6 Flexural Component Design

7/29/2019 6 Flexural Component Design

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PCI 6th Edition

Flexural Component Design


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Presentation Outline

Whats new to ACI 318 Gravity Loads

Load Effects

Concrete Stress Distribution Nominal Flexural Strength

Flexural Strength Reduction Factors

Shear Strength

Torsion

Serviceability Requirements


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New to ACI 31802

Load Combinations Stress limits

Member Classification

Strength Reduction factor is a function ofreinforcement strain

Minimum shear reinforcement requirements

Torsion Design Method


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Load Combinations

U = 1.4 (D + F)

U = 1.2 (D + F + T) + 1.6 (L + H) + 0.5 (Lror S or R)

U = 1.2D + 1.6 (Lror S or R) + (1.0L or 0.8W)

U = 1.2D + 1.6W + 1.0L + 0.5(Lror S or R) U = 1.2D + 1.0E + 1.0L + 0.2S

U= 0.9D + 1.6W + 1.6H

U= 0.9D + 1.0E + 1.6H


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Comparison of Load Combinations

U=1.2D + 1.6 L 2002

U= 1.4D + 1.7L 1999

If L=.75D

i.e. a 10% reduction in required strength

Ratio 1.2D 1.6 .75D 1.4D 1.7 .75D

0.90


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Classifications

No Bottom Tensile Stress Limits Classify Members Strength Reduction

Factor Tension-Controlled

Transition

Compression Controlled

Three Tensile Stress Classifications Class U Un-cracked

Class T Transition

Class C Cracked


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Copied from ACI 318 2002, ACI 318-02 table R18.3.3


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Class C Members

Stress Analysis Based on Cracked SectionProperties

No Compression Stress limit

No Tension Stress limit Increase awareness on serviceability

Crack Control

Displacements

Side Skin Reinforcement


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0

50

100

150

200

0 5,000 10,000 15,000 20,000

Concrete Strength, f'c, psi

Minimum Shear Reinforcing

1999

2002

Avfy

bws


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System Loads

Gravity Load Systems Beams

Columns

Floor Member Double Tees, Hollow Core

Spandrels

Tributary Area Floor members, actual top area

Beams and spandrels

Load distribution Load path

Floor members spandrels or beams Columns


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Live Loads can be reduced based on:

Where:

KLL = 1

Lo = Unreduced live load andAt = tributary area

Live Load Reduction

L Lo

0.25 15

KLL A t


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Live Load Reduction

Or the alternative floor reduction shall notexceed

or

Where:

R = % reduction 40%r = .08

R r (At

150)

R 23.1 1 D

Lo


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Member Shear and Moment

Shear and moments on members can befound using statics methods and beam tables

from Chapter 11


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Strength Design

Strength design is based using the rectangular stressblock

The stress in the prestressing steel at nominalstrength, fps, can be determined by straincompatibility or by an approximate empirical equation

For elements with compression reinforcement, thenominal strength can be calculated by assuming thatthe compression reinforcement yields. Then verified.

The designer will normally choose a section and

reinforcement and then determine if it meets thebasic design strength requirement:

Mn

Mu


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Concrete Stress Distribution

Parabolic distribution

Equivalent rectangular distribution


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Stress Block Theory

Stress-Strain

relationship

is not constantE(f 'c )

E(f 'c )

fc=3,000 psi

fc=6,000 psi


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Stress Block Theory

Stress-Strain relationship Stress-strain can be modeled by:

fc

2 f ''

c (

)

1 ( )2

Where :strain at max. stress 1.71 f '

c

Ec

and :max stressf ''c .9 f 'c


18/127

Stress Block Theory

The Whitney stress block is a simplifiedstress distribution that shares the same

centroid and total force as the real stress

distribution

=


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Equivalent Stress Blockb1 Definition

b1 = 0.85

when fc < 3,000 psi

b1 = 0.65

when fc > 8,000 psi

a b1c

b1

1.05 05 f '

c

1,000psi


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Design Strength

Mild Reinforcement Non - Prestressed

Prestress Reinforcement


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Strength Design Flowchart

Figure 4.2.1.2page 4-9

Non-Prestressed

Path Prestressed Path


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Non-Prestressed Members

Find depth of compression block


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Depth of Compression Block

Where:

As is the area of tension steel

As is the area of compression steel

fy is the mild steel yield strength

a A

s f

yA '

s f '

y

.85 f 'c b

Assumes

compressionsteel yields


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Flanged Sections

Checked to verify that the compression block is truly

rectangular


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Compression Block Area

If compression block is rectangular, the flanged

section can be designed as a rectangular beam

Acomp

a b

= =


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Compression Block Area

If the compression block is not rectangular (a> hf),

=

To find a

Af

(b bw) h

f

Aw

Acomp

Af

a A

w

bw


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Determine Neutral Axis

From statics and strain compatibility

c a / b


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Check Compression Steel

Verify that compression steel has reached yield using

strain compatibility

3 'c d


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Compression Comments

By strain compatibility, compression steel yields if:

If compression steel has not yielded, calculation for amust be revised by substituting actual stress for yieldstress

Non prestressed members should always be tensioncontrolled, therefore c / dt < 0.375

Add compression reinforcement to create tesnioncontrolled secions

c 3 d'


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Moment Capacity

2 equations rectangular stress block in the flange section

rectangular stress block in flange and stemsection


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Strength Design Flowchart

Figure 4.2.1.2page 4-9Non- PrestressedPathPrestressed Path


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This portion of theflowchart is dedicated todetermining the stress in

the prestressreinforcement


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Stress in Strand

fse - stress in the strand after losses

fpu - is the ultimate strength of the strand

fps - stress in the strand at nominal strength


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Stress in Strand

Typically the jacking force is 65% or

greater

The short term losses at midspan are

about 10% or less

The long term losses at midspan are

about 20% or less

fse 0.5 fpu

S i S


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Stress in Strand

Nearly all prestressed concrete is bonded

S i S d


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Stress in Strand

Prestressed Bonded reinforcement

p = factor for type of prestressing strand, see ACI 18.0

= .55 for fpy/fpu not less than .80

= .45 for fpy

/fpu

not less than .85

= .28 for fpy/fpu not less than .90 (Low Relaxation Strand)

rp = prestressing reinforcement ratio

fps

fpu

1

p

b1

rp

fpu

f 'c

d

dp

'

D i C i Bl k


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Determine Compression Block


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Fl S ti Ch k


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Flange Sections Check

C i St l Ch k


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Compression Steel Check

Verify that compression steel has reached yield usingstrain compatibility

3 'c d

M t C it


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Moment Capacity

2 Equations rectangular stress block in flange section

rectangular stress block in flange and stem

section

Fl l St th R d ti F t


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Flexural Strength Reduction Factor

Based on primary reinforcement strain Strain is an indication of failure

mechanism

Three Regions

M b Cl ifi ti


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Member Classification

On figure 4.2.1.2

C i C t ll d


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Compression Controlled

< 0.002 at extreme

steel tension fiber or

c/dt > 0.600

= 0.70 with spiral ties

= 0.65 with stirrups

T i C t ll d


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Tension Controlled

> 0.005 at extremesteel tension fiber, or

c/dt < 0.375

= 0.90 with spiral tiesor stirrups

Transition Zone


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Transition Zone

0.002 < < 0.005 at extremesteel tension fiber, or

0.375 < c/dt < 0.6

= 0.57 + 67() or

= 0.48 + 83() with spiralties

= 0.37 + 0.20/(c/dt) or

= 0.23 + 0.25/(c/dt) with

stirrups

Strand Slip Regions


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Strand Slip Regions

ACI Section 9.3.2.7

where the strand embedment length is

less than the development length

=0.75

Limits of Reinforcement


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To prevent failure immediately upon cracking,Minimum As is determined by:

As,min is allowed to be waived if tensilereinforcement is 1/3 greater than required by

analysis

As,min

3 f '

c

fy b

w

d 200 b

w d

fy



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The flexural member must also have adequate

reinforcement to resist the cracking moment

Where Mn

1.2Mcr

Mcr Sbc

P

A Pe

Sb

fr

Mnc

Sbc

Sb

1

Section after compositehas been applied,including prestress

forces

Correction forinitial stresses on

non-composite,prior to toppingplacement

Critical Sections


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Critical Sections

Horizontal Shear


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Horizontal Shear

ACI requires that the interface betweenthe composite and non-composite, be

intentionally roughened, clean and free of

laitance Experience and tests have shown that

normal methods used for finishing precast

components qualifies as intentionallyroughened

Horizontal Shear F Positive Moment Region


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Horizontal Shear, Fh Positive Moment Region

Based on the force transferred in topping (page 4-53)

Horizontal Shear F Negative Moment Region


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Horizontal Shear, Fh Negative Moment Region

Based on the force transferred in topping (page 4-53)

Unreinforced Horizontal Shear


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Unreinforced Horizontal Shear

Fh

80 bv

lvh

Where 0.75

bv width of shear arealvh - length of the member subject to shear, 1/2 the

span for simply supported members

Reinforced Horizontal Shear


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Reinforced Horizontal Shear

Where

0.75

rv

- shear reinforcement ratio

Acs - Area of shear reinforcement

me - Effective shear friction coefficient

Fh (260 0.6 rv fy ) bv l vh

Acs

F

h

me fy

Shear Friction Coefficient


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Shear Friction Coefficient

me

1000 A

cr m

Vu

F

h

Shear Resistance by Non-Prestressed Concrete


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Shear Resistance by Non-Prestressed Concrete

Shear strength for

non-prestressed

sections

Vc

2 f 'c

bw

d

Prestress Concrete Shear Capacity


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Where:

ACI Eq 11-9

Effective prestress must be 0.4fpu Accounts for shear combined with moment

May be used unless more detail is required

Vc

0.6 f 'c

700V

u d

Mu

b

w

d

Vu

d

Mu

1



59/127


Concrete shear strength is minimum is

Maximum allowed shear resistance fromconcrete is:

Vc

2 f 'c

bw

d

Vc 5 f 'c bw d

Shear Capacity Prestressed


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Shear Capacity, Prestressed

Resistance by concrete when diagonal cracking is aresult of combined shear and moment

Vci

0.6 f 'c

bw

d Vd

V

i M

cr

Mmax

Where:Vi and Mmax - factoredexternally appliedloadse.g. no self weightVd - is un-factored dead

load shear

Shear Capacity Prestressed


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Shear Capacity, Prestressed

Resistance by concrete when diagonal cracking is aresult of principal tensile stress in the web is in excess of

cracking stress.

Vcw

3.5 f 'c

0.3 fpc bw d Vp

Where:Vp = the verticalcomponent of effectiveprestress force (harpedor draped strand only)

V


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Vcmax

Shear capacity is the minimum of Vc, or if a

detailed analysis is used the minimum of Vci

or Vcw

Shear Steel


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Shear Steel

If:

Then:

Vu

Vc

v

sV

nV

cor v

s

Vu

V

c


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Torsion


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Torsion

Current ACI Based on compact sections

Greater degree of fixity than PC can provide

Provision for alternate solution Zia, Paul and Hsu, T.C., Design for Torsion and

Shear in Prestressed Concrete, Preprint 3424,American Society of Civil Engineers, October,1978. Reprinted in revised form in PCI JOURNAL,

V. 49, No. 3, May-June 2004.

Torsion


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Torsion

For members loaded two sides, such as invertedtee beams, find the worst case condition with

full load on one side, and dead load on the

other

1.0D 1.2D+1.6L

Torsion


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Torsion

In order to neglect Torsion

Where:

Tu(min) minimum torsional strength providedby concrete

Tu

Tu(min)


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Prestress Factor,


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Prestress Factor,

For Prestressed Members

Where:

fpc level of prestress after losses

1 10fpc

f`c

Maximum Torsional Strength


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Maximum Torsional Strength

Avoid compression failures due to overreinforcing

Where:

Tn(max)

1

3K

t f

cx2y

1 K

tV

t

30 C t Tu

2

Tn(max)

T

u

Kt

12 10fpc

fc

Ct

b

w d

x2y

Maximum Shear Strength


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u S e S e g

Avoid compression failures due to over

reinforcing

Vn(max) 10 f`

c bw d

1 30 C

t T

u

Kt

Vt

2

Vn(max)

Vu

Torsion/Shear Relationship


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p

Determine the torsion carried by the concrete

Where:

Tcand Vc - concrete resistance under puretorsion and shear respectively

Tc and Vc - portions of the concrete resistanceof torsion and shear

Tc

T '

c

1 T '

cT

u

V 'c Vu

2

Torsion/Shear Relationship


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p

Determine the shear carried by the concrete

Vc

V '

c

1 V 'c VuT '

cT

u

2

Torsion Steel Design


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g

Provide stirrups for torsionmoment - in addition toshear

Wherex and y - short and longdimensions of the closedstirrup

A t

Tu

T

c

s

t

x1

y1

fy

t

0.66 0.33 y1

x1

1.5

Torsion Steel Design


75/127

g

Minimum area of closed stirrups is

limited by

Av

2At min 50 b

w sfy

()2 200 bw sfy


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Longitudinal Steel limits


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g

Al

400 x

fy

T

u

Tu

V

u

3 Ct

2 A t

s

x1

y1

The factor in

the second equation need not exceed

2 At

s

50 bwfy

1 12 fpcf

c

50 bwfy

Detailing Requirements, Stirrups


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g q , p

135 degree hooks are required unless sufficient

cover is supplied

The 135 degree stirrup hooks are to be anchored

around a longitudinal bar Torsion steel is in addition to shear steel

Detailing Requirements, Longitudinal Steel


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Placement of the bars should be around theperimeter

Spacing should spaced at no more than 12 inches

Longitudinal torsion steel must be in addition to

required flexural steel (note at ends flexural demandreduces)

Prestressing strand is permitted (@ 60ksi)

The critical section is at the end of simply supportedmembers, therefore U-bars may be required to meet

bar development requirements

Serviceability Requirements


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y q

Three classifications for prestressedcomponents

Class U: Uncracked

Class T: Transition

Class C: Cracked

t

7.5 f 'c

7.5 f 'c

t

12 f 'c

t 12 f '

c

Stress

Uncracked Section


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Table 4.2.2.1 (Page 4.24) Easiest computation

Use traditional mechanicsof materials methods to

determine stresses, grosssection and deflection.

No crack control or sideskin reinforcementrequirements

Transition Section


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Table 4.2.2.1 (Page 4.24) Use traditional mechanics

of materials methods todetermine stresses only.

Use bilinear crackedsection to determinedeflection

No crack control or sideskin reinforcement

requirements

Cracked Section


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Table 4.2.2.1 (Page 4.24) Iterative process

Use bilinear crackedsection to determine

deflection and todetermine memberstresses

Must use crack controlsteel per ACI 10.6.4

modified by ACI 18.4.4.1and ACI 10.6.7

Cracked Section Stress Calculation


84/127

Class C member require stress to be

check using a Cracked Transformed

Section The reinforcement spacing

requirements must be adhered to

Cracked TransformedSection Property Calculation Steps


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Section Property Calculation Steps

Step 1 Determine if section is crackedStep 2 Estimate Decompression Force in Strand

Step 3 Estimate Decompression Force in mildreinforcement (if any)

Step 4 Create an equivalent force in topping if presentStep 5 Calculate transformed section of all elements

and modular ratios

Step 6 Iterate the location of the neutral axis until thenormal stress at this level is zero

Step 7 Check Results with a a moment and forceequilibrium set of equations

Steel Stress


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fdc decompression stress

stress in the strand when the

surrounding concrete stress is zero

Conservative to use, fse (stress after

losses) when no additional mild steel is

present.

Simple Example


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Page 4-31


88/127


89/127


90/127


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Deflection Calculation

Bilinear Cracked Section


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Deflection before themember has cracked iscalculated using thegross (uncracked)

moment of inertia, Ig

Additional deflectionafter cracking iscalculated using themoment of inertia of the

cracked section Icr

Effective Moment of Inertia


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Alternative method

Ie

M

cr

Ma

3

Ig

1 M

cr

Ma

3

Icr

or based on stress

Mcr

Ma

1 ftl

fr

fl

Where:ftl = final stressfl = stress due to live loadfr = modulus of rupture

Prestress Losses


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Prestressing losses Sources of total prestress loss (TL)

TL = ES + CR + SH + RE

Elastic Shortening (SH)

Creep (CR)

Shrinkage (SH)

Relaxation of tendons (RE)

Elastic Shortening


95/127

Caused by the prestressed force in the precast member

Where:Kes = 1.0 for pre-tensioned members

Eps = modulus of elasticity of prestressing tendons (about28,500 ksi)

Eci = modulus of elasticity of concrete at time prestress isapplied

fcir= net compressive stress in concrete at center of gravity ofprestressing force immediately after the prestress has beenapplied to the concrete

ES Kes

Eps

fcir

Eci

fcir


96/127

Where:Pi = initial prestress force (after anchorage seating loss)

e = eccentricity of center of gravity of tendons with respect tocenter of gravity of concrete at the cross sectionconsidered

Mg = bending moment due to dead weight of prestressed

member and any other permanent loads in place at time ofprestressing

Kcir= 0.9 for pretensioned members

fcir

Kcir

Pi

Ag

Pi e

2

Ig

Mg e

Ig

Creep


97/127

Creep (CR)

Caused by stress in the concrete

Where:Kcr= 2.0 normal weight concrete

= 1.6 sand-lightweight concrete

fcds = stress in concrete at center of gravity of

prestressing force due to all uperimposedpermanent dead loads that are applied tothe member after it has been prestressed

CR Kcr

Eps

Eci fcir fcds

fcds


98/127

Where:

Msd = moment due to all superimposed permanent

dead and sustained loads applied after prestressing

fcds

M

sd e

Ig

Shrinkage


99/127

Volume change determined by section andenvironment

Where:

Ksh = 1.0 for pretensioned members

V/S = volume-to-surface ratio

R.H. = average ambient relative humidityfrom map

SH 8.2 106 Ksh Eps 1 0.06 V S 100 R.H.

Relative Humidity


100/127

Page 3-114 Figure 3.10.12

Relaxation


101/127

Relaxation of prestressing tendons is based on thestrand properties

Where:

Kre and J - Tabulated in the PCI handbook

C - Tabulated or by empirical equations in the PCI

handbook

RE Kre

J SH CR ES C

Relaxation Table


102/127

Values for Kre and J

for given strand

Table 4.7.3.1

page 4-85

Relaxation Table Values for C


103/127

fpi = initial stress inprestress strand

fpu = ultimate stress

for prestress strand

Table 4.7.3.2

(Page 4-86)

Prestress Transfer Length


104/127

Transfer lengthLength when the stress

in the strand is applied

to the concrete

Transfer length is notused to calculate

capacity

t se bl f 3 d

lt

fse

3 db

Prestress Development Length


105/127

Development length -length required todevelop ultimate strandcapacity

Development length is

not used to calculatestresses in the member

ld

lt

fps

fse

ld

fse

3 db fps fse

Beam Ledge Geometry


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Beam Ledge Design


107/127

For Concentrated loads where s > bt + hl, findthe lesser of:

Vn 3 f 'c hl 2 bl b b t hl

Vn

f 'c

hl

2 bl

b

b

t

hl

2 de

Beam Ledge Design


108/127

For Concentrated loads where s < bt + hl, findthe lesser of:

Vn 1.5 f 'c hl 2 bl b bt hl s

Vn

f 'c

hl

bl

b

bt

hl

2

d

e

s

Beam Ledge Reinforcement


109/127

For continuous loads or closely spaced concentratedloads:

Ledge reinforcement should be provided by 3 checks

As, cantilevered bending of ledge

Al, longitudinal bending of ledge

Ash, shear of ledge

Vn

24 hl f '

c

Beam Ledge Reinforcement


110/127

Transverse (cantilever) bending reinforcement, As

Uniformly spaced over width of 6hl on either side of thebearing

Not to exceed half thedistance to the next load

Bar spacing should not

exceed the ledge depth,hl, or 18 in

As

1

fy

Vu

a

d

Nu h

l

d

0.2 N

u

Vdl

Longitudinal Ledge Reinforcement


111/127

Placed in both the top and bottom of the ledgeportion of the beam:

Where:

dl - is the depth of steel

U-bars or hooked bars may

be required to develop

reinforcement at the end

of the ledge

Al

200 b

l b dl

fy

Hanger Reinforcement


112/127

Required for attachment of the ledge to the web

Distribution and spacingof Ash reinforcementshould follow the sameguidelines as for As

Ash

V

u

fy

m

Hanger (Shear) Ledge Reinforcement


113/127

Ash is not additive to shear and torsion

reinforcement

m is a modification factor which can be

derived, and is dependent on beam section

geometry. PCI 6th edition has design aids

on table 4.5.4.1

Dap Design


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(1) Flexure (cantilever bending) and axial tension in the

extended end. Provide flexural reinforcement, Af, plusaxial tension reinforcement, An.

Dap Design


115/127

(2) Direct shear at the junction of the dap and the main body

of the member. Provide shear friction steel, composed ofAvf+ Ah, plus axial tension reinforcement, An

Dap Design


116/127

(3) Diagonal tension emanating from the re-entrant

corner. Provide shear reinforcement, Ash

Dap Design


117/127

(4) Diagonal tension in the extended end. Provide shear

reinforcement composed of Ah and Av

Dap Design


118/127

(5) Diagonal tension in the undapped portion. This is

resisted by providing a full development length for Asbeyond the potential crack.

Dap Reinforcement


119/127

5 Main Areas of Steel

Tension - As Shear steel - Ah

Diagonal cracking Ash,Ash Dap Shear Steel - Av


120/127

Shear Steel

Ah


121/127

The potential vertical crack (2) is resisted by a

combination of As and Ah

Ah

2 V

u

3 fy

me

An

ShearSteel

Ah


122/127

Note the development ld of Ah beyond the

assumed crack plane. Ah is usually a U-barsuch that the bar is developed in the dap


123/127

Dap Shear Steel

Av


124/127

Additional reinforcement for Crack (4) is

required in the extended end, such that:

Vn

Av

fy

Ah

fy

2 b d f 'c

Dap Shear Steel

Av


125/127

At least one-half of the reinforcement

required in this area should be placed

vertically. Thus:

Av

1

2 fy

V

u

2 b d f '

c

Dap Limitations and Considerations


126/127

Design Condition as a dap if any of thefollowing apply The depth of the recess exceeds 0.2H or 8 in.

The width of the recess (lp) exceeds 12 in.

For members less than 8 in. wide, less than one-half of the main flexural reinforcement extends tothe end of the member above the dap

For members 8 in. or more wide, less than one-third of the main flexural reinforcement extends to

the end of the member above the dap


127/127

Documents

6 Flexural Component Design