Design Report

Engineer Design Project (EDP) (200 points) Due 1600 Lesson 36 (SA 27) Annex A - Preliminary Design Requirements

CE403001

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UNITED STATES MILITARY ACADEMY

ENGINEERING DESIGN PROBLEM

WEST POINT POST EXCHANGE PRELIMINARY DESIGN

CE403: STRUCTURAL ANALYSIS

SECTION I1

LTC ALBERT BLEAKLEY

BY

TEAM GLAD-IT'S-FINISHED: CADET MATT SPEARS, 00 CO B4 CADET CHRIS WAGAR, 00 CO G 1

CADET DAN WALLESTAD, 00 CO B4

WEST POINT, NEW YORK

30 NOVEMBER 1999

Table of Contents

Table of Contents

List Tables and Figures .................................................................................... ii

List of Symbols .............................................................................................. iii

Introduction ............................................................................................... ...... 1

Assumptions .. ............................................................................................... 1

General Approach .................................................. : ..................................... 1-2

Results and Discussion .................................................................................. 2-17 Loads ................................................. ............................................................ 2-6 Tributary Area ................................................................................................... 6 Load Case Combinations ... . , ............................................................................. 6-8 Design ................................................ .............................................................. 8 Bar Joist Design ............................................................................................. 8-11 N-S Bay Design ............................................................................................ 12-16

Conclusions ....................................................................................... ....... 16-17

Documentation .... ................... , ............. , ........ , ................ , .... " .. , " .. " .................. 18

Appendix A .................................................................................................. 19 Preliminary PX Plan (Top View Drawing)

Appendix B ................................................................................................ 20-21 Visual Analysis Drawing of Bar Joist

Appendix c. ............. '" ............................................. '" ..................... '" ........... 22 Designed Bar Joist Drawing

Appendix D ....................................................................... '" ....................... 23-26 Bar Joist Hand Calculations/Comparison With Visual Analysis Output

Appendix E. '" ....................................................................... '" ...... '" ......... 27-28 Visual Analysis Drawing of N-S Bay

Appendix F .. ... '" ........................................................................................ 29-34 Designed N-S Bay Drawing (North Elevation and West Elevation)

N-S Bay Hand Calculations/Comparison With Visual Analysis Output

Appendix G .... '" ...................... , .................... '" ......................... '" ........ 35 Bill of Materials

i

Report Body

List of Tables and Figures

FIGURES

Figurel: Roof Section ........ .............................................................. '" .... '" ........ 2

Figure 2: Wind Loads ......................................................................................... 5

Figure 3: Tributary Area-Roof ............................................................................ 6

TABLES

Table 1: First-Run Bar Joist Design Results ............................................................ l0

Table 2: Bar Joist Deflection ..... .......................................................................... 11

Table 3: N-S Bay Internal Forces ......................................................................... 13

Table 4: N-S Bay Member Deflections .................................................................... 15

Table 5: N-S Bay Incremental Analysis ................................................................... 16

Table 6: Final Design Specifications ..................................................................... 16

ii

List of Symbols

B base (width of building)

D dead load

E earthquake load

F load due to fluids with well-defined pressures and max heights

GCp product of external pressure coefficient and gust effect factor

GCpi product of internal pressure coefficient and gust effect factor

H load due to the weight and lateral pressure of soil and water in soil

L live load

L length of member

LCC load case combination

lbs pounds

L,. rooflive load

M" nominal capacity

M" factored moment

psf pounds per square foot

q velocity pressure

qh velocity pressure evaluated at mean roof height

R rain load

S snow load

T self-straining force

W wind load

reduction factor for design

P" design axial strength

P" factored axial internal force

iii

West Point Post Exchange Preliminary Design

INTRODUCTION

The Glad-It's-Finished Team of the Messrs. Goebel and Wannabee structural engineering firm

has been tasked to design the main structural frame and structural roofing system for the

upcoming construction of the new West Point Post Exchaqge (PX). Specifically, our team will

design:

1. One roof bar joist using LCC#3. 2. One entire interior North-South (N-S) bay of the building frame. 3. Beams in the N-S bay. 4. Columns in the N-S bay. 5. Bracing in the N-S bay against lateral loads.

Enclosed in Appendix A is the preliminary plan drawing for the new PX. The circled N-S bay is

the one that we will design.

ASSUMPTIONS

In order to simplify the structure for design and analysis, we made the following

assumptions:

l. The wind loads on the 2' parapet will be transferred as a point load to the top of the ~ exterior columns.

2. Bracing will be included in the N-S bay; however, there is no bracing in the interior bays v-of the real frame.

3. Since we do not design the beams in the East-West (E-W) direction, we will assume that those beams weigh the same as the bar joists, and therefore transmit the same load onto the N-S columns as the bar joists do onto the N-W beams.

4. Even though each beam and column of the frame will need to carry a different load, we ./' will design all members for the worst possible load.

5. Bar joists are braced against buckling because the top chord is anchored to the roof deck. ~

GENERAL APPROACH

Our approach to designing the assigned N-S bay was from top to bottom. We began with

the roof, and the associated loads, then transferred them down to the bar joists, to the N-S beams,

and finally into the columns.

1


To design the bar joists, we used the Load Case Combination #3. Once the bar joists

were complete, we re-worked them using LCC#4 so that we could transfer the loads to the beams

in the N-S bay. Along with the point loads from the bar joists, we applied the wind loads to the

exterior columns and analyzed the bay in its entirety.

RESULTS AND DISCUSSION

Loads

Using this approach, we first calculated the loads on the roof. Figure I below shows a

cross section of the roof.

Roof Membrane

1+1-- Roof Boanl InsuL,tion

H--- Roof Board Insulation

~---------------

--+--Galvnruze(l Metal Deck ~---------------

Figure 1: Roof Section

Included in that section is a liquid applied waterproofing roof membrane (I psf), 2 layers of 2"

urethane foam roof board insulation (.5 psf per I" thickness), and a 20 gauge-I '12 inch deep wide

rib galvanized metal deck with 6 inch rib spacing (2.5 psf). Additionally, on the inside of the PX,

the roof will support the lights, AC and heating ducts, plumbing, etc. (4 psf) and an interior

suspended steel channel system ceiling (2 psf). Combined, these loads are the dead loads (D) that

act on the roof.

D = ~+(220.5)+2.5+4+2]= 11.5psJ /

2


There are also roof live loads (L,) which act on the roof. The roof live loads are

calculated using the equation:

Where: RJ = 1.2 - 0.001 A, if the tributary area (A,) that-the roof supports is > 200 sq ft and> 600 sq ft

Rz = 1 because we have a flat roof

L, = 19.78 psf ./"'

Another type of load which acts on the roof is the snow load. The snow load equation is:

Pf = 0.7C,C,IPg Where: C, = exposure factor = I

C, = thermal factor = 0.9 I = importance factor = 1.1 Pg = ground snow loads = 30 psf

PI = 20.79 psf .--The final load that we will consider in our design acts on both the roof, and the columns

of the frame: wind loads (W). Below are the steps we took to calculate the wind loads.

v = Basic Wind Speed = 110 mph The PX is located in Exposure Class C ----1= 1.15 (PX is category II/)-z = 18ft high roof ----k I ,#; . h . h 9 (.9-.S5}(20-IS) : = ve OClty pressure exposure eOejjlClent at elg t z =. - ( ) = 0.88

20-15 k" = topographic factor =1

From the above information, we calculated the velocity pressure, and design pressure for the wind

loads on the windward, side, and leeward walls, and the roof.

Velocity Pressure, qJ8 = (0.00256)(.88)( 1)(1. 15)( lJd) = 31.35 psf

Design Pressure, p = qGCp q"GCpi

Where: q = velocity pressure GCp = product of external pressure coefficient and gust effect

factor GCpi = product of internal pressure coefficient and gust effect

Factor q" = velocity pressure evaluated at mean roof height

3

Windward Wall:

Leeward Wall:

Side Walls

Roof


PO-15 = (.00256)!O-85)( 1)( 1.15)( 1 ](1)(0.8) 4.797 psf ../'

= 19.426 psf or 29.017 psf

PI5.'O = (.00256)(0.90)(1)(1.15)(110'(0.8) 4.797psf

= 20.85 psf or 30.445 psf

VB = 0.483, so Cp = -0.5 ----

p = (31.35 psf)(-O. 5)(0. 8) 4.797 psf ---

= -7.743 psfor-17.337 /'"

Cp = 0.7

p = (31.35)(-0.7)(0.8) 4.797psf= ---

PO-18 = (31.35 psf)(-O. 9)(0. 8) = -22.57 psf ----

P18-36

P36./md

= (31.35 psf)(-0.5)(0.8) = -12.54 psf

= (31.35 psf)(-0.3)(0.8) = -7.524 psf

With the wind loads for the leeward and windward walls, we converted the loads from pounds per

square foot (psf), into pounds per inch Ob/in) so that we could apply the loads to the exterior

columns of the frame. Each E-W bay is 38', or 456" long. Each exterior column supports the

loads on half of the area of the wall to the left and right of the columns. Therefore, the area that

the exterior columns support is:

Height 0-15 ft. (456" long)( 180" high) = 82,080 in' ----

Height 15-18 ft (456" long)(36" high) = 16,416 in'

But there is also the 2 foot parapet that extends past the height of the roof. The area from the

parapet that the exterior column supports is:

(456" iong)(24 " high) = 1 0,944 in'

For the roof, the areas are as follows:

Length 0-18 ft

Length 18-36 ft

Length 36-end

(216" long)(456" wide) = 98,496 in'

98,496 in'

(1568.04 in2)(456") = 715,026.24 in'

4


We can multiply the areas by the loads in psf to get the total load in pounds and simply divide by

the length of the column to get the load in pounds per inch. However, the wind loads on the 2

foot parapet will be transferred to the exterior columns as a point load at the top of the column (as

stated in our assumptions).

North-side column:

0-15 ft (S2,OSO in'l 144 in')(29.017 psf)11S0" = 91.SSIh!in ----

15-1Sft ./' (16,416 in'l 144 in')(30.44 psf)l36 " = 96.411blin

Parapet (10,944 in' 1144)(30.445 psf) = 2313.Slb point load ----

South-side column:

Roof:

O-IS ft [(S2,OSOin' +16,416 in')I(144in')](-17.337psf)l216"=-54.9Iblin ~

Parapet (10,944 in' 1144)( -17.337 psf) = 13 17.6 lb point load ---

Length O-ISjt (9S,496 in'I144 in])(-22.57 psf)l216" = -71.47Iblill ----

Length IS-36ft (98,496 in'I144 in2)(-/2.54 psf)l216 " = -39.71lblill

Length 36-end (715,026.24 in'I144 in')(-7.524 psf)ll568 " = -23.83Iblin

With the above wind loads, we drew the diagram in Figure 2 below, depicting the direction of the

wind loads-a negative wind load means that the load acts away from the building, and a positive

load acts towards the building. The loads shown are the worst case loads, which we will use in

our design.

"16" "'16" 1 - 1- 1 1668" I

71.471 rfT rm ?{ lbiin 23.831biin ~I 2313.8Ib, -1317.6Ib 96 .4llblin L ~ C" 180" 91.881biin C" 54.9 lblin

t::: ... N S ...

Figure 2: Wmd Loads

5


[(33'4")16](38') = 211.J2tt =30,401.28 in'

Load Case Combinations

In the following section we calculated the Load Case Combinations (LCC) for the roof.

We calculated the worst case possible for each of the LCCs. To accomplish the worst cases for

-------

each load case, we used 0 for the wind loads, and used the greatest load in cases with an 'or.'

This is because on the roof, the wind loads act upwards (negative) and therefore actually decrease

the overall load on the roof. So when the wind is not blowing, there is the largest force acting

downward on the roof.

In an actual design, we would consider all load cases, and pick the worst possible case

---out of all LCCs. However, for this design, we will used LCC#3 to design the bar joists, and ~

LCC#4 to design the beams and columns. For LCC #3 we showed all possible combinations as

an example of what would be done for each load case in an actual design. ,:; oOD /

Load Case Combinations (LCC):

1. I.4D 1.4( 11.5) = 16.1 pst

2. 1.2(D + F + T) + 1.6(L + H) + 0.5(L, or S or R) 1.2(11.5 + 0 + 0) + 1.6(0 +0) + 0.5(20.79) = 24.195 pst

3. 1.2D + 1.6(L, or S or R) + (0.5L or 0.8W) 1.2D + 1.6(L,) + (0.5L) 1.2( 11.5) + 1.6(19.78) + 0.8(0) = 45.448 pst

1.2D + 1.6(L,) + (O.8W) 1.2( 11.5) + 1.6(19.78) + 0.8(0) = 45.448 pst

1.2D + 1.6(L,) + (O.8W) 1.2( 11.5) + 1.6(19.78) + 0.8(-22.57) = 27.392 pst

1.2D + 1.6(L,) + (O.8W) 1.2(11.5) + 1.6(19.78) + 0.8(-12.54) = 35.416 pst

1.2D + 1.6(L,) + (O.8W) 1.2( 11.5) + 1.6( 19.78) + 0.8(-7.524) = 39.4288 pst

1.2D + 1.6(S) + (0.5L) 1.2( 11.5) + 1.6(20.79) + 0.8(0) = 47.064 pst

7


I.2D + I.6(S) + (O.8W) 1.2(11.5) + 1.6(20.79) + 0.8(0) = 47.064 psf

I.2D + I.6(S) + (O.8W) I.2( 11.5) + 1.6(20.79) + 0.8(-22.57) = 29.008 psf

I.2D + I.6(S) + (O.8W) 1.2(11.5) + 1.6(20.79) + 0.8(-12.54) = 37.032 psf

I.2D + I.6(S) + (O.8W) I.2( I 1.5) + 1.6(20.79) + 0.8(-7.524) = 41.0448 psf

I.2D + I.6(R) + (0.5L) 1.2( Il.5) + 1.6(0) + 0.5(0) = 13.8 psf

1.2D + 1.6(R) + (O.8W) 1.2(11.5) + 1.6(0) + 0.8(0) = 13.8 psf

1.2D + I.6(R) + (O.8W) 1.2( I 1.5) + 1.6(0) + 0.8(-22.57) = -4.256psf

1.2D + 1.6(R) + (O.8W) 1.2(11.5) + 1.6(0) + 0.8(-12.54) = 3. 768psf

1.2D + 1.6(R) + (O.8W) I.2( I 1.5) + 1.6(0) + 0.8(-7.524) = 7.7808 psf

4. 1.2D + I.3W + 0.5L +0.5(L, orS or R) 1.2( I 1.5) + 1.3(0) + 0.5(0) + 0.5(20.79) = 24.195 psf

, 1- :>dC?UC/7) liVl!.(.iJDI tVf"''' - T}IT$ IS wfMT MV:>iE.6 /~JfL 5. 1.2D + I.OE + 0.5L + 0.2S LOt'r[)S biJ T7f;< FMItfE .

1.2(11.5) + 0.5(30.445) + 0.2(20.79) = 33.1805 psf

6. 0.9D + (1.3W or I.OE) 0.9( 11.5) + 1.3(0) = 10.35 psf

Design

In order to design the members of the frame. the bracing, and the bar joist truss, we used

the following criteria.

1. !{iPn ~ p", where !{i is the reduction factor, 0.9, Pn is the design axial strength, and p" is the factored axial internal force.

-----2. Deflections must be less than Ll360 inches. r

3. For the beams and columns, !{iM n ~ M" ' where !{iM n is the nominal capacity of the section times the reductions factor and M" is the factored moment.

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AlbertPolygon

AlbertText BoxYou only need to show the largest total for each LCC


Bar Joist Design

The bar joists are made by pairs of angles for the upper and lower chords with continuous

bar stock woven in between. For simplification in design, we used the same angles for the top

and bottom chords. Additionally, we did not design the kicker braces between bar joists even

though they are actually used in real design. ~ Our first step in designing the bar joists was to determine the total load that one bar joist

...-'

would support. From LCC#3, the worst case combination was 47.064 psf, and the tributary area

of one bar joist was 211.12 fe. Therefore, one bar joist supports a total load of:

(211. 12 )(47.064 ) =9,936.15 lbs

Because the bar joist is a truss, we cannot analyze the total load as it is in reallife--a distributed

load. Rather, we divide the total load by the number of nodes in the truss, and then apply the total

load as point loads at the nodes of the bar joist. Therefore, we had to determine how many nodes

we wanted in the bar joist. We decided to make 20 nodes in the truss with 24" spacing. We

based our decision upon the visual inspection of the actual PX, and pictures shown in our

textbook. We also decided to make the truss equilateral-all members of the truss are the same

length. Using this method, the bar joist is 20.78" deep. Refer to Appendix B for a Visual

Analysis drawing of the bar joist, to include member names and the deflected shape. Since the

two end nodes of the bar joist are supports, there are 18 nodes on which to distribute the load of

the roof. /

(9936.15Ibs)l18 = 552.011bs per node

Using our Visual Analysis software, we built the bar joist and applied the nodal loads. For the

first run analysis, we did not include the member weights. Table 1 below show the maximum

internal forces for each member of the bar joist, and the displacement of the top nodes.

9


Table 1: First-Run Bar Joist Design Results

Using the information from Table 1, we applied our design criteria. First, we designed the top

and bottom chords of the bar joist for axial strength. The greatest internal force in the top and

bottom chords was in members M28 and M29-28,689.64 Ibs.

P"i/I 2 P"

(A) (36,000psi)' (0.9)2 28,689.64Ibs

A 2 0.8855in'

The double angle L2 x 2 x 118 provides the smallest area for double angle members--O.960 in',

so we used that angle for both the upper and lower chords.

We followed the same steps to design the diagonal members of the bar joist. The greatest

internal force in the diagonals was in members M38, M56, M57, and M75-5736.97 Ibs.

(A) (36,000psi). (0.9)2 5,736.97lbs

A 2 0.1770in',//

10

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The diagonals are made from bar stock. The closest area was al"x1!4" rectangular section with

an area of 0.188 in2

Next, we designed the entire truss for deflection, based upon the deflection of the nodes

and the total length of the truss. The maximum deflection was at nodes N 10, N 11, and N 12-

0.12 inches, which is less than the maximum allowable deflection, U360l)(:= 456"/360l(= 1.267".

Using the results from the first run design, we went back into Visual Analysis and

changed the top and bottom chords to L2x2x1l8 and the diagonals to 1 "x 1/4" rectangular

sections. We also included the member weights. This time, the bar joist did not meet the

specifications for deflection-the truss deflected -1.65 inches. With the included member

weights, however, the bar joist still met the specifications for axial strength. Therefore, we only

had to design the bar joist for deflection from here. We used a trial and error method to

determine the members of the truss. Table 2 shows the results, with the corresponding deflection

at node NIO (where the max deflection occurs).

Table 2: Bar Joist Deflection

We chose the double angle member L2x2xO.1875 and a rectangular bar stock diagonal with area

of 0.5 in' because it gave us the smallest deflection. Using those members, the bar joist still met

the design requirements for axial strength. Refer to Appendix C for a hand drawing of the

designed bar joist. Appendix D contains hand calculations of the internal forces of the bar joist.

In order to check the calculations by hand, we re-ran Visual Analysis and excluded the member

weights. A comparison the computer output and hand calculations is also in Appendix D. We

found that there were not any significant differences between the two forms of calculation-the

largest percent error was 0.289%. 5iJPt/.~

11

../


N-S Bay Design

BEAMS: Continuing with our top-down approach, we transferred the loads of the bar

joists to the beams in the N-S bay. As the provided diagram in Appendix A shows, there are five

bar joists in each 3S'x33'4" section of the frame. Additionally, there are beams in the E-W

direction that the N-S columns support. From the bar joist design above, the support reactions at

the ends of the bar joists were 496S.09 lbs. But this value was calculated using LCC#3. For the

beams, we design them using LCC#4. We went back into Visual Analysis and analyzed the bar ~()"D

joist using LCC#4. We changed the value of the nodal loads to 276.011bs, which we calculated

the same way as we did for LCC#3 in the Bar Joist Design section above. The end reactions were

2729 lbs. Because we are designing a bay in the middle of the frame, there are actually ten bar

joists that each beam supports-five on each side. Therefore, the applied point loads to the N-S

beams will be (2)(2729) = 510S.05 lbs ............

COLUMNS: The columns of the N-S bay are steel pipes and are pinned at the bottom.

----We used moment connections to connectthe tops of the columns with the beams. We did this

~ because the moment connection is more efficient. Only the exterior columns of the bay have

applied horizontal loads, as shown in the wind load calculation section above.

BRACING: For the bracing, the real PX structure uses rectangular tubes. We also used

rectangular tubes. We placed bracing in the first two and last two sections of the bay, in an 'A'

shape. We did not put bracing in the middle section because is did not provide significant support

to the structure when we ran test runs in Visual Analysis.

We built the frame in Visual Analysis as shown in Appendix E. From the Visual

Analysis output, we made the following table with shows the maximum internal force and

bending moment for each member, excluding member weights.

12

AlbertText BoxUse the largest LCC to design the beams and columns, not necessarily LCC #4

AlbertPolygon


Table 3: NS Bay Internal Forces

Using the same method as we did to design the members of the bar joist, we designed the beams,

columns, and bracing for axial strength. We also designed these members for internal bending.

The greatest internal force for the beams was 16,095.7 lbs, in member M 1.

(A) (36,000 psi) (0.9)2 16,095.71bs

A 2 0.4968in'

The W-Shape WI0x12 provides a cross-sectional area of 3.54 in', which is the smallest W-

Shape. However, we also have to design them for bending. The maximum internal bending

moment was 1,186,196.59 in Ibs in member M5. We used the design tables in the Steel

Reference Manual to get the rpM" term.

rpM" 2 1,186,196 in Ibs

13


The W -Shape W 12x26 meets the design requirements. Because this beam is larger than the one

designed for axial strength, the bending criteria is more critical and we will therefore use the

W12x26 member for all beams in the N-S bay."'-/

The greatest internal force in the columns was 30,581.1 lbs in member M9. Because we

are using moment connections at the tops of the columns, we have to find the effective length of

the columns. The effective length is equal to kL, where k = 0.7 because the tops of the column

are moment connections. Therefore, the effective length of the columns is 12.6 ft, rounding up is

13 ft. Designing the column for axial strength:

Pnl/l2: Pu

(A) (36,000 psi). (0.9)2: 30581.11bs

A 2: 0.9439in 2

The circular pipe PX1.5 meets the design requirements with the smallest area available-1.07 in2

We would have designed the member for bending, but the Load and Resistance Factor Design

book does not have design tables for bending of columns.

The greatest internal force in the bracing members was 8,041.23 lbs in member M13.

Pnl/l2: Pu

(A) (36,000 psi). (0.9) 2: 8041.231bs

A 2: 0.2482in 2

The structural bracing TS1.5x1.5x.1875 has the smallest area available, 0.894 in2. We had the

same problem designing the bracing for bending as we did for the columns.

We went back to Visual Analysis and changed each member to the specifications that we

designed above. We then ran the analysis which provided the us results for the member

deflections. Table 4 below summarizes those results.

14


We can see that the maximum deflection for the beams occurred in member MS, for columns was

in member M6, and for bracing in member M12. Next we gradually increased the size of the

beam, column, and bracing members until we found a combination that would meet the deflection

restrictions. The deflection restriction is that the member cannot deflect more than L/360". For

each type of member, the restriction is:

Beams:

Columns:

Bracing:

400"/360 = 1.11 "

216"/360 = 0.6"

454.8"/360 = 1.263" /

Table S below shows the results from our incremental analysis. The member sizes are shown in

areas (in2) and the maximum deflection is shown for each type of member, regardless of what

member it actually was.

15


Table 5: NS Bay Incremental Analysis

As shown in Table 5, we chose members PX12 for the columns, W 16x57 for the beams, and

TS5x4x0.375 for the bracing members. Team Glad-It's-Finished recommends that Messrs.

Goebel and Wannabee structural engineering firm use these members in their final design report

to the constructors.

Some hand calculations for the N-S Bay are in Appendix F.

CONCLUSIONS

Table 6 below shows the final specifications of the designed N-S bay and the bar joist

trusses.

We did have some problems designing the columns and bracing in the N-S bay for bending

because we did not find design tables in the Load & Resistance Factor Design book. However,

we feel that the columns and bracing are sufficient for bending based upon the results from

analyzing the beams for bending and deflection, and the bar joist for deflection. For both the /'

beam and the bar joist, the deflection requirement was the critical design characteristic. In other

words, if we would have designed the bar joist and beams strictly for deflection, we would have

----met the requirements for bending and axial strength-we would have the same results.

16


Therefore, the fact that we did not design the columns and bracing for internal bending should not

discredit our results.

As a final requirement for our design, we calculated the total cost of the frame using the

members that we designed. We do note, however, that we did not include the E-W beams in our

calculation because we did not design these members. Additionally, we only included one set of

bracing in the calculation. Steel currently costs $0.75 per lb. To calculate the total cost, we

determined how much the frame weighed, based upon the unit weight and length of the members.

Appendix G summarizes our calculation of the total cost. We found that the frame of the post

exchange will cost $284,254

17

NOTES

None.

1 ======-----;:---18 ___ ===1

Appendix A: Preliminary Plan

I i

L-____________________ ~A~p~p~e~nd~cr_~A~:_P_r~e~Um~in~a~ry~P~w~n~ ____________________ ~I

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19

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Appendix B: Visual Analysis Drawing oj Bar Joist

-552.01 Ih,,--;Io

-552.01 Ih.,,"---,~~1!l

-552.01 IbS'-~~l -552.01 IbS'-~~Jl

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~

VisualAnalysis (version 3.12.STUDENT) - Result View Bar Joists, Mon Nov 29 23:16:031999 Matt Spears, For Educational Purposes Only Nodal Loads Inches, Pounds, Degrees

~ co

Appendix C: Designed Bar Joist Drawing

----.~----------------L---------------L------------------r_--

10.t

) , ~o

Appendix D: Bar Joist Hand Calculations

. _ .. _. ____ ... _ .. __ ____ ~ _______ ... _________ ~:!fIlSr ~ji!I'!Q !!!!~e;/I?ly7(I"'_~_. ____ .. _

* /ltl #f;r1lt8]e f't9f>e 20

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M3 6056.7 6059.6 4.79E04 M4 ~ 1.65 75E M4 I: 112.64 1916.4~ 2.00E03 M4 --31E '.21 318".02 ;.96E 15 M5 1593.8' 11;89.21 M4, 2549.77 ~ M6 4157.5 1.18E03 M4' 1912.32 1912.2 ~ 15 M7 .6'~ M45 1274.88 1274.79 ,--06E' 15 M8 Il288.1 8:'0 I.' 7.40E04 M46 637.44 637.38 ~

~ 17 0 0.026 O. ~ ~ !1 ~ 6.72E( 18E '.44 637.38 9. ~ '9 14.88 1; 14.79 7.06E ~ l28! 12 ~ 10 12.32 1 12.2

., N 19. '7 14':~ ~ E03 M, : 187.!1

. 15 '.26 M 116 :.6' 117 6.7

M18 M20 5737.93

=B ~:! - M 3 191

M24 22, M25 M26 2677; M27 28052.1

=i=:fi1 ---r:ii31 m 'M32 ,2' 864.36

M33 !231,.17 M34 9121i.43 M35 .1 M36 M37 '.9 M38 5 36.9

19' '.47 M5 60 1.6 4.7 E04 M5

~19 10840.15 1.70E04 M56 5736.66 2.21 E04 M57 5736.97

. :1 E04 18

. :2E04 19 ,,04 22~","'f 11:9'---+--::::.;:: .. 2 E04 24, 1.8 :.2 E04

;771 2, 1459 2, 21

9 26771

2:!3m 19 I! 2; 17 1!;2 74 lOll:< ,1 573Ei.66 573EI,66

E04 E114 E04

2.2IE04

!:~~4 :.2 E04 '.2 E04 :,2 E04

26

M6 M6 M 13 M 14

15 M66 M6 M6 M6

170 '1

17: M7: M7 M75 M76

3187. ~549.

1, 74.88 6 '.44

63 '.44 12~ 1.88 19' :,32 ;:549.77

Appendix E: Visual Analysis of N-S Bay

2554.03 Ibs 5108.05Ib 5108.05Ibs 5108.05 Ibs 5108.05Ibs 5108.05Ibs 5108.05Ibs 5108.05Ibs 5108.05Ib 5108.05Ib 5108.05Ib 5108.05 Ibs 5108.05Ib 5108.05Ib 5108.05 Ibs 5108.05Ibs 5108.05 Ibs 5108.05Ibs 5108.05Ibs 5108.05Ibs 5108.05Ib

:s: 5108.05Ibs

.91 5108.05Ibs

>

:6 5108.05Ibs

W

050) 5108.05Ibs

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w ..

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VisualAnalysis (version 3.12.STUDENT) - Result View N-S Beam Analysis, Mon Nov 29 23:17:031999 Matt Spears, For Educational Purposes Only Beam Loads Inches, Pounds, Degrees

,Q ,Q

" r-

ro oi '"

0)

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M3 M4 .M5.

"';toE 13979.91bs

,Q ,Q ,Q ,Q

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t .,' f" ~ .~.-I

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FI = 7!,oc..a I> of.

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1/J.,w l

"Joioc II: 7-i/41l .J.Gv T 4, -'41::"'~~--\!"-----

Appendix G: Bill of Materials I I

Appendix G: Bill of Materials

1176 751.53

78 4

I: .. TotalCostl;C 'I $ 284,254.29 I "This cost does not include the price of the East-West beams because we did not design them for this project. Furthermore. we only included the one set of bracing that We designed in this price.

35

Documents

Design Report