146 0.00170 0.00232 0.00170

0.00170 0.00232 0.00170 0.00146 0.00234 0.00273

L / dp 60.00 51.39 40.00 65.45 48.00 65.45 40.00 51.39 60.00

fps (ksi) 186.25 187.27 191.59 189.22 186.60 189.22

(ACI 318, 18.7.2, b & c) 189.22 200.98 189.22 191.59 187.27 186.25

Aps (in2 / ft) 0.131 0.131 0.131 0.153 0.153 0.153

Actual Area 0.153 0.15 0.15 0.131 0.131 0.131 d (in) 6.63 6.75 6.63 6.75 6.63 6.75 6.63 6.75 6.63

a (in) 0.19 0.36 0.60 0.83 0.57 0.540.54 0.58 0.83 0.60 0.36 0.19

Required As (in2 / ft) 0.000 0.000 0.006 0.000

Bottom Bars, Each Way Not ReqD Not ReqD # 4 @ 378in. o.c. Not ReqD

Actual As (in2 / ft) 0.000 0.000 0.006 0.000

Required As' (in2) 0.000 2.683 6.705 2.683 0.000

Top Bars at Column Not ReqD 9 # 5 22 # 5 9 # 5 Not ReqD

Actual As' (in2) 2.170 2.790 6.820 2.790 2.170

φ Mn (ft-k / ft) -9.28 8.27 -16.22 -21.74 11.35 -18.36

Actual Capacity -18.36 12.02 -21.74 -16.22 8.27 -9.28


pt, (ACI 318, 18.8.1) 0.0469 0.0284 0.0271 0.0187 0.0203 0.0302

c(dp - a / β1) / (a / β1) 0.0302 0.0199 0.0187 0.0271 0.0284 0.0469




RDL (k) 35.99 90.01 108.00 90.01 35.99 RLL (k) 22.49 56.26 67.50 56.26 22.49

RSec (k) -26.99 -70.21 -86.40 -70.21 -26.99

Vu =1.2 RDL + 1.6 RLL + 1.0 Rsec 52.18 127.81 151.21 127.81 52.18

Required b0d, (ACI 318, 11-36) 219.34 527.85 613.74 527.85 219.34Required d, (ACI 318, 11.0) 3.04 4.90 5.57 4.90 3.04

For φ Vn > Vu, the required 0.00 0.00 0.00 0.00 0.00

column cap thickness, tcap (in) Not ReqD Not ReqD Not ReqD Not ReqD Not ReqD












fpy = 243 ksi

fse = 175 ksi

dia = 1 / 2 in

Aps = 0.153 in2


Location Left Mid of L1 Support Mid of L2 Support Mid of L3 Right

Span (ft) 20 30 20DL (psf) 120 120 120LL (psf) 75 75 75

Balanced DL (60%-80% suggested) 75% 80% 75% dCGS (in, from bottom) 4 3.33 7.5 2.5 7.5 3.33 4REQD EFFECTIVE PT (k / ft) 22.31 25.92 22.31 Total if banded (kips) 669.4 777.6 669.4

Tendons 1 / 2 in Dia @ 14 in o.c. 1 / 2 in Dia @ 12 in o.c. 1 / 2 in Dia @ 14 in o.c. Total Number if banded 26 30 26Top Bars at Column 7 # 5 , L = 3.03 ft 15 # 5 , L = 9.39 ft 15 # 5 , L = 9.39 ft 7 # 5 , L = 3.03 ftBot Bars, Cont., E. Way Not ReqD # 4 @ 58in. o.c. Not ReqDRequired column cap thk, (in) Not ReqD Not ReqD Not ReqD Not ReqD



MDL (ft-k / ft) 0.00 1.96 -8.08 5.42 -8.08 1.96 0.00 MLL (ft-k / ft) 0.00 1.23 -5.05 3.39 -5.05 1.23 0.00

Balanced Load (psf, uplift) -90 -96 -90

Balanced MBal (ft-k / ft) 0.00 -1.32 6.37 -4.43 6.37 -1.32 0.00

Required Effective PT (k / ft) 22.31 25.92 22.31 Tendon Spacing (in) 14 12 14

Primary MFe (ft-k / ft) 0.00 -1.25 6.51 6.51 -1.25 0.00

7.56 -3.24 7.56

Secondary MSec (ft-k / ft) 0.00 0.07 0.14 0.14 0.07 0.00

1.19 1.19 1.19



A (in2 / ft) 96 96 96 96 96 96 96 S (in3 / ft) 128 128 128 128 128 128 128

F / A (ksi) 0.232 0.232 0.232 0.232 0.232 0.2320.270 0.270 0.270


M / S + F / A , (ksi) 0.232 0.293 0.072 0.072 0.293 0.232for load combination (DL + PT) 0.110 0.363 0.110 - M / S + F / A , (ksi) 0.232 0.172 0.393 0.393 0.172 0.232for load combination (DL + PT) 0.430 0.177 0.430


M / S + F / A , (ksi) 0.232 0.408 -0.401 -0.401 0.408 0.232 for load (DL + LL + PT) -0.363 0.681 -0.363 - M / S + F / A , (ksi) 0.232 0.057 0.866 0.866 0.057 0.232 for load (DL + LL + PT) 0.903 -0.141 0.903


DanielTian Li




Max. Nc (k / ft), (ACI 318, 18.0) 0.000 1.158 0.000

As (in2 / ft), (ACI 318, 18.9.3.2) 0.000 0.000 0.000

Bottom Bars, Each Way Not ReqD Not ReqD Not ReqD

Max. Acf (in2), (ACI 318, 18.0) 2880 2880 2880 2880

As' (in2), (ACI 318, 18.9.3.2) 2.160 2.160 2.160 2.160

Top Bars at Column 7 # 5 7 # 5 7 # 5 7 # 5 L (ft), (ACI 318, 18.9.4.1) 3.03 9.39 9.39 3.03



Factored Mu (ft-k / ft) 0.00 4.39 -17.63 -17.63 4.39 0.00

1.2 MDL + 1.6 MLL + 1.0 MSec -16.58 13.12 -16.58

dp (in) 4.00 4.67 7.50 5.50 7.50 4.67 4.00

ρp 0.00273 0.00234 0.00146 0.00146 0.00234 0.00273

0.00170 0.00232 0.00170

L / dp 60.00 51.39 40.00 65.45 40.00 51.39 60.00

fps (ksi) 186.25 187.27 191.59 191.59 187.27 186.25

(ACI 318, 18.7.2, b & c) 189.22 200.98 189.22

Aps (in2 / ft) 0.131 0.131 0.131 0.131 0.131 0.131

Actual Area 0.153 0.15 0.15 d (in) 6.63 6.75 6.63 6.75 6.63 6.75 6.63

a (in) 0.19 0.33 0.66 0.66 0.33 0.190.61 0.65 0.61

Required As (in2 / ft) 0.000 0.041 0.000

Bottom Bars, Each Way Not ReqD # 4 @ 58in. o.c. Not ReqD

Actual As (in2 / ft) 0.000 0.041 0.000

Required As' (in2) 0.000 4.367 4.367 0.000

Top Bars at Column Not ReqD 15 # 5 15 # 5 Not ReqD

Actual As' (in2) 2.170 4.650 4.650 2.170

φ Mn (ft-k / ft) -9.28 8.30 -17.90 -17.90 8.30 -9.28

Actual Capacity -20.03 13.13 -20.03


pt, (ACI 318, 18.8.1) 0.0469 0.0314 0.0241 0.0241 0.0314 0.0469

c(dp - a / β1) / (a / β1) 0.0267 0.0173 0.0267




RDL (k) 35.99 90.01 90.01 35.99 RLL (k) 22.49 56.26 56.26 22.49

RSec (k) -26.99 -70.21 -70.21 -26.99

Vu =1.2 RDL + 1.6 RLL + 1.0 Rsec 52.18 127.82 127.82 52.18

Required b0d, (ACI 318, 11-36) 219.34 527.86 527.86 219.34Required d, (ACI 318, 11.0) 3.04 4.90 4.90 3.04

For φ Vn > Vu, the required 0.00 0.00 0.00 0.00

column cap thickness, tcap (in) Not ReqD Not ReqD Not ReqD Not ReqD












fpy = 243 ksi

fse = 175 ksi

dia = 1 / 2 in

Aps = 0.153 in2


Location Left Mid of L1 Support Mid L2 Right

Span (ft) 20 20DL (psf) 120 120LL (psf) 75 75

Balanced DL (60%-80% suggested) 75% 75% dCGS (in, from bottom) 4 3.33 7.5 3.33 4REQD EFFECTIVE PT (k / ft) 22.31 22.31 Total if banded (kips) 669.4 669.4

Tendons 1 / 2 in Dia @ 14 in o.c. 1 / 2 in Dia @ 14 in o.c. Total Number if banded 26 26Top Bars at Column 7 # 5 , L = 3.03 ft 7 # 5 , L = 6.06 ft 7 # 5 , L = 3.33 ftBot Bars, Cont., E. Way # 4 @ 73in. o.c. # 4 @ 73in. o.c.Required column cap thk, (in) Not ReqD Not ReqD Not ReqD



MDL (ft-k / ft) 0.00 3.50 -5.00 3.50 0.00 MLL (ft-k / ft) 0.00 2.19 -3.13 2.19 0.00

Balanced Load (psf, uplift) -90 -90

Balanced MBal (ft-k / ft) 0.00 -2.63 3.75 -2.63 0.00

Required Effective PT (k / ft) 22.31 22.31 Tendon Spacing (in) 14 14

Primary MFe (ft-k / ft) 0.00 -1.25 6.51

6.51 -1.25 0.00

Secondary MSec (ft-k / ft) 0.00 1.38 2.76

2.76 1.38 0.00



A (in2 / ft) 96 96 96 96 96 S (in3 / ft) 128 128 128 128 128

F / A (ksi) 0.232 0.232 0.2320.232 0.232 0.232


M / S + F / A , (ksi) 0.232 0.314 0.115for load combination (DL + PT) 0.115 0.314 0.232 - M / S + F / A , (ksi) 0.232 0.150 0.350for load combination (DL + PT) 0.350 0.150 0.232


M / S + F / A , (ksi) 0.232 0.520 -0.178 for load (DL + LL + PT) -0.178 0.520 0.232 - M / S + F / A , (ksi) 0.232 -0.055 0.643 for load (DL + LL + PT) 0.643 -0.055 0.232


DanielTian Li




Max. Nc (k / ft), (ACI 318, 18.0) 0.250 0.250

As (in2 / ft), (ACI 318, 18.9.3.2) 0.000 0.000

Bottom Bars, Each Way Not ReqD Not ReqD

Max. Acf (in2), (ACI 318, 18.0) 2880 2880 2880

As' (in2), (ACI 318, 18.9.3.2) 2.160 2.160 2.160

Top Bars at Column 7 # 5 7 # 5 7 # 5 L (ft), (ACI 318, 18.9.4.1) 3.03 6.06 3.33



Factored Mu (ft-k / ft) 0.00 9.08 -8.24 9.08 0.00

1.2 MDL + 1.6 MLL + 1.0 MSec -8.24

dp (in) 4.00 4.67 7.50 4.67 4.00

ρp 0.00273 0.00234 0.00146

0.00146 0.00234 0.00273

L / dp 60.00 51.39 32.00 51.39 60.00

fps (ksi) 186.25 187.27 214.46

(ACI 318, 18.7.2, b & c) 191.59 187.27 186.25

Aps (in2 / ft) 0.131 0.131 0.131

Actual Area 0.131 0.131 0.131 d (in) 6.63 6.75 6.63 6.75 6.63

a (in) 0.19 0.52 0.260.27 0.52 0.19

Required As (in2 / ft) 0.033 0.033

Bottom Bars, Each Way # 4 @ 73in. o.c. # 4 @ 73in. o.c.

Actual As (in2 / ft) 0.033 0.033

Required As' (in2) 0.000 0.000 0.000

Top Bars at Column Not ReqD 0 # 5 Not ReqD

Actual As' (in2) 2.170 2.170 2.170

φ Mn (ft-k / ft) -9.28 9.08 -17.66

Actual Capacity -16.00 9.08 -9.28


pt, (ACI 318, 18.8.1) 0.0469 0.0186 0.0669

c(dp - a / β1) / (a / β1) 0.0648 0.0186 0.0469




RDL (k) 35.99 72.02 35.99 RLL (k) 22.50 45.01 22.50

RSec (k) -26.99 -54.01 -26.99

Vu =1.2 RDL + 1.6 RLL + 1.0 Rsec 52.19 104.42 52.19

Required b0d, (ACI 318, 11-36) 219.36 438.91 219.36Required d, (ACI 318, 11.0) 3.04 4.19 3.04

For φ Vn > Vu, the required 0.00 0.00 0.00

column cap thickness, tcap (in) Not ReqD Not ReqD Not ReqD





INPUT DATA & DESIGN SUMMARYSPECIAL INSPECTION ( 0=NO, 1=YES ) 0 No, (reduced fm' by 0.5)TYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiSEISMIC PERFORMANCE CATEGORY D Seismic D

( C,D,E, 0=WIND, 5=GRAVITY)

SERVICE AXIAL LOAD P = 4.9 kSERVICE SHEAR LOAD V = 6 kSERVICE MOMENT LOAD M = 0 ft-kTHICKNESS OF WALL t = 8 inLENGTH OF WALL w = 4.67 ft

EFFECTIVE HEIGHT of WALL hw = 10.67 ftVERT. REINF. AT EACH END (As) 1 # 5 => DIST. FR BAR'S CENT. TO END a = 8 inWALL HORIZ. REINF. (Ash) 1 # 4 @ 16 in o.c.WALL VERT. REINF. (Asv) 1 # 4 @ 16 in o.c.

ANALYSISREINF. AREA AT EACH END As = 0.31 in2 MODULAR RATIO n = 21.48GROSS WALL AREA Ag = 448 in2 REINFORCEMENT RATIO ρ = 0.0008EFFECTIVE LENGTH OF WALL d = 48 in ALLOWABLE STRESS FACTOR SF = 0.667EFFECTIVE THICKNESS OF WALL bw = 7.63 in REQD MIN. HORIZ. REINF. Ash,min = 0.128 in2/ft

MASONRY ELASTICITY MODULUS Em = 1350 ksi Ssh,max = 19 inSTEEL ELASTICITY MODULUS Es = 29000 ksi REQD MIN. VERT. REINF. Asv,min = 0.128 in2/ft

Ssv,max = 19 inTHE ALLOWABLE STRESS DUE TO FLEXURE IS THE ALLOWABLE REINF. STRESS DUE TO FLEXURE IS

ksi 32.00 ksi

THE TOTAL AXIAL FORCE ACTING AT BOTTOM IS THE TOTAL MOMENT ACTING AT BOTTOM IS

9.4 kips ft-kips

THE NEUTRAL AXIS DEPTH FACTOR IS THE SHEAR STRESS IN MASONRY IS

= 0.173 = 14 psi


Pm (k) 4.9

Mm (ft-k)

64.02

Masonry Shear Wall Design Based on ACI 530-02

0.330

DanielTian Li

[THE WALL DESIGN IS ADEQUATE.]

vw w

Vf

b l=

( )( )'0.33b mSF fF = =

T wM V hM = + =

( )22k n n nρ ρ ρ= + −

( )( )& 241.33S wind seismic onlyF = =

( )T P wall weightP = + =

0

20

40

60

80

100

120

140

160

0 20 40 60 80 100

THE CODE SEC. 2.3.2.2.1 PERMITS COMPRESSION FORCES TO BE RESISTED BY COMPRESSION REINFORCEMENT

ONLY IF THE LATERAL SUPPORT REQUIREMENTS OF CODE SEC. 2.1.6.5 ARE MET. SINCE IT IS VIRTUALLY IMPOSSIBLE

TO MEET THESE PROVISIONS IN WALLS, THE CONTRIBUTION OF REINFORCING STEEL TO COMPRESSIVE FORCE

MUST BE NEGLECTED.

THE MAXIMUN DESIGN AXIAL LOAD STRENGTH IS P m = t Lw Fb = 147.9456 kips. THE DESIGN MOMENT CAPACITYAT MAXIMUM AXIAL LOAD STRENGTH IS 0 ft-kips.THE DESIGN AXIAL AND MOMENT CAPACITIES AT THE WALL CRACKED BUT STEEL STRESS ZERO ARE 85 kipsAND 85 ft-kips.

FOR THE BALANCED STRAIN CONDITION UNDER COMBINED FLEXURE AND AXIAL LOAD, THE MAXIMUM STRAIN IN

THE MASONRY AND IN THE TENSION REINFORCEMENT MUST SIMULTANEOUSLY REACH THE VALUES ASεm = Fb / Em and εs = Fs / Es . THE DESIGN AXIAL AND MOMENT CAPACITIES AT THE BALANCED STRAIN

CONDITION ARE 60 kips AND 87 ft-kips.


Pm (kips) Mm (ft-kips)AT AXIAL LOAD ONLY = 148 0

AT LARGE AXIAL LOAD = 116 58

AT 0 % TENSION = 85 85

AT 25 % TENSION = 77 87

AT 50 % TENSION = 71 88

AT BALANCED STRAIN CONDITION = 60 87

AT SMALL AXIAL LOAD = 17 69

AT FLEXURE ONLY = 0 59

THE DESIGN FORCES P & M ARE ALSO PLOTTED ON THE INTERACTION DIAGRAM. FROM THE INTERACTION DIAGRAM,THE ALLOWABLE MOMENT AT AN AXIAL LOAD P IS

Mm = 65 ft-kips. > M [Satisfactory]

CHECK SHEAR CAPACITYTHE ALLOWABLE SHEAR STRESS IS GIVEN BY

Fv, without reinf. =

= 23.33 psi > fv [Satisfactory]

( ) '( , 35) , 1.0Tm

MSF MIN forfVd

≥

( ) '1 454 , 80 , 1.03

T TTm

MM MSF MIN forfVd Vd Vd

− − <

Fv, max. =


CHECK THE MINIMUM AREA OF SHEAR REINFORCEMENT REQUIRED :

0.05 in2 / ft < 0.15 in2 / ft (No shear reinf. reqd.)

Techincal References: 1. "Masonry Designers' Guide, Third Edition" (MDG-3), The Masonry Society, 2001. 2. Alan Williams: "Structural Engineering Reference Manual", Professional Publications, Inc, 2001.

vAs

=s

VdF

=

( ) '(1.5 , 75) , 1.0Tm

MSF MIN forfVd

≥

( ) '1 454 , 120 , 1.02

T TTm


− − <



INPUT DATA & DESIGN SUMMARYSPECIAL INSPECTION ( 0=NO, 1=YES ) 1 YesTYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiLATERAL PERFORMANCE CATEGORY 4 Zone 4

( 0=Wind,1=Zone 1, 2=Zone 2, 3=Zone 3, 4=Zone 4, 5=Gravity Only)

SERVICE AXIAL LOAD P = 245 kSERVICE SHEAR LOAD V = 160 kSERVICE MOMENT LOAD M = 8550 ft-kTHICKNESS OF WALL t = 10 inLENGTH OF WALL w = 29 ft

EFFECTIVE HEIGHT of WALL hw = 10 ftVERT. REINF. AT EACH END (As) 2 # 9 => DIST. FR BAR'S CENT. TO END a = 4 inWALL HORIZ. REINF. (Ash) 2 # 4 @ 16 in o.c.WALL VERT. REINF. (Asv) 2 # 9 @ 8 in o.c.

ANALYSISREINF. AREA AT EACH END As = 2.00 in2 MODULAR RATIO n = 25.78GROSS WALL AREA Ag = 3480 in2 REINFORCEMENT RATIO ρ = 0.0006EFFECTIVE LENGTH OF WALL d = 344 in ALLOWABLE STRESS FACTOR SF = 1.333EFFECTIVE THICKNESS OF WALL bw = 9.63 in REQD MIN. HORIZ. REINF. Ash,min = 0.300 in2/ft

MASONRY ELASTICITY MODULUS Em = 1125 ksi Ssh,max = 16 inSTEEL ELASTICITY MODULUS Es = 29000 ksi REQD MIN. VERT. REINF. Asv,min = 0.100 in2/ft

Ssv,max = 40 inTHE ALLOWABLE STRESS DUE TO FLEXURE IS THE ALLOWABLE REINF. STRESS DUE TO FLEXURE IS

ksi 32.00 ksi

THE TOTAL AXIAL FORCE ACTING AT BOTTOM IS THE TOTAL MOMENT ACTING AT BOTTOM IS

277.6 kips ft-kips

THE NEUTRAL AXIS DEPTH FACTOR IS THE SHEAR STRESS IN MASONRY IS

= 0.162 = 76 psi, (UBC 2107.1.7)(The factor 1.5 for zone 3 & 4 only)


9

Pm (k)

Mm (ft-k)

10150.00


Masonry Shear Wall Design Based on UBC 97

0.660

DanielTian Li

1.5v

ww

Vf

jb l=

( )( )'0.33b mSF fF = =

T wM V hM = + =

( )22k n n nρ ρ ρ= + −


( )T P wall weightP = + =

0

500

1000

1500

2000

2500

0 2000 4000 6000 8000 10000 12000

THE ACI 530-02 SEC. 2.3.2.2.1 PERMITS COMPRESSION FORCES TO BE RESISTED BY COMPRESSION REINFORCEMENT

ONLY IF THE LATERAL SUPPORT REQUIREMENTS OF ACI 530-02 SEC. 2.1.6.5 ARE MET. SINCE IT IS VIRTUALLY IMPOSSIBLE

TO MEET THESE PROVISIONS IN WALLS, THE CONTRIBUTION OF REINFORCING STEEL TO COMPRESSIVE FORCE

MAY BE NEGLECTED.

THE MAXIMUN DESIGN AXIAL LOAD STRENGTH IS P m = t Lw Fb = 2296.8 kips. THE DESIGN MOMENT CAPACITYAT MAXIMUM AXIAL LOAD STRENGTH IS 0 ft-kips.THE DESIGN AXIAL AND MOMENT CAPACITIES AT THE WALL CRACKED BUT STEEL STRESS ZERO ARE 1514 kipsAND 7484 ft-kips.

FOR THE BALANCED STRAIN CONDITION UNDER COMBINED FLEXURE AND AXIAL LOAD, THE MAXIMUM STRAIN IN

THE MASONRY AND IN THE TENSION REINFORCEMENT MUST SIMULTANEOUSLY REACH THE VALUES ASεm = Fb / Em and εs = Fs / Es . THE DESIGN AXIAL AND MOMENT CAPACITIES AT THE BALANCED STRAIN

CONDITION ARE 1028 kips AND 9235 ft-kips.


Pm (kips) Mm (ft-kips)AT AXIAL LOAD ONLY = 2297 0

AT LARGE AXIAL LOAD = 1905 4710

AT 0 % TENSION = 1514 7484

AT 25 % TENSION = 1379 8061

AT 50 % TENSION = 1256 8509

AT BALANCED STRAIN CONDITION = 1028 9235

AT SMALL AXIAL LOAD = 270 11334

AT FLEXURE ONLY = 0 10918

THE DESIGN FORCES P & M ARE ALSO PLOTTED ON THE INTERACTION DIAGRAM. FROM THE INTERACTION DIAGRAM,THE ALLOWABLE MOMENT AT AN AXIAL LOAD P IS

Mm = 11410 ft-kips. > M [Satisfactory]

CHECK SHEAR CAPACITYTHE ALLOWABLE SHEAR STRESS IS GIVEN BY

Fv, without reinf. =

= 46.67 psi < fv (Shear reinf. reqd to carry full shear load.)

( ) '( , 35) , 1.0Tm

MSF MIN forfVd

≥

( ) '1 454 , 80 , 1.03

T TTm


− − <

Fv, max. =



0.17 in2 / ft < 0.30 in2 / ft [Satisfactory]

Techincal References: 1. "Masonry Designers' Guide, Third Edition" (MDG-3), The Masonry Society, 2001. 2. Alan Williams: "Structural Engineering Reference Manual", Professional Publications, Inc, 2001. 3. "Seismic Design of Masonry Using the 1997 UBC", Concrete Masonry Association of NV & CA, 2003.

vAs

=s

VdF

=

( ) '(1.5 , 75) , 1.0Tm

MSF MIN forfVd

≥

( ) '1 454 , 120 , 1.02

T TTm


− − <



INPUT DATA & DESIGN SUMMARYSPECIAL INSPECTION ( 0=NO, 1=YES ) 0 NoTYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiSERVICE GRAVITY LOAD P = 400 lbs / ftSERVICE LATERAL LOAD w1 = 25 plf / ftSERVICE PARAPET LOAD w2 = 45 plf / ftTHICKNESS OF WALL t = 8 inPARAPET HEIGHT hp = 4 ftWALL HEIGHT h = 20 ftECCENTRICITY e = 6 inWALL VERT. REINF. (Asv) 2 # 6 @ 16 in o.c. (at each face)

[THE WALL DESIGN IS ADEQUATE.]ANALYSISREINF. AREA AT EACH SIDE As = 0.33 in2 MODULAR RATIO n = 21.48EFFECTIVE DEPTH d = 5.26 in REINFORCEMENT RATIO ρ = 0.0052WIDTH OF SECTION bw = 12.00 in ALLOWABLE STRESS FACTOR SF = 0.667EFFECTIVE THICKNESS te = 7.63 in THE NEUTRAL AXIS DEPTH FACTOR ISMASONRY ELASTICITY MODULUS Em = 1350 ksiSTEEL ELASTICITY MODULUS Es = 29000 ksiTHE ALLOWABLE STRESS DUE TO FLEXURE IS THE ALLOWABLE REINF. STRESS DUE TO FLEXURE IS

psi 32000 psi

THE DISTANCE FROM BOTTOM TO M1 IS THE GOVERNING MOMENTS AND AXIAL FORCES ARE

= 10.0 ft = 1250 ft-lbs/ft

= lbs / ft

THE GOVERNING SHEAR FORCES ARE= 360 ft-lbs/ft

= 250 lbs / ft

= lbs / ft

= 250 lbs / ft THE GOVERNING SHEAR STRESS IN MASONRY IS

= 180 lbs / ft 23 = 2.73 psi

DETERMINE THE REGION FOR FLEXURE AND AXIAL LOAD (MDG Tab 12.2.1, Fig 12.2-12 & 13, page 12-25).

1. Wall is in compression and not cracked. 2. Wall is cracked but steel is in compression. 3. Wall is cracked and steel is in tension.

REGION 3 APPLICABLE FOR (M1, P1)


Masonry Bearing Wall Design Based on ACI 530-02

330

1613

747

= 0.37489

Tian LiDaniel

( )1 2 3, ,v

e w

MAX V V Vf

t b=

( )( )'0.33b mSF fF = =

( )22k n n nρ ρ ρ= + −


6eM t

Pd d≤

12 3

eM tPd d

≤ −

12 3

eM tPd d

> −

( )2

1 2 21 2

1

122 pwPe h hM

w h = + −

( )2

12p

p

h h PeS h h

h hw

+ = + − −

( )1 P wall weightP = +2

22

2pw h

M =

( )2 P wall weightP = +( ) ( )2

11 1 2

pp

h h w PehV h w

h h

+= + − +

2 1 1hV w V= −

3 2pV h w=

(cont'd)CHECK REGION 1 CAPACITY

3031 ft-lbs / ft > M1 [Not applicable]

3123 ft-lbs / ft > M2 [Not applicable]CHECK REGION 2 CAPACITY

476 ft-lbs / ft < M1 [Not applicable]

230 ft-lbs / ft < M2 [Not applicable]CHECK REGION 3 CAPACITY (The moment maybe limited by either the masonry compression or steel tension, MDG page 12-25).

1301 ft-lbs / ft > M1 [Satisfactory]


THE ALLOWABLE SHEAR STRESS IS GIVEN BY



2

6 6w e e

m bb t tPM F

= − =

222 3e

mbw

t PPMb F

= − =

1,

2 3 2 3 2 3e e

m b s swkd kd kdt tMIN kd d P d d PbM F A F

= − − − − + −

( ) ( )' , 50v mSF MIN fF =

=



INPUT DATA & DESIGN SUMMARYTYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiSERVICE DEAD LOAD PDL = 625 lbs / ftLATERAL LOAD (E/1.4 or W) w1 = 26.9 plf / ftLATERAL LOAD (E/1.4 or W) w2 = 26.9 plf / ftTHICKNESS OF WALL t = 8 inPARAPET HEIGHT hp = 0 ftWALL HEIGHT h = 23 ftECCENTRICITY e = 8 inWALL VERT. REINF. (Asv) 1 # 5 @ 16 in o.c. (at middle)SEISMIC COEFFICIENT Ca = 0.484 [THE WALL DESIGN IS ADEQUATE.]IMPORTANCE FACTOR I = 1

ANALYSISREINF. AREA AT EACH SIDE As = 0.23 in2/ft EFFECTIVE THICKNESS te = 7.63 in

EFFECTIVE DEPTH d = 3.82 in MASONRY ELASTICITY MODULUS Em = 1125 ksi

WIDTH OF SECTION bw = 12.00 in STEEL ELASTICITY MODULUS Es = 29000 ksi

GROSS MOMENT OF INERTIA Ig = 444 in4/ft MODULAR RATIO n = 25.78

CHECK REINFORCING RATIO (UBC 2108.2.4.2)

ρ = As / d bw = < 0.5ρb = 0.50.85(0.85)fm'(0.003Es) / [fy(0.003Es+fy)] =

[Satisfactory]CHECK WALL SLENDERNESS REQUIRED BY EQ(8-19)

(Pw + Pf) / Ag = 16.5 psi < 0.04 fm' = 60 psi [Satisfactory]

where Pw = (0.5 h + hp)(115 psf) t = 882 lbs / ft , Pf = 625 lbs / ft

THE CRACKING MOMENT STRENGTH IS

Mcr = S fr = (bw te2 / 6)(4 fm'0.5) = 1503 ft-lbs/ft

CHECK CAPACITY OF LOAD COMBINATION UBC(12-6), 0.9D+Eh

Pu = 0.9 (PDL + Pw) = lbs/ft

THE DEPTH OF THE COMPRESSIVE STRESS BLOCK IS THE DEPTH OF NEUTRAL AXIS IS

a = (Pu + As fy) / (0.85 fm' bw) = 1.00 in c = a/ 0.85 = 1.18 in

THE EFFECTIVE AREA OF REINFORCING STEEL IS THE CRACKED MOMENT OF INERTIA IS

Ase = (Pu + As fy) / fy = 0.26 in2/ft Icr = n Ase(d-c)2 + bc3 / 3 = 52 in4/ft

THE MOMENT AND DEFLECTION AT THE MID-HEIGHT OF THE WALL ARE GIVEN BY 1st Cycle 2nd Cycle 3rd Cycle Final

∆u = 5Mcrh2/(48EmIg) + 5(Mu - Mcr)h

2/(48EmIcr) = 0 2.187 2.591 2.683 in

Mu = wuh2/8 + Puf e/2 + Pu∆u = 2678 > Mcr 2927 2974 2984 ft-lbs/ft

[Satisfactory] => Eq (8-29) ApplicableTHE MOMENT CAPACITY OF THE WALL IS

φMn = φ[Asefy(d-a/2) - Pu(d-te/2)] = <= Not applicable

φMn = φAsefy(d-a/2) = 3385 ft-lbs/ft

where φ = 0.8, (SEC 2108.1.4.2.1)

CHECK DEFLECTION LIMITATION BY SEC. 2108.2.4.61st Cycle 2nd Cycle 3rd Cycle Final

∆s = 5Mcrh2/(48EmIg) + 5(Mser - Mcr)h

2/(48EmIcr) = 0 1.069 1.289 1.346 in

Mser = wh2/8 + Pf e/2 + P∆s = 1987 > Mcr 2123 2151 2158 ft-lbs/ft

[Satisfactory] => Eq (8-29) Applicable

0.007 h = 1.93 in > ∆s [Satisfactory]

Mu

Masonry Bearing Wall Design Based on UBC 97

0.005 0.005

DanielTian Li

1370

[Satisfactory]>

CHECK CAPACITY OF LOAD COMBINATION UBC(12-5), (1.2+0.5CaI)D+0.5L+Eh (cont'd)

Pu = (1.2+0.5CaI)(PDL + Pw) = 2195 lbs/ft

THE DEPTH OF THE COMPRESSIVE STRESS BLOCK IS THE DEPTH OF NEUTRAL AXIS IS

a = (Pu + As fy) / (0.85 fm' bw) = 1.06 in c = a/ 0.85 = 1.24 in

THE EFFECTIVE AREA OF REINFORCING STEEL IS THE CRACKED MOMENT OF INERTIA IS

Ase = (Pu + As fy) / fy = 0.27 in2/ft Icr = n Ase(d-c)2 + bc3 / 3 = 51 in4/ft

THE MOMENT AND DEFLECTION AT THE MID-HEIGHT OF THE WALL ARE GIVEN BY

wu = 1.4 w1 = 37.7 plf / ft

1st Cycle 2nd Cycle 3rd Cycle Final

∆u = 5Mcrh2/(48EmIg) + 5(Mu - Mcr)h

2/(48EmIcr) = 0 2.413 3.142 3.457 in

Mu = wuh2/8 + Puf e/2 + Pu∆u = 2791 > Mcr 3232 3365 3423 ft-lbs/ft

[Satisfactory] => Eq (8-29) ApplicableTHE MOMENT CAPACITY OF THE WALL IS

φMn = φ[Asefy(d-a/2) - Pu(d-te/2)] = <= Not applicable

φMn = φAsefy(d-a/2) = 3538 ft-lbs/ft

where φ = 0.8, (SEC 2108.1.4.2.1)

CHECK DEFLECTION LIMITATION BY SEC. 2108.2.4.61st Cycle 2nd Cycle 3rd Cycle Final

∆s = 5Mcrh2/(48EmIg) + 5(Mser - Mcr)h

2/(48EmIcr) = 0 1.069 1.289 1.346 in

Mser = wh2/8 + Pf e/2 + P∆s = 1987 > Mcr 2123 2151 2158 ft-lbs/ft

[Satisfactory] => Eq (8-29) Applicable

0.007 h = 1.93 in > ∆s [Satisfactory]

CHECK SHEAR CAPACITY BY SEC. 2108.2.4.5

φVn = φ 2 Amv (fm')0.5 = 4255 lbs/ft > Vu = 1.4 [w1h/2 + w2(h + 0.5hP)hP/h + PDLe/h] = 458 lbs/ft

where φ = 0.6 [Satisfactory]CHECK PARAPET BENDING CAPACITY

φMn > Mu [Not applicable]

Techincal References: 1. "Masonry Designers' Guide, Third Edition" (MDG-3), The Masonry Society, 2001. 2. "Seismic Design of Masonry Using the 1997 UBC", Concrete Masonry Association of NV & CA, 2003.

> Mu [Satisfactory]



INPUT DATA & DESIGN SUMMARYSPECIAL INSPECTION ( 0=NO, 1=YES ) 1 YesTYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiGIRDER SERVICE LOAD PG = 37 kipsECCENTRICITY e = 3 in

BEARING LENGTH Lbr = 20 in

SERVICE LATERAL LOAD w1 = 25 psfSERVICE PARAPET LOAD w2 = 45 psfTHICKNESS OF WALL t = 8 inPARAPET HEIGHT hp = 2 ftWALL HEIGHT h = 15 ft

WALL VERT. REINF. (Asv) 2 # 7@ 16 in o.c. (at each face)


ANALYSISEFFECTIVE THICKNESS te = 7.63 in

EFFECTIVE LENGTH (ACI 530, 2.1.9.1)

Le = 4te + Lbr = in

LOAD DISTRIBUTION P = PG / Le = 8789 lbs / ft

CHECK BEARING CAPACITY (ACI 530, 2.1.9.3)

fbr = PG / [(t - 3) Lbr] = 370 psi

< 0.25 fm' = 375 psi

[Satisfactory]REINF. AREA AT EACH SIDE As = 0.45 in2

EFFECTIVE DEPTH d = 5.19 in REINFORCEMENT RATIO ρ = 0.0072WIDTH OF SECTION bw = 12.00 in ALLOWABLE STRESS FACTOR SF = 1.333MASONRY ELASTICITY MODULUS Em = 1350 ksi THE NEUTRAL AXIS DEPTH FACTOR ISSTEEL ELASTICITY MODULUS Es = 29000 ksiMODULAR RATIO n = 21.48THE ALLOWABLE STRESS DUE TO FLEXURE IS THE ALLOWABLE REINF. STRESS DUE TO FLEXURE IS

psi 32000 psi

THE DISTANCE FROM BOTTOM TO M1 IS THE GOVERNING MOMENTS AND AXIAL FORCES ARE

= 13.2 ft = 2186 ft-lbs/ft

= lbs / ft

Design for Girder at Masonry Wall Based on ACI 530-02

DanielTian Li

50.52

660

9116

= 0.42309

( )( )'0.33b mSF fF = =

( )22k n n nρ ρ ρ= + −


( )2

1 2 21 2

1

122 pwPe h hM

w h = + −

( )2

12p

p

h h PeS h h

h hw

+ = + − −

( )1 P wall weightP = +

(cont'd)

THE GOVERNING SHEAR FORCES ARE= 90 ft-lbs/ft

= 331 lbs / ft

= lbs / ft

= 44 lbs / ft THE GOVERNING SHEAR STRESS IN MASONRY IS

= 90 lbs / ft = 3.61 psi

DETERMINE THE REGION FOR FLEXURE AND AXIAL LOAD (MDG Tab 12.2.1, Fig 12.2-12 & 13, page 12-25).

1. Wall is in compression and not cracked. 2. Wall is cracked but steel is in compression. 3. Wall is cracked and steel is in tension.



CHECK REGION 1 CAPACITY5438 ft-lbs / ft > M1 [Not applicable]

5454 ft-lbs / ft > M2 [Satisfactory]CHECK REGION 2 CAPACITY


2286 ft-lbs / ft > M2 [Not applicable]CHECK REGION 3 CAPACITY (The moment maybe limited by either the masonry compression or steel tension, MDG page 12-25).

45



THE ALLOWABLE SHEAR STRESS IS GIVEN BY



8962

( )1 2 3, ,v

e w

MAX V V Vf

t b=

6eM t

Pd d≤

12 3

eM tPd d

≤ −

12 3

eM tPd d

> −

2

6 6w e e

m bb t tPM F

= − =

222 3e

mbw

t PPMb F

= − =

1,

2 3 2 3 2 3e e

m b s swkd kd kdt tMIN kd d P d d PbM F A F

= − − − − + −

( ) ( )' , 50v mSF MIN fF =

22

22

pw hM =

( )2 P wall weightP = +( ) ( )2

11 1 2

pp

h h w PehV h w

h h

+= + − +

2 1 1hV w V= −

3 2pV h w=

=



INPUT DATA & DESIGN SUMMARYSPECIAL INSPECTION ( 0=NO, 1=YES ) 0 No, (reduced fm' by 0.5)TYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiSERVICE SHEAR LOAD V = 4.56 kSERVICE MOMENT LOAD M = 13.68 ft-kWIDTH b = 8 inEFFECTIVE DEPTH d = 45 inCLEAR SPAN Lc = 12 ftLOAD TYPE (1=SEISMIC, 0=WIND, 5=GRAVITY) 1 Seismic [THE BEAM DESIGN IS ADEQUATE.]VERTICAL REINF. 1 # 4 @ 8 in o.c.TENSION REINFORCEMENT 2 # 6

ANALYSISALLOWABLE STRESS FACTOR SF =

ALLOWABLE REINF. STRESS (1.33wind & seismic only)Fs = 32 ksiALLOWABLE MASONRY STRESS Fb=(SF)(0.33fm') = 0.33 ksiMASONRY ELASTICITY MODULUS Em = 1350 ksi, (Sec. 1.8.2.2.1)STEEL ELASTICITY MODULUS Es = 29000 ksiEFFECTIVE WIDTH bw = 7.63 in [Satisfactory, Lc < 32 bw]MODULAR RATIO n = 21.48TENSION REINFORCEMENT RATIO ρ = 0.003

THE NEUTRAL AXIS DEPTH FACTOR IS THE LEVER-ARM FACTOR IS

= 0.281 = 0.906

THE TENSILE STRESS IN REINFORCEMENT DUE TO FLEXURE IS

= 4.574 ksi < Fs [SATISFACTORY]

THE COMPRESSIVE STRESS IN THE EXTREME FIBER DUE TO FLEXURE IS

= 0.08 ksi < Fb [SATISFACTORY]

THE SHEAR STRESS IN MASONRY IS

< 15.1 25.82 psi

= 13 psi [SATISFACTORY]

(Sec. 2.3.5.2.1) < 77.46 psi [SATISFACTORY]


0.04 in2 / ft < 0.30 in2 / ft ( No shear reinf. Reqd )

0.667

Tian LiDaniel

Masonry Beam Design Based on ACI 530-02

( )22k n n nρ ρ ρ= + − 13k

j = −

ss

Mf

jdA=

2

2b

w

Mf

jkb d=

vw

Vf

db=

( ) '( , 50)v mSF MIN fF = =

( ) '(3 , 150)v mSF MIN fF = =

vAs

=s

VdF

=



INPUT DATA & DESIGN SUMMARYSPECIAL INSPECTION ( 0=NO, 1=YES ) 1 Yes, (Sec. 2107.1.2)TYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiSERVICE SHEAR LOAD V = 15.1 kSERVICE MOMENT LOAD M = 83 ft-kWIDTH b = 12 inEFFECTIVE DEPTH d = 40 inCLEAR SPAN Lc = 12 ftLOAD TYPE (1=SEISMIC, 0=WIND, 5=GRAVITY) 5 Gravity Only [THE BEAM DESIGN IS ADEQUATE.]VERTICAL REINF. 1 # 4 @ 8 in o.c.TENSION REINFORCEMENT 2 # 7

ANALYSISALLOWABLE STRESS FACTOR SF =

ALLOWABLE REINF. STRESS (1.33wind & seismic only)Fs = 24 ksiALLOWABLE MASONRY STRESS Fb=(SF)(0.33fm') = 0.495 ksiMASONRY ELASTICITY MODULUS Em = 750 fm' = 1125 ksi, (Eq. 6-4. 2106.2.12.1)STEEL ELASTICITY MODULUS Es = 29000 ksiEFFECTIVE WIDTH bw = 11.63 in [Satisfactory, Lc < 32 bw]MODULAR RATIO n = 25.78TENSION REINFORCEMENT RATIO ρ =

THE NEUTRAL AXIS DEPTH FACTOR IS THE LEVER-ARM FACTOR IS

= 0.304 = 0.899

THE TENSILE STRESS IN REINFORCEMENT DUE TO FLEXURE IS

= 23.09 ksi < Fs [SATISFACTORY]

THE COMPRESSIVE STRESS IN THE EXTREME FIBER DUE TO FLEXURE IS

= 0.39 ksi < Fb [SATISFACTORY]

THE SHEAR STRESS IN MASONRY IS

< 38.73 psi

= 36 psi [SATISFACTORY]

(Sec. 2107.2.17) < 116.19 psi [SATISFACTORY]



1.000

0.0026

Masonry Beam Design Based on UBC 97T. Li

Daniel

( )22k n n nρ ρ ρ= + − 13k

j = −

ss

Mf

jdA=

2

2b

w

Mf

jkb d=

vw

Vf

djb=

( ) '( , 50)v mSF MIN fF = =

( ) '(3 , 150)v mSF MIN fF = =

vAs

=s

VdF

=



INPUT DATA & DESIGN SUMMARYSPECIAL INSPECTION ( 0=NO, 1=YES ) 0 No, (reduced fm' by 0.5)TYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiLOAD TYPE ( 1=SEISMIC, 0=WIND, 5=GRAVITY) 1 SeismicSERVICE AXIAL LOAD P = 50 kSERVICE SHEAR LOAD V = 9.5 kMOMENT AT MIDHEIGHT M = 8.2 ft-kEFFECTIVE WIDTH b = 15.63 inEFFECTIVE DEPTH d = 15.63 in DISTANCE BETWEEN COL. REINF. a = 10.37 inEFFECTIVE HEIGHT h = 15 ftVERTICAL REINF. (EACH SIDE) 2 # 6HORIZ. TIES 2 legs # 4 @ 8 in o.c. [THE COLUMN DESIGN IS ADEQUATE.]

ANALYSISTOTAL REINFORCEMENT AREA As = 1.76 in2 MODULAR RATIO n = 21.48EFFECTIVE COLUMN AREA An = 244 in2 REINFORCEMENT RATIO ρ = 0.009NET EFFECTIVE MOMENT OF INERTIA In = 4973 in4 ALLOWABLE STRESS FACTOR SF = 0.667RADIUS OF GYRATION r = 4.51 in MAX. TIES SPACING Smax = 12 inMASONRY ELASTICITY MODULUS Em = 1350 ksi THE TRANSFORMED COLUMN AREA ISSTEEL ELASTICITY MODULUS Es = 29000 ksi 333 in2

CHECK VERTICAL REINFORCEMENT LIMITATION (ACI 530, 2.1.6.4)

As = 1.76 in2> 0.0025An = 0.61 in2

[Satisfactory]

< 0.04An = 9.77 in2[Satisfactory]

THE ALLOWABLE STRESS DUE TO AXIAL LOAD ONLY IS THE AXIAL STRESS AT MIDHEIGHT OF THE COLUMN IS

0.230 ksi 0.155 ksi

[for h/r < 99] < Fa, [Satisfactory]THE ALLOWABLE STRESS DUE TO FLEXURE IS THE ALLOWABLE REINF. STRESS DUE TO FLEXURE IS

ksi 32.0 ksi

THE TOTAL MOMENT ACTING AT MIDHEIGHT IS THE TRANSFORMED MOMENT OF INERTIA IS

11.5 ft-kips 6959 in4

THE STRESS IN THE EXTREME FIBER DUE TO MT IS THE MAX.STRESS COMBINED AXIAL & FLEXURE IS

0.154 ksi 0.310 ksi< fa, [Satisfactory, the section is uncracked] < Fb, [Satisfactory]

THE MAX. REINF. STRESS COMBINED AXIAL & FLEXURE IS THE AXIAL LOAD AT BASE OF THE COLUMN IS

11.1 ksi 53.435 k< Fs, [Satisfactory]

THE ALLOWABLE AXIAL LOAD FOR AXIAL COMPRESSION ONLY IS

89.75 k > Pt, [Satisfactory][for h/r < 99]

THE SHEAR DESIGN MAY BE DETERMIND FROM THE FOLLOWING EXPRESSION

> 25.82 psi

= 39 psi (Shear reinf. reqd to carry full shear load.)

(Sec. 2.3.5.2.1) < 106 77.46 psi [Satisfactory]

0.23 in2 / ft < 0.60 in2 / ft [Satisfactory]


0.330

Masonry Column Design Based on ACI 530-02

DanielTian Li

( )( )2

'0.25 1.0140a m

hSF fF

r

= − =

( )( )1 2 1t n nA A ρ= + − =

v

Vf

bd=

( ) '(3 , 150)v mSF MIN fF = =

vAs

=s

VdF

=

( ).a

t

P half col weightf

A

+= =

( )( )'0.33b mSF fF = = ( )( )& 241.33S wind seismic onlyF = =

( )0.12

TPd

MM = + =

2T

bt

dMfI

= =

( )2

2 12t n sa

nI I A = + − =

m a bf f f= + =

2 bS a

afnf f

d = + =

( ) '( , 50)v mSF MIN fF = =

( ).t P full col weightP = + =

( )( )2

'0.25 0.65 1.0140a n S Sm

hSF fP A F A

r

= + − =



INPUT DATA & DESIGN SUMMARYSPECIAL INSPECTION ( 0=NO, 1=YES ) 1 Yes, (Sec. 2107.1.2)TYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiLOAD TYPE 4 Zone 4 ( 1=Zone 1, 2=Zone 2, 3=Zone 3, 4=Zone 4, 0=WIND, 5=GRAVITY)

SERVICE AXIAL LOAD P = 11.5 k, @ top of col.MAX SHEAR LOAD V = 20 kMOMENT AT MIDHEIGHT M = 106 ft-k, @ mid of colEFFECTIVE WIDTH b = 23.63 inEFFECTIVE DEPTH c = 23.63 in DISTANCE BETWEEN COL. REINF. a = 17.37 inEFFECTIVE HEIGHT h = 29 ftVERTICAL REINF. (EACH SIDE) 3 # 8HORIZ. TIES 2 legs # 4 @ 8 in o.c. [THE COLUMN DESIGN IS ADEQUATE.]

ANALYSISREINFORCEMENT AREA AT ONE SIDE As = 2.37 in2 REINFORCEMENT RATIO ρ = ρ' = 0.005EFFECTIVE COLUMN AREA An = 558 in2 DISTANCE d' = 3.13 , d = 20.50 inNET EFFECTIVE MOMENT OF INERTIA In = 25982 in4 ALLOWABLE STRESS FACTOR SF = 1.333RADIUS OF GYRATION r = 6.82 in MAX. TIES SPACING Smax = 8 inMASONRY ELASTICITY MODULUS Em = 1125 ksi NEUTRAL AXIS DEPTH FACTORSTEEL ELASTICITY MODULUS Es = 29000 ksi k = [nρ+(2n-1)ρ' ]2+2[nρ+(2n-1)ρ(d'/d)]2 - [nρ+(2n-1)ρ' ]MODULAR RATIO n = 25.78 = 0.310

TRANSFORMED AREA At = An(1-ρ+nρ+nρ' ) = 696 in2LEVER-ARM FACTOR j = 1-k/3 = 0.897

CHECK VERTICAL REINFORCEMENT LIMITATION (UBC 2107.2.13.1)

As,total = 4.74 in2> 0.005Ae = 0.005bd = 2.42 in2

[Satisfactory]

< 0.04Ae = 0.04bd = 19.38 in2[Satisfactory]

THE AXIAL LOAD AT MIDDLE OF THE COLUMN IS THE TOTAL MOMENT ACTING AT MIDHEIGHT IS

19.090 k ft-kips

CHECK IF THERE IS TENSILE STRESS IN CROSS SECTION THE AXIAL LOAD AT BASE OF THE COLUMN ISPMid / A = 34 psi < MMid / (bc2/6) = 585 psi

(tensile exist) 26.681 kTHE ALLOWABLE AXIAL LOAD FOR AXIAL COMPRESSION ONLY IS

245.71 k > Pt, [Satisfactory][for h/r < 99]

THE ALLOWABLE REINF. STRESS DUE TO FLEXURE IS THE ALLOWABLE STRESS DUE TO FLEXURE IS

32.0 ksi 0.622 ksi

THE CORRESPONDING STRAIN IN THE TENSILE BARS IS THE STRAIN IN THE EXTREME COMPRESSION FIBER IS

0.0011 =(steel governs)

THE STRESS IN THE EXTREME FIBER DUE TO MT IS THE MOMENT DUE TO THE MASONRY IS

0.558 ksi < Fb', [Satisfactory] ft-kips

THE STRAIN IN THE COMPRESSION BARS IS THE STRESS IN THE COMPRESSION BARS IS

14.602 ksi

107.1

0.0004960

0.0002518

64.19

DanielT. Li

Masonry Column Design Based on UBC 97

( )'' 0.33Midb m

a

PSF fFP

= − =


( )0.12

MidPd

MM = + =

mmbf e E= =

( ).t P full col weightP = + =

( )2

'0.25 0.65 1.0140a n S stm

hfP A F A

r

= + − =

'

, bm s

m

kd FMINe ed kd E

= −

SS

S

Fe

E= =

' '2 S SSf eE= ='

'S m

kd de e

kd− = =

1( )

2 3m b

kdbkd dfM = − =

( ).Mid P half col weightP = + =

(cont'd)THE MOMENT DUE TO THE COMPRESSION BARS IS THE ALLOWABLE BENDING MOMENT IS

ft-kips ft-kips> Mt, [Satisfactory]

THE SHEAR DESIGN MAY BE DETERMIND FROM THE FOLLOWING EXPRESSION

< 51.64 psi

= 40 psi [Satisfactory]

(Sec. 2107.2.17) < 154.92 psi [Satisfactory]


Techincal References: 1. "Masonry Designers' Guide, Third Edition" (MDG-3), The Masonry Society, 2001. 2. "Seismic Design of Masonry Using the 1997 UBC", Concrete Masonry Association of NV & CA, 2003.

50.09 114.28

v

Vf

jbd=

( ) '(3 , 150)v mSF MIN fF = =

vAs

=s

VdF

=

( ) '( , 50)v mSF MIN fF = =

' '' ( )S SS df dM A= − = 'S mM M M+ ==


JOB NO. : DATE : REVIEW BY : Design for Bending Post at Top of Wall, Based on ACI 530-02 & UBC 97

INPUT DATA & DESIGN SUMMARYMASONRY STRENGTH fm' = 1.5 ksi

REBER YIELD STRESS fy = 60 ksi

SERVICE BENDING LOAD M = 0.6 ft-kipsSERVICE SHEAR LOAD V = 0.116 kipsWALL THICKNESS T = 8 inANCHORAGE REBARS 2 # 4 @ middle of wall 4

BASE PLATE YIELD STRESS Fy = 36 ksi

BASE PLATE WIDTH a = 6 inBASE PLATE LENGTH b = 10 in (REQUIRED BASE PLATE THK. t = 3/8 in & MIN. REBAR SPLICE LENGTH Ls = 24 in)

[THE ANCHORAGE DESIGN IS ADEQUATE.]

ANALYSISDETERMINE BASE PLATE THICKNESS (AISC-ASD F2.1, page 5-48)

3/8 in

where (4/3) is seismic/wind factor, typical.

CHECK MASONRY BEARING CAPACITY (ACI 530 2.1.9.3 & UBC 2107.2.10)

7.20 kips

480 psi < 500 psi [SATISFACTORY]

where 0.25 is for ACI 530, while 0.26 shall be used for UBC.

CHECK REBAR CAPACITY (ACI 530 2.3.2.1 & UBC 2107.2.11)

T = 7.20 kips < (4/3) Fs As = (4/3) (24 ksi) As = 12.80 kips [SATISFACTORY]

CHECK SHEAR CAPACITY (ACI 530 2.1.4.2.3 & UBC 2107.1.5.3)

1.49 kips > V [SATISFACTORY]

where n = 2 (rebars numbers)

Lbe = 3.56 in (rebar edge distance)

F = [MIN(Lbe , 12φ) -1"] / (12φ - 1") = 0.513 (edge reduction factor. Here 1" is for ACI 530, 1 1/2" shall be used for UBC.)

DETERMINE LAP SPLICE LENGTH (ACI 530 2.1.10.6.1.1, UBC 2107.2.2.3)

L d = 0.002 d fs = 48 d = 24.00 in

where for UBC, 50% increase may be used if rebar stress > 80%. (UBC 2107.2.12)

DanielTian Li

( )6

40.75

3 y

M

b Ft = =

( )20.5a

Tf

b a= =

( )2

20.5

3

MT

a= =

'40.25

3 mf =

( )'4350 , 0.12b ballow m yn MAX F f fV A A= =


JOB NO. : DATE : REVIEW BY : Group Fasteners in Combined Stresses Based on ACI 530-02 & CBC 2001

INPUT DATA & DESIGN SUMMARYMASONRY STRENGTH

fm' = 1.5 ksi

FASTENER YIELD STRESS

fy = 60 ksi

TENSION STRESS, ASD

ba = 0.42 kips / ft 5

SHEAR STRESS, VERTICAL

bv,V = 1 kips / ft

SHEAR STRESS, HORIZONTAL

bv,H = 0.85 kips / ft

WALL THICKNESSb = 8 in

FASTENER DIAMETER φ = 1/2 in

EFFECTIVE EMBEDMENT Lb = 5 in [THE ANCHORAGE DESIGN IS ADEQUATE.]

EDGE DISTANCE TO WALL TOP Lbe = 16 in

FASTENER SPACING S = 12 in, o.c.LOAD TYPE (1=SEISMIC, 0=WIND, 5=GRAVITY) 1 Seismic

ANALYSISCHECK MIN. EMBEDMENT (ACI 530 2.1.4.2.1)

Lb,min = MIN[ 4φ , 2] = 2.00 in < Lb [SATISFACTORY]

CHECK TENSION CAPACITY FOR A FASTENER (ACI 530 2.1.4.2.2 & CBC 2107A.1.5.2)

Ba = MIN[ 0.5Ap(fm')0.5 , 0.2Abfy] = 1.52 kips / fastener

> k S ba [SATISFACTORY]

Where L = MIN[ Lb , Lbe] = 5.00 in

θ = COS-1(0.5S / L) = 0.00 rad

Aseg = L2 [ θ - 0.5 SIN(2θ)] = 0.00 in2

Ap = π L2 - 2Aseg = 78.54 in2

Ab = π φ2 / 4 = 0.20 in2

k = 1.4 (3/4) = 1.05 Seismic , (IBC 2003 1620.2 & CBC 1633.2.8.1)

CHECK SHEAR CAPACITY (ACI 530 2.1.4.2.3 & CBC 2107A.1.5.3)

Bv = MIN[ (F)350(Abfm')1/4 , 0.12Abfy] = 1.41 kips / fastener

> S bv,V , Gravity only [SATISFACTORY]

> k S bv , Combined shear [SATISFACTORY]

Where F = [MIN(Lbe , 12φ) -1] / (12φ - 1) = 1.000

bv = (bv,V + bv,H)0.5 = 1.312 kips / ft

CHECK COMBINED SHEAR AND TENSION CAPACITY (ACI 530 2.1.4.2.4 & CBC 2107A.1.5.4)

S ba / Ba + S bv / Bv = 1.20 < 4/3 [SATISFACTORY]

DanielTian Li


JOB NO. : DATE : REVIEW BY : Double Fasteners in Tension & Shear Based on ACI 530-02 & CBC 2001


FASTENER YIELD STRESS fy = 60 ksi

SERVICE TENSION LOAD ba = 0.91 kips / 2 fasteners

SERVICE SHEAR LOAD bv = 0.728 kips / 2 fasteners

WALL THICKNESS b = 8 inFASTENER DIAMETER φ = 3/4 in

EFFECTIVE EMBEDMENT Lb = 7 in 0.42

FASTENER SPACING S = 6 inLOAD TYPE 1 Seismic (1=SEISMIC, 0=WIND, 5=GRAVITY)




CHECK TENSION CAPACITY (ACI 530 2.1.4.2.2 & CBC 2107A.1.5.2)

Ba = 2 MIN[ 0.5Ap(fm')0.5 , 0.2Abfy] = 1.40 kips / 2 fasteners

> k ba [SATISFACTORY]

Where Lbe = 3.44 in

L = MIN[ Lb , Lbe] = 3.44 in

θ = COS-1(0.5S / L) = 0.51 rad

Aseg = L2 [ θ - 0.5 SIN(2θ)] = 1.00 in2

Ap = π L2 - Aseg = 36.18 in2

Ab = π φ2 / 4 = 0.44 in2

k = 3/4 Seismic


Bv = 2 MIN[ (F)350(Abfm')1/4 , 0.12Abfy] = 1.08 kips / 2 fasteners

> k bv [SATISFACTORY]

Where F = [MIN(Lbe , 12φ) -1] / (12φ - 1) = 0.305


ba / Ba + bv / Bv = 1.32 < 4/3 [SATISFACTORY]

DanielTian Li


JOB NO. : DATE : REVIEW BY : Group Fasteners in Tension & Shear Based on ACI 530-02 & CBC 2001


FASTENER YIELD STRESS fy = 60 ksi

UPLIFT STRESS, ASD ba = 0.9 kips / ft

SHEAR STRESS IN WALL DIR bv = 0.045 kips / ft

WALL THICKNESS b = 8 inFASTENER DIAMETER φ = 1/2 in

EFFECTIVE EMBEDMENT Lb = 5 in

FASTENER SPACING S = 12 in, o.c.LOAD TYPE 1 Seismic (1=SEISMIC, 0=WIND, 5=GRAVITY)




CHECK TENSION CAPACITY FOR A FASTENER (ACI 530 2.1.4.2.2 & CBC 2107A.1.5.2)

Ba = MIN[ 0.5Ap(fm')0.5 , 0.2Abfy] = 0.77 kips / fastener

> k S ba [SATISFACTORY]

Where Lbe = 3.57 in

L = MIN[ Lb , Lbe] = 3.57 in

θ = COS-1(0.5S / L) = 0.00 rad

Aseg = L2 [ θ - 0.5 SIN(2θ)] = 0.00 in2

Ap = π L2 - 2Aseg = 39.93 in2

Ab = π φ2 / 4 = 0.20 in2

k = 3/4 Seismic


Bv = MIN[ (F)350(Abfm')1/4 , 0.12Abfy] = 0.74 kips / fastener

> k S bv [SATISFACTORY]

Where F = [MIN(Lbe , 12φ) -1] / (12φ - 1) = 0.513


S ba / Ba + S bv / Bv = 1.22 < 4/3 [SATISFACTORY]

DanielTian Li



INPUT DATA & DESIGN SUMMARYSPECIAL INSPECTION ( 0=NO, 1=YES ) 0 NoTYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUMASONRY STRENGTH fm' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiSERVICE LATERAL LOAD w = 45 psfTHICKNESS OF WALL t = 8 inWALL HEIGHT h = 24 ftPILASTER SPACING L = 10 ftPILASTER SIZE c1 = 24 in

c2 = 24 in [THE WALL DESIGN IS ADEQUATE.]WALL HORIZ. REINF. (Asv) 1 # 4 @ 48 in o.c. (at middle)

ANALYSISDESIGN CRITERIA 1. Pilaster spacing less than one half the unsupported vertical span of out-of-plane wall. (MDG-3,page 11-8)

L = 10 ft < 0.5 h = 12 ft [Satisfactory] 2. Pilaster stiffness greater than that the tributary area of wall. (CBC 2106A.1.7)

beff = c1 + 12 t = 120 in, (ACI 530, 1.7.5.1 & 1.9.4.2.3)y = 7.4 inE Ipilaster = 52809 E ,(ACI 530, 1.9.2)E Iwall = 4096 E ,(ACI 530, 1.9.2) 10

E Ipilaster > E Iwall [Satisfactory]

CHECK WALL HORIZONTAL BENDING CAPACITYd = 3.82 in, effective depth Em = 1350 ksi

bw = 12 in Es = 29000 ksite = 7.63 in, effective thickness n = 21.48 , modular ratio

As = 0.05 in2 / ft ρ = 0.0011 , reinforcement ratioSF = 0.667 , allowable stress factor

0.1944

330 psi psi

437 ft-lbs/ft, (MDG-3, page 11-3)

360 ft-lbs/ft < M allowable [Satisfactory]


26 psi > f v = 0.5 (L - c 1 ) w / (b w d) = 4 psi[Satisfactory]

Techincal References: 1. "Masonry Designers' Guide, Third Edition" (MDG-3), The Masonry Society, 2001.

Masonry Wall Design at Horizontal Bending, Based on ACI 530-02

32000

Tian LiDaniel

( )( )'0.33b mSF fF = =

( )22k n nnρ ρρ= + − =

( )1.33 24000SF = =

1,

2 3 3allowable b s swkd kd

MIN kd d dbM F A F = − − =

( ) ( )' , 50v mSF MIN fF = =

( )21

max 8w L c

M−

= =



TENSION DEVELOPMENT

Ld = 0.002 Fs db = 48 db = 68 in

CBC 2107.2.2.3where Bar size # 11

db = 1.41 in

fy = 60 ksi

Fs = 24 ksi, CBC 2107.2.11

COMPRESSION DEVELOPMENT

Ld = 0.0015Fs db = 36 db = 51 in


db = 1.41 in

fy = 60 ksi

Fs = 24 ksi, CBC 2107.2.11

ANCHORAGE OF FLEXURAL REINFORCEMENT

Lanchor = MAX ( h , 12 db ) = 34 db = 48 in, CBC 2106.3.4


db = 1.41 in

h = 48 in

TENSION SPLICE WITH MORE 80% STRESS

Ls = 1.5 MAX( Ld , 40 db ) = 102 in

CBC 2107.2.2.6 & 2107.2.12

( DSA : Ls = 1.5 MAX( Ld , 48 db ) = 102 in )

CBC 2107A.2.2.6 & 2107A.2.12where Bar size # 11

db = 1.41 in

fy = 60 ksi

Fs = 24 ksi, CBC 2107.2.11

Ld = 68 in, CBC 2107.2.2.3

TENSION SPLICE WITH 80% STRESS OR LESS 5

Ls = MAX( Ld , 40 db ) = 68 in

CBC 2107.2.2.6

( DSA : Ls = MAX( Ld , 48 db ) = 68 in )

CBC 2107A.2.2.6where Bar size # 11

db = 1.41 in

fy = 60 ksi

Fs = 24 ksi, CBC 2107.2.11

Ld = 68 in, CBC 2107.2.2.3

COMPRESSION SPLICE

Ls = MAX( Ld , 30 db ) = 51 in

CBC 2107.2.2.6

( DSA : Ls = MAX( Ld , 36 db ) = 51 in )

CBC 2107A.2.2.6where Bar size # 11

db = 1.41 in

fy = 60 ksi

Fs = 24 ksi, CBC 2107.2.11

Ld = 51 in, CBC 2107.2.2.3

Development & Splice of Reinforcement in Masonry Based on CBC 2001, ASD

DanielTian Li



TENSION OR COMPRESSION DEVELOPMENT

Ld = MAX (Lde / φ , 12) = 65 db = 92 in


db = 1.41 in

fy = 60 ksi

fm' = 2.5 ksi

K = 3 db = 4.23 in

Lde = MIN[0.15 db2 fy / (K fm' 0.5 ) , 52 db] = 73 in

φ = 0.8 , CBC 2108.1.4.6.1

( DSA : Ld = MAX (Lde / φ , 12) = 75 db = 106 in

Lde = 0.15 db2 fy / (K fm' 0.5 ) = 85 in, CBC 8A-13 )


Lanchor = MAX ( h , 12 db ) = 34 db = 48 in, CBC 2106.3.4


db = 1.41 in

h = 48 in

TENSION OR COMPRESSION SPLICE

Ls = MAX (Lde / φ , 12) = 65 db = 92 in


db = 1.41 in

fy = 60 ksi

fm' = 2.5 ksi 48

K = 3 db = 4.23 in

Lde = MIN[0.15 db2 fy / (K fm' 0.5 ) , 52 db] = 73 in

φ = 0.8 , CBC 2108.1.4.6.2

( DSA : Ls = MAX (Lde / φ , 12) = 75 db = 106 in

Lde = 0.15 db2 fy / (K fm' 0.5 ) = 85 in, CBC 8A-13 )

Development & Splice of Reinforcement in Masonry Based on CBC 2001, SD

DanielTian Li



TENSION DEVELOPMENT

Ld = MAX( 0.0015Fs db , 12 ) = 36 db = 23 in

ACI 530-02 2.1.10.2where Bar size # 5

db = 0.625 in

fy = 60 ksi

Fs = 24 ksi, ACI 530-02 2.3.2.1

COMPRESSION DEVELOPMENT

Ld = 0.0015Fs db = 36 db = 23 in

ACI 530-02 2.1.10.2where Bar size 5

db = 0.625 in

fy = 60 ksi

Fs = 24 ksi, ACI 530-02 2.3.2.1


Lanchor = MAX ( d , 12 db ) = 64 db = 40 in, ACI 530-02 2.1.10.3.1.3

where Bar size # 5

db = 0.625 in

d = 40 in

TENSION OR COMPRESSION SPLICE

Ld = 0.002Fs db = 48 db = 30 in

ACI 530-02 2.1.10.6.1.1

where Bar size # 5

db = 0.625 in

fy = 60 ksi

Fs = 24 ksi, ACI 530-02 2.3.2.1

Development & Splice of Reinforcement in Masonry Based on ACI 530-02

DanielTian Li




MASONRY STRENGTH fm' = 1.5 ksi


THICKNESS OF WALL t = 8 inHORIZ. BENDING SPAN L = 10.17 ft

SEISMIC COFFICIENT Ca = 0.44 (CBC Tab 16A-Q)

IMPORTACE FACTOR IP = 1.15 (CBC Tab 16A-K)

ELEVATOR CAR WEIGHT Wcar = 2.5 kips

ELEVATOR RATED LOAD Wrat = 3 kips

WALL HORIZ. REINF. (As) 1 # 6 @ 8 in o.c. (at middle)

[THE WALL DESIGN IS NOT ADEQUATE, SEE BELOW.]

ANALYSIS

DESIGN LOADS

0.482 Wp = 37 psf , ASD

Where : ap = 1.0 Rp = 3.0 Wp = 76.7 psf hx = hr

(CBC Tab.16A-O) (CBC Tab.16A-O) ( attachement height, CBC 1633A.2,

horizontal bending, hx dynamic)

1.85 kip / ft , ASD

Where : Factor 2/3 , 0.5 , & 40% from CBC 1633A.2.13.1b = 8.0 in, CBC 2106A.3.8

5.18 ft-kip / ft , ASD (simple beam since wall cracked)

CHECK HORIZONTAL BENDING CAPACITY

2.21 ft-kip / ft , ASD

< M max [Unsatisfactory]

Where : bw = 12 in, te = 8 in, d = 4 in

As = 0.66 in2 / ft ρ = 0.0144 n = 25.8

k = 0.567 Fb = 660 psi Fs = 32000 psi

Elevator Masonry Wall Design Based on CBC 2001 Chapter ATian LiDaniel

1,

2 3 3allowable b s swkd kd

MIN kd d dbM F A F = − − =

11.4

pp a p xp

p r

a C WI hF

R h

= + =

( )2 12"0.5 40%

3 car ratP W Wb

= + =

2

max 8 4p PLF L

M = + =


JOB NO. : DATE : REVIEW BY : Wood Joist Design Based on NDS 2001, ICC PFC-4354 & PFC-5803

INPUT DATA & DESIGN SUMMARY AVAILABLE MINIMUM Douglas Fir-Larch SIZESJOIST SPAN L = 35 ftDEAD LOAD DL = 22 psf, (w/o self Wt) AVAILABLE MINIMUM TJI SIZESLIVE LOAD / SNOW LL = 15.5 psf 30" TJI/L65 28" TJI/L90 26" TJI/H90JOIST SPACING S = 24 in o.c. AVAILABLE MINIMUM SSI SIZESDURATION FACTOR C D = 1.25 (NDS Tab. 2.3.2) 28" SSI 42MX 26" SSI 43LXREPETITIVE FACTOR C r = 1.15 (NDS 4.3.9. For DSA, 1.0)DEFLECTION LIMIT OF LIVE LOAD ∆ LL = L / 360 ( L / 360 , 1.2 in ) DEFLECTION LIMIT OF LONG-TERM LOAD ∆ 1.5(DL+0.33LL) = L / 480 ( L / 480 , 0.9 in ) DEFLECTION LIMIT OF TOTAL LOAD ∆ (DL+LL) = L / 240 ( L / 240 , 1.8 in )

ANALYSISJOIST PROPERTIES & ALLOWABLE MOMENT & SHEAR

2x No. 2, Douglas Fir-Larch ( ASD Supplements, Tab. 5.4a) 2x No. 1, Douglas Fir-Larch (from WoodBeam.xls)Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106 Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106

(CF included) (in2-lbs) (CF included) (in2-lbs)

4 1.00 4130 630 9 4 1.00 4594 630 96 2.00 8850 990 33 6 2.00 9831 990 358 2.00 14200 1310 76 8 2.00 15769 1305 81

10 3.00 21200 1670 158 10 3.00 23530 1665 16812 4.00 28500 2030 285 12 4.00 31641 2025 303

2x Structural, Douglas Fir-Larch (ASD Supplements, Tab. 5.4a)Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106

(CF included) (in2-lbs) Where:4 1.00 6890 630 10 1. ASD Supplements, Tab. 5.4a is from American Wood Council, 2001.6 2.00 14700 990 40 2. Assume that the joist top is fully lateral supported by diaphragm. (CL = 1.0)8 2.00 23700 1310 91 3. WoodBeam.xls is at www.engineering-international.com

10 3.00 35300 1670 18812 4.00 47500 2030 338

TJI/L65 ( from Trusjoist # 1062, page 5) SSI 32MX ( from ICC PFC-5803, page 5 & 6)Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106 Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106 C x 106

(in2-lbs) (in2-lbs) (in2-lbs)11 7/8 3.30 6750 1925 450 11 7/8 3.10 5391 2115 460 9.39

14 3.60 8030 2125 666 14 3.30 6570 2330 667 10.9916 3.90 9210 2330 913 16 3.60 7684 2530 900 12.5018 4.20 10380 2535 1205 18 3.90 8800 2735 1170 14.0220 4.40 11540 2740 1545 20 4.10 9918 2935 1478 15.5522 4.70 12690 2935 1934 22 4.40 11038 3135 1824 17.0824 5.00 13830 3060 2374 24 4.70 12159 3335 2211 18.6226 5.30 14960 2900 2868 26 5.00 13279 3540 2638 20.1528 5.50 16085 2900 3417 28 5.20 14401 3740 3106 21.6830 5.80 17205 2900 4025 30 5.50 15524 3940 3616 23.21



14 4.50 11430 2125 913 14 4.10 9274 2350 924 11.1516 4.70 13115 2330 1246 16 4.30 10863 2620 1246 12.6818 5.00 14785 2535 1635 18 4.60 12456 2895 1617 14.2220 5.30 16435 2740 2085 20 4.90 14051 3165 2040 15.7722 5.60 18075 2935 2597 22 5.10 15649 3440 2514 17.3224 5.80 19700 3060 3172 24 5.40 17248 3710 3042 18.8726 6.10 21315 2900 3814 26 5.70 18849 3985 3622 20.4228 6.40 22915 2900 4525 28 6.00 20450 4255 4257 21.9730 6.60 24510 2900 5306 30 6.20 22052 4530 4948 23.53

TJI/H90 ( from Trusjoist # 1062, page 5) SSI 43L ( from ICC PFC-5803, page 5 & 6)Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106 Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106 C x 106


14 4.90 13090 2125 1015 14 4.90 12081 2260 1031 7.9116 5.20 15065 2330 1389 16 5.20 14251 2425 1394 8.9718 5.40 17010 2535 1827 18 5.40 16269 2590 1944 10.0520 5.70 18945 2740 2331 20 5.70 18419 2755 2454 11.1322 6.00 20855 2935 2904 22 5.90 20573 2920 3026 12.2124 6.30 22755 3060 3549 24 6.20 22730 3090 3661 13.3026 6.50 24645 2900 4266 26 6.40 24889 3255 4358 14.3928 6.80 26520 2900 5059 28 6.70 27050 3420 5119 15.4730 7.10 28380 2900 5930 30 7.00 29212 3585 5944 16.56

DanielT. Li

(cont'd)DESIGN EQUATIONS

( from Trusjoist # 1062, page 21)

( from ICC PFC-5803, page 2)

CHECK JOIST CAPACITIES & DEFLECTIONS

2x No. 2, Douglas Fir-Larch 2x No. 1, Douglas Fir-Larch

Deep (in) M (ft-lbs) V (lbs) ∆LL ∆LT ∆D+L CHECK Deep (in) M (ft-lbs) V (lbs) ∆LL ∆LT ∆D+L CHECK

(in) (in) (in) (in) (in) (in)4 8096 925 116.30 310.80 285.12 N.G. 4 8096 925 116.30 310.80 285.12 N.G.6 8202 937 31.72 86.30 78.78 N.G. 6 8202 937 29.91 81.37 74.28 N.G.8 8202 937 13.77 37.47 34.21 N.G. 8 8202 937 12.92 35.16 32.10 N.G.

10 8309 950 6.62 18.34 16.67 N.G. 10 8309 950 6.23 17.25 15.68 N.G.12 8415 962 3.67 10.35 9.36 N.G. 12 8415 962 3.45 9.73 8.80 N.G.

2x Structural, Douglas Fir-Larch

Deep (in) M (ft-lbs) V (lbs) ∆LL ∆LT ∆D+L CHECK

(in) (in) (in)4 8096 925 104.67 279.72 256.61 N.G.6 8202 937 26.17 71.20 65.00 N.G.8 8202 937 11.50 31.29 28.57 N.G.

10 8309 950 5.57 15.42 14.01 N.G.12 8415 962 3.10 8.73 7.89 N.G.

TJI/L65 SSI 32MXDeep (in) M (ft-lbs) V (lbs) ∆LL ∆LT ∆D+L CHECK Deep (in) M (ft-lbs) V (lbs) ∆LL ∆LT ∆D+L CHECK

(in) (in) (in) (in) (in) (in)11 7/8 8341 953 2.40 6.68 6.06 N.G. 11 7/8 8319 951 2.32 6.45 5.85 N.G.

14 8373 957 1.63 4.57 4.14 N.G. 14 8341 953 1.61 4.48 4.07 N.G.16 8405 961 1.20 3.38 3.05 N.G. 16 8373 957 1.20 3.36 3.04 N.G.18 8437 964 0.92 2.59 2.34 N.G. 18 8405 961 0.93 2.61 2.36 N.G.20 8458 967 0.72 2.04 1.85 N.G. 20 8426 963 0.74 2.08 1.88 N.G.22 8490 970 0.58 1.65 1.49 N.G. 22 8458 967 0.60 1.70 1.54 N.G.24 8522 974 0.48 1.37 1.23 N.G. 24 8490 970 0.50 1.42 1.28 N.G.26 8554 978 0.40 1.15 1.03 N.G. 26 8522 974 0.42 1.20 1.08 N.G.28 8575 980 0.34 0.97 0.88 N.G. 28 8543 976 0.36 1.03 0.93 N.G.30 8607 984 0.29 0.84 0.75 o.k. 30 8575 980 0.31 0.89 0.80 N.G.

TJI/L90 SSI 42MXDeep (in) M (ft-lbs) V (lbs) ∆LL ∆LT ∆D+L CHECK Deep (in) M (ft-lbs) V (lbs) ∆LL ∆LT ∆D+L CHECK


14 8468 968 1.21 3.43 3.10 N.G. 14 8426 963 1.17 3.31 2.99 N.G.16 8490 970 0.89 2.55 2.30 N.G. 16 8447 965 0.88 2.48 2.24 N.G.18 8522 974 0.69 1.97 1.78 N.G. 18 8479 969 0.68 1.93 1.74 N.G.20 8554 978 0.54 1.57 1.41 N.G. 20 8511 973 0.54 1.55 1.40 N.G.22 8586 981 0.44 1.28 1.15 N.G. 22 8532 975 0.44 1.27 1.14 N.G.24 8607 984 0.37 1.06 0.95 N.G. 24 8564 979 0.37 1.06 0.96 N.G.26 8639 987 0.31 0.90 0.80 N.G. 26 8596 982 0.31 0.90 0.81 N.G.28 8671 991 0.26 0.77 0.69 o.k. 28 8628 986 0.27 0.78 0.70 o.k.30 8692 993 0.23 0.66 0.59 o.k. 30 8650 989 0.23 0.68 0.60 o.k.

TJI/H90 SSI 43LDeep (in) M (ft-lbs) V (lbs) ∆LL ∆LT ∆D+L CHECK Deep (in) M (ft-lbs) V (lbs) ∆LL ∆LT ∆D+L CHECK


14 8511 973 1.09 3.13 2.82 N.G. 14 8511 973 1.07 3.07 2.77 N.G.16 8543 976 0.81 2.32 2.09 N.G. 16 8543 976 0.80 2.31 2.07 N.G.18 8564 979 0.62 1.79 1.61 N.G. 18 8564 979 0.58 1.68 1.51 N.G.20 8596 982 0.49 1.43 1.28 N.G. 20 8596 982 0.47 1.36 1.22 N.G.22 8628 986 0.40 1.16 1.04 N.G. 22 8618 985 0.38 1.11 1.00 N.G.24 8660 990 0.33 0.97 0.87 N.G. 24 8650 989 0.32 0.94 0.84 N.G.26 8682 992 0.28 0.82 0.73 o.k. 26 8671 991 0.27 0.80 0.71 o.k.28 8713 996 0.24 0.70 0.63 o.k. 28 8703 995 0.23 0.69 0.62 o.k.30 8745 999 0.21 0.61 0.54 o.k. 30 8735 998 0.20 0.60 0.54 o.k.

2

8 D r

wLMC C

=2 D r

wLV

C C=

4 2

5

22.5 2.26

10TJI

w wL LEI d

= +∆ ×

4 25384

SSIw wL L

EI C= +∆

45384

DFLwL

EI=∆


JOB NO. : DATE : REVIEW BY : Double Joist Design for Mechanical Equipment Based on NDS 2001, ICC PFC-4354 & PFC-5803

INPUT DATA & DESIGN SUMMARY AVAILABLE MINIMUM Douglas Fir-Larch SIZESJOIST SPAN L = 30 ftDEAD LOAD DL = 26 psf, (w/o self Wt) AVAILABLE MINIMUM TJI SIZESJOIST SPACING S = 24 in o.c. 30" TJI/L65 22" TJI/L90 20" TJI/H90DURATION FACTOR C D = 1.33 (NDS Tab. 2.3.2) AVAILABLE MINIMUM SSI SIZES

24" SSI 42MX 20" SSI 43LDEFLECTION LIMITATION ∆ (DL+E) = L / 240 ( L / 240 , 1.5 in )

EQUIPMENT WEIGHT W = 3 kipsHEIGHT OF MASS CENTER H = 3 ft, 2/3 total heightEQUIPMENT LENGTH D = 4 ftEQUIPMENT WIDTH B = 6 ft, double joist spacing

SEISMIC LOADS, (CBC 1632.2)FH = Fp = MAX 0.7CaIpW , MIN[ apCaIp(1+3hx/hr)/Rp W , 4CaIpW ]

= MAX 0.39W , MIN[ 0.74W , 2.23W ] = 0.74 W , (SD)= 0.53 W , (ASD) = 1.59 kips

FV = Fp / 3 = 0.53 kips, up & down, CBC Tab. 16A-O footnote 20

wE = (0.5 FH H / B + 0.25 FV) / L = 133 plf / joistat middle of span

where C a = 0.484

I p = 1.15

a p = 1 (CBC Tab 16A-O)

R p = 3 (CBC Tab 16A-O)

h x = hr ft

h r = 36 ft

GRAVITY LOADSwR = 0.5(B + S) DL = 104 plf / joist, full span

wD = 0.25 W / D = 188 plf / joist, at middle of span

ANALYSISDESIGN EQUATIONS

6

JOIST PROPERTIES & ALLOWABLE MOMENT & SHEAR2x No. 2, Douglas Fir-Larch ( ASD Supplements, Tab. 5.4a) 2x No. 1, Douglas Fir-Larch (from WoodBeam.xls)

Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106 Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106

(CF included) (in2-lbs) (CF included) (in2-lbs)

4 1.00 4130 630 9 4 1.00 4594 630 96 2.00 8850 990 33 6 2.00 9831 990 358 2.00 14200 1310 76 8 2.00 15769 1305 81

10 3.00 21200 1670 158 10 3.00 23530 1665 16812 4.00 28500 2030 285 12 4.00 31641 2025 303

2x Structural, Douglas Fir-Larch (ASD Supplements, Tab. 5.4a)Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106

(CF included) (in2-lbs) Where:4 1.00 6890 630 10 1. ASD Supplements, Tab. 5.4a is from American Wood Council, 2001.6 2.00 14700 990 40 2. Assume that the joist top is fully lateral supported by diaphragm. (CL = 1.0)8 2.00 23700 1310 91 3. WoodBeam.xls is at www.engineering-international.com

10 3.00 35300 1670 18812 4.00 47500 2030 338



14 3.60 8030 2125 666 14 3.30 6570 2330 667 10.9916 3.90 9210 2330 913 16 3.60 7684 2530 900 12.5018 4.20 10380 2535 1205 18 3.90 8800 2735 1170 14.0220 4.40 11540 2740 1545 20 4.10 9918 2935 1478 15.5522 4.70 12690 2935 1934 22 4.40 11038 3135 1824 17.0824 5.00 13830 3060 2374 24 4.70 12159 3335 2211 18.6226 5.30 14960 2900 2868 26 5.00 13279 3540 2638 20.1528 5.50 16085 2900 3417 28 5.20 14401 3740 3106 21.6830 5.80 17205 2900 4025 30 5.50 15524 3940 3616 23.21

DanielTian Li

( )( ) 345

384 48D ER

DL E

Dw w Lw LEI EI

++

= +∆

( )2 2

8 4 8R

D ED D D

LDw L DM w wC C C

= + + +

( )2

R D E

D

L Dw w wVC

+ +=

(cont'd)TJI/L90 ( from Trusjoist # 1062, page 5) SSI 42MX ( from ICC PFC-5803, page 5 & 6)

Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106 Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106 C x 106


14 4.50 11430 2125 913 14 4.10 9274 2350 924 11.1516 4.70 13115 2330 1246 16 4.30 10863 2620 1246 12.6818 5.00 14785 2535 1635 18 4.60 12456 2895 1617 14.2220 5.30 16435 2740 2085 20 4.90 14051 3165 2040 15.7722 5.60 18075 2935 2597 22 5.10 15649 3440 2514 17.3224 5.80 19700 3060 3172 24 5.40 17248 3710 3042 18.8726 6.10 21315 2900 3814 26 5.70 18849 3985 3622 20.4228 6.40 22915 2900 4525 28 6.00 20450 4255 4257 21.9730 6.60 24510 2900 5306 30 6.20 22052 4530 4948 23.53

TJI/H90 ( from Trusjoist # 1062, page 5) SSI 43L ( from ICC PFC-5803, page 5 & 6)Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106 Deep (in) Wt (lbs/ft) M (ft-lbs) V (lbs) EI x 106 C x 106


14 4.90 13090 2125 1015 14 4.90 12081 2260 1031 7.9116 5.20 15065 2330 1389 16 5.20 14251 2425 1394 8.9718 5.40 17010 2535 1827 18 5.40 16269 2590 1944 10.0520 5.70 18945 2740 2331 20 5.70 18419 2755 2454 11.1322 6.00 20855 2935 2904 22 5.90 20573 2920 3026 12.2124 6.30 22755 3060 3549 24 6.20 22730 3090 3661 13.3026 6.50 24645 2900 4266 26 6.40 24889 3255 4358 14.3928 6.80 26520 2900 5059 28 6.70 27050 3420 5119 15.4730 7.10 28380 2900 5930 30 7.00 29212 3585 5944 16.56

CHECK JOIST CAPACITIES & DEFLECTIONS

2x No. 2, Douglas Fir-Larch 2x No. 1, Douglas Fir-Larch

Deep (in) M (ft-lbs) V (lbs) ∆(DL+E) CHECK Deep (in) M (ft-lbs) V (lbs) ∆(DL+E) CHECK

(in) (in)4 16581 1665 348.85 N.G. 4 16581 1654 348.85 N.G.6 16666 1677 95.14 N.G. 6 16666 1654 89.70 N.G.8 16666 1677 41.31 N.G. 8 16666 1654 38.76 N.G.

10 16751 1688 19.87 N.G. 10 16751 1654 18.69 N.G.12 16835 1699 11.02 N.G. 12 16835 1654 10.36 N.G.

2x Structural, Douglas Fir-Larch

Deep (in) M (ft-lbs) V (lbs) ∆(DL+E) CHECK

(in)4 16581 1665 313.97 N.G.6 16666 1677 78.49 N.G.8 16666 1677 34.50 N.G.

10 16751 1688 16.70 N.G.12 16835 1699 9.29 N.G.

TJI/L65 SSI 32MX


(in) (in)11 7/8 16776 1691 6.98 N.G. 11 7/8 16759 1689 6.83 N.G.

14 16801 1695 4.71 N.G. 14 16776 1691 4.71 N.G.16 16827 1698 3.44 N.G. 16 16801 1695 3.49 N.G.18 16852 1702 2.61 N.G. 18 16827 1698 2.68 N.G.20 16869 1704 2.03 N.G. 20 16844 1700 2.12 N.G.22 16894 1707 1.62 N.G. 22 16869 1704 1.72 N.G.24 16920 1711 1.32 N.G. 24 16894 1707 1.42 N.G.26 16945 1714 1.09 N.G. 26 16920 1711 1.19 N.G.28 16962 1716 0.92 N.G. 28 16937 1713 1.01 N.G.30 16987 1720 0.78 o.k. 30 16962 1716 0.87 N.G.

TJI/L90 SSI 42MX


(in) (in)11 7/8 16852 1702 5.06 N.G. 11 7/8 16818 1697 4.93 N.G.

14 16877 1705 3.44 N.G. 14 16844 1700 3.40 N.G.16 16894 1707 2.52 N.G. 16 16861 1703 2.52 N.G.18 16920 1711 1.92 N.G. 18 16886 1706 1.94 N.G.20 16945 1714 1.51 N.G. 20 16911 1709 1.54 N.G.22 16970 1717 1.21 o.k. 22 16928 1712 1.25 N.G.24 16987 1720 0.99 o.k. 24 16954 1715 1.03 o.k.26 17013 1723 0.82 o.k. 26 16979 1718 0.87 o.k.28 17038 1726 0.69 o.k. 28 17004 1722 0.74 o.k.30 17055 1729 0.59 o.k. 30 17021 1724 0.63 o.k.

TJI/H90 SSI 43L


(in) (in)11 7/8 16886 1706 4.57 N.G. 11 7/8 16886 1706 4.44 N.G.

14 16911 1709 3.09 N.G. 14 16911 1709 3.05 N.G.16 16937 1713 2.26 N.G. 16 16937 1713 2.25 N.G.18 16954 1715 1.72 N.G. 18 16954 1715 1.62 N.G.20 16979 1718 1.35 o.k. 20 16979 1718 1.28 o.k.22 17004 1722 1.08 o.k. 22 16996 1721 1.04 o.k.24 17030 1725 0.88 o.k. 24 17021 1724 0.86 o.k.26 17047 1727 0.74 o.k. 26 17038 1726 0.72 o.k.28 17072 1731 0.62 o.k. 28 17064 1730 0.61 o.k.30 17097 1734 0.53 o.k. 30 17089 1733 0.53 o.k.



INPUT DATADL = 40 lbs / ft MEMBER SIZE = 6 x 8 No. 2, Douglas Fir-LarchLL = 40 lbs / ftSELF = 9 lbs / ft MEMBER SPAN = 8 ft Camber =>0 in (1.5 DL deflection)TL = 89 lbs / ftPLL = 2000 lbs (Non concurrent point live load, see IBC Tab 1607.1 or UBC Tab16-A) Code DesignationDoes member support plaster ? (See IBC Tab 1604.3 or UBC Tab 16-D) 1 Select Structural, Douglas Fir-Larch (1= yes, 0= no) 1 Yes 2 No. 1, Douglas Fir-Larch

3 No. 2, Douglas Fir-Larch

Code Duration Factor, CD Condition Choice => 31 0.90 Dead Load Does member have continuous lateral support ?2 1.00 Occupance Live Load (1= yes, 0= no) 0 No3 1.15 Snow Load

4 1.25 Construction Load CD CM Ct Ci CL CF

5 1.60 Wind/Earthquake Load 1.25 1.00 1.00 1.00 1.00 1.00

6 2.00 Impact Load Cf CV Cc Cr

1.00 1.00 1.00 1.00Choice => 4 Construction Load

DESIGN SUMMARYL = 8.0 ft 1. Check Bending: M/S < Fb w = 49 lbs/ft 1,022 psi < 1,089 psi ok

PLL = 2000 lbs <= Case 2 Governs ! R = 2,196 lbs 2. Check Shear: 1.5 V/A < Fv

79 psi < 213 psi ok

= 2,165 lbs2000

3. Check Live Load Deflection:= 4,392 ft-lbs

0.03 in < 0.27 in ok

= 0.03 in4. Check Long-term Deflection:

= 0.04 in 0.04 in < 0.40 in ok

ANALYSIS 6 x 8 Properties Maximum length permitted by:

b = 5.50 in KbE = 0.439

d = 7.50 in E = 14.9 ft

A = 41.3 in2RB = 6.661 < 50 Bending : = 20.5 ft (case 1)

S = 51.6 in3E'y = 1300 ksi

I = 193 in4FbE = 12861 psi

E = Ex = 1300 ksi Fb* = 1093.75 psi = 8.5 ft (case 2)

Fb = 875 psi F = FbE / Fb* = 11.76

Fv = 170 psi, (CH included to comply with NDS 97)

E' = 1,300 ksi Shear : = 132.7 ft (case 1)

Fb' = 1,089 psi

Fv' = 213 psi= 158.3 ft (case 2)

CASE 1: DL+LL Governs => CASE 2: DL+PLL

Wood Beam Design Base on NDS 2001

DanielT. Li

2

8 4llw LL PM = +

8 btl

tl

Sw FLw

=

22

1.5V

tl

AFL dw

= +

4 35384 48

llllll

w L P LEI EI

= +∆

4 3(0.5 )

(0.5 )5

384 48dl ll ll

dl llw L P L

EI EI+

+ = +∆

( 0 .5 )2 4 0

d l l lL

+ ≤∆

3 6 0l lL≤∆

2ll

LV d w P

= − +

2 8ll b lldl

dl

SwP F PLw

+ −=

2 221.5

V ll

dl dl

AF PL dw w

= + −

SUBDIAPHRAGM CHORD DESIGN

AXIAL LOAD P = 3.5 kips

THE ALLOWABLE COMPRESSIVE STRESS ISFc' = Fc CD CP CF = 754 psi

Where Fc = 600 psi

CD = 1.33

CF = 1.00 (Lumber only)

CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0.5 = 0.944

Fc* = Fc CD CF = 798 psi

Le = Ke L = 1.0 L = 96 ind = 7.5 inSF = slenderness ratio = 12.8 < 50 [Satisfies NDS 2001 Sec. 3.7.1.4]

FcE = KcE E / SF2 = 3317 psi

KcE = 0.418

F = FcE / Fc* = 4.156c = 0.8

THE ACTUAL COMPRESSIVE STRESS ISfc = P / A = 85 psi < Fc' [Satisfactory]

THE ALLOWABLE FLEXURAL STRESS ISFb' = 1164 psi, [ for CD = 1.33 ]

THE ACTUAL FLEXURAL STRESS ISfb = (M + Pe) / S = 1277 psi > Fb' [Unsatisfactory]

CHECK COMBINED STRESS [NDS 2001 Sec. 3.9.2]

(fc / Fc' )2 + fb / [Fb' (1 - fc / FcE)] = 1.138 < 4 / 3 [Satisfactory]

Hence the beam design is inadequate.

Note: This is straight beam. The top of the beam is continuously braced by the plywood sheathing and nonslenderness factor, CL , adjustment is necessary for gravity load. For axial load , the only strong axis factor, CP , should be consided.



INPUT DATA DESIGN SUMMARYMEMBER TYPE: 3 1

2 STUD USE: 2 - 2'' x 6'' DOUGLAS FIR-LARCH No. 13 KING STUD

GEOMETRY DATA: 1. CHECK VERTICAL LOADS : fc < Fc' ?HEIGHT h = 12.67 ft 364 psi < 599 psi okUNBRACED LENGTH Le x-x (H) = 12.7 ft

Le y-y (B) = 0 ft 2. CHECK BENDING LOADS : fb < Fb' ?LOAD DATA: 876 psi < 1300 psi ok

DEAD LOAD 3000 lbsLIVE LOAD 3000 lbsTOTAL 6,000 lbs 3. CHECK INTERACTION :LATERAL LOAD x-x 55 plf

M= 1104 ft-lbsV= 348 lbs 1.848 > 1 NG

LOAD DURATION 2 OCCUPANCY LIVE LOAD

DESIGN CRITERIA: 4. CHECK SHEAR LOADS : fv < Fv' ?SECTION 2 pcs, B = 2 in 32 psi < 180 psi ok

H = 6 inSPECIES DOUGLAS FIR-LARCH 5. DEFLECTIONGRADE No. 1 3 ∆ = 5wh4 / (384EI) + 2.4wh2 / (Ebd) = 4/8 inLUMBER GRADING TYPE 1 VISUALLY GRADED ( h / 330 )SHAPE TYPE 1 SAWN LUMBERWET / DRY USE ? 1 DRY

ANALYSISCOLUMN BASIC DESIGN STRESSES:

COMPRESSIVE STRESS Fc = 1500 psiMODULUS OF ELASTICITY E = 1700 ksi

BENDING STRESS (X-Axis) Fbx = 1000 psi

BENDING STRESS (Y-Axis) Fby = 1000 psi

SHEAR STRESS (X-Axis) Fv = 180 psi

COLUMN PROPERTIES:COLUMN SECTION X-Dir dx = 5.50 in

Y-Dir dy = 1.50 inAREA A = 16.5 in2

SECTION MODULI Abt. xx Sx = 15.13 in3

Ix = 41.59 in4

Abt. yy Sy = 4.13 in3

LENGTH-DEPTH RATIO Le x-x / dx = 27.6Le y-y / dy = 0.0

ADJUSTMENT FACTORS: Fbx' Fby' Fc' Fv' E'

DURATION FACTOR CD 1.00 1.00 1.00 1.00

MOISTURE FACTOR CM 1.00 1.00 1.00 1.00 1.00 COLUMN PARAMETER c = 0.80

TEMPERATURE FACTOR Ct 1.00 1.00 1.00 1.00 1.00 EULER BUCKLING COEFFICIENT

INCISING FACTOR Ci 1.00 1.00 1.00 1.00 1.00 KcE = 0.300

SIZE FACTOR CF 1.30 1.30 1.10 1.00 CRITICAL EULER BUCKLING VALUES

FLAT USE FACTOR Cfu 1.15 FcE = 667 psi

COLUMN STABILITY CP 0.363 Fc* = 1650 psi

REPETITIVE MEMBER Cr 1.00 1.00

BEAM STABILITY CL 1.00 1.00

ADJUSTED PROPERTIES:

MODULUS OF ELASTICITY E' = 1700 ksi AXIAL STRESS Fc' = 599 psi

BENDING STRESS (X-Axis) Fbx' = 1300 psi SHEAR STRESS Fv' = 180 psi

BENDING STRESS (Y-Axis) Fby' = 1495 psi

ACTUAL STRESSES:

AXIAL STRESS fc = 363.6 psi SHEAR STRESS fv = 32 psi

BENDING STRESSES fbx = 875.6 psi

Wood Column Design

POST

DanielTian Li

2

' '

11 ?

1bxc

c cEx bxc

fffF F F

+ ≤ −



INPUT DATALATERAL FORCE ON DIAPHRAGM: vdia, WIND = 350 plf,for wind

vdia, SEISMIC = 350 plf,for seismic

GRAVITY LOADS ON THE ROOF: wDL = 262 plf,for dead load

wLL = 0 plf,for live load

DIMENSIONS: Lw = 8 ft , h = 16 ftL = 7 ft , hp = 0 ft

PANEL GRADE ( 0 or 1) = 1 <= Sheathing and Single-FloorMINIMUM NOMINAL PANEL THICKNESS = 15/32 in 262COMMON NAIL SIZE ( 0=6d, 1=8d, 2=10d ) 1 8dSPECIFIC GRAVITY OF FRAMING MEMBERS 0.5EDGE STUD SECTION 2 pcs, b = 2 in , h = 6 in

DESIGN SUMMARYBLOCKED 15/32 SHEATHING WITH 8d COMMON NAILS @ 4 in O.C. BOUNDARY & ALL EDGES / 12 in O.C. FIELD,5/8 in DIA. x 10 in LONG ANCHOR BOLTS @ 34 in O.C.

HOLD-DOWN FORCES: TL = 4.02 k , TR = 3.87 k (USE PHD5-SDS3 SIMPSON HOLD-DOWN)DRAG STRUT FORCES: F = -0.35 kEDGE STUD: 2 - 2'' x 6'' DOUGLAS FIR-LARCH No. 1, CONTINUOUS FULL HEIGHT.SHEAR WALL DEFLECTION: ∆ = 0.97 in

ANALYSISTHE MAX SHEAR WALL DIMENSION RATIO L / B = 2.0 < 2 [Satisfactory]THE UNIT SHEAR FORCE vb = 306 plf, ( 1 Side Diaphragm Required, the Max. Nail Spacing = 4 in )

THE SHEAR CAPACITIES PER UBCTable 23-II-I-1 :Min. Min.

Common PenetrationThicknessNail (in) (in) 6 4 3 28d 1 1/2 15/32 260 380 490 640

Note: The indicated shear numbers have reduced by specific gravity factor per note 1 of the table.

THE DRAG STRUT FORCE: F = (L-Lw) MAX( vdia, WIND, Ω0vdia, SEISMIC ) = -0.35 k ( Ω0 = 1 ) (Sec. 1633.2.6)THE MAX SPACING OF 5/8" DIA ANCHOR BOLT (Tab.11E, NDS 2001, Page 85) S = 34 in

THE HOLD-DOWN FORCES:vdia Safty Holddown(plf) Factors SIMPSON

Left 0.9 TL = 3922Right 0.9 TR = 3716Left 2/3 TL = 4024

Right 2/3 TR = 3871

THE MAXIMUM SHEAR WALL DEFLECTION: ( Section 4.3, ASD MANUAL SUUP, Page SW-17)

= 0.971 in

Where: vb = 306 plf Lw = 8 ft E = 1.7E+06 psiA = 16.50 in2 h = 16 ft G = 9.0E+04 psit = 0.298 in en = 0.037 in da = 0.15 in

CHECK EDGE STUD CAPACITYFc = 1500 psi CD = 1.6 CP = 0.14 A = 16.5 in2

E = 1700 ksi CF = 1.10 Fc' = 365 psi > fc = 201 psi

[Satisfactory] Techincal References: 1. "National Design Specification, NDS", 2001 Edition, AF&AP, AWC, 2001. 2. Alan Williams: "Structuiral Engineering Reference Manual", Professional Publications, Inc, 2001.

Shear Wall Design

Blocked Nail SpacingBoundary & All Edges

Sheathing and Single-Floor

Panel Grade

205

Net Uplift(lbs)

OverturningMoments (ft-lbs)

Wall Seismicat mid-story (lbs)

ResistingMoments (ft-lbs)

12349

40838

39200

10515

DanielTian Li

PHD5-SDS3350

350

SEISMIC

WIND

1234910515

380.75b b a

Bending Shear Nail slip Chord splice slip nw w

h hv h v dheEA GtL L

∆ = + + + = + + +∆ ∆ ∆ ∆



INPUT DATALATERAL FORCE ALONG L SIDE: wL, WIND = 256 plf,for wind

wL, SEISMIC = 224 plf,for seismic

LATERAL FORCE ALONG B SIDE: wB, WIND = 256 plf,for wind

wB, SEISMIC = 343 plf,for seismic

DIMENSIONS: L = 240 ft , B = 110 ftB1 = 45 ft , B2 = 40 ft

PANEL GRADE ( 0 or 1) = 1 <= Sheathing and Single-FloorMINIMUM NOMINAL FRAMING WITH ( 2 or 3) = 3 inMINIMUM NOMINAL PANEL THICKNESS = 15/32 inCOMMON NAIL SIZE ( 0=6d, 1=8d, 2=10d ) 1 8dSPECIFIC GRAVITY OF FRAMING MEMBERS 0.43

DESIGN SUMMARYA1: (2) - 16 ft x 110 ft

BLOCKED 15/32 SHEATHING WITH 8d COMMON NAILS @ 4 in O.C. BOUNDARY / 6 in O.C. EDGES / 12"O.C. FIELD.

A2: (2) - 14 ft x 110 ftBLOCKED 15/32 SHEATHING WITH 8d COMMON NAILS @ 6" O.C. BOUNDARY & EDGES / 12"O.CFIELD.

A3: (1) - 180.00 ft x 110 ftUNBLOCKED 15/32 SHEATHING WITH 8d COMMON NAILS @ 6" O.C. ALL EDGES / 12"O.CFIELD.

THE CHORD FORCES: TL = CL = 16.76 k , TB = CB = 2.16 kTHE DRAG STRUT FORCES: F1 = 6.84 k , F2 = 27.37 kTHE MAXIMUM DIAPHRAGM DEFLECTION: ∆ = 3.09 in

ANALYSISTHE DIAPHRAGM IS CONSIDED FLEXIBLE IF ITS MAXIMUM LATERAL DEFORMATION IS MORE THAN TWO TIMES THE AVERAGE SHEARWALL DEFLECTION OF THE ASSOCIATED STORY. WITHOUT FURTHER CALCULATIONS, ASSUME A FLEXIBLE DIAPHRAGM HERE.

FROM THE TABLE 3.1 IN ASD MANUAL SUPP, PAGE SP-12, THE PANEL BENDING STRENGTH CAPACITY IS 355 in-lbs/ft, THAT IS THEDIAPHRAGM CAN RESISTS 59 psf GRAVITY LOADS (DL+LL) AT 2'-0" o.c. SPACING SUPPORTS.

THE MAX DIAPHTAGM DIMENSION RATIO L / B = 2.2 < 3, [satisfactory]THE MAX SHEAR FORCE ALONG B SIDE vL = 279 plf, ( Boundary Spacing = 4 in, Edges ReqD = 6 in )THE MAX SHEAR FORCE ALONG L SIDE vB = 79 plf, ( Required Boundary/Edges Nail Spacing for Case 3 = 6 in )THE ALLOWABLE SHEAR FORCE FOR CASE 1 @ 6 in NAIL SPACING v1 = 246 plf, L1 = 105.7 ftTHE MAX ALLOWABLE UNBLOCKED SHEAR FORCE FOR CASE 1 v1 = 217 plf, L2 = 93.4 ft

THE SHEAR CAPACITIES PER UBC Table 23-II-H :Min. Min. Member

Common PenetrationThickness WidthNail (in) (in) (in) 6 / 6 4 / 6 2.5 / 4 2 / 3 Case 1 Others8d 1 1/2 15/32 3 246 328 492 554 217 164


THE CHORD FORCES: TL = CL = ( wLL2 ) / ( 8B ) = 16.76 k TB = CB = ( wBB2 ) / ( 8L ) = 2.16 k

THE DRAG STRUT FORCES: F1 = 0.5 (B-2B1) MAX( v1, WIND, Ω0v1, SEISMIC ) = 6.84 k (UBC 1633.2.6, F2 = B2 MAX( v1, WIND, Ω0v1, SEISMIC ) = 27.37 k or IBC 1620.1.6)

THE MAXIMUM DIAPHRAGM DEFLECTION: ( Section 3.3, ASD MANUAL SUUP, Page SW-12)

= 3.092 in

Where: vL = 279 plf L = 240 ft E = 1.7E+06 psiA = 21.75 in2 B = 110 ft G = 9.0E+04 psi,(UBC97 Page3-421)

t = 0.298 in,(UBC97 Page3-420) en = 0.037 in,(UBC97 Page3-422) Σ(Dcx) = 45.00 inNote: The deflection, ∆, above is based on completely blocked. For unblocked diaphragm, 2.4∆ should be used.

Techincal References: 1. "National Design Specification, NDS", 2001 Edition, AF&AP, AWC, 2001. 2. Alan Williams: "Structuiral Engineering Reference Manual", Professional Publications, Inc, 2001.


Blocked Nail SpacingBoundary / Other Edges

UnblockedPanel Grade

Diaphragm Design

DanielT. Li

Ω0 = 2.8

( )350.188

8 4 2CL L

Bending Shear Nail slip Chord splice slip n

xL Dv vL LeEAB Gt B

∆ = + + + = + + +∆ ∆ ∆ ∆



INPUT DATALength L = 60 ftWidth B = 46 ftRoof Height hr = 28 ftParapet Height hP = 4 ftWall Weight WP = 112 psfCoefficient SDS = 0.54Importance Factor Iw = 1Diaphragm Shear Capacity vallowable = 720 plf

ANALYSIS The subdiaphragms comply with 2.5:1 of max. length-to-width rario. (Sec.1620.2.1,IBC 2000) The wall anchor force is given by IBC Sec. 1620.1.7, 1620.2.1 & 1604.8.2 as

885 plf

Where : Fmin = 200 plf(IBC Sec.1604.8.2)

Note: the cofficient in Eq 16-64 was revised from 1.2 to 0.8 as a result of voting during the 2001 publichearing on the 2000 IBC.

Wood subdiaphragm shear :

= 412 plf, for ASD< vallowable

Satisfactory to use diaphragm nailing for subdiaphragm.

Chord force :

= 8.66 k, (Indicated force has NOT been reduced for ASD)

Reaction force :

= 26.54 k(Indicated force has NOT been reduced for ASD)

Subdiaphragm Design Based on IBC 2000T. Li

Daniel

0.51.4

PLFvB

=

2

8PF LT CB

= =

2PLFR =

( )2

min0.8 , 400 ,2

P E EDS p DSr

h hr pMAX S W SF I I F

h

+ = =



INPUT DATALength L = 60 ftWidth B = 46 ftRoof Height hr = 28 ftParapet Height hP = 4 ftWall Weight WP = 112 psfCoefficient Ca = 0.44Seismic Zone (2A, 2B, 3, 4) 4Importance Factor IP = 1Diaphragm Capacity vallowable = 720 plf (ASD)

ANALYSIS

Rp = 3ap = 1.5

The subdiaphragms comply with 2.5:1 max. length-to-width rario. (Sec.1633.2.9,UBC 97)

99 psf

Check minimum wall-roof anchorage force : (Sec.1633.2.8.1 & 1611.4, UBC97)

1802 plf > 420 plf

Thus, w = 1802 plf

Wood subdiaphragm shear : (Sec.1633.2.8.1 item 5, UBC97)

= 714 plf, for ASD< 720 plf

Use diaphragm nailing for subdiaphragm is adequate.

Chord force :

= 17.63 kips, (for SD) = 12.59 kips, (for ASD)

Steel tie/reaction force : (Sec.1633.2.8.1 item 4, UBC97)

75.69 kips, (SD) = 54.07 kips, (for ASD)

Subdiaphragm Design Based on UBC 97

DanielT. Li

41 3p pp a p a pxp p

p pr

a C a C wI IhwF

hR R

= + = =

( )2

2p

r

h hr pw F

h

+= =

0.50.85 , 3 4

1.40.5

, 1 21.4

wLfor zone or

BvwL

for zone orB

=

2

8wLT C

B= =

1.4 , 3 42

, 1 22

wLfor zone or

RwL

for zone or

= =



INPUT DATA & DESIGN SUMMARYDIAPHRAGM CHORD FORCE RESISTED BY THE TOP PLATE T = C = 15 kNAIL TYPE ( 0=Common Wire, 1=Box, 2=Sinker ) 0 Common Wire NailNAIL PENNY-WEIGHT ( 12d, 16d, 20d ) 16dLUMBER TYPE ( 0=Douglas Fir-Larch, 1=Douglas Fir-Larch(N), 0 Douglas Fir-Larch, G=0.5

2=Hem-Fir(N), 3=Hem-Fir, 4=Spruce-Pine-Fir) LUMBER GRADE ( 0=Select Structural, 1=No.1 & Btr, 2=No.1, 3=No.2, 2 No.1

4=No.3, 5=Stud, 6=Construction, 7=Standard, 8=Utility) TOP PLATE SIZE Double 2 x 6 No.1, Douglas Fir-Larch, G=0.5LOAD DURATION FACTOR ( Tab 2.3.2, NDS 2001, Page 9 ) CD = 1.6

WET SERVICE FACTOR ( Tab 10.3.3, NDS 2001, Page 59 ) CM = 1.0

TEMPERATURE FACTOR ( Tab 10.3.4, NDS 2001, Page 59 ) Ct = 1.0

Use CMSTC16 with 32-16d sinkers, Each Side.

ANALYSISDESIGN VALUE FOR TENSION (Tab 4A, NDS 2001 SUPP, Page 31) Ft = 675 psiAREA OF CROSS SECTION FOR ONE 2 x 6 MEMBER A = 8.25 in2

SIZE FACTOR (Tab 4A, NDS 2001 SUPP, Page 30) CF = 1.30

ALLOWABLE TENSION CAPACITY FOR ONE 2 x 6 ONLY T' = AFtCDCMCtCF = 11.58 k< T, SIMPSON STRAP REQUIRED.

NAIL LENGTH L = 3 1/2 inSIDE MEMBER THICKNESS ts = 1 1/2 inTHE PENETRATION OF THE NAIL INTO THE MAIN MEMBER p = 1.99 inNAIL DIAMETER D = 0.162 inTHE PENETRATION FACTOR (Note 3, Tab 11N, NDS 2001, Page 97) Cd = 1.00

THE NOMINAL DESIGN VALUE FOR SINGLE SHEAR IS TABULATED IN NDS 2001 TABLE 11N, PAGE 97, ASZ = 141 lbf

THE ALLOWABLE LATERAL DESIGN VALUE FOR THE ONE NAIL IS

Z' = ZCDCMCtCd = 226 lbf

THE NUMBER OF NAILS REQUIRED IS n = T / Z' = 66.5 = 67 NailsTHE MIN. FORCE RESISTED BY THE SIMPSON STRAP T - T' = 3.42 k

Use CMSTC16 with 32-16d sinkers, Each Side.


Top Plate Connection

DanielT. Li



INPUT DATA & DESIGN SUMMARYAXIAL TENSILE FORCE (ASD) T = 5.3 kNUMBER OF BOLTS n = 3BOLT DIAMETER φ = 3/4BOLT SPACING S = 3 in

END DISTANCE OF WOOD En = 4 in

END DISTANCE OF STEEL En,s = 1.5 in

LUMBER TYPE ( 0=Douglas Fir-Larch, 1=Douglas Fir-Larch(N), 0 Douglas Fir-Larch, G=0.5 2=Hem-Fir(N), 3=Hem-Fir, 4=Spruce-Pine-Fir)

LUMBER SIZE ( 2 ) - 2 thk. x 6 widthSTRAP SIZE 5 width x 1/4 thk.LOAD DURATION FACTOR ( Tab 2.3.2, NDS 2001, Page 9 ) CD = 1.6WET SERVICE FACTOR ( Tab 10.3.3, NDS 2001, Page 59 ) CM = 1.0TEMPERATURE FACTOR ( Tab 10.3.4, NDS 2001, Page 59 ) Ct = 1.0

THE CONNECTION DESIGN IS ADEQUATE.

11.8

ANALYSISCHECK STEEL STRAP CAPACITIES (AISC-ASD)

Ag = 1.25 in2, yielding critierion Fy = 36.00 ksi

Tallow = 0.6 Fy Ag = 27.00 k > T [Satisfactory]

An = 1.03 in2, fracture critierion Fu = 58.00 ksi

Tallow = 0.5 Fu An = 29.91 k > T [Satisfactory]

Av = 1.33 in2, block shear

Tallow = 0.3 Fu Av + 0.5 Fu (0.5 An) = 38.06 k > T [Satisfactory]

rmin = t / (12)0.5 = 0.072 in L = Max (En , S ) = 4 in

L / rmin = 55 < 300 [Satisfactory] (AISC-ASD, D2, page 5-40)

CHECK EDGE, END, & SPACING DISTANCE REQUIREMENTS (NDS 2001, Table 11.5.1A, Table 11.5.1B, & Table 11.5.1C)

Eg = 2.75 in > 1.5 D [Satisfactory]

En = 4 in > 3.5 D [Satisfactory]

S = 3 in > 3 D [Satisfactory]

CHECK WOOD CAPACITY

C∆ = Min (C∆1 , C∆2 , C∆3) = 0.762 , (geometry factor, NDS 2001, 11.5.1, page 76)

where C∆1 = (actual end distance) / (min end distance for full design value) = En / 7D = 0.762

C∆2 = (actual shear area) / (min shear area for full design value) = 1.000

C∆3 = (actual spacing) / (min spacing for full design value) = S / 4D = 1.000

0.988 , (group action factor, NDS 2001, 10.3.6, page 60)

where n = 3 REA = Min [(EsAs/EmAm) , (EmAm/EsAs)] = 0.616

EsAs = 3.8E+07 lbs, (NDS 2001, Table 10.3.6C) γ = 180000 D1.5 = 116913

tm = 3 in u = 1+γ S/2 [1 / EmAm + 1 / EsAs] = 1.012

EmAm = 2.3E+07 lbs, (NDS 2001, Table 10.3.6C) m = u - (u2 - 1)0.5 = 0.855

Z'II = n ZII CD CM Ct Cg C∆ = 5.309 kips > T [Satisfactory]

where ZII = 1470 lbs / bolt, (interpolated from NDS 2001, Table 11B, Page 82)


Bolt Connection Design Based on NDS 2001

DanielTian Li

( )( )( )

2

2

1 111 1 1

EA

EA

n

n n

m m RC g mn mm mR

− + = = − + + − +



INPUT DATA & DESIGN SUMMARYAXIAL TENSILE FORCE (ASD) T = 8 kNUMBER OF STUDS n = 6THREAD STUD DIAMETER φ = 3/4THREAD STUD SPACING S = 12 in

END DISTANCE OF WOOD En = 4 in

NAILER TYPE ( 0=Douglas Fir-Larch, 1=Douglas Fir-Larch(N), 0 Douglas Fir-Larch, G=0.5 2=Hem-Fir(N), 3=Hem-Fir, 4=Spruce-Pine-Fir)

NAILER SIZE 4 thk. x 6 width

SECTION AREA OF STEEL MEMBER As = 11.8 in2

LOAD DURATION FACTOR ( Tab 2.3.2, NDS 2001, Page 9 ) CD = 1.6WET SERVICE FACTOR ( Tab 10.3.3, NDS 2001, Page 59 ) CM = 1.0TEMPERATURE FACTOR ( Tab 10.3.4, NDS 2001, Page 59 ) Ct = 1.0

GROUP ACTION FACTOR APPLY? ( 0 = No, 1 = Yes) 1 Yes, ( Cg = 0.719 , NDS 2001, 10.3.6)

(If T is drag/collector force or stud spacing less than 12" o.c., the group action factor Cg must apply.)


ANALYSISCHECK EDGE, END, & SPACING DISTANCE REQUIREMENTS (NDS 2001, Table 11.5.1A, Table 11.5.1B, & Table 11.5.1C)

Eg = 2.75 in > 1.5 D [Satisfactory]

En = 4 in > 3.5 D [Satisfactory]

S = 12 in > 3 D [Satisfactory]

CHECK WOOD CAPACITY

C∆ = Min (C∆1 , C∆2 , C∆3) = 0.762 , (geometry factor, NDS 2001, 11.5.1, page 76)

where C∆1 = (actual end distance) / (min end distance for full design value) = En / 7D = 0.762

C∆2 = (actual shear area) / (min shear area for full design value) = 1.000

C∆3 = (actual spacing) / (min spacing for full design value) = S / 4D = 4.000

5 0.719 , (group action factor, NDS 2001, 10.3.6, page 60)

where n = 6 REA = Min [(EsAs/EmAm) , (EmAm/EsAs)] = 0.079

EsAs = 3.4E+08 lbs γ = 180000 D1.5 = 116913

tm = 3.5 in u = 1+γ S/2 [1 / EmAm + 1 / EsAs] = 1.028

EmAm = 2.7E+07 lbs, (Em fr NDS, Tab.10.3.6C) m = u - (u2 - 1)0.5 = 0.789

Z'II = n ZII CD CM Ct Cg C∆ = 8.629 kips > T [Satisfactory]

where ZII = 1640 lbs / stud, (interpolated from NDS 2001, Table 11E, Page 85)


Nailer Connection Design Based on NDS 2001

DanielTian Li

( )( )( )

2

2

1 111 1 1

EA

EA

n

n n

m m RC g mn mm mR

− + = = − + + − +



INPUT DATALATERAL FORCE ON DIAPHRAGM: vdia, WIND = 268 plf,for wind

(SERVICE LOADS) vdia, SEISMIC = 350 plf,for seismic

DIMENSIONS: L1 = 4 ft , L2 = 16 ft , L3 = 4 ft

H1 = 4 ft , H2 = 6 ft , H3 = 6.5 ft

KING STUD SECTION 2 pcs, b = 2 in , h = 6 in

EDGE STUD SECTION 1 pcs, b = 4 in , h = 6 in

PANEL GRADE ( 0 or 1) = 1 <= Sheathing and Single-Floor

MINIMUM NOMINAL PANEL THICKNESS = 15/32 in

COMMON NAIL SIZE ( 0=6d, 1=8d, 2=10d ) 1 8d

SPECIFIC GRAVITY OF FRAMING MEMBERS 0.5

DESIGN SUMMARYBLOCKED 15/32 SHEATHING, EACH SIDE, WITH 8d COMMON NAILS

@ 2 in O.C. BOUNDARY & ALL EDGES / 12 in O.C. FIELD,

5/8 in DIA. x 10 in LONG ANCHOR BOLTS @ 48 in O.C.

HOLD-DOWN FORCES: TL = 5.99 k , TR = 5.99 k (USE HDQ8-SDS3 SIMPSON HOLD-DOWN)MAX STRAP FORCE: F = 4.68 k (USE SIMPSON CMSTC16 OVER WALL SHEATHING WITH FLAT BLOCKING)KING STUD: 2 - 2'' x 6'' DOUGLAS FIR-LARCH No. 1, CONTINUOUS FULL HEIGHT.EDGE STUD: 1 - 4'' x 6'' DOUGLAS FIR-LARCH No. 1, CONTINUOUS FULL HEIGHT.SHEAR WALL DEFLECTION: ∆ = 1.58 in

4

DanielTian Li

Wood Shear Wall with an Opening

cont'dANALYSISCHECK MAX SHEAR WALL DIMENSION RATIO h / w = 1.5 < 2 [Satisfactory]

THE FORCES OF FREE-BODY INDIVIDUAL PANELS OF WALL ARE GIVEN BY FIGURE ABOVE AND TABLES AS

INDIVIDUAL PANEL W (ft) H (ft) MAX SHEAR STRESS (plf) NO. FORCE (lbf) NO. FORCE (lbf)

1 4.00 6.50 27 F1 108 F13 3325

2 8.00 6.50 512 F2 4092 F14 3325

3 8.00 6.50 512 F3 108 F15 6475

4 4.00 6.50 27 F4 3325 F16 3150

5 4.00 3.00 1050 F5 4200 F17 3150

6 4.00 3.00 1050 F6 4092 F18 6475

7 4.00 3.00 1050 F7 4092 F19 4682

8 4.00 3.00 1050 F8 4200 F20 4682

9 4.00 4.00 -121 F9 175 F21 2668

10 8.00 4.00 667 F10 3150 F22 -482

11 8.00 4.00 667 F11 3150 F23 4682

12 4.00 4.00 -121 F12 175 F24 -482

THE UNIT SHEAR FORCE vb = 1050 plf, ( 2 Sides Diaphragm Required, the Max. Nail Spacing = 2 in )

THE SHEAR CAPACITIES PER UBC Table 23-II-I-1 :Min. Min.

Common Penetration ThicknessNail (in) (in) 6 4 3 28d 1 1/2 15/32 260 380 490 640


THE MAX SPACING OF 5/8" DIA ANCHOR BOLT (Tab.11E, NDS 2001, Page 85) S = 48 in

THE HOLD-DOWN FORCES:vdia Safty Holddown

(plf) Factors SIMPSON

Left 0.9 TL = 5993

Right 0.9 TR = 5993

Left 2/3 TL = 4422

Right 2/3 TR = 4422

THE MAXIMUM SHEAR WALL DEFLECTION: ( Section 4.3, ASD MANUAL SUUP, Page SW-17)

= 1.581 in

Where: vb = 1050 plf Lw = 8 ft E = 1.7E+06 psiA = 16.50 in2 h = 17 ft G = 9.0E+04 psit = 0.298 in en = 0.037 in da = 0.15 in

CHECK KING STUD CAPACITY

Fc = 1500 psi CD = 1.6 CP = 0.13 A = 16.5 in2


[Satisfactory]

CHECK EDGE STUD CAPACITY

Fc = 1500 psi CD = 1.6 CP = 0.13 A = 19.25 in2


[Satisfactory]


1. "National Design Specification, NDS", 2001 Edition, AF&AP, AWC, 2001.

2. Alan Williams: "Structuiral Engineering Reference Manual", Professional Publications, Inc, 2001.

3. "2000 IBC Structuiral/Seismic Design Manual - Volume 1, Code Application Examples", Structural Engineers

Association of California, 2001.

Blocked Nail SpacingBoundary & All Edges


Panel Grade

Wall Seismicat mid-story (lbs)

ResistingMoments (ft-lbs)

Net Uplift(lbs)

OverturningMoments (ft-lbs)

SEISMIC

WIND

0

0

0

143827

106128

0634

HDQ8-SDS3350

268

380.75b b a

Bending Shear Nail slip Chord splice slip nw w

h hv h v dheEA GtL L

∆ = + + + = + + +∆ ∆ ∆ ∆



INPUT DATA & DESIGN SUMMARYNAIL TYPE ( 0=Common Wire, 1=Box, 2=Sinker ) 0 Common Wire NailNAIL PENNY-WEIGHT ( 6d, 7d, 8d, 10d, 12d, 16d, 20d, 30d, 40d, 50d, 60d ) 10dLUMBER SPECIES ( 0=Douglas Fir-Larch, 1=Douglas Fir-Larch(N), 0 Douglas Fir-Larch, G=0.5

2=Hem-Fir(N), 3=Hem-Fir, 4=Spruce-Pine-Fir) LOAD DURATION FACTOR ( Tab 2.3.2, NDS 2001, Page 9 ) CD = 1.6

WET SERVICE FACTOR ( Tab 10.3.3, NDS 2001, Page 59 ) CM = 1.0

TEMPERATURE FACTOR ( Tab 10.3.4, NDS 2001, Page 59 ) Ct = 1.0

THE ALLOWABLE LATERAL DESIGN VALUE FOR THE TOE-NAIL = 157 lbf

ANALYSISTOE-NAIL FACTOR ( Sec 11.5.4, NDS 2001, Page 78 ) Ctn = 0.83

NAIL LENGTH L = 3 inTHE SIDE MEMBER THICKNESS IS TAKEN TO BE EQUAL TO ts = L / 3 = 1 inTHE PENETRATION OF THE NAIL INTO THE MAIN MEMBER p = L(cos30o) -ts = 1.60 inNAIL DIAMETER D = 0.148 inTHE PENETRATION FACTOR (Note 3, Tab 11N, NDS 2001, Page 97) Cd = 1.00

THE NOMINAL DESIGN VALUE FOR SINGLE SHEAR IS TABULATED IN NDS 2001 TABLE 11N, PAGE 97, ASZ = 118 lbf

THE ALLOWABLE LATERAL DESIGN VALUE FOR THE TOE-NAIL IS

Z' = ZCDCMCtCdCtn = 157 lbf

Typical Nail Dimensions (Appendix L, NDS 2001, Page 168)

6d 7d 8d 10d 12d 16d 20d 30d 40d 50d 60d10 6 7 8 10 12 16 20 30 40 50 60

Length 2 2 1/4 2 1/2 3 3 1/4 3 1/2 4 4 1/2 5 5 1/2 6

Diameter 0.113 0.113 0.131 0.148 0.148 0.162 0.192 0.207 0.225 0.244 0.263

Head 0.266 0.266 0.281 0.312 0.312 0.344 0.406 0.438 0.469 0.500 0.531

Length 2 2 1/4 2 1/2 3 3 1/4 3 1/2 4 4 1/2 5

Diameter 0.099 0.099 0.113 0.128 0.128 0.135 0.148 0.148 0.162

Head 0.266 0.266 0.297 0.312 0.312 0.344 0.375 0.375 0.406

Length 1 7/8 8 1/8 2 3/8 2 7/8 3 1/8 3 1/4 3 3/4 4 1/4 4 3/4 5 3/4

Diameter 0.092 0.099 0.113 0.120 0.135 0.148 0.177 0.192 0.207 0.244

Head 0.234 0.250 0.266 0.281 0.312 0.344 0.375 0.406 0.438 0.500


Box

Sinker

Penny-Weight

Common

Type

Toe-Nail Connection

DanielT. Li



INPUT DATAGRAVITY LOAD D+L = 250 plfSERVICE ANCHORAGE FORCE Fanch = 420 plf

SERVICE DIAPHRAGM SHEAR FORCE V = 640 plfLEDGER SPECIES ( 0=Douglas Fir-Larch, 1=Douglas Fir-Larch(N), 2=Hem-Fir(N), 3=Hem-Fir, 4=Spruce-Pine-Fir) 0 Douglas Fir-Larch, G=0.5LEDGER GRADE 2 No.1

( 0=Select Structural, 1=No.1 & Btr, 2=No.1, 3=No.2, 4=No.3, 5=Stud, 6=Construction, 7=Standard, 8=Utility)

LEDGER SIZE 4 x 8( No.1, Douglas Fir-Larch, G=0.5 )

ANCHOR BOLT DIAMETER φ = 3/4 inWET SERVICE FACTOR ( Tab 10.3.3, NDS 2001, Page 59 ) CM = 1.0TEMPERATURE FACTOR ( Tab 10.3.4, NDS 2001, Page 59 )Ct = 1.0PURLIN ANCHORS SPACING S = 36 inTYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUTHICKNESS OF WALL t = 8 inSPECIAL INSPECTION ( 0=NO, 1=YES ) 0 NoMASONRY STRENGTH fM' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiWALL HORIZ. REINF. 1 # 5 @ 16 in. o. c.

DESIGN SUMMARY4 x 8 LEDGER with 3/4 in DIA. A.B.'s @ 36 in o.c.SIMPSON PLURLIN ANCHOR PAI23 w/ 9-10d x 1 1/2 Nails @ 36 in o.c.

ANALYSISTHE DIAPHRAGM SHEAR TRANSFERS INTO THE TOP OF THE LEDGER BY NAILS THOUGH PLY WOOD AND THE LEDGER BOLTS WOULDRESIST BOTH THE DIAPHRAGM SHEAR BY PARALLEL TO GRAIN BEARING AND THE GRAVITY LOAD BY PERPENDICULAR TO GRAINBEARING. LATER FORCES IN THE OTHER DIRECTION WOULD BE RESIST BY SIMPSON PLURLIN ANCHORS.CHECK LEDGER CAPACITYLOAD DURATION FACTOR ( Tab 2.3.2, NDS 2001, Page 9 ) (CD)Lat = 1.6 (for wind/seismic loads)

(CD)D+L = 1.0 (for gravity loads)GROUP ACTION FACTOR (Sec 10.3.6, NDS 2001, Page 60) Cg = 1.0GEOMETRY FACTOR ( Sec 11.5.1, NDS 2001, Page 76 ) C∆ = 1.0 ( All dimensions conform to the specified minimums.)

DESIGN VALUE FOR THE BOLT BEARING PERPENDICULAR TO GRAIN (Tab 11E, NDS 2001, Page 85) Z = 880 lbfDESIGN VALUE FOR THE BOLT BEARING PARALLEL TO GRAIN (Tab 11E, NDS 2001, Page 85) Z = 1640 lbfTHE ALLOWABLE DESIGN VALUE FOR THE BOLT BEARING PERPENDICULAR TO GRAINZ' = Z(CD)D+LCMCtCgC = 880 lbf

THE ALLOWABLE DESIGN VALUE FOR THE BOLT BEARING PARALLEL TO GRAIN Z' = Z(CD)LATCMCtCgC = 2624 lbf

THE ANGLE BETWEEN DIRECTION OF COMBINED LOAD AND DIRECTION OF GRAIN θ = 21.34o

THE ALLOWABLE DESIGN VALUE FOR THE BOLT BEARING AT THE ANGLE TO GRAIN Z' = Z'Z' /(Z'cos2θ+Z||'sin2θ) = 2079 lbf

THE MAXIMUM ALLOWABLE BOLT SPACING SHALL BE CALCULATED AS FOLLOWS :

= 36.30 in ( Use 36 in )

BASIC DESIGN VALUE FOR SHEAR (Tab 4A, NDS 2001 SUPP, Page 31) Fv = 180 psiTHE ALLOWABLE DESIGN VALUE FOR SHEAR Fv' = Fv(CD)D+LCMCt = 180 psiDEPTH FROM THE UNLOADED EDGE OF THE LEDGER TO THE CENTER OF THE BOLTde = 3.63 inTHE ALLOWABLE DESIGN SHEAR IS GIVEN BY NDS 2001 Eq. 3.4-6, Page 17, AS

= 761 lbf > (D+L)S = 756 lbf [Satisfactory]

T. LiConnection from Diaphragm, Ledger to CMU Wall

Daniel

( )

' '

12 2

,Z ZMINSD L D L V

θ⊥ = + + +

2' '2

3e

vdbdV Fd

=

DESIGN PURLIN ANCHORS cont'dSELECT SIMPSON PAI23 w/ 9-10d x 1 1/2 Nails @ 36 in o.c.DESIGN FORCE FOR STEEL ANCHOR (UBC 1633.2.8.1, IBC 2003 1620.2) :

1.4FanchS = 1764 lbf < 1880 lbf, allowableDESIGN FORCE FOR WOOD MEMBER (UBC97SEC. 1633.2.8.1, ITEM 5) :

1071 lbf < 1128 lbf, allowable [Satisfactory]

CHECK WALL CAPACITY TO SPAN LATERALLY S SPACINGALLOWABLE MASONRY STRESS FACTOR : SF = 0.667Allowable reinf. stress Fs = 32000 psi Modular ratio n = 21.48Allowable stress Fb=(SF)(0.33fm') = 330 psi Wall reinf. area As = 0.23 in / ftMasonry elasticity modulus Em = 1350 ksi Tension reinf. ratio ρ = 0.005Steel elasticity modulus Es = 29000 ksi The neutral axis depth factor k = 0.36Effective width bw = 12 in The lever-arm factor j = 0.88The tensile stress in reinforcement is The compressive stress in the extreme fiber is

6938 psi < Fs 185 psi < Fb



0.85FanchS =

2

8anch

ss

SFfjdA

= =2

24anch

bw

SFfjkb d

= =



INPUT DATAGRAVITY LOAD D+L = 250 plfSERVICE ANCHORAGE FORCE Fanch = 420 plf

SERVICE DIAPHRAGM SHEAR FORCE V = 640 plfLEDGER SPECIES ( 0=Southern Pine with 2"-4" Wide, 1=Southern Pine with 5"-6" Wide) 0 Southern PineLEDGER GRADE 2 Non-Dense Select Structural

( 0=Dense Select Structural, 1=Select Structural, 2=Non-Dense Select Structural, 3=No.1 Dense,

4=No.1, 5=No.1 Non-Dense, 6=No.2 Dense, 7=No.2, 8=No.2 Non-Dense)

LEDGER SIZE 4 x 10( Non-Dense Select Structural, Southern Pine )

ANCHOR BOLT DIAMETER φ = 3/4 inWET SERVICE FACTOR ( Tab 10.3.3, NDS 2001, Page 59 ) CM = 1.0TEMPERATURE FACTOR ( Tab 10.3.4, NDS 2001, Page 59 )Ct = 1.0PURLIN ANCHORS SPACING S = 36 inTYPE OF MASONRY ( 1=CMU, 2=BRICK ) 1 CMUTHICKNESS OF WALL t = 8 inSPECIAL INSPECTION ( 0=NO, 1=YES ) 0 NoMASONRY STRENGTH fM' = 1.5 ksiREBAR YIELD STRESS fy = 60 ksiWALL HORIZ. REINF. 1 # 5 @ 16 in. o. c.

DESIGN SUMMARY4 x 10 LEDGER with 3/4 in DIA. A.B.'s @ 37 in o.c.SIMPSON PLURLIN ANCHOR PAI23 w/ 9-10d x 1 1/2 Nails @ 36 in o.c.

ANALYSISTHE DIAPHRAGM SHEAR TRANSFERS INTO THE TOP OF THE LEDGER BY NAILS THOUGH PLY WOOD AND THE LEDGER BOLTS WOULDRESIST BOTH THE DIAPHRAGM SHEAR BY PARALLEL TO GRAIN BEARING AND THE GRAVITY LOAD BY PERPENDICULAR TO GRAINBEARING. LATER FORCES IN THE OTHER DIRECTION WOULD BE RESIST BY SIMPSON PLURLIN ANCHORS.CHECK LEDGER CAPACITYLOAD DURATION FACTOR ( Tab 2.3.2, NDS 2001, Page 9 ) (CD)Lat = 1.6 (for wind/seismic loads)

(CD)D+L = 1.0 (for gravity loads)GROUP ACTION FACTOR (Sec 10.3.6, NDS 2001, Page 60) Cg = 1.0GEOMETRY FACTOR ( Sec 11.5.1, NDS 2001, Page 76 ) C∆ = 1.0 ( All dimensions conform to the specified minimums.)

DESIGN VALUE FOR THE BOLT BEARING PERPENDICULAR TO GRAIN (Tab 11E, NDS 2001, Page 85) Z = 950 lbfDESIGN VALUE FOR THE BOLT BEARING PARALLEL TO GRAIN (Tab 11E, NDS 2001, Page 85) Z = 1680 lbfTHE ALLOWABLE DESIGN VALUE FOR THE BOLT BEARING PERPENDICULAR TO GRAINZ' = Z(CD)D+LCMCtCgC = 950 lbf

THE ALLOWABLE DESIGN VALUE FOR THE BOLT BEARING PARALLEL TO GRAIN Z' = Z(CD)LATCMCtCgC = 2688 lbf

THE ANGLE BETWEEN DIRECTION OF COMBINED LOAD AND DIRECTION OF GRAIN θ = 21.34o

THE ALLOWABLE DESIGN VALUE FOR THE BOLT BEARING AT THE ANGLE TO GRAIN Z' = Z'Z' /(Z'cos2θ+Z||'sin2θ) = 2164 lbf

THE MAXIMUM ALLOWABLE BOLT SPACING SHALL BE CALCULATED AS FOLLOWS :

= 37.79 in ( Use 37 in )

BASIC DESIGN VALUE FOR SHEAR (Tab 4B, NDS 2001 SUPP, Page 37) Fv = 175 psiTHE ALLOWABLE DESIGN VALUE FOR SHEAR Fv' = Fv(CD)D+LCMCt = 175 psiDEPTH FROM THE UNLOADED EDGE OF THE LEDGER TO THE CENTER OF THE BOLT de = 4.63 inTHE ALLOWABLE DESIGN SHEAR IS GIVEN BY NDS 2001 Eq. 3.4-6, Page 17, AS

= 944 lbf > (D+L)S = 787 lbf [Satisfactory]

DESIGN PURLIN ANCHORSSELECT SIMPSON PAI23 w/ 9-10d x 1 1/2 Nails @ 36 in o.c.DESIGN FORCE FOR STEEL ANCHOR (UBC97,SEC 1633.2.8.1, ITEM 4) : 1764 lbf < 1880 lbf, allowableDESIGN FORCE FOR WOOD MEMBER (UBC97SEC. 1633.2.8.1, ITEM 5) : 1071 lbf < 1128 lbf, allowable

[Satisfactory]

Connection from Diaphragm, Ledger to CMU Wall

DanielT. Li

1.4FanchS =0.85FanchS =

( )

' '

12 2

,Z ZMINSD L D L V

θ⊥ = + + +

2' '2

3e

vdbdV Fd

=

CHECK WALL CAPACITY TO SPAN LATERALLY S SPACING cont'dALLOWABLE MASONRY STRESS FACTOR : SF = 0.667Allowable reinf. stress Fs = 32000 psi Modular ratio n = 21.48Allowable stress Fb=(SF)(0.33fm') = 330 psi Wall reinf. area As = 0.23 in / ftMasonry elasticity modulus Em = 1350 ksi Tension reinf. ratio ρ = 0.005Steel elasticity modulus Es = 29000 ksi The neutral axis depth factor k = 0.36Effective width bw = 12 in The lever-arm factor j = 0.88The tensile stress in reinforcement is The compressive stress in the extreme fiber is

6938 psi < Fs 185 psi < Fb



2

8anch

ss

SFfjdA

= =2

24anch

bw

SFfjkb d

= =



INPUT DATATOTAL SHEAR FORCE (ASD) Fp = 15.16 kips

NUMBER OF SEGMENTS n = 5

Segment 1 2 3 4 5Length, ft 17.5 22.67 13.83 35.5 10

Shear Wall ? Yes No Yes No Yes

ANALYSISTOTAL DRAG LENGTH Ldrag = 99.5 ft

TOTAL SHEAR WALL LENGTH Lwall = 41.33 ft

DIAPHRAGM SHEAR STRESS vdiaphragm = Fp / vdrag = 152 plf

SHEAR WALL SHEAR STRESS vshear wall = Fp / vwall = 367 plf

Section Point 0 1 2 3 4 5Distance, ft 0 17.5 40.17 54 89.5 99.5Axial Force 0 3.75 0.30 3.26 -2.14 0.00

DRAG / COLLECTOR FORCE DIAGRAM

SHEAR WALL & DRAG ELEVATION

DanielTian Li

Drag / Collector Force Diagram Generator

-3-2-1012345

Distance, ft

Dra

g A

xial

For

ce, k

ips


JOB NO. : DATE: REVIEW BY : Joist Design Based on AISI 2001 & ICBO ER-4943P


SECTION & SPACING 600S162-54 @ 24 in o.c( 50 ksi )

JOIST SPAN L = 11.5 ft

DEAD LOAD DL = 20 psfLIVE LOAD LL = 20 psf

LATERAL SUPPORTED ? 1 compression flange (0=No, 1=compression flange, 2=tension flange)DEFLECTION LIMITATION FOR LIVE LOAD ? 1 L /240 (0=No., 1= L /240, 2= L /360, 3= L /180, 4= L /120) THE DESIGN IS ADEQUATE.

ANALYSISSECTION PROPERTIES OF EACH METAL STUD (SSMA page 7 & 8)

t = 0.0566 in Fy = 50 ksi Ixx = 2.86 in4Mn/Ωb = 27.76 in-kips

h = 6 in Wt = 0.556 lb/ft Sxx = 0.927 in3Vn/Ωv = 2708 lbs

A = 0.556 in2rx = 2.267 in ry = 0.57 in xo = -1.072 in

J = 0.000594 in4Cw = 1.318 in6

CHECK MAX WEB DEPTH-TO-THICKNESS RATIO (AISI B1.2)h / t = 106.01 < 200 [Satisfactory]

CHECK FLEXURAL CAPACITY (AISI C3.1)

Mn/ΩΩΩΩb = 27.76 in-kips > M [Satisfactory]

Where M = [(DL+LL) S + Wt ] L2 / 8 = 15.98 in-kipsS = 24 in o.c., spacing as gaven.

CHECK SHEAR CAPACITY (AISI C3.2)

Vn/ΩΩΩΩV = 2708 lbs > V [Satisfactory]Where V = [(DL+LL) S + Wt ] L / 2 = 463 lbs

CHECK LATERAL-TORSIONAL BUCKLING (AISI C3.1.2) CASE 1: BOTH TOP & BOTTOM FLANGES UNSUPPORTED <== Does not apply.

9.0 ksi < 2.78 Fy = 139.0 ksi

< 0.56 Fy = 28.0 ksi

Where Cb = 1.0

ro = (rx2 + ry

2 + xo2)0.5 = 2.572 in

Sf = 0.95 in34.967 ksi

E = 29500 ksi (AISI pg 18)G = 11300 ksi (AISI pg 21)

Ky = 1.0

Kt = 1.0 7.305 ksi

Ly = 138 in

Lt = 138 in

= 9.0 ksi

DanielTian Li

AC robFe ey tS fσ σ= =

( )2

2/

Eey

K L ry y y

πσ = =

( )21

2 2ECwGJt

Aro K Lt t

πσ = + =

, 2.78

10101 , 2.78 0.56

9 36

, 0.56

forF F Fy e e

F y forF F F Fc y e eFe

forF F Fe e e

≥

= − > ≥ ≤

(cont'd)

Mn/ΩΩΩΩb = 5.16 in-kips < M [Satisfactory]

Where Sc = 0.95 in3 from SSMA page 7 & 8

Ωb = 1.67

Mn = ScFc = 8.61 in-kips

M = [(DL+LL) S/12 + Wt ] L2 / 8 = 15.98 in-kips

CASE 2: BOTTOM FLANGE SUPPORTED ONLY <== Does not apply.Mn/ΩΩΩΩb = 16.65 in-kips > M [Satisfactory]

Where Se = 0.93 in3, from Sxx

Ωb = 1.67R = 0.60

Mn = RSeFy = 27.81 in-kips

CHECK CAPACITY COMBINED BENDING & SHEAR AT ANY SAME SECTION (AISI C3.3.1)

= 0.3606 < 1.0 [Satisfactory]

Where M = 15.98 in-kipsV = 463 lbs

Vn/Ωv = 2708 lbs

Ωb = 1.67

Mn = MIN( Bending , Buckling) = 46.359 in-kips

Mn/Ωb = 27.76 in-kips, from SSMA page 7 & 8 for bending only.

= 0.58 > 0.5 = 0.17 < 0.7


CHECK DEFLECTION

0.19 in < L /240 = 0.58 in [Satisfactory]

DETERMINE SCREWS AT EACH LEG OF CONNECTION (SSMA page 48)

Vmax = 463 lbs

vallow = 344 lbs / screw, for # 8 screws. = = > (2)- # 8 screws required.370 lbs / screw, for # 10 screws. = = > (2)- # 10 screws required.384 lbs / screw, for # 12 screws. = = > (2)- # 12 screws required.

Techincal References: 1. AISI STANDARD, 2001 Edition. American Iron and Steel Institute. 2. SSMA, Product Technical Information, ICBO ER-4943P, Steel Stud Manufactures Association, 2001.

22M Vb v

VM n n

Ω Ω+

0.6M Vb v

VM n n

Ω Ω+

MbM n

Ω

VvV n

Ω

( ) 45

384

LL S LLL

EI xx= =∆


JOB NO. : DATE: REVIEW BY :


VERT. MEMBERS 3 x 1000S250-68 (TOTAL SECTION: 8 x 10 , 50 ksi )

SPAN L = 10 ft

DEAD LOAD DL = 0.2 kips / ftLIVE LOAD LL = 0.36 kips / ft

COMPRESSION FLANGE SUPPORTED ? (0=No, 1=Yes) 0 No.

DEFLECTION LIMITATION FOR LIVE LOAD ? 1 L /240 (0=No., 1= L /240, 2= L /360, 3= L /180, 4= L /120) THE DESIGN IS ADEQUATE.

ANALYSISSECTION PROPERTIES OF EACH VERTICAL STUD (SSMA page 7 & 8)

t = 0.0713 in Fy = 50 ksi Ixx = 15.751 in4Mn/Ωb = 79.94 in-kips

h = 10 in Wt = 1.121 lb/ft Sxx = 2.67 in3Vn/Ωv = 3209 lbs


J = 0.001899 in4Cw = 15.726 in6

CHECK MAX WEB DEPTH-TO-THICKNESS RATIO (AISI B1.2)h / t = 140.25 < 200 [Satisfactory]


Mn/ΩΩΩΩb = 19.99 ft-kips > M [Satisfactory]

Where M = (DL + LL + Wt ) L2 / 8 = 7.04 ft-kips


Vn/ΩΩΩΩV = 9.63 kips > V [Satisfactory]Where V = (DL + LL + Wt ) L / 2 = 2.82 kips

CHECK LATERAL-TORSIONAL BUCKLING (AISI C3.1.2.2)

1.38 ft < L

Where Cb = 1.0

Sf = 9.45 in3 (total vertical studs, SSMA page 7 & 8.)

E = 29500 ksi (AISI page 18)G = 11300 ksi (AISI page 21)

Iy = 25.213 in4 (neglecting top & bottom tracks conservatively.)

J = 0.006 in4

19.2 ksi < 2.78 Fy = 139.0 ksi

< 0.56 Fy = 28.0 ksi

Where Ky = 1.0

Ly = 120 in

= 19.2 ksi

Box Beam Design Based on AISI 2001 & ICBO ER-4943P

DanielTian Li

Cb EGJF Ie ySK Ly y f

π= =

, 2.78

10101 , 2.78 0.56

9 36

, 0.56

forF F Fy e e

F y forF F F Fc y e eFe

forF F Fe e e

≥

= − > ≥ ≤

0.36Cb EGJL Iu ySF y f

π= =

(cont'd)


Where Sc = 9.45 in3 (total vertical studs, SSMA page 7 & 8.)

Ωb = 1.67

Mn = ScFc = 181.16 in-kips

M = (DL + LL + Wt ) L2 / 8 = 7.04 ft-kips



Where M = 7.04 ft-kipsV = 2.82 kips

Vn/Ωv = 9.63 kips

Ωb = 1.67

Mn = MIN( Bending , Buckling) = 15.10 ft-kips

Mn/Ωb = 19.99 ft-kips, for bending only.

= 0.78 > 0.5 = 0.29 < 0.7


CHECK DEFLECTION

0.06 in < L /240 = 0.50 in [Satisfactory]


22M Vb v

VM n n

Ω Ω+

0.6M Vb v

VM n n

Ω Ω+

MbM n

Ω

VvV n

Ω

( ) 45384LL L

LL EI= =∆


JOB NO. : DATE: REVIEW BY : Wall Stud Design Based on AISI 2001 & ICBO ER-4943P


SECTION & SPACING 600S162-97 @ 16 in o.c( 50 ksi )

WALL HEIGHT h = 16 ft

hp = 4 ft

SERVICE GRAVITY LOAD P = 1500 lbs / ft

SERVICE LATERAL LOAD w1 = 25 psf

SERVICE PARAPET LOAD w2 = 45 psfECCENTRICITY e = 0 in

DEFLECTION LIMITATION ? 1 h /240 (0=No., 1= h /240, 2= h /360, 3= h /180, 4= h /120)

THE DESIGN IS ADEQUATE.ANALYSISSECTION PROPERTIES OF EACH METAL STUD (SSMA page 7 & 8)

thk = 0.1017 in Fy = 50 ksi Ixx = 4.797 in4Mn/Ωb = 56.73 in-kips

t = 6 in Wt = 0.966 lb/ft Sxx = 1.599 in3Vn/Ωv = 11124 lbs


J = 0.003329 in4Cw = 2.093 in6

CHECK MAX WEB DEPTH-TO-THICKNESS RATIO (AISI B1.2)t / (thk) = 59.00 < 200 [Satisfactory]


Mn/ΩΩΩΩb = 56.73 in-kips / stud > M [Satisfactory]

Where 7.50 ft

11.25 in-kips / stud

S = 1.333 ft o.c., spacing as gaven.

5.76 in-kips / stud

M = 0.75 MAX( M1 , M2) = 8.44 in-kips / stud, (0.75 for wind/seismic, from AISI App. A4.1.2, typical)


Vn/ΩΩΩΩV = 11124 lbs / stud > V [Satisfactory]

Where 250 lbs / stud

V2 = h w1 S - V1 = 283 lbs / stud

V3 = hp w2 S = 240 lbs / stud

V = 0.75 MAX( V1 , V2 , V3) = 213 lbs / stud

CHECK CAPACITY COMBINED BENDING & SHEAR AT ROOF/FLOOR SECTION (AISI C3.3.1)


Where M = 4.32 in-kips / stud, (0.75 included)V = 213 lbs / stud, (0.75 included)

Vn/Ωv = 11124 lbs / stud, from SSMA page 7 & 8.

Mn/Ωb = 56.73 in-kips / stud, from SSMA page 7 & 8.

DanielTian Li

22M Vb v

VM n n

Ω Ω+

( )2

12p

p

h h Pex h h

h hw

+ = + − − =

( )2

1 2 21 2

1

122 pwPe Sh hM

w h = + − =

22

22

pw h SM = =

( ) ( )21

1 1 2p

p

h h w Peh SV h w

h h

+ = + − + =

(cont'd)

= 0.08 < 0.5 = 0.02 < 0.7


CHECK COMPRESSION CAPACITY WITH, AT LEAST, ONE FLANGE THOUGH-FASTENED TO SHEATHING (AISI C4.6)

Pn/ΩΩΩΩc = 9.12 kips / stud > P [Satisfactory]

Where Ωc = 1.8

Pn = C1C2C3 AE / 29500 = 16.42 kips / studC1 = (0.79 x + 0.54) = 0.949C2 = (1.17 α t + 0.93) = 1.049C3 = α (2.5b - 1.63d) + 22.8 = 17.070E = 29500 ksi (AISI pg 18)P = 2.30 kips / stud (included wall weight, 18psf.)

CHECK CAPACITY COMBINED AXIAL LOAD & BENDING (AISI C5.2.1)


Where M = 8.44 in-kips / stud, (0.75 included)P = 2.30 kips / stud

Pn/Ωc = 9.12 kips / stud

Mn/Ωb = 56.73 in-kips / stud

Cm = 1.0

37.89 kips / stud

0.891

CHECK DEFLECTION

0.35 in < h /240 = 0.80 in [Satisfactory]

NOTE : 1. STUD FLANGES SHALL BE FASTENED TO SHEATHING AT EACH SIDE OF WALL BEFORE VERICAL LOAD ADDED.2. THE LATERAL LOADS MAY BE REDUCED BY 0.75 PER AISI APPENDIX A.4.1.2 .


0.6M Vb v

VM n n

Ω Ω+

MbM n

Ω

VvV n

Ω

( ) 45 1384

Sw LEI xx

∆ = =

MP Cbc mP Mn nα

ΩΩ + =

( )2

2E I xPEx

K Lx x

π= =

1Pc

PExα Ω= − =


JOB NO. : DATE: REVIEW BY :


VERT. MEMBERS 3 x 600S162-68 ( 50 ksi ) (TOTAL SECTION: 6 x 5 , INSIDE 6 in THK. WALL)

0.36HEIGHT h = 9 ft

SERVICE GRAVITY LOAD P = 6.8 kipsSERVICE LATERAL LOAD w = 0.33 kips / ft

DEFLECTION LIMITATION ? 1 h /240 (0=No., 1= h /240, 2= h /360, 3= h /180, 4= h /120) THE DESIGN IS ADEQUATE.

ANALYSISSECTION PROPERTIES OF EACH VERTICAL STUD (SSMA page 7 & 8)

thk = 0.0713 in Fy = 50 ksi Ixx = 3.525 in4Mn/Ωb = 39.46 in-kips

t = 6 in Wt = 0.693 lb/ft Sxx = 1.164 in3Vn/Ωv = 5468 lbs


J = 0.001174 in4Cw = 1.596 in6

CHECK MAX WEB DEPTH-TO-THICKNESS RATIO (AISI B1.2)t / (thk) = 84.15 < 200 [Satisfactory]



Where M = 0.75 w h2 / 8 = 2.51 ft-kips, (0.75 for wind/seismic, from AISI App. A4.1.2, typical)


Vn/ΩΩΩΩV = 16.40 kips > V [Satisfactory]Where V = 0.75 w L / 2 = 1.11 kips

CHECK COMPRESSION CAPACITY WITH, AT LEAST, ONE FLANGE THOUGH-FASTENED TO SHEATHING (AISI C4.6)

Pn/ΩΩΩΩc = 18.97 kips > P [Satisfactory]

Where Ωc = 1.8

Pn = C1C2C3 AE / 29500 = 34.14 kipsC1 = (0.79 x + 0.54) = 0.949C2 = (1.17 α t + 0.93) = 1.013C3 = α (2.5b - 1.63d) + 22.8 = 17.070E = 29500 ksi (AISI pg 18)P = 6.82 kips (included studs weight.)



Where M = 1.88 ft-kips, (0.75 included)V = 0.84 kips, (0.75 included)

Vn/Ωv = 16.40 kips

Mn/Ωb = 9.87 ft-kips

= 0.19 < 0.5 = 0.05 < 0.7


Jamb/Column Design Based on AISI 2001 & ICBO ER-4943P

DanielTian Li

22M Vb v

VM n n

Ω Ω+

0.6M Vb v

VM n n

Ω Ω+

MbM n

Ω

VvV n

Ω

(cont'd)CHECK CAPACITY COMBINED AXIAL LOAD & BENDING (AISI C5.2.1)


Where M = 1.88 ft-kips, (0.75 included)P = 6.82 kips

Pn/Ωc = 18.97 kips

Mn/Ωb = 9.87 in-kips

Cm = 1.0

263.97 kips

0.954

CHECK DEFLECTION

0.16 in < h /240 = 0.45 in [Satisfactory]

NOTE : 1. STUD FLANGES SHALL BE FASTENED TO SHEATHING AT EACH SIDE OF WALL BEFORE VERICAL LOAD ADDED.2. THE LATERAL LOADS MAY BE REDUCED BY 0.75 PER AISI APPENDIX A.4.1.2 .


45384

wLEI

∆ = =

MP Cbc mP Mn nα

ΩΩ + =

( )2

2E I xPEx

K Lx x

π= =

1Pc

PExα Ω= − =


JOB NO. : DATE: REVIEW BY : Brace Design Based on AISI 2001 & ICBO ER-4943P


SECTION & SPACING 350S125-43 @ 60 in o.c

BRACE LENGTH L = 12 ftBRACE SLOPE 12 / 12

WALL LATERAL LOAD, ASD Fp = 5 psfWALL HEIGHT H = 16 ft

THE DESIGN IS ADEQUATE.

ANALYSIS

Check Brace Compression Capacity (AISI C4.6)P = Fp ( 0.5 H) S Cos α = 0.283 kips / bracePn/ΩΩΩΩc = 0.31 kips / brace > P [Satisfactory]

Where Ωc = 1.8

Ae = 0.272 in2 (SSMA page 6-7, ICBO ER-4943P)ry = 0.41 in (SSMA page 6-7, ICBO ER-4943P)

E = 29500 ksi (AISI pg 18)Fe = π2 E / (KL / ry)

2 = 2 ksiFy = 33 ksi

λC = (Fy / Fe)0.5 = 3.74

2.1 ksi

Pn = Ae Fn = 0.56 kips / brace

NOTE : THE LATERAL LOADS MAY BE REDUCED BY 0.75 PER AISI APPENDIX A.4.1.2 .


DanielTian Li

2

2

0.658 , 1.5

0.877, 1.5

yc c

ny c

c

forFF

forF

λ λ

λλ

≤= = >


JOB NO. : DATE: REVIEW BY : Connection Design of Jamb to Track Based on AISI 2001 & ICBO ER-4943P


JAMB MEMBERS 3 x 600S250-43 (TOTAL SECTION: 6 x 8 , INSIDE 6 in THK. WALL)

TOP TRACK 600T200-54 ( 16 GA , 50 ksi )

JAMB LATERAL LOAD F = 0.2 kips 11.5


ANALYSISSECTION PROPERTIES OF TOP TRACK (SSMA page 10 & 11)

t = 2 in, leg length

Fy = 50 ksi

thk = 0.0566 in, metal thicknesswall = 6 in, wall thickness < track width

[Satisfactory]CHECK BENDING CAPACITY OF TRACK LEG

d = 8 in, jamb widthb = d + 2 t (tan 60o) = 14.9 in, effective widthM = F (t + 1/4") / 2 = 0.2 in-kipsS = b (thk)2 / 6 = 0.0080 in3

fb = M / S = 28 ksi < (4/3) Fy [Satisfactory]

(If jamb lateral load have reduced 0.75, the factor 4/3 does not apply. AISI App. A4.1.2)


DanielTian Li


JOB NO. : DATE: REVIEW BY : Seismic Design for Ordinary Concentrically Braced Frames Based on IBC & AISC Seismic


BRACE SECTION (Tube or Pipe) = > HSS5X5X3/8 Tube A rmin t hBRACE AXIAL LOAD AT SERVICE LEVEL D = 20 kips 6.18 1.87 0.35 5.00

L = 10 kipsBRACE AXIAL LOAD AT HORIZ. SEISMIC QE = 50 kips (IBC 1617.1)

SEISMIC PARAMETER SDS = 0.533 (IBC 1615.1.3) THE DESIGN IS ADEQUATE.UNBRACED LENGTH OF THE BRACE = 14.142 ftBUILDING LIMITATION FOR SDC D or E H = 10 ft (< 35 ft, IBC 2003)REQUIRED CONNECTION = > ( 1/2 in Gusset Plate with 17 in Length, 4 leg, 1/4 in Fillet Weld. Cover Plate 3/4 x 4 at Each Sides.)

CHECK LIMITING WIDTH THICKNESS RATIO λps FOR COMPRESSION ELEMENT, LOCAL BUCKLING (AISC Seismic 02 Tab. I-8-1)

D / t = 0.044 Es / Fy = 36.46 , for Pipe

( D / t = 1300 / Fy for AISC-Seismic 97, Tab. 1-9-1)

h / t = 0.64 (Es / Fy)0.5 = 16.07 , for Tube

[ h / t = 110 / (Fy)0.5 for AISC-Seismic 97, Tab. 1-9-1]

Where Fy = 46 ksi

Es = 29000 ksi

CHECK LIMITING SLENDERNESS RATIO FOR V OR INVERTED-V CONFIGURATIONS (AISC Seismic 02 Sec. 14.2)4.23 (Es / Fy) = 106.2 > K / r = 90.6 [Satisfactory]

[ 720 / (Fy)0.5 for AISC-Seismic 97, Sec. 14.2]

Where K = 1.0

DETERMINE FACTORED DESIGN LOADS (IBC 1617.1.2, AISC Seiemic 02 Tab. C-I-4.1)

Put = 0.9D - Ω0QE - 0.2SDSD = -84.13 kips (Tension)

Puc = 1.2D + f1L + Ω0QE + 0.2SDSD = 131.13 kips (Compression, Governs)

Where Ω0 = 2 (IBC Tab. 1617.6, AISC Seiemic Tab. I-4.1, Pg. 7)f1 = 0.5 (IBC 1605.4)

(Note: the special seismic load combinations above must be used to determine all member and connection forces.)

CHECK DESIGN STRENGTH IN COMPRESSION (LRFD Sec.E2)

φcPn = φcAgFcr = 139.17 kips > Puc [Satisfactory]

Where φc = 0.85

λc = K / (rπ) (Fy / E)0.5 = 1.15 (0.658λc^2 )Fy = 26.49 kis, for λc <1.5

0.877 / (λc2 )Fy = N/A kis, for λc >1.5

DETERMINE CONNECTION DESIGN FORCE (AISC Seismic 02 & 97 Sec. 14.2)

Put = RyFyAg = 369.56 kips (Tension)

Where Ry = 1.3 (AISC Seiemic 02 & 97 Tab. I-6-1)( 1.4 for Pipe)

DETERMINE BEST FILLET WELD SIZE (LRFD Sec.J2.2b) 50> wMIN = 0.1875 in< wMAX = 0.25 in

[Satisfactory]DETERMINE REQUIRED WELD LENGTH (LRFD Sec.J2.4)

L = Put / [(4) φ Fw (0.707 w)] = 369.56 / [(4) 0.75 (0.6x70)(0.707x1/4)] = 16.59 in

( USE 17 in )CHECK DESIGN SHEAR RUPTURE CAPACITY OF SLOTED BRACE (LRFD Sec.J4.1)

φPn = φ(0.6Fu)Anv = 619.41 kips > PutWhere φ = 0.75 [Satisfactory]

Fu = 58 ksi (LRFD Tab.1-4, Pg. 1-21)

Anv = 4 t L = 4 x 0.349 x 17 = 23.73 in2

DETERMINE REQUIRED THICKNESS OF GUSSET PLATE (LRFD Tab. J2.4)

tg = 1/2 inCHECK SHEAR RUPTURE CAPACITY OF GUSSET PLATE (LRFD Sec.J4.1)

φPn = φ(0.6Fu)Anu = 443.70 kips > Put [Satisfactory]Where φ = 0.75

Fu = 58 ksi (A36 Steel)

Anu = 2 tg L = 2 x 1/2 x 17 = 17.00 in2

w = 1/4 in

Fcr =

> Actual [Satisfactory]

DanielTian Li

(cont'd)CHECK TENSION CAPACITY AT SLOTED BRACE (LRFD Sec.D1).

φPn = φ Fu Ae = 222.30 kips < Put [Cover Plate Required]Where φ = 0.75

Fu = 58 ksi (LRFD Tab.1-4, Pg. 1-21)x = 3 h / 8 = 1.88 , for Tube (HSS Spectification 2.1-4)

D / π = 1.59 , for Pipe (HSS Spectification 2.1-3)U = MIN( 1 - x / L , 0.9 ) = 0.89 ,(LRFD B3.)

An = Ag - 2 (tg + 1/8) t = 5.74 in2

Ae = U An = 5.11 in2

Try Cover Plate 3/4 x 4 , at Each Sides.

Region x 0.5 An x A

HSS 1.88 2.87 5.38 x = 14.01 / 5.87 = 2.39Cover Plate 2.88 3.00 8.63 U = MIN( 1 - x / L , 0.9 ) = 0.86

Σ 5.87 14.01 An = 5.74 + 6.00 = 11.74 in2

Ae = U An = 10.10 in2

Thus, φPn = φ Fu Ae = 439.16 kips > Put [Satisfactory]

Where Fu = 58 ksi, use plate value

Techincal References: 1. Rafael Sabelli: "Structuiral Engineering Review Workshop", BYA publications, 2005. 2. Alan Williams: "Seismic and Wind Forces, Structural Design Examples", International Code Council, 2003. 3. SEAOC: "2000 IBC Structural/Seismic Design Manual - Volume 3", International Code Council, 2003. 4. AISC: "Seismic Provisions for Structural Steel Buildings", American Institute of Steel Construction, May 1, 2002.


JOB NO. : DATE : REVIEW BY : Seismic Design for Ordinary Concentrically Braced Frames Based on CBC 2001

INPUT DATA & DESIGN SUMMARYBRACE SECTION (Tube or Pipe) = > HSS8X8X5/8 Tube A rmin t hBRACE AXIAL LOAD AT SERVICE LEVEL D = 24 kips 16.40 2.98 0.58 8.00

L = 11 kips

BRACE AXIAL LOAD AT HORIZ. SEISMIC Eh = 40 kips (CBC 30A-1) THE BRACE DESIGN IS ADEQUATE.SEISMIC COEFFICIENT Ca = 0.44 (CBC Tab. 16A-Q)IMPORTANCE FACTOR I = 1.15 (CBC Tab. 16A-K)REDUNDANCY FACTOR ρ = 1.5UNBRACED LENGTH OF THE BRACE = 18.5 ftCHEVRON BRACING ? = > Yes (CBC 2213A.8.4.1)SUM OF HORIZ. FORCES EXCEED 70% ? = > No (CBC 2213A.8.2.3)

REQUIRED CONNECTION = > ( 1/2 in Gusset Plate with 5 in Length, 4 leg, 5/16 in Fillet Weld.)

CHECK LIMITING WIDTH THICKNESS RATIO FOR COMPRESSION ELEMENT, LOCAL BUCKLING (CBC Sec. 2213A.8.2.5)

D / t = 1300 / Fy = 37.14 , for Pipe

h / t = 110 / (Fy)0.5 = 16.22 , for Tube

[Note: For building of two stories or less, this condition need not be met if ΩoPc < (1.7)Pc,allow. (CBC 2213A.8.5) ]

CHECK LIMITING SLENDERNESS RATIO (CBC 2213A.8.2.1)720 / (Fy)

0.5 = 106.2 > K / r = 74.4 [Satisfactory]

DETERMINE ALLOWABLE DESIGN LOADS (CBC 1612A.3.1, 2213A.8.2, & 2213A.8.4.1)

Pt = 0.9D - f (ρEh + 0.5CaID) / 1.4 = -49 kips (Tension)

Pc = D + f 0.75[L + (ρEh + 0.5CaID) / 1.4] = 89 kips (Compression)

Pc = D + f (ρEh + 0.5CaID) / 1.4 = 95 kips (Compression, Governs)

Where f = Ωo x 1.5 = 1.50 (1.5 for chevron bracing or force not exceed 70%, CBC 2213A.8.4.1)

Ωo = 1 (If force not exceed 70%, 1.0 should apply, CBC 2213A.8.2.3)

CHECK STRENGTH IN COMPRESSION (ASD Sec.E2)

Pc,allow = A(BFa) = 234.03 kips > Pc [Satisfactory]

Where K = 1.0Fy = 46 ksi (1-F2/2)Fy / (5/3+3F/8-F3/8) = 19.03 kis, for Cc > (K/r)

Es = 29000 ksi 12π2Es/[23(K/r)2] = N/A kis, for Cc < (K/r)

Cc = (2π2Es/Fy)0.5 = 112

K / r = 74

F = (K / r) / Cc = 0.67B = 1/( 1 + F/2) = 0.75 (If force exceed 70%, 1.0 should apply, CBC 2213A.8.2.3)

DETERMINE CONNECTION DESIGN FORCE (CBC Sec. 2213A.8.3.1)

Pconn= MAX[1.4Pc, MIN(FyA , D+L+ΩoEh)] = 133 kips (Compression)

Where Ωo = 2.2 (CBC Tab. 16A-N)

DETERMINE BEST FILLET WELD SIZE (ASD Sec.J2.2b)> wMIN = 0.1875 in 28< wMAX = 0.4375 in

[Satisfactory]DETERMINE REQUIRED WELD LENGTH (ASD Sec.J2.4 )

L = Pconn / [(1.7)(4) (0.3) Fu (0.707 w)] = 132.71 / [(1.7)(4) (0.3) (70) (0.707x5/16)] = 4.21 in

( USE 5 in )

DanielTian Li

w = 5/16 in

Fa =


(cont'd)CHECK SHEAR RUPTURE CAPACITY OF SLOTED BRACE (ASD Sec.J4)

Pt,rup,brace =1.7(0.3Fu)Anu = 343.72 kips > Pconn[Satisfactory]

Where Fu = 58 ksi (LRFD Tab.1-4, Pg. 1-21)

Anu = 4 t L = 4 x 0.581 x 5 = 11.62 in2

DETERMINE REQUIRED THICKNESS OF GUSSET PLATE (ASD Tab. J2.4)

tg = 1/2 inCHECK SHEAR RUPTURE CAPACITY OF GUSSET PLATE (ASD Sec.J4 & CBC 2213A.4.2)

Pt,rup,gusset =1.7(0.3Fu)Anv = 147.9 kips > Pconn [Satisfactory]

Where Fu = 58 ksi (plate value)

Anv = 2 tg L = 2 x 1/2 x 5 = 5.00 in2

CHECK TENSION CAPACITY AT SLOTED BRACE (ASD Sec.D1). CHECK GUSSET BLOCK SHEAR CAPACITY (ASD J4)Pt,brace = 1.7(0.5FuUAn) = 590.28 kips > Pconn Ps,guss = 1.7[0.3FuAnv + 0.5FuAgt] =

[Satisfactory] = 148 +1.7[0.5FuAgt]

Where U = 0.75 (ASD Sec.B3) > Pconn = 132.7 [Satisfactory]

An = A - 2 t tg = 15.964 in2

Techincal References: 1. ICBO: "2001 California Building Code, Title 24, Part 2, Volume 2", 2002. 2. SEAOC: "Seismic Design Manual - Volume 3", International Code Council, 2000. 3. AISC: "Manual of Steel construction 9th", American Institute of Steel Construction, 1990.



INPUT DATA & DESIGN SUMMARYBEAM SECTION = > W16X77 = > A d tw bf tf SxBEAM DISTRIBUTED SERVICE LOADS D = 1 kips / ft 22.9 16.5 0.46 10.30 0.76 136

L = 0.5 kips / ft Ix rx ry Zx kBEAM LENGTH = 28 ft 1120 6.99 2.45 152 1.47

BEAM YIELD STRESS Fy = 50 ksi

LATERALLY UNBRACED LENGTH b = 14 ft THE BEAM DESIGN IS ADEQUATE.

CHECK LOCAL BUCKLING LIMITATION (ASD Tab. B5.1)

bf / (2tf ) = 6.78 < 65 / (Fy)0.5 = 9.19 [Satisfactory]

d / tw = 36.26 < 640 / (Fy)0.5 = 90.51 [Satisfactory]

THE BEAM SHOULD BE DESIGNED AS IF BRACES DO NOT EXIST (CBC Sec.2213A.8.4.1-3)M = (D + L) 2/ 8 = ft-kips

DETERMINE GOVERNING UNBALANCED SEGMENT LENGTH (AISC-ASD F1)c = MIN[76bf/(Fy)

0.5 , 20000/(d/Af)Fy] = 9.23 ft

u = MAX[rT(102000Cb/Fy)0.5 , 12000Cb/(d/Af)0.6Fy] = 15.81 ft

3 = rT(510000Cb/Fy)0.5 = 22.04 ft

Where (d/Af) = 2.11 in-1

rT = 2.62

Cb = 1.00

CHECK ALLOWABLE BENDING STRESS

= 0.66Fy = N/A ksi, for Lb @ [0, Lc]

= 0.60Fy = 30.00 ksi, for Lb @ (Lc, Lu]

= MAX(Fb1, Fb3) = N/A ksi, for Lb @ (Lu, L3]

= MAX(Fb2, Fb3) = N/A ksi, for Lb @ (L3, Larger)

Where Fb1 = MIN[2/3 - Fy(L/rT)2/(1530000Cb)]Fy , 0.6Fy = 26.61 ksi

Fb2 = MIN[170000Cb/(L/rT)2, Fy/3] = 16.67 ksi

Fb3 = MIN[12000Cb/(Ld/Af), 0.6Fy] = 30.00 ksi

CHECK FLEXURAL CAPACITY

Mallow = FbSx = 340 ft-kips > M [Satisfactory]


DanielTian Li

147.0

Fb =



INPUT DATA & DESIGN SUMMARYCOLUMN SECTION = > W12X50 = > A d tw bf tf SxCOLUMN AXIAL SERVICE LOADS D = 80 kips 14.6 12.2 0.37 8.08 0.64 64.2

L = 30 kips Ix rx ry Zx k

COLUMN AXIAL LOAD AT HORIZ. SEISMIC Eh = 110 kips (CBC 1630A.1.1) 391 5.18 1.96 72 1.14UNBRANCED COLUMN LENGTH = 14 ft

COLUMN YIELD STRESS (36 or 50) Fy = 50 ksi THE COLUMN DESIGN IS ADEQUATE.

DETERMINE DESIGN LOADS (CBC 2213A.5.1)

Pt = 0.85D - Ω0Eh = -174 kips (Tension)

Pc = D + 0.7L + Ω0Eh = 343 kips (Compression, Governs)

Where Ω0 = 2.2 (CBC Tab. 16A-N)

CHECK COMPRESSION CAPACITY (CBC 2213A.4.2 & ASD E2)

Pc,allow = 1.7FaA = 443.8 kips > Pc [Satisfactory]

Where K = 1.0Es = 29000 ksi (1-F2/2)Fy / (5/3+3F/8-F3/8) = 17.88 kis, for Cc > (K/r)

Cc = (2π2Es/Fy)0.5 = 107 12π2Es/[23(K/r)2] = N/A kis, for Cc < (K/r)

MAX(K x/rx, K y/ry) = 85.55 < 200 [Satisfactory]

F = (K / r) / Cc = 0.80

Techincal References: 1. ICBO: "2001 California Building Code, Title 24, Part 2, Volume 2", 2002. 1.5 2. SEAOC: "Seismic Design Manual - Volume 3", International Code Council, 2000. 3. AISC: "Manual of Steel construction 9th", American Institute of Steel Construction, 1990.

DanielTian Li

Fa =


JOB NO. : DATE : REVIEW BY : Seismic Design for Special Concentrically Braced Frames Based on IBC & AISC Seismic


BRACE SECTION (Tube or Pipe) = > HSS10.750X0.500 Pipe A rmin t DBRACE AXIAL LOAD AT SERVICE LEVEL D = 29 kips 15.00 3.64 0.47 10.75

L = 19 kipsBRACE AXIAL LOAD AT HORIZ. SEISMIC QE = 204 kips (IBC 1617.1)

SEISMIC PARAMETER SDS = 1 (IBC 1615.1.3) THE BRACE DESIGN IS ADEQUATE.UNBRACED LENGTH OF THE BRACE = 18.8 ftREDUNDANCY FACTOR ρ = 1.26

LENGTH OF END BRACE TO JUNCTION Lg = 17 inREQUIRED CONNECTION = > ( 1 in Gusset Plate with 14 in Length, 4 leg, 1/4 in Fillet Weld.)

CHECK LIMITING WIDTH THICKNESS RATIO FOR COMPRESSION ELEMENT, LOCAL BUCKLING (AISC Seismic 02 Tab. I-8-1)

D / t = 0.044 Es / Fy = 36.46 , for Pipe

( D / t = 1300 / Fy for AISC-Seismic 97, Tab. 1-9-1)

h / t = 0.64 (Es / Fy)0.5 = 16.07 , for Tube

[ h / t = 110 / (Fy)0.5 for AISC-Seismic 97, Tab. 1-9-1]

Where Es = 29000 ksi

Fy = 35 ksi

CHECK LIMITING SLENDERNESS RATIO FOR V OR INVERTED-V CONFIGURATIONS (AISC Seismic 02 Sec. 13.2a)5.87 (Es / Fy)

0.5 = 169.0 > K / r = 61.9 [Satisfactory]

[ 1000 / (Fy)0.5 for AISC-Seismic 97, Sec. 13.2a]

Where K = 1.0

DETERMINE FACTORED DESIGN LOADS (IBC1605.2 & AISC Seiemic 02 Tab. C-I-4.1, Pg. 78)

Put = 0.9D - ρQE - 0.2SDSD = -236.74 kips (Tension)

Puc = 1.2D + f1L + ρQE + 0.2SDSD = 307.14 kips (Compression, Governs)

Where f1 = 0.5

CHECK DESIGN STRENGTH IN COMPRESSION (LRFD Sec.E2)

φcPn = φcAgFcr = 366.69 kips > Puc [Satisfactory]

Where φc = 0.85 (0.658λc^2 )Fy = 28.76 kis, for λc <1.5

λc = K / (rπ) (Fy / E)0.5 = 0.68 0.877 / (λc2 )Fy = N/A kis, for λc >1.5

DETERMINE CONNECTION DESIGN FORCE (AISC Seismic Sec. 13.3a)

Put = MIN(RyFyAg , Pmax) = 307.14 kips (Tension)

Where Ry = 1.4 (AISC Seiemic Tab. I-6-1, Pg. 8)

Pmax = 307.14 kips, (the max force, indicated by analysis, that can be transferredto the brace by the system.)

DETERMINE BEST FILLET WELD SIZE (LRFD Sec.J2.2b)> wMIN = 0.1875 in< wMAX = 0.3125 in


L = Put / [(4) φ Fw (0.707 w)] = 307.14 / [(4) 0.75 (0.6x70)(0.707x1/4)] = 13.79 in

( USE 14 in )CHECK SHEAR RUPTURE CAPACITY OF SLOTED BRACE

(LRFD Sec.J4.1) 14

φPn = φ(0.6Fu)Anu = 703.08 kips > PutWhere φ = 0.75 [Satisfactory]

Fu = 60 ksi (LRFD Tab.1-4, Pg. 1-21)

Anu = 4 t L = 4 x 0.465 x 14 = 26.04 in2

DETERMINE REQUIRED THICKNESS OF GUSSET PLATE (LRFD Tab. J2.4)tg = 1 in

DanielTian Li

w = 1/4 in

Fcr =


(cont'd)CHECK SHEAR RUPTURE CAPACITY OF GUSSET PLATE (LRFD Sec.J4.1)

φPn = φ(0.6Fu)Anv = 730.80 kips > Put [Satisfactory]Where φ = 0.75


Anv = 2 tg L = 2 x 1 x 14 = 28.00 in2

CHECK GUSSET BLOCK SHEAR CAPACITY (LRFD J4-3b) CHECK GUSSET COMPRESSION CAPACITY (LRFD E2)φRn = φ(0.6Fu)Anv + φFyAgt = 730.80 + φFyAgt φcPn = φcFcrLwtg = 633.22 kips > Put

> Put = 307.14 Where φc = 0.85 [Satisfactory][Satisfactory] K = 1.2

rg = tg / (12)0.5 = 0.29 in

CHECK GUSSET TENSION YIELDING CAPACITY (LRFD D1.a) K Lg / rg < 200 [Satisfactory]

φtPn = φtFyLwtg = 872.07 kips > Put λc = 0.793 (LRFD E2-4, Pg 6-47)

Where φt = 0.9 [Satisfactory] Fcr = 27.677 ksi (LRFD Sec.E2, Pg 6-47)

Fy = 36 ksi (plate value)

Lw = D +2 tan30o = 26.916 in

CHECK SHEAR LAG FRACTURE OF BRACE (LRFD Sec.D.1)

φPn = φ Fu Ae = 474.45 kips > Put [Satisfactory]Where φ = 0.75



An = Ag - 2 (tg + 1/8) t = 13.95 in2

Ae = U An = 10.54 in2

Try Cover Plate 1/2 x 5 , at Each Sides.

Region x 0.5 An x A


Σ 9.48 36.54 An = 13.95 + 5.00 = 18.95

Ae = U An = 13.73 in2

Thus, φPn = φ Fu Ae = 597.44 kips > Put [Satisfactory]





INPUT DATA & DESIGN SUMMARYBEAM SECTION = > W27X307 = > A d tw bf tf SxBEAM DISTRIBUTED SERVICE LOADS D = 0.2 kips / ft 90.4 29.6 1.16 14.40 2.09 887

L = 0.35 kips / ft Ix rx ry Zx kBEAM LENGTH L = 25 ft 13100 12.04 3.41 1030 2.88TOP FLANGE CONTINUOUSLY BRACED ? 0 NoBEAM YIELD STRESS Fy = 50 ksi

THE BEAM DESIGN IS ADEQUATE.DETERMINE FACTORED AXIAL LOAD ON THE BEAM (AISC Seiemic 02 & 97 Sec. 13.4a)

Pu = 0.5 (RyFyAg + 0.3φcPn) cos α = 280.92 kips

Where Ry = 1.4 (AISC Seiemic Tab. I-6-1, see brace sheet)

α = 48.33 o

Fy = 35 ksi (see brace sheet)

Ag = 15.00 in2 (see brace sheet)

φcPn = 366.69 kips (see brace sheet)

CHECK LOCAL BUCKLING LIMITATION (AISC Seiemic 02 Tab. I-8-1)bf / (2tf ) = 3.44 < 0.3 (Es / Fy)

0.5 = 7.22 [Satisfactory][ 52 / (Fy)

0.5 for AISC Seismic 97, Tab. I-9-1]


3.14(Es/Fy)0.5(1-1.54Pu/φbPy) = 67.58 , for Pu/φbPy < 0.125

[ 520 / (Fy)0.5(1-1.54Pu/φbPy) for AISC Seismic 97, Tab. I-9-1]

1.12(Es/Fy)0.5(2.33-Pu/φbPy) = N/A , for Pu/φbPy > 0.125

MAX[ 191 / (Fy)0.5(2.33-Pu/φbPy) , 253 / (Fy)

0.5] for AISC Seismic 97, Tab. I-9-1

[Satisfactory] Where φb = 0.9 , Py = FyA = 4520 kips

DETERMINE UNBALANCED VERTICAL FORCE ON BEAM (AISC Seiemic 02 & 97 Sec. 13.4a)Qb = (RyFyAg - 0.3φcPn) sin α = 466.83 kips (Vertical)

DETERMINE FACTORED MOMENT ON THE BEAMMnt = (1.2D + 0.5L) L 2/ 8 + Qb l/ 4 = ft-kips

1DETERMINE UNBALANCED SEGMENT LENGTH ABOUT X - AND Y - AXES

L x = 25 ft

L y = 12.5 ft ( AISC Seiemic Sec.13.4a-4 ,lateral supported at the intersection of chevorn braces

with axial load 0.02 Fy bf tf .)CHECK COMPRESSION CAPACITY (LRFD E2)

φcPn = φcFcrA = 3334.61 kips > Pu [Satisfactory]Where φc = 0.85

K = 1.0MAX(KL x/rx, KL y/ry) = 44.01 < 200 [Satisfactory]λc = 0.582 (LRFD E2-4, Pg 6-47)

Fcr = 43.40 ksi (LRFD E2.a or b, Pg 6-47)

DETERMINE FLEXURAL DESIGN STRENGTH (AISC-LRFD F1)L b = 12.50 ft

L p = 1.76 ry (E / Fyf)0.5 = 12.04 ft

L r = ry X1 [1 + (1 + X2 FL2 )0.5 ]0.5 / FL = 46.65 ft

M p = MIN( Fy Zx , 1.5 Fy Sx) = 4291.7 ft-kips

M r = FL Sx = 2956.7 ft-kips

M cr = Cb Sx ry X1 (2 + X12 X2 ry

2 / Lb2)0.5 / Lb = 23961 ft-kips

Where X 1 = π (0.5 E G J A)0.5 / Sx = 4312.8

X 2 = 4 Cw [Sx / (G J)]2 / Iy = 0.0005 Iy G J Cw

Fr = 10.00 ksi 1050 11200 101 199000

FL = MIN( Fyf - Fr , Fyw) = 40.00 ksi

C b = 1.30 , (AISC-LRFD Table 4.1)

Mp = N/A ft-kips, for L b @ [0 , L p]

M n = MINCb [Mp - (Mp - Mr) (Lb - Lp) / (Lr - Lp)] , Mp = 4291.7 ft-kips, for L b @ (L p , L r]

MIN(Mcr , Mp) = N/A ft-kips, for L b @ (L r , Larger)

φ b M n = 0.9 Mn = 3862.5 ft-kips

DanielTian Li

h / tw = 20.55 <

2950.11

(cont'd)

CHECK FLEXURAL CAPACITY (LRFD C1)Mu = B1Mnt = 2970.14 ft-kips < φbMnx = MIN(φbFyZ, φMxCb) = 3863 ft-kips

Where Pe1 = π2EsIx / (KL x)2 = 41661 kips [Satisfactory]

Cm = 1.0 (LRFD C1-3, Pg 6-41) Where φb = 0.9

B1 = Cm/(1 - Pu/Pe1) = 1.007 Cb = 1.67 (LRFD Tab 4-1, Pg 4-9)

CHECK INTERACTION CAPACITY (LRFD H1.1)For Pu/φcPn>0.2, Pu / φcPn + 8 / 9 (Mux / φbMnx) = N/A

For Pu/φcPn<0.2, Pu / (2φcPn) + Mux / φbMnx = 0.81


< 1 [Satisfactory]



INPUT DATA & DESIGN SUMMARYCOLUMN SECTION = > W14X132 = > A d tw bf tf SxCOLUMN AXIAL SERVICE LOADS D = 135 kips 38.8 14.7 0.65 14.70 1.03 209


COLUMN AXIAL LOAD AT HORIZ. SEISMIC QE = 274 kips (IBC 1617.1) 1530 6.28 3.76 234 1.63UNBRANCED COLUMN LENGTH = 14 ft


DETERMINE FACTORED DESIGN LOADS (IBC 1617.1.2, AISC Seiemic 02 Tab. C-I-4.1)

Put = 0.9D - ρQE - 0.2SDSD = -251 kips (Tension)

Puc = 1.2D + f1L + ρQE + 0.2SDSD = 551 kips (Compression, Governs)Where ρ = 1.26

f1 = 0.5 (IBC 1605.4)

CHECK LOCAL BUCKLING LIMITATION (AISC Seiemic 02 Tab. I-8-1)

bf / (2tf ) = 7.14 < 0.3 (Es / Fy)0.5 = 7.22 [Satisfactory]

[ 52 / (Fy)0.5 for AISC Seismic 97, Tab. I-9-1]


3.14(Es/Fy)0.5(1-1.54Pu/φbPy) = N/A , for Pu/φbPy < 0.125


1.12(Es/Fy)0.5(2.33-Pu/φbPy) = 54.33 , for Pu/φbPy > 0.125

MAX[ 191 / (Fy)0.5(2.33-Pu/φbPy) , 253 / (Fy)


[Satisfactory] Where φc = 0.9

Py = FyA = 1940 kips

CHECK COMPRESSION CAPACITY (LRFD E2)

φcPn = φcFcrA = 1424.83 kips > Pu [Satisfactory]

Where φc = 0.85K = 1.0MAX(K x/rx, K y/ry) = 44.70 < 200 [Satisfactory]

λc = 0.591 (LRFD E2-4, Pg 6-47)


CHECK AMPLIFIED SEISMIC LOAD EFFECTS FOR Pu / φ Pn > 0.4 (AISC Seismic 02 8.3, AISC Seismic 97 8.2 )

Puc / φ Pn = 0.39 < 0 [Amplified Seismic Load Do Not Need to Checke]

Puc = 1.2D + f1L + Ω0QE + 0.2SDSD = 754 kips < φcPn [Satisfactory]

Where Ω0 = 2


DanielTian Li

h / tw = 17.74 <


JOB NO. : DATE : REVIEW BY : Seismic Design for Special Concentrically Braced Frames Based on CBC 2001

INPUT DATA & DESIGN SUMMARYBRACE SECTION (Tube or Pipe) = > HSS8X8X5/8 Tube A rmin t hBRACE AXIAL LOAD AT SERVICE LEVEL D = 24 kips 16.40 2.98 0.58 8.00

L = 11 kips

BRACE AXIAL LOAD AT HORIZ. SEISMIC Eh = 228 kips (CBC 30A-1) THE BRACE DESIGN IS ADEQUATE.SEISMIC COEFFICIENT Ca = 0.44 (CBC Tab. 16A-Q)IMPORTANCE FACTOR I = 1.15 (CBC Tab. 16A-K)REDUNDANCY FACTOR ρ = 1.5UNBRACED LENGTH OF THE BRACE = 18.5 ft

LENGTH OF END BRACE TO JUNCTION Lg = 17 in

REQUIRED CONNECTION = > ( 1 in Gusset Plate with 17 in Length, 4 leg, 5/16 in Fillet Weld.)

THE LIMITING WIDTH THICKNESS RATIO FOR COMPRESSION ELEMENT, LOCAL BUCKLING (CBC Sec. 2213A.9.2.4)

D / t = 1300 / Fy = 37.14 , for Pipe

h / t = 110 / (Fy)0.5 = 16.22 , for Tube

Where Fy = 46 ksi

THE LIMITING SLENDERNESS RATIO FOR V OR INVERTED-V CONFIGURATIONS (CBC Sec. 2213A.9.2.1)1000 / (Fy)

0.5 = 147.4 > K / r = 74.4 [Satisfactory]Where K = 1.0

THE ALLOWABLE DESIGN LOADS (CBC Sec. 1612A.3.1)

Pt = 0.9D - (ρEh + 0.5CaID) / 1.4 = -227 kips (Tension)

Pc = D + 0.75[L + (ρEh + 0.5CaID) / 1.4] = 219 kips (Compression)

Pc = D + (ρEh + 0.5CaID) / 1.4 = 273 kips (Compression, Governs)THE DESIGN STRENGTH IN COMPRESSION (ASD Sec.E2)

Pc,allow = AFa = 312.07 kips > Pc [Satisfactory]


Cc = (2π2Es/Fy)0.5 = 112 (1-F2/2)Fy / (5/3+3F/8-F3/8) = 19.03 kis, for Cc > (K/r)

K / r = 74 12π2Es/[23(K/r)2] = N/A kis, for Cc < (K/r)

F = (K / r) / Cc = 0.67

THE CONNECTION DESIGN FORCE (CBC Sec. 2213A.9.3.1)

Pconn = MAX(1.4Pc , MIN(FyA , D+L+ΩoEh) = 537 kips (Tension)

Where Ωo = 2.2 (CBC Tab. 16A-N)THE BEST FILLET WELD SIZE (ASD Sec.J2.2b)

> wMIN = 0.1875 in< wMAX = 0.4375 in

[Satisfactory]THE REQUIRED WELD LENGTH (ASD Sec.J2.4 & CBC 2213A.4.2)

L = Pt / [(4) 1.7 (0.3) Fu (0.707 w)] = 536.60 / [(4) 1.7 (0.3) (70) (0.707x5/16)] = 17.01 in

( USE 17 in )THE DESIGN SHEAR RUPTURE CAPACITY OF SLOTED BRACE

(ASD Sec.J4 & CBC 2213A.4.2)Pt,rup,brace =1.7(0.3Fu)Anu = 1168.65 kips > Pt

[Satisfactory]Where Fu = 58 ksi (LRFD Tab.1-4, Pg. 1-21)

Anu = 4 t L = 4 x 0.581 x 17 = 39.51 in2

THE REQUIRED THICKNESS OF GUSSET PLATE (ASD Tab. J2.4)

tg = 1 in

DanielTian Li


w = 5/16 in

Fa =

(cont'd)THE DESIGN SHEAR RUPTURE CAPACITY OF GUSSET PLATE (ASD Sec.J4 & CBC 2213A.4.2)

Pt,rup,gusset =1.7(0.3Fu)Anv = 1005.7 kips > Pconn [Satisfactory]


Anv = 2 tg L = 2 x 1 x 17 = 34.00 in2

THE TENSION CAPACITY AT SLOTED BRACE (ASD Sec.D1). THE GUSSET BLOCK SHEAR CAPACITY (ASD J4)Pt,brace = 1.7(0.5FuUAn) = 751.23 kips > Pconn Ps,guss = 1.7[0.3FuAnv + 0.5FuAgt] =


Where U = 1 (ASD Sec.B3) > Pconn = 536.6 [Satisfactory]

An = A - 2 t tg = 15.238 in2

THE GUSSET TENSION YIELDING CAPACITY (CBC 2213A.4.2) THE GUSSET COMPRESSION CAPACITY (CBC 2213A.4.2)Pt,guss = FyLwtg = 994.68 kips > Pconn Pc,guss = 1.7FaLwtg = 768 kips > Pconn

[Satisfactory] [Satisfactory]Where Fy = 36 ksi (plate value) Where K = 1.2

Lw = D +2 tan30o = 27.63 in rg = tg / (12)0.5 = 0.29 in

K Lg / rg < 200 (ASD B7) [Satisfactory]

Cc = 126 (ASD E2, Pg 5-42)

Fa = 16.361 ksi (ASD E2, Pg 5-42)




INPUT DATA & DESIGN SUMMARYBEAM SECTION = > W24X279 = > A d tw bf tf SxBEAM DISTRIBUTED SERVICE LOADS D = 1 kips / ft 82 26.7 1.16 13.30 2.09 718

L = 0.5 kips / ft Ix rx ry Zx kBEAM LENGTH = 28 ft 9600 10.82 3.17 835 2.59TOP FLANGE CONTINUOUSLY BRACED ? 1 Yes


THE BEAM DESIGN IS ADEQUATE.THE UNBALANCED AXIAL LOAD ON THE BEAM

Pc = 0.5[FyAg + 1.7(0.3Pc,allow)] cos α = 345.7 kips

Where α = 40.82 o

Fy = 46 ksi (see brace sheet)

Ag = 16.40 in2 (see brace sheet)

Pc,allow = 312.07 kips (see brace sheet)

THE LOCAL BUCKLING LIMITATION (ASD Tab. B5.1) < = does not apply for top flange continuously braced.

bf / (2tf ) = 3.18 < 65 / (Fy)0.5 = N/A [Satisfactory]

640 (1 - 3.74 Pc / Py) / (Fy)0.5 = N/A , for Pc / Py < 0.16

257 / (Fy)0.5 = N/A , for Pc / Py > 0.16

[Satisfactory] Where Py = FyA = 4100 kips

THE UNBALANCED VERTICAL FORCE ON BEAM (CBC Sec.2213A.9.4.1)

Pb = [FyAg - 1.7(0.3Pc,allow)] sin α = 389.1 kips (Vertical)

THE FACTORED MOMENT ON THE BEAM (CBC Sec.2213A.9.4.1-3)(1.2D + 0.5L) 2/ 8 + Pb / 4 = ft-kips

(0.9D) 2/ 8 - Pb / 4 = -2636 ft-kips,not govern.

THE UNBALANCED SEGMENT LENGTH ABOUT X - AND Y - AXES x = 28 ft

y = 14 ft ( CBC Sec.2213A.9.4.1-4 ,lateral supported at the intersection of chevorn braces.)

THE COMPRESSION CAPACITY (CBC 2213A.4.2 & ASD E2)


Where K = 1.0Es = 29000 ksi (1-F2/2)Fy / (5/3+3F/8-F3/8) = 23.87 ksi, for Cc > (K/r)

Cc = (2π2Es/Fy)0.5 = 107 12π2Es/[23(K /r)2] = N/A ksi, for Cc < (K/r)


F = (K / r) / Cc = 0.50

THE FLEXURAL CAPACITY (CBC 2213A.4.2)

Ms = FyZx = 3479 ft-kips > Mb [Satisfactory]

THE INTERACTION CAPACITY (ASD N4, Pg 5-95)

Pc / Pc,allow + Cm Mb / [(1 - Pc / Ps) Ms] = 0.963 < 1.0 [Satisfactory]

Pc / (A Fy) + Mb / (1.18Ms) = 0.782 < 1.0 [Satisfactory]

Where Cm = 1.00 (ASD H1)

Fe = 12π2Es/[23(K/r)2] = 53.1 ksi

Ps = (23/12)Fe A = 8346 kips


DanielTian Li

2865.9Mb =

Fa =

d / tw = 23.02 <



INPUT DATA & DESIGN SUMMARYCOLUMN SECTION = > W12X50 = > A d tw bf tf SxCOLUMN AXIAL SERVICE LOADS D = 80 kips 14.6 12.2 0.37 8.08 0.64 64.2


COLUMN AXIAL LOAD AT HORIZ. SEISMIC Eh = 110 kips (CBC 1630A.1.1) 391 5.18 1.96 72 1.14UNBRANCED COLUMN LENGTH = 14 ft


THE LOCAL BUCKLING LIMITATION (CBC 2213A.9.5 ,CBC 2213A.7.3, & ASD N7 Pg 5-96)

bf / (2tf ) = 6.31 < 7.00 [Satisfactory]

THE DESIGN LOADS (CBC 2213A.5.1)

Pt = 0.85D - Ω0Eh = -174 kips (Tension)



THE COMPRESSION CAPACITY (CBC 2213A.4.2 & ASD E2)





F = (K / r) / Cc = 0.80

Techincal References: 1. ICBO: "2001 California Building Code, Title 24, Part 2, Volume 2", 2002. 18.5 2. SEAOC: "Seismic Design Manual - Volume 3", International Code Council, 2000. 3. AISC: "Manual of Steel construction 9th", American Institute of Steel Construction, 1990.

DanielTian Li

Fa =


JOB NO. : DATE : REVIEW BY : Seismic Design for Ecconcentrically Braced Frames Based on IBC & AISC Seismic

INPUT DATA & DESIGN SUMMARYLINK SECTION = > W16X77 = > A d tw bf tf Sx

MAX SERVICE LOADS AT LINK END VDL = 1.8 kips 22.9 16.5 0.46 10.30 0.76 136

PDL = 7.4 kips Ix rx ry Zx kMDL = 14.4 ft-kips 1120 6.99 2.45 152 1.47VLL = 1.3 kipsPLL = 5.3 kips THE LINK DESIGN IS ADEQUATE.MLL = 9.6 ft-kips ( USE 1/2 x 4-3/4 @ 16 in o.c. INTERMEDIATE

MAX HORIZ. SEISMIC LOADS AT LINK END VE = 84 kips (QE, IBC 1617.1) & END STIFFENERS WITH 1/4" FILLET WELD.)

PE = 5.5 kips (QE, IBC 1617.1)

ME = 168 ft-kips (QE, IBC 1617.1)LINK LENGTH e = 4 ft

LINK YIELD STRESS Fy = 50 ksiREDUNDANCY FACTOR ρ = 1.19SEISMIC PARAMETER SDS = 1 (IBC 1615.1.3)BEAM LENGTH BETWEEN COL. CENTERS L = 30 ft (inclding link)STORY HEIGHT h = 12.5 ftMAXIMUM INELASTIC STORY DRIFT δ = 0.7 in (IBC eq.16-46)

DETERMINE FACTORED DESIGN LOADS AT LINK END (IBC1605.2 & AISC Seiemic 02 Tab. C-I-4.1)

Vu = (1.2 + 0.2SDS)VDL + f1VLL + ρVE = 103.13 kips

Pu = (1.2 + 0.2SDS)PDL + f1PLL + ρPE = 19.56 kips

Mu = (1.2 + 0.2SDS)MDL + f1MLL + ρME = 224.88 ft-kips

Where f1 = 0.5 (IBC 1605.4)CHECK LOCAL BUCKLING LIMITATION (AISC Seiemic 02 Tab. I-8-1)







MAX[ 191 / (Fy)0.5(2.33-Pu/φbPy) , 253 / (Fy)


[Satisfactory] Where φb = 0.9 , Py = FyA = 1145 kipsCHECK SHEAR CAPACITY (AISC Seiemic 02 & 97, Sec. 15.2)

φVn = φ MIN(Vp, 2Mp/e) = 184.0 kips > Vu [Satisfactory]

Where φ = 0.9 (Ignored axial force effect since Pu < 0.15 Py = 0.15 Fy Ag, AISC Seiemic 15.2)

Aw = (d - 2tf)tw = 6.82 in2

Vp = 0.6FyAw = 204.5 kips

Mp = FyZ = 633.3 ft-kipsCHECK FLEXURAL CAPACITY (LRFD F1.1)

φbMp = 570.0 > Mu [Satisfactory]

Where φb = 0.9

CHECK ADDITIONAL SHEAR CAPACITY REQUIREMENT FOR Pu>0.15Py ONLY (AISC Seiemic 15.2) < = DOES NOT APPLY.

φVna = φ MIN(Vpa, 2Mpa/e) = 184.0 kips > Vu [Satisfactory]

Where φ = 0.9

Vpa = Vp[1 - (Pu/Py)2 ]0.5 = 204.4 kips

Mpa = 1.18 Mp(1 - Pu/Py) = 734.6 ft-kips

CHECK ADDITIONAL LINK LENGTH REQUIREMENT FOR Pu>0.15Py ONLY (AISC Seiemic 15.2) < = DOES NOT APPLY.

[1.15 - 0.5ρ' (Aw/Ag)](1.6Mp/Vp) = N/A ft, for ρ' (Aw/Ag) > 0.3

(1.6Mp/Vp) = 4.96 ft, for ρ' (Aw/Ag) < 0.3

[Satisfactory] Where ρ' = Pu / Vu = 0.19

Aw / Ag = 0.30

DanielTian Li

e <

h / tw = 29.80 <

(cont'd)CHECK LINK ROTATION ANGLE LIMITATION (AISC Seiemic 02 & 97, Sec. 15.2)

γp = L δ / (h e) = 0.04 rad < γp,allowable = 0.080 rad [Satisfactory]

Where γp,allowable == 0.08 rad for e < 1.6Mp/Vp;

= 0.02 rad for e > 2.6Mp/Vp;

= linear interpolation [0.02, 0.08] by e value.

1.6Mp/Vp = 4.96 ft, 2.6Mp/Vp = 8.05 ft

CHECK LINK STIFFENER REQUIREMENT (AISC Seiemic 15.3)

bst = (bf - 2tw) /2 = 4.70 in

tst = MAX (0.75 tw , 3/8 ) = 0.375 in

USE 3/8 x 4-3/4 END STIFFENERS AT EACH SIDE.

s = see table following = 17.9 inProvide 2 stiffeners to give s = 16.0 in

Where 1.6 Mp / Vp = 4.96 ft

2.6 Mp / Vp = 8.05 ft

5.0 Mp / Vp = 15.49 ft

γp = 0.04 rad

e = 4 ftd = 16.5 in

tst = MAX (tw , 3/8 ) = 0.455 in

USE 1/2 x 4-3/4 @ 16 in o.c. INTERMEDIATESTIFFENERS AT EACH SIDE.

γpe [0 ~ 0.02] (0.02 ~ 0.08) 0.08

[0~1.6Mp/Vp] 52tw-d/5 178tw/3-d/5-1100γptw/3 30tw-d/5

(1.6Mp/Vp~2.6Mp/Vp] MIN(52tw-d/5 , bf) Min(178tw/3-d/5-1100γptw/3 , 1.5bf) MIN(30tw-d/5 , 1.5bf)

(2.6Mp/Vp~5Mp/Vp) 1.5bf 1.5bf 1.5bf

[5.0Mp/Vp~Greater] Not ReqD Not ReqD Not ReqD

The best fillet weld size (LRFD Sec.J2.2b)> wMIN = 0.1875 in< wMAX = 0.3125 in

[Satisfactory]The required weld length between A36 stiffener and web(LRFD Sec.J2.4 & AISC Seiemic 02 & 97, Sec. 15.3)

Lw = AstFy / [(2) φ Fw (0.707 w)] = (3/8 x 4-3/4) x 36 / [(2) 0.75 (0.6x70)(0.707x1/4)] = 4.07 in

< (d - 2k), [Satisfactory]The required weld length between A36 stiffener and flange (LRFD Sec.J2.4 & AISC Seiemic 02 & 97, Sec. 15.3)

Lf = 0.25AstFy / [(2) φ Fw (0.707 w)] = 0.25(3/8 x 4-3/4) x 36 / [(2) 0.75 (0.6x70)(0.707x1/4)] = 1.02 in

< (bst - k), [Satisfactory]

CHECK COMBINED LINK CAPACITY (SEAOC Design Manual Vol. 3, page 104)

f = Pu,link / (2Af) + Mu,link / Zf = 42.3 < Fy [Satisfactory]

Where Pu,link = Ω Pu = 38.8 kips

Mu,link = Vp (e/2) = 409.0 ft-kips

Zf = (d - tf) bf tf = 123.2 in3

Ω = Vn / Vu = 1.98

Af = bf tf = 7.83 in2

Techincal References: 1. Alan Williams: "Seismic and Wind Forces, Structural Design Examples", International Code Council, 2003. 2. SEAOC: "2000 IBC Structural/Seismic Design Manual - Volume 3", International Code Council, 2003. 3. AISC: "Seismic Provisions for Structural Steel Buildings", American Institute of Steel Construction, May 1, 2002.

w = 1/4 in




BRACE SECTION (Tube or Pipe) = > HSS10X10X5/8 Tube A rmin t h

MAX SERVICE LOADS PDL = 11.8 kips 21.00 3.80 0.58 10.00

PLL = 8.3 kips

UNBRACED LENGTH OF THE BRACE Lb = 18 ft(SEE LINK DESIGN SPREADSHEET FOR BALANCE OF INPUT DATA)

THE BRACE DESIGN IS ADEQUATE.REQUIRED CONNECTION = > ( 5/8 in Gusset Plate with 21 in Length, 4 leg, 5/16 in Fillet Weld.)

DETERMINE LIMITING WIDTH THICKNESS RATIO FOR COMPRESSION ELEMENT, LOCAL BUCKLING (LRFD Tab. B5.1)D / t = NA , for Pipe

h / t = 190 / (Fy)0.5 = 28.01 , for Tube

Where Fy = 46 ksi

Es = 29000 ksi

DETERMINE FACTORED DESIGN LOADS (AISC Seiemic 02 & 97Sec.15.6)

Pu = (1.2 + 0.2SDS)PDL + f1PLL + PE = 572.8 kips

Where PE = 1.25 Ry [Vn L Lb / (L-e) h] = 552.1 kips

Ry = 1.3 (1.4 for Pipe.)

DETERMINE DESIGN STRENGTH IN COMPRESSION (LRFD Sec.E2)

φcPn = φcAgFcr = 661.06 kips > Pu [Satisfactory]

φc = 0.85

λc = K / (rπ) (Fy / E)0.5 = 0.72 (0.658λc^2 )Fy = 37.03 kis, for λc <1.5

K = 1.0 0.877 / (λc2 )Fy = N/A kis, for λc >1.5

DETERMINE CONNECTION DESIGN FORCE

Put = Pu = 572.76 kips (Tension)

DETERMINE BEST FILLET WELD SIZE (LRFD Sec.J2.2b)> wMIN = 0.1875 in< wMAX = 0.4375 in


L = Put / [(4) φ Fw (0.707 w)] = 572.8 / [(4) 0.75 (0.6x70)(0.707x5/16)] = 20.57 in

( USE 21 in )CHECK DESIGN SHEAR RUPTURE CAPACITY OF SLOTED BRACE

(LRFD Sec.J4.1)φRn = φ(0.6Fu)Anu = 1273.78 kips > PutWhere φ = 0.75 [Satisfactory]

Fu = 58 ksi (LRFD Tab.1-4, Pg. 1-21)

Anu = 4 t L = 4 x 0.581 x 21 = 48.80 in2


tg = 5/8 inCHECK SHEAR RUPTURE CAPACITY OF GUSSET PLATE (LRFD Sec.J4.1)

φRn = φ(0.6Fu)Anv = 685.13 kips > Put [Satisfactory]Where φ = 0.75

Fu = 58 ksi (plate value)

Anv = 2 tg L = 2 x 5/8 x 21 = 26.25 in2

CHECK TENSION CAPACITY AT SLOTED BRACE (LRFD Sec.D1). THE GUSSET BLOCK SHEAR CAPACITY (LRFD J4-3b)φtPn = φtRyFuUAn = 1146.48 kips > Put φRn = φ(0.6Fu)Anv + φFyAgt = 685.13 + φFyAgt

Where φt = 0.75 [Satisfactory] > Put = 572.76U = 1 (LRFD Sec.B3.2d) [Satisfactory]

An = A - 2 t tg = 20.274 in2


DanielTian Li


w = 5/16 in

Fcr =



INPUT DATA & DESIGN SUMMARYMAX SERVICE LOADS AT OUTSIDE OF LINK VDL = 6.8 kips

PDL = 1 kipsMDL = 17 ft-kipsVLL = 4.8 kipsPLL = 0.7 kips 12.5MLL = 11.3 ft-kips

SEISMIC LOADS AT OUTSIDE OF LINK VE = 8.7 kips (QE, IBC 1617.1)

PE = 105 kips (QE, IBC 1617.1)

ME = 113 ft-kips (QE, IBC 1617.1)

(SEE LINK DESIGN SPREADSHEET FOR BALANCE OF INPUT DATA) THE DESIGN IS INADEQUATE, SEE ANALYSIS BELOW

DETERMINE FACTORED DESIGN LOADS AT SECTION OF LINK AND BEAM (AISC Seiemic 02 & 97, Sec. 15.6 & Tab. C-I-4.1)

Vu = (1.2 + 0.2SDS)VDL + f1VLL + ρVE = 42.4 kips

Pu = (1.2 + 0.2SDS)PDL + f1PLL + PE = 298.7 kips

Mu = (1.2 + 0.2SDS)MDL + f1MLL + ME = 524.3 ft-kips

Where f1 = 0.5 (IBC 1605.4)

Ry = 1.1 (AISC Seiemic Tab. I-6-1)

Vn = 204.5 kips (from link design)

Mn = Vn e / 2 = 408.95 ft-kips

VE = (1.1Ry Vn / VE , link) VE = 25.6 kips

PE = 1.1Ry Vn L / 2h = 296.9 kips

ME = 1.1Ry Mn = 494.8 ft-kips








MAX[ 191 / (Fy)0.5(2.33-Pu/φbPy) , 253 / (Fy)



CHECK UNBALANCED SEGMENT LENGTH 1 = (L - e - dc) / 2 = 12.41 ft, (top & bottom flange bracing with a design strength greater than below will be provided

Brace Load : Pb,link = 0.06Ry Fy bf tf = 25.8 kips, [AISC Seiemic Sec.15.5] at each end of the link segment.)

2 = 1 / 2 = 6.20 ft, ( lateral supported at middle of beam outside of link with following design strength.)

Brace Load : Pb,mid = 0.02Fy bf tf = 7.8 kips, [AISC Seiemic Sec.15.6.(2)]

Mb,mid =0.02Fy bf tf d = 10.8 ft-kips, [AISC Seiemic Sec.15.6.(2)]



Where φc = 0.85K = 1.0MAX(K 1/rx, K 2/ry) = 30.33 < 200 [Satisfactory]

λc = 0.401 (LRFD E2-4, Pg 6-47)


DanielTian Li

h / tw = 29.80 <

(cont'd)

DETERMINE FLEXURAL DESIGN STRENGTH (AISC-LRFD F1)

L b = 6.20 ft

L p = 1.76 ry (E / Fyf)0.5 = 8.67 ft

L r = ry X1 [1 + (1 + X2 FL2 )0.5 ]0.5 / FL = 25.40 ft




2 / Lb2)0.5 / Lb = 6373.8 ft-kips

Where X 1 = π (0.5 E G J A)0.5 / Sx = 2767.7 A Iy tf ry Sx

X 2 = 4 Cw [Sx / (G J)]2 / Iy = 0.0025 22.9 138 0.76 2.45 136

Fr = 10.00 ksi E G J Cw Zx

FL = MIN( Fyf - Fr , Fyw) = 40.00 ksi 29000 11200 3.86 8570 152.0

C b = 1.30 , (AISC-LRFD Table 4.1)

Mp = 633.33 ft-kips, for L b @ [0 , L p]

M n = MINCb [Mp - (Mp - Mr) (Lb - Lp) / (Lr - Lp)] , Mp = N/A ft-kips, for L b @ (L p , L r]


φ b M n = 0.9 Mn = 570 ft-kips

CHECK FLEXURAL CAPACITY (LRFD C1)

Mux = B1Mu = 524.28 ft-kips < φbMnx = Min(Ry φbFyZ, φbMn) = 570 ft-kips

Where Pe1 = π2EsI x / (K x)2 = 14459 kips [Satisfactory]

Cm = 0.6 (LRFD C1-3, Pg 6-41) Where φb = 0.9

B1 = Cm/(1 - Pu/Pe1) = 1.000

CHECK INTERACTION CAPACITY (LRFD H1.1)

For Pu/φcPn>0.2, Pu / φcPn + 8 / 9 (Mux / φbMnx) = 1.15

For Pu/φcPn<0.2, Pu / (2φcPn) + Mux / φbMnx = N/A


> 1 [Unsatisfactory]



INPUT DATA & DESIGN SUMMARYCOLUMN SECTION = > W14X99 = > A d tw bf tf Sx

COLUMN AXIAL SERVICE LOADS PDL = 151 kips 29.1 14.2 0.49 14.60 0.78 157

PLL = 46 kips Ix rx ry Zx kNUMBER OF STORIES n = 4 1110 6.18 3.72 173 1.38

COLUMN YIELD STRESS (36 or 50) Fy = 50 ksi

THE DESIGN IS INADEQUATE, SEE ANALYSIS BELOW

UNBRANCED COLUMN LENGTH = h = 14 ft

DETERMINE COLUMN AXIAL SEISMIC LOAD (AISC Seiemic 02, & 97, Sec. 15.8)PE = (n - 1) 1.1 Ry Vn = 742.25 kips

DETERMINE FACTORED DESIGN LOADS (IBC1605.2 & AISC Seiemic 02 Tab. C-I-4.1, Pg. 78)

Pu,t = (0.9 - 0.2SSD ) PDL - PE = -637 kips (Tension)

Pu,c = (1.2 + 0.2SDS)PDL + f1PLL + PE = 977 kips (Compression)

Where f1 = 0.5 (IBC 1605.4)


bf / (2tf ) = 9.36 > 0.3 (Es / Fy)0.5 = 7.22 [Unsatisfactory]






MAX[ 191 / (Fy)0.5(2.33-Pu/φbPy) , 253 / (Fy)


[Satisfactory] Where φc = 0.9

Py = FyA = 1455 kips



Where φc = 0.85K = 1.0MAX(K x/rx, K y/ry) = 45.20 < 200 [Satisfactory]

λc = 0.597 (LRFD E2-4, Pg 6-47)



DanielTian Li

h / tw = 23.59 <


JOB NO. : DATE : REVIEW BY : Seismic Design for Ecconcentrically Braced Frames Based on CBC 2001


LINK SECTION = > W16X77 = > A d tw bf tf Sx

MAX SERVICE LOADS AT LINK END VDL = 1.8 kips 22.9 16.5 0.46 10.30 0.76 136PDL = 7.4 kips

MDL = 14.4 ft-kips Ix rx ry Zx kVLL = 1.3 kips 1120 6.99 2.45 152 1.47PLL = 5.3 kipsMLL = 9.6 ft-kips

MAX HORIZ. SEISMIC LOADS AT LINK END VE = 84 kips (CBC 30A-1) THE LINK DESIGN IS ADEQUATE.PE = 5.5 kips (CBC 30A-1) ( USE 1/2 x 4-3/4 @ 16 in o.c. INTERMEDIATE

ME = 168 ft-kips (CBC 30A-1) & END STIFFENERS WITH 1/4" FILLET WELD.)LINK LENGTH e = 4 ft

LINK YIELD STRESS Fy = 50 ksiREDUNDANCY FACTOR ρ = 1.19IMPORTANCE FACTOR I = 1.15 (CBC Tab. 16A-K)SEISMIC COEFFICIENT Ca = 0.57 (CBC Tab. 16A-Q)BEAM LENGTH BETWEEN COL. CENTERS L = 30 ft (inclding link)STORY HEIGHT h = 14 ftMAXIMUM INELASTIC STORY DRIFT δM = 0.7 in (CBC eq.30A-17)

DETERMINE ALLOWABLE DESIGN LOADS (CBC Sec. 1612A.3.1)

LC1 = D + 0.75[L + (ρEh + 0.5CaID) / 1.4]

LC2 = D + (ρEh + 0.5CaID) / 1.4

V = MAX(LC1 , LC2) = 73.6 kipsP = MAX(LC1 , LC2) = 14.9 kipsM = MAX(LC1 , LC2) = 160.6 ft-kips

CHECK LOCAL BUCKLING LIMITATION (CBC Sec. 2213A.10.2, ASD Tab. B5.1)


640 (1 - 3.74 Pc / Py) / (Fy)0.5 = 86.11 , for Pc / Py < 0.16

257 / (Fy)0.5 = N/A , for Pc / Py > 0.16


CHECK FLEXURAL CAPACITY (CBC 2213A.10.3)

Mrs = Zx(fy - fa) = 621.3 ft-kips > M [Satisfactory]

Where fa = P / (2bftf) = 0.95 ksi

Af = bf tf = 7.83 in2

CHECK SHEAR CAPACITY (CBC 2213A.10.3, 2213A.10.5)

Ω = Vs / V = 2.80 > Ωmin = 1.0 / 0.8 = 1.25 [Satisfactory]

Where Vs = MIN( 0.55Fydtw , 2Mrs/e ) = 206.5 kips

CHECK COMBINED LINK CAPACITY USING LOADS ANTICIPATED TO YIELD LINK (SEAOC Design Manual Vol. 3)

f = Plink / (2Af) + Mlink / Zf = 42.9 < Fy [Satisfactory]

Where Plink = Ω P = 41.7 kips

Mlink = Vs (e/2) = 412.9 ft-kips

Zf = (d - tf) bf tf = 123.2 in3

CHECK LINK ROTATION ANGLE LIMITATION (CBC 2213A.10.4)

γp = L δM / (h e) = 0.031 rad < γp,allowable = 0.090 rad [Satisfactory]

Where γp,allowable == 0.09 rad for e < 1.6Ms / Vs;

= 0.03 rad for e > 2.6Ms / Vs;

= linear interpolation [0.03, 0.09] by e value.

1.6Mrs / Vs = 4.81 ft, 3.0Mrs / Vs = 9.03 ft

DanielTian Li

d / tw = 36.26 <

(cont'd)CHECK LINK STIFFENER REQUIREMENT (CBC 2213A.10.7 to 2213A.10.11)

bst = (bf - 2tw) /2 = 4.70 in

tst = MAX (0.75 tw , 3/8 ) = 0.375 in

USE 3/8 x 4-3/4 END STIFFENERS AT EACH SIDE.

s = see table following = 22.0 ins

γp 2Mrs/e > 0.45Fydtw 2Mrs/e < 0.45Fydtw [0 ~ 0.03] 56tw-d/5

(0.03 ~ 0.09) 65tw-d/5-300γptw0.09 38tw-d/5

Provide 2 pair stiffeners to give s = 16.0 in

Where γp = 0.03 rad

e = 4 ft

tst = MAX (tw , 3/8 ) = 0.455 in

USE 1/2 x 4-3/4 @ 16 in o.c. INTERMEDIATE

The best fillet weld size (LRFD Sec.J2.2b)> wMIN = 0.1875 in 0.45< wMAX = 0.3125 in

[Satisfactory]The required weld length between A36 stiffener and web (ASD Sec.J2.4 & CBC 2213A.4.2)

Lw = Ast Fy / [(2)(1.7)(0.3Fu)(0.707 w)] = (3/8 x 4-3/4) x 36 / [(2)(1.7)(0.3x70)(0.707x1/4)] = 3.59 in

< (d - 2k), [Satisfactory]The required weld length between A36 stiffener and flange (ASD Sec.J2.4 & CBC 2213A.4.2)

Lf = 0.25Ast Fy / [(2)(1.7)(0.3Fu)(0.707 w)] = 0.25(3/8 x 4-3/4) x 36 / [(2)(1.7)(0.3x70)(0.707x1/4)] = 0.90 in

< (bst - k), [Satisfactory] Techincal References: 1. ICBO: "2001 California Building Code, Title 24, Part 2, Volume 2", 2002. 2. SEAOC: "Seismic Design Manual - Volume 3", International Code Council, 2000. 3. AISC: "Manual of Steel construction 9th", American Institute of Steel Construction, 1990.

w = 1/4 in

Not ReqD




BRACE SECTION (Tube or Pipe) = > HSS10X10X5/8 Tube A rmin t h

MAX SERVICE LOADS PDL = 11.8 kips 21.00 3.80 0.58 10.00

PLL = 8.3 kips

UNBRACED LENGTH OF THE BRACE Lb = 18 ft(SEE LINK DESIGN SPREADSHEET FOR BALANCE OF INPUT DATA)

THE DESIGN IS INADEQUATE, SEE ANALYSISREQUIRED CONNECTION = > ( 5/8 in Gusset Plate with 16 in Length, 4 leg, 5/16 in Fillet Weld.)

CHECK LIMITING WIDTH THICKNESS RATIO FOR COMPRESSION ELEMENT, LOCAL BUCKLING (CBC Sec. 2213A.8.2.5)

D / t =1300 / Fy = 37.14 , for Pipe

h / t = 110 / (Fy)0.5 = 16.22 , for Tube

Where Fy = 46 ksi

CHECK LIMITING SLENDERNESS RATIO (CBC 2213A.8.2.1)720 / (Fy)

0.5 = 106.2 > K / r = 56.8 [Satisfactory]

Where K = 1.0


LC1 = D + 0.75[L + (Eh + 0.5CaID) / 1.4]

LC2 = D + (Eh + 0.5CaID) / 1.4P = MAX(LC1 , LC2) = 288.0 kips

Where PE = 1.3 [Vs L Lb / (L-e) h] = 382.9 kips

Vs = 206.5 (see LINK sheet)

DETERMINE DESIGN STRENGTH IN COMPRESSION (ASD Sec.E2)

Pc,allow = AFa = 456.76 kips > P [Satisfactory]


Cc = (2π2Es/Fy)0.5 = 112 (1-F2/2)Fy / (5/3+3F/8-F3/8) = 21.75 kis, for Cc > (K/r)

K / r = 57 12π2Es/[23(K/r)2] = N/A kis, for Cc < (K/r)

F = (K / r) / Cc = 0.51

DETERMINE CONNECTION DESIGN FORCE (AISC Seismic Sec. 13.3a)

Pconn = P = 288.03 kips (Tension)

DETERMINE BEST FILLET WELD SIZE (ASD Sec.J2.2b)> wMIN = 0.1875 in< wMAX = 0.4375 in

[Satisfactory]DETERMINE REQUIRED WELD LENGTH (ASD Sec.J2.4 )

L = Pconn / [(4)(0.3Fw)(0.707 w)] = 288.0 / [(4)(0.3x70)(0.707x5/16)] = 15.52 in

( USE 16 in )

CHECK DESIGN SHEAR RUPTURE CAPACITY OF SLOTED BRACE (ASD Sec.J4)

Pt,rup,brace =(0.3Fu)Anu = 647.00 kips > Pconn[Satisfactory]


Anu = 4 t L = 4 x 0.581 x 16 = 37.18 in2

DETERMINE REQUIRED THICKNESS OF GUSSET PLATE (ASD Tab. J2.4)

tg = 5/8 in

CHECK SHEAR RUPTURE CAPACITY OF GUSSET PLATE (ASD Sec.J4 & CBC 2213A.4.2)

Pt,rup,gusset =(0.3Fu)Anv = 348.00 kips > Pconn [Satisfactory]


Anv = 2 tg L = 2 x 5/8 x 16 = 20.00 in2

CHECK TENSION CAPACITY AT SLOTED BRACE (ASD Sec.D1). THE GUSSET BLOCK SHEAR CAPACITY (ASD J4)Pt,brace = (0.5FuUAn) = 511.51 kips > Pconn Ps,guss = [0.3FuAnv + 0.5FuAgt] =

[Satisfactory] = 348.00 +[0.5FuAgt]

Where U = 0.87 (ASD Sec.B3) > Pconn = 288.03 [Satisfactory]

An = A - 2 t tg = 20.274 in2


DanielTian Li

w = 5/16 in

Fa =

< Actual [Unsatisfactory]



INPUT DATA & DESIGN SUMMARYMAX SERVICE LOADS AT OUTSIDE OF LINK VDL = 6.8 kips

PDL = 1 kipsMDL = 17 ft-kipsVLL = 4.8 kips 168PLL = 0.7 kipsMLL = 11.3 ft-kips

SEISMIC LOADS AT OUTSIDE OF LINK VE = 8.7 kips (CBC 30A-1)PE = 105 kips (CBC 30A-1)ME = 113 ft-kips (CBC 30A-1) THE BEAM DESIGN IS ADEQUATE.

DETERMINE DESIGN LOADS AT SECTION OF LINK AND BEAM (CBC 2213A.10.13)

LC1 = D + 0.75[L + (Eh + 0.5CaID) / 1.4]

LC2 = D + (Eh + 0.5CaID) / 1.4

V = MAX(LC1 , LC2) = 30.5 kipsP = MAX(LC1 , LC2) = 229.5 kipsM = MAX(LC1 , LC2) = 447.0 ft-kips

Where Vs = 229.4 kips (from link design)

Ms = Vn e / 2 = 458.8 ft-kips

VE = (1.3 Vs / VE , link) VE = 30.9 kips

PE = MAX[1.3(Vs L / 2h) , PE]= 319.5 kips

ME = MAX[1.3 Mn , ME] = 596.4 ft-kips

CHECK LOCAL BUCKLING LIMITATION (CBC Sec. 2213A.10.2, ASD Tab. B5.1)


640 (1 - 3.74 Pc / Py) / (Fy)0.5 = N/A , for Pc / Py < 0.16

257 / (Fy)0.5 = 36.35 , for Pc / Py > 0.16


CHECK UNBALANCED SEGMENT LENGTH. (CBC 2213A.10.18) 1 = (L - e) / 2 = 13 ft, (top & bottom flange bracing with a design strength greater than below will be provided

Where Pb,link = 0.06Fy bf tf = 23.5 kips, (brace axial load.) at each end of the link segment.)

2 = 1 / 2 = 6.5 ft, ( lateral supported at middle of beam outside of link with following design strength.)

< max = 76 bf / (Fy)0.5 = 9.23 ft [Satisfactory]

Where Pb,mid = 0.01Ry Fy bf tf = 3.9 kips, (brace axial load.)

Mb,mid = 0.01Ry Fy bf tf d = 5.4 kips, (brace bending load.)



Where K = 1.0Es = 29000 ksi (1-F2/2)Fy / (5/3+3F/8-F3/8) = 26.93 ksi, for Cc > (K/r)

Cc = (2π2Es/Fy)0.5 = 107 12π2Es/[23(K /r)2] = N/A ksi, for Cc < (K/r)

MAX(K 1/rx, K 2/ry) = 31.77 < 200 [Satisfactory]

F = (K / r) / Cc = 0.30

CHECK FLEXURAL CAPACITY (CBC 2213A.4.2)

Ms = FyZx = 633 ft-kips > M [Satisfactory]

CHECK INTERACTION CAPACITY (ASD N4, Pg 5-95)

Pc / Pc,allow + Cm M / [(1 - Pc / Ps) Ms] = 0.951 < 1.0 [Satisfactory]

Pc / (A Fy) + M / (1.18Ms) = 0.906 < 1.0 [Satisfactory]

Where Cm = 1.00 (ASD H1)

Fe = 12π2Es/[23(K/r)2] = 147.9 ksi

Pe = (23/12)Fe A = 6492 kips


DanielTian Li

d / tw = 36.26

Fa =

<



INPUT DATA & DESIGN SUMMARYCOLUMN SECTION = > W14X211 = > A d tw bf tf Sx

COLUMN AXIAL SERVICE LOADS PDL = 151 kips 62 15.7 0.98 15.80 1.56 338

PLL = 46 kips Ix rx ry Zx kNUMBER OF STORIES n = 4 2660 6.55 4.08 390 2.16

COLUMN YIELD STRESS (36 or 50) Fy = 50 ksi


UNBRANCED COLUMN LENGTH = h = 14 ft

DETERMINE COLUMN AXIAL SEISMIC LOAD (CBC 2213A.10.14)PE = (n - 1) 1.25 Vs = 860.23 kips


Pt = 0.85PD - ΩoPE = -2280 kips (Tension)

Pc = 1.0PD + 0.7PL + ΩoPE = 2592 kips (Compression)

Where Ωo = 2.8 (CBC Tab. 16A-N)






F = (K / r) / Cc = 0.39


DanielTian Li

Fa =


JOB NO. : DATE : REVIEW BY : Seismic Design for Intermediate/Ordinary Moment Resisting Frames Based on AISC Seismic - LRFD

INPUT DATA & DESIGN SUMMARYCOLUMN SECTION = > W14X132

A d tw bf tf Sx Ix rx ry Zx k38.8 14.7 0.65 14.70 1.03 209 1530 6.28 3.76 234 1.63

BEAM SECTION = > W21X122


STRUCTURAL STEEL YIELD STRESS Fy = 50 ksi

THE FACTOR AXIAL LOAD ON THE COLUMN Pu = 200 kipsBEAM LENGTH BETWEEN COL. CENTERS L = 28 ftAVERAGE STORY HEIGHT OF ABOVE & BELOW h = 12 ft

THE DESIGN IS ADEQUATE.(Continuity column stiffeners 1 x 7 with 7/16" fillet weld to web & CP to flanges.A doubler plate is required with thickness of 13/16 in. )

ANALYSISTHE SEISMIC DESIGN FACTOR COMPARISON

FRAME TYPE R Ωo Cd L / dSMRF 8 3 5 1/2 7IMRF 6 3 5 5OMRF 4 3 3 1/2 5

CHECK BEAM LOCAL BUCKLING LIMITATION (AISC Seiemic 02 Tab. I-8-1)




h / tw = 31.30 < 2.45 (Es / Fy)0.5 = 59.00 [Satisfactory]

[ 418 / (Fy)0.5 for FEMA Sec. 3.3.1.2]

CHECK COLUMN LOCAL BUCKING LIMITATION (AISC Seiemic 02 Tab. I-8-1)






MAX[ 191 / (Fy)0.5(2.33-Pu/φbPy) , 253 / (Fy)



CHECK CONTINUITY PLATE REQUIREMENT ((AISC Seismic 02 Sec. 10.5, AISC Seismic 97 Sec. 10.3, FEMA Sec. 3.3.3.1)

tcf = MIN bbf / 6 , 0.4[1.8bbf tbf (FybRyb) / (FybRyb)] 0.5 = 1.85 in > actual tcf(The continuity plates required.)

tst = tbf for interior connection, or (tbf /2) for exterior connection = 0.96 in,USE 1.00 in, ( 1 in )

bst = 7 in < 1.79 (Es / Fyst )0.5 tst = 50.80 in, (LRFD Sec. K1.9)

[Satisfactory]

φcPn,st = φcFcr A = 703.6 kips

Where φc = 0.85 hst = dc - 2k = 11.44

K = 0.75 K hst / rst < 200 (LRFD B2) [Satisfactory]

I = tst (2bst + twc) 3 / 12 = 259 in4 λc = 0.029 (LRFD E2-4, Pg 6-47)

A = 2bsttst + 25(twc) 2 = 23 in2

Fcr = 35.99 ksi (LRFD Sec.E2, Pg 6-47)

rst = ( I / A )0.5 = 3.36 in Fyst = 36 kips, plate yield stress

Pu,st = Ryb Fyb bfb tfb = 654.7 kips < φcPn,st [Satisfactory]

The best fillet weld size (LRFD Sec.J2.2b)> wMIN = 0.25 in

< wMAX = 0.5625 in

[Satisfactory]The required weld length between A36 continuity plates and column web (FEMA Fig 3-6)

Lw = 0.6tstLnstFy / [(2) φ Fw (0.707 w)] = (1 x 8.4) x 36 / [(2) 0.75 (0.6x70)(0.707x7/16)] = 7.12 in

Where Lnet = dc - 2(k + 1.5) = 8.4 < 2(Lnet -0.5) [Satisfactory]

(Use complete joint penetration groove welds between continuity plates & column flanges.)

DanielTian Li

w = 7/16 in

h / tw = 17.74 <

(cont'd)

CHECK PANEL ZONE THICKNESS REQUIREMENT (FEMA Sec. 3.3.3.2, No additional requirement for AISC Seismic.)

tReqD = MAX (t1, t2) = 1.41 in

t1 = Cy Mpr (h - db ) / [0.9 (0.6) Fyc Ryc dc (db - tfb) h] = 1.41 in

Where Cy = Sb / (Cpr Zb) = 0.77

Cpr = 1.15 (FEMA Sec. 3.5.5.1)


Sb = 2Ib / db = 273 in2

Ib = Ix = 2960 in4

Mpr = Nb CprRyFyb Zb = 1618 ft-kips 12

Nb = 1 , (if double side connection of beams, input 2)

t2 = (dz + wz ) / 90 = (db -2tst + dc - 2k) / 90 = 0.35 in

Since twc = 0.65 in < tReqD , a doubler plate isrequired with thickness of 13/16 in.

Techincal References: 1. FEMA 350: "Recommended Seismic Design Criteria for New Steel Moment-frame Buildings.", SAC Joint Venture, 2000. 2. Alan Williams: "Seismic and Wind Forces, Structural Design Examples", International Code Council, 2003. 3. SEAOC: "2000 IBC Structural/Seismic Design Manual - Volume 3", International Code Council, 2003. 4. AISC: "Manual of Steel construction, LRFD, 2th", American Institute of Steel Construction, 1998. 5. AISC: "Seismic Provisions for Structural Steel Buildings", American Institute of Steel Construction, May 1, 2002.


JOB NO. : DATE : REVIEW BY : Seismic Design for Ordinary Moment Resisting Frames Based on CBC 2001 - ASD


A d tw bf tf Sx Ix rx ry Zx k62 15.7 0.98 15.80 1.56 338 2660 6.55 4.08 390 2.16



STRUCTURAL STEEL YIELD STRESS Fy = 50 ksi

THE GRAVITY LOAD (1.2D + 0.5 L) ON THE BEAM wg = 2 klfBEAM LENGTH BETWEEN COL. CENTERS = 28 ft

COLUMN AXIAL LOAD AT SERVICE LEVEL D = 113 kipsL = 75 kips

COLUMN AXIAL LOAD AT HORIZ. SEISMIC Eh = 5.5 kips (CBC 30A-1)

SEISMIC COEFFICIENT Ca = 0.44 (CBC Tab. 16A-Q)IMPORTANCE FACTOR I = 1.15 (CBC Tab. 16A-K)REDUNDANCY FACTOR ρ = 1.25AVERAGE STORY HEIGHT OF ABOVE & BELOW h = 12 ft

THE OMRF DESIGN IS ADEQUATE.(Continuity column stiffeners 13/16 x 7 with 1/2" fillet weld to web & CP to flanges.A doubler plate is not required. )

DETERMINE ALLOWABLE COLUMN DESIGN LOADS (CBC 1612A.3.1, 2213A.8.2, & 2213A.8.4.1)

Pt = 0.9D - (ρEh + 0.5CaID) / 1.4 = 76 kips (Compression)

Pc = D + 0.75[L + (ρEh + 0.5CaID) / 1.4] = 188 kips (Compression, Governs)

Pc = D + (ρEh + 0.5CaID) / 1.4 = 138 kips (Compression)

CHECK BEAM - COLUMN RATIO REQUIREMENT (SEAOC Design Manual, Vol. III, Pg 182, & CBC2213A.7.5)

ΣMpc* / (ΣMpb* ) = 2.09 > 1.00 [Satisfactory]

( > 1.25 The column flanges need to be laterally supported only at the beam top flange.

Where ΣMpc* = Nc Zc (Fyc - Pc / Ag ) = 3053 ft-kips CBC 2213A.7.7.1)

Nc = 2 , (if only one column below, input 1)

ΣMpb* = Nb (Mpr + Mv) = 1459 ft-kips


Mv = 0.5wg(L -dc)(0.5dc) = 17 ft-kips

Mpr = Fyb Zb = 1442 ft-kips

DETERMINE COLUMN DESIGN LOADS (CBC 2213A.5.1)

Pt = 0.85D - Ω0Eh = 81 kips (Compression)



CHECK COLUMN COMPRESSION CAPACITY (CBC 2213A.4.2 & ASD E2)




MAX(Kx/rx, Ky/ry) = 67.10 < 200 [Satisfactory]

F = (K / r) / Cc = 0.63

CHECK CONTINUITY PLATE REQUIREMENT (SEAOC Design Manual, Vol. III, Pg 182, FEMA Sec. 3.3.3.1 & CBC 2213A.7.4)

tst = tbf for interior connection, or (tbf /2) for exterior connection = 0.76 in,USE 0.81 in, ( 13/16 in )

bst = 7 in > bfb /3 - twc /2 = 3.01 in, (ASD Sec. K1.8) [Satisfactory]

Ast = [Pbf - Fyc twc(tfb + 5kc)] / Fyst = 4.2156 in2, (ASD K1-9) < tst bst [Satisfactory]

Where Pbf = 1.8bfbtfbFyb = 718.2 kips, (CBC 2213A.7.4)

Fyst = 36 kips, plate yield stress

The best fillet weld size (ASD Sec.J2.2b)> wMIN = 0.25 in

< wMAX = 0.6875 in

[Satisfactory]

Fa =

w = 1/2 in

DanielTian Li

(Cont'd)

The required weld length between A36 continuity plates and column web (FEMA Fig 3-6)

Lw = Pbf / [(2) 1.7 (0.3Fu )(0.707 w)] = 718.2 / [(2) 1.7 (0.3x70)(0.707x1/2)] = 21.46 in

< 2(Lnet) [Satisfactory]

Where Lnet = dc - 2(k) = 11.4


CHECK PANEL ZONE THICKNESS REQUIREMENT (SEAOC Design Manual, Vol. III, Pg 182, & CBC 2213A.7.2)


t1 = Vz / (0.55Fycdc) - 3bfctfc2 / (dcdb) = 0.62 in, (CBC 13A-1)

Where Vz = Nb [(0.8) ΣMf / (db - tfb/2) - (0.8)Mf / h] = 374.3 kips(SEAOC Design Manual, Vol. III, Pg 182)

t2 = (dz + wz ) / 90 = (db -2tst + dc - 2kc) / 90 = 0.44 in5.5

Since twc = 0.98 in > tReqD , a doubler plate isnot required.

Techincal References: 1. FEMA 350: "Recommended Seismic Design Criteria for New Steel Moment-frame Buildings.", SAC Joint Venture, 2000. 2. Alan Williams: "Seismic and Wind Forces, Structural Design Examples", International Code Council, 2003. 3. SEAOC: "2000 IBC Structural/Seismic Design Manual - Volume 3", International Code Council, 2003. 4. CBC : "2001 California Building Code, Volume 2", International Conference of Building Officials, 2001. 5. AISC: "Seismic Provisions for Structural Steel Buildings", American Institute of Steel Construction, May 1, 2002.


JOB NO. : DATE : REVIEW BY : Seismic Design for Special Moment Resisting Frames Based on AISC Seismic - LRFD





STRUCTURAL STEEL YIELD STRESS Fy = 50 ksi THE SMRF DESIGN IS ADEQUATE.THE FACTOR GRAVITY LOAD ON THE BEAM wu = 0.25 klf (Continuity column stiffeners 1 x 8

THE FACTOR AXIAL LOAD ON THE COLUMN Pu = 800 kips with 3/4" fillet weld to web & CP to flanges.BEAM LENGTH BETWEEN COL. CENTERS L = 30 ft A doubler plate is not required. )AVERAGE STORY HEIGHT OF ABOVE & BELOW h = 12 ft

REDUCED SECTION DIMENSIONS a = 7 in, [0.5~0.75bf]

b = 25 in, [0.65~0.85db]

c = 2.5 in, [ < 0.25bf]

ANALYSIS r = (4c2 + b2 ) / 8c = 32.5 in

Sh = dc/2 + a + b/2 = 28.9 in

CHECK BEAM LOCAL BUCKLING LIMITATIONS (AISC Seiemic 02 Tab. I-8-1)




h / tw = 52.03 < 2.45 (Es / Fy)0.5 = 59.00 [Satisfactory]

[ 418 / (Fy)0.5 for FEMA Sec. 3.3.1.2]

CHECK COLUMN LOCAL BUCKING LIMITATIONS (AISC Seiemic 02 Tab. I-8-1)






MAX[ 191 / (Fy)0.5(2.33-Pu/φbPy) , 253 / (Fy)



CHECK BEAM - COLUMN RATIO REQUIREMENT (AISC Seiemic 02 Sec. 9.6, Pg. 16)


Where ΣMpc* = Nc Zc (Fyc - Pu / Ag ) = 6315 ft-kips


ΣMpb* = Nb (MRBS + Mv) = 2623 ft-kips, at center of column


Mv = VRBS Sh = [2MRBS /(L-2Sh)+wu(L-2Sh)/2] Sh= 427 ft-kips

MRBS = CprRyFyb ZRBS = 2196 ft-kips


ZRBS = Zb - 2c tf (d - tf) = 417 in3

Cpr = 1.15 (FEMA Sec. 3.5.5.1)

CHECK BENDING MOMENT AT THE COLUMN FACE (FEMA Sec. 3.5.5.1)

Mf = MRBS + [2MRBS / (L - 2Sh) + wu(L - 2Sh)/2] (a + b/2)

= 2485 ft-kips < RyFyb Zb = 2663 ft-kips [Satisfactory]

CHECK CONTINUITY PLATE REQUIREMENT (FEMA Sec. 3.3.3.1)

tcf = MIN bbf / 6 , 0.4[1.8bbf tbf (FybRyb) / (FybRyb)] 0.5 = 1.80 in < actual tcf(The continuity plates may not be required.)

tst = tbf for interior connection, or (tbf /2) for exterior connection = 0.94 in, USE 1.00 in, ( 1 in )

bst = 8 in < 95 / (Fyst )0.5 tst = 15.83 in, (LRFD Sec. K1.9)

[Satisfactory]

φcPn,st = φcFcr A = 788.2 kips

Where φc = 0.85 hst = dc - 2k = 11.44

K = 0.75 K hst / rst < 200 (LRFD B2) [Satisfactory]

I = tst (2bst + twc) 3 / 12 = 383 in4 λc = 0.025 (LRFD E2-4, Pg 6-47)

A = 2bsttst + 25(twc) 2 = 26 in2

Fcr = 35.99 ksi (LRFD Sec.E2, Pg 6-47)

rst = ( I / A )0.5 = 3.86 in

Pu,st = Ryb Fyb bfb tfb = 620.4 kips < φcPn,st [Satisfactory]

Daniel

6.09 <

Tian Li

h / tw =

(Cont'd)

The best fillet weld size (LRFD Sec.J2.2b)> wMIN = 0.3125 in

< wMAX = 0.9375 in


Lw = 0.6tstLnstFy / [(2) φ Fw (0.707 w)] = (1 x 8.4) x 36 / [(2) 0.75 (0.6x70)(0.707x3/4)] = 3.86 in

Where Lnet = dc - 2(kc + 1.5) = 8.4 < 2(Lnet -0.5) [Satisfactory]


CHECK PANEL ZONE THICKNESS REQUIREMENT (AISC Sec. 9.3 & FEMA Sec. 3.3.3.2)


t1 = Cy Mc (h - db ) / [0.9 (0.6) Fyc Ryc dc (db - tfb) h] = 1.08 in

Where Cy = Sb / (Cpr ZRBS) = 0.88 12

Sb = 2Ib / db = 424 in2

Ib = Ix - (2 c tfb)(0.5db - 0.5tfb)2 = 7604 in4

Mc = ΣMpb* = 2623 ft-kips

t2 = (dz + wz ) / 90 = (db -2tst + dc - 2k) / 90 = 0.50 in


Techincal References: 1. FEMA 350: "Recommended Seismic Design Criteria for New Steel Moment-frame Buildings.", SAC Joint Venture, 2000. 2. AISC: "Manual of Steel construction, LRFD, 2th", American Institute of Steel Construction, 1998. 3. Alan Williams: "Seismic and Wind Forces, Structural Design Examples", International Code Council, 2003. 4. SEAOC: "2000 IBC Structural/Seismic Design Manual - Volume 3", International Code Council, 2003. 5. AISC: "Seismic Provisions for Structural Steel Buildings", American Institute of Steel Construction, May 1, 2002.

w = 3/4 in


JOB NO. : DATE : REVIEW BY : Seismic Design for Special Moment Resisting Frames Based on CBC 2001 - ASD





STRUCTURAL STEEL YIELD STRESS Fy = 50 ksi THE SMRF DESIGN IS ADEQUATE.THE GRAVITY LOAD (1.2D + 0.5 L) ON THE BEAM wg = 2 klf (Continuity column stiffeners 13/16 x 7BEAM LENGTH BETWEEN COL. CENTERS = 28 ft with 1/2" fillet weld to web & CP to flanges.COLUMN AXIAL LOAD AT SERVICE LEVEL D = 113 kips A doubler plate is not required. )

L = 75 kips

COLUMN AXIAL LOAD AT HORIZ. SEISMIC Eh = 5.5 kips (CBC 30A-1)

SEISMIC COEFFICIENT Ca = 0.44 (CBC Tab. 16A-Q)IMPORTANCE FACTOR I = 1.15 (CBC Tab. 16A-K)REDUNDANCY FACTOR ρ = 1.25

AVERAGE STORY HEIGHT OF ABOVE & BELOW h = 12 ft

REDUCED SECTION DIMENSIONS a = 6 in, [0.5~0.75bf]

b = 24 in, [0.65~0.85db]

c = 2.25 in, [ < 0.25bf]

ANALYSIS r = (4c2 + b2 ) / 8c = 33.1 in

Sh = dc/2 + a + b/2 = 25.9 in

CHECK BEAM LOCAL BUCKLING LIMITATION (CBC 2213A.7.3)


d / tw = 54.68 < 640 / (Fy)0.5 = 90.51 [Satisfactory]

CHECK BEAM BRACING LIMITATION (CBC 2213A.7.8)

Smax = 96ry = 17.17 ftHence, both flanges of beam shall be braced at maximum spacing 17.17 ft.

CHECK COLUMN LOCAL BUCKLING LIMITATION (CBC 2213A.7.3 & ASD N7, Pg 5-96)

bf / (2tf ) = 5.06 < 7.00 [Satisfactory]

DETERMINE ALLOWABLE COLUMN DESIGN LOADS (CBC 1612A.3.1, 2213A.8.2, & 2213A.8.4.1)

Pt = 0.9D - (ρEh + 0.5CaID) / 1.4 = 76 kips (Compression)

Pc = D + 0.75[L + (ρEh + 0.5CaID) / 1.4] = 188 kips (Compression, Governs)

Pc = D + (ρEh + 0.5CaID) / 1.4 = 138 kips (Compression)

CHECK BEAM - COLUMN RATIO REQUIREMENT (CBC2213A.7.5)


( If > 1.25 The column flanges need to be laterally supported only at the beam top flange.

Where ΣMpc* = Nc Zc (Fyc - Pc / Ag ) = 3053 ft-kips CBC 2213A.7.7.1)


ΣMpb* = Nb (MRBS + Mv) = 1509 ft-kips, at center of column


Mv = VRBS Sh = [2MRBS /(L-2Sh)+wg(L -2Sh)/2] Sh= 275 ft-kips

MRBS = β Fyb ZRBS = 1233 ft-kips

ZRBS = Zb - 2c tf (d - tf) = 247 in3

β = 1.2 (FEMA Sec. 7.5.2.2)

CHECK BENDING MOMENT AT THE COLUMN FACE (CBC2213A.4.2)

Mf = MRBS + [2MRBS / (L - 2Sh) + wg(L - 2Sh)/2] (a + b/2)

= 1425 ft-kips < Fyb Zb = 1442 ft-kips [Satisfactory]

DETERMINE COLUMN DESIGN LOADS (CBC 2213A.5.1)

Pt = 0.85D - Ω0Eh = 81 kips (Compression)



DanielTian Li

(Cont'd)

CHECK COLUMN COMPRESSION CAPACITY (CBC 2213A.4.2 & ASD E2)




MAX(Kx/rx, Ky/ry) = 35.33 < 200 [Satisfactory]

F = (K / r) / Cc = 0.33

CHECK CONTINUITY PLATE REQUIREMENT (FEMA Sec. 3.3.3.1 & CBC 2213A.7.4)

tst = tbf for interior connection, or (tbf /2) for exterior connection = 0.76 in, USE 0.81 in, ( 13/16 in )

bst = 7 in > bfb /3 - twc /2 = 3.01 in, (ASD Sec. K1.8) [Satisfactory]

Ast = [Pbf - Fyc twc(tfb + 5kc)] / Fyst = 4.2156 in2, (ASD K1-9) < tst bst [Satisfactory]

Where Pbf = 1.8 bfb tfb Fyb = 718.2 kips, (CBC 2213A.7.4)

Fyst = 36 kips, plate yield stress

The best fillet weld size (ASD Sec.J2.2b)> wMIN = 0.25 in

< wMAX = 0.6875 in


Lw = Pbf / [(2) 1.7 (0.3Fu )(0.707 w)] = 718.2 / [(2) 1.7 (0.3x70)(0.707x1/2)] = 21.46 in

< 2(Lnet) [Satisfactory]

Where Lnet = dc - 2(kc) = 11.4


CHECK PANEL ZONE THICKNESS REQUIREMENT (CBC 2213A.7.2)


t1 = Vz / (0.55Fycdc) - 3bfctfc2 / (dcdb) = 0.61 in, (CBC 13A-1)

Where Vz = Nb [(0.8) ΣMf / (db - tfb/2) - (0.8)Mf / h] = 370.0 kips 28(SEAOC Design Manual, Vol. III, Pg 182)

t2 = (dz + wz ) / 90 = (db -2tst + dc - 2kc) / 90 = 0.44 in


Techincal References: 1. FEMA 350: "Recommended Seismic Design Criteria for New Steel Moment-frame Buildings.", SAC Joint Venture, 2000. 2. Alan Williams: "Seismic and Wind Forces, Structural Design Examples", International Code Council, 2003. 3. SEAOC: "2000 IBC Structural/Seismic Design Manual - Volume 3", International Code Council, 2003. 4. CBC : "2001 California Building Code, Volume 2", International Conference of Building Officials, 2001. 5. AISC: "Seismic Provisions for Structural Steel Buildings", American Institute of Steel Construction, May 1, 2002.

Fa =

w = 1/2 in


JOB NO. : DATE : REVIEW BY : Steel Beam Design with Gravity Loading Based on AISC-ASD 9th


BEAM SECTION = > W18X40 = > Ix Sx rT bf tf tw612 68.4 1.52 6.02 0.53 0.32

SLOPED DEAD LOADS wDL,1 = 1.15 kips / ftwDL,2 = 0.5 kips / ft

PROJECTED LIVE LOADS wLL,1 = 0.8 kips / ftwLL,2 = 0.5 kips / ft

CONCENTRATED LOADS PDL = 3 kipsPLL = 3 kips

BEAM SPAN LENGTH L 1 = 30 ft

CANTILEVER LENGTH L 2 = 10 ft, (0 for no cantilever)

BEAM SLOPE 4 : 12 ( θ = 18.43 0 )



ANALYSISDETERMINE REACTIONS, MOMENTS & SHEARS

= 50.17 kips

= 26.47 kips

X 1 = 13.16 ft

X 2 = 13.16 ft

X 3 = 3.69 ft

111.4 ft-kips

174.1 ft-kips

V max = 39.89 kips, at R2 left.

CHECK M Min BENDING CAPACITY (AISC-ASD, F1.3, page 5-46)

10.00 ft, unbraced length

1.75 , since M1 is 0

r T = 1.52 in , A f = 3.16 in2

5.39 ft

10.30 ft

16.92 ft

DanielTian Li

∆

∆

( ),1 ,2,1 ,2

2 1 22 1 1 2

1 1

0.5 0.5cos cos

DL DL

LL LL

w w L L LPw wR L L LL Lθ θ

+ = + + + + +

,1 ,2,1 ,21 1 2 2cos cos

DL DL

LL LL

w w Pw wR L L Rθ θ = + + + + −

,2

,222 20.5

cosDL

LLMinw PwM L Lθ

= + + =

( ),1,1

21 2

cos 8DL

LLMaxX Xw

wM θ+ = + =

( )76 20000

,/

fc

f yy

bMINL

d A FF

= =

( )102000 12000

,0.6 /

Tb b

uy f y

C CMAXL rdF A F

= =

3510000

Tb

y

CL r

F= =

21

2

11.75 1.05 0.32

bMM

CMM

= + + =

( )2 3,l Max L X= =

(cont'd)

27.5 ksi

16.7 ksi

30.0 ksi

30.0 ksi

f b = M Min / S x = 19.5 ksi < F b [Satisfactory]

CHECK LOCAL BUCKLING (AISC-ASD Tab. B5.1)

bf / (2tf ) = 5.73 < 65 / (Fy)0.5 = 9.19

[Satisfactory]d / tw = 56.83 < 640 / (Fy)

0.5 = 90.51

[Satisfactory]

CHECK M Max BENDING CAPACITY (AISC-ASD, F1.3, page 5-46)

f b = M Max / S x = 30.5 ksi < F b = 0.66 F y = 33.0 ksi [Satisfactory]

CHECK SHEAR CAPACITY (AISC-ASD, F4, page 5-49)f v = V Max / t w d = 7.1 ksi < F v = 0.4 F y = 20.0 ksi [Satisfactory]

DETERMINE CAMBER AT DEAD LOAD CONDITIONL = L 1 / cos θ = 31.62 ft, beam sloped span

a = L 2 / cos θ = 10.54 ft, beam sloped cantilever length

P = P DL cos θ = 2.85 kips, perpendicualr to beam

w 1 = w DL,1 cos θ = 1.09 klf, perpendicualr to beam

w 2 = w DL,2 cos θ = 0.47 klf, perpendicualr to beam

-0.69 in, uplift perpendicular to beam.

USE C = 3/4" AT CANTILEVER.

0.25 in, downward perpendicular at middle of beam.

USE C = 1/4" AT MID BEAM.

CHECK DEFLECTION AT LIVE LOAD CONDITIONP = P LL cos θ = 2.85 kips, perpendicualr to beam

w 1 = w LL,1 cos 2 θ = 0.72 klf, perpendicualr to beam

w 2 = w LL,2 cos 2 θ = 0.45 klf, perpendicualr to beam

3 -0.19 in, uplift to vertical direction.

< 2L 2 / 240 = 1.00 in [Satisfactory]

0.14 in, downward to vertical direction.

< L 1 / 240 = 1.50 in [Satisfactory]

( )2

1/2

, 0.63 1530000

Tyb y y

b

lF rMINF F F

C

= − =

( )2 2

170000,

3/ T

ybb

FCMINFl r

= =

( )312000

, 0.6/

bb y

f

CMINF Fl d A

= =

( )( )

1 3 3

2 3 3

0.66 ,0.6 ,

, ,

, ,

y c

y c ub

b b u

b b

for lF Lfor lF L L

FMAX for lF F L L

MAX for lF F L

≤ < <= = ≤ < ≥

( ) ( )212 33 4 33 24 24End

P L a L aaa w aw LEI EI EI

+ += − + =∆

1 222 4 2

16 384 32MidPa w w aL L L

EI EI EI= − + − =∆

( ) ( )212 33 4 3

cos3 24 24End

P L a L aaa w aw LEI EI EI

θ + +

= − + =∆

1 222 4 2

cos16 384 32MidPa w w aL L L

EI EI EIθ

= − + − =∆


JOB NO. : DATE : REVIEW BY : Simply Supported Beam Design with Torsional Loading Based on AISC-ASD 9th

INPUT DATA & DESIGN SUMMARYBEAM SECTION = > W10X54 = > A d rx ry Ix SxGRAVITY DISTRIBUTED LOAD w = 1.15 kips / ft 15.8 10.1 4.38 2.55 303 60

LATERAL POINT LOAD AT MID F = 5 kips Iy Sy λλλλ tw bf tfTORSION AT MID SPAN T = 5.1 ft-kips 103 20.6 0.0174 0.37 10.00 0.62AXIAL LOAD P = 96 kipsBEAM LENGTH L = 15 ft


VERTICAL BENDING UNBRACED LENGTH L b = 15 ft

AXIAL VERTICAL UNBRACED LENGTH L x = 15 ft

AXIAL HORIZONTAL UNBRACED LENGTH L y = 7.5 ft 7.5

ANALYSISCHECK LOCAL BUCKLING (AISC-ASD Tab. B5.1)

bf / (2tf ) = 8.13 < 65 / (Fy)0.5 = 9.19

[Satisfactory]d / tw = 27.30 < 640 / (Fy)

0.5 = 90.51 THE BEAM DESIGN IS ADEQUATE.[Satisfactory]

DETERMINE GOVERNING MOMENTS AT MIDDLE OF SPAN

Mx = w L 2/ 8 = 32.3 ft-kips

My = F L/ 4 = 18.8 ft-kips

M0 = T L/ (4d) = 22.7 ft-kips 0.584 ,(Philip page 101)

MT = βM0 = 13.3 ft-kips

DETERMINE GOVERNING UNBALANCED SEGMENT LENGTH (AISC-ASD F1)

L c = MIN[76bf/(Fy)0.5 , 20000/(d/Af)Fy] = 8.96 ft

L u = MAX[rT(102000Cb/Fy)0.5 , 12000Cb/(d/Af)0.6Fy] = 20.30 ft

L 3 = rT(510000Cb/Fy)0.5 = 22.39 ft

Where (d/Af) = 1.64 in-1

rT = 2.66

Cb = 1.00

DETERMINE ALLOWABLE BENDING STRESSES (AISC-ASD F1)

= 0.66Fy = N/A ksi, for L b @ [0, L c]

= 0.60Fy = 30.00 ksi, for L b @ (L c, L u]

= MAX(Fb1, Fb3) = N/A ksi, for L b @ (L u, L 3]

= MAX(Fb2, Fb3) = N/A ksi, for L b @ (L 3, Larger)

Where Fb1 = MIN[2/3 - Fy(L /rT)2/(1530000Cb)]Fy , 0.6Fy = 25.85 ksi

Fb2 = MIN[170000Cb/(L /rT)2, Fy/3] = 16.67 ksi

Fb3 = MIN[12000Cb/(L d/Af), 0.6Fy] = 30.00 ksi

CHECK VERTICAL FLEXURAL CAPACITY (AISC-ASD F & Philip page 100)

fbx / Fbx = 0.73 < 1.00 [Satisfactory]

Where fbx = Mx / Sx + 2MT / Sy = 21.93 ksi

DanielTian Li

Fbx =

24 sinh

2sinh

L

L L

λ

β λ λ

= =

(cont'd)CHECK COMPRESSION CAPACITY (AISC-ASD E2)

fa / Fa = 0.24 < 1.33 [Satisfactory]

Where fa = P / A = 6.08 ksi

K = 1.0Es = 29000 ksi (1-F2/2)Fy / (5/3+3F/8-F3/8) = 25.68 kis, for Cc > (K/r)

Cc = (2π2Es/Fy)0.5 = 107 12π2Es/[23(KL /r)2] = N/A kis, for Cc < (K/r)

KL /r = MAX(KL x/rx, KL y/ry) = 41.10 < 200 [Satisfactory]

F = (KL / r) / Cc = 0.38

CHECK COMBINED STRESS (AISC-ASD H1)

fa / Fa = 0.24 > 0.15

1.33 < 1.33

Where Cm = 1.00

fby = My / Sy = 10.92 ksi

Fby = 0.75 Fy = 37.50 ksi

88.39 ksi 120.18 ksi

1.22 < 1.33

1.26 < 1.33 <== Not applicable.

[Satisfactory]

Techincal References: 1. AISC: "Manual of Steel construction 9th", American Institute of Steel Construction, 1990. 2. Philip H. Lin: "Simplified Design for Torsional Loading of Rolled Steel Members", Engineering Journal, AISC, 1977.

Fa =

1 1' '

fCf f myC bymxa bxfF a fa aFbx FbyFex Fey

+ + =

− −

212'2

23

EFex

K l xr x

π =

=212'

223

EFey

K l yr y

π =

=

0.6

ff f bya bxF F Fbx byy

+ + =

ff f bya bxF F Fa bx by

+ + =


JOB NO. : DATE : REVIEW BY : Simply Supported Beam Design with Torsional Loading Based on AISC-LRFD 3rd

INPUT DATA & DESIGN SUMMARYBEAM SECTION = > W10X54 = > A d rx ry Ix Sx

GRAVITY DISTRIBUTED LOAD wu = 1.725 kips / ft 15.8 10.1 4.38 2.55 303 60

LATERAL POINT LOAD AT MID Fu = 7.5 kips Iy Sy λλλλ tw bf tfTORSION AT MID SPAN Tu = 3.213 ft-kips 103 20.6 0.0174 0.37 10.00 0.62

AXIAL LOAD Pu = 144 kips E G J Cw Zx ZyBEAM LENGTH L = 15 ft 29000 11200 1.82 2310 66.6 31.3


VERTICAL BENDING UNBRACED LENGTH L b = 15 ft

AXIAL VERTICAL UNBRACED LENGTH L x = 15 ft

AXIAL HORIZONTAL UNBRACED LENGTH L y = 7.5 ft

ANALYSISCHECK LOCAL BUCKLING (AISC-LRFD Tab. B5.1)

bf / (2tf ) = 8.13 < 0.38 (E / Fy)0.5 = 9.15 96

[Satisfactory]d / tw = 27.30 < 3.76 (E / Fy)

0.5 = 90.55

[Satisfactory]DETERMINE GOVERNING MOMENTS AT MIDDLE OF SPAN

Mu,x = wu L 2/ 8 = 48.5 ft-kips THE BEAM DESIGN IS ADEQUATE.Mu,y = Fu L/ 4 = 28.1 ft-kips

M0,u = Tu L / (4d) = 14.3 ft-kips 0.584 ,(Philip page 101)

Mu,T = βM0,u = 8.4 ft-kips

DETERMINE GOVERNING UNBALANCED SEGMENT LENGTH (AISC-LRFD F1)

L p = 1.76 ry (E / Fyf)0.5 = 9.02 ft

L r = ry X1 [1 + (1 + X2 FL2 )0.5 ]0.5 / FL = 30.08 ft

Where X 1 = π (0.5 E G J A)0.5 / Sx = 3578.1

X 2 = 4 Cw [Sx / (G J)]2 / Iy = 0.0008

Fr = 10.00 ksi

FL = MIN( Fyf - Fr , Fyw) = 40.00 ksi

DETERMINE FLEXURAL DESIGN STRENGTH (AISC-LRFD F1)




2 / Lb2)0.5 / Lb = 578.8 ft-kips

Where C b = 1.14 , (AISC-LRFD Table 4.1)

Mp = N/A ft-kips, for L b @ [0 , L p]

M n = MINCb [Mp - (Mp - Mr) (Lb - Lp) / (Lr - Lp)] , Mp = 277.50 ft-kips, for L b @ (L p , L r]


φ b M n,x = 0.9 Mn = 249.75 ft-kips > M u,x [Satisfactory]

φ b M n,y = 0.9 MIN( Fy Zy , 1.5 Fy Sy) = 115.88 ft-kips > M u,y [Satisfactory]

CHECK COMPRESSION CAPACITY (AISC-LRFD E2)

φ P n = 0.85 Ag Fcr = 593.5 ft-kips > P u [Satisfactory]

Where KL /r = MAX(KL x/rx, KL y/ry) = 41.10 < 200 [Satisfactory]

λc = (KL /r) (Fy / E)0.5 / π = 0.54K = 1.0

[0.658(λc λc)] Fy = 44.19 ksi, for λ c 1.5

(0.877 / λc2) Fy = N/A ksi, for λ c > 1.5

= F cr

DanielTian Li

24 sinh

2sinh

L

L L

λ

β λ λ

= =

(cont'd)

CHECK COMBINED BENDING AND AXIAL CAPACITY (AISC-LRFD H1)

Pu / φPn + (8/9) (Mu,x / φbMn,x + Mu,y / φbMn,y) = 0.63 , for Pu / φPn 0.2

Pu / 2φPn + (Mu,x / φbMn,x + Mu,y / φbMn,y) = N/A , for Pu / φPn < 0.2

CHECK TORSION STRESS AND COMBINED BENDING, AXIAL FORCE (AISC-LRFD H2 & Philip page 100)

f un = Pu / A + Mu,x / Sx + (Mu,y + 2Mu,T) / Sy = 44.94 ksi < φ fn = 0.9 Fy = 45.00 ksi[Satisfactory]

Techincal References: 1. AISC: "Manual of Steel construction, LRFD 3rd", American Institute of Steel Construction, 2001. 2. Philip H. Lin: "Simplified Design for Torsional Loading of Rolled Steel Members", Engineering Journal, AISC, 1977.

[Satisfactory]< 1.00


JOB NO. : DATE : REVIEW BY : Plate Girder Design Based on AISC-ASD 9th, Chapter G


STEEL YIELD STRESS F y = 50 ksi

SIMPLY SUPPORTED SPAN S = 72 ftSUPERIMPOSED UNIFORM DEAD LOAD DL = 1 kips / ftUNIFORM LIVE LOAD LL = 1 kips / ftPOINT DEAD LOAD P DL = 60 kips

POINT LIVE LOAD P LL = 60 kips

TOP FLANGE WIDTH b f,top = 16 in

TOP FLANGE THICKNESS t f,top = 1.5 in

BOTTOM FLANGE WIDTH b f,bot = 16 in

BOTTOM FLANGE THICKNESS t f,bot = 1.5 in

WEB THICKNESS t w = 0.375 inBEAM DEPTH d = 53 in 1/4 x 7 STIFFENER, E. SIDES, AT SUPPORTS & POINT LOAD.DISTANCE POINT LOAD TO END c = 24 ft FLANGE TO WEB WELDING USE 1/4 in - 24 in @ 32 in o.c.UNBRACED LENGTH l = 8 ft

THE GIRDER DESIGN IS ADEQUATE.

ANALYSIS

DETERMINE Fb (AISC-ASD, F1.3, page 5-46)

1.02

r T = 4.34 in , A f = 24.00 in2

14.33 ft

16.54 ft

36.98 ft

30.0 ksi

16.7 ksi

30.0 ksi

33.0 ksi

CHECK WEB SLENDERNESS (AISC-ASD, G1, page 5-51)

a = 47.4 ft, the max clear distance between stiffeners.

h / t w = 133.33 > 132.30 [Satisfactory]

< 282.84

[Satisfactory]

DanielTian Li

( )76 20000

,/

fc

f yy

bMINL

d A FF

= =

( )102000 12000

,0.6 /

Tb b

uy f y

C CMAXL rdF A F

= =

3510000

Tb

y

CL r

F= =

( )2

1/2

, 0.63 1530000

Tyb y y

b

lF rMINF F F

C

= − =

( )2 2

170000,

3/ T

ybb

FCMINFl r

= =

( )312000

, 0.6/

bb y

f

CMINF Fl d A

= =

( )( )

1 3 3

2 3 3

0.66 ,0.6 ,

, ,

, ,

y c

y c ub

b b u

b b

for lF Lfor lF L L

FMAX for lF F L L

MAX for lF F L

≤ < <= = ≤ < ≥

22 2 21

22

2 24 411.75 1.05 0.3 1.75 1.05 0.322

bM l SM l S

CMM S S

− −= + + ≈ + + =

760

bF=

( )14000

, 1.516.5

2000, 1.5

yf yf

yf

for a hF F

for a hF

> + = ≤

(cont'd)DETERMINE ALLOWABLE FLEXURAL STRESS (AISC-ASD G2, pg 5-51)

A w = 18.75 in2 , α = 0.6 F yw / F b = 0.91

0.375 1.000

0.999

32.94 ksi

DETERMINE ALLOWABLE SHEAR STRESS (F4-2, pg 5-49)

h = d - t f,top - t f,bot = 50 in , h / t w = 133

= 5.37 = 0.47

8.08 ksi

TOTAL SUPERIMPOSED GRAVITY LOAD w = DL + LL = 2.000 kips / ft , P = PDL + PLL = 120.00 kips

CHECK EACH SECTION CAPACITIES

Section Left 0.06 S 0.11 S 0.17 S 0.22 S 0.28 S Point 0.44 S 0.56 S 0.67 S 0.78 S 0.89 S RightDistance 0 4.00 8.00 12.00 16.00 20.00 24.00 32.00 40.00 48.00 56.00 64.00 72.00

d (in) 53 53 53 53 53 53 53 53 53 53 53 53 53y (in) 27 27 27 27 27 27 27 27 27 27 27 27 27I (in4) 35742 35742 35742 35742 35742 35742 35742 35742 35742 35742 35742 35742 35742

Wt (plf) 227.1 227.1 227.1 227.1 227.1 227.1 227.1 227.1 227.1 227.1 227.1 227.1 227.1V (kips) 160.18 151.27 142.36 133.45 124.54 115.63 0.00 31.09 48.91 66.73 84.54 102.36 120.18M (ft-k) 0 623 1210 1762 2278 2758 3203 3025 2705 2243 1638 890 0f v (ksi) 8.06 7.61 7.16 6.71 6.27 5.82 0.00 1.56 2.46 3.36 4.25 5.15 6.05

f b (ksi) 0.00 5.54 10.77 15.67 20.27 24.54 28.50 26.92 24.07 19.95 14.57 7.92 0.00

f b,max = 28.50 ksi @ 24.00 ft, from heel.

< F' b = 32.94 ksi[Satisfactory]

'b PGbF F R Re= =

, 0.42.89

yvyv

C FMIN FF

= =

( )

( )

2

2

5.344.0 , / 1.0

/

4.05.34 , / 1.0

/

v

for a ha h

for a ha h

k

+ ≤

= + >

( )245000

, 0.8/

190, 0.8

/

v

wv

v

yw

k for Cvh t

k for Cvh t F

C

≤

= >

7601 0.0005 , 1.0w

PGf b

hAMINRtA F

= − − =

( )312 3

, 1.0

12 2

w

fe

w

f

AA

MINRAA

α α

+ − = = +

BENDING STRESS

-35.00

-30.00

-25.00

-20.00

-15.00

-10.00

-5.00

0.00

Length

fb (k

si) &

F'b

(ksi

)

SHEAR STRESS

-10.00

-5.00

0.00

Length

fv (k

si) &

Fv

(ksi

)

(cont'd)

f v,max = 8.06 ksi @ 0.00 ft, from heel.

< F v = 8.08 ksi [Satisfactory]

DETERMINE DEFLECTION AT MID SPAN

1.40 in ( L / 619 ) (for camber, self Wt included.)

where E = 29000 ksi w = 1.227 kips / ftI = 35742 in4 P = 60 kipsb = 24.6 ft L = 72.0 ft

1.26 in ( L / 684 )

where P = 60 kips w = 1.000 kips / ft

DETERMINE FLANGE TO WEB WELDING

w = 1/4 inw min = 3/16 in, < w

w max = 5/16 in, > w

V max = 160.18 kips

Q = A f (d - y - 0.5 t f,top ) = 618 in3

v max = V max Q / I = 2.77 kips / in

A = 24 in 32 in. o.c.

USE 1/4 in - 24 in @ 32 in o.c.

DESIGN STIFFENERS

1. BEARING STIFFENERS ARE REQUIRED AT EACH END SUPPORT. (AISC-ASD, K1.8, page 5-82)

2. CHECK LOCAL WEB YIELDING FOR THE CONCENTRATED LOAD. (AISC-ASD, K3, page 5-81)R = P = 120.00 kipsN = 0 in, bearing length, point.

k = t f,top + w = 1.75 in

36.57 > 0.66F y [Unsatisfactory]

(BEARING STIFFENERS MUST BE PROVIDED.)

3. CHECK WEB CRIPPLING FOR THE CONCENTRATED LOAD. (AISC-ASD, K4, page 5-81)

134.24 > P [Satisfactory]

(Note : If item 2, local web yielding is Satisfactory, this item does not need to be checked.)

4. CHECK SIDESWAY WEB BUCKING FOR THE CONCENTRATED LOAD. (AISC-ASD, K5, page 5-81) d c = d - 2k = 49.50 in

(d c / t w ) / (l / b f ) = 22.00

120.00 > P [Satisfactory]

(Note : If item 2, local web yielding is Satisfactory, this item does not need to be checked.)

( )4

3/ 2225 0.06415384DL

w PbLbL

EI EIL= + − =∆

( )4

3/ 2225 0.06415384LL

w PbLbL

EI EIL= + − =∆

( )( )0.3 0.707

max

w AFuBv

= =

( )

( )

,5

,2.5

Rfor c d

N kt wR

for c dN kt w

> +

≤ +

=

1.52

1.52

67.5 1 3 , 0.5

34 1 3 , 0.5

yw fww

f w

yw fww

f w

tFN t for c dtd t t

RtFN t for c dt

d t t

+ ≥ = + <

=

33

33

6800 / /0.4 , 1.7

/ /

6800 / /1 0.4 , 1.7 2.3

/ /

/, 2.3

/

w c w c w

f f

w c w c w

f f

c w

f

t d t d tforh l lb b

t d t d tR forh l lb b

d tP forl b

<

= + ≤ < ≥

=

(cont'd)

5. DETERMINE STIFFENER SIZE.

t w = 5/8 in , b st = 7 in

b st / t w = 11.20 < 95 / Fy0.5 , AISC-ASD, B5.1

[Satisfactory]A eff = 10.44 in2 , I = 155 in4

f a = 15.3 ksi

E s = 29000 ksiK / r = 0.75 h / ( I / A eff ) 0.5 = 9.7

C c = (2 π 2 E s / F y ) 0.5 = 107

29.3 ksi, (AISC-ASD, E2, page 5-42)

> f a [Satisfactory]

Techincal Reference: 1. AISC: "Manual of Steel construction 9th", American Institute of Steel Construction, 1990.

( )

( ) ( )

( )

2

2

3

3

2

2

/1

2,

3 / /53 8 8

12,

23 /

yc

c

ac c

c

kl rF

C klfor C

rkl r kl rF

C C

E klfor C

rkl r

π

− ≤= = + − >


JOB NO. : DATE : REVIEW BY : Web Tapered Girder Design Based on AISC-ASD 9th, Appendix F

INPUT DATA & DESIGN SUMMARYSTEEL YIELD STRESS F y = 36 ksi

SIMPLY SUPPORTED SPAN S = 84.67 ftSUPERIMPOSED DEAD LOAD DL = 0.48 kips / ftLIVE LOAD LL = 0.288 kips / ft

FLANGE WIDTH b f = 12 in

FLANGE THICKNESS t f = 0.625 in

WEB THICKNESS t w = 0.3125 in

HEEL DEPTH d 0 = 22 in

MID-SPAN DEPTH d L = 60 inDISTANCE BETWEEN STIFFENERS a = 7.2 ft

UNBRACED LENGTH / PURLIN SPACING L = 7.2 ft(Diaphragm is not bracing member. L is different with " l " in F1.3, pg 5-47)


ANALYSISTOTAL SUPERIMPOSED GRAVITY LOAD w = DL + LL = 0.768 kips / ft

ALLOWABLE FLEXURAL STRESS (APP. F7.14, pg 5-103)

= 21.60 ksi

where A f = 7.50 in2

γ = MIN[(dL - d0) / d0 , 0.268 L/d0, 6.0] = 1.05

A To = tf bf + d0 tw / 6 = 8.65 in2

I To = (tf bf3 + d0 tw

3 / 6) / 12 = 90 in4

3.23 in 34.18 ksi

1.39 in 192.93 ksi

1.11 in 1.37

ALLOWABLE SHEAR STRESS (F4, pg 5-49)

= 5.63 ksi

where h = d L - 2 t f = 59 in

h / t w = 188 > 380 / F y0.5 = 63

= 0.45

= 7.19

DanielTian Li

ToTo

To

IA

r = =

01.0 0.0230f

LdA

hs γ= + =

1.0 0.00385To

Lhw rγ= + =

12000/s

fLh d As oF γ = =

( )2170000

/w

ToLhwF

rγ = =

2 2

2 2

21.0 0.60 , / 3

3 6

, / 3

b ys wb

s w b y

F y forF F F Fy yB F F

B forF F F F

Fγ

γ γγ

γ γ γ

− ≤ > += + ≤

1.751.0 0.25

Bγ

= =+

0.4 , / 380

0.4 , / 3802.89

yw

v yvy yw

for h tF Fy

C Ffor h tF F

F ≤

= ≤ >

( )

( )

2

2

5.344.0 , / 1.0

/

4.05.34 , / 1.0

/

v

for a ha h

for a ha h

k

+ ≤

= + >

( )245000

, 0.8/

190, 0.8

/

v

wv

v

yw

k for Cvh t

k for Cvh t F

C

≤

= >

(cont'd)CHECK EACH SECTION CAPACITIES

Section HEEL 1/24 S 1/12 S 1/8 S 1/6 S 5/24 S 1/4 S 7/24 S 1/3 S 3/8 S 5/12 S 11/24 S MIDDistance 0 3.53 7.06 10.58 14.11 17.64 21.17 24.70 28.22 31.75 35.28 38.81 42.34

d (in) 22 25 28 32 35 38 41 44 47 51 54 57 60I (in4) 2092 2790 3603 4535 5592 6778 8099 9559 11163 12917 14826 16893 19125

Wt (plf) 73.1 76.5 79.8 83.2 86.6 89.9 93.3 96.7 100.0 103.4 106.8 110.1 113.5V (kips) 36.46 33.49 30.51 27.51 24.50 21.48 18.45 15.40 12.34 9.28 6.20 3.10 0.00M (ft-k) 0 123 236 339 430 511 582 642 691 729 756 772 778f v (ksi) 5.30 4.26 3.45 2.79 2.26 1.82 1.44 1.12 0.83 0.59 0.37 0.17 0.00

f b (ksi) 0.00 6.68 11.15 14.11 16.01 17.13 17.68 17.79 17.57 17.09 16.42 15.59 14.64


< F b = 21.60 ksi[Satisfactory]




2.10 in ( Span / 484 ) 1.05 in ( Span / 971 ) (for camber, self Wt included.)

where E = 29000 ksiSection HEEL 1/24 S 1/12 S 1/8 S 1/6 S 5/24 S 1/4 S 7/24 S 1/3 S 3/8 S 5/12 S 11/24 S MIDDistance 0.00 3.53 7.06 10.58 14.11 17.64 21.17 24.70 28.22 31.75 35.28 38.81 42.34I S (in4) 2092 2790 3603 4535 5592 6778 8099 9559 11163 12917 14826 16893 19125

mDL (ft-k) 0 82 157 226 287 341 388 428 461 487 505 516 520mLL (ft-k) 0 41 79 113 143 170 194 213 229 242 251 256 258

munit 0.00 1.76 3.53 5.29 7.06 8.82 10.58 12.35 14.11 15.88 17.64 19.40 21.17∆DL,S 0.00 0.00 0.02 0.04 0.07 0.09 0.10 0.11 0.12 0.12 0.13 0.13 0.12∆LL,S 0.00 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.06 0.06 0.06 0.06 0.06


BENDING STRESS

-Fb

-25.00

-20.00

-15.00

-10.00

-5.00

0.00

Length

-fb

(ksi

)

SHEAR STRESS

-Fv

-6.00-5.00

-4.00-3.00

-2.00-1.00

0.00

Length

-fv

(ksi

)

0

SDL unit

DLs

m m dsE I

= =∆0

SLL unit

LLs

m m dsE I

= =∆


JOB NO. : DATE : REVIEW BY : Web Tapered Girder Design Based on AISC-ASD 9th, Appendix F


STEEL YIELD STRESS F y = 36 ksi

SIMPLY SUPPORTED SPAN S = 90 ftNUMBER OF CONCENTRATED LOADS n = 8 ( @ 10.0 ft o.c.)SUPERIMPOSED DEAD LOAD P D = 4.8 kips

LIVE LOAD P L = 2.88 kips

FLANGE WIDTH b f = 12 in

FLANGE THICKNESS t f = 0.625 in

WEB THICKNESS t w = 0.3125 in

HEEL DEPTH d 0 = 22 in

MID-SPAN DEPTH d L = 60 inDISTANCE BETWEEN STIFFENERS a = 7.2 ft

UNBRACED LENGTH / PURLIN SPACING L = 7.2 ft(Diaphragm is not bracing member. L is different with " l " in F1.3, pg 5-47)


ANALYSISTOTAL SUPERIMPOSED GRAVITY LOAD P = PD + PL = 7.680 kips @ 10.0 ft o.c. 0.288

ALLOWABLE FLEXURAL STRESS (APP. F7.14, pg 5-103)

= 21.60 ksi

where A f = 7.50 in2

γ = MIN[(dL - d0) / d0 , 0.268 L/d0, 6.0] = 1.05

A To = tf bf + d0 tw / 6 = 8.65 in2

I To = (tf bf3 + d0 tw

3 / 6) / 12 = 90 in4

3.23 in 34.18 ksi

1.39 in 192.93 ksi

1.11 in 1.37

ALLOWABLE SHEAR STRESS (F4, pg 5-49)

= 5.63 ksi

where h = d L - 2 t f = 59 in

h / t w = 188 > 380 / F y0.5 = 63

= 0.45

= 7.19

DanielTian Li

ToTo

To

IA

r = =

01.0 0.0230f

LdA

hs γ= + =

1.0 0.00385To

Lhw rγ= + =

12000/s

fLh d As oF γ = =

( )2170000

/w

ToLhwF

rγ = =

2 2

2 2

21.0 0.60 , / 3

3 6

, / 3

b ys wb

s w b y

F y forF F F Fy yB F F

B forF F F F

Fγ

γ γγ

γ γ γ

− ≤ > += + ≤

1.751.0 0.25

Bγ

= =+

0.4 , / 380

0.4 , / 3802.89

yw

v yvy yw

for h tF Fy

C Ffor h tF F

F ≤

= ≤ >

( )

( )

2

2

5.344.0 , / 1.0

/

4.05.34 , / 1.0

/

v

for a ha h

for a ha h

k

+ ≤

= + >

( )245000

, 0.8/

190, 0.8

/

v

wv

v

yw

k for Cvh t

k for Cvh t F

C

≤

= >

(cont'd)CHECK EACH SECTION CAPACITIES

Section HEEL 1/18 S 2/18 S 3/18 S 4/18 S 5/18 S 6/18 S 7/18 S 8/18 S MIDDistance 0 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00

d (in) 22 26 30 35 39 43 47 52 56 60

I (in4) 2092 3048 4211 5592 7203 9056 11163 13536 16186 19125Wt (plf) 73.1 77.6 82.1 86.6 91.1 95.6 100.0 104.5 109.0 113.5V (kips) 34.92 34.54 26.46 26.04 17.92 17.45 9.28 8.77 0.56 0.00M (ft-k) 0 174 345 477 606 694 780 825 868 869f v (ksi) 5.08 4.22 2.78 2.40 1.47 1.30 0.63 0.54 0.03 0.00

f b (ksi) 0.00 8.96 14.98 17.73 19.62 19.83 19.85 18.86 17.94 16.361 2 4 6 9 12 16 20 25 30


< F b = 21.60 ksi[Satisfactory]




2.21 in ( Span / 489 ) 1.79 in ( Span / 605 ) (for camber, self Wt included.)

where E = 29000 ksiSection HEEL 1/18 S 2/18 S 3/18 S 4/18 S 5/18 S 6/18 S 7/18 S 8/18 S MID

Distance 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00I S (in4) 2092 3048 4211 5592 7203 9056 11163 13536 16186 19125

mDL (ft-k) 0 116 230 304 375 406 435 422 407 351mLL (ft-k) 0 58 115 173 230 288 346 403 461 518

munit 0.00 2.50 5.00 7.50 10.00 12.50 15.00 17.50 20.00 22.50∆DL,S 0.00 0.01 0.06 0.10 0.14 0.16 0.17 0.17 0.16 0.14∆LL,S 0.00 0.01 0.03 0.06 0.08 0.11 0.13 0.15 0.16 0.18


BENDING STRESS

-25.00

-20.00

-15.00

-10.00

-5.00

0.00

Length

-fb

(ksi

)

SHEAR STRESS

-6.00

-5.00

-4.00

-3.00

-2.00

-1.00

0.00

Length

-fv

(ksi

)

0

SDL unit

DLs

m m dsE I

= =∆0

SLL unit

LLs

m m dsE I

= =∆


JOB NO. : DATE : REVIEW BY : Composite Beam Design with Verco Floor Deck Based on AISC-ASD

INPUT DATA & DESIGN SUMMARYBEAM SECTION = > W12X16 = > A d Ix Sx

4.71 12.0 103 17.1FLOOR DECK TYPE = > W3-5" NW

RIBS PERPENDICULAR TO BEAM ? Yes (perpendicular)BEAM SPAN L = 22.17 ftBEAM SPACING (DECK SPAN) B = 8.25 ft, o.c.SUPERIMPOSED LOAD w s = 80 lbs / ft2

BEAM YIELD STRESS F y = 50 ksi

CONCRETE STRENGTH f c ' = 3 ksi

SHEAR STUD DIA. (1/2, 5/8, 3/4) φ = 5/8 in

NUMBER OF STUD IN ONE RIB N r = 1

( Total 9 - 3/4 x 4.5" Shear Studs Required)


ANALYSISCHECK DIMENSION REQUIREMENTS (AISC-ASD I5.1, page 5-61)

t 0 = 2 in > 2 in [Satisfactory]

h r = 3 in < 3 in [Satisfactory]

φ = 3/4 in < 3/4 in [Satisfactory]

H s = h r + 1.5 = 4.5 in > 3 in [Satisfactory]s = 33 in o.c. < 36 in o.c. [Satisfactory]

w r = 6 in > 2 in [Satisfactory]

DETERMINE COMPOSITE PROPERTIESb = MIN ( L / 4 , B ) = 66.51 in, (AISC-ASD I1.1, page 5-56)

9.29 , (ACI 8.5.1)

A ctr = b t 0 / n = 14.3 in2

13.5 in, from steel bottom.

462 in4

34 in3, refered to steel bottom.

133 in3, refered to concrete top.

CHECK BENDING & SHEAR CAPACITIESw = w s + w wt = 80.00 + 42.30 = 122.30 lbs / ft2 (total gravity loads)

62 ft-kips, (changeable). 11 kips, (changeable per actual).

Bottom: 22 ksi < 0.9 F y = 45 ksi, (non-shored, AISC-ASD I2.2, page 5-57)[Satisfactory]

Top: 1.024 ksi < 0.45 f c ' = 1.35 ksi, (AISC-ASD I2.2, page 5-57)

[Satisfactory]

Shear: 0.143 ksi < 12.386 ksi [Satisfactory](neglecting steel capacity conservativly)

DanielTian Li

En

Ec= =

( )00.5 0.5ctr

ctr

rb

d Adh tAyAA

+ + += =

+

( ) ( )2 22 0

00.5 0.512ctr

tr x ctr rb btAd d yyA tI I A h− + −= + + + =+

trtr

b

IS

y= =

( )0

trt

r b

IS

yd h t= =

+ + −

maxb

tr

Mf

S= =

max1.7c

t

Mf

nS= =

max

0

1.7v

Vf

bt= = ' '

0 02 2(0.85)c cb bf ft tφ = =

2

max8

wBLM = = max

2

wBLV = =

(cont'd)CHECK SHEAR CONNECTOR CAPACITY

117.75 kips, (AISC-ASD I4-1 & I4-2, page 5-58)

S eff = M max / (0.66 F y ) = 23 in3, refered to steel bottom.

29.44 kips, (AISC-ASD I2-1 & I4, page 5-57 & 58)

205

0.850 < 1.0 , (AISC-ASD I5-1, page 5-61)

q' = ρ q = 9.78 kips, (AISC-ASD I5.2, page 5-61)

Allowable Horizontal Shear Load for One Connector (q, kips)(AISC-ASD Table I4.1 with cofficient Table I4.2, page 5-59) 6778

Dia. φ min. Hs( in ) ( in ) 3.0 3.5 4.0 or Larger

0.500 1/2 2 5.1 5.5 5.90.625 5/8 2 1/2 8.0 8.6 9.20.750 3/4 3 11.5 12.5 13.3

2 N 1 ' = 2 V h ' / q' = 7 , total number on the beam for partical composite action.

2 N 1 = 2 V h / q' = 25 , total number on the beam for full composite action.

n = MAX(2N 1 ' , 2N 1 /4) = 9 , total number required on the beam, (AISC-ASD I4, page 5-59)

Techincal References: 1. AISC: "Manual of Steel construction 9th", American Institute of Steel Construction, 1990. 2. Alan Williams Ph.D., S.E., C.Eng.: "Structural Steel Design - Volume 1: ASD", ICBO, 2001.

Concrete fc'

( )'0.85 / 2 / 2,c y sh cfMINV A F A= =

2' , 0.25

eff sh h

tr s

S SMAXV V

S S

− = = −

0.851.0 ,

0.6 1.0 ,

w H sr for Perpendicularh hN r rr

w H sr for Parallelh hr r

ρ

− = =

−


JOB NO. : DATE : REVIEW BY : Composite Beam Design with Formed Steel Deck Based on AISC-ASD

INPUT DATA & DESIGN SUMMARYBEAM SECTION = > W21X62 = > A d Ix SxBEAM SPAN L = 40 ft 18.3 21.0 1330 127BEAM SPACING (DECK SPAN) B = 11 ft, o.c.TOTAL GRAVITY LOAD (DL+LL) w = 205 lbs / ft2

RIBS PERPENDICULAR TO BEAM ? yes (perpendicular)BEAM YIELD STRESS F y = 50 ksi

CONCRETE STRENGTH f c ' = 4.5 ksi

TOPPING CONCRETE THICK. t 0 = 3 in

SHEAR STUD DIA. (1/2, 5/8, 3/4) φ = 5/8 in

NOMINAL RIB HEIGHT h r = 2 in 3/4

AVERAGE WIDTH OF RIB w r = 3 in

NUMBER OF STUD IN ONE RIB N r = 1

( Total 34 - 5/8 x 3.5" Shear Studs Required)


ANALYSISCHECK DIMENSION REQUIREMENTS (AISC-ASD I5.1, page 5-60)

t 0 = 3 in > 2 in [Satisfactory]

h r = 2 in < 3 in [Satisfactory]

φ = 5/8 in < 3/4 in [Satisfactory]

H s = h r + 1.5 = 3.5 in > 2.5 in [Satisfactory]s = 15 in o.c. < 36 in o.c. [Satisfactory]

w r = 3 in > 2 in [Satisfactory]

DETERMINE COMPOSITE PROPERTIESb = MIN ( L / 4 , B ) = 120 in, (AISC-ASD I1.1, page 5-56)

7.58 , (ACI 8.5.1)

A ctr = b t 0 / n = 47.5 in2

20.6 in, from steel bottom.

3954 in4

192 in3, refered to steel bottom.

733 in3, refered to concrete top.

CHECK BENDING & SHEAR CAPACITIES

451 ft-kips, (changeable). 45 kips, (changeable per actual).

Bottom: 28 ksi < 0.9 F y = 45 ksi, (non-shored, AISC-ASD I2.2, page 5-57)[Satisfactory]

Top: 1.655 ksi < 0.45 f c ' = 2.025 ksi, (AISC-ASD I2.2, page 5-57)

[Satisfactory]

Shear: 0.213 ksi < 41.054 ksi [Satisfactory](neglecting steel capacity conservativly)

DanielTian Li

29000'57

En

Ec f c

= = =

( )00.5 0.5ctr

ctr

rb

d Adh tAyAA

+ + += =

+

( ) ( )2 22 0

00.5 0.512ctr

tr x ctr rb btAd d yyA tI I A h− + −= + + + =+

trtr

b

IS

y= =

( )0

trt

r b

IS

yd h t= =

+ + −

maxb

tr

Mf

S= =

max1.7c

t

Mf

nS= =

max

0

1.7v

Vf

bt= = ' '

0 02 2(0.85)c cb bf ft tφ = =

2

max8

wBLM = = max

2

wBLV = =

(cont'd)CHECK SHEAR CONNECTOR CAPACITY

457.5 kips, (AISC-ASD I4-1 & I4-2, page 5-58)

S eff = M max / (0.66 F y ) = 164 in3, refered to steel bottom.

148.62 kips, (AISC-ASD I2-1 & I4, page 5-57 & 58)

0.956 < 1.0 , (AISC-ASD I5-1, page 5-61)

q' = ρ q = 8.80 kips, (AISC-ASD I5.2, page 5-61)

Allowable Horizontal Shear Load for One Connector (q, kips), (AISC-ASD Table I4.1, page 5-59)Dia. φ min. Hs( in ) ( in ) 3.0 3.5 4.0 or Larger

0.500 1/2 2 5.1 5.5 5.90.625 5/8 2 1/2 8.0 8.6 9.20.750 3/4 3 11.5 12.5 13.3

2 N 1 ' = 2 V h ' / q' = 34 , total number on the beam for partical composite action.

2 N 1 = 2 V h / q' = 104 , total number on the beam for full composite action.

n = MAX(2N 1 ' , 2N 1 /4) = 34 , total number required on the beam, (AISC-ASD I4, page 5-59)

Techincal References: 1. AISC: "Manual of Steel construction 9th", American Institute of Steel Construction, 1990. 2. Alan Williams Ph.D., S.E., C.Eng.: "Structural Steel Design - Volume 1: ASD", ICBO, 2001.

Concrete fc'

( )'0.85 / 2 / 2,c y sh cfMINV A F A= =

0.851.0sr

r rr

w Hh hN

ρ = − =

2' , 0.25

eff sh h

tr s

S SMAXV V

S S

− = = −


JOB NO. : DATE : REVIEW BY : WF Base Plate Design Based on AISC-ASD 9th Edition

INPUT DATA & DESIGN SUMMARYAXIAL LOAD OF COMPRESSION P = 600 kipsCONCRETE STRENGTH fc' = 3 ksiCOLUMN SIZE => W21X73BASE PLATE SIZE N = 19 in

B = 16 inAREA OF CONCRETE SUPPORT A2 = 1156 in2

(geometrically similar to and concentric with the loaded area.)

USE19 x 16

2-1/4 in thick plate

ANALYSISCHECK BEARING PRESSURE [Page 3-107]:

P / A1 > 0.35 f'c < = Not govern( 1.97 ) ( 1.05 )

Where A1 = 304 in2, actual area of base plate.

P / A1 < MIN [ 0.35 f'C (A2 / A1)0.5, 0.7 f'C ] [Satisfactory]

( 1.97 ) ( 2.05 )

DETERMINE THE VALUE OF m AND n :m = 0.5 (N - 0.95 d) = -0.57 inn = 0.5 (B - 0.8 bf ) = 4.68 inWhere d = 21.20 in, depth of column section.

bf = 8.30 in, flange width of column section.

DETERMINE THICKNESS OF BASE PLATE [Page 3-108]:

tp = 2 c (P / A1 Fy)0.5 = 2.19 in

Where c = MAX ( m, n, λn' ) = 4.68 inλ = 1.0

n' = ( d bf )0.5 / 4 = 3.32 in

DanielT. Li


JOB NO. : DATE : REVIEW BY : TS Base Plate Design Based on AISC-ASD 9th Edition

INPUT DATA & DESIGN SUMMARYAXIAL LOAD OF COMPRESSION P = 100 kipsCONCRETE STRENGTH fc' = 3 ksiCOLUMN SIZE => HSS8X6X3/16BASE PLATE SIZE N = 16 in



USE16 x 14

1 in thick plate


P / A1 < 0.35 f'c < = Not govern( 0.45 ) ( 1.05 )



( 0.45 ) ( 2.10 )

DETERMINE THE VALUE OF m AND n :m = 0.5 (N - 0.95 d) = 4.20 inn = 0.5 (B - 0.95 b) = 4.15 inWhere d = 8.00 in, depth of column section.

b = 6.00 in, width of column section.


tp = 2 c (P / A1 Fy)0.5 = 0.94 in

Where c = MAX ( m, n, λn' ) = 4.20 inλ = 1.0n' = d / 4 = 2.00 in

DanielT. Li


JOB NO. : DATE : REVIEW BY : Pipe Base Plate Design Based on AISC-ASD 9th Edition

INPUT DATA & DESIGN SUMMARYAXIAL LOAD OF COMPRESSION P = 100 kipsCONCRETE STRENGTH fc' = 3 ksiCOLUMN SIZE => HSS7.500X0.312BASE PLATE SIZE N = 16 in



USE16 x 16



P / A1 < 0.35 f'c < = Not govern( 0.39 ) ( 1.05 )



( 0.39 ) ( 1.97 )

DETERMINE THE VALUE OF m AND n :m = 0.5 (N - 0.80 d) = 5.00 inn = 0.5 (B - 0.80 d) = 5.00 inWhere d = 7.50 in, depth of column section.


tp = 2 c (P / A1 Fy)0.5 = 1.04 in

Where c = MAX ( m, n, λn' ) = 5.00 inλ = 1.0n' = d / 4 = 1.88 in

DanielT. Li


JOB NO. : DATE : REVIEW BY : WF Base Plate Design Based on AISC-LRFD 2nd Edition

INPUT DATA & DESIGN SUMMARYFACTORED AXIAL FORCE Ru = 1100 kips

CONCRETE STRENGTH fc' = 3 ksiCOLUMN SIZE => W33X354BASE PLATE SIZE N = 28 in



USE28 x 26



Ru / A1 < φc ( 0.85 f'c ) < = Not govern( 1.51 ) ( 1.53 )


φc = 0.6

Ru / A1 < φc 0.85 f'C MIN[(A2 / A1)0.5, 2] [Satisfactory]

( 1.51 ) ( 1.70 )

DETERMINE THE VALUE OF m AND n :m = 0.5 (N - 0.95 d) = -2.91 inn = 0.5 (B - 0.8 bf ) = 6.56 inWhere d = 35.60 in, depth of column section.

bf = 16.10 in, flange width of column section.


tp = l ( 2Ru / 0.9 A1 Fy)0.5 = 2.00 in

Where l = MAX ( m, n, λn' ) = 6.56 inλ = 1.0

n' = ( d bf )0.5 / 4 = 5.99 in

DanielT. Li


JOB NO. : DATE : REVIEW BY : Beam Connection Based on AISC-ASD 9th Edition

INPUT DATA & DESIGN SUMMARYWF BEAM SECTION = > W21x44GRAVITY SERVICE LOAD P = 37.57 kipsLATERAL TENSION LOAD, ASD T = 180 kipsPLATE THICKNESS t = 0.5 inPLATE MATERIAL (A36, A529-42, A529-50,A572-65) ASTM = A36TRIAL WELD SIZE w = 0.437 in ( 7/16 in)BOLT DIAMETER φ = 1 in ( 1 in)BOLT MATERIAL (A307, A325, A490) ASTM = A325HOLE TYPE (STD, NSL, OVS, SSL, LSL) = > STD

STD = Standard round holes ( d + 1/16 " )NSL = Long or short-slotted hole normal to load direction OVS = Oversize round holesSSL = Short-slotted holes holesLSL = Long-slotted holes USE PLATE 18" x 6 1/2" x 1/2" WITH WELD 7/16" EACH SIDE TO

CONNECTION TYPE (SC, N, X) = > SC COLUMN AND 2 ROW OF TOTAL (11) - 1" BOLTS AT BEAM END.SC = Slip critical connectionN = Bearing-type connection with threads included in the shear plane

X = Bearing-type connection with threads excluded from the shear planeIS TOP FLANGE COPED ? (1=Yes, 0=No,) = > 1 Yes

ANALYSISDATA FOR ROLLED SECTION CHOSEN

A d tw bf tf Sx13 20.7 0.35 6.5 0.45 81.6

CHECK CAPACITY OF BOLTSAllow shear per bolt = 13.4 kips / bolt3/4(P2 + T2 )0.5 = 138 kipsNo. of bolts required = 10.3Bolt spacing required = 3.00 in Bolt spacing used = 3.0 inEdge spacing required = 1.50 in Edge spacing used = 1.5 inBolt group capacity = 147 kips > 3/4(P2 + T2 )0.5 = 138 kips

> P = 38 kips [Satisfactory]CHECK CAPACITY OF WELDING (70XX)

e = 3.5 inPlate thickness = 0.50 inWeld size,w = 0.44 inMin allowable weld = 0.19 in [Satisfactory]Max allowable weld = 0.44 in [Satisfactory]te = 0.31 inD = 18.0 inI = 2 ( te D3 / 12 ) = 300.3 in4

Vertical shear = P / Aw = P / 2 D te = 3.4 ksiBending stress = P e / I = 0.4 ksi

Tension stress = T / Aw = T / 2 D te = 16.2 ksi

Resultant Stress = 3/4 [ (P/Aw)2 + (P e/ I + T/Aw)2 ]0.5 = 12.7 ksi

Resultant Stress = [ (P/Aw)2 + (P e/ I )2 ]0.5 = 3.4 ksiAllow shear Fv = 0.3 x 70 = 21.0 ksi > 12.7 ksi [Satisfactory]

CHECK PLATE FOR SHEAR CAPACITYP / A = 4.2 ksi < 0.4 x 36 = 14.4 ksi [Satisfactory]

CHECK PLATE FOR TENSION CAPACITY(3/4) T / A = 15.0 ksi < 0.6 x 36 = 21.6 ksi [Satisfactory]

CHECK NET SHEAR FRACTUREFu = 58 ksi

Pallow = 0.3 Fu [ D - n (ds + 1/8 ) ] t = 98 kips > 37.57 kips [Satisfactory]

CHECK NET TENSION FRACTUREFu = 58 ksi

Tallow = (4/3) 0.5 Fu [ D - n (ds + 1/8 ) ] t = 218 kips > 180 kips [Satisfactory]

CHECK BLOCK SHEAR ( WEB TEAR-OUT)( Applies only if top flange is coped for P)

lh = 3.0 in

lv = 11.0 in

Fu = 58 ksi

Rbs,P = 0.3 Av Fu + 0.5 At Fu = (0.3 lv + 0.5 lh) tw Fu = 97 kips> P = 37.57 kips [Satisfactory]

Rbs,T = 4/3 (0.5 lv + (2) 0.3 lh) tw Fu = 198 > T = 180 kips [Satisfactory]

DanielT. Li


JOB NO. : DATE : REVIEW BY : Brace Connection Design Based on AISC-ASD 9th

INPUT DATA & DESIGN SUMMARYBRACE SECTION (Tube or Pipe) = > HSS8X8X5/8 Tube A rmin t hBRACE AXIAL LOAD AT SERVICE LEVEL T = 450 kips 16.40 2.98 0.58 8.00

(For seismic T = Ω0 x ASD / 1.7 suggested. CBC 2213.4.2)

ANGLE BETWEEN BRACE & COLUMN θ = 50 0

COLUMN INTERFACE DIMENSION 2 β = 24 in

COLUMN CENTER TO INTERFACE ec = 7.33 in

BEAM CENTER TO INTERFACE eb = 14.8 in

( 5/8" Gusset Plate with 5/16" Fillet Weld, 4 leg x 25" Length at Brace, and 2 leg x 24" at Column Interface, 2 leg x 50" at Beam Interface. )


ANALYSISDETERMINE BEST FILLET WELD SIZE PER BRACE THICKNESS (ASD Sec.J2.2b)

> wMIN = 0.1875 in< wMAX = 0.4375 in

[Satisfactory]

DETERMINE REQUIRED WELD LENGTH AT BRACE (ASD Sec.J2.4 )

L = T / [(4) (0.3) Fu (0.707 w)] = 450.00 / [(4) (0.3) (70) (0.707x5/16)] = 24.25 in

( USE 25 in )

CHECK SHEAR RUPTURE CAPACITY OF SLOTED BRACE (ASD Sec.J4)

Tt,rup,brace =(0.3Fu)Anu = 1010.94 kips > T[Satisfactory]


Anu = 4 t L = 4 x 0.581 x 25 = 58.10 in2

DETERMINE REQUIRED THICKNESS OF GUSSET PLATE PER Tt,rup,brace ABOVE (ASD Tab. J2.4)

tg = 5/8 in

CHECK SHEAR RUPTURE CAPACITY OF GUSSET PLATE (ASD Sec.J4 & CBC 2213A.4.2)

Tt,rup,gusset = (0.3Fu)Anv = 543.8 kips > T [Satisfactory]

Where Fu = 58 ksi (A36 Steel)

Anv = 2 tg L = 2 x 5/8 x 25 = 31.25 in2

CHECK TENSION CAPACITY AT SLOTED BRACE (ASD Sec.D1). CHECK GUSSET BLOCK SHEAR CAPACITY (ASD J4)Tt,brace = (0.5FuUAn) = 458.75 kips > T Ts,guss = [0.3FuAnv + 0.5FuAgt] =


Where U = 1 (ASD Sec.B3) > T = 450.0 [Satisfactory]

An = A - 2 t tg = 15.819 in2

DETERMINE CONNECTION INTERFACE FORCES, USING THE UNIFORM FORCE METHOD & TO REMAIN FREE OF INTERFACE MOMENTS

β = 12.00 in, as given

α = (eb + β) tanθ - ec = 24.61 in

r = [(eb + β)2 + (ec + α)2] 0.5 = 41.69 in

Vc = (β / r) T = 129.5 kips

Hc = (ec / r) T = 79.1 kips 50

Vb = (eb / r) T = 159.7 kips

Hb = (α / r) T = 265.6 kips

CHECK WELD CAPACITY AT INTERFACES (ASD Sec.J2.4 )fVc = Vc / (4 β 0.707 w) = 12.21 ksi

fHc = Hc / (4 β 0.707 w) = 7.46 ksi

fVb = VB / (4 α 0.707 w) = 7.34 ksi

fHb = HB / (4 α 0.707 w) = 12.21 ksi

fv,c = [(fvc)2 + (fHc)

2] 0.5 = 14.31 ksi < 0.3 Fu = 21.00 ksi [Satisfactory]

fv,b = [(fvb)2 + (fHb)2] 0.5 = 14.25 ksi < 0.3 Fu = 21.00 ksi [Satisfactory]

in

DanielTian Li

w = 5/16

β

α

θ

θ

β

α


JOB NO. : DATE : REVIEW BY : Brace Connection Design Based on AISC-LRFD 2th & 3th

INPUT DATA & DESIGN SUMMARYBRACE SECTION (Tube or Pipe) = > HSS8X8X5/8 Tube A rmin t h

BRACE AXIAL LOAD AT STRENGTH LEVEL Tu = 980 kips, SD 16.40 2.98 0.58 8.00

(For OCBF Tu = Ry Fy Ag suggested. AISC Seismic 14.2)

ANGLE BETWEEN BRACE & COLUMN θ = 50 0

COLUMN INTERFACE DIMENSION 2 β = 24 in

COLUMN CENTER TO INTERFACE ec = 7.33 in

BEAM CENTER TO INTERFACE eb = 14.8 in

( 5/8" Gusset Plate with 5/16" Fillet Weld, 4 leg x 36" Length at Brace, and 2 leg x 24" at Column Interface, 2 leg x 50" at Beam Interface. Cover Plate 3/4 x 7 at Each Sides.)


ANALYSISDETERMINE BEST FILLET WELD SIZE PER BRACE THICKNESS (LRFD Sec.J2.2b)

> wMIN = 0.1875 in< wMAX = 0.4375 in

[Satisfactory]

DETERMINE REQUIRED WELD LENGTH AT BRACE (LRFD Sec.J2.4)

L = Tu / [(4) φ Fw (0.707 w)] = 980.00 / [(4) 0.75 (0.6 x 70) (0.707x5/16)] = 35.20 in

( USE 36 in )

CHECK DESIGN SHEAR RUPTURE CAPACITY OF SLOTED BRACE (LRFD Sec.J4.1)

φTn = φ(0.6Fu)Anv = 2183.63 kips > Tu [Satisfactory]Where φ = 0.75

Fu = 58 ksi (LRFD Tab.1-4, Pg. 1-21)

Anu = 4 t L = 4 x 0.581 x 0.1875 = 83.66 in2


tg = 5/8 in

CHECK SHEAR RUPTURE CAPACITY OF GUSSET PLATE (LRFD Sec.J4.1)

φTn = φ(0.6Fu)Anv = 1174.5 kips > Tu [Satisfactory]

Where φ = 0.75


Anv = 2 tg L = 2 x 5/8 x 36 = 45.00 in2

CHECK TENSION CAPACITY AT SLOTED BRACE (LRFD Sec.D1).

φTn = φ Fu Ae = 607.94 kips < Tu [Cover Plate Required]Where φ = 0.75



An = Ag - 2 (tg + 1/8) t = 15.53 in2

Ae = U An = 13.98 in2

7.33Try Cover Plate 3/4 x 7 , at Each Sides.

Region x 0.5 An x A


Σ 13.01 46.26 An = 15.53 + 10.50 = 26.03 in2

Ae = U An = 23.43 in2

Thus, φTn = φ Fu Ae = 1019.02 kips > Tu [Satisfactory]

DanielTian Li

w = 5/16 in

β

α

θ

(cont'd)

DETERMINE CONNECTION INTERFACE FORCES, USING THE UNIFORM FORCE METHOD & TO REMAIN FREE OF INTERFACE MOMENTS

β = 12.00 in, as given

α = (eb + β) tanθ - ec = 24.61 in

r = [(eb + β)2 + (ec + α)2] 0.5 = 41.69 in

Vc = (β / r) Tu = 282.1 kips

Hc = (ec / r) Tu = 172.3 kips

Vb = (eb / r) Tu = 347.9 kips

Hb = (α / r) Tu = 578.4 kips

CHECK WELD CAPACITY AT INTERFACES (ASD Sec.J2.4 )fVc = Vc / (4 β 0.707 w) = 26.60 ksi

fHc = Hc / (4 β 0.707 w) = 16.25 ksi

fVb = VB / (4 α 0.707 w) = 16.00 ksi

fHb = HB / (4 α 0.707 w) = 26.60 ksi

fv,c = [(fvc)2 + (fHc)

2] 0.5 = 31.17 ksi < φ Fw = 31.50 ksi [Satisfactory]

fv,b = [(fvb)2 + (fHb)2] 0.5 = 31.04 ksi < φ Fw = 31.50 ksi [Satisfactory]

θ

β

α



INPUT DATA & DESIGN SUMMARYWF SECTION = > W24x76 OPENING DIMENSIONS b = 48 in

MOMENT @ ABCD SECTION MABCD = 90 ft-k h = 10 in

MOMENT @ EFGH SECTION MEFGH = 75 ft-k OPENING LOCATION e = 8 in

MAX SHEAR @ OPENING V = 60 kips PLATE SIZE @ EACH SIDE t = 0.75 inSTEEL MATERIALS (A36, A50) ASTM = A50 L = 4.25 inTRIAL WELD SIZE w = 0.25 in

USE (4) - 3/4" x 4-1/4" x 6' -10" PLATES, WITH WELD 1/4" AT EACH SIDES, TOP & BOTTOM.

ANALYSISDATA FOR ROLLED SECTION CHOSEN

A d tw bf tf Sx Fb = 33.00 ksi

22.4 23.9 0.44 8.99 0.68 176 Fv = 20.00 ksi

PROPERTIES OF OPENING SECTION

t L y1t y1b y2t y2b A top A bott Y1 Y2 I top I bott I total0.75 4.25 4.12 3.88 2.88 3.04 15.7 14.8 12.2 11.7 173 89 2,402

CHECK BENDING STRESSESV top V bott Ms top Ms bott34.48 25.52 69.0 51.0

MAIN BENDING STRESSES SECONDARY BENDING STRESSES TOTAL BENDING STRESSES

σ1 A = -5.50 ksi σ2 A = -19.14 ksi ( σ1+σ2 ) A = -24.64 ksiσ1 B = -1.91 ksi σ2 B = 19.14 ksi ( σ1+σ2 ) B = 17.23 ksiσ1 C = 2.59 ksi σ2 C = -20.29 ksi ( σ1+σ2 ) C = -17.70 ksiσ1 D = 5.25 ksi σ2 D = 20.29 ksi ( σ1+σ2 ) D = 25.54 ksiσ1 E = -4.59 ksi σ2 E = 19.14 ksi ( σ1+σ2 ) E = 14.55 ksiσ1 F = -1.59 ksi σ2 F = -19.14 ksi ( σ1+σ2 ) F = -20.73 ksiσ1 G = 2.16 ksi σ2 G = 20.29 ksi ( σ1+σ2 ) G = 22.44 ksiσ1 H = 4.37 ksi σ2 H = -20.29 ksi ( σ1+σ2 ) H = -15.91 ksi

Max fb = Max ( σ1+σ2 ) = 25.54 ksi < 33.00 ksi [Satisfactory]

DETERMINE STIFFENER EXTENSIONSMax bending stress fb @ stiffener = 22.4 ksiForce, F = 8.5 x 0.75 x 22.44 = 143 kAllow stress in web = 20.0 ksiExtension = 143.1 / ( 0.44 x 20.00 ) = 16.3 in

Say => 17.0 inCHECK WELDINGWeld width, w = 0.25 in Min weld = 0.25 in Max weld = 0.69 in

V top Q top I top q top q critical = 6.38 k/in34.48 23.5 173 4.69 te = 0.707w = 0.18 in

q / 4 te = 9.03 ksi < Fv = 20.00 ksi [Satisfactory]V bott Q bott I bott q bott25.52 22.4 89 6.38

T. LiCheck Reinforcement of WF Beam at Opening

Daniel



INPUT DATA & DESIGN SUMMARYBOLT DIAMETER φ = 0.875 in ( 7/8 in)BOLT MATERIAL (A307, A325, A490) ASTM = A325HOLE TYPE (STD, NSL, OVS, SSL, LSL) = > STD

STD = Standard round holes ( d + 1/16 " )NSL = Long or short-slotted hole normal to load direction OVS = Oversize round holesSSL = Short-slotted holes holesLSL = Long-slotted holes

LOADING (S=single shear,S=double shear) DCONNECTION TYPE (SC, N, X) = > N

SC = Slip critical connectionN = Bearing-type connection with threads included in the shear planeX = Bearing-type connection with threads excluded from the shear plane

PLATE THICKNESS t1 = 1 inPLATE MATERIAL (A36, A441-46, A472-50,A514-90) ASTM = A36NUMBER OF ROWS = > 3

NUMBER OF BOLTS / ROW = > 3 Pallowable = 161 kipsBOLT SPACING s = 3 inEDGE DISTANCE e = 1.25 in

ANALYSISMIN.THICKNESS OF SIDE PLATES t2 = 0.5 in PLATE YIELD STRENGTH Fy = 36.0 kips

PLATE WIDTH D = 8.50 in PLATE TENSILE STRENGTH Fu = 58.0 kips

Bolt φ P (kips) CHECK MIDDLE PLATE FOR SHEAR : 0.875 25.30 228

CHECK MIDDLE PLATE FOR BEARING DUE TO BOLT SPACING : 0.875 60.90 5483 φ 1.5 φ

CHECK MIDDLE PLATE FOR BEARING DUE TO EDGE DISTANCE : 3.00 1.50 0.875 36.3 327

CHECK MIDDLE PLATE FOR TENSION : 0.6 Fy Ag = 184 0.5 Fu U An = 161 <=gov 161

kips / bolt

Capacity of Bolts in Bearing Connection Based on AISC-ASD 9th Edition

DanielTian Li



INPUT DATA & DESIGN SUMMARYBOLT DIAMETER φ = 0.75 in ( 3/4 in)BOLT MATERIAL (A307, A325, A490) ASTM = A325NUMBER OF BOLTS n = 6DISTANCE FROM BOLT CENTERLINE TO EDGE a = 1.778 inDISTANCE FROM BOLT CENTERLINE TO STEM FACE b = 1.792 inSMALLER THICKNESS OF FLANGES t = 0.695 inLENGTH OF FLANGE TRIBUTARY TO EACH BOLT p = 4.5 in

YIELD STRENGTH OF THE FLANGE MATERIAL Fy = 50 ksi

Pallowable = 93.0 kips

ANALYSIS (AISC-ASD, page 4-89 to 4-95)

d' = φ + 1/16 = 0.813 in B = 19.4 kips / bolt, (from table below, AISC-ASD page 4-3 )b' = b - 0.5 φ = 1.417 ina' = a + 0.5 φ = 2.153 in 0.819

0.9890.753

15.5 kips / bolt

TENSIONALLOWABLE LOADS ( kips )

TENSION ON GROSS ( NOMINAL ) AREA

ASTM DESIGNATION Ft 5/8 3/4 7/8 1 1 1/8 1 1/4 1 3/8 1 1/2

( ksi )0.3068 0.4418 0.6013 0.7854 0.9940 1.2270 1.4850 1.7670

A307 20 6.1 8.8 12.0 15.7 19.9 24.5 29.7 35.3A325 44 13.5 19.4 26.5 34.6 43.7 54.0 65.3 77.7A490 54 16.6 23.9 32.5 42.4 53.7 66.3 80.2 95.4

DanielTian Li

AREA BASED ON NOMINAL DIAMETER ( in2 )

Tensile Capacity of Bolts Connection Based on AISC-ASD 9th Edition

NOMINAL BOLT DIAMETER, φ ( in )

'8c

y

Bbt

pF= =

'

1 dp

δ = − =

2'

'

'

11

1

cttb

a

αδ

= − = +

( )

( )

2

'

2

' '

'

1 , 1

1 , 0 1

, 0

c

allowc

tB for

t

tB forT

t

B for

δ α

δα α

α

+ >

= + ≤ ≤ =

<



INPUT DATA & DESIGN SUMMARYTHICKER PART JOINTED t = 0.75 inWELD SIZE w = 0.25 inECCENTRICITY TO EDGE e = 30 in

WELD LENGTH, DA L1 = 4 in

WELD LENGTH, DC L2 = 18 in

WELD LENGTH, CB L3 = 2 in

Pallowable = 10.7 kips

ANALYSISMIN WELD SIZE wmin = 0.25 in

MAX WELD SIZE wmax = 0.69 in

EFFECTIVE THROAT THICHNESS te = 0.707 w = 0.18 in

CENTROID OF WELD GROUP X0 = Σ Xi Ai / Σ Ai = 0.42 in

Y0 = Σ Yi Ai / Σ Ai = 9.75 in

CENTRODIAL MOMENT OF INERTIA Ix = Σ (b h3/ 12 + A d2 ) = 169 in4

Iy = Σ (b h3/ 12 + A d2 ) = 4 in4

TOTAL ECCENTRICITY e total = e + L1 - X0 = 33.6 in

FACTOR α = e total / ( Ix + Iy ) = 0.194

ALLOWABLE STRESS Fv = 0.3 Fu = 21.0 ksi

DIRECT SHEAR FORCE @ POINTS A, B, C, D

f i y = Py / Aw = 0.24 P

f i x = Px / Aw = 0.00 P

THE VERTICAL & HORIZONTAL COMPONENTS OF SHEAR FORCE @ POINTS A, B, C, D

Point A : Ryi = α xi P = 0.70 P Rxi = α yi P = 1.60 P

Point B : Ryi = α xi P = 0.31 P Rxi = α yi P = 1.89 P

Point C : Ryi = α xi P = 0.08 P Rxi = α yi P = 1.89 P

Point D : Ryi = α xi P = 0.08 P Rxi = α yi P = 1.60 P

THE RESULTANT SHEAR FORCE @ POINTS A, B, C, D

Point A : Ri = [ ( 0.24 + 0.70 ) 2 + ( 0.00 + 1.60 ) 2 ] 0.5= 1.85 P

Point B : Ri = [ ( 0.24 + 0.31 ) 2 + ( 0.00 + 1.89 ) 2 ] 0.5= 1.97 P

Point C : Ri = [ ( 0.24 + 0.08 ) 2 + ( 0.00 + 1.89 ) 2 ] 0.5= 1.92 P

Point D : Ri = [ ( 0.24 + 0.08 ) 2 + ( 0.00 + 1.60 ) 2 ] 0.5= 1.63 P

Ri,max = 1.97 P

THE ALLOWABLE LOAD

Pallowable = Fv / Ri,max = 10.66 kips

Weld Capacity of Eccentric Connection Based on AISC-ASD 9th Edition

DanielT. Li



INPUT DATA & DESIGN SUMMARYTHICKER PART JOINTED t = 0.75 inWELD LENGTH L = 10 in

KL = 4 inECCENTRICITY TO EDGE e = 8 in

FACTORED LOAD Pu = 50 kips

USE w = 3/8 in WELD

ANALYSISMAX WELD SIZE wmax = 0.69 in

MIN WELD SIZE wmin = 0.25 inTHICKER PART JOINED MIN SIZE OF FILLET WELDTo 1/4 " inclusive 1/8Over 1/4 to 1/2 3/16Over 1/2 to 3/4 1/4Over 3/4 5/16

k = 0.40x = 0.80xL = 8.0aL = 4.00a = 0.40C = 0.88 (from AISC-LRFD Table 8-42)C1 = 1.00 (E70 electrodes, from AISC-LRFD Table 8-37)D = no. of sixteenths of an inch fillet weld size required

D = Pu / ( C C1 L ) = 5.69 / 16 inSAY => 6 / 16 in

GOVERNS => 6 / 16 in

DanielT. Li

Weld Size of Eccentric Connection Based on AISC-LRFD 2nd Edition



INPUT DATA & DESIGN SUMMARYTHICKER PART JOINTED t = 0.75 inWELD SIZE w = 0.375 inECCENTRICITY TO EDGE x = 6 inWELD LENGTH D = 10 inSERVICE LEVEL LOAD P = 29.8 kips

ANALYSIS

MIN WELD SIZE w min = 0.25 in

MAX WELD SIZE w max = 0.69 in

EFFECTIVE THROAT THICHNESS t e = 0.707 w = 0.27 in

CENTRODIAL MOMENT OF INERTIA I x = 2 (t e D3 / 12 ) = 44.2 in4

DIRECT SHEAR STRESS f y = P / 2 D t e = 5.6 ksi

BENDING STRESS f x = D P x / 2 I x = 20.2 ksi

RESULTANT STRESS f = ( f x 2 + f y

2 ) 0.5 = 21.0 ksi

ALLOWABLE STRESS Fv = 0.3 F u = 21.0 ksi > f [Satisfactory]

Weld Capacity of Eccentric Connection Based on AISC-ASD 9th Edition

DanielT. Li



INPUT DATA & DESIGN SUMMARYSERVICE LEVEL LOAD P = 81 kipsTHICKER PART JOINTED t = 0.5 inWELD SIZE w = 0.1875 in

WELD LENGTH AT END L2 = 5 inCENTROID OF ANGLE SECTION d = 1.75 in

Pallowable = 81.0 kipsL 1 = 7.7 inL 3 = 16.4 in

ANALYSIS

MIN WELD SIZE w min = 0.1875 in

MAX WELD SIZE w max = 0.44 in

EFFECTIVE THROAT THICHNESS t e = 0.707 w = 0.13 in

ALLOWABLE STRESS Fv = 0.3 Fu = 21.0 ksi

THE LOAD CARRIED BY END WELD R 2 = Fv L2 t e = 13.92 ksi

THE LOAD CARRIED BY BOTTOM WELD R 1 = P ( 1 - d / L2 ) - R2 / 2 = 45.69 ksi

THE LOAD CARRIED BY TOP WELD R 3 = P - R1 - R2 = 21.39 ksi

REQUIRED WELD LENGTH AT BOTTOM L 1 = R1 / (Fv t e ) = 16.41 in

REQUIRED WELD LENGTH AT TOP L 3 = R3 / (Fv t e ) = 7.68 in

THE ALLOWABLE LOAD Pallowable = R1 + R2 + R3 = 81.00 in

Balance Weld Length & Compute Capacity ased on AISC-ASD 9th Edition

DanielT. Li



INPUT DATA & DESIGN SUMMARYNO. OF SPANS (1,2 or 3) n = 2 USE:DECK VERT. SPAN LENGTH = 9 ft 1 1/2" x 20 GA. VERCO PLB-36/HSB-36 GALVANIZED ROOF DECK

GAGE (22,20,18,16) ? => == > 20 GA ( 2 SPANS MINS)DEAD LOAD DL = 20 psf 5 -1/2 "Ø PUDDLE WELDS PER SHEET, EACH SUPPORT.LIVE LOAD LL = 20 psf 1/2"Ø PUDDLE WELD @ 12" O.C. EACH PARALLEL SUPPORT.DIAPHRAGM HORIZ SPAN L = 150 ft SIDELAP TOP SEAM WELD (TSW) @ 12" O.C.DIAPHRAGM HORIZ DEPTH d = 50 ftTHE MAX DIAPHRAGM SHEAR v = 680 plf (THE DIAPHRAGM DEFLECTION, 0.69 in, AT MIDDLE SPAN.)

NO. OF SUPPORT WELD (4, 5 or 7) == > 5 per sheetSPACING OF PUDDLE WELD == > 12 in o.c.SIDE LAP TYPE (0=BP, 1=TSW) 1 Top Seam WeldSPACING OF SIDELAP CONNECTION 12 in o.c.

ANALYSISPLB & HSB SECTION PROPERTIES (ER-2078P, Table 4, page 3) PUDDLE WELDS ALLOWABLE DIAPHRAGM SHEAR (ER-2078P, Table 1, page 2)

GAGE thk, in I, in4/ft +S, in3/ft -S, in3/ft Wt, psf GAGE 6" o.c. 9" o.c. 12" o.c. 18" o.c.

16 0.0598 0.377 0.411 0.417 3.5 16 4186 2791 2093 1395

18 0.0478 0.302 0.322 0.335 2.9 18 3346 2231 1673 1115

20 0.0359 0.216 0.235 0.248 2.3 20 2513 1675 1257 838

22 0.0299 0.175 0.187 0.198 1.9 22 2093 1395 1047 698

4 5 6 7 8 9 10 11 12HSB-36 ALLOWABLE DIAPHRAGM SHEAR, q (plf), AND FLEXIBILITY FACTORS, F (ER-2078P, Table 20, page 40-47)

SUPPORT GAGE BP 4'-0" 5'-0" 6'-0" 7'-0" 8'-0" 9'-0" 10'-0" 11'-0" 12'-0"

36/5 20 24 q 690 675 590 516 447 405 361 334 `

36/5 20 24 F 4.0+91R 5.1+73R 6.5+61R 8.2+52R 10.4+45R 12.8+40R 15.9+36R 19.0+33R 23.1+30R

36/5 20 12 q 762 733 656 568 501 450 408 374 346

36/5 20 12 F 3.9+91R 4.9+73R 6.1+61R 7.7+52R 9.5+45R 11.7+40R 14.3+36R 17.2+33R 20.6+30R

TSW

36/5 20 24 q 991 872 730 679 598 572 518 504 465

36/5 20 24 F 20.5+43R 18.2+34R 23.7+28R 21.5+24R 26.2+21R 24.1+19R 28.3+17R 26.3+16R 30.2+14R

36/5 20 18 q 1084 941 788 727 680 608 583 561 518

36/5 20 18 F 13.4+43R 13.4+34R 17.2+28R 16.6+24R 16.3+21R 19.4+19R 18.8+17R 18.5+16R 21.1+14R

36/5 20 12 q 1169 1006 895 816 756 709 672 641 615

36/5 20 12 F 10.0+43R 10.6+34R 11.0+28R 11.4+24R 11.8+21R 12.1+19R 12.4+17R 12.6+16R 12.8+14R

36/5 20 6 q 1469 1293 1174 1088 1023 972 931 818 688

36/5 20 6 F 5.4+43R 5.5+34R 5.6+28R 5.7+24R 5.8+21R 5.9+19R 5.9+17R 6.0+16R 6.0+14R

CHECK VERTICAL BENDING CAPACITY

20.72 ksi, (Vero PunchLok Book, page 5.)

< Fb [Satisfactory]

Where w = (DL + Wt) + LL = 42 psfFb = 22.8 ksi, (Vero PunchLok Book, page 4.)

CHECK VERTICAL DEFLECTION

0.19 in, (Vero PunchLok Book, page 5.)

< / 240 = 0.45 in [Satisfactory]

Where wLL = 20 psf, as givenCd = 1728 , (Vero PunchLok Book, page 5.)

E = 29500 ksi, , (from ER-2078P, page 6)

CHECK HORIZONTAL DIAPHRAGM SHEAR CAPACITY

v = 680 plf, as given < vallow = 709 plf, (for 36/5 & TSW from table above.) [Satisfactory]< vallow = 1257 plf, ( for Puddle Weld from table above.) [Satisfactory]

DETERMINE HORIZONTAL DIAPHRAGM DEFLECTION

0.0600 + 0.6316 = 0.6916 in, (from ER-2078P, page 6)

Where w = 2 d v / L = 453 plfR = (4 - n) / 3 = 0.667 , (from ER-2078P, page 39 footnotes.)F = 12.1+19R = 24.77 , for TSW connection.A = 16.2 in2, steel chord member area. 50= 0.5 A d2 = 2916000 in2, (ER-2078 page 6)

Design of 1 1/2" Type "B" Roof Deck Based on ICBO ER-2078P

DanielTian Li

20.125,

20.125,

20.1,

wl for Simple SpanS

wlf for Double Spansb S

wl for Triple SpansS

+−=

−−

−

=

4

4

4

0.013,

0.0054,

0.0069,

LL

LL

LL

w l Cd for Simple SpanEI

w l Cd for Double SpansLL EI

w l Cd for Triple SpansEI

=∆

=

( )( )

34 2

65 12

384 8 10f w

w w FL LEI d

∆ = + = +∆ ∆×

=


JOB NO. : DATE: REVIEW BY : DEPRESSED FLOOR DECK CAPACITY USING STEEL PROPERTIES ONLY (NON COMPOSITE)

DEAD LOAD DL = 70 psf (including partitions)LIVE LOAD LL = 50 psf (non-reduceable)SPAN LENGTH L = 8 ftGAGE (22, 20, 18, 16) GA = 18DECK TYPE (B, W2, W3, N) TYPE = W2

THE DEPRESSED FLOOR DECK IS ADEQUATE.SECTION PROPERTIES (See Tables below)

Type Gage I (in4/ft) +S (in3/ft) -S (in3/ft) Fb

W2 18 0.555 0.51 0.511 22.8

For one span,

Mmax = 0.125 (DL+LL) L2 = 0.960 ft-kips

Mallowable = Fb (+S) = 0.969 ft-kips ok

∆max,Total = 5 (DL+LL) L4 / (384 EI) = 0.7 in ( L / 140 )

∆max,LL = 5 (LL) L4 / (384 EI) = 0.3 in ( L / 335 )

For two spans, uniform LL on one span

Mmax = 0.0070 (DL) L2 + 0.096 (LL) L2 = 0.621 ft-kips

Mallowable = Fb (+S) = 0.969 ft-kips ok

∆max,Total = (0.00541DL + 0.0092LL) L4 / (EI) = 0.4 in ( L / 260 )

∆max,LL = ( 0.0092LL) L4 / (EI) = 0.2 in ( L / 475 )

For two spans, all spans loaded

-Mmax = -0.125 (DL+LL) L2 = -0.960 ft-kips

-Mallowable = Fb (-S) = -0.969 ft-kips ok

∆max,Total = 0.00541(DL + LL) L4 / (EI) = 0.3 in ( L / 336 )

∆max,LL = ( 0.0054LL) L4 / (EI) = 0.1 in ( L / 807 )

For three spans, all spans loaded

- Mmax = - 0.100 (DL+LL) L2 = -0.768 ft-kips

- Mallowable = Fb (-S) = -0.969 ft-kips ok

∆max,Total = 0.0069 (DL+LL) L4 / (EI) = 0.4 in ( L / 264 )

∆max,LL = 0.0069 (LL) L4 / (EI) = 0.2 in ( L / 633 )

B SECTION PROPERTIES (Verco Page 34 & 35) N SECTION PROPERTIES (Verco Page 88 & 89)Gage I (in4/ft) +S (in3/ft) -S (in3/ft) Gage I (in4/ft) +S (in3/ft) -S (in3/ft)

16 0.377 0.411 0.417 16 1.542 0.851 0.91418 0.302 0.322 0.335 18 1.146 0.664 0.73720 0.216 0.235 0.248 20 0.780 0.466 0.54822 0.175 0.187 0.198 22 0.613 0.361 0.446

W2 SECTION PROPERTIES (Verco Page 52 & 53) W3 SECTION PROPERTIES (Verco Page 70 & 71)Gage I (in4/ft) +S (in3/ft) -S (in3/ft) Gage I (in4/ft) +S (in3/ft) -S (in3/ft)

16 0.694 0.639 0.639 16 1.509 0.960 0.96018 0.555 0.510 0.511 18 1.203 0.767 0.76720 0.423 0.361 0.370 20 0.896 0.534 0.56422 0.340 0.283 0.287 22 0.718 0.418 0.444

DanielT. Li



DEAD LOAD = 20 psfLIVE LOAD = 20.02 psf TRIAL CODE 1 THRU 9 ==> 5TOTAL LOAD = 40.02 psf

JOIST SPACING = 8 ft USE 40LH12 @ 8 ft o.c.JOIST SPAN = 73.667 ft

1. STRENGTH CHECK ACTUAL < ALLOW

PLASTER CEILING ? 0 no (TOTAL LOAD, lbs / ft) ( 1 = YES, 0 = NO ) 345 < 382 ok

DEAD LOAD = 160.0 lbs / ft 2. DEFLECTION CHECK ∆ < L / 360LIVE LOAD = 160.2 lbs / ft (LIVE LOAD, lbs / ft)SELF LOAD = 25.0 lbs / ft 160 < 197 okTOTAL LOAD = 345 lbs / ft

STANDARD LOAD TABLES / OPEN WEB STEEL JOISTS LH-SERIESThe black figures give the TOTAL load capacity;

The blue figures are the LIVE load that will give a deflection of L/360;

JOIST WT DEPTH CLEAR SPAN IN FEETID lbs/ft in

65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 8040LH08 16 40 254 247 241 234 228 222 217 211 206 201 196 192 187 183 178 174

150 144 138 132 127 122 117 112 108 104 100 97 93 90 86 8340LH09 21 40 332 323 315 306 298 291 283 276 269 263 256 250 244 239 233 228

196 188 180 173 166 160 153 147 141 136 131 126 122 118 113 10940LH10 21 40 367 357 347 338 329 321 313 305 297 290 283 276 269 262 255 249

216 207 198 190 183 176 169 162 156 150 144 139 134 129 124 11940LH11 22 40 399 388 378 368 358 349 340 332 323 315 308 300 293 286 279 273

234 224 215 207 198 190 183 176 169 163 157 151 145 140 135 13040LH12 25 40 486 472 459 447 435 424 413 402 392 382 373 364 355 346 338 330

285 273 261 251 241 231 222 213 205 197 189 182 176 169 163 15740LH13 30 40 573 557 542 528 514 500 487 475 463 451 440 429 419 409 399 390

334 320 307 295 283 271 260 250 241 231 223 214 207 199 192 18540LH14 35 40 656 638 620 603 587 571 556 542 528 515 502 490 478 466 455 444

383 367 351 336 323 309 297 285 273 263 252 243 233 225 216 20940LH15 36 40 734 712 691 671 652 633 616 599 583 567 552 538 524 511 498 486

427 408 390 373 357 342 328 315 302 290 279 268 258 248 239 23040LH16 42 40 808 796 784 772 761 751 730 710 691 673 655 638 622 606 591 576

469 455 441 428 416 404 387 371 356 342 329 316 304 292 282 271

40LH Series Truss Joist / Vulcraft or Equial

Daniel

T. Li



DEAD LOAD = 20 psfLIVE LOAD = 20.02 psf TRIAL CODE 1 THRU 9 ==> 1TOTAL LOAD = 40.02 psf

JOIST SPACING = 6 ft USE 36LH07 @ 6 ft o.c.JOIST SPAN = 60 ft

1. STRENGTH CHECK ACTUAL < ALLOW

PLASTER CEILING ? 0 no (TOTAL LOAD, lbs / ft) ( 1 = YES, 0 = NO ) 256 < 266 ok

DEAD LOAD = 120.0 lbs / ft 2. DEFLECTION CHECK ∆ < L / 360LIVE LOAD = 120.1 lbs / ft (LIVE LOAD, lbs / ft)SELF LOAD = 16.0 lbs / ft 120 < 153 okTOTAL LOAD = 256 lbs / ft




57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 7236LH07 16 36 292 283 274 266 258 251 244 237 230 224 218 212 207 201 196 191

177 168 160 153 146 140 134 128 122 117 112 107 103 99 95 9136LH08 18 36 321 311 302 293 284 276 268 260 253 246 239 233 227 221 215 209

194 185 176 168 160 153 146 140 134 128 123 118 113 109 104 10036LH09 21 36 411 398 386 374 363 352 342 333 323 314 306 297 289 282 275 267

247 235 224 214 204 195 186 179 171 163 157 150 144 138 133 12736LH10 21 36 454 440 426 413 401 389 378 367 357 347 338 328 320 311 303 295

273 260 248 236 225 215 206 197 188 180 173 165 159 152 146 14036LH11 23 36 495 480 465 451 438 425 412 401 389 378 368 358 348 339 330 322

297 283 269 257 246 234 224 214 205 196 188 180 173 166 159 15336LH12 30 36 593 575 557 540 523 508 493 478 464 450 437 424 412 400 389 378

354 338 322 307 292 279 267 255 243 232 222 213 204 195 187 17936LH13 36 36 697 675 654 634 615 596 579 562 546 531 516 502 488 475 463 451

415 395 376 359 342 327 312 298 285 273 262 251 240 231 222 21336LH14 36 36 768 755 729 706 683 661 641 621 602 584 567 551 535 520 505 492

456 434 412 392 373 356 339 323 309 295 283 270 259 247 237 22836LH15 36 36 809 795 781 769 744 721 698 677 656 637 618 600 583 567 551 536

480 464 448 434 413 394 375 358 342 327 312 299 286 274 263 252

Daniel


T. Li



DEAD LOAD = 40 psfLIVE LOAD = 20 psf TRIAL CODE 1 THRU 9 ==> 4TOTAL LOAD = 60 psf

JOIST SPACING = 8 ftJOIST SPAN = 34.667 ft USE 24LH06 @ 8 ft o.c.

PLASTER CEILING ? 0 no 1. STRENGTH CHECK ACTUAL < ALLOW ( 1 = YES, 0 = NO ) (TOTAL LOAD, lbs / ft)

496 < 555 okDEAD LOAD = 320.0 lbs / ftLIVE LOAD = 160.0 lbs / ft 2. DEFLECTION CHECK ∆ < L / 360SELF LOAD = 16.0 lbs / ft (LIVE LOAD, lbs / ft)TOTAL LOAD = 496 lbs / ft 160 < 356 ok




33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 4824LH03 11 24 342 339 336 323 307 293 279 267 255 244 234 224 215 207 199 191

235 226 218 204 188 175 162 152 141 132 124 116 109 102 96 9024LH04 12 24 419 398 379 360 343 327 312 298 285 273 262 251 241 231 222 214

288 265 246 227 210 195 182 169 158 148 138 130 122 114 107 10124LH05 13 24 449 446 440 419 399 380 363 347 331 317 304 291 280 269 258 248

308 297 285 264 244 226 210 196 182 171 160 150 141 132 124 11724LH06 16 24 604 579 555 530 504 480 457 437 417 399 381 364 348 334 320 307

411 382 356 331 306 284 263 245 228 211 197 184 172 161 152 14224LH07 17 24 665 638 613 588 565 541 516 491 468 446 426 407 389 373 357 343

452 421 393 367 343 320 297 276 257 239 223 208 195 182 171 16124LH08 18 24 707 677 649 622 597 572 545 520 497 475 455 435 417 400 384 369

480 447 416 388 362 338 314 292 272 254 238 222 208 196 184 17324LH09 21 24 832 808 785 764 731 696 663 632 602 574 548 524 501 480 460 441

562 530 501 460 424 393 363 337 313 292 272 254 238 223 209 19624LH010 23 24 882 856 832 809 788 768 737 702 668 637 608 582 556 533 511 490

596 559 528 500 474 439 406 378 351 326 304 285 266 249 234 22024LH011 25 24 927 900 875 851 829 807 787 768 734 701 671 642 616 590 567 544

624 588 555 525 498 472 449 418 388 361 337 315 294 276 259 243


Daniel

T. Li



DEAD LOAD = 20 psf TRIAL CODE 1 THRU 8 ==> 8LIVE LOAD = 20.02 psfTOTAL LOAD = 40.0 psf

USE 24K12 @ 6 ft ocJOIST SPAN = 60 ftJOIST SPACING = 6 ft

1. STRENGTH CHECK: ACTUAL ALLOWPLASTER CEILING ? 0 no TOTAL LOAD, # / ft =>256.12 < 325 ok ( 1 = YES, 0 = NO )

2. DEFLECTION CHECK ACTUAL ALLOWDEAD LOAD = 120 lbs / ft LIVE LOAD, lbs / ft => 120 < 153 ( ∆ = L / 360 ) okLIVE LOAD = 120 lbs / ftSELF LOAD = 16.0 lbs / ftTOTAL LOAD = 256 lbs / ft

STANDARD LOAD TABLES / OPEN WEB STEEL JOISTS K-SERIESThe black figures give the TOTAL load capacity;

The blue figures are the LIVE load that will give a deflection of L/360;CODE 1 2 3 4 5 6 7 8JOIST 24K4 24K5 24K6 24K7 24K8 24K9 24K10 24K12DEPTH 24 24 24 24 24 24 24 24

WT,lbs/ft 8.4 9.3 9.7 10.1 11.5 12 13.1 16SPAN

1 24 520 550 550 550 550 550 550 550 12 516 544 544 544 544 544 544 544 23 25 479 540 550 550 550 550 550 550 34 456 511 520 520 520 520 520 520 45 26 442 499 543 550 550 550 550 550 56 405 453 493 499 499 499 499 499 67 27 410 462 503 550 550 550 550 550 78 361 404 439 479 479 479 479 479 89 28 381 429 467 521 550 550 550 550 9

10 323 362 393 436 456 456 456 456 1011 29 354 400 435 485 536 550 550 550 1112 290 325 354 392 429 436 436 436 1213 30 331 373 406 453 500 544 550 550 1314 262 293 319 353 387 419 422 422 1415 31 310 349 380 424 468 510 550 550 1516 237 266 289 320 350 379 410 410 1617 32 290 327 357 397 439 478 549 549 1718 215 241 262 290 318 344 393 393 1819 33 273 308 335 373 413 449 532 532 1920 196 220 239 205 289 313 368 368 2021 34 257 290 315 351 388 423 502 516 2122 179 201 218 242 264 286 337 344 2223 35 242 273 297 331 366 399 473 501 2324 164 184 200 221 242 262 308 324 2425 36 229 258 281 313 346 377 447 487 2526 150 169 183 203 222 241 283 306 2627 37 216 244 266 296 327 356 423 474 2728 138 155 169 187 205 222 260 290 2829 38 205 231 252 281 310 338 401 461 2930 128 143 156 172 189 204 240 275 3031 39 195 219 239 266 294 320 380 449 3132 118 132 144 159 174 189 222 261 3233 40 185 208 227 253 280 304 361 438 3334 109 122 133 148 161 175 206 247 3435 41 176 198 216 241 266 290 344 427 3536 101 114 124 137 150 162 191 235 3637 42 168 189 206 229 253 276 327 417 3738 94 106 115 127 139 151 177 224 3839 43 160 180 196 219 242 263 312 406 3940 88 98 107 118 130 140 165 213 4041 44 153 172 187 209 231 251 298 387 4142 82 92 100 110 121 131 154 199 4243 45 146 164 179 199 220 240 285 370 4344 76 86 93 103 113 122 144 185 4445 46 139 157 171 191 211 230 272 354 4546 71 80 87 97 106 114 135 174 4647 47 133 150 164 183 202 220 261 339 4748 67 75 82 90 99 107 126 163 4849 48 128 144 157 175 194 211 250 325 4950 63 70 77 85 93 101 118 153 50

24K Series Truss Joist / Vulcraft or Equial

T. LiDaniel


JOB NO. : DATE : REVIEW BY : Joist Girder / Vulcraft or Equial

JOIST SPACING = 6.67 ftDEAD LOAD = 15 psf PANEL LOAD = 9.3 kips (USE 10.0 kips )LIVE LOAD = 20 psf GIRDER SELF LOAD = 111 lbs / ftTOTAL LOAD = 35 psf

JOIST SPAN = 40 ftNO. OF JOIST SPACES = 15

Where 72 G 15 N 10 KGIRDER DEPTH = 72 in Depth No. of spaces Load on each panel pointGIRDER SPAN = 100 ft

USE 72G 15N 10K JOIST GIRDER

DanielT. Li


JOB NO. : DATE : REVIEW BY : Steel Stair Design Based on AISC-ASD 9th

INPUT DATA & DESIGN SUMMARYCOLUMN SECTION (Tube or Pipe) = > HSS4X4X1/4

Tube A rmin t h3.37 1.52 0.23 4.00

FLOOR BEAM - 1 = > W16X26A d Ix Sx

7.68 15.70 301.00 38.40STRINGER - 1 (Channel or Tube) = > MC8X8.5

Channel A d Ix Sx2.50 8.00 23.30 5.82

STRINGER - 2 (Channel or Tube) = > C12X20.7Channel A d Ix Sx

6.08 12.00 129.00 21.50LANDING BEAM - 1 (Channel or Tube) = > C8X18.75

Channel A d Ix Sx5.51 8.00 43.90 11.00

LANDING BEAM - 2 (Channel or Tube) = > C8X18.75Channel A d Ix Sx

5.51 8.00 43.90 11.00

DIMENTIONS H = 16 ft, story HtL1 = 12 ftL2 = 6 ftL3 = 10 ft

NUMBER OF STORIES n = 2

GRAVITY LOAD DL = 50 psfLL = 100 psf

THE STAIR DESIGN IS ADEQUATE.

ANALYSIS

STRINGER - 1

θ = 33.69 deg, from horizontalw = 0.25 (DL / Cos θ + LL) L3 = 400 plf , projectedR = 0.5 w L1 = 2.40 kips

M = w L12 / 8 = 7.20 ft-kips

fb = M / Sx = 14.85 ksi

Fb = 0.6 Fy = 21.60 ksi > fb [Satisfactory]E = 29000 ksi∆LL = 5 (wLL Cos θ) (L1 / Cos θ)4 / (384 E I) = 0.46 in

< (L1 / Cos θ) / 240 = 0.72 in [Satisfactory]

LANDING BEAM - 1

w = 0.5 (DL + LL) L2 = 450 plfP = 2.40 kips, from STRINGER - 1R = 0.5 w L3 + P = 4.65 kips

M = w L32 / 8 + P L3 / 2 = 17.63 ft-kips

fb = M / Sx = 19.23 ksi

Fb = 0.6 Fy = 21.60 ksi > fb [Satisfactory]

∆LL = 5 wLL L34 / (384 E I) + PLL L3

3 / (24 E I) = 0.14 in

< L3 / 240 = 0.50 in [Satisfactory]

DanielTian Li

(cont'd)STRINGER - 2

w = 400 plf , projected, from STRINGER - 1P = 4.65 kips, from LANDING BEAM - 1RL = [w L1 (0.5 L1 + L2) + P L2] / (L1 + L2) = 4.75 kipsRR = [w L1 (0.5 L1) + P L1] / (L1 + L2) = 4.70 kipsX = RL / w = 11.87 ft, from left

Mmax = RL X - (0.5 w X2 ) = 28.21 ft-kips

fb = Mmax / Sx = 15.75 ksi


∆LL = 5 wLL (L1 + L2)4 / (384 E I) + PLL (L1 +L2)

3 / (48 E I) = 0.24

< (L1 + L2) / 240 = 0.90 in [Satisfactory]

LANDING BEAM - 2

w = 450 plf, from LANDING BEAM - 1R = 0.5 w L3 = 2.25 kips

M = w L32 / 8 = 5.63 ft-kips

fb = M / Sx = 6.14 ksi


∆LL = 5 wLL L34 / (384 E I) = 0.05 in

< L3 / 240 = 0.50 in [Satisfactory]

FLOOR BEAM - 1

L = 12 ftw = 600 plf, floor gravity loadP = P1 + P2 = 4.80 kips, total point loads, from STRINGER - 1R = 0.5 w L + 0.5 P = 6.00 kipsM = w L2 / 8 + P L3 / 4 = 25.21 ft-kips

fb = M / Sx = 7.88 ksi


∆LL = 5 wLL L4 / (384 E I) + PLL L

3 / (48 E I) = 0.04 in

< L / 240 = 0.60 in [Satisfactory]

COLUMN

P = 6.95 kipsR = n P = 13.90 kipsKL = H = 16 ftK = 1.0

Fy = 46 ksi

Cc = (2π2Es/Fy)0.5 = 112

K / r = 126

F = (K / r) / Cc = 1.13

(1-F2/2)Fy / (5/3+3F/8-F3/8) = N/A kis, for Cc > (K/r)

12π2Es/[23(K/r)2] = 9.38 kis, for Cc < (K/r)

12 Rallow = A Fa = 31.60 kips > R [Satisfactory]

Fa =



INPUT DATAExposure category (A,,B, C or D) = BImportance factor, pg 55, (0.87, 1.0 or 1.15) I = 1.00 Category IIBasic wind speed V = 90 mphTopographic factor (Sec.6.5.7.2, pg 29 & 39) Kzt = 1 FlatBuilding height to eave he = 35 ftBuilding height to ridge hr = 39 ftBuilding length L = 100 ftBuilding width B = 60 ftEffective area of components A = 50 ft2

DESIGN SUMMARYMax horizontal force normal to building length, L, face = 39.00 kipsMax horizontal force normal to building length, B, face = 22.20 kipsMax total horizontal force at wind with any angle to building = 39.00 kipsMax total upward force = 60.48 kips

ANALYSISVelocity pressure

qh = 0.00256 Kh Kzt Kd V2 I = 13.08 psf

where: qh = velocity pressure at mean roof height, h. (Eq. 6-13, page 30)

Kh = velocity pressure exposure coefficient evaluated at height, h, (Tab. 6-5, Case 1,pg 60) = 0.74Kd = wind directionality factor. (Tab. 6-6, for building, page 61) = 0.85h = mean roof height = 37.00 ft

< 60 ft, [Satisfactory]Design pressures for MWFRSp = qh [(G Cpf )-(G Cpi )]where: p = pressure in appropriate zone. (Eq. 6-16, page 32).

G Cp f = product of gust effect factor and external pressure coefficient, see table below. (Fig. 6-4, page 43)

G Cp i = product of gust effect factor and internal pressure coefficient.(Tab. 6-7, Enclosed Building, page 62)

= 0.18 or -0.18a = width of edge strips, Fig 6-4, note 7, page 43, (IBC 1609.6.3), MAX[ MIN(0.1B, 0.4h), 0.04B,3] = 6.00 ft

Net Pressures (psf), Case A Net Pressures (psf), Case B7.59 0.00

(+GCp i ) (-GCp i )

(+GCp i ) (-GCp i ) (+GCp i ) (-GCp i ) 1 -0.45 -8.24 -3.531 0.42 3.17 7.88 0.40 2.88 7.59 2 -0.69 -11.38 -6.672 -0.69 -11.38 -6.67 -0.69 -11.38 -6.67 3 -0.37 -7.19 -2.483 -0.39 -7.44 -2.73 -0.37 -7.19 -2.48 4 -0.45 -8.24 -3.534 -0.31 -6.46 -1.76 -0.29 -6.15 -1.44 5 0.40 2.88 7.59

1E 0.64 6.05 10.76 0.61 5.62 10.33 6 -0.29 -6.15 -1.442E -1.07 -16.35 -11.64 -1.07 -16.35 -11.64 1E -0.48 -8.63 -3.923E -0.56 -9.65 -4.94 -0.53 -9.29 -4.58 2E -1.07 -16.35 -11.644E -0.47 -8.45 -3.74 -0.43 -7.98 -3.27 3E -0.53 -9.29 -4.58

4E -0.48 -8.63 -3.925E 0.61 5.62 10.336E -0.43 -7.98 -3.27

Wind Analysis for Low-rise Building, Based on ASCE 7-98

Surface

Roof angle θ = Roof angle θ =

G Cp fNet Pressure withNet Pressure with

G Cp f

Net Pressure withG Cp fSurface

DanielT. Li

Case A @ (a) CORNER I Case A @ (b) CORNER IIArea Area

(ft2) (+GCp i ) (-GCp i ) (ft2) (+GCp i ) (-GCp i )

1 3080 9.77 24.27 1 1790 5.15 13.582 2663 -30.30 -17.76 2 2421 -27.55 -16.153 2663 -19.82 -7.28 3 2421 -17.42 -6.024 3080 -19.91 -5.41 4 1790 -11.01 -2.58

1E 420 2.54 4.52 1E 430 2.42 4.442E 363 -5.94 -4.23 2E 605 -9.90 -7.053E 363 -3.50 -1.79 3E 605 -5.62 -2.774E 420 -3.55 -1.57 4E 430 -3.43 -1.40

Horiz. 34.06 34.06 Horiz. 22.00 22.00Vert. -59.04 -30.79 Vert. -59.95 -31.70

10 psf min. Horiz. 39.00 39.00 10 psf min. Horiz. 22.20 22.20Sec. 6.1.4.1 Vert. -60.00 -60.00 Sec. 6.1.4.1 Vert. -60.00 -60.00

Case B @ (a) CORNER I Case B @ (b) CORNER IIArea Area

(ft2) (+GCp i ) (-GCp i ) (ft2) (+GCp i ) (-GCp i )

1 3080 -25.38 -10.88 1 1790 -14.75 -6.322 2663 -30.30 -17.76 2 2421 -27.55 -16.153 2663 -19.16 -6.62 3 2421 -17.42 -6.024 3080 -25.38 -10.88 4 1790 -14.75 -6.325 2008 5.78 15.23 5 3290 9.47 24.966 2008 -12.34 -2.89 6 3290 -20.22 -4.73

1E 420 -3.63 -1.65 1E 430 -3.71 -1.692E 363 -5.94 -4.23 2E 605 -9.90 -7.053E 363 -3.37 -1.66 3E 605 -5.62 -2.774E 420 -3.63 -1.65 4E 430 -3.71 -1.695E 212 1.19 2.19 5E 210 1.18 2.176E 212 -1.69 -0.69 6E 210 -1.68 -0.69



Design pressures for components and claddingp = qh[ (G Cp) - (G Cpi)]where: p = pressure on component. (Eq. 6-18, pg 32)

pmin = 10 psf (Sec. 6.1.4.2).

G Cp = external pressure coefficient.

see table below. (Fig. 6-5, page 44 ~ 47)

EffectiveArea (ft2) GCP - GCP GCP - GCP GCP - GCP GCP - GCP GCP - GCP

Comp. 50 0.23 -0.93 0.23 -1.31 0.23 -1.61 0.79 -0.88 0.79 -1.04(Walls reduced 10 %, Fig. 6-5A note 5.)

Positive Negative Positive Negative Positive Negative Positive Negative Positive Negative

10.00 -14.52 10.00 -19.50 10.00 -23.43 12.67 -13.85 12.67 -15.93

Pressure (k) withPressure (k) with

Σ

Σ

Surface

Surface

Σ

Surface

Pressure( psf )

Pressure (k) with

Σ

Zone 1 Zone 2

Comp. & Cladding

Zone 3 Zone 4 Zone 5

Zone 1 Zone 2 Zone 3 Zone 4 Zone 5

SurfacePressure (k) with



INPUT DATAExposure category (A,,B, C or D) = BImportance factor (0.87, 1.0 or 1.15) I = 1.00 Category II, page 55Basic wind speed V = 90 mphTopographic factor (Sec.6.5.7.2) Kzt = 1 Flat, page 29 & 39Building height to roof H = 108 ftParapet height HP = 0 ftBuilding length L = 100 ftBuilding width B = 50 ftEffective area of mullion AM = 55 ft2

Effective area of panel AP = 27 ft2

DESIGN SUMMARYMax building horizontal force normal to building length, L, face = 187.7 kipsMax base moment at wind normal to building length, L, face = 11595.0 ft - kipsMax building horizontal force normal to building length, B, face = 77.5 kipsMax base moment at wind normal to building length, B, face = 5234.5 ft - kipsMax building upward force = 78.7 kipsMax building torsion force = 586.6 ft - kips

ANALYSISVelocity pressures

qz = 0.00256 Kz Kzt Kd V2 I

where: qz = velocity pressure at height, z. (Eq. 6-13, page 30)

Kz = velocity pressure exposure coefficient evaluated at height, z. (Tab. 6-5, Case 2, page 60)

Kd = wind directionality factor. (Tab. 6-6, for building, page 61) = 0.85z = hetght above ground

z (ft) 0 - 15 20 25 30 40 50 60 70 80 90 100 108

Kz 0.57 0.62 0.66 0.70 0.76 0.81 0.85 0.89 0.93 0.96 0.99 1.01

qz (psf) 10.05 10.93 11.63 12.34 13.40 14.28 14.98 15.69 16.39 16.92 17.45 17.80

z (ft) 108 108 108 108 108 108 108 108 108 108

Kz 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.01

qz (psf) 17.80 17.80 17.80 17.80 17.80 17.80 17.80 17.80 17.80 17.80

Design pressures for MWFRSp = q G Cp - qh (G Cpi)where: p = pressure on surface for rigid building with all h. (Eq. 6-15, page 31).

q = qz for windward wall at height z above the ground, see table above.

G Cp i = internal pressure coefficient. (Tab. 6-7, Enclosed Building, page 62) = 0.18 or -0.18

qh = qz value at mean roof height, h, for leeward wall, side walls, and roof.

G = gust effect factor for rigid (Sec. 6.5.8.1, Page 29) = 0.85 Fig. 6-3 fo θ < 10o, page 42Cp = external pressure coefficient, see right down tables. Roof h / B Distance Cp

To L Face 2.16 50 -1.04To L Face 2.16 50To L Face 2.16 50To L Face 2.16 50

Roof h / L Distance CpTo B Face 1.08 54 -1.04To B Face 1.08 100 -0.70To B Face 1.08 100To B Face 1.08 100

Fig. 6-3, page 42Wall Direction L / B Cp

Windward Wall All All 0.80

q G Cp Figure for Gable, Hip Roof, page 41 Leedward Wall To L Dir 0.50 -0.50

Leedward Wall To B Dir 2.00 -0.30Side Wall All All -0.70

Wind Analysis for Building with h > 60 ft, Based on ASCE 7-98

DanielT. Li

(cont'd)Hence, MWFRS Net Pressures are given by following tables (Sec. 6.5.12.3, Page 32)

Surface z (ft) GCPi - GCPi Surface z (ft) GCPi - GCPi

0 - 15 3.63 10.04 Side Wall All -13.80 -7.3920 4.23 10.6425 4.71 11.1130 5.19 11.5940 5.90 12.31 Surface z (ft) GCPi - GCPi Surface z (ft) GCPi - GCPi

50 6.50 12.91 Leeward All -10.77 -4.36 Leeward All -7.74 -1.3460 6.98 13.3970 7.46 13.8780 7.94 14.3590 8.30 14.71 Surface Dist. (ft) GCPi - GCPi Surface Dist. (ft) GCPi - GCPi

100 8.66 15.07 0 - 50 -18.94 -12.53 0 - 54 -18.94 -12.53108 8.90 15.31 100 -13.80 -7.39

Figure 6-9, page 54

Case 1 Case 2 Case 1 Case 2 Case 3 Case 4

VBase (kips) 188 164 78 68 152 133

MBase (ft - kips) 11595 10149 5235 4581 9541 8336 Fig. 6-9

TBase (ft - kips) 0 587 0 242 0 475 Page 54

FUpward (kips) 79 69 67 58 77 68

Vmin (kips) 108 108 54 54 122 121 Sec. 6.1.4.1

FUp,min (kips) 50 50 50 50 50 50 ( 10 psf )

Design pressures for components and cladding

p = q (G Cp) - qi (G Cpi)where: p = pressure on component for building with h > 60 ft. (Eq. 6-19, page 32).

pmin = 10 psf (Sec. 6.1.4.2, page 23).



G Cp i = internal pressure coefficient. (Tab. 6-7) = 0.18 or -0.18a = Zone width = MAX[ MIN(0.1B, 0.1L), 3] = 5.0 ft, (Fig 6-8 note 8, pg 53)G Cp = external pressure coefficient. (Fig. 6-8, page 53)

WallComp. GCP - GCP GCP - GCP

Mullion 0.81 -0.84 0.81 -1.55Panel 0.87 -0.88 0.87 -1.73

Positive Negative Positive Negative Positive Negative Positive Negative

0 - 15 10.00 -18.11 10.00 -30.77 10.57 -18.89 10.57 -33.9220 10.77 -18.11 11.50 -30.77 11.50 -18.89 11.50 -33.9225 11.47 -18.11 12.24 -30.77 12.24 -18.89 12.24 -33.9230 12.16 -18.11 12.98 -30.77 12.98 -18.89 12.98 -33.9240 13.20 -18.11 14.09 -30.77 14.09 -18.89 14.09 -33.9250 14.07 -18.11 15.02 -30.77 15.02 -18.89 15.02 -33.9260 14.77 -18.11 15.76 -30.77 15.76 -18.89 15.76 -33.9270 15.46 -18.11 16.50 -30.77 16.50 -18.89 16.50 -33.9280 16.16 -18.11 17.24 -30.77 17.24 -18.89 17.24 -33.9290 16.68 -18.11 17.80 -30.77 17.80 -18.89 17.80 -33.92100 17.20 -18.11 18.36 -30.77 18.36 -18.89 18.36 -33.92108 17.55 -18.11 18.73 -30.77 18.73 -18.89 18.73 -33.92

Zone 5Zone 4

Zone 4 Zone 5Mullion Pressure (psf) Panel Pressure (psf)

Zone 4 Zone 5z (ft)

27

Base Forces

Actual EffectiveArea ( ft2 )

55

Wind with Angle

Normal to B Face

P (psf) with P (psf) with

Normal to L Face Normal to B Face ASCE-7

P (psf) with

Normal to B Face P (psf) with

Win

dwar

d W

all

Normal to L Face P (psf) with


Roof Roof

cont'dZone 1 Zone 2 Zone 3 - GCP - GCP - GCP

0 -1.40 -2.30 -3.20

10 -1.40 -2.30 -3.20

59 -1.17 -1.98 -2.79

108 -1.10 -1.87 -2.65

157 -1.05 -1.81

206 -1.01 -1.76

255 -0.99 -1.72

304 -0.96 -1.69

353 -0.94 -1.66

402 -0.93 -1.64

451 -0.91

500 -0.90

3600 -0.90

3600 400 75

Zone 1 Zone 2 Zone 30 -28.13 -44.15 -60.17

10 -28.13 -44.15 -60.17

59 -24.09 -38.49 -52.90

108 -22.71 -36.57 -50.42

157 -21.86 -35.38

206 -21.24 -34.51

255 -20.76 -33.83

304 -20.36 -33.27

353 -20.02 -32.80

402 -19.72 -32.38

451 -19.46

500 -19.23

3600 -19.23

Effective

Com

pone

nts

and

Cla

ddin

g

Net Pressure (psf)

Area ( ft2 )

Com

pone

nts

and

Cla

ddin

g

Roof

RoofEffectiveArea ( ft2 )



INPUT DATAExposure category (A,,B, C or D) = BImportance factor, pg 73, (0.87, 1.0 or 1.15) I = 1.00 Category IIBasic wind speed V = 90 mphTopographic factor (Sec.6.5.7.2, pg 30 & 47) Kzt = 1 FlatBuilding height to eave he = 35 ft 68Building height to ridge hr = 39 ftBuilding length L = 100 ftBuilding width B = 60 ftEffective area of components A = 50 ft2

DESIGN SUMMARYMax horizontal force normal to building length, L, face = 39.00 kipsMax horizontal force normal to building length, B, face = 22.20 kipsMax total horizontal torsional load = 422.21 ft-kipsMax total upward force = 60.00 kips

ANALYSISVelocity pressure

qh = 0.00256 Kh Kzt Kd V2 I = 13.08 psf

where: qh = velocity pressure at mean roof height, h. (Eq. 6-15, page 31)

Kh = velocity pressure exposure coefficient evaluated at height, h, (Tab. 6-3, Case 1,pg 75) = 0.74

Kd = wind directionality factor. (Tab. 6-4, for building, page 76) = 0.85h = mean roof height = 37.00 ft

< 60 ft, [Satisfactory]Design pressures for MWFRSp = qh [(G Cpf )-(G Cpi )]where: p = pressure in appropriate zone. (Eq. 6-18, page 32).

G Cp f = product of gust effect factor and external pressure coefficient, see table below. (Fig. 6-10, page 55 & 56)

G Cp i = product of gust effect factor and internal pressure coefficient.(Fig. 6-5, Enclosed Building, page 49) = 0.18 or -0.18

a = width of edge strips, Fig 6-0, note 9, page 56, MAX[ MIN(0.1B, 0.4h), 0.04B,3] = 6.00 ft(IBC Fig.1609.6.2.2, footnote 5)

Net Pressures (psf), Basic Load Cases Net Pressures (psf), Torsional Load Cases7.59 0.00 7.59

(+GCp i ) (-GCp i ) (+GCp i ) (-GCp i ) (+GCp i ) (-GCp i )1 0.42 3.17 7.88 0.40 2.88 7.59 1T 0.42 0.79 1.972 -0.69 -11.38 -6.67 -0.69 -11.38 -6.67 2T -0.69 -2.84 -1.673 -0.39 -7.44 -2.73 -0.37 -7.19 -2.48 3T -0.39 -1.86 -0.684 -0.31 -6.46 -1.76 -0.29 -6.15 -1.44 4T -0.31 -1.62 -0.44

1E 0.64 6.05 10.76 0.61 5.62 10.33 0.002E -1.07 -16.35 -11.64 -1.07 -16.35 -11.643E -0.56 -9.65 -4.94 -0.53 -9.29 -4.58 (+GCp i ) (-GCp i )4E -0.47 -8.45 -3.74 -0.43 -7.98 -3.27 1T 0.40 0.72 1.905 -0.45 -8.24 -3.53 -0.45 -8.24 -3.53 2T -0.69 -2.84 -1.676 -0.45 -8.24 -3.53 -0.45 -8.24 -3.53 3T -0.37 -1.80 -0.62

4T -0.29 -1.54 -0.36

DanielTian Li

Net Pressure with

Surface

Roof angle θ =

G Cp fNet Pressure with

Surface

Roof angle θ =

G Cp f

Surface

Roof angle θ = Roof angle θ =

G Cp fNet Pressure withNet Pressure with

G Cp f

Wind Analysis for Low-rise Building, Based on ASCE 7-02

Basic Load Cases in Transverse Direction Basic Load Cases in Longitudinal DirectionArea Area

(ft2) (+GCp i ) (-GCp i ) (ft2) (+GCp i ) (-GCp i )1 3080 9.77 24.27 1 1790 5.15 13.582 2663 -30.30 -17.76 2 2421 -27.55 -16.153 2663 -19.82 -7.28 3 2421 -17.42 -6.024 3080 -19.91 -5.41 4 1790 -11.01 -2.58

1E 420 2.54 4.52 1E 430 2.42 4.442E 363 -5.94 -4.23 2E 605 -9.90 -7.053E 363 -3.50 -1.79 3E 605 -5.62 -2.774E 420 -3.55 -1.57 4E 430 -3.43 -1.40



Torsional Load Cases in Transverse Direction Torsional Load Cases in Longitudinal DirectionArea Area

(ft2) (+GCp i ) (-GCp i ) (+GCp i ) (-GCp i ) (ft2) (+GCp i ) (-GCp i ) (+GCp i ) (-GCp i )1 1330 4.22 10.48 93 231 1 680 1.96 5.16 17 462 1150 -13.09 -7.67 -38 -22 2 1816 -20.66 -12.11 68 403 1150 -8.56 -3.14 25 9 3 1816 -13.06 -4.51 -43 -154 1330 -8.60 -2.33 189 51 4 680 -4.18 -0.98 37 9

1E 420 2.54 4.52 112 199 1E 430 2.42 4.44 58 1062E 363 -5.94 -4.23 -35 -25 2E 605 -9.90 -7.05 33 233E 363 -3.50 -1.79 20 10 3E 605 -5.62 -2.77 -19 -94E 420 -3.55 -1.57 156 69 4E 430 -3.43 -1.40 82 341T 1750 1.39 3.45 -35 -86 1T 1110 0.80 2.10 -12 -312T 1513 -4.30 -2.52 14 8 2T 2421 -6.89 -4.04 -46 -273T 1513 -2.82 -1.03 -9 -3 3T 2421 -4.35 -1.50 29 104T 1750 -2.83 -0.77 -71 -19 4T 1110 -1.71 -0.40 -25 -6

422 422 180.3 180.3

Design pressures for components and claddingp = qh[ (G Cp) - (G Cpi)]where: p = pressure on component. (Eq. 6-22, pg 33)

pmin = 10 psf (Sec. 6.1.4.2).

G Cp = external pressure coefficient. see table below. (Fig. 6-11, page 57~60)

EffectiveArea (ft2) GCP - GCP GCP - GCP GCP - GCP GCP - GCP GCP - GCP

Comp. 50 0.36 -0.83 0.36 -1.61 0.36 -1.61 0.79 -0.88 0.79 -1.04(Walls reduced 10 %, Fig. 6-11A note 5.)

Positive Negative Positive Negative Positive Negative Positive Negative Positive Negative

10.00 -13.21 10.00 -23.42 10.00 -23.42 12.67 -13.85 12.67 -15.93

Pressure (k) withTorsion (ft-k)

Surface

Σ

SurfacePressure (k) withPressure (k) with

Total Horiz. Torsional Load, MT

SurfacePressure (k) with

Surface

Zone 5

Zone 3 Zone 4Zone 1 Zone 2

Comp. & Cladding Zone 1 Zone 2

Σ

Torsion (ft-k)

Total Horiz. Torsional Load, MT

Pressure( psf )

Zone 5

Zone 3 Zone 4



INPUT DATAExposure category (B, C or D) = BImportance factor (0.87, 1.0 or 1.15) I = 1.00 Category II, page 73Basic wind speed V = 90 mphTopographic factor (Sec.6.5.7.2) Kzt = 1 Flat, page 30 & 47Building height to roof H = 68 ftParapet height HP = 4 ftBuilding length L = 300 ftBuilding width B = 180 ftEffective area of mullion AM = 55 ft2

Effective area of panel AP = 27 ft2

DESIGN SUMMARYMax building horizontal force normal to building length, L, face = 337.1 kipsMax overturning moment at wind normal to building length, L, face = 44925.2 ft - kipsMax building horizontal force normal to building length, B, face = 179.0 kipsMax overturning moment at wind normal to building length, B, face = 31505.9 ft - kipsMax building upward force = 605.9 kipsMax building torsion force = 11377.5 ft - kips

ANALYSISVelocity pressures

qz = 0.00256 Kz Kzt Kd V2 I

where: qz = velocity pressure at height, z. (Eq. 6-15, page 31)

Kz = velocity pressure exposure coefficient evaluated at height, z. (Tab. 6-3, Case 2, page 75)

Kd = wind directionality factor. (Tab. 6-4, for building, page 76) = 0.85z = hetght above ground

z (ft) 0 - 15 20 25 30 40 50 60 70 72 72 72 72

Kz 0.57 0.62 0.66 0.70 0.76 0.81 0.85 0.89 0.90 0.90 0.90 0.90

qz (psf) 10.05 10.93 11.63 12.34 13.40 14.28 14.98 15.69 15.83 15.83 15.83 15.83

z (ft) 72 72 72 72 72 72 72 72 72 72

Kz 0.90 0.90 0.90 0.90 0.90 0.90 0.90 0.90 0.90 0.90

qz (psf) 15.83 15.83 15.83 15.83 15.83 15.83 15.83 15.83 15.83 15.83

Design pressures for MWFRSp = q G Cp - qh (G Cpi)where: p = pressure on surface for rigid building with all h. (Eq. 6-17, page 32).


G Cp i = internal pressure coefficient. (Fig. 6-5, Enclosed Building, page 49) = 0.18 or -0.18


G = gust effect factor for rigid (Sec. 6.5.8.1, Page 30) = 0.85 Fig. 6-6 fo θ < 10o, page 50Cp = external pressure coefficient, see right down tables. Roof h / B Distance Cp

To L Face 0.40 36 -0.90To L Face 0.40 72 -0.90To L Face 0.40 144 -0.50To L Face 0.40 180 -0.30

Roof h / L Distance CpTo B Face 0.24 36 -0.90To B Face 0.24 72 -0.90To B Face 0.24 144 -0.50

60 To B Face 0.24 300 -0.30

Fig. 6-6, page 50Wall Direction L / B Cp

Windward Wall All All 0.80

q G Cp Figure for Gable, Hip Roof, page 50 Leedward Wall To L Dir 0.60 -0.50

Leedward Wall To B Dir 1.67 -0.37Side Wall All All -0.70

Wind Analysis for Building with h > 60 ft, Based on ASCE 7-02

DanielTian Li

(cont'd)Hence, MWFRS Net Pressures are given by following tables (Sec. 6.5.12.3, Page 32)

Surface z (ft) GCPi - GCPi Surface z (ft) GCPi - GCPi

0 - 15 3.98 9.68 Side Wall All -12.27 -6.5720 4.58 10.2825 5.06 10.7630 5.54 11.2440 6.26 11.96 Surface z (ft) GCPi - GCPi Surface z (ft) GCPi - GCPi

50 6.86 12.56 Leeward All -9.58 -3.88 Leeward All -7.78 -2.0860 7.34 13.0470 7.82 13.5272 7.91 13.61

Surface Dist. (ft) GCPi - GCPi Surface Dist. (ft) GCPi - GCPi

0 - 36 -14.96 -9.26 0 - 36 -14.96 -9.2672 -14.96 -9.26 72 -14.96 -9.26

144 -9.58 -3.88 144 -9.58 -3.88180 -6.89 -1.19 300 -6.89 -1.19

Figure 6-9, page 54

Case 1 Case 2 Case 1 Case 2 Case 3 Case 4

VBase (kips) 337 253 179 134 387 215

MBase (ft - kips) 44925 33694 31506 23629 57323 30893 Fig. 6-9

MT (ft - kips) 0 11378 0 3625 0 11262 Page 54

FUpward (kips) 450 338 357 268 606 324

Vmin (kips) 216 216 130 130 259 252 Sec. 6.1.4.1

FUp,min (kips) 540 540 540 540 540 540 ( 10 psf )

Design pressures for components and cladding

p = q (G Cp) - qi (G Cpi)where: p = pressure on component for building with h > 60 ft. (Eq. 6-23, page 33).

pmin = 10 psf (Sec. 6.1.4.2, page 23).



G Cp i = internal pressure coefficient. (Fig. 6-5) = 0.18 or -0.18a = Zone width = MAX[ MIN(0.1B, 0.1L), 3] = 18.0 ft, (Fig 6-17 note 8, pg 67)G Cp = external pressure coefficient. (Fig. 6-17, page 67)

WallComp. GCP - GCP GCP - GCP

Mullion 0.81 -0.84 0.81 -1.55Panel 0.87 -0.88 0.87 -1.73

Positive Negative Positive Negative Positive Negative Positive Negative

0 - 15 10.00 -16.10 10.00 -27.36 10.57 -16.80 10.57 -30.1620 10.77 -16.10 11.50 -27.36 11.50 -16.80 11.50 -30.1625 11.47 -16.10 12.24 -27.36 12.24 -16.80 12.24 -30.1630 12.16 -16.10 12.98 -27.36 12.98 -16.80 12.98 -30.1640 13.20 -16.10 14.09 -27.36 14.09 -16.80 14.09 -30.1650 14.07 -16.10 15.02 -27.36 15.02 -16.80 15.02 -30.1660 14.77 -16.10 15.76 -27.36 15.76 -16.80 15.76 -30.1670 15.46 -16.10 16.50 -27.36 16.50 -16.80 16.50 -30.1672 15.60 -16.10 16.65 -27.36 16.65 -16.80 16.65 -30.16

P (psf) with

Normal to B Face P (psf) with

Win

dwar

d W

all



Roof Roof

Wind with Angle

Normal to B Face

P (psf) with P (psf) with

Normal to L Face Normal to B Face ASCE-7

z (ft)

27

Base Forces

Actual EffectiveArea ( ft2 )

55

Zone 5Mullion Pressure (psf) Panel Pressure (psf)

Zone 4 Zone 5

Zone 5Zone 4

Zone 4

cont'dZone 1 Zone 2 Zone 3 - GCP - GCP - GCP

0 -1.40 -2.30 -3.20

10 -1.40 -2.30 -3.20

59 -1.17 -1.98 -2.79

108 -1.10 -1.87 -2.65

157 -1.05 -1.81 -2.57

206 -1.01 -1.76 -2.50

255 -0.99 -1.72 -2.45

304 -0.96 -1.69 -2.41

353 -0.94 -1.66 -2.38

402 -0.93 -1.64 -2.35

451 -0.91 -1.62 -2.32

500 -0.90 -1.60 -2.30

38016 -0.90 -1.60 -2.30

38016 4104 972

Zone 1 Zone 2 Zone 30 -25.01 -39.25 -53.50

10 -25.01 -39.25 -53.50

59 -21.42 -34.23 -47.03

108 -20.19 -32.51 -44.83

157 -19.44 -31.45 -43.47

206 -18.89 -30.68 -42.48

255 -18.46 -30.08 -41.70

304 -18.10 -29.58 -41.06

353 -17.80 -29.16 -40.52

402 -17.54 -28.79 -40.05

451 -17.30 -28.47 -39.63

500 -17.09 -28.17 -39.25

38016 -17.09 -28.17 -39.25

Net Pressure (psf)

Area ( ft2 )

Com

pone

nts

and

Cla

ddin

g

Roof

RoofEffectiveArea ( ft2 )

Effective

Com

pone

nts

and

Cla

ddin

g



INPUT DATAExposure category = 2 Exp. C

Importance category = 2 Hazardous

Basic wind speed V = 100 mphBuilding height to eave he = 14 ftBuilding height to ridge hr = 28 ftBuilding length L = 150 ftBuilding width B = 110 ft

DESIGN SUMMARYHorizontal pressure = 43.88 psf ,for roof

= 40.73 psf ,for wallUpward pressure = 23.63 psfTotal horizontal force on building length = 134.91 kipsTotal horizontal force on building width = 65.15 kipsTotal upward force = 389.83 kipsOverturning moment on building length = 2384.00 ft-kipsOverturning moment on building width = 959.94 ft-kips

ANALYSISImportance factor Iw = 1.15Mean roof height h = 21.0 ftHgt-exp-gust factor Ce = 1.14 ,for roof

1.06 ,for wallStagnation pressure qs = 25.7 psfPressure coefficients Cq = 1.30 horizontal any direction (Method 2, Table 16-H)

= 0.70 upward

DanielT. Li

Wind Analysis Based on UBC 1997

ELEMENT & COMPONENT LOADS

Exposure category 2 Exp. CBasic wind speed V = 100 mph

Building height to eave he = 14 ft

Building height to ridge hr = 28 ftBuilding length L = 150 ftBuilding width B = 110 ftWidth of edge strips (CBC 1622A)

a = MIN(0.1B, 10ft) = 10.00 ft

Importance factor Iw = 1.15Mean roof height h = 21 ft

Hgt-exp-gust factor Ce = 1.142 ,for roof

Stagnation pressure qs = 25.7 psfRoof slope 4 :12

Pressure coefficients , for continuity Cq,10 = 1.3 outward ,(CBC Tab.16A-H)

for discontinuity Cq,10 = 2.6 outward

for overhang Cq,10 = 3.1 outward ( 2.3 @ top + 0.8 @ bottom )

Cq = Cq,10 - 0.2 + (100-TA) / 450 , (CBC Tab.16A-H, footnote 2)

Cq = Cq,10 - 0.8 + (100-TA) / 112.5 , (only at discontinuities with slope less than 7:12)Design wind pressures (CBC 1620A)

P = Ce Cq qs Iw (CBC 20A-1)

Continuity Discontinuity Overhangoutward outward outward

Cq P (psf) Cq P (psf) Cq P (psf)<10 1.30 44 2.60 88 3.10 10510 1.30 44 2.60 88 3.10 10515 1.29 44 2.56 86 3.06 10320 1.28 43 2.51 85 3.01 10225 1.27 43 2.47 83 2.97 10030 1.26 42 2.42 82 2.92 9935 1.24 42 2.38 80 2.88 9740 1.23 42 2.33 79 2.83 9645 1.22 41 2.29 77 2.79 9450 1.21 41 2.24 76 2.74 9355 1.20 41 2.20 74 2.70 9160 1.19 40 2.16 73 2.66 9065 1.18 40 2.11 71 2.61 8870 1.17 39 2.07 70 2.57 8775 1.16 39 2.02 68 2.52 8580 1.14 39 1.98 67 2.48 8485 1.13 38 1.93 65 2.43 8290 1.12 38 1.89 64 2.39 8195 1.11 38 1.84 62 2.34 79

100 1.10 37 1.80 61 2.30 78>100 1.10 37 1.80 61 2.30 78

>1000

Trib

utar

y A

rea

(ft2 )

use primary frame values

PROJECT : PAGE : CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :

SEISMIC ANALYSIS BASED ON IBC 2000 (Equivalent Lateral-Force Procedure, Sec.1617.4)

INPUT DATA DESIGN SUMMARYTypical floor height h = 9.0 ft Total base shearTypical floor weight wx = 780 k V = 0.04 W, (SD) = 453 k, (SD)Number of floors (20 max) n = 15 = 0.03 W, (ASD) = 323 k, (ASD)Seismic factor (1616.2) IE = 1.00 Seismic design category = DBuilding location Zip Code = 89107 Latitude: 36.170034Site class (A, B, C, D, E, F) = D (If no soil report, use D) Longitude: -115.20822

The cofficient (1617.4.2.1) Ct = 0.03 SS = 61.812 %g,Sms = 0.807 %g, Fa = 1.306

The cofficient(Tab1617.6) R = 5.50 S1 = 18.366 %g,Sm1 = 0.379 %g, Fv = 2.065

SDS = 0.538 %g , SD1 = 0.253 %g

hn = 135.0 ft k = 1.34

W = 11,700 k Σwxhk = Ta = Ct (hn)

3/4 = 1.19

VERTICAL DISTRIBUTION OF LATERAL FORCESLevel Level Floor to floor Heigth Weight Lateral force @ each level Diaphragm forceNo. Name Height hx wx wxhx

k Cvx Fx Vx O. M. ΣFi ΣW i Fpx

ft ft k k k k-ft k k k15 Roof 135.0 780 569,408 0.145 65.6 65.6 780 66

9.00 65.614 15th 126.0 780 518,980 0.132 59.8 590 125.3 1,560 63

9.00 125.313 14th 117.0 780 469,778 0.119 54.1 1,718 179.4 2,340 63

9.00 179.412 13th 108.0 780 421,861 0.107 48.6 3,333 228.0 3,120 63

9.00 228.011 12th 99.0 780 375,300 0.095 43.2 5,385 271.2 3,900 63

9.00 271.210 11th 90.0 780 330,175 0.084 38.0 7,826 309.2 4,680 63

9.00 309.29 10th 81.0 780 286,578 0.073 33.0 10,609 342.2 5,460 63

9.00 342.28 9th 72.0 780 244,619 0.062 28.2 13,689 370.4 6,240 63

9.00 370.47 8th 63.0 780 204,430 0.052 23.5 17,022 393.9 7,020 63

9.00 393.96 7th 54.0 780 166,174 0.042 19.1 20,567 413.1 7,800 63

9.00 413.15 6th 45.0 780 130,058 0.033 15.0 24,285 428.0 8,580 63

9.00 428.04 5th 36.0 780 96,357 0.025 11.1 28,137 439.1 9,360 63

9.00 439.13 4th 27.0 780 65,457 0.017 7.5 32,089 446.7 10,140 63

9.00 446.72 3rd 18.0 780 37,956 0.010 4.4 36,109 451.0 10,920 63

9.00 451.01 2nd 9.0 780 14,951 0.004 1.7 40,169 452.8 11,700 63

9.00 452.8 Ground 0.0 44,244

3,932,081

DanielTian Li



Base Shear (Derived from IBC 1617.4.1, Pg.360)

V = MAX MIN[SD1IE / (RT) , SDS IE / R ] , 0.044SDS IE , 0.5S1 IE / R W

= MAX MIN[ 0.33W , 0.17W ] , 0.05W , 0.05W ^= 0.17 W, (SD) (E or F) & S1>=0.6g only)= 0.12 W, (ASD)

Where SDS = 0.826 (IBC 1615.1.3)

SD1 = 0.469 (IBC 1615.1.3)

R = 6 (IBC Tab 1617.6, Pg. 365 )

IE = 1.25 (IBC Tab 1604.5, Pg. 297 )

S1 = 0.457 (IBC 1615.1)

hn = 36 ft

CT = 0.02 (0.035 for steel MRF, 0.03 for concrete MRF & steel EBF)

T = CT (hn)(3/4) = 0.294 sec, (IBC 16-39, Pg. 361)

Vertical Distribution of Force (IBC 1617.4.3, Pg.361)

Level Wx hxk Wxhx

k

Roof W3 36k36kW3 V(36kW3) / ΣWxhx

k

3RD W2 24k24kW2 V(24kW2) / ΣWxhx

k

2ND W1 12k12kW1 V(12kW1) / ΣWxhx

k

ΣW ΣWxhxk

Where k = 1 for T <= 0.5

k = 0.5 T + 0.75 for T @ (0.5 , 2.5)

k = 2 for T >= 2.5

Diaphragm Force for D, E, & F (IBC 1620.3.3, Pg.372), (Note: see 1620.1.5, Pg. 372,for B & C)

Level Wx ΣΣΣΣWx Fx ΣΣΣΣFx

Roof W3 W3 F3 F3 MAX[Fmin , MIN(ΣFx W3 / ΣWx , Fmax )]

3RD W2 W2+W3 F2 F2+F3 MAX[Fmin , MIN(ΣFx W2 / ΣWx , Fmax )]

2ND W1 W1+W2+W3 F1 F1+F2+F3 MAX[Fmin , MIN(ΣFx W1 / ΣWx , Fmax )]

ΣW V

Where Fmin = 0.15 SDS IE Wx

Fmax = 0.3 SDS IE Wx

Story Drift Determination (IBC 16-46, Pg.362)δx = Cd δxe / IE

Where Cd = IBC Tab. 1617.6, Pg.365

Daniel

V

Fpx (16-65)

Fx (16-41 & 1642)

Seismic Analysis Based on IBC 2000Tian Li


JOB NO. : DATE : REVIEW BY : Redundancy Factor, ρρρρ, Based on IBC 2000

(IBC 1617.2.2)

where ρ = apply for SDC D,E, & F of this direction of entire building

rmax,i = MAX ( hi / Vi )

hi = the horizontal shear of full height level of the building.

Vi = the total i story shear.

Ai = floor area of the diaphragm level immediately above the story.

BRACED FRAME STRUCTURE

ρ = [1.0 , 1.5]

ρ = max(ρ3rd, ρ2nd, ρ1st)

ρ3rd =2-20 / [max( Pe, Pf, P5, P6) cos α / V3rd ] (A2nd)0.5

ρ2nd =2-20 / [max( Pa, Pb, P1, P2) cos α / V2nd ] (A1st)0.5

ρ1st =2-20 / [max( Pc, Pd, P3, P4) cos α / V1st ] (Aground)0.5

MOMENT FRAME STRUCTURE

ρ = [1.0 , 1.25] ,for SDC Dρ = [1.0 , 1.1] ,for SDC E & F

ρ = max[ 2 - 20 / (r2nd A1st0.5) , 2 - 20 / (r1st Aground

0.5) ]

r2nd = max[ (Va+0.7Vb) , 0.7(Vb+Vc) , (Vc+0.7Vd) , (V1+0.7V2) , (0.7V2+V3)] / V2nd

r1st = max[ (Ve+0.7Vf) , 0.7(Vf+Vg) , (Vg+0.7Vh) , (V4+0.7V5) , (0.7V5+V6)] / V1st

BUILDING FRAME SYSTEM WITH SHEAR WALL

ρ = [1.0 , 1.5]


0.5) ]

r2nd = max[ 10Va / (V2nd La ) , 10V1 / (V2nd L1 )]

r1st = max[ 10Vb / (V1st Lb ) , 10Vc / (V1st Lc ) , 10V2 / (V1st L2 )]

DUAL SYSTEM

ρ = 0.8 ρcalcs > 1.0< 1.5


0.5) ]

r2nd = max[ 10Va / (V2nd La ) , Px cos α / V2nd , (V1+V2) / V2nd , (V5+V6) / V2nd]

r1st = max[ 10Vb / (V1st Lc ) , Py cos α / V1st , (V3+V4) / V1st , (V7+V8) / V1st]

VERTICAL COMBINATION OF STRUCTURAL SYSTEMS

ρ2nd , ρ1st = [1.0 , 1.5]

ρ2nd = ρU

ρ1st = max( ρL , RLρU / RU)

Note: 1. One direction is bearing wall , the R, ΩΩΩΩ 0 , and C d used for the orthogonal should use their direction values. (IBC 1617.6.3)2. When a combination of structural systems is used in the same direction , the R, ΩΩΩΩ 0 , and C d used be not greater than the least value. (IBC 1617.6.2)

Tian LiDaniel

( )max,

20max 2i

i ir Aρ ρ= = −

PROJECT : PAGE : CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :

SEISMIC ANALYSIS BASED ON IBC 2003 (Equivalent Lateral-Force Procedure, ASCE 9.5.5)

INPUT DATA DESIGN SUMMARYTypical floor height h = 9.0 ft Total base shearTypical floor weight wx = 780 k V = 0.06 W, (SD) = 679 k, (SD)Number of floors (20 max) n = 15 = 0.04 W, (ASD) = 485 k, (ASD)Seismic factor (IBC Tab.1604.5) IE = 1.00 Seismic design category = DBuilding location Zip Code 89107 Latitude: 36.170034Site class (A, B, C, D, E, F) D (If no soil report, use D) Longitude: -115.20822

The cofficient (ASCE 9.5.5.3.2) Ct = 0.02 SS = 61.812 %g,Sms = 0.807 %g, Fa = 1.306

The cofficient (ASCE 9.5.5.3.2) x = 0.75 S1 = 18.366 %g,Sm1 = 0.379 %g, Fv = 2.065

The cofficient(IBC Tab. 1617.6.2) R = 5.50 SDS = 0.538 %g , SD1 = 0.253 %g

hn = 135.0 ft k = 1.15

W = 11,700 k Σwxhk = Ta = Ct (hn)x = 0.79

VERTICAL DISTRIBUTION OF LATERAL FORCESLevel Level Floor to floor Heigth Weight Lateral force @ each level Diaphragm forceNo. Name Height hx wx wxhx

k Cvx Fx Vx O. M. ΣFi ΣW i Fpx

ft ft k k k k-ft k k k15 Roof 135.0 780 215,558 0.133 90.6 90.6 780 91

9.00 90.614 15th 126.0 780 199,170 0.123 83.7 816 174.4 1,560 87

9.00 174.413 14th 117.0 780 182,953 0.113 76.9 2,385 251.3 2,340 84

9.00 251.312 13th 108.0 780 166,917 0.103 70.2 4,647 321.5 3,120 84

9.00 321.511 12th 99.0 780 151,075 0.094 63.5 7,540 385.0 3,900 84

9.00 385.010 11th 90.0 780 135,442 0.084 56.9 11,005 442.0 4,680 84

9.00 442.09 10th 81.0 780 120,037 0.074 50.5 14,983 492.4 5,460 84

9.00 492.48 9th 72.0 780 104,880 0.065 44.1 19,415 536.5 6,240 84

9.00 536.57 8th 63.0 780 89,997 0.056 37.8 24,244 574.4 7,020 84

9.00 574.46 7th 54.0 780 75,423 0.047 31.7 29,413 606.1 7,800 84

9.00 606.15 6th 45.0 780 61,201 0.038 25.7 34,868 631.8 8,580 84

9.00 631.84 5th 36.0 780 47,391 0.029 19.9 40,554 651.7 9,360 84

9.00 651.73 4th 27.0 780 34,081 0.021 14.3 46,420 666.1 10,140 84

9.00 666.12 3rd 18.0 780 21,414 0.013 9.0 52,414 675.1 10,920 84

9.00 675.11 2nd 9.0 780 9,676 0.006 4.1 58,490 679.1 11,700 84

9.00 679.1 Ground 0.0 64,602

1,615,216

DanielTian Li



Base Shear (Derived from ASCE 7-02 Sec. 9.5.5, Pg.146)

V = MAX MIN[SD1IE / (RT) , SDS IE / R ] , 0.044SDS IE , 0.5S1 IE / R W

= MAX MIN[ 0.33W , 0.17W ] , 0.05W , 0.05W ^= 0.17 W, (SD) (E or F only)= 0.12 W, (ASD)

Where SDS = 0.826 (IBC 1615.1.3)

SD1 = 0.469 (IBC 1615.1.3)

R = 6 (IBC Tab 1617.6.2, Pg. 334 )

IE = 1.25 (IBC Tab 1604.5, Pg. 272 )

S1 = 0.457 (IBC 1615.1)

hn = 36 ft

Ct = 0.02 (0.028 for steel MRF, 0.016 for concrete MRF, & 0.03 steel EBF)

x = 0.75 (0.8 for steel MRF, 0.9 for concrete MRF & steel EBF)

T = Ct (hn)x = 0.294 sec, (ASCE 7-02 Sec. 9.5.5.3.2, Pg. 147) D

Vertical Distribution of Force (ASCE 7-02, Sec. 9.5.5.4, Pg.148)

Level Wx hxk Wxhx

k

Roof W3 36k36kW3 V(36kW3) / ΣWxhx

k

3RD W2 24k24kW2 V(24kW2) / ΣWxhx

k

2ND W1 12k12kW1 V(12kW1) / ΣWxhx

k

ΣW ΣWxhxk

Where k = 1 for T <= 0.5

k = 0.5 T + 0.75 for T @ (0.5 , 2.5)

k = 2 for T >= 2.5

Diaphragm Force for D, E, & F (IBC 1620.4.3, Pg. 341), (Note: see 1620.2.5, Pg. 339,for B & C)


Roof W3 W3 F3 F3 MAX[Fmin , MIN(ΣFx W3 / ΣWx , Fmax )]

3RD W2 W2+W3 F2 F2+F3 MAX[Fmin , MIN(ΣFx W2 / ΣWx , Fmax )]

2ND W1 W1+W2+W3 F1 F1+F2+F3 MAX[Fmin , MIN(ΣFx W1 / ΣWx , Fmax )]

ΣW V

Where Fmin = 0.2 SDS IE Wx

Fmax = 0.4 SDS IE Wx

Story Drift Determination (ASCE 7-02 Sec. 9.5.5.7.1, Pg.149)δx = Cd δxe / IE

Where Cd = ASCE 7-02 Tab. 9.5.2.2, Pg.135

Daniel

V

Fpx (16-63)

Fx (16-41 & 1642)

Seismic Analysis Based on IBC 2003Tian Li


JOB NO. : DATE : REVIEW BY : Redundancy Factor, ρρρρ, Based on IBC 2003

(IBC 1617.2.2.2)

where ρ = apply for SDC D,E, & F of this direction of entire building

rmax,i = MAX ( hi / Vi )

hi = the horizontal shear of full height level of the building.


Ai = floor area of the diaphragm level immediately above the story.


ρ = [1.0 , 1.5]

ρ = max(ρ3rd, ρ2nd, ρ1st)

ρ3rd =2-20 / [max( Pe, Pf, P5, P6) cos α / V3rd ] (A2nd)0.5

ρ2nd =2-20 / [max( Pa, Pb, P1, P2) cos α / V2nd ] (A1st)0.5

ρ1st =2-20 / [max( Pc, Pd, P3, P4) cos α / V1st ] (Aground)0.5


ρ = [1.0 , 1.25] ,for SDC Dρ = [1.0 , 1.1] ,for SDC E & F


0.5) ]

r2nd = max[ (Va+0.7Vb) , 0.7(Vb+Vc) , (Vc+0.7Vd) , (V1+0.7V2) , (0.7V2+V3)] / V2nd

r1st = max[ (Ve+0.7Vf) , 0.7(Vf+Vg) , (Vg+0.7Vh) , (V4+0.7V5) , (0.7V5+V6)] / V1st


ρ = [1.0 , 1.5]


0.5) ]

r2nd = max[ 10Va / (V2nd La ) , 10V1 / (V2nd L1 )]

r1st = max[ 10Vb / (V1st Lb ) , 10Vc / (V1st Lc ) , 10V2 / (V1st L2 )]

DUAL SYSTEM

ρ = 0.8 ρcalcs > 1.0< 1.5


0.5) ]

r2nd = max[ 10Va / (V2nd La ) , Px cos α / V2nd , (V1+V2) / V2nd , (V5+V6) / V2nd]

r1st = max[ 10Vb / (V1st Lc ) , Py cos α / V1st , (V3+V4) / V1st , (V7+V8) / V1st]

VERTICAL COMBINATION OF STRUCTURAL SYSTEMS

ρ2nd , ρ1st = [1.0 , 1.5]

ρ2nd = ρU

ρ1st = max( ρL , RLρU / RU)

Note: 1. One direction is bearing wall or other system, the R, ΩΩΩΩ 0 , and C d used for the orthogonal should use their direction values. (IBC 1617.6.2.3)2. When a combination of structural systems is used in the same direction , the R, ΩΩΩΩ 0 , and C d used be not greater than the least value. (IBC 1617.6.2.2)

Tian LiDaniel

( )max,

20max 2i

i ir Aρ ρ= = −

PROJECT : PAGE :

CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :

INPUT DATASeismic zone (1, 2A, 2B, 3, or 4) = 4Coefficient, R = 4.5Occupancy Category(1, 2, 3, or 4) = 4Soil type (SA, SB, SC, SD, or SE ) = SDNo. of Stories = 10Typical story height = 10 ftTypical story weight = 1500 kipsCoefficient, Ct = 0.02Seismic source type ( A, B, or C ) = AClosest distance to seismic source = 8.6 km

VERTICAL DISTRIBUTION ANALYSISStory H (ft) W (k) Fi (k) V (k) Fpx (k) Coefficient, Ca = 0.484Roof 10 1500 733 726 Coefficient, Cv = 0.87010th 10 1500 660 733 697 Period (method A) = 0.639th 10 1500 587 1,393 660 Period (method B) =8th 10 1500 513 1,980 623 Zone factor, Z = 0.4007th 10 1500 440 2,493 587 Importance factor, I = 1.006th 10 1500 367 2,933 550 V = ( Cv I / R T ) W = 4,587 k5th 10 1500 293 3,300 513 V = ( 2.5 Ca I / R ) W = 4,033 k (max)4th 10 1500 220 3,593 477 V = 0.11 Ca I W = 799 k (min)3rd 10 1500 147 3,813 440 V = ( 0.8 Z Nv I / R ) W = 1,451 k(minZ=4)2nd 10 1500 73 3,960 403 Whip effect force, Ft = 0 k

Grnd 4,033 Base shear, V = 4,033 kWhip force, Ft = 0 k

V = 0.269 w, (SD)= 0.192 w, (ASD)

Seismic Analysis Based on UBC 1997

DanielT. Li



Base Shear (Derived from CBC 1630A.2.1)

V = MAX MIN[ 2.5Ca I W / R , Cv I W / (R T) ] , 0.11Ca I W , 0.8Z Nv I W / R

= MAX MIN[ 0.25W , 0.75W ] , 0.06W , 0.08W ^= 0.25 W, (SD) (Z 4 only)= 0.18 W, (ASD)

Where Ca = 0.484 (CBC Tab 16A-Q)

I = 1.15 (CBC Tab 16A-K)

R = 5.5 (CBC Tab 16A-N)

Cv = 0.768 (CBC Tab 16A-R)

Z = 0.4 (CBC Tab 16A-I)

hn = 36 ft

Ct = 0.02 (0.035 for steel MRF, 0.03 for concrete MRF)

Nv = 1.2 (CBC Tab 16A-S)

T = Ct (hn)(3/4) / (I Nv) = 0.213 sec, (CBC 30A-8)

Vertical Distribution of Force (CBC 1630A.5)

Level Wx hx Wxhx

Roof W3 36 36W3 Ft+(V-Ft)36W3 / ΣWxhx

3RD W2 24 24W2 (V-Ft)24W2 / ΣWxhx

2ND W1 12 12W1 (V-Ft)12W1 / ΣWxhx

ΣW ΣWxhx

Where Ft = MIN(0.07TV , 0.25V) , for T > 0.7 (CBC 1630A.5)

Diaphragm Force (CBC 1633A.2.9)


Roof W3 W3 F3 F3 MAXFmin , MIN[(Ft + ΣFx)W3 / ΣWx , Fmax]

3RD W2 W2+W3 F2 F2+F3 MAXFmin , MIN[(Ft + ΣFx)W2 / ΣWx , Fmax]

2ND W1 W1+W2+W3 F1 F1+F2+F3 MAXFmin , MIN[(Ft + ΣFx)W1 / ΣWx , Fmax]

ΣW V

Where Fmin = 0.5Ca I Wx

Fmax = 1.0Ca I Wx

The Maximum Inelastic Response Displacement (CBC 1630A.9.2)∆M = 0.7 R ∆s , (CBC 30A-17)

Daniel

V

Fpx (33A-1)

Fx (30A-15)

Seismic Analysis Based on CBC 2001/UBC97T. Li



The seismic requirements of CBC Division VI-R apply to RETROFIT OF SCHOOL BUILDING (DSA), andRETROFIT / REHABILITATION OF HOSPITAL BUILDING (OSHPD 1 & 4). (CBC 1640A)

RETROFIT: All design and construction work undertaken to construct any new or to repair or strengthenany existing structural or nonstructural elements required by the evaluation and designof the building. (CBC 1641A)

REHABILITATION: The evaluation and retrfit of an existing nonconforming building or a school buildingconforming to earlier code requirements to bring the building into conformance with thesafty standards of the currently effective regulations. (CBC 1641A)

OSHPD 1: General acute-care hospitals. (CBC 1601A.1)OSHPD 4: Correctional Treatment Centers. (CBC 1601A.1)

Base Shear (Derived from CBC 1644A.4.1)

V = MAX MIN[ 2.5H Ca I W / R , H Cv I W / (R T) ] , 0.11H Ca I W , 0.8H Z Nv I W / R

= MAX MIN[ 0.23W , 0.75W ] , 0.06W , 0.08W ^= 0.23 W, (SD) (Z 4 only)= 0.16 W, (ASD)

Where H = 1.0 (OSHPD, CBC 1643A.8.1 & 1643A.8.2)1.2 (DSA, CBC 1643A.8.3)

Ca = 0.44 (CBC Tab 16A-Q)I = 1.15 (school, CBC Tab 16A-K)

1.5 (hospital, CBC Tab 16A-K)R = 5.5 (1976 or later, CBC 1644A.3.1.2 & Tab 16A-N)

5.5 (Building Frame System, CBC 1644A.3.1.1)4.5 (CBC 1644A.3.1)

Cv = 0.768 (CBC Tab 16A-R)

Z = 0.4 (CBC Tab 16A-I)

hn = 36 ft

Ct = 0.02 (0.035 for steel MRF, 0.03 for concrete MRF)

Nv = 1.2 (CBC Tab 16A-S)

T = Ct (hn)(3/4) / (I Nv) = 0.213 sec, (CBC 30A-8)

Vertical Distribution of Force (CBC 1630A.5)

Level Wx hx Wxhx

Roof W3 36 36W3 Ft+(V-Ft)36W3 / ΣWxhx

3RD W2 24 24W2 (V-Ft)24W2 / ΣWxhx

2ND W1 12 12W1 (V-Ft)12W1 / ΣWxhx

ΣW ΣWxhx

Where Ft = MIN(0.07TV , 0.25V) , for T > 0.7 (CBC 1630A.5)

Daniel

V

Fx (30A-15)

Seismic Analysis Based on CBC Division VI-RT. Li

(cont'd)Diaphragm Force (CBC 1646A.2.9.2)


Roof W3 W3 F3 F3 ββββ MAXFmin , MIN[(Ft + ΣFx)W3 / ΣWx , Fmax]

3RD W2 W2+W3 F2 F2+F3 ββββ MAXFmin , MIN[(Ft + ΣFx)W2 / ΣWx , Fmax]

2ND W1 W1+W2+W3 F1 F1+F2+F3 ββββ MAXFmin , MIN[(Ft + ΣFx)W1 / ΣWx , Fmax]

ΣW V

Where β = 1.0 (CBC 1646A.2.9.2)V = (R / 4) V (masonry or concrete wall, CBC 1646A.2.9.4)

Fmin = 0.5 β H I Wx (no Ca, code print wrong? Anyway governs, CBC 1646A.2.9.2)

Fmax = 1.0 β H Ca I Wx (CBC 1646A.2.9.2)

Lateral Force on Elements, Nonstructural Components, and Equipments (CBC 1644A.1.3)

Fp = β H Ca Ip Wp (CBC 44A-14)

Alternatively,

Fp = β ap H Ca Ip (1 + 3 hx / hr) Wp / Rp (CBC 44A-15)

> 0.7 β H Ca Ip Wp

< 4 β H Ca Ip Wp (CBC 44A-16)

Use Fp = β H Ca Ip Wp

Where β = 4.0 (CBC 1645A)

SD Load Combinations (CBC 1644A.4.1.1)

φ Cn = 1.05D + 0.25 L + βE (CBC 44A-5)

φ Cn = βE - 0.9D (CBC 44A-6)

Where β = 1.0 (concrete, CBC 1645A.3 & Tab 16A-R-1)

ASD Load Combinations (CBC 1644A.4.1.1)

Cw = D + L + βE/1.4 (CBC 44A-7)

Cw = βE/1.4 - 0.9D (CBC 44A-8)

Where β = 2.5 (masonry, CBC 1645A.4)1.5 (steel, CBC 1645A.5)2.0 (wood, CBC 1645A.6 & Tab 16A-R-2)

Zone 3 & 4 Load Combinations (CBC 1644A.9.2)

φ Cn = D + 0.8L + Ω0 β E (CBC 44A-9)

φ Cn = Ω0 β E - 0.9D (CBC 44A-10)

Where Ω0 β < R

Fpx (33A-1)


JOB NO. : DATE : REVIEW BY : Redundancy Factor, ρρρρ, Based on UBC 97

(UBC 1630.1.1)

where ρ = apply for this direction of entire building

rmax = MAX ( hi / Vi )

hi = the horizontal shear at or below the 2/3 height level of the building.


AB = ground floor area of the structure.


ρ = [1.0 , 1.5]

rmax = max[ max( Pa, Pb, P1, P2) cos α / V2nd ,

max( Pc, Pd, P3, P4) cos α / V1st ]


ρ = [1.0 , 1.25]

rmax = max max[ (Va+0.7Vb) , 0.7(Vb+Vc) , (Vc+0.7Vd)] / V2nd ,

max[ (Ve+0.7Vf) , 0.7(Vf+Vg) , (Vg+0.7Vh)] / V1st ,

max[(V1+0.7V2) , (0.7V2+V3)] / V2nd , max[(V4+0.7V5) , (0.7V5+V6)] / V1st


ρ = [1.0 , 1.5]

rmax = max[ 10Va / (V2nd La ) , 10Vb / (V2nd Lb ) , 10V1 / (V2nd L1 ) ,

10Vc / (V1st Lc ) , 10Vd / (V1st Ld ) , 10V2 / (V1st L2 )]

(Note: For light-framed construction of CBC 2001 OSHPD 1 & 4, 10 / Lw < 1.0 )

DUAL SYSTEM

ρ = 0.8 ρcalcs > 1.0< 1.5

rmax = max[ 10Va / (V2nd La ) , Px cos α / V2nd , (V1+V2) / V2nd , (V5+V6) / V2nd ,

10Vb / (V1st Lc ) , Py cos α / V1st , (V3+V4) / V1st , (V7+V8) / V1st]

COMBINATION OF STRUCTURAL SYSTEMS

ρ = 1.0 ρcalcs > 1.0< 1.5

Note: 1. One direction is bearing wall , the R value used for the orthogonal can not be greater than that for the bearing wall system. (UBC 1630.4.3)2. When a combination of structural systems is used in the same direction , the R used be not greater than the least value. (UBC 1630.4.4)

T. LiDaniel

max

202

Br Aρ = −



INPUT DATAWALL THICKNESS t = 8 inPARAPET HEIGHT hp = 4 ftWALL HEIGHT h = 14 ftTOTAL WALL DENSITY ρ = 150 pcfSEISMIC PARAMETER SDS = 0.54 (IBC Sec.1615.1.3)

SEISMIC DESIGN CATEGORY SDC = CDIAPHRAGM FLEXIBLE ? (0=no, 1=yes) 1 Yes

IMPORTANCE FACTOR Iw = 1 (IBC Tab. 1604.5)

BASIC WIND SPEED V = 90 mphEXPOSURE CATEGORY (B, C, D) = C 90TOPOGRAPHIC FACTOR KZt = 1 Flat, (ASCE Eq.6-1)

DESIGN SUMMARYOut-of-plane force for wall design w1 = 23.9 psf (Wind governs)Out-of-plane force for parapet design w2 = 61.7 psf (Wind governs)Out-of-plane force for anchorage design Fanch = 450 plf (Horizontal direction)

(The governing seismic forces have been reduced by 0.7 for ASD)

WIND ANALYSISOut-of-plane wind force for wall design (ASCE 7-98, Eq.6-18)

= 23.9 psf

Where : Kh = 0.86 , Kd = 0.85 , GCp = -1.40 , GCpi = 0.18(mean roof h = 16 ft, changeable) (ASCE Tab. 6-6) (corner ? Yes , TA = 10 ft2 ) (ASCE Tab. 6-7)

(ASCE Tab. 6-5) (ASCE Fig. 6-5A)Out-of-plane wind force for parapet design (ASCE 7-98, Eq.6-18)

= 61.7 psf

Where : Kh = 0.88 , Kd = 0.85 , GCp = -1.40 , GCp = -2.80 GCpi = 0.18(ASCE Tab. 6-5) (ASCE Tab. 6-6) = 1.00 (roof, ASCE 7-02 Fig. 6-11B) (roof, ASCE Fig. 6-5)

(wall, ASCE 7-02 Fig. 6-11A)Out-of-plane wind force for anchorage design

= 450 plf (Horizontal)

SEISMIC ANALYSISOut-of-plane seismic force for wall design (IBC 2000, Sec.1620.1.7)

= 0.22 Wp = 21.6 psf

Where : Wp = 100.0 psf , IE = 1.0(IBC Sec.1621.1.6)

Out-of-plane seismic force for parapet design (IBC 2000, Sec. 1621.1.4)

= 0.65 Wp = 64.8 psf

Where : ap = 2.5 , Ip = 1.0 , Rp = 2.5(IBC Tab.1621.2) (IBC Sec.1621.1.6) (IBC Tab.1621.2)

Out-of-plane seismic force for anchorage designFor seismic design category A & B, both flexible & rigid diaphragm (IBC 2000, Sec.1604.8.2 & 1620.1.7)

0.54 2.50 Wp =250 plf (Horizontal)

(Not applicable)Where : Fmin = 200 plf

(IBC Sec.1604.8.2)

For seismic design category C and above, flexible diaphragm (IBC 2000, Sec.1604.8.2 ,1620.1.7, & 1620.2.1)

5.00 Wp =500 plf (Horizontal)

(Applicable)For seismic design category C and above, rigid diaphragm (IBC 2000, Sec.1604.8.2 ,1620.1.7, & 1621.1.4)

= 3.00 Wp = 300 plf (Horizontal) (Not applicable)Where : ap = 1.0

(IBC Tab.1621.2)

DanielTian Li

Lateral Force for One-Story Wall Based on IBC 2000

( ) ( ) ( ) ( ) ( )21, 0.00256 h Zt d wwind P Pi P Pih G G G Gqw C C V C CK K K I= − = −

( )1, 0.4 ,0.1Eseismic DS p pMAXw S W WI=

( )2

, min1.2

0.3 , , 1.6 , 400 ,2

pp DSanch seismic p p EDS DS p DS

p

h h pa S IMAX MAX MINS S W SF I I I FhR

+ =


, 1, 2,12 2

panch wind wind p wind

h hw h wF

h = + +

2,1.2

0.3 , , 1.6pp DS pp pseismic DS p DS p

p

a S WIMAX MINw S W S WI IR

=

( ) ( )2 2

, min0.4 , 0.1 , 400 ,2 2

anch seismic E EDS p p DS

h hh hp pMAX S W W SF I I F

h h

+ + = =

( ) ( )2 2

, min0.8 , 0.1 , 400 ,2 2



h h

+ + = =



INPUT DATAWALL THICKNESS t = 8 inPARAPET HEIGHT hp = 4 ftWALL HEIGHT h = 14 ftTOTAL WALL DENSITY ρ = 150 pcfSEISMIC PARAMETER SDS = 0.54 (IBC Sec.1615.1.3)

SEISMIC DESIGN CATEGORY SDC = CDIAPHRAGM FLEXIBLE ? (0=no, 1=yes) 1 Yes

IMPORTANCE FACTOR Iw = 1 (IBC Tab. 1604.5)

BASIC WIND SPEED V = 90 mphEXPOSURE CATEGORY (B, C, D) = CTOPOGRAPHIC FACTOR KZt = 1 Flat, (ASCE Eq.6-3)

DESIGN SUMMARYOut-of-plane force for wall design w1 = 23.9 psf (Wind governs)Out-of-plane force for parapet design w2 = 61.7 psf (Wind governs)Out-of-plane force for anchorage design Fanch = 450 plf (Horizontal direction)

(The governing seismic forces have been reduced by 0.7 for ASD)

WIND ANALYSISOut-of-plane wind force for wall design (ASCE 7-02, Eq.6-22)

= 23.9 psf

Where : Kh = 0.86 , Kd = 0.85 , GCp = -1.40 , GCpi = 0.18(mean roof h = 16 ft, changeable) (ASCE Tab. 6-4) (corner ? Yes , TA = 10 ft2 ) (ASCE Fig. 6-5)

(ASCE Tab. 6-3) (ASCE Fig. 6-11A)Out-of-plane wind force for parapet design (ASCE 7-02, Eq.6-24)

= 61.7 psf, (ASCE7-02,6.5.12.4.4)

Where : Kp = 0.88 , Kd = 0.85 , GCp = -1.40 , GCp = -2.80 GCpi = 0.18(ASCE Tab. 6-3) (ASCE Tab. 6-4) = 1.00 (roof, ASCE Fig. 6-11B) (roof, ASCE Fig. 6-5)

(wall, ASCE Fig. 6-11A)

Out-of-plane wind force for anchorage design


SEISMIC ANALYSISOut-of-plane seismic force for wall design (IBC 2003, Sec.1620.2.7)

= 0.22 Wp = 21.6 psf

Where : Wp = 100.0 psf , IE = 1.0(IBC Sec.1604.5)

Out-of-plane seismic force for parapet design (ASCE 7-02, Sec. 9.6.1.3)

= 0.65 Wp = 64.8 psf

Where : ap = 2.5 , Ip = 1.0 , Rp = 2.5(ASCE Tab.9.6.2.2) (ASCE Sec. 9.6.1.5) (ASCE Tab.9.6.2.2)

Out-of-plane seismic force for anchorage designFor seismic design category A & B, both flexible & rigid diaphragm (IBC 2003, Sec.1604.8.2 & 1620.2.7)


(Not applicable)Where : Fmin = 280 plf

(IBC Sec.1604.8.2)

For seismic design category C and above, flexible diaphragm (IBC 2003, Sec.1604.8.2 ,1620.2.7, & 1620.3.1)


(Applicable)For seismic design category C and above, rigid diaphragm (IBC 2003, Sec.1604.8.2 ,1620.2.7, & ASCE Sec. 9.6.1.3)

= 3.00 Wp = 300 plf (Horizontal) (Not applicable)Where : ap = 1.0

(ASCE Tab.9.6.2.2)

DanielTian Li

Lateral Force for One-Story Wall Based on IBC 2003


( )1, 0.4 ,0.1Eseismic DS p pMAXw S W WI=

( )2

, min1.2

0.3 , , 1.6 , 400 ,2

pp DSanch seismic p p EDS DS p DS

p

h h pa S IMAX MAX MINS S W SF I I I FhR

+ =

( ) ( ) ( ) ( ) ( )22, 0.00256 h Zt d wwind P Pi P Pip G G G Gqw C C V C CK K K I= − = −

, 1, 2,12 2

panch wind wind p wind

h hw h wF

h = + +

2,1.2

0.3 , , 1.6pp DS pp pseismic DS p DS p

p

a S WIMAX MINw S W S WI IR

=

( ) ( )2 2

, min0.4 , 0.1 , 400 ,2 2



h h

+ + = =

( ) ( )2 2

, min0.8 , 0.1 , 400 ,2 2



h h

+ + = =



INPUT DATAWALL THICKNESS t = 12 inPARAPET HEIGHT hp = 0 ftWALL HEIGHT h = 44 ftTOTAL WALL DENSITY ρ = 58 pcfSEISMIC ZONE (1, 2A, 2B, 3, or 4) 4 Zone 4SEISMIC COFFICIENT Ca = 0.44 (UBC Tab. 16-Q)

IMPORTANCE FACTOR Ip = 1 (UBC Tab. 16-K)

BASIC WIND VELOCITY V = 70 mphEXPOSURE TYPE (B, C, D) B

DESIGN SUMMARYOut-of-plane force for wall design w1 = 18.5 psf (Seismic governs)Out-of-plane force for parapet design w2 = 60.8 psf (Seismic governs)Out-of-plane force for anchorage design Fanch = 802 plf (Horizontal direction)

(The governing seismic forces have been divided by 1.4 for ASD)

WIND ANALYSISOut-of-plane wind force for wall design (UBC 97 Sec.1620)

= 16.1 psf

Where : Ce = 0.86 , Cq = 1.49 , qs = 12.60 psf , Iw = 1.00(UBC Tab.16-G) (UBC Tab.16-H) (UBC Tab.16-F) (UBC Tab.16-K)

Mean roof h = 44 ft Corner ? Yes (1.5 Yes, 1.2 No)TA = 16 ft2 (10 ft2 default)[Cq - 0.2 + (100-TA)/450 ] for TA @ [10,100]

Out-of-plane wind force for parapet design (UBC 97 Sec.1620)= 16.3 psf

Where : Ce = 0.86 , Cq = 1.50 , qs = 12.60 psf , Iw = 1.00(UBC Tab.16-G) TA = 8 ft2 (UBC Tab.16-F) (UBC Tab.16-K)

Out-of-plane wind force for anchorage design


Where : Ce = 0.86 , Cq = 1.49 , qs = 12.60 , Iw = 1.00(UBC Tab.16-G) (UBC Tab.16-H) (UBC Tab.16-F) (UBC Tab.16-K)

SEISMIC ANALYSISOut-of-plane seismic force for wall design (UBC 97 Sec.1632.2)

= 0.45 Wp = 25.9 psf

Where : ap = 1.0 , Rp = 3.0 , Wp = 58.0 psf(UBC Tab.16-O) (UBC Tab.16-O)

Out-of-plane seismic force for parapet design (UBC 97 Sec.1632.2)

= 1.47 Wp = 85.1 psf

Where : ap = 2.5 , Rp = 3.0 , Wp = 58.0 psf(UBC Tab.16-O) (UBC Tab.16-O)

Out-of-plane seismic force for anchorage design (UBC 97 Sec.1632.2 & 1633.2.8.1)

= 0.88 Wp =1123 plf (Horizontal)

Where : ap = 1.5 , Rp = 3.0 , Wp = 1276 plf (Horizontal) , Fmin = 420 plf(UBC Tab.16-O) (UBC Tab.16-O) (UBC 1633.2.8.1)

DanielTian Li

Lateral Force for One-Story Wall Based on UBC 97

1, wwind e q sqw C C I=

2, wwind e q sqw C C I=

1,41

0.7 , , 42

pp a pp pseismic a p a p

p

a C WIMAX MINw C W C WI IR

= +

2,4

0.7 , , 4pp a pp pseismic a p a p

p

a C WIMAX MINw C W C WI IR

=

0.7 , , 4pp a pp pa p a p

p

a C WIMAX MINC W C WI IR

, min4

0.7 , , 4 ,pp a panch seismic p pa p a p

p

a C WIMAX MINC W C WF I I FR

=

( )2

,2

anch wind we q s

hh pqC CF I

h

+=


JOB NO. : DATE : REVIEW BY : Guardrail Design Based on AISC-ASD & ACI 318-02

INPUT DATA & DESIGN SUMMARYGUARDRAIL SECTION = > PIPE-1 1/2 = > A I S t dGUARDRAIL YIELD STRESS F y = 35 ksi 0.80 0.31 0.33 0.15 1.90BALUSTER SECTION = > PIPE-1 1/4 = > A I S t dBALUSTER YIELD STRESS F y = 35 ksi 0.67 0.20 0.24 0.14 1.66

GUARDRAIL SPAN L = 48 inBALUSTER HEIGHT H = 36 inBALUSTER SLEEVE DEPTH D = 3 inEDGE DISTANCE TO SLEEVE c = 1 in

CONCRETE STRENGTH f c' = 3 ksiHORIZ. LOAD PERP. TO GUARDRAIL w = 50 plf

(UBC Tab.16-B, IBC 1607.7.1)

THE BRACE DESIGN IS ADEQUATE.

ANALYSISCHECK GUARDRAIL CAPACITIES (AISC-ASD, 5-48 & 5-49)

100 ft-lbs 100 lbs

3.68 ksi < (4/3) 0.66 F y = 30.80 ksi [Satisfactory]

(where 4/3 from IBC 1607.7.1.3, Typical.)


CHECK BALUSTER CAPACITIES (AISC-ASD, 5-48 & 5-49)

600 ft-lbs 200 lbs



CHECK CONCRETE BREAKOUT STRENGTH AT BALUSTER SLEEVE (ACI 318-02 Appendix D)

= 0.562 kips > Vu [Satisfactory]


A V /A Vo and ψ6 terms are 1.0 for single shear sleeve not influenced by more than one free edge. l is load bearing length of the anchor for shear, not to exceed 8d.

V u = 1.4 V = 0.280 kips

CHECK CONCRETE PRYOUT STRENGTH AT BALUSTER SLEEVE (ACI 318-02 Appendix D)


where : ψ3 term is 1.0 for location where concrete cracking is likely to occur.

k cp = 2.0 for D > 2.5 in.

A n = 3 D (1.5D + c) = 49.50 in2

DanielTian Li

2

8wLM = =

2wL

V = =

bM

FS

= =

vV

FA

= =

M wLH= = V wL= =

bM

FS

= =

vV

FA

= =

0.2' 1.5

6 7 6 7 7V Vcb b c

Vo Vo

lA A d fV V cdA A


( ) ( )' 1.52 3 32

0.30.7 24

9 1.5N N

cp cp b cp cNo

cA A fV k N k DDA D

φ φ φψ ψ ψ = = +


JOB NO. : DATE : REVIEW BY : Guardrail Design Based on AISC-ASD & ACI 318-02

INPUT DATA & DESIGN SUMMARYGUARDRAIL SECTION = > PIPE-1 1/2 = > A I S t dGUARDRAIL YIELD STRESS F y = 35 ksi 0.80 0.31 0.33 0.15 1.90BALUSTER SECTION = > PIPE-1 1/4 = > A I S t dBALUSTER YIELD STRESS F y = 35 ksi 0.67 0.20 0.24 0.14 1.66

GUARDRAIL SPAN L = 48 inBALUSTER HEIGHT H = 36 inBALUSTER SLEEVE DEPTH D = 3 inEDGE DISTANCE TO SLEEVE c = 1 in

CONCRETE STRENGTH f c' = 3 ksiPOINT LOAD PERP. TO GUARDRAIL P = 200 lbs 35

(UBC Tab.16-B, IBC 1607.7.1)

THE BRACE DESIGN IS ADEQUATE.

ANALYSISCHECK GUARDRAIL CAPACITIES (AISC-ASD, 5-48 & 5-49)

200 ft-lbs, (P @ middle) 200 lbs, (P @ end)


(where 4/3 from IBC 1607.7.1.3, Typical.)


CHECK BALUSTER CAPACITIES (AISC-ASD, 5-48 & 5-49)

600 ft-lbs 200 lbs



CHECK CONCRETE BREAKOUT STRENGTH AT BALUSTER SLEEVE (ACI 318-02 Appendix D)



A V /A Vo and ψ6 terms are 1.0 for single shear sleeve not influenced by more than one free edge. l is load bearing length of the anchor for shear, not to exceed 8d.

V u = 1.4 V = 0.280 kips

CHECK CONCRETE PRYOUT STRENGTH AT BALUSTER SLEEVE (ACI 318-02 Appendix D)

= 4.800 kips > Vu [Satisfactory] 3

where : ψ3 term is 1.0 for location where concrete cracking is likely to occur.

k cp = 2.0 for D > 2.5 in.

A n = 3 D (1.5D + c) = 49.50 in2

DanielTian Li

4PL

M = = V P= =

bM

FS

= =

vV

FA

= =

M PH= = V P= =

bM

FS

= =

vV

FA

= =

0.2' 1.5

6 7 6 7 7V Vcb b c

Vo Vo

lA A d fV V cdA A


( ) ( )' 1.52 3 32

0.30.7 24

9 1.5N N

cp cp b cp cNo

cA A fV k N k DDA D

φ φ φψ ψ ψ = = +


JOB NO. : DATE : REVIEW BY : Sign Design Based on AISC-ASD, ACI, and IBC 1805.7 / UBC1806.8

INPUT DATA & DESIGN SUMMARYCOLUMN SECTION (Tube or Pipe) HSS8X8X1/2 Tube A rmin t h S

COLUMN YIELD STRESS F y = 46 ksi 13.50 3.04 0.47 8.00 31.20DIMENSIONS L = 20 ft

H S = 8 ft

H C = 10 ft

SIGN GRAVITY LOAD ( lbs / ft2 ) D = 10 psfSIGN LATERAL LOAD ( lbs / ft2 ) W = 25 psf, ASDCOLUMN LATERAL LOAD ( plf ) F = 8 plf, ASDDIAMETER OF POLE FOOTING b = 2 ftALLOW SOIL PRESSURE Q a = 1 ksf

LATERAL SOIL CAPACITY P P = 0.266 ksf / ftRESTRAINED @ GRADE ?(1=yes,0=no) 0 No

Use 2 ft dia x 8.64 ft deep pole footing, unrestrained @ ground level.


ANALYSISCHECK COMPRESSION AND BENDING CAPACITY OF COLUMN

0.79 < 4/3 [Satisfactory] (AISC-ASD, H1, page 5-54)

Where f a = (D H S L + col wt ) / A = 0.1 ksi

E s = 29000 ksiK / r = 71

C c = (2 π 2 E s / F y ) 0.5 = 112



C m = 0.6

M max = 56.40 ft-kips, at base of column

V max = 4.08 kips, at base of column

f b = M max / S = 21.692 ksi

F b = 0.6 F y = 27.6 ksi, (AISC-ASD, F3, page 5-48) > 3/4 f b [Satisfactory]

29.64 ksi

DESIGN POLE FOOTING (IBC Sec.1805.7 / UBC Sec.1806.8)By trials, use pole depth, d = 8.639 ftLateral bearing @ bottom, S 3 = 2 P P d = 4.60 ksfLateral bearing @ d / 3, S 1 = 0.33 S 3 = 1.53 ksfRequire Depth is given by

8.639 ft [Satisfactory]

1

Where P = V max = 4.08 kips

A = 2.34 P / (b S 1 ) = 3.15

h = M max / V max = 13.82 ft

CHECK VERTICAL SOIL BEARING CAPACITY (ACI, Sec. 15.2.2)q soil = (D H S L + col wt ) / ( π b 2 / 4) = 0.55 ksf, (net weight of pole footing included.)

< Q a [Satisfactory]

DanielTian Li

'

, 0.15

1arg , 0.15

0.6

a b a

a b a

ma b

a ab a

ea

a b

y b

f f ffor

F F Ff fC

fFF f

L er of forFFf f

F F

+ ≤ +

= − > +

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2

2

3

3

2

2

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2,

3 / /53 8 8

12,

23 /

yc

c

ac c

c

kl rF

C klfor C

rkl r kl rF

C C

E klfor C

rkl r

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23 /e

EF

kl r

π= =

3

4.361 1

2

4.25

A hfor nonconstrained

Ad

Phfor constrained

bS

+ +

= =


JOB NO. : DATE : REVIEW BY : Sign Design Based on AISC-ASD, ACI, and IBC 1805.7 / UBC1806.8


COLUMN SECTION W8X40 A rx ry d Sx bf rT d / Af

COLUMN YIELD STRESS F y = 50 ksi 11.70 3.53 2.05 8.25 35.50 8.07 2.21 1.83DIMENSIONS L = 20 ft

H S = 8 ft

H C = 10 ft

SIGN GRAVITY LOAD ( lbs / ft2 ) D = 10 psfSIGN LATERAL LOAD ( lbs / ft2 ) W = 25 psf, ASDCOLUMN LATERAL LOAD ( plf ) F = 8 plf, ASDDIAMETER OF POLE FOOTING b = 3 ftALLOW SOIL PRESSURE Q a = 1 ksf

LATERAL SOIL CAPACITY P P = 0.266 ksf / ftRESTRAINED @ GRADE ?(1=yes,0=no) 0 No

Use 3 ft dia x 5.75 ft deep pole footing, unrestrained @ ground level.


ANALYSISCHECK COMPRESSION AND BENDING CAPACITY OF ONE COLUMN

0.33 < 4/3 [Satisfactory] (AISC-ASD, H1, page 5-54)

Where f a = (0.5 D H S L + col wt ) / A = 0.1 ksi

E s = 29000 ksi

K / ry = 105

C c = (2 π 2 E s / F y ) 0.5 = 107



C m = 0.6

M max = 28.40 ft-kips, at base of column

V max = 2.08 kips, at base of column

f b = M max / S x = 9.6 ksi

C b = 1.0

7.23 ft

18.26 ft

18.60 ft

17.7 ksi

DanielTian Li

'

, 0.15

1arg , 0.15

0.6

a b a

a b a

ma b

a ab a

ea

a b

y b

f f ffor

F F Ff fC

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= =

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b

LF rMINF F F

C

= − =

(cont'd)

16.7 ksi

16.7 ksi

30.0 ksi, (AISC-ASD, F1.3, page 5-46)10 > 3/4 f b [Satisfactory]

39.9 ksi

DESIGN POLE FOOTING (IBC Sec.1805.7 / UBC Sec.1806.8)By trials, use pole depth, d = 5.75 ftLateral bearing @ bottom, S 3 = 2 P P d = 3.06 ksfLateral bearing @ d / 3, S 1 = 0.33 S 3 = 1.02 ksfRequire Depth is given by

5.75 ft [Satisfactory]

Where P = V max = 2.08 kips

A = 2.34 P / (b S 1 ) = 1.60

h = M max / V max = 13.65 ft

CHECK VERTICAL SOIL BEARING CAPACITY (ACI, Sec. 15.2.2)q soil = (0.5 D H S L + col wt ) / ( π b 2 / 4) = 0.13 ksf, (net weight of pole footing included.)

< Q a [Satisfactory]

( )2

'2

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23 /e

x

EF

kl r

π= =

3

4.361 1

2

4.25

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Ad

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MAX forF F L L

≤ < <= = ≤ < ≥


JOB NO. : DATE : REVIEW BY : Snow Load Analysis Based on UBC 97

INPUT DATA & DESIGN SUMMARY (UBC Sec. 1614)

TOTAL SNOW LOAD S = 75 psf wsnow = S - Rs(α - 20) = 61.3 psf

PITCH DEGREE α = 30 o Where S/40 - 1/2 = 1.375 psf, (for α > 20o)

N/A (for α < 20o)

INPUT DATA & DESIGN SUMMARY (UBC / CBC Appendix Chapter 16)

BASIC GROUND SNOW LOAD Pg = 35 psf

SNOW EXPOSURE FACTOR Ce = 0.9 (Tab. A-16-A) Pf, roof = 31.50 psf

IMPORTANCE FACTOR I = 1 (Tab. A-16-B) Pf, overhang = 63.00 psfOBSTRUCTED SLIPPERY SURFACE ON ROOF ? (1=Yes, 0=No) 0 Unobstructed Pf, valley = 59.72 psfHEATED GREENHOUSES ? (1=Yes, 0=No) = = > 0 Unheated Pf, parapet = 51.38 psf

ROOF SLOPE α = 18 o Wd = 4.29 ft

PARAPET HEIGHT (DRIFT CORNER HEIGHT) hr = 4 ft, see fig. belowLENGTH FROM PARAPET TO ROOF EDGE W = 35 ft, see fig. below

THE ROOF SNOW LOADS (CBC 40-1-1)Pf = Ce I Pg = 31.50 psf

THE APPLICABLE ROOF SNOW LOADS (CBC Sec. 1640)Pf, roof = Cs Pf = 31.50 psf

Where CsPf is derived from CBC Sec.1640 as following table

Pg [0, 20] (20, 70] (70, 100] (100, greater]

Unobstructed Max(Cs,1Pf, Pg) Max(Cs,1Pf, 20) Max(Cs,1Pf, 70CeI) Max(Cs,5Pf, 70CeI)

Obstructed Max(Cs,2Pf, Pg) Max(Cs,2Pf, 20) Max(Cs,2Pf, 70CeI) Max(Cs,5Pf, 70CeI)

Unobstructed Max(Cs,3Pf, Pg) Max(Cs,3Pf, 20) Max(Cs,3Pf, 70CeI) Max(Cs,5Pf, 70CeI)

Obstructed Max(Cs,4Pf, Pg) Max(Cs,4Pf, 20) Max(Cs,4Pf, 70CeI) Max(Cs,5Pf, 70CeI)

Cs,1 = MIN [ 1-(α - 30) / 40, 1.0] , CBC (40-2-1)

Cs,2 = MIN [ 1-(α - 45) / 25, 1.0] , CBC (40-2-2)

Cs,3 = MIN [ 1-(α - 15) / 55, 1.0] , CBC (40-2-3)

Cs,4 = MIN [ 1-(α - 30) / 40, 1.0] , CBC (40-2-4)

Cs,5 = MIN [ 1-(α - 20)(Pf - 20) / (40Pf), 1.0] , CBC (40-2-5)

THE SNOW LOADS AT OVERHANG, VALLEY, AND PARAPET CORNER

Pf, overhang = 2 Pf, roof = 63.00 psf, (CBC Fig. A-16-10)

Pf, valley = Cv Pf, roof = 59.72 psf, (CBC Fig. A-16-12)

Where θ = 90 o, roof intersection angle

Cv = 1.90 (CBC Fig. A-16-11)

Pf, parapet = Cd Pf, roof = Pm = MIN[D(hd + hb), Dhr]= 51.38 psf, (CBC Eq. 44-4)

Where D = MIN(0.13Pg+ 14, 35) = 18.55 pcf, (44-2)

hb = Pf, roof / D = 1.70 ft, (44-3)

(hr - hb) / hb = 1.36 > 0.2, (44-3)(Drift load need be considered)

Wb = MIN[ W , 50] = 35.00 ft, (Sec.1644.5)

hd = 0.5 [0.43(Wb)1/3(Pg+ 10)1/4 - 1.5]= 1.07 ft,(1644.5 & 44-1)

Cd = Pm/ Pf, roof = 1.63

Wd = MIN[4(hr - hb), 4hd] = 4.29 ft, (Sec.1644.2)

Note : Where design snow loads exceed 30 psf, the seismic dead load shall include 25% design snow load. (UBC 1630.1.1)

DanielTian Li

Rs =

Unheated

Heated


JOB NO. : DATE : REVIEW BY : Snow Load Analysis Based on ASCE 7-98


BASIC GROUND SNOW LOAD Pg = 75 (ASCE page 69)

SNOW EXPOSURE FACTOR Ce = 1 (Tab. 7-2, pg 82) Pf, roof = 45.82 psf

THERMAL FACTOR Ct = 1.2 (Tab. 7-3, pg 83) Pf, overhang = 91.64 psf

IMPORTANCE FACTOR I = 1 (Tab. 7-4, pg 83) Pf, valley = 126.00 psf

ROOF SLOPE αroof = 30 o Pf, parapet = 95.00 psf

PARAPET HEIGHT (DRIFT CORNER HEIGHT) hr = 4 ft, see fig. below Wd = 8.34 ft

LENGTH OF THE ROOF UPWIND OF THE DRIFT Lu = 35 ft, see fig. belowOBSTRUCTED SLIPPERY SURFACE ON ROOF(Sec. 7.4.0, pg70) ? (1=Yes, 0=No) 0 Unobstructed

ANALYSISTHE FLAT SNOW LOADS (Sec 7.3, pg 69)

Pf = 0.7CeCt I Pg = 63.00 psf

Where Pf, min = 0.00 psf, (Sec. 7.3.4, pg 69)

THE ROOF SNOW LOADS (Sec. 7.4, pg 70)Ps = Cs Pf = 45.82 psf, (Eq.7-2)

Where Cs = 0.727 , Derived from Fig 7-2, page 76, as following table

Ct [0, 1.0] (1.0, 1.2) [1.2, greater]

Unobstructed Cs,1 Cs,1+5(Ct-1)(Cs,3-Cs,1) Cs,3

Obstructed Cs,2 Cs,2+5(Ct-1)(Cs,4-Cs,2) Cs,4

Cs,1 = MIN [ (70 - α) / 65, 1.0] , Fig. 7-2a dash line

Cs,2 = MIN [ (70 - α) / 40, 1.0] , Fig. 7-2a solid line

Cs,3 = MIN [ (70 - α) / 55, 1.0] , Fig. 7-2b dash line

Cs,4 = MIN [ (70 - α) / 26, 1.0] , Fig. 7-2b solid line


Pf, overhang = 2 Ps = 91.64 psf, (Sec.7.4.5, pg 70)

Pf, valley = Cv Pf = 126.00 psf, (Sec.7.6.3, pg 71)

Where Cv = 2 / Ce = 2.00 , (Fig. 7-6, pg 80)

Pf, parapet = Cd Ps = MIN[γ(hd + hb), γhr]= 95.00 psf, (Sec.7.7.1, pg 72)

Where γ = MIN(0.13Pg+ 14, 30) = 23.75 pcf, (7-4)

hb = Ps / γ = 1.93 ft, (Sec. 7.1)

hc = hr - hb = 2.07 ft, (Fig. 7.8)

hc / hb = 1.07 > 0.2, (Sec.7.7.1)(Drift load need be considered)

hd = 0.75 [0.43(Lu)1/3(Pg+ 10)1/4 - 1.5] = 2.08 ft, (Sec.7.8 & Fig.7-9)

Cd = 2.07 , (see fig. right)

4hd = N/A ft, for hd<hc

4hd2 / hc = 8.34 ft, for hd>hc

(Sec.7.7.1, pg 72)

Wd,max = 8 hc = 16.57 ft

Note : Where flat roof snow loads exceed 30 psf, the seismic dead load shall include 20% design snow load. (IBC 2000 1617.4.1)

DanielTian Li

Wd =


JOB NO. : DATE : REVIEW BY : Snow Load Analysis Based on ASCE 7-02


BASIC GROUND SNOW LOAD Pg = 5 (ASCE page 77)

SNOW EXPOSURE FACTOR Ce = 1 (Tab. 7-2, pg 90) Pf, roof = 5.00 psf

THERMAL FACTOR (0.85 , 1.0 , 1.1 , 1.2) Ct = 1.2 (Tab. 7-3, pg 91) Pf, overhang = 10.00 psf

IMPORTANCE FACTOR I = 1 (Tab. 7-4, pg 91) Pf, valley = 10.00 psf

ROOF SLOPE αroof = 14 o Pf, parapet = 18.93 psf

PARAPET HEIGHT (DRIFT CORNER HEIGHT) hr = 4 ft, see fig. below Wd = 3.80 ft

LENGTH OF THE ROOF UPWIND OF THE DRIFT Lu = 35 ft, see fig. belowOBSTRUCTED SLIPPERY SURFACE ON ROOF(Sec. 7.4.0, pg78) ? (1=Yes, 0=No) 0 Unobstructed

ANALYSISTHE FLAT SNOW LOADS (Sec 7.3, pg 77)

Pf = 0.7CeCt I Pg = 5.00 psf

Where Pf, min = 5.00 psf, (Sec. 7.3.4, pg 77)

THE ROOF SNOW LOADS (Sec. 7.4, pg 78)Ps = Cs Pf = 5.00 psf, (Eq.7-2)

Where Cs = 1.000 , Derived from Fig 7-2, page 84, as following table

Ct 1.1 1.2

Unobstructed Cs,1 Cs,3 Cs,5

Obstructed Cs,2 Cs,4 Cs,6

Cs,1 = MIN [ (70 - α) / 65, 1.0] , Fig. 7-2a dash line

Cs,2 = MIN [ (70 - α) / 40, 1.0] , Fig. 7-2a solid line

Cs,3 = MIN [ (70 - α) / 60, 1.0] , Fig. 7-2b dash line

Cs,4 = MIN [ (70 - α) / 32.5, 1.0] , Fig. 7-2b solid line

Cs,5 = MIN [ (70 - α) / 55, 1.0] , Fig. 7-2c dash line

Cs,6 = MIN [ (70 - α) / 26, 1.0] , Fig. 7-2c solid line


Pf, overhang = 2 Ps = 10.00 psf, (Sec.7.4.5, pg 78)

Pf, valley = Cv Pf = 10.00 psf, (Sec.7.6.3, pg 79)

Where Cv = 2 / Ce = 2.00 , (Fig. 7-6, pg 88)

Pf, parapet = Cd Ps = MIN[γ(hd + hb), γhr]= 18.93 psf, (Sec.7.7.1, pg 79)

Where γ = MIN(0.13Pg+ 14, 30) = 14.65 pcf, (7-4)

hb = Ps / γ = 0.34 ft, (Sec. 7.1)

hc = hr - hb = 3.66 ft, (Fig. 7.8)

hc / hb = 10.72 > 0.2, (Sec.7.7.1)(Drift load need be considered)

hd = 0.75 [0.43(Lu)1/3(Pg+ 10)1/4 - 1.5] = 0.95 ft, (Sec.7.8 & Fig.7-9)

Cd = 3.79 , (see fig. right)

4hd = 3.80 ft, for hd<hc

4hd2 / hc = N/A ft, for hd>hc

(Sec.7.7.1, pg 79)

Wd,max = 8 hc = 29.27 ft

Note : Where flat roof snow loads exceed 30 psf, the seismic dead load shall include 20% design snow load. (IBC 2003 1617.5.1)

DanielTian Li

Wd =

0.85 or 1.0


JOB NO. : DATE : REVIEW BY : Live Load Reduction Based on IBC 2003

INPUT DATA & DESIGN SUMMARYMEMBER TYPE (0=Beam, 1=Column) 1 Column

ROOF TRIBUTARY AREA SUPPORTED BY THE MEMBER A r = 500 ft2, (if no roof, input 0.)

ROOF SLOPE 4 / 12NUMBER OF FLOORS n = 1

TOTAL FLOOR TRIBUTARY AREA SUPPORTED BY THE MEMBER A f = 700 ft2

FLOOR LIVE LOAD (IBC Table 1607.1) L = 100 psfGROUP A OCCUPANCIES ? (0=No, 1=Yes) 0 No

UNIFORM ΣTHE MINIMUM ROOF LIVE LOAD 14.00 psf 7.00 kipsTHE MINIMUM FLOOR LIVE LOAD 53.35 psf 37.34 kipsTOTAL LOAD SUPPORTED BY THE COLUMN 44.34 kips

Note: Live loads are horizontal projected loads.

ANALYSIS

MINIMUM ROOF LIVE LOAD (IBC 1607.11.2)L r = 20 R1 R2 = 14.00 psf

Where R1 = 0.70

TRIBUTARY AREA [ 0 ~ 200] ( 200 ~ 600) [ 600 ~ over)R1 1 1.2 -0.001 A f 0.6

R2 = 1.00

ROOF SLOPE, F / 12 [ 0 ~ 4] / 12 ( 4 ~ 12) / 12 [ 12 ~ over) / 12R2 1 1.2 -0.05 F 0.6

L r = [ 12 , 20 ]

MINIMUM FLOOR LIVE LOAD BY INFLUENCE AREA METHOD(IBC 1607.9.1)L = L0 [ 0.25 + 15 / (A I)

0.5 ] = 53.35 psf < = = note: 1. Min reduced ,L, 50% for one level only, 40% for others.

Where L0 = 100 psf

KLL = 4 ,(IBC Table 1607.9.1)

A I = KLL A f = 2800 ft2

MINIMUM FLOOR LIVE LOAD BY TRIBUTARY AREA METHOD(IBC 1607.9.2)L = (100 psf ) x (100 - R) / 100 = 60.00 psf

Where r = 0.08R = r ( A f - 150 ) = 40.0 < = = note: 1. For floor live loads exceeding 100 psf, no reduction

shall be made, except that design live loads on columnsmay be reduced 20 percent.

2. Max reduction ,R, 40% for one level only, 60% for others.3. No reduction permitted in Group A occupancies.

DanielTian Li


JOB NO. : DATE : REVIEW BY : Live Load Reduction Based on CBC 2001

INPUT DATA & DESIGN SUMMARYMEMBER TYPE (0=Beam, 1=Column) 1 Column

ROOF TRIBUTARY AREA SUPPORTED BY THE MEMBER A r = 500 ft2, (if no roof, input 0.)

ROOF SLOPE 4 / 12NUMBER OF FLOORS n = 1

TOTAL FLOOR TRIBUTARY AREA SUPPORTED BY THE MEMBER A f = 700 ft2

FLOOR LIVE LOAD (CBC Table 16-A, or Table 16A-A) L = 100 psfDSA or OSHPD PROJECT ? (0=No, 1=Yes) 0 No

UNIFORM ΣTHE MINIMUM ROOF LIVE LOAD 12.64 psf 6.32 kipsTHE MINIMUM FLOOR LIVE LOAD 53.35 psf 37.34 kipsTOTAL LOAD SUPPORTED BY THE COLUMN 43.66 kips

Note: Live loads are horizontal projected loads. 4

ANALYSIS

MINIMUM ROOF LIVE LOAD (CBC Table 16-C) METHOD METHOD 1 METHOD 2

TRIBUTARY AREA UNIFORM REDUCTION MAXIMUM

ROOF SLOPE [ 0 ~ 200] ( 200 ~ 600] ( 600 ~ over) L r (psf) r R (%)[0~4) / 12 20 16 12 20 0.08 40

[4~12) / 12 16 14 12 16 0.06 25[12~over) / 12 12 12 12 12 0 0

16 0.06 25 MIINIMUM L r 14 psf 12.64 psf

MINIMUM FLOOR LIVE LOAD BY TRIBUTARY AREA METHOD(CBC 1607.5)L = (100 psf ) x (100 - R) / 100 = 60.00 psf

Where r = 0.08R = r ( A f - 150 ) = 40.0 < = = note: 1. For storage loads exceeding 100 psf, no reduction

shall be made, except that design live loads on columnsmay be reduced 20 percent.

2. Max reduction ,R, 40% for one level only, 60% for others.

MINIMUM FLOOR LIVE LOAD BY INFLUENCE AREA METHOD(CBC 1607.6)L = L0 [ 0.25 + 15 / (A I)

0.5 ] = 53.35 psf < = = note: 1. This method Not adopted by DSA or OSHPD.

Where L0 = 100 psf 2. Min reduced ,L, 50% for one level only, 40% for others.

A I = 2800 ft2

DanielTian Li



INPUT DATALATERAL DIAPHRAGM FORCE : Fp = 380 plf

DIMENSIONS: X1 = 72 ft , X2 = 24 ft , X3 = 16 ft

Y1 = 20 ft , Y2 = 20 ft , Y3 = 20 ft

DESIGN SUMMARYTHE MAXIMUM STRAP FORCE, T = C = 4.66 kips

THE MAXIMUM SHEAR STRESS, v x = 466 plf

v y = 480 plf

ANALYSIS

16

DanielTian Li

Flexible Diaphragm with an Opening

(cont'd)REACTIONS : VL = 21.280 kips , VR = 21.280 kips

HATCHED AREA : a = X1 + 0.5 X2 = 84.00 ft , b = 0.5 Y2 + Y3 = 30.00 ft

Y = Y1 + Y2 + Y3 = 60.00 ft , w = Fp b / Y = 190.0 plf

M = a VL - 0.5 Fpx a2 = 446.9 ft-k, total moment at middle opening , V = b VL / Y = 10.640 kips

F2 = M / Y = 7448 lbf , F9 = F2 = 7448 lbf

F5 = ( F2 b - 0.5 w a2 ) / a = -5320 kips , F15 = w a - V + F5 = 0 kips

INDIVIDUAL PANEL X (ft) Y (ft) vx (plf) vy (plf) NO. FORCE (lbf) NO. FORCE (lbf)

1 72.00 20.00 138 480 F1 9956 F15 0

2 12.00 20.00 209 266 F2 7448 F16 0

3 12.00 20.00 323 380 F3 3572 F17 -2512

4 16.00 20.00 223 299 F4 3040 F18 -1034

5 72.00 10.00 103 103 F5 -5320 F19 -4655

6 16.00 10.00 466 466 F6 7600 F20 1108

7 72.00 10.00 103 103 F7 2508 F21 -2508

8 16.00 10.00 466 466 F8 -3876 F22 3876

9 72.00 20.00 138 480 F9 7448 F23 3040

10 12.00 20.00 209 266 F10 7448 F24 5320

11 12.00 20.00 323 380 F11 2512 F25 7600

12 16.00 20.00 223 299 F12 1034 F26 9956

F13 4655 F27 3572

F14 -1108


1. Kelly E. Cobeen: "Structuiral Engineering Review Workshop", BYA publications, 2005.


JOB NO. : DATE : REVIEW BY : Formulas of Moment Resisting Frame at Bottom Fixed Condition

INPUT DATA & DESIGN SUMMARYBEAM SECTION W16X26 I b = 301 in4

COLUMN SECTION W10X33 I c = 171 in4

BEAM LENGTH BETWEEN COL. CENTERS L = 30 ftSTORY HEIGHT h = 14 ft

GRAVITY LOAD w = 1.8 klfHA = HD = 10.3 kips MB = MC = 95.7 ft-kipsVA = VD = 27.0 kips Mmax = 106.8 ft-kips

MA = MD = 47.8 ft-kips

4

LATERAL LOAD P = 15 kipsHA = HD = 7.5 kips MB = MC = 43.6 ft-kipsVA = VD = 2.9 kips ∆H = 0.90 in

MA = MD = 61.4 ft-kips KH = 16.7 kips / in(KH is frame stiffness.)

GRAVITY ANALYSIS

0.821 27.0 kips

10.3 kips 47.8 ft-kips


LATERAL ANALYSIS

5.82 ft 8.18 ft 7.5 kips


29000 ksi 43.6 ft-kips

0.90 in 16.7 kips / in

DanielTian Li

b

c

hInLI

= =2A D

wLV V= = =

( )2

4 2A D

wLH H

h n= = =

+ ( )2

12 2A D

wLM M

n= = =

+

( )( )

2

max

2 324 2

w nLM

n

+= =

+ ( )2

6 2B C

wLM M

n= = =

+

36 1

na h

n = = +

( )31 6A DnPh

V VL n

= = =+

2A D

PH H= = =

( )( )

1 32 1 6

A D

Ph nM M

n

+= = =

+

32

26

c c bH

a LbP aI I I

E

+ + = =∆

3 16 1n

b hn

+ = = +

( )3

2 1 6B C

nPhM M

n= = =

+E =

HH

PK = =

∆


JOB NO. : DATE : REVIEW BY : Formulas of Moment Resisting Frame at Bottom Pinned Condition

INPUT DATA & DESIGN SUMMARYBEAM SECTION W16X26 I b = 301 in4

COLUMN SECTION W10X33 I c = 171 in4


GRAVITY LOAD w = 1.8 klfHA = HD = 2.4 kips MB = MC = 32.9 ft-kipsVA = VD = 18.0 kips Mmax = 57.1 ft-kips

LATERAL LOAD P = 15 kipsHA = HD = 7.5 kips MB = MC = 105.0 ft-kipsVA = VD = 10.5 kips ∆H = 3.36 in

KH = 4.5 kips / in(KH is frame stiffness.)

GRAVITY ANALYSIS

1.232 18.0 kips


57.1 kips

LATERAL ANALYSIS

7.5 kips 29000 ksi


3.36 in 4.5 kips / in

DanielTian Li

b

c

hInLI

= =2A D

wLV V= = =

( )2

4 3 2A D

wLH H

h n= = =

+

( )( )2

max

2 18 2 3

w nLM

n

+= =

+

( )2

4 3 2B C

wLM M

n= = =

+

A DPh

V VL

= = =

2A D

PH H= = =

( )3 1 212

Hc

P nhnEI

+= =∆

2B CPh

M M= = =

E =

HH

PK = =

∆


JOB NO. : DATE : REVIEW BY : Formulas of Concentrically Braced Frame

INPUT DATA & DESIGN SUMMARYBRACE SECTION HSS6X6X1/2 EA = 282460 kips , Tube


GRAVITY LOAD w = 1.8 klfHA = HD = 15.8 kips MBE = MEC = 28.5 ft-kipsVA = VD = 27.0 kips ME = 50.6 ft-kips

NAE = NDE = 23.1 kips

LATERAL LOAD P = 15 kipsHA = HD = 7.5 kips NAE = NDE = 10.3 kipsVA = VD = 7.0 kips ∆H = 0.01 in 14

KH = 1226.2 kips / in , (KH is frame stiffness.)

GRAVITY ANALYSIS (Neglected axial deformations.)

43.03 0 27.0 kips



LATERAL ANALYSIS

7.5 kips 7.0 kips

10.3 kips

0.01 in 1226.2 kips / in

DanielTian Li

1tan0.5

hL

α − = = 2A D

wLV V= = =

5tan

16A D

wLH H α= = =

29512BE ECwL

M M= = =

2

32E

wLM = =

A DPh

V VL

= = =2

A DP

H H= = =

34 cosH

PLEA α

= =∆ HH

PK = =

∆

516cosAE DE

wLN N α

= = =

2cosAE DEP

N N α= = =


JOB NO. : DATE : REVIEW BY : Formulas of Ecconcentrically Braced Frame

INPUT DATA & DESIGN SUMMARYBRACE SECTION HSS6X6X1/2 EA = 282460 kips , Tube

BEAM SECTION W16X26 EI = 8729000 kips-in2

BEAM LENGTH BETWEEN COL. CENTERS L = 30 ftSTORY HEIGHT h = 14 ftLINK LENGTH e = 4 ft

GRAVITY LOAD w = 1.8 klfHA = HD = 16.0 kips MBE = MFC = 25.8 ft-kipsVA = VD = 27.0 kips ME = MF = 24.5 ft-kips

NAE = NDF = 23.5 kips

LATERAL LOAD P = 15 kipsHA = HD = 7.5 kips NAE = NDF = 11.0 kipsME = MF = 14.0 ft-kips ∆H = 0.04 in

QEF = VA = VD = 7.0 kips KH = 384.5 kips / in(QEF is link shear force.) (KH is frame stiffness.)

GRAVITY ANALYSIS (Neglected axial deformations.)

47.12 0 24.5 ft-kips , (by moment distribution procedure.)

27.0 kips 16.0 kips


LATERAL ANALYSIS

7.5 kips 7.0 kips


0.04 in 384.5 kips / in

DanielTian Li

1tan0.5 0.5

hL e

α − = = −

2A DwL

V V= = = ( )22

8E

A D

w eLMH H

h h

−= = + =

( )2

32 2E

BE FC

w L e MM M

−= = − =

E FM M= =

A DEF

PhQ V V

L= = = =

2A D

PH H= = =

2 2

22 sin 12cosH

Ph Ph eEA EILα α

= + =∆ HH

PK = =

∆

cosA

AE DFH

N N α= = =

2cosAE DFP

N N α= = =

2E F

PehM M

L= = =

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American Structural Design Calculation