CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT...

CES 4743C Structural Design

FINAL PROJECT SUBMITTAL

Team Vandelay:

Jonathon Ambar

Paul Heagney

Dominick Tota

David Lutz

Due Date: November 25, 2014

Instructor: Professor Justin Quillen

TABLE OF CONTENTS Pg 1

Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL

TABLE OF CONTENTS:

TABLE OF CONTENTS…………………………………………………………………... 1

EXECUTIVE SUMMARY…………………….…..……………………………………… 3

SCHEDULE………………………………………………………………………………... 4

TABULATION OF SPECIFIED LOADS………………………………………………… 7

MWFRS CALCULATIONS……….……………………………………………………… 9

COMPONENTS & CLADDING CALCULATIONS………………………..……………. 12

ROOF DECK DESIGN………………...………………………………………………….. 17

ROOF JOIST DESIGN…………………………………………………………………......20

ROOF JOIST GIRDER DESIGN…………………………………………………………..23

ROOF PERIMETER BEAM DESIGN…………………………………………………..... 25

COMPOSITE FLOOR DECK DESIGN…………...……………………………………… 35

FLOOR BEAM DESIGN……………………………………………………………...…... 40

STEEL COLUMN DESIGN……………..…………………………………………………83

TWO-WAY SLAB DESIGN……….……………………………………………………… 95

CONCRETE CONTINUOUS L-BEAM DESIGN…………………………..……………. 143

CONCRETE COLUMN DESIGN……..………………………………………………….. 156

SIMPLE SHEAR CONNECTIONS…….………………………………………………......158

ROOF CONNECTIONS...…………………………………………………………………..160

DIAPHRAGM DESIGN…...……….……………………………………………………… .162

LATERAL STEEL ANALYSIS…………………….………………………..……………..164

APPROXIMATE SECOND ORDER ANALYSIS…………………………………………180

BRACED FRAME DESIGN……………………………………………………………......187

TABLE OF CONTENTS Pg 2


SHEAR WALL DESIGN……….…………………………………………………………..204

FOUNDATION DESIGN…………………………..………………………........................ 208

APPENDIX……………..……………………………………………………………...…... 214

AUTOCAD………………..……………..………………………………………………… 229

REFERENCES…………….……….……………………………………………………… 230

EXECUTIVE SUMMARY Pg 3


EXECUTIVE SUMMARY

Vandelay Industries is pleased to present this final product for our Structural Design (CES

4743C) project. This paper is entirely stand-alone and includes all of the work completed for

this project throughout the 2014 Fall Semester.

The structural design of the building in this project is for a six-story reinforced concrete and

structural steel building located in Orlando, FL. The ground floor features a slab on grade and

was not included in this project’s scope. The second and third floors feature a thirteen (13) inch

two-way reinforced concrete slab and the green roof features a similar twelve (12) inch slab. The

fourth, fifth, and sixth floors feature a steel frame implementing various wide-flange steel

sections and a 3VLI19 composite steel deck with normal-weight concrete fill and a slab depth of

5.5 inches. The roof features a 1.5B22 steel deck with no composite fill and is supported by

30K11 joists and 50G10KN13.6K joist girders.

In addition, 2L6x6x1/2x3/8 braces were used in the steel frame’s main wind resisting force

system (MWRFS). Moment frames were not implemented in the design because moment frames

are expensive, difficult to construct and design, require more detailed inspections, and often

include costly complete or partial penetration joint welds. Bolted/Welded double angle simple

shear connections are used to connect all steel beams to columns and beams to beams. In order

for consistency, there are no bolted connections in any of the steel columns.

The two-way concrete slabs are supported by 24-inch x 24-inch reinforced concrete columns

with a reinforcement ratio of 0.01. The end of each slab is also supported by an L-shaped

reinforced concrete beam; there are no interior beams featured in the slab design. There are also

four twenty-five feet long reinforced concrete shear walls that are used to resist the main wind

forces. All columns are supported by square foundations of various sizes.

In general, this project was designed using mostly allowable strength design (ASD) where

applicable. ASD was used in order to maintain consistency with the steel deck and also to act as

a learning experience because the engineer’s involved had limited experience with ASD. ASD

also makes the design more conservative because it is based on a member’s yielding strength

instead of its ultimate strength.

The main codes used to design this building were the 2010 Florida Building Code, the American

Institute of Steel Construction’s Specification for Structural Steel Buildings June 22, 2010, the

14th

edition of the AISC Steel Construction Manual, the American Society of Civil Engineers’

Minimum Design Loads for Buildings and Other Structures (7-10), the 2012 International

Building Code, and the American Concrete Institute’s Structural Concrete Building Code (ACI

318-11) and Commentary.

Vandelay Industries would like to thank all parties involved with the execution this project.

Jonathon Ambar Paul Heagney November 25, 2014

David Lutz Dominick Tot

SCHEDULE Pg 4


SCHEDULE:

Below is the project schedule as of November 25, 2014. This schedule shows the hours worked

by each group member by task and also shows dates of completion as well as target dates.

SCHEDULE Pg 5


SCHEDULE Pg 6


TABULATION OF SPECIFIED LOADS Pg 7


TABULATION OF SPECIFIED LOADS

The following specified loads were researched and tabulated using ASCE 7-10, the 2010 FBC,

and reliable Internet sources.

SUMPERIMPOSED DEAD LOADS

Note: Self weight must also be considered in addition to these loads

Item Load (psf)

2ND FLOOR

Mixed-use 30

Total: 30

GREEN ROOF

Growing Medium 25

Two layers 3/4" plywood 5

Water storage/filter 10

Single-ply waterproofing membrane 3

Vegetation 3

Miscellaneous 5

1" insulation (3rd floor exterior)

Total: 50.6

OFFICE

Carpet 2

Acousitcal dropped ceiling 1

Fire sprinklers 3

Mechanical Allowance 2

Miscellaneious (3rd interior and above) 2

Total: 9.5

ROOF

Acousical dropped ceiling 1

3.25 inch rigid insulation 5

Roof board and cover board (5/8" gypsum each) 5

Three-ply ready roofing with single-ply waterproofing membrance 12

Fire sprinklers 3

Mechanical Allowance 2

TABULATION OF SPECIFIED LOADS Pg 8


Miscellaneious (top of building steel roof) 2

Total: 30

EXTERIOR WALLS

Steel stud framing with gypsum board each side 11

7/8" stucco 10

Miscellaneous 2

Total: 23

CONCRETE WALLS

Finishes 15

Total: 15

FIRST FLOOR SLAB ON GRADE

4" thick concrete weight 150

Total: 150

LIVE LOADS Note: These loads are not reduced

Item Load (psf)

MIXED-USE AREAS

*Assembly Loading (other) 100

Total: 100

GREEN ROOF

*Assembly (other) 100

Total: 100

OFFICE

*Office 50

*Partition 15

Total: 65

ROOF

*Minimum roof live load 20

Total: 20

FIRST FLOOR SLAB ON GRADE

*Assembly (lobbies) 100

Total: 100

*Denotes load found in 2010 FBC

MWRFS CALCULATIONS Pg 9


MWRFS CALCULATIONS

The Main Wind Resisting Force System calculations were completed in Excel. The equations

used by this excel program are as follows:

𝑞𝑧 = 0.00256𝐾𝑧𝐾𝑧𝑡𝐾𝑑𝑉2

𝑝 = 𝑞𝐺𝐶𝑝 − 𝑞𝑖𝐺𝐶𝑝𝑖

Note: 𝐾𝑧 was found from the tables and not calculated

The Excel results are displayed below.

Building Conditions:

Enclosed

Topography factor

Kzt= 1

Rigid Structure

Exposure Factor B

Risk category 2

Assumed Enclosed

MWFRS Calcs

Design Wind speed based on ASCE 7-10 = 140 mph

Directionality Factor Kd= 0.85

Gust Effect Factor G= 0.85 Internal Pressure

Coefficient (GCpI)= 0.18 or -0.18

Windward wall Ext. Pressure Coefficient Cp (Use with qz )

Leeward Wall Ext. Pressure Coef Cp (Use

with qz)

Side Wall Cp (Use with qh)

0.8 -0.5 -0.7

Roof Cp (Use qh)

Distance From Edge Case A Case B

0 to h/2 -0.9 -0.18

h/2 to h -0.9 -0.18

h to 2h -0.5 -0.18

> 2h -0.3 -0.18



Wind Pressure p = qGCp – qi(GCpi):

Height Windward p Leeward p Side p

15 5.2 11.4 -10.4 -4.2 -13.3 -7.1

30 7.1 13.2 -10.4 -4.2 -13.3 -7.1

40 7.9 14.1 -10.4 -4.2 -13.3 -7.1

48 8.6 14.8 -10.4 -4.2 -13.3 -7.1

60 9.2 15.4 -10.4 -4.2 -13.3 -7.1

70 9.8 16.0 -10.4 -4.2 -13.3 -7.1

90 10.8 17.0 -10.4 -4.2 -13.3 -7.1

98 11.3 17.4 -10.4 -4.2 -13.3 -7.1

Roof Wind Pressure p = qGCp – qi(GCpi):

Distance From Edge Case A Case B

0 to h/2 -16.2 -10.0 -

5.7 0.5

h/2 to h -16.2 -10.0 -

5.7 0.5

h to 2h -10.4 -4.2 -

5.7 0.5

> 2h -7.5 -1.3 -

5.7 0.5



C&C CALCULATIONS Pg 12


C&C CALCULATIONS:

These calculations were originally completed for Homework #3 and incorrectly used a wind

velocity of 150 mph. They have now been updated with the correct 140 mph and to reflect our

revised roof design. We have also updated the calculations to include values of 𝐾𝑧 at every floor

height.

For the roof, find:

a. The effective wind area of one joist.

b. The effective wind area of the roof deck. Hint: actual area is span*1 ft. Use the greater

of the actual area or the minimum effective wind area.

c. Using the effective wind area for the joist and deck from part 1a and 1b, find the GCP

value that is associated with each, for zones 1,2 and 3. Use figure 30.6-1. You should

have one set of values for the joist and one set of values for the deck.

SOLUTIONS:

1) From the AutoCAD drawings, the span of a joist is 50 feet and spacing is 5 feet. Because

the spacing is less than one third the span, use one-third the span to find the effective

area. Therefore,

𝐴𝑒𝑓𝑓,𝑗𝑜𝑖𝑠𝑡 = 50 𝑓𝑡 ∗50

3 𝑓𝑡 = 833.33 𝑓𝑡2

2) From the AutoCAD drawings, the span of the deck is 5 feet. Therefore the actual area is

5 square feet and the effective wind area is 5 feet multiplied by one-third the span (or 5/3

feet), which equals 8.33 square feet. Therefore, use 8.33 square feet.

3) From figure ASCE 7-10 30.6-1, the GCP values are as follows:

Zone GCP for Joist GCP for Deck

Zone 1 -0.9 -1.40

Zone 2 -1.60 -2.30

Zone 3 -2.30 -3.20

For the concrete walls on level 1 and 2, find:

a. The effective wind area of the wall. Hint: Keep in mind we design walls based on the

unit length of the wall, so the actual tributary area is wall height from support to

support*unit length.

b. Using that effective area, find the GCP value that is associated with the wall from 2a for

zones 4 and 5. Use figure 30.6-1.

SOLUTIONS:



1) From the AutoCAD file, the wall height on level 1 is 20 feet and level 2 is 18 feet.

Therefore the actual tributary area of the wall on level 1 is 20 square feet per foot and on

level 2 is 18 square feet per foot. The effective wind area is found by:

𝐿𝑒𝑣𝑒𝑙 1: 20 𝑓𝑡 ∗ 20 𝑓𝑡

3 = 133.33 𝑓𝑡2

𝐿𝑒𝑣𝑒𝑙 2: 18 𝑓𝑡 ∗18 𝑓𝑡

3= 108 𝑓𝑡2

2) Finding the values from the specified figure gives:

Wall GCp values Level Zone 4 Zone 5

Level 1 0.7 -0.79 0.7 -1.35

Level 2 0.75 -0.8 0.75 -1.4

Using equation (30.6-1), find the design wind pressures associated with the GCP values from

problems 1 and 2. Use your project information for values and information not given here (wind

speed, height, etc.) Calculate the positive 𝑞𝑧 at the max roof height (not the MRH = h).

SOLUTIONS:

1) From project specifications, use risk category 2.

2) From ASCE 7 figure 26.5-1A, 𝑉 = 140 𝑚𝑝ℎ

3) Determining directionality factor from section 26.6 gives: 𝐾𝑑 = 0.85

From project specifications, use exposure category B.

From project specifications, the topographic factor is 𝐾𝑧𝑡 = 1.0

From table 26.11-1, the internal pressure coefficient is (𝐺𝐶𝑝𝑖) = {0.18−0.18

4) From table 26.9-1, 𝛼 = 7.0 and 𝑧𝑔 = 1200 𝑓𝑡.

From table 30.3-1:

𝐾𝑧 = 2.01 (𝑧

𝑧𝑔)

2𝛼

= 2.01 (98 𝑓𝑡

1200 𝑓𝑡)

27.0= 0.98

Using Excel to calculate kz values for values of z at every floor height of the building

gives:



z (ft) Kz

98 0.98

83 0.94

68 0.89

53 0.82

38 0.75

20 0.62

5) The velocity pressure is found by:

𝑞𝑧 = 0.00256𝐾𝑧𝐾𝑧𝑡𝐾𝑑𝑉2 = 0.00256(0.98)(1)(. 85)(140 𝑚𝑝ℎ)2 = 41.91 𝑝𝑠𝑓

Using Excel to find the velocity pressure at each value of kz gives:

z (ft) Kz qz (psf)

98 0.98 41.91

83 0.94 39.96

68 0.89 37.75

53 0.82 35.16

38 0.75 31.97

20 0.62 26.61

6) Applying the equation 𝑞𝑧(𝐺𝐶𝑝) − 𝑞𝑧(𝐺𝐶𝑃𝑖) and taking the highest negative value:

For the joist, use the height of 98 ft:

Zone 1: 𝑝 = {41.91𝑝𝑠𝑓(−0.9) − 41.91𝑝𝑠𝑓(. 18) = −𝟒𝟓. 𝟑𝒑𝒔𝒇

41.91𝑝𝑠𝑓(−0.9) − 41.91𝑝𝑠𝑓(−.18) = −30.2𝑝𝑠𝑓

Using Excel to ease the calculations gives:

Zone Load with GCP+ (psf) Load with GCP- (psf) Design loads (psf)

1 -45.3 -30.2 -45.3 16

2 -74.6 -59.5 -74.6 16

3 -103.9 -88.8 -103.9 16

For the deck, use the height of 98 ft:

Zone Load with GCP+ (psf)

Load with GCP- (psf) Design loads (psf)

1 -66.2 -51.1 -66.2 16

2 -103.9 -88.8 -103.9 16

3 -141.6 -126.6 -141.6 16

For the concrete walls:

Level 1, use a height of 20 ft:





Zone 4&5 13.8 23.4 N/A 23.4

Zone 4 -25.8 -16.2 -25.8 N/A

Zone 5 -40.7 -31.1 -40.7 N/A

Level 2, use a height of 38 ft:



Zone 4&5 18.2 29.7 N/A 29.7

Zone 4 -31.3 -19.8 -31.3 N/A

Zone 5 -50.5 -39.0 -50.5 N/A

Fill out the following tables for the tributary areas and zones shown. Show plus and minus

values. Keep these numbers for the semester. You will use them to design members exposed to

wind and will need to place these tables on your design drawings.

SOLUTIONS:

1) Using ASCE 7, the following information can be found:

Components and cladding: Roof GCp

Tributary area (sq. ft) Zone 1 Zone 2 Zone 3

10 -1.4 -2.3 -3.2

20 -1.22 -2.18 -3.05

50 -1.2 -2 -2.85

100 -1.1 -1.9 -2.68

Components and cladding: Wall GCp

Tributary area (sq. ft) Zone 4 Zone 5

10 0.9 -0.9 0.9 -1.8

20 0.9 -0.9 0.9 -1.8

50 0.8 -0.85 0.8 -1.57

100 0.75 -0.8 0.75 -1.4

200 0.7 -0.78 0.7 -1.21

500 0.6 -0.7 0.6 -1



2) Calculating the pressures in excel in the same manner as problem #3 gives the following

design pressures:

Components and cladding: Roof wind pressures (psf)

Tributary area (sq. ft) Zone 1 Zone 2 Zone 3

10 -66.2 16.0 -103.9 16.0 -141.6 16.0

20 -58.7 16.0 -98.9 16.0 -135.4 16.0

50 -57.8 16.0 -91.4 16.0 -127.0 16.0

100 -53.6 16.0 -87.2 16.0 -119.8 16.0

Components and cladding: Wall wind pressures (psf) at level 1


10 -28.7 28.7 -52.7 28.7

20 -28.7 28.7 -52.7 28.7

50 -27.4 26.1 -46.6 26.1

100 -26.1 24.7 -42.0 24.7

200 -25.5 23.4 -37.0 23.4

500 -23.4 20.8 -31.4 20.8

Components and cladding: Wall wind pressures (psf) at level 2


10 -34.5 34.5 -63.3 34.5

20 -34.5 34.5 -63.3 34.5

50 -32.9 31.3 -55.9 31.3

100 -31.3 29.7 -50.5 29.7

200 -30.7 28.1 -44.4 28.1

500 -28.1 24.9 -37.7 24.9

ROOF DECK DESIGN Pg 17


ROOF DECK DESIGN

Use a span of 5 feet and a two span condition. Do not use less than 22 gauge for safer/easier

construction. Use the highest value from all zones for the wind data for easier

design/construction and standardization (this is zone 3).

KNOWN/GIVEN VALUES:

𝐿𝑟 = 20 𝑝𝑠𝑓

𝑊 = −141.6 𝑝𝑠𝑓, 16 𝑝𝑠𝑓

𝐴𝑇 = 5 𝑓𝑡 ∗ 50 𝑓𝑡 = 250 𝑓𝑡2

𝐷 = 30 𝑝𝑠𝑓 (𝑛𝑜𝑡 𝑖𝑛𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡)

LIVE LOAD REDUCTIONS:

Use ASCE 7-10 Section 4.8

𝐿𝑟 = 𝐿0𝑅1𝑅2

12 ≤ 𝐿𝑟 = 20 𝑝𝑠𝑓 ≤ 20 𝑝𝑠𝑓 → 𝑂𝐾

𝑅1 = 1.2 − 0.001𝐴𝑇 = 1.2 − 0.001(250 𝑓𝑡2) = .95 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 200 𝑓𝑡2 < 𝐴𝑇 < 400 𝑓𝑡

2

𝐹 = .25 𝑖𝑛𝑐ℎ𝑒𝑠 𝑜𝑓 𝑟𝑖𝑠𝑒/𝑓𝑜𝑜𝑡 𝑅2 = 1

Therefore, 𝐿𝑟 = (20 𝑝𝑠𝑓)(. 95)(1) = 19 𝑝𝑠𝑓

DESIGN:

Assume a dead self-weight of 2.5 psf.

Therefore, 𝐷 = 30 𝑝𝑠𝑓 + 2.5 𝑝𝑠𝑓 = 32.5 𝑝𝑠𝑓

Finding ASD load combinations from an Excel File:

[LOAD COMBINATIONS FOUND ON NEXT PAGE]



Using Vulcraft manual, try a 1.5B22. This meets the max SDI Construction Span of 6’-11” and

the allowable load of 100 psf and deflection limit live load of 213 psf. Check using the actual

self-weight of 1.78 psf:

The deck’s capacity still meets the required load using the actual self-weight.

CHECK UPLIFT:



𝐹𝑏 = 0.6𝐹𝑦 = 0.6(33 𝑘𝑠𝑖) = 19.8 𝑘𝑠𝑖

𝑀 =𝑤𝑙2

8=65.89 𝑝𝑠𝑓 ∗ (5 𝑓𝑡)2

8= 2.47

𝑘𝑖𝑝 ∗ 𝑖𝑛

𝑓𝑡

𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 =𝑀

𝐹𝑏=2.47

𝑘𝑖𝑝 ∗ 𝑖𝑛𝑓𝑡

19.8 𝑘𝑠𝑖= 0.125

𝑖𝑛3

𝑓𝑡

This meets the section modulus of a 1.5B22 (𝑆𝑛 = .192𝑖𝑛3

𝑓𝑡).

CHECK UPLIFT DEFLECTION:

∆𝑚𝑎𝑥=𝐿

240=5 𝑓𝑡 ∗ 12 𝑖𝑛

240= 0.25 𝑖𝑛

∆= 0.0052𝑤𝑙4

𝐸𝐼

𝐼𝑚𝑖𝑛 = 0.0052𝑤𝑙4

𝐸∆𝑚𝑎𝑥= 0.0052

65.89 𝑙𝑏𝑓𝑡2

∗1 𝑘𝑖𝑝1000 𝑙𝑏

∗ (𝑓𝑡12 𝑖𝑛)

2

(5 𝑓𝑡 ∗ 12 𝑖𝑛)4

29000 𝑘𝑠𝑖 ∗ .25 𝑖𝑛= 0.004

𝑖𝑛4

𝑓𝑡

This minimum I meets the I of the deck, which is 0.184 𝑖𝑛4

𝑓𝑡→ 𝑂𝐾

THEREFORE USE A 1.5B22 ROOF DECK.

ROOF JOIST DESIGN Pg 20


ROOF JOIST DESIGN

Use a span of 50 ft. and a spacing of 5 ft. Use ASD and do not design for uplift as this will be

completed by the joist engineer. Use zone 3 from the C&C calculations for standardization,

repetition, and easier construction.

KNOWN/GIVEN VALUES:

𝐿𝑟 = 20 𝑝𝑠𝑓 ∗ 5 𝑓𝑡 = 100 𝑙𝑏/𝑓𝑡

𝑊 = −103.9 𝑝𝑠𝑓 ∗ 5 𝑓𝑡 = −519.5𝑙𝑏

𝑓𝑡, 16 𝑝𝑠𝑓 ∗ 5 𝑓𝑡 = 80 𝑙𝑏/𝑓𝑡

𝐴𝑇 = 5 𝑓𝑡 ∗ 50 𝑓𝑡 = 250 𝑓𝑡2

𝐷 = 30 𝑝𝑠𝑓 + 1.78 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 𝑑𝑒𝑐𝑘 = 31.78 𝑝𝑠𝑓 ∗ 5𝑓𝑡= 158.9 𝑙𝑏/𝑓𝑡 (𝑛𝑜𝑡 𝑖𝑛𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡)




12 ≤ 𝐿𝑟 = 20 𝑝𝑠𝑓 ≤ 20 𝑝𝑠𝑓 → 𝑂𝐾


2


Therefore, 𝐿𝑟 = (20 𝑝𝑠𝑓)(. 95)(1) = 19 𝑝𝑠𝑓 ∗ 5𝑓𝑡 = 95 𝑙𝑏/𝑓𝑡

DESIGN:

Assume the joist has a self-weight of 7.5 lb/ft.

Using an Excel file to find the ASD load combinations gives a dead weight of 158.9 lb/ft + 7.5

lb/ft =166.4 lb/ft.

[LOAD COMBINATIONS FOUND ON NEX PAGE]



From the Vulcraft manual, consider a 30K11 because it is the lightest joist that can handle these

loads.

Factoring in the actual self-weight of a 30K11 in ASD load combinations gives a new dead load

of 158.9 lb/ft + 16.4 lb/ft = 175.3 lb/ft and the following results:




The 30K11’s capacity of 333 lb/ft is greater than the 282.55 lb/ft from the load combinations.

Also, the deflection causing load maximum is 190 lb/ft which is greater than the 95 lb/ft.

BRIDGING REQUIREMENTS:

From page 16 in the Vulcraft manual, a 30K11 requires two rows of top chord diagonal bridging.

THEREFORE USE 30K11.

ROOF JOIST GIRDER DESIGN Pg 23


ROOF JOIST GIRDER DESIGN

Use a span of 50 ft. and a spacing of 50 ft. Use ASD and zone 3 C&C values for the joist for the

sake of standardization, repetition, and easier construction.

KNOWN/GIVEN VALUES:

𝐴𝑇 =(50 𝑓𝑡 + 50 𝑓𝑡)

2∗50

10= 250 𝑓𝑡2

𝐿𝑟 = 20 𝑝𝑠𝑓 ∗ 250 𝑓𝑡2 = 5 𝑘𝑖𝑝

𝑊 = −103.9 𝑝𝑠𝑓 ∗ 250 𝑓𝑡2 = −26 𝑘𝑖𝑝, 16 𝑝𝑠𝑓 ∗ 250 𝑓𝑡2 = 4 𝑘𝑖𝑝

𝐷 = 30 𝑝𝑠𝑓 + 1.78 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 𝑑𝑒𝑐𝑘

= ((31.78 𝑝𝑠𝑓 ∗ 50𝑓𝑡) + 16.4𝑙𝑏

𝑓𝑡 𝑗𝑜𝑖𝑠𝑡 𝑤𝑒𝑖𝑔ℎ𝑡) ∗ 5𝑓𝑡 = 8.03 𝑘𝑖𝑝




12 ≤ 𝐿𝑟 = 20 𝑝𝑠𝑓 ≤ 20 𝑝𝑠𝑓 → 𝑂𝐾


2


Therefore, 𝐿𝑟 = (20 𝑝𝑠𝑓)(. 95)(1) = 19 𝑝𝑠𝑓 ∗ 250𝑓𝑡2 = 4.75 𝑘𝑖𝑝

DESIGN:

Choose a depth of 50 inches because the span is 50 feet.

There are 10 spaces so use 10N.

Using an Excel file to find the ASD load combinations gives:


ROOF JOIST GIRDER DESIGN Pg 24


Do not design for uplift since this will be taken care of by the joist engineer. Therefore, the Joist

Girder designation is 50G10N13.6K.

Check the joist girder’s deflection:

𝐼 = 0.027𝑁𝑃𝐿𝑑 = 0.027(10)(13.6)(50)(50) = 9180 𝑖𝑛4

i c ki Ni PiNi

1 0.5 0.1 0.0370 0.5032

2 0.5 0.2 0.0710 0.9656

3 0.5 0.3 0.0990 1.3464

4 0.5 0.4 0.1180 1.6048

5 0.5 0.5 0.1250 1.7000

6 0.5 0.6 0.1180 1.6048

7 0.5 0.7 0.0990 1.3464

8 0.5 0.8 0.0710 0.9656

9 0.5 0.9 0.0375 0.5032

Σ(PiNi)= 10.54

𝐷 = 1.15∑(𝑃𝑖𝑁𝑖)𝑙

3

6𝐸𝐼= 1.15

10.54(50 𝑓𝑡 ∗ 12 𝑖𝑛)3

6(29000 𝑘𝑠𝑖)(9180 𝑖𝑛4)= 1.43 𝑖𝑛

𝐷𝑚𝑎𝑥 =𝐿

360=50 𝑓𝑡 ∗ 12 𝑖𝑛

360= 1.667 𝑖𝑛 > 1.43 𝑖𝑛 → 𝑂𝐾

THEREFORE USE 50G10N13.6K. THIS JOIST GIRDER HAS AN APPROXIMATE

SELF-WEIGHT OF 78 LB/FT.

ROOF PERIMETER BEAM DESIGN Pg 25


ROOF PERIMETER BEAM DESIGNS:

Use a wide flange that is braced in the top flange every 5 ft by the joists and is completely

unbraced in the bottom flange for the east to west beams. Use simply supported beams with a

length of 25 feet each.

Use a wide flange that is fully braced in the top flange by the roof deck and is completely

unbraced in the bottom flange for the north to south beams. Use simply supported beams with a

length of 25 feet each.

For both beam directions, use the following maximum deflections:

𝐷𝑒𝑎𝑑 + 𝐿𝑖𝑣𝑒 +𝑊𝑖𝑛𝑑 =𝐿

240

𝐿𝑖𝑣𝑒 +𝑊𝑖𝑛𝑑 𝑜𝑛𝑙𝑦 =𝐿

360𝑜𝑟 1.5"

Also, use AISC Manual Table 2-4 ASTM A992 for material properties:

𝐹𝑦 = 50 𝑘𝑠𝑖, 𝐹𝑢 = 65 𝑘𝑠𝑖

GIVEN/KNOWN VALUES:

For the east to west beams:

𝐴𝑇 =(50 𝑓𝑡)

2∗50

10= 125 𝑓𝑡2

𝐿𝑟 = 20 𝑝𝑠𝑓 ∗ 125𝑓𝑡2 = 2.5 𝑘𝑖𝑝

𝑊 = −103.9 𝑝𝑠𝑓 ∗ 125 𝑓𝑡2 = −13 𝑘𝑖𝑝, 16 𝑝𝑠𝑓 ∗ 125 𝑓𝑡2 = 2 𝑘𝑖𝑝

𝐷 = [(30 𝑝𝑠𝑓 + 1.78 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 𝑑𝑒𝑐𝑘) (50 𝑓𝑡

2) + (16.4

𝑙𝑏

𝑓𝑡 𝑓𝑟𝑜𝑚 𝑗𝑜𝑖𝑠𝑡𝑠)] (5 𝑓𝑡)

= 4.05 𝑘 (𝑒𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡)

For the north to south beams:

𝐴𝑇 =(5 𝑓𝑡)

2∗100

4= 62.5 𝑓𝑡2

𝐿𝑟 = 20 𝑝𝑠𝑓 ∗5

2𝑓𝑡 = .05 𝑘/𝑓𝑡

𝑊 = −103.9 𝑝𝑠𝑓 ∗ 2.5 𝑓𝑡 = −.26𝑘

𝑓𝑡, 16 𝑝𝑠𝑓 ∗ 2.5 𝑓𝑡2 = .04 𝑘/𝑓𝑡

𝐷 = [(30 𝑝𝑠𝑓 + 1.78 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 𝑑𝑒𝑐𝑘) (5 𝑓𝑡

2)] = 0.08

𝑘

𝑓𝑡(𝑒𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡)




12 ≤ 𝐿𝑟 = 20 𝑝𝑠𝑓 ≤ 20 𝑝𝑠𝑓 → 𝑂𝐾

𝑅1 = 1 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝐴𝑇 < 200 𝑓𝑡2




Therefore, 𝐿𝑟 = (20 𝑝𝑠𝑓)(1)(1) = 20 𝑝𝑠𝑓 ∗ 2.5 𝑓𝑡2 = .05 𝑘/𝑓𝑡

DESIGN OF NORTH-SOUTH BEAMS:

Visual Analysis was used to analyze the north to south 25 ft. beams. The following load cases

were used:

Dead load with an assumed self-weight of 100lb/ft.:

Live roof load:

Wind in the positive direction:

Wind in the negative direction (uplift):

ASD load combination 𝐷 + .75(𝐿 + .6𝑊 + 𝐿𝑟) gives the maximum positive moment:



ASD load combination . 6𝐷 + .6𝑊 gives the largest negative moment:

Check uplift first because this seems to control the design, and try a W12x40 (increasing the size

from the previous W10x22). Checking LTB:

𝐹𝑐𝑟 =𝐶𝑏𝜋

2𝐸

(𝐿𝑏𝑟𝑡𝑠)2√1 + .078

𝐽𝑐

𝑆𝑥ℎ0(𝐿𝑏𝑟𝑡𝑠)2

=1.14𝜋2(29000)

(25 ∗ 122.21 )

2√1 + .078

. 906 ∗ 1

51.5 ∗ 11.4(25 ∗ 12

2.21)2

= 31.8 𝑘𝑠𝑖 𝑀𝑛Ω=𝐹𝑐𝑟𝑆𝑥Ω

=31.8 ∗ 51.5

1.67= 81.7 𝑘 ∗ 𝑓𝑡 > 3.75 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾

Finding the max positive moment using its actual self-weight of 40 lb/ft and the controlling ASD

load combinations mentioned previously in visual analysis gives:



The beam’s Zone I capacity of 𝑀𝑛

Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗57𝑖𝑛3

1.67= 142.2 𝑘 ∗ 𝑓𝑡 > 13.7 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾

Finding the max negative moment using its actual self-weight of 40 lb/ft and the controlling

ASD load combinations mentioned previously in visual analysis gives:

The beam’s Zone III capacity meets the demand:

𝑀𝑛Ω=𝐹𝑐𝑟𝑆𝑥Ω

=31.8 ∗ 51.5

1.67= 81.7 ∗ 𝑓𝑡 > 6.6 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾

Running visual analysis to find the maximum shear values gives:

These values meet the beam’s shear capacity found in the AISC manual pg. 3-26 of 70.2k:

𝑉𝑛Ω= 70.2𝑘 > 2.2𝑘 → 𝑂𝐾

𝑉𝑛Ω= 70.2𝑘 > 1.05𝑘 → 𝑂𝐾

Finding the deflections from Visual Analysis:

For 𝐷 + 𝐿𝑟 +𝑊:



These deflections meet the requirements:

𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > 0.09 𝑖𝑛 → 𝑂𝐾

𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > 0.17 𝑖𝑛 → 𝑂𝐾

For 𝐿𝑟 +𝑊:



These deflections meet the requirements: 𝐿

360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= 0.833 𝑖𝑛 > 0.21 𝑖𝑛 → 𝑂𝐾

𝐿

360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= 0.833 𝑖𝑛 > 0.09 𝑖𝑛 → 𝑂𝐾

THEREFORE USE A W12X40 FOR THE NORTH TO SOUTH PERIMETER BEAMS.

FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 2.2K.

DESIGN OF EAST-WEST BEAMS:

Visual Analysis was used to analyze the east to west 25 ft. beams. The following load cases

were used:


Live roof load:

Wind in the positive direction:



Wind in the negative direction (uplift):

ASD load combination 𝐷 + .75(𝐿 + .6𝑊 + 𝐿𝑟) gives the maximum positive moment:

ASD load combination . 6𝐷 + .6𝑊 gives the largest negative moment:

The top flange is braced at every 5 feet. Because a W10 is too small, try a W12x87.

Checking uplift:

𝐿𝑝 = 10.8 𝑓𝑡 < 𝐿𝑏 = 25 𝑓𝑡 < 𝐿𝑟 = 43.1 𝑓𝑡 → 𝑍𝑜𝑛𝑒 𝐼𝐼 Checking LTB:

32.9 𝑘𝑠𝑖 𝑀𝑛Ω= 𝐶𝑏 [

𝑀𝑝

Ω−𝐵𝐹

Ω(𝐿𝑏 − 𝐿𝑝)] = 1.0[329 − 3.81(25 − 10.8)] = 275 𝑘 ∗ 𝑓𝑡 > 76.1 𝑘 ∗ 𝑓𝑡

→ 𝑂𝐾





The beam’s Zone I capacity:

𝑀𝑛

Ω=𝐹𝑦𝑍𝑥


1.67= 329 𝑘 ∗ 𝑓𝑡 > 110 𝑘 ∗ 𝑓𝑡 → 𝐺𝑜𝑜𝑑

Finding the max negative moment using its actual self-weight of 78 lb/ft and the controlling

ASD load combinations mentioned previously in visual analysis gives:

The beam’s Zone II capacity meets the demand:

𝑀𝑛Ω= 𝐶𝑏 [

𝑀𝑝

Ω−𝐵𝐹

Ω(𝐿𝑏 − 𝐿𝑝)] = 1.0[329 − 3.81(25 − 10.8)] = 275 𝑘 ∗ 𝑓𝑡 > 76.1 𝑘 ∗ 𝑓𝑡

Checking Deflection:

For Lr+W:



Check the selected beam for deflection:

𝐿

360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .833 𝑖𝑛 > .832 𝑖𝑛 → 𝐺𝑜𝑜𝑑

𝐿

360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .833 𝑖𝑛 > .36 𝑖𝑛 → 𝐺𝑜𝑜𝑑

For D+Lr+W:

Check the selected beam for deflection:



𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > .48 𝑖𝑛 → 𝐺𝑜𝑜𝑑

𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > .71 𝑖𝑛 → 𝐺𝑜𝑜𝑑

Check the selected beam for maximum shear:

These values meet the beam’s shear capacity found in the AISC manual pg. 3-24 of 129k: 𝑉𝑛Ω= 129𝑘 > 21.6𝑘 → 𝑂𝐾

𝑉𝑛Ω= 129𝑘 > 15.5𝑘 → 𝑂𝐾

THEREFORE USE A W12X87 FOR THE EAST TO WEST PERIMETER BEAMS.

FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 21.6 K

COMPOSITE FLOOR DECK DESIGN Pg 35


COMPOSITE FLOOR DECK DESIGN

Note: All designs for the floor are the same for the 4th

, 5th

, and 6th

floors.

FOR THE AREAS WITHOUT THE HOLES FOR THE STAIR WELLS/ELEVATORS:

Design a composite deck with normal weight concrete with a span of 12.5 ft and a two span

condition.

GIVEN/KNOWN VALUES:

𝐷 = 9.5 𝑝𝑠𝑓 (𝑛𝑜𝑡 𝑖𝑛𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 − 𝑤𝑒𝑖𝑔ℎ𝑡) 𝐿 = 65 𝑝𝑠𝑓

𝐴𝑇 = 12.5 𝑓𝑡 ∗ 25 𝑓𝑡 = 312.5 𝑓𝑡2


From ASCE 7-10 Table 4-2, 𝐾𝐿𝐿 = 1

Therefore, 𝐴𝑇𝐾𝐿𝐿 = 312.5𝑓𝑡2 ∗ 1 < 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

DESIGN:

Factoring the loads using ASD and finding the superimposed load:

From page 54 in the Vulcraft Manual, choose a total slab depth of 5.5 inch and deck type

3VLI19. This deck satisfies the following checks:

𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦: 74.50 𝑝𝑠𝑓 < 90 𝑝𝑠𝑓 → 𝑂𝐾

𝑀𝑎𝑥 𝑆𝐷𝐼 𝑆𝑝𝑎𝑛: 12.5 𝑓𝑡 < 13 𝑓𝑡 8 𝑖𝑛 → 𝑂𝐾



For future reference, this deck’s self weight is 2.5 psf from the steel and 51 psf from the

concrete, which totals to 53.5 psf.

THEREFORE US A 3VLI19 WITH A SELF WEIGHT OF 53.5 PSF AND TOTAL SLAB

DEPTH OF 5.50 INCHES.

FOR SOME OF THE AREA AFFECTED BY THE ELEVATOR:

Design a composite deck with normal weight concrete with a span of 7.75 ft and a one span

condition.

GIVEN/KNOWN VALUES:


𝐴𝑇 = 7.75 𝑓𝑡 ∗ 25𝑓𝑡 = 193.75 𝑓𝑡2



Therefore, 𝐴𝑇𝐾𝐿𝐿 = 193.75 𝑓𝑡2 ∗ 1 < 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

DESIGN:


From page 54 in the Vulcraft Manual, check to see if the previously chosen deck works. If this

deck works, although overdesigned, construction will be simplified for the contractor. Also,



money should not be much of an issue because there are not many areas which will be

overdesigned like this and labor will not be escalated for the installation of different sized decks.

Therefore, for a 3VLI19 with a span of 7.75 ft and by interpolation:

𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 270 − (270 − 244)7.75 − 7.5

8 − 7.5= 257 𝑝𝑠𝑓

𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦: 74.50 𝑝𝑠𝑓 < 257 𝑝𝑠𝑓 → 𝑂𝐾

𝑀𝑎𝑥 𝑆𝐷𝐼 𝑆𝑝𝑎𝑛: 5 𝑓𝑡 < 11 𝑓𝑡 4 𝑖𝑛 → 𝑂𝐾 For future reference, this deck’s self-weight is 2.5 psf from the steel and 51 psf from the




FOR THE REST OF THE AREA AFFECTED BY THE ELEVATOR:

Design a composite deck with normal weight concrete with a span of 4.75 ft and a two span

condition.

GIVEN/KNOWN VALUES:


𝐴𝑇 = 4.75 𝑓𝑡 ∗ 16𝑓𝑡 = 76 𝑓𝑡2



Therefore, 𝐴𝑇𝐾𝐿𝐿 = 76 𝑓𝑡2 ∗ 1 < 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

DESIGN:







money should not be much of an issue because there are not many areas that will be


Therefore, for a 3VLI19 with a span of 4.75 ft and by extrapolation:

𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 302 − (302 − 0)7 − 4.75

7 − 0= 204.9 𝑝𝑠𝑓

𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦: 74.50 𝑝𝑠𝑓 < 204.9 𝑝𝑠𝑓 → 𝑂𝐾

𝑀𝑎𝑥 𝑆𝐷𝐼 𝑆𝑝𝑎𝑛: 5 𝑓𝑡 < 13 𝑓𝑡 8 𝑖𝑛 → 𝑂𝐾 For future reference, this deck’s self weight is 2.5 psf from the steel and 51 psf from the




FOR AREA AFFECTED BY THE STAIRWELL:

Design a composite deck with normal weight concrete with a span of 8.85 ft and a one span

condition.

GIVEN/KNOWN VALUES:


𝐴𝑇 = 8.85 𝑓𝑡 ∗ 25𝑓𝑡 = 221.25 𝑓𝑡2





Therefore, 𝐴𝑇𝐾𝐿𝐿 = 221.15 𝑓𝑡2 ∗ 1 < 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

DESIGN:




money should not be much of an issue because there are not many areas that will be


Therefore, for a 3VLI19 with a span of 8.85 ft and by interpolation:

𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 222 − (222 − 203)8.85 − 8.5

9 − 8.5= 208.7 𝑝𝑠𝑓

𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦: 74.50 𝑝𝑠𝑓 < 208.7 𝑝𝑠𝑓 → 𝑂𝐾

𝑀𝑎𝑥 𝑆𝐷𝐼 𝑆𝑝𝑎𝑛: 5 𝑓𝑡 < 11 𝑓𝑡 4 𝑖𝑛 → 𝑂𝐾 For future reference, this deck’s self weight is 2.5 psf from the steel and 51 psf from the




FLOOR BEAM DESIGN Pg 40


FLOOR BEAM DESIGN

FOR THE OUTSIDE STAIRWELL FLOOR BEAM:

Use a wide flange that is fully braced in the top flange by the floor deck and is completely

unbraced in the bottom flange. Use a simply supported beam with a length of 25 feet.

Use the following maximum deflections:


240


360𝑜𝑟 1.5"



GIVEN/KNOWN VALUES:

𝐴𝑇 =(8.85 𝑓𝑡)

2∗ 25 𝑓𝑡 = 110.625 𝑓𝑡2

𝐿 = 65 𝑝𝑠𝑓 ∗8.85 𝑓𝑡

2= .287625 𝑘/𝑓𝑡

𝐷 = [(9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟 𝑑𝑒𝑐𝑘 + 8 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑤𝑎𝑙𝑙 )]8.85 𝑓𝑡

2= .314175 𝑘/𝑓𝑡 (𝑒𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡)



Therefore, 𝐴𝑇𝐾𝐿𝐿 = 110.625 𝑓𝑡2 ∗ 2 = 221.25 < 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

DESIGN OF BEAM:

Visual Analysis was used to analyze the beam. The following load cases were used:


Live load:



ASD load combination 𝐷 + 𝐿 gives the maximum positive moment:

The top flange is fully braced, so use the plastic moment to find the required section modulus:

𝑍𝑟𝑒𝑞 =𝑀𝑢Ω

𝐹𝑦=(54.828 𝑘 ∗ 𝑓𝑡)(1.67) ∗ 12𝑖𝑛

50 𝑘𝑠𝑖= 21.98 𝑖𝑛3

From the AISC manual, table 3-2, try a W12X19.


Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗24.7𝑖𝑛3

1.67= 61.63 𝑘 ∗ 𝑓𝑡 > 54.9 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾

Test the beam using its self-weight of 19lb/ft instead of the assumed weight as before. The max

moment found in visual analysis is:

The beam’s Zone I capacity meets the demand:

𝑀𝑛Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖 ∗ 24.7𝑖𝑛3

1.67= 61.63 𝑘 ∗ 𝑓𝑡 > 48.5 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾




These values meet the beam’s shear capacity found in the AISC manual pg. 3-26 of 57.3k: 𝑉𝑛Ω= 57.3𝑘 > 7.76𝑘 → 𝑂𝐾


For 𝐷 + 𝐿:

These deflections do not meet the requirements:

𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 < 1.4472 𝑖𝑛 → 𝑁𝑜𝑡 𝑔𝑜𝑜𝑑

Retry with a W14X26.




Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗40.2𝑖𝑛3

1.67= 100.3 𝑘 ∗ 𝑓𝑡 > 49.06 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾






For 𝐷 + 𝐿:


𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > 0.777 𝑖𝑛 → 𝑂𝐾

For 𝐿:


𝐿

360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .883 𝑖𝑛 > 0.356 𝑖𝑛 → 𝑂𝐾

THEREFORE USE A W14X26 FOR THE OUTSIDE STAIRWELL BEAM.




FOR THE INSIDE STAIRWELL BEAM:





240


360𝑜𝑟 1.5"



GIVEN/KNOWN VALUES:

𝐴𝑇 =(8.85 + 12.5𝑓𝑡)

2∗ 25 𝑓𝑡 = 266.875 𝑓𝑡2

𝐿 = 65 𝑝𝑠𝑓 ∗(8.85 𝑓𝑡 + 12.5 𝑓𝑡)

2= .694 𝑘/𝑓𝑡

𝐷 = [(9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟 𝑑𝑒𝑐𝑘)](8.85 𝑓𝑡 + 12.5𝑓𝑡)




Therefore, 𝐴𝑇𝐾𝐿𝐿 = 266.875 𝑓𝑡2 ∗ 2 = 533.75 > 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

𝐿 = 𝐿0 (0.25 +15

√𝐾𝐿𝐿𝐴𝑇)

𝐿 = .694 (0.25 +15

√2 ∗ 266.875) = .624

𝑘

𝑓𝑡

. 624𝑘

𝑓𝑡> .5 ∗ .694 = .347 → 𝑂𝐾

DESIGN OF BEAM:



Live load:






𝐹𝑦=(109.14 𝑘 ∗ 𝑓𝑡)(1.67) ∗ 12𝑖𝑛

50 𝑘𝑠𝑖= 43.74 𝑖𝑛3



Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗47.3𝑖𝑛3

1.67= 118.01 𝑘 ∗ 𝑓𝑡 > 109.14 𝑘 ∗ 𝑓𝑡 →

𝑂𝐾 Test the beam using its self-weight of 30lb/ft instead of the assumed weight as before. The max



𝑀𝑛Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖 ∗ 47.3𝑖𝑛3

1.67= 118.01 𝑘 ∗ 𝑓𝑡 > 103.68 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For 𝐷 + 𝐿:



These deflections do not meet the requirements:

𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 < 1.3822 𝑖𝑛 → 𝑁𝑜𝑡 𝑔𝑜𝑜𝑑

Retry with a W16X31.




Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗54 𝑖𝑛3

1.67= 134.7 𝑘 ∗ 𝑓𝑡 > 103.8 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾




For 𝐷 + 𝐿:




𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > 1.07 𝑖𝑛 → 𝑂𝐾

For 𝐿:


𝐿

360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .883 𝑖𝑛 > 0.504 𝑖𝑛 → 𝑂𝐾

THEREFORE USE A W16X31 FOR THE INSIDE STAIRWELL BEAM.


FOR THE INSIDE MOST ELEVATOR BEAM:





240


360𝑜𝑟 1.5"



GIVEN/KNOWN VALUES:

𝐴𝑇 =(4.75 + 4.75𝑓𝑡)

2∗ 16 𝑓𝑡 = 76 𝑓𝑡2



𝐿 = 65 𝑝𝑠𝑓 ∗(4.75 𝑓𝑡 + 4.75 𝑓𝑡)

2= .30875 𝑘/𝑓𝑡

𝐷 = [(9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟 𝑑𝑒𝑐𝑘)](4.75 𝑓𝑡 + 4.75 𝑓𝑡)




Therefore, 𝐴𝑇𝐾𝐿𝐿 = 76 𝑓𝑡2 ∗ 2 = 152 𝑓𝑡2 < 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

DESIGN OF BEAM:



Live load:




𝐹𝑦=(22.656 𝑘 ∗ 𝑓𝑡)(1.67) ∗ 12𝑖𝑛

50 𝑘𝑠𝑖= 8.92 𝑖𝑛3



Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗20.1𝑖𝑛3

1.67= 50.15 𝑘 ∗ 𝑓𝑡 > 22.66 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾

Test the beam using its self-weight of 16 lb/ft instead of the assumed weight as before. The max





𝑀𝑛Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖 ∗ 20.1𝑖𝑛3

1.67= 50.15 𝑘 ∗ 𝑓𝑡 > 19.97 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For 𝐷 + 𝐿:

This deflections meets the requirement:

𝐿

240=16 𝑓𝑡 ∗ 12 𝑖𝑛

240= .8 𝑖𝑛 > .308 𝑖𝑛 → 𝑂𝐾

For L:

This deflections meets the requirement:

𝐿

360=16 𝑓𝑡 ∗ 12 𝑖𝑛

360= .533 𝑖𝑛 > .152 𝑖𝑛 → 𝑂𝐾



𝑉𝑛Ω= 52.8𝑘 > 4.99𝑘 → 𝑂𝐾

THEREFORE USE A W12X16 FOR THE MOST INSIDE ELEVATOR BEAM.




FOR THE SMALL GIRDER BY THE ELEVATOR:

Use a wide flange that is fully braced in the top flange by the steel deck. Use a simply supported

beam with a length of 9.5 feet.



240


360𝑜𝑟 1.5"



KNOWN/GIVEN VALUES:

𝐿 = 65 𝑝𝑠𝑓 ∗ 8 𝑓𝑡 = .52 𝑘/𝑓𝑡

𝐴𝑇 =(16 𝑓𝑡)

2∗ 9.5 𝑓𝑡 = 76 𝑓𝑡2

𝐷 = [(9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟 𝑑𝑒𝑐𝑘)](8 𝑓𝑡) = .504 𝑘/𝑓𝑡 (𝑒𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡)

𝑃𝐷 = ((9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓) ∗ 8 𝑓𝑡 ∗ 4.75 𝑓𝑡) + 16𝑙𝑏

𝑓𝑡∗ 8𝑓𝑡 = 2.52 𝑘𝑖𝑝

𝑃𝐿 = 65 𝑝𝑠𝑓 ∗ 8 𝑓𝑡 ∗ 4.75 𝑓𝑡 = 2.47 𝑘𝑖𝑝



Therefore, 𝐴𝑇𝐾𝐿𝐿 = 76𝑓𝑡2 ∗ 2 = 152 𝑓𝑡2 < 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

DESIGN OF BEAM:


Dead load with an assumed self-weight of 100 lb/ft.:

Live load:






𝐹𝑦=(24.5 𝑘 ∗ 𝑓𝑡)(1.67) ∗ 12𝑖𝑛

50 𝑘𝑠𝑖= 9.82 𝑖𝑛3

From the AISC manual, table 3-2, try a W18X55. Using the actual self-weight gives:


Ω=𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗112 𝑖𝑛3

1.67= 279.4 𝑘 ∗ 𝑓𝑡 > 24 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For 𝐷 + 𝐿:



This deflection meets the requirements:

𝐿

240=9.5 𝑓𝑡 ∗ 12 𝑖𝑛

240= .475 𝑖𝑛 > .01 𝑖𝑛 → 𝑂𝐾

For 𝐿:

This deflection meets the requirement: 𝐿

360=9.5 𝑓𝑡 ∗ 12 𝑖𝑛

360= .32 𝑖𝑛 > .007 𝑖𝑛 → 𝐺𝑜𝑜𝑑





THEREFORE USE A W18X55 FOR THE SMALL ELEVATOR GIRDER. FOR THE

CONNECTION FORCE, USE GRAVITY SHEAR AT 7.62K.

FOR THE MIDDLE ELEVATOR BEAMS:





240


360𝑜𝑟 1.5"



GIVEN/KNOWN VALUES:

𝐴𝑇 =(4.75 + 7.75𝑓𝑡)

2∗ 25 𝑓𝑡 = 156.25 𝑓𝑡2

𝐿 = 65 𝑝𝑠𝑓 ∗(4.75 𝑓𝑡 + 7.75 𝑓𝑡)

2= .41 𝑘/𝑓𝑡



𝑃𝐷 = 3.92 𝑘 𝑓𝑟𝑜𝑚 𝑔𝑖𝑟𝑑𝑒𝑟 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛

𝑃𝐿 = 3.71 𝑘 𝑓𝑟𝑜𝑚 𝑔𝑖𝑟𝑑𝑒𝑟 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛



Therefore, 𝐴𝑇𝐾𝐿𝐿 = 156.25 𝑓𝑡2 ∗ 2 = 312.5 𝑓𝑡2 < 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

DESIGN OF BEAM:



Live load:






𝐹𝑦=(109.6 𝑘 ∗ 𝑓𝑡)(1.67) ∗ 12𝑖𝑛

50 𝑘𝑠𝑖= 43.9 𝑖𝑛3



Ω==

𝐹𝑦𝑍𝑥


1.67= 134.7 𝑘 ∗ 𝑓𝑡 > 109.6 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾




𝑀𝑛Ω=𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖 ∗ 54𝑖𝑛3

1.67= 134.7 𝑘 ∗ 𝑓𝑡 > 104.5 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾




For 𝐷 + 𝐿:

This deflection meets the requirement:

𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > 1.03 𝑖𝑛 → 𝑂𝐾

For L:


𝐿

360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .833 𝑖𝑛 > .503 𝑖𝑛 → 𝑂𝐾



𝑉𝑛Ω= 87.5𝑘 > 15.4𝑘 → 𝑂𝐾

THEREFORE USE A W16X31 FOR THE MIDDLE ELEVATOR BEAMS.




FOR THE OUTSIDE ELEVATOR BEAMS:





240


360𝑜𝑟 1.5"



GIVEN/KNOWN VALUES:

𝐴𝑇 =(7.75 + 12.5𝑓𝑡)

2∗ 25 𝑓𝑡 = 253.125 𝑓𝑡2

𝐿 = 65 𝑝𝑠𝑓 ∗(7.75 𝑓𝑡 + 12.5 𝑓𝑡)

2= .658125 𝑘/𝑓𝑡





Therefore, 𝐴𝑇𝐾𝐿𝐿 = 253.125 𝑓𝑡2 ∗ 2 = 506.25 𝑓𝑡2 > 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

𝐿 = 𝐿0 (0.25 +15


𝐿 = .658125 (0.25 +15

√2 ∗ 253.125) = .6034

𝑘

𝑓𝑡

. 6034𝑘

𝑓𝑡> .5 ∗ .658125 = .329 → 𝑂𝐾

DESIGN OF BEAM:





Live load:




𝐹𝑦=(104.79 𝑘 ∗ 𝑓𝑡)(1.67) ∗ 12𝑖𝑛

50 𝑘𝑠𝑖= 42 𝑖𝑛3



Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗47.3𝑖𝑛3

1.67= 118 𝑘 ∗ 𝑓𝑡 > 104.79 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾




𝑀𝑛Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖 ∗ 47.3𝑖𝑛3

1.67= 118 𝑘 ∗ 𝑓𝑡 > 99.3 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For 𝐷 + 𝐿:

This deflection does not meet the requirement:



𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 < 1.3242 𝑖𝑛 → 𝑁𝑜𝑡 𝑔𝑜𝑜𝑑

Retry with a W16X31. Visual Analysis gives the following maximum moment considering a

self-weight of 31 lb/ft:


𝑀𝑛Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖 ∗ 54𝑖𝑛3

1.67= 134.7 𝑘 ∗ 𝑓𝑡 > 99.41 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For D+L:


𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > 1.03 𝑖𝑛 → 𝑂𝐾

For L:


𝐿

360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .833 𝑖𝑛 > .488 𝑖𝑛 → 𝑂𝐾





𝑉𝑛Ω= 87.5𝑘 > 15.91𝑘 → 𝑂𝐾

THEREFORE USE A W16X31 FOR THE OUTSIDE ELEVATOR BEAMS.


FOR THE GIRDERS NORTH OF THE ELEVATOR:

Use a wide flange that is continuously braced in the top flange by the floor deck. Use a simply

supported beam with a length of 25 feet.



240


360𝑜𝑟 1.5"



KNOWN/GIVEN VALUES:

𝐴𝑇 = (12.5𝑓𝑡)(25 𝑓𝑡) + (12.5𝑓𝑡)(7.75𝑓𝑡) = 409.3𝑓𝑡2

𝐿1 = 65 𝑝𝑠𝑓(25 𝑓𝑡) = 1.63 𝑘/𝑓𝑡 𝐿2 = 65 𝑝𝑠𝑓(12.5 𝑓𝑡) = .81 𝑘/𝑓𝑡

𝐷1 = [(9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟 𝑑𝑒𝑐𝑘)](25 𝑓𝑡)= 1.58 𝑘/𝑓𝑡 (𝑒𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡)

𝐷2 = [(9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟 𝑑𝑒𝑐𝑘)](12.5 𝑓𝑡)= .79 𝑘/𝑓𝑡 (𝑒𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡)

𝑃𝐷 = 7.9 𝑘 𝑓𝑟𝑜𝑚 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑑𝑒𝑠𝑖𝑔𝑛

𝑃𝐿 = 7.5 𝑘 𝑓𝑟𝑜𝑚 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑑𝑒𝑠𝑖𝑔𝑛

𝑃𝐷2 = (9.5 + 53.5)(10.125 ∗ 25) + 35(12.5) + 31(12.5) = 16.8 𝑘

𝑃𝐿2 = 65(10.125 ∗ 25) = 16.5 𝑘


Do not consider live load reductions because of the change in areas/loading.

DESIGN OF BEAM:





Live load:



Checking the maximum moment using the actual self-weight of 97 lb/ft:




Ω=𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗211 𝑖𝑛3

1.67= 526 𝑘 ∗ 𝑓𝑡 > 494 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For 𝐷 + 𝐿:


𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > 1.01 𝑖𝑛 → 𝑂𝐾

For 𝐿:




360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .833 𝑖𝑛 > .50 𝑖𝑛 → 𝐺𝑜𝑜𝑑



THEREFORE USE A W18X97 FOR THE GIRDERS NORTH OF THE ELEVATOR.


FOR THE GIRDERS SOUTH OF THE ELEVATOR:

Use a wide flange that is continuously braced in the top flange by the floor deck. Use a simply

supported beam with a length of 25 feet.



240


360𝑜𝑟 1.5"



KNOWN/GIVEN VALUES:

𝐴𝑇 = (25 𝑓𝑡)(25 𝑓𝑡) = 625 𝑓𝑡2

𝐿 = 65 𝑝𝑠𝑓(25 𝑓𝑡) = 1.63 𝑘/𝑓𝑡 𝐷 = [(9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟 𝑑𝑒𝑐𝑘)](25 𝑓𝑡) = 1.58 𝑘/𝑓𝑡 (𝑒𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡) 𝑃𝐷 = 6.8 𝑘 𝑓𝑟𝑜𝑚 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑑𝑒𝑠𝑖𝑔𝑛

𝑃𝐿 = 6.5 𝑘 𝑓𝑟𝑜𝑚 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑑𝑒𝑠𝑖𝑔𝑛

𝑃𝐷2 = (9.5 + 53.5)(10.125 ∗ 25) + 35(12.5) + 31(12.5) = 16.8 𝑘

𝑃𝐿2 = 65(10.125 ∗ 25) = 16.5 𝑘




Do not consider live load reductions because of the point loads – they may complicate the

design.

DESIGN OF BEAM:



Live load:







Ω=𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗211 𝑖𝑛3

1.67= 526 𝑘 ∗ 𝑓𝑡 > 499 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For 𝐷 + 𝐿:


𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > 1.02 𝑖𝑛 → 𝑂𝐾

For 𝐿:


360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .833 𝑖𝑛 > .50 𝑖𝑛 → 𝐺𝑜𝑜𝑑





THEREFORE USE A W18X97 FOR THE GIRDERS SOUTH OF THE ELEVATOR.


FOR THE GIRDERS NOT AFFECTED BY THE STAIRS/ELEVATOR:

Use a wide flange that is fully braced in the top flange by the steel deck. Use simply supported

beams with a length of 25 feet each spaced accordingly.



240


360𝑜𝑟 1.5"



GIVEN/KNOWN VALUES:

𝐴𝑇 = 25𝑓𝑡 ∗ 25𝑓𝑡 = 625𝑓𝑡2

𝐿 = 65 𝑝𝑠𝑓 ∗ 25𝑓𝑡2 = 1.625 𝑘/𝑓𝑡 𝐷 = [(9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟 𝑑𝑒𝑐𝑘)](25 𝑓𝑡)

= 1.575 𝑘/𝑓𝑡 (𝑒𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡)

𝑃𝐷 = (12.5 𝑓𝑡 ∗ (9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓) + 35𝑙𝑏

𝑓𝑡) 25 𝑓𝑡 = 20.6 𝑘

𝑃𝐿 = (12.5 𝑓𝑡)(25 𝑓𝑡)(65 𝑝𝑠𝑓) = 20.3 𝑘



Therefore, 𝐴𝑇𝐾𝐿𝐿 = 625 𝑓𝑡2 ∗ 2 = 1250 𝑓𝑡2 > 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

𝐿 = 𝐿0 (0.25 +15




𝐿 = 1.625 (0.25 +15

√2 ∗ 625) = 1.1

𝑘

𝑓𝑡

1.1 > .5 ∗ 1.625 = .8125 → 𝑂𝐾

Visual Analysis was used to design the 25 ft. beams. The following load cases were used:


Live load:




𝐹𝑦=(472.42 𝑘 ∗ 𝑓𝑡)(1.67) ∗ 12𝑖𝑛

50 𝑘𝑠𝑖= 189 𝑖𝑛3







Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗211 𝑖𝑛3

1.67= 526 𝑘 ∗ 𝑓𝑡 > 472 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾




For 𝐷 + 𝐿:




𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > .93 𝑖𝑛 → 𝑂𝐾


For 𝐿:


360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .883 𝑖𝑛 > 0.42 𝑖𝑛 → 𝑂𝐾

THEREFORE USE SECTION W18X97 FOR ANY INTERNAL SECTION SPACED 25ft CENTER TO CENTER. FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT

55.1 K.

FOR THE BEAMS NOT AFFECTED BY THE STAIRS/ELEVATOR:

Use a wide flange that is fully braced on the top flange and is completely unbraced in the bottom

flange for the interior beams. Use simply supported beams with a length of 25 feet each spaced

accordingly.



240


360𝑜𝑟 1.5"





GIVEN/KNOWN VALUES:

𝐴𝑇 = 25𝑓𝑡 ∗ 12.5𝑓𝑡 = 312.5 𝑓𝑡2

𝐿 = 65 𝑝𝑠𝑓 ∗ 12.5𝑓𝑡2 = .8125 𝑘/𝑓𝑡 𝐷 = [(9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟 𝑑𝑒𝑐𝑘)](12.5 𝑓𝑡)

= .7875 𝑘/𝑓𝑡 (𝑒𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡)



Therefore, 𝐴𝑇𝐾𝐿𝐿 = 312.5 𝑓𝑡2 ∗ 2 = 625 𝑓𝑡2 > 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

𝐿 = 𝐿0 (0.25 +15


𝐿 = .8125 (0.25 +15

√2 ∗ 312.5) = .6907

𝑘

𝑓𝑡

. 6907 > .5 ∗ .8125 = .406 → 𝑂𝐾

Visual Analysis was used to design the 25 ft. beams. The following load cases were used:


Live load:






𝐹𝑦=(123.3 𝑘 ∗ 𝑓𝑡)(1.67) ∗ 12𝑖𝑛

50 𝑘𝑠𝑖= 49.42 𝑖𝑛3





Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗66.5𝑖𝑛3

1.67= 166𝑘 ∗ 𝑓𝑡 > 118.3 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾




For 𝐷 + 𝐿:




𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > .8995 𝑖𝑛 → 𝑂𝐾


For 𝐿:


360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .883 𝑖𝑛 > 0.4106 𝑖𝑛 → 𝑂𝐾

THEREFORE USE SECTION W18X35 FOR ANY INTERNAL SECTION SPACED 12.5ft CENTER TO CENTER. FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT

18.92 K.

FOR THE GIRDERS NEXT TO THE STAIRWELLS:

Use a wide flange that is fully braced in the top flange by the steel deck. Use a simply supported

beam with a length of 25 feet.



240


360𝑜𝑟 1.5"



KNOWN/GIVEN VALUES:

𝐴1 = (25 𝑓𝑡 ∗ 12.5 𝑓𝑡) = 312.5 𝑓𝑡2

𝐴2 = (12.5 𝑓𝑡 ∗ 8.85 𝑓𝑡) = 110.625 𝑓𝑡2

𝐿1 = 65 𝑝𝑠𝑓(12.5 𝑓𝑡) = .8125 𝑘/𝑓𝑡 𝐿2 = .8125 ∗ 2 = 1.625 𝑘/𝑓𝑡 𝐷1 = (9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓)(12.5 𝑓𝑡) = .7875 𝑘/𝑓𝑡 𝐷2 = 2(. 7875) = 1.575 𝑘/𝑓𝑡




None of the tributary areas are greater than 400 sq. ft., therefore reductions are not permitted.

DESIGN OF BEAM:



Live load:







Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗112 𝑖𝑛3

1.67= 279.4 𝑘 ∗ 𝑓𝑡 > 161.3 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For 𝐷 + 𝐿:


𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > .70 𝑖𝑛 → 𝑂𝐾

For 𝐿:




360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .833 𝑖𝑛 > .28 𝑖𝑛 → 𝐺𝑜𝑜𝑑



THEREFORE USE A W18X55 FOR THE GIRDERS NEAR THE STAIRWELL. FOR

THE CONNECTION FORCE, USE GRAVITY SHEAR AT 28.07K.

FLOOR PERIMETER BEAM DESIGNS:

FOR THE NORTH TO SOUTH DIRECTION:

Use a wide flange that is fully braced in the top flange by the floor deck and is completely braced

in the bottom flange by the exterior wall. Use a simply supported beam with a length of 25 feet.



240


360𝑜𝑟 1.5"



GIVEN/KNOWN VALUES:

𝐴𝑇 =(12.5𝑓𝑡)

2∗ 25 𝑓𝑡 = 156.25 𝑓𝑡2

𝐿 = 65 𝑝𝑠𝑓 ∗(12.5 𝑓𝑡)

2= .40625 𝑘/𝑓𝑡

𝐷 = [(9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟 𝑑𝑒𝑐𝑘)](12.5 𝑓𝑡)

2+ (23 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑤𝑎𝑙𝑙)(15 𝑓𝑡)






Therefore, 𝐴𝑇𝐾𝐿𝐿 = 156.25 𝑓𝑡2 ∗ 2 = 312.5 𝑓𝑡2 < 400 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

DESIGN OF BEAM:



Live load:




𝐹𝑦=(97.266 𝑘 ∗ 𝑓𝑡)(1.67) ∗ 12𝑖𝑛

50 𝑘𝑠𝑖= 38.98 𝑖𝑛3



Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗40.2𝑖𝑛3

1.67= 100.3 𝑘 ∗ 𝑓𝑡 > 97.27 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾






𝑀𝑛Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖 ∗ 40.2𝑖𝑛3

1.67= 100.3 𝑘 ∗ 𝑓𝑡 > 91.5 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For 𝐷 + 𝐿:


𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 < 1.45 𝑖𝑛 → 𝑁𝑜𝑡 𝑔𝑜𝑜𝑑

Retry with a W14X30. Visual Analysis gives the following maximum moment considering a

self-weight of 30 lb/ft:


𝑀𝑛Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖 ∗ 47.3𝑖𝑛3

1.67= 118 𝑘 ∗ 𝑓𝑡 > 91.8 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For D+L:


𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > 1.23 𝑖𝑛 → 𝑂𝐾

For L:




𝐿

360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .833 𝑖𝑛 > .423 𝑖𝑛 → 𝑂𝐾



𝑉𝑛Ω= 74.5𝑘 > 14.7𝑘 → 𝑂𝐾



FOR THE EAST TO WEST DIRECTION:

Use a wide flange that is fully braced in the top flange by the floor deck. Use a simply supported

beam with a length of 25 feet.



240


360𝑜𝑟 1.5"



GIVEN/KNOWN VALUES:

𝐴𝑇 = 12.5 𝑓𝑡 ∗ 25 𝑓𝑡 = 300 𝑓𝑡2

𝐿 = 65 𝑝𝑠𝑓 ∗ 12.5 𝑓𝑡 = .813 𝑘/𝑓𝑡 𝐷 = [(9.5 𝑝𝑠𝑓 + 53.5 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟 𝑑𝑒𝑐𝑘)](12.5 𝑓𝑡)






Therefore, 𝐴𝑇𝐾𝐿𝐿 = 156.25 𝑓𝑡2 ∗ 2 = 300 𝑓𝑡2 > 600 𝑓𝑡2 → 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑒𝑑

𝐿 = 𝐿0 (0.25 +15


𝐿 = .813 (0.25 +15

√2 ∗ 300) = .7

𝑘

𝑓𝑡

. 7 > .5 ∗ .813 = .41 → 𝑂𝐾

DESIGN OF BEAM:



Live load:








𝑀𝑛Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖 ∗ 112𝑖𝑛3

1.67= 279 𝑘 ∗ 𝑓𝑡 > 120 𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For 𝐷 + 𝐿:


𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > .53 𝑖𝑛 → 𝑂𝐾

For L only:

This deflection does meet the requirement:

𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= .833 𝑖𝑛 > .24 𝑖𝑛 → 𝐺𝑜𝑜𝑑






FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 19.2k.

FOR THE NORTH TO SOUTH DIRECTION NOT AFFECTED BY STAIRWELL:

Use a wide flange that is fully braced in the top flange by the floor deck and is completely braced

in the bottom flange by the exterior wall. Use a simply supported beam with a length of 25 feet.



240


360𝑜𝑟 1.5"



GIVEN/KNOWN VALUES:

𝐴𝑇 =(16.15𝑓𝑡)

2∗ 25 𝑓𝑡 = 201.872 𝑓𝑡2

𝐷 = (23 𝑝𝑠𝑓 𝑓𝑟𝑜𝑚 𝑤𝑎𝑙𝑙)(15 𝑓𝑡) = .345 𝑘/𝑓𝑡 (𝑒𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡)

DESIGN OF BEAM:





ASD load combination 𝐷 gives the maximum positive moment:



𝐹𝑦=(34.77 𝑘 ∗ 𝑓𝑡)(1.67) ∗ 12𝑖𝑛

50 𝑘𝑠𝑖= 13.9𝑖𝑛3



Ω==

𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖∗20.1𝑖𝑛3

1.67= 50.15𝑘 ∗ 𝑓𝑡 > 34.77𝑘 ∗ 𝑓𝑡 → 𝑂𝐾

Test the beam using its self-weight of 16lb/ft instead of the assumed weight as before. The max



𝑀𝑛Ω=𝐹𝑦𝑍𝑥

Ω=50 𝑘𝑠𝑖 ∗ 20.1𝑖𝑛3

1.67= 50.15 𝑘 ∗ 𝑓𝑡 > 28.21𝑘 ∗ 𝑓𝑡 → 𝑂𝐾


For 𝐷:


𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛 > 1.062 𝑖𝑛 → 𝐺𝑜𝑜𝑑





𝑉𝑛Ω= 52.8𝑘 > 4.513𝑘 → 𝑂𝐾



STEEL COLUMN DESIGNS Pg 83


STEEL COLUMN DESIGN:

FINDING THE LOADS FOR EACH OF THE COLUMNS FROM THE ROOF:

For the columns on the corners of the building:

𝐿𝑟 = 20 𝑝𝑠𝑓(12.5 )(12.5) = 3.1 𝑘

𝐷 = (30 + 1.78 𝑝𝑠𝑓)(12.5 𝑓𝑡)(12.5 𝑓𝑡) + 40𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡) + 87

𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡)

+ 16.4𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡)(2) = 7.0 𝑘

𝑊 = {−103.9 𝑝𝑠𝑓(12.5 𝑓𝑡)(12.5 𝑓𝑡) = −16.2 𝑘

16 𝑝𝑠𝑓(12.5 𝑓𝑡)(12.5 𝑓𝑡) = 2.5 𝑘

For the columns in the east to west direction on the perimeter:

𝐿𝑟 = 20 𝑝𝑠𝑓(25 𝑓𝑡)(25 𝑓𝑡) = 12.5 𝑘

𝐷 = (30 + 1.78 𝑝𝑠𝑓)(25 𝑓𝑡)(25 𝑓𝑡) + 87𝑙𝑏

𝑓𝑡(25 𝑓𝑡) + 16.4

𝑙𝑏

𝑓𝑡(25 𝑓𝑡)(5) = 24.1 𝑘

𝑊 = {−103.9(25 𝑓𝑡)(25 𝑓𝑡) = −64.9 𝑘

16(25 𝑓𝑡)(25 𝑓𝑡) = 10 𝑘

For the columns supporting the joist girder:

𝐿𝑟 = 20 𝑝𝑠𝑓(50 𝑓𝑡)(50 𝑓𝑡) = 50 𝑘

𝐷 = (30 + 1.78 𝑝𝑠𝑓)(50 𝑓𝑡)(50 𝑓𝑡) + 16.4𝑙𝑏

𝑓𝑡(50 𝑓𝑡)(11) + 78

𝑙𝑏

𝑓𝑡(50 𝑓𝑡) = 92.4 𝑘

𝑊 = {−103.9 𝑝𝑠𝑓(50 𝑓𝑡)(50 𝑓𝑡) = −259.8 𝑘

16 𝑝𝑠𝑓(50 𝑓𝑡)(50 𝑓𝑡) = 40 𝑘

For the columns in the north to south direction on the perimeter:

𝐿𝑟 = 20 𝑝𝑠𝑓(25 𝑓𝑡)(25 𝑓𝑡) = 12.5 𝑘

𝐷 = (30 + 1.78 𝑝𝑠𝑓)(25 𝑓𝑡)(25 𝑓𝑡) + 16.4𝑙𝑏

𝑓𝑡(25 𝑓𝑡)(5) + 78

𝑙𝑏

𝑓𝑡(25 𝑓𝑡) = 23.9 𝑘

𝑊 = {−103.9 𝑝𝑠𝑓(25 𝑓𝑡)(25 𝑓𝑡) = −64.9 𝑘

16 𝑝𝑠𝑓(25 𝑓𝑡)(25 𝑓𝑡) = 10 𝑘

FINDING THE LOADS FOR EACH OF THE COLUMNS FROM THE FLOOR:

For columns E3 and A9:

𝐿 = 65 𝑝𝑠𝑓(12.5 𝑓𝑡)(12.5 𝑓𝑡) = 10.2 𝑘

𝐷 = (9.5 + 53.5 𝑝𝑠𝑓)(12.5 𝑓𝑡)(12.5 𝑓𝑡) + 30𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡) + 55

𝑙𝑏

𝑓𝑡 (12.5 𝑓𝑡)

+ 35𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡) + 23 𝑝𝑠𝑓(15 𝑓𝑡)(25 𝑓𝑡) = 20𝑘



For the columns E4 to E7 and A5 to A8:

𝐿 = 65 𝑝𝑠𝑓(25 𝑓𝑡)(12.5 𝑓𝑡) = 20.3 𝑘

𝐷 = (9.5 + 53.5 𝑝𝑠𝑓)(12.5 𝑓𝑡)(25 𝑓𝑡) + 55𝑙𝑏

𝑓𝑡(25 𝑓𝑡) + 35

𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡)(3)

+ 23 𝑝𝑠𝑓(15 𝑓𝑡)(25 𝑓𝑡) = 31 𝑘

For the columns E9 and A3:

𝐷 =30𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡) + 55

𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡) + 23 𝑝𝑠𝑓(15 𝑓𝑡)(25 𝑓𝑡) = 9.69 𝑘

For the columns A4 and E8:

𝐿 = 65 𝑝𝑠𝑓(12.5 𝑓𝑡)(12.5 + 8.85 𝑓𝑡) = 17.35 𝑘

𝐷 = (9.5 + 53.5 𝑝𝑠𝑓)(12.5 + 8.85 𝑓𝑡) (12.5 𝑓𝑡) +31𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡) +

26𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡)

+55𝑙𝑏

𝑓𝑡(25 𝑓𝑡) +

35𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡) + 23 𝑝𝑠𝑓(15 𝑓𝑡)(25 𝑓𝑡) = 30 𝑘

For the columns D4 to D7, C4, C5, C7, C8, B5, B7, and B8:

𝐿 = 65 𝑝𝑠𝑓(25 𝑓𝑡)(25 𝑓𝑡) = 40.6 𝑘

𝐷 = (9.5 + 53.5 𝑝𝑠𝑓)(25 𝑓𝑡)(25 𝑓𝑡) + 35𝑙𝑏

𝑓𝑡(3)(25 𝑓𝑡) + 97

𝑙𝑏

𝑓𝑡(25 𝑓𝑡) = 44.4 𝑘

For the column B6:

𝐿 = 65 𝑝𝑠𝑓(25 𝑓𝑡)(25 𝑓𝑡) = 40.6 𝑘

𝐷 = (9.5 + 53.5 𝑝𝑠𝑓)(25 𝑓𝑡)(25 𝑓𝑡) + 35𝑙𝑏

𝑓𝑡(3)(12.5 𝑓𝑡) + 31

𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡)(4)

+ 16𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡) + 97

𝑙𝑏

𝑓𝑡(25 𝑓𝑡) = 44.9𝑘

For the column C6:

𝐿 = 65 𝑝𝑠𝑓[(25 𝑓𝑡)(25 𝑓𝑡) − (9.5 𝑓𝑡)(9 𝑓𝑡)] = 35.1 𝑘

𝐷 = (9.5 + 53.5 𝑝𝑠𝑓)[(25 𝑓𝑡)(25 𝑓𝑡) − (9.5 𝑓𝑡)(9 𝑓𝑡)] + 35𝑙𝑏

𝑓𝑡(3)(12.5 𝑓𝑡)

+ 31𝑙𝑏

𝑓𝑡 (12.5 𝑓𝑡)(4) + 16

𝑙𝑏

𝑓𝑡(12.5 − 8 𝑓𝑡) + 55

𝑙𝑏

𝑓𝑡(9.5 𝑓𝑡) + 97

𝑙𝑏

𝑓𝑡(25 𝑓𝑡)

= 39.9𝑘

For the columns B3 and D9:

𝐿 = 65 𝑝𝑠𝑓(12.5 𝑓𝑡)(12.5 𝑓𝑡) = 10.2 𝑘

𝐷 = (9.5 + 53.5 𝑝𝑠𝑓)(12.5 𝑓𝑡)(12.5 𝑓𝑡) + 30𝑙𝑏

𝑓𝑡(25 𝑓𝑡) + 35

𝑙𝑏

𝑓𝑡 (12.5 𝑓𝑡) + 97

𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡)

+ 23 𝑝𝑠𝑓(15 𝑓𝑡)(25 𝑓𝑡) = 20.9𝑘



For the columns D3, C3, C9 and B9:

𝐿 = 65 𝑝𝑠𝑓(12.5 𝑓𝑡)(25 𝑓𝑡) = 20.3 𝑘

𝐷 = (9.5 + 53.5 𝑝𝑠𝑓)(12.5 𝑓𝑡)(25 𝑓𝑡) + 30𝑙𝑏

𝑓𝑡(25 𝑓𝑡) + 35

𝑙𝑏

𝑓𝑡 25 + 97

𝑙𝑏

𝑓𝑡(12.5 𝑓𝑡)

+ 23 𝑝𝑠𝑓(15 𝑓𝑡)(25 𝑓𝑡) = 31.2𝑘

INTERNAL COLUMN DESIGN

Use a wide flange that is braced each floor deck. Model the column as a 15 foot span with

summed axial loading on the top connection. Use a pin-pin connection.



GIVEN/KNOWN VALUES:

Use the following maximum loading for internal beams per floor:

𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 = 44𝑘

𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 = 92.4𝑘

𝐿𝑖𝑣𝑒 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 = 41𝑘

𝐿𝑖𝑣𝑒 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 = 50𝑘

𝑊𝑖𝑛𝑑 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 = 40𝑘 DESIGN OF COLUMN:

For pin-pin connection k=1.0

kL=(1.0)(15)= 15ft

Each internal column reaching the roof will have the loading from the roof and the three floors

that it spans. Therefore multiply 44k and 41k by 3 to have the total loading to factor. Negative

wind loading is neglected because this will cause tension and not add to the compression.

𝑇𝑜𝑡𝑎𝑙 𝐷𝑒𝑎𝑑 = 𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 + 𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 ∗ 3 = 224.4 𝑘

𝑇𝑜𝑡𝑎𝑙 𝐿𝑖𝑣𝑒 = 𝐿𝑖𝑣𝑒 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 ∗ 3 = 123 𝑘

ASD Load combinations are found on the next page.



Using Table 4.1 in the AISC manual and finding a value over 447k for a KL=15ft, choose a

W14x61 for the internal columns reaching the roof: 𝑃𝑛Ω= 361𝑘 > 347𝑘 → 𝑂𝐾

THEREFORE USE A W14X61 FOR THE INTERNAL COLUMNS. THESE COLUMNS

ARE SPECIFIED ON THE FLOOR LAYOUT AS D4 TO D8, C4 TO C8, AND B4 TO B8.

*FROM THE FUTURE SECOND ORDER ANALYSIS, USE A W14X109 INSTEAD.*

CORNER COLUMNS BY THE STAIRWELL DESIGN:


summed axial loading on the top connection. Use a pin-pin support.





GIVEN/KNOWN VALUES:

Use the following maximum loading for corner beams per floor:

𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 = 7𝑘

𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 = 9.7𝑘

𝐿𝑖𝑣𝑒 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 = 0𝑘

𝐿𝑖𝑣𝑒 𝑟𝑜𝑜𝑓 = 3.1𝑘

𝑊𝑖𝑛𝑑 = 2.5𝑘 DESIGN OF COLUMN:


kL=(1.0)(15)= 15ft

Each column reaching the roof will have the loading from the roof and the three floors that it

spans. Therefore multiply 9.7k and 0k by 3 to have the total loading to factor. Negative wind

loading is neglected because this will cause tension and not compression.

𝑇𝑜𝑡𝑎𝑙 𝐷𝑒𝑎𝑑 = 𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 + 𝐷𝑒𝑎𝑑𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 ∗ 3 = 36.1𝑘 𝑇𝑜𝑡𝑎𝑙 𝐿𝑖𝑣𝑒 = 𝐿𝑖𝑣𝑒 ∗ 3 = 0𝑘

ASD load combinations are as follows:




From Table 4.1 in AISC manual, choose W14X48. This meets the factored loads:

𝑃𝑛Ω= 221𝑘 > 36.1𝑘 → 𝑂𝐾

THEREFORE USE A W14X48 FOR THE CORNER COLUMNS NEAR THE

STAIRWELLS. THESE COLUMNS ARE SPECIFIED ON THE FLOOR LAYOUT AS

E9 AND A3.


PERIMETER COLUMN DESIGN:







GIVEN/KNOWN VALUES:

Use the following maximum loading:


𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 = 31.2𝑘 𝐿𝑖𝑣𝑒 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 = 20.3𝑘


𝑊𝑖𝑛𝑑 = 10𝑘 DESIGN OF COLUMN:


kL=(1.0)(15)= 15ft

Each perimeter column reaching the roof will have the loading from the roof and the three floors

that it spans. Therefore multiply 31.2k and 20.3k by 3 to have the total loading to factor.

Negative wind loading is neglected.

𝑇𝑜𝑡𝑎𝑙 𝐷𝑒𝑎𝑑 = 𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 + 𝐷𝑒𝑎𝑑𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 ∗ 3 = 117.7𝑘 𝑇𝑜𝑡𝑎𝑙 𝐿𝑖𝑣𝑒 = 𝐿𝑖𝑣𝑒 ∗ 3 = 60.9𝑘

The ASD load combinations are found on the following page:




Checking Table 4.1 in the AISC manual for 178.6k choose a W14X48 for perimeter columns

reaching the roof: 𝑃𝑛Ω= 221𝑘 > 178.6𝑘 → 𝑂𝐾

THEREFORE USE A W14X48 FOR ALL PERIMETER COLUMNS. THESE

COLUMNS ARE SPECIFIED ON THE FLOOR LAYOUT AS D3, C3, B3, D9, C9, B9, E4

TO E8 AND A4 TO A9.


ELEVATOR COLUMN DESIGN:







GIVEN/KNOWN VALUES:

Use the following maximum loading:

𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 = 0𝑘

𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 = 44.9𝑘 𝐿𝑖𝑣𝑒 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 = 40.6𝑘

𝐿𝑖𝑣𝑒 𝑟𝑜𝑜𝑓 = 0𝑘

𝑊𝑖𝑛𝑑 = 0𝑘 DESIGN OF BEAM:

For a pin-pin connection k=1.0

kL=(1.0)(15)= 15ft

Each perimeter column reaching the roof will have the loading from the roof and the three floors

that it spans. Therefore multiply 44.9k and 40.6k by 3 to have the total loading to factor.

𝑇𝑜𝑡𝑎𝑙 𝐷𝑒𝑎𝑑 = 𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 + 𝐷𝑒𝑎𝑑𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 ∗ 3 = 134.7𝑘 𝑇𝑜𝑡𝑎𝑙 𝐿𝑖𝑣𝑒 = 𝐿𝑖𝑣𝑒 ∗ 3 = 121.8 𝑘

ASD load combinations in Excel are as follows:




Checking Table 4.1 in the AISC manual for 256.6k choose a W14X61 for elevator columns: 𝑃𝑛Ω= 361𝑘 > 256.6𝑘 → 𝑂𝐾

THEREFORE USE A W14X61 FOR THE ELEVATOR COLUMNS. THESE COLUMNS

ARE SPECIFIED ON THE FLOOR LAYOUT AS C6 AND B6.


STAIRWELL COLUMN DESIGN







GIVEN/KNOWN VALUES:

Use the following maximum loading for beams per floor:


𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 = 30𝑘 𝐿𝑖𝑣𝑒 𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 = 17.4𝑘


𝑊𝑖𝑛𝑑 = 10𝑘 DESIGN OF COLUMN:

For a pin-pin connection k=1.0

kL=(1.0)(15)= 15ft

Each column reaching the roof will have the loading from the roof and the three floors that it

spans. Therefore multiply 30k and 17.4k by 3 to have the total loading to factor.

𝑇𝑜𝑡𝑎𝑙 𝐷𝑒𝑎𝑑 = 𝐷𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑟𝑜𝑜𝑓 + 𝐷𝑒𝑎𝑑𝑓𝑟𝑜𝑚 𝑓𝑙𝑜𝑜𝑟𝑠 ∗ 3 = 114.1𝑘 𝑇𝑜𝑡𝑎𝑙 𝐿𝑖𝑣𝑒 = 𝐿𝑖𝑣𝑒 ∗ 3 = 52.2𝑘

ASD load combinations from Excel are shown on the next page:




Checking Table 4.1 in the AISC manual and finding a value under 166.3k for a KL=15 ft, choose

a W14x48 for the stairwell columns:

𝑃𝑛Ω= 221𝑘 > 166.3𝑘 → 𝑂𝐾

THEREFORE USE A W14X48 FOR THE STAIRWELL COLUMNS. THESE

COLUMNS ARE SPECIFIED ON THE FLOOR LAYOUT AS A3, A4, B3, E8, E9, AND

D9.


TWO-WAY SLAB DESIGN Pg 95


TWO-WAY SLAB DESIGN:

3rd

Floor Two-Way Slab- Green Roof:

Initial slab thickness: h=12in per ACI Table 9.5c

Superimposed dead load: DLsup= 50.6psf

Concrete density: γ=150pcf Initial deal load: DL= h* γ=12in*150pcf=150psf

Load Combinations:

[COMBINATIONS PROVIDED ON NEXT PAGE]



LRFD LOAD COMBINATIONS

Loads: Combinations:

Dead (D) 200.60 psf 1) 1.4D

Live (L) 0.00 psf 2) 1.2D + 1.6L + .5(Lr or S or R)

Live Roof (Lr) 100.00 psf 3) 1.2D + 1.6(Lr or S or R) + (.5L or .5W)

Snow (S) 0.00 psf 4) 1.2D + W+.5L+.5(Lr or S or R)

Rain ( R) 0.00 psf 5) 1.2D + E + .5L + .2S

Positive Wind (W) 23.60 psf 6) 0.9D + W

Negative Wind (W) -34.10 psf 7) 0.9D + 1E

Seismic (E) 0.00 psf

Results:

Positive: Negative: DESIGN VALUES (psf):

1) 280.84 psf - psf Positive: 412.52

2) 290.72 psf - psf Negative: 146.44

3) 412.52 psf 400.72

psf

4) 314.32 psf - psf

5) 240.72 psf psf

6) 204.14 psf 146.44 psf

7) 180.54 psf - psf

LRFD Max Positive Moment: Mmax positive= 412.52 psf

LRFD Max Negative Moment: Mmax negative= 146.44 psf

Column cross section dimensions (square): Column Dimensions (CD) = 24in x 24in

L1=L2= 25ft

Ln= L1-CD=22.917ft

hmin=Ln/33=(22.917*12)/33=8.333in Check: h > hmin 12in > 8.33in OK!

qu=Mmax positive=412.52psf

Mo =𝑞𝑢 ∗ 𝐿1 ∗ 𝐿𝑛2

8=412.52 ∗ 25 ∗ 22.9172

8= 451.36𝑘𝑖𝑝 ∗ 𝑓𝑡



Green Roof Slab design Tables Location along Span Coefficient Moment

Interior Span M- 0.65 443.2656


End span interior M- 0.7 477.363

End span M+ 0.5 340.9736

End Span Exterior M- 0.3 204.5841

Location along Span Coefficient Moment




End span M+ 0.5 184.1257


Area C Area C

1 10045.44 1 19005.44

2 4861.44 2 1538.19

Sum 14906.88 Sum 20543.63

Variable Value

Ib 23930.1818

Is 23328

Alpha F1=Ib/Is 1.025813692

Bt 0.440321288

Interior Slab Panel Factored Moments k-ft

Location Column Strip Middle Strip


End Span M+ 204.5841375 136.389425

End Span Interior 358.0222406 119.3407469

Interior Span M- 332.4492234 110.8164078

Interior Span M+ 179.0111203 59.67037344



Exterior Slab Panel Factored Moments k-ft



End Span M+ 138.0942928 46.03143094


Interior Span M- 179.5225807 32.22200166

Interior Span M+ 96.66600497 32.22200166

Exterior slab panel column strip beam and slab factored moments k-ft

Location Beam Moments from slab

Slab moment


End Span M+ 127.0467494 57.07897436


Interior Span M- 165.1607742 74.20266667

Interior Span M+ 88.93272457 39.95528205

Exterior Slab Panel Column Strip Beam moments k-ft

Location From Slab From Perimeter wall and beam stem weight

Total beam design moment

End Span Exterior M- 76.22804963 44.24363636 120.471686

End Span M+ 127.0467494 9.618181818 136.6649312

End Span Interior M- 177.8654491 48.668 226.5334491

Interior Span M- 165.1607742 44.24363636 209.4044106

Interior Span M+ 88.93272457 30.4175 119.3502246

Interior slab panel column strip beam and slab factored

moments k-ft


Slab moment


End Span M+ 235.2717581 105.7018044


Interior Span M- 305.8532856 137.4123457

Interior Span M+ 164.6902307 73.99126306



Interior Slab Panel Column Strip Beam moments k-ft




End Span M+ 235.2717581 44.24363636 279.5153945

End Span Interior 329.3804614 10.58 339.9604614

Interior Span M- 305.8532856 44.24363636 350.0969219

Interior Span M+ 164.6902307 30.4175 195.1077307

Finding the centroid:

B=12in

HIb=24in

hw= HIb-h=24in-12in=12in

Centroid (from the top) =

𝐻𝑙𝑏2 ∗ 𝐵 ∗ 𝐻𝑙𝑏 +

ℎ2 ∗ 𝑥 ∗ ℎ

𝐵 ∗ 𝐻𝑙𝑏 + 𝑥 ∗ ℎ=

242 ∗ 12 ∗ 24 +

122 ∗ 12 ∗ 12

12 ∗ 24 + 12 ∗ 12= 10𝑖𝑛

𝐼𝑏 =1

3[(𝐵 + 𝑥)(𝐶𝑒𝑛𝑡𝑟𝑜𝑖𝑑2) + (𝐵(𝐻𝑙𝑏 − 𝐶𝑒𝑛𝑡𝑟𝑜𝑖𝑑)

3) + (𝑥(𝐻𝑙𝑏 − 𝐶𝑒𝑛𝑡𝑟𝑜𝑖𝑑)3)] = 19008𝑖𝑛4

Check: Centroid < h 10in < 12 in Good!

𝐼𝑠 =1

12[(𝐶𝐷

2+𝐿12) ℎ3 =

1

12[(24

2+25

2) 123 = 23328𝑖𝑛4

𝛼𝐹1 = (𝐼𝑏𝐼𝑠) =

19008

23328= 0.815

Check: αF1 > 0.80 0.815 > 0.80 Good! We may now use direct design method!

Reinforcement:

Exterior Column Panel Negative Steel Reinforcement:

Φ1=0.90

db1 = diameter of steel reinforcement = 0.875in

fc’= 5.5ksi

fy=60ksi

ECSW= CSW/2 +CD/2 = 87in

Critical Section: 𝑑1 = ℎ − 0.75𝑖𝑛 − 0.50 ∗ 𝑑𝐼𝑏 = 12 − .75 − 0.5(0.875) = 10.8125𝑖𝑛

𝑀𝑛1 =𝐸𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝐶𝑜𝑙𝑢𝑚𝑛 𝑃𝑎𝑛𝑒𝑙 𝑁𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑀𝑜𝑚𝑒𝑛𝑡𝑚𝑎𝑥Φ1 ∗ Exterior Column Panel Width ∗ 𝑑 21

=199.33𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 87𝑖𝑛 ∗ 10.8125𝑖𝑛 2

= 261.30𝑝𝑠𝑖



ρ1 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .2613ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))

= 4.48(10−3)

As = ρ1 ∗ Exterior Column Panel Width ∗ d1 = 4.48(10−3)(87in)(10.8125in) = 4.218in2

Top Steel: Number of bars 8 #7 bar

Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 8 (

𝜋

4) (. 875𝑏1

2 ) = 4.81 > 4.218 → 𝐺𝑜𝑜𝑑!

Check: As= 8(.60in2) = 4.8in

2 > 4.218 in

2 Good!

Exterior Column Panel Positive Steel Reinforcement:

Φ1=0.90

db2= 0.75in

fc’= 5.50ksi

fy=60ksi

ECSW= 87in

Critical Section: 𝑑2 = ℎ − 0.75𝑖𝑛 − 0.50 ∗ 𝑑𝑏2 = 12 − 0.75 − 0.50(0.75𝑖𝑛) = 10.875𝑖𝑛

𝑀𝑛2 =𝐸𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝐶𝑜𝑙𝑢𝑚𝑛 𝑃𝑎𝑛𝑒𝑙 𝑃𝑜𝑖𝑡𝑖𝑣𝑒 𝑀𝑜𝑚𝑒𝑛𝑡𝑚𝑎𝑥Φ1 ∗ Exterior Column Panel Width ∗ 𝑑 22

=204.60 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 87𝑖𝑛 ∗ 10.875𝑖𝑛 2

= 265.135 𝑝𝑠𝑖

ρ2 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .265135ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))

= 4.56(10−3)


Bottom Steel: Number of bars 10 #6 bar

Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 10 (

𝜋

4) (. 75𝑏1

2 ) = 4.42𝑖𝑛2 > 4.32𝑖𝑛2 → 𝐺𝑜𝑜𝑑!

Check: As= 10(.44in2) = 4.4in

2 > 4.32 in

2 Good!

Interior Column Panel Negative Steel Reinforcement:



Φ1=0.90

db1= 0.875in

fc’= 5ksi

fy=60ksi

ICSW= 2*min (.25*L1, .25*L2) = 150in


𝑀𝑛1 =𝐼𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝐶𝑜𝑙𝑢𝑚𝑛 𝑃𝑎𝑛𝑒𝑙 𝑁𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑀𝑜𝑚𝑒𝑛𝑡𝑚𝑎𝑥Φ1 ∗ Interior Column Strip Width ∗ 𝑑 21

=358.02 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 150𝑖𝑛 ∗ 10.8125𝑖𝑛 2

= 272.21𝑝𝑠𝑖

ρ1 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .27221ksi

. 85 ∗ 5𝑘𝑠𝑖) 12))

= 4.69(10−3)

As = ρ1 ∗ Interior Column Panel Width ∗ d1 = 4.69(10−3)(150in)(10.8125in) = 7.61in2


Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 13 (

𝜋

4) (. 875𝑏1

2 ) = 7.813 > 7.61 → 𝐺𝑜𝑜𝑑!

Check: As= 13(0.60in2) = 7.80 in

2 > 7.61 in

2 Good!

Interior Column Panel Positive Steel Reinforcement:

Φ1=0.90

db2= 0.75in

fc’= 5ksi

fy=60ksi

CSW= 150in


𝑀𝑛2 =𝐼𝑛𝑛𝑒𝑟 𝐶𝑜𝑙𝑢𝑚𝑛 𝑃𝑎𝑛𝑒𝑙 𝑃𝑜𝑖𝑡𝑖𝑣𝑒 𝑀𝑜𝑚𝑒𝑛𝑡𝑚𝑎𝑥Φ1 ∗ Interior Column Panel Width ∗ 𝑑 22

=143.20 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 150𝑖𝑛 ∗ 10.875𝑖𝑛 2

= 107.63𝑝𝑠𝑖

ρ2 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .10763ksi

. 85 ∗ 5𝑘𝑠𝑖) 12))

= 1.823(10−3)





Check: 𝑛𝜋

4> 𝐴𝑠 → 7 (

𝜋

4) (. 75𝑏1

2 ) = 3.09𝑖𝑛2 > 2.964𝑖𝑛2 → 𝐺𝑜𝑜𝑑!

Check: As= 7(0.44in2) = 3.08in

2 > 2.964in

2 Good!

Middle Column Panel Negative Steel Reinforcement:

Φ1=0.90

db1= 0.75in

fc’= 5ksi

fy=60ksi

MSW= L2-CSW = (25ft)(12in) -150in = 150in

Critical Section: 𝑑1 = ℎ − 0.75𝑖𝑛 − 0.50 ∗ 𝑑𝐼𝑏 = 12 − .75 − 0.5(0.75) = 10.875 𝑖𝑛

𝑀𝑛1 =𝑀𝑖𝑑𝑑𝑙𝑒 𝐶𝑜𝑙𝑢𝑚𝑛 𝑃𝑎𝑛𝑒𝑙 𝑁𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑀𝑜𝑚𝑒𝑛𝑡𝑚𝑎𝑥Φ1 ∗ Middle Column Panel Width ∗ 𝑑 21

=119.30 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 150𝑖𝑛 ∗ 10.875𝑖𝑛 2

= 89.67𝑝𝑠𝑖

ρ1 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .08967ksi

. 85 ∗ 5𝑘𝑠𝑖) 12))

= 1.51(10−3) As = ρ1 ∗ Middle Column Panel Width ∗ d1 = 1.51(10−3)(150in)(10.875in) = 2.464in2


Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 6 (

𝜋

4) (. 75𝑏1

2 ) = 2.65 > 2.464 → 𝐺𝑜𝑜𝑑!

Check: As= 6(0.44in2) = 2.64 in

2 > 2.464 in

2 Good!

Middle Column Panel Positive Steel Reinforcement:

Φ1=0.90

db2= 0.875in

fc’= 5ksi

fy=60ksi

MSW= 150 in

Critical Section: 𝑑2 = ℎ − 0.75𝑖𝑛 − 0.50 ∗ 𝑑𝑏2 = 12 − 0.75 − 0.50(0.875𝑖𝑛) = 10.8125 𝑖𝑛



𝑀𝑛2 =𝐶𝑜𝑙𝑢𝑚𝑛 𝑃𝑎𝑛𝑒𝑙 𝑃𝑜𝑖𝑡𝑖𝑣𝑒 𝑀𝑜𝑚𝑒𝑛𝑡𝑚𝑎𝑥

Φ1 ∗ Middle Column Panel Width ∗ 𝑑 22=136.40 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 150𝑖𝑛 ∗ 10.8125𝑖𝑛 2= 103.71𝑝𝑠𝑖

ρ2 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .10371ksi

. 85 ∗ 5𝑘𝑠𝑖) 12))

= 1.75(10−3)

As = ρ2 ∗ Middle Column Panel Width ∗ d2 = 1.75(10−3)(150in)(10.8125in) = 2.84in2


Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 5 (

𝜋

4) (. 875𝑏1

2 ) = 3.01𝑖𝑛2 > 2.85𝑖𝑛2 → 𝐺𝑜𝑜𝑑!

Check: As= 5(0.60in2) = 3 in

2 > 2.85in

2 Good!

Green Roof Slab Shear:

Case A Edge

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 412.52 psf

Slab

thickness 12 inch

db 1.128 inch

qdu 200.6 lbs

qlu 100 lbs

One way Shear Check

𝑑 = 𝑆𝑙𝑎𝑏 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 − .75 − .5(𝑑𝑏) = 10.686 𝑖𝑛

𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 265.2375 𝑓𝑡

2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 109.4158k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 356.6𝑘

Bw=l2



Case A Edge

d 10.686 in

At 265.2375 ft^2

Vu 109.4158 kip

(F)Vc 356.6 kip OK

Two-way Slab Shear

𝐴𝑡 = (𝑙12+𝑐

2) ∗ 𝑙2 − 𝑏1 ∗ 𝑏2 = 330.4𝑓𝑡^2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 136.3𝑘 b=1 , as=30

𝑏1 = 𝑐 +𝑑

2= 29.41𝑖𝑛 , 𝑏2 = 𝑐 + 𝑑 = 34.686

𝑏0 = 2 ∗ 𝑏1 + 𝑏2 = 93.372

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=222k

Case A Edge

At 330.3909 ft^2

Vu 136.2929 kip

b 1

as 30

b1 29.40625 in

b2 34.686

b0 93.372

(F)Vc 222 kip OK

Munb=full slab moment (col+mid) for and edge column=End span exterior M- = 199.3k-ft

𝛾𝐹 =1

1 +23 ∗√𝑏1𝑏2

= .62

𝛾𝐹 ∗ 𝑀𝑢𝑛𝑏 = 123.54𝑘 ∗ 𝑓𝑡 Effective Width = 3h+c = 5ft

𝐴𝑐 = (2𝑏1 + 𝑏2) ∗ 𝑑 = 1054.9𝑖𝑛^2

𝐽

𝑐= 𝑏1 ∗ 𝑑(𝑏1 + 6𝑏2) + 𝑑

3

6= 12613𝑖𝑛^3

𝛾𝑣 = 1 − 𝛾𝐹 = .38

𝑉𝑚𝑎𝑥 =𝑉𝑢

𝐴𝑐+ 𝛾𝑣 ∗ 𝑀𝑢𝑛𝑏 ∗ 12

𝐽/𝑐= 201.3𝑘/𝑖𝑛2

(𝜙)𝑉𝑐 = .75 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘



Case A Edge

Munb 199.3 k-ft

YF 0.620071

Effective

Width 5 ft

Min Munb 46.2875 k-ft

Ac 1054.9 in^2

J/c 12613 in^3

Yv 0.379929

Vmax 201.3 Kips/in^2

(F)Vc 222.486 kips OK

Case B:

Case B Interior

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 412.52 psf

Slab thickness 12 inch

db 1.128 inch

qdu 200.6 lbs

qlu 100 lbs One way Shear Check


𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 265.2375 𝑓𝑡

2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 109.4158k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 356.6𝑘

Bw=l2

Case B Interior d 10.8125 in

At 265.24 ft^2 Vu 109.4158 kip (F)Vc 356.62 kip OK

Two-way Slab Shear



𝐴𝑡 = (𝑙1) ∗ 𝑙2 − 𝑏1 ∗ 𝑏2 = 616.645𝑓𝑡2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 254.384𝑘

b=1 , as=40

𝑏1 = 𝑐 + 𝑑 = 34.8125𝑖𝑛 , 𝑏2 = 𝑐 + 𝑑 = 34.686𝑖𝑛

𝑏0 = 𝑏1 + 𝑏2 = 138.744𝑖𝑛

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=329.862

Case B Interior At 616.645 ft^2

Vu 254.3784 kip b 1 as 40 b1 34.686 in b2 34.686 in b0 138.744 in (F)Vc 329.862 kip OK

Munb=25.571 k-ft

𝛾𝐹 =1

1 +23 ∗√𝑏1𝑏2

= .6


𝐴𝑐 = 2(𝑏1 + 𝑏2) ∗ 𝑑 = 1482.618𝑖𝑛^2

𝐽

𝑐= 𝑏1 ∗ 𝑑(𝑏1 + 3𝑏2) + 𝑑

3

3= 17548.8𝑖𝑛^3

𝛾𝑣 = 1 − 𝛾𝐹 = .4



𝐽/𝑐= 184.2𝑘/𝑖𝑛2

(𝜙)𝑉𝑐 = .75 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘

Case B Interior

Munb 25.571 k-ft

Yf 0.6

Yf*Munb check 15.3426

Effective Width 5 ft

Min Munb 46.2875 k-ft

Ac 1482.618 in^2

J/c 17548.78 in^3

Yv 0.4

Vmax 184.2345 Kips/in^2

(F)Vu 222.486 kips



Case C:

Case C Edge

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 412.52 psf


db 1.128 inch

qdu 200.6 lbs

qlu 100 lbs

One way Shear Check


𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 265.2375 𝑓𝑡

2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 109.4158k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 356.6𝑘

Bw=l2

Case C Edge d 10.686 in

At 265.2375 ft^2 Vu 109.4158 kip

(F)Vc 356.6227 kip OK

Two-way Slab Shear

𝐴𝑡 = (𝑙12+𝑐

2) ∗ 𝑙2 − 𝑏1 ∗ 𝑏2 = 330.4𝑓𝑡^2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 136.3𝑘 b=1 , as=30

𝑏1 = 𝑐 +𝑑

2= 29.41𝑖𝑛 , 𝑏2 = 𝑐 + 𝑑 = 34.78125

𝑏0 = 2 ∗ 𝑏1 + 𝑏2 = 93.372

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=222k



Case C Edge At 330.432 ft^2

Vu 136.3098 kip b 1

as 30 b1 29.343 in b2 34.686 in b0 93.372 in (F)Vc 221.9905 kip OK

𝛾𝐹 =1

1 +23 ∗√𝑏1𝑏2

= .62


𝐴𝑐 = (𝑏1 + 2𝑏2) ∗ 𝑑 = 997.77𝑖𝑛^2

𝑗

𝑐=2 ∗ 𝑏1

2 ∗ 𝑑(𝑏1 ∗ 2𝑏2) + 𝑑3(2𝑏1 + 𝑏2)

6𝑏1= 10964.82𝑖𝑛^3

𝛾𝑣 = 1 − 𝛾𝐹 = .38



𝐽/𝑐= 219.5𝑘/𝑖𝑛2

(𝜙)𝑉𝑐 = .75 ∗ $ ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘

Case C Edge Munb 199.3 k-ft

Yf 0.619896 Yf*Munb check 123.5452 Effective Width 5 ft Min Munb 46.2875 k-ft Ac 997.7732 in^2 J/c 10964.82 in^3 Yv 0.380104 Vmax 219.5207 Kips/in^2 (F)Vu 222.486 kips OK



Case D:

Case D Corner

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 412.52 psf


db 1.128 inch

qdu 200.6 lbs

qlu 100 lbs


𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 265.2375 𝑓𝑡

2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 109.4158k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 356.6𝑘

Bw=l2

Case D Corner d 10.686 in


Two-way Slab Shear

𝐴𝑡 = (𝑙12) ∗𝑙22− (𝑐 +

𝑑

2)2

= 150.27𝑓𝑡^2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 62𝑘 b=1 , as=30

𝑏1 = 𝑐 +𝑑

2= 29.41𝑖𝑛 , 𝑏2 = 𝑐 +

𝑑

2= 29.41𝑖𝑛

𝑏0 = 𝑏1 + 𝑏2 = 58.69𝑖𝑛

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=139.53k



Case D Corner At 150.2708 ft^2


Munb = 199.3k-ft

𝛾𝐹 =1

1 +23 ∗√𝑏1𝑏2

= .6

Effective Width = 3h+c = 5ft

𝛾𝐹 ∗ 𝑀𝑢𝑛𝑏 = 119.58𝑘 ∗ 𝑓𝑡 𝐴𝑐 = (𝑏1 + 𝑏2) ∗ 𝑑 = 627.12𝑖𝑛

2

𝐽

𝑐= 𝑏12 ∗ 𝑑(𝑏1 + 4𝑏2) + 𝑑

3(𝑏1 + 𝑏2)

6𝑏1= 21664.01𝑖𝑛3

𝛾𝑣 = 1 − 𝛾𝐹 = .4



𝐽/𝑐= 143𝑘/𝑖𝑛2

(𝜙)𝑉𝑐 = .75 ∗ $ ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘

Case D Corner Munb 199.3 k-ft

Yf 0.6 Yf*Munb check 119.58 Effective Width 5 ft Min Munb 46.2875 k-ft Ac 627.1186 in^2 J/c 21664.01 in^3 Yv 0.4 Vmax 143.0065 Kips/in^2 (F)Vu 222.486 kips OK

3rd

Floor Two-Way Slab:

Initial slab thickness: h=13in

Per ACI Table 9.5c



Superimposed dead load: DLsup=9.50psf

Concrete density: γ=150pcf Initial deal load: DL= h* γ=13in*150pcf=162.5psf

Load Combinations:



Dead (D) 172.00 psf 1) 1.4D




Rain ( R) 0.00 psf 5) 1.2D + E + .5L + .2S


Negative Wind (W) 0.00 psf 7) 0.9D + 1E


Results:


1) 240.80 psf - psf Positive: 310.40

2) 310.40 psf - psf Negative: 154.80 3) 238.90 psf 238.90 psf

4) 238.90 psf -

psf

5) 238.90 psf psf

6) 154.80 psf 154.80 psf

7) 154.80 psf - psf



Column cross section dimensions (square): Column Dimensions (CD)= 24in x 24in

L1=L2= 25ft

Ln= L1-CD=22.917ft




8=310.40𝑝𝑠𝑓 ∗ 25𝑓𝑡 ∗ 22.917𝑓𝑡2

8= 509.43𝑘𝑖𝑝 ∗ 𝑓𝑡



3rd Floor Slab design tables Location along Span Coefficient Moment




End span M+ 0.5 256.565






End span M+ 0.5 138.5451


Area C Area C

1 19005.44 1 12310.52

2 1781.19 2 4024.057

Sum 20786.63 Sum 16334.58

Variable Value

Ib 18647.85

Is 17968.5

Alpha F1=Ib/Is 1.03780783

Bt 0.578418621






End Span M+ 153.939 102.626

End Span Interior M- 269.39325 89.79775

Interior Span M- 250.150875 83.383625

Interior Span M+ 107.7573 71.8382




End Span M+ 103.908825 34.636275


Interior Span M- 135.0814725 24.2453925

Interior Span M+ 72.7361775 24.2453925

Exterior slab panel column strip beam and slab factored moments k-ft


Slab moment


End Span M+ 95.596119 42.948981


Interior Span M- 124.2749547 55.8336753

Interior Span M+ 66.9172833 30.0642867




Location From Slab

From Perimeter wall and beam stem weight


End Span Exterior M-

57.3576714 44.24363636 101.6013078

End Span M+ 95.596119 44.24363636 139.8397554


Interior Span M- 124.2749547 44.24363636 168.5185911

Interior Span M+ 66.9172833 30.4175 97.3347833

Interior slab panel column strip beam and slab factored moments k-ft


Slab moment


End Span M+ 177.02985 79.53515 End Span Interior 247.84179 111.34921 Interior Span M- 230.138805 103.395695 Interior Span M+ 123.920895 55.674605


Location From Slab

From Perimeter wall and beam stem weight


End Span Exterior M- 106.21791

44.24363636 150.4615464

End Span M+ 177.02985 44.24363636 221.2734864


Interior Span M- 230.138805 44.24363636 274.3824414

Interior Span M+ 123.920895 30.4175 154.338395




B=12in

HIb=24in




ℎ2 ∗ 𝑥 ∗ ℎ

𝐵 ∗ 𝐻𝑙𝑏 + 𝑥 ∗ ℎ=

242 ∗ 12 ∗ 24 +

112 ∗ 11 ∗ 11

12 ∗ 24 + 11 ∗ 11= 10.18𝑖𝑛

𝐼𝑏 =1


3) + (𝑥(𝐻𝑙𝑏 − 𝐶𝑒𝑛𝑡𝑟𝑜𝑖𝑑)3)]

= 18647.85𝑖𝑛4 Check: Centroid < h 10.18in < 12 in Good!

𝐼𝑠 =1

12[(𝐶𝐷

2+𝐿12) ℎ3 =

1

12[(24

2+25

2) 123 = 17968.5𝑖𝑛4


18647.85

17968.5= 1.038

Check: αF1 > 0.80 1.038 > 0.80 Good!

Reinforcement:


Φ1=0.90


fc’= 5.5ksi

fy=60ksi




=147.94𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 87𝑖𝑛 ∗ 11.8125𝑖𝑛 2

= 162.49𝑝𝑠𝑖

ρ1 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .16249ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))

= 2.76(10−3)



Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 5 (

𝜋

4) (. 875𝑏1

2 ) = 3.01𝑖𝑛2 > 2.83𝑖𝑛2 → 𝐺𝑜𝑜𝑑!

Check: As= 6(0.60in2) = 3.0in

2 > 2.83in

2 Good!




Φ1=0.90

db2= 0.75in

fc’= 5.5ksi

fy=60ksi

ECSW= 87in


𝑀𝑛2 =𝐸𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝐶𝑜𝑙𝑢𝑚𝑛 𝑃𝑎𝑛𝑒𝑙 𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑀𝑜𝑚𝑒𝑛𝑡𝑚𝑎𝑥Φ1 ∗ Exterior Column Panel Width ∗ 𝑑 22

= 153.94 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 87𝑖𝑛 ∗ 11.875𝑖𝑛 2

= 167.303 𝑝𝑠𝑖

ρ2 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .1673ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))

= 2.84(10−3) As = ρ2 ∗ Exterior Column Panel Width ∗ d2 = 2.861(10−3)(87in)(11.875in) = 2.93in2


Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 7 (

𝜋

4) (. 75𝑏1

2 ) = 3.09𝑖𝑛2 > 2.93𝑖𝑛2 → 𝐺𝑜𝑜𝑑!

Check: As= 7(0.44in2) = 3.08in

2 > 2.93 in

2 Good!


Φ1=0.90

db1= 0.875in

fc’= 5.5ksi

fy=60ksi

ICSW= 2*min (.25*L1, .25*L2) = 150in



=269.39 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 150𝑖𝑛 ∗ 11.8125𝑖𝑛 2

= 171.61𝑝𝑠𝑖

ρ1 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

4𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .17161ksi

. 85 ∗ 4𝑘𝑠𝑖) 12))

= 2.91(10−3)





Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 9 (

𝜋

4) (. 875𝑏1

2 ) = 5.41 > 5.16 → 𝐺𝑜𝑜𝑑!

Check: As= 9(0.6in2) = 5.40 in

2 > 5.16 in

2 Good!


Φ1=0.90

db2= 0.75in

fc’= 5.5ksi

fy=60ksi

CSW= 150in



=107.76 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 150𝑖𝑛 ∗ 11.875𝑖𝑛 2

= 67.93𝑝𝑠𝑖

ρ2 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .06793ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))

= 1.14(10−3)


Bottom Steel: Number of bars 5#6 bar

Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 3 (

𝜋

4) (. 75𝑏1

2 ) = 2.21𝑖𝑛2 > 2.03𝑖𝑛2 → 𝐺𝑜𝑜𝑑!

Check: As= 5(0.44in2) = 2.20in

2 > 2.03in

2 Good!


Φ1=0.90

db1= 0.75in

fc’= 5.5ksi

fy=60ksi






=89.8 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 150𝑖𝑛 ∗ 11.875𝑖𝑛 2

= 56.61𝑝𝑠𝑖

ρ1 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .8967ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))



Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 4 (

𝜋

4) (. 75𝑏1

2 ) = 1.77 𝑖𝑛2 > 1.69in2 → 𝐺𝑜𝑜𝑑!

Check: As= 4(0.44in2) = 1.76 in

2 > 1.69 in

2 Good!


Φ1=0.90

db2= 0.875in

fc’= 5.5ksi

fy=60ksi

MSW= 150 in




0.90 ∗ 150𝑖𝑛 ∗ 11.8125𝑖𝑛 2= 65.38𝑝𝑠𝑖

ρ2 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .06538ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))

= 1.10(10−3)



Check: 𝑛𝜋

4(𝑑𝑏12 ) < 𝐴𝑠 → 4 (

𝜋

4) (. 875𝑏1

2 ) = 2.41𝑖𝑛2 > 1.94𝑖𝑛2 → 𝐺𝑜𝑜𝑑!

Check: As= 4(0.6in2) = 2.40 in

2 > 1.94 in

2 Good!



3rd

floor Slab Shear:

Case A Edge

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 310.4 psf


db 1.128 inch

qdu 172 lbs

qlu 65 lbs

One way Shear Check


𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 263.2 𝑓𝑡

2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 81.7k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 390𝑘

bw=l2

Case A Edge One Way Shear d 11.686 in


Two-way Slab Shear

𝐴𝑡 = (𝑙12+𝑐

2) ∗ 𝑙2 − 𝑏1 ∗ 𝑏2 = 330.1𝑓𝑡^2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 102.46𝑘 b=1 , as=30

𝑏1 = 𝑐 +𝑑

2= 29.91𝑖𝑛 , 𝑏2 = 𝑐 + 𝑑 = 35.686𝑖𝑛

𝑏0 = 2 ∗ 𝑏1 + 𝑏2 = 95.372𝑖𝑛

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=247.96k



Case A Edge Two Way Shear At 330.1043 ft^2


Munb= 147.94k-ft

𝛾𝐹 =1

1 +23 ∗√𝑏1𝑏2

= .62

𝛾𝐹 ∗ 𝑀𝑢𝑛𝑏 = 91.9𝑘 ∗ 𝑓𝑡 Effective Width = 3h+c = 5.25ft

𝐴𝑐 = (2𝑏1 + 𝑏2) ∗ 𝑑 = 1182.8𝑖𝑛^2

𝐽

𝑐= 𝑏1 ∗ 𝑑(𝑏1 + 6𝑏2) + 𝑑

3

6= 14445.92𝑖𝑛^3

𝛾𝑣 = 1 − 𝛾𝐹 = .38



𝐽/𝑐= 133.17𝑘/𝑖𝑛2

(𝜙)𝑉𝑐 = .75 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘

Case A Edge Munb 147.94 k-ft

Yf 0.621253 Yf*Munb 91.90814 Effective Width 5.25 ft Min Munb 30.08688 k-ft Ac 1182.798 in^2 J/c 14445.9 in^3 Yv 0.378747 Vmax 133.1736 Kips/in^2 (F)Vu 222.486 kips OK



Case B Interior

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 310.4 psf


db 1.128 inch

qdu 172 lbs

qlu 65 lbs

One way Shear Check


𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 263.2 𝑓𝑡

2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 81.7k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 390𝑘

bw=l2

Case B Interior One Way Shear d 11.686 in


Two-way Slab Shear

𝐴𝑡 = (𝑙1) ∗ 𝑙2 − 𝑏1 ∗ 𝑏2 = 616.15𝑓𝑡2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 191.25𝑘

b=1 , as=40

𝑏1 = 𝑐 + 𝑑 = 35.686𝑖𝑛 , 𝑏2 = 𝑐 + 𝑑 = 35.686𝑖𝑛

𝑏0 = 𝑏1 + 𝑏2 = 142.744𝑖𝑛

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=371.13k



Case B Interior Two Way Shear At 616.1563 ft^2


Munb=19.24 k-ft

𝛾𝐹 =1

1+2

3∗√𝑏1𝑏2

= .6


𝐴𝑐 = 2(𝑏1 + 𝑏2) ∗ 𝑑 = 1668.101𝑖𝑛^2

𝐽

𝑐= 𝑏1 ∗ 𝑑(𝑏1 + 3𝑏2) + 𝑑

3

3= 20374.35𝑖𝑛^3

𝛾𝑣 = 1 − 𝛾𝐹 = .4



𝐽/𝑐= 121.74𝑘/𝑖𝑛2

(𝜙)𝑉𝑐 = .75 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘

Case B Interior Munb 19.24 k-ft

Yf 0.6 Yf*Munb 11.544 Effective

Width 5.25 ft Min Munb 30.08688 k-ft Ac 1668.106 in^2 J/c 20374.64 in^3 Yv 0.4 Vmax 121.742 Kips/in^2 (F)Vu 222.486 kips OK



Case C:

Case C Edge

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 310.4 psf


db 1.128 inch

qdu 172 lbs

qlu 65 lbs

One way Shear Check


𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 263.2 𝑓𝑡

2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 81.7k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 390𝑘

bw=l2



Two-way Slab Shear

𝐴𝑡 = (𝑙12+𝑐

2) ∗ 𝑙2 − 𝑏1 ∗ 𝑏2 = 330.06𝑓𝑡^2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 102.46𝑘 b=1 , as=30

𝑏1 = 𝑐 +𝑑

2= 29.84𝑖𝑛 , 𝑏2 = 𝑐 + 𝑑 = 35.686𝑖𝑛

𝑏0 = 2 ∗ 𝑏1 + 𝑏2 = 95.372𝑖𝑛

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=247.96k



Case C Edge Two Way Shear At 330.1043 ft^2


Munb = 147.4k-ft

𝛾𝐹 =1

1 +23 ∗√𝑏1𝑏2

= .62


𝐴𝑐 = (𝑏1 + 2𝑏2) ∗ 𝑑 = 1114.52𝑖𝑛^2

𝑗

𝑐=2 ∗ 𝑏1

2 ∗ 𝑑(𝑏1 ∗ 2𝑏2) + 𝑑3(2𝑏1 + 𝑏2)

6𝑏1= 12616.1𝑖𝑛^3

𝛾𝑣 = 1 − 𝛾𝐹 = .38



𝐽/𝑐= 145.1𝑘/𝑖𝑛2

(𝜙)𝑉𝑐 = .75 ∗ $ ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘





Case D:

Case D Corner

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 310.4 psf


db 1.128 inch

qdu 172 lbs

qlu 65 lbs

One way Shear Check


𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 263.2 𝑓𝑡

2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 81.7k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 390𝑘

bw=l2

Case D Corner One Way Shear d 11.686 in


Two-way Slab Shear

𝐴𝑡 = (𝑙12) ∗𝑙22− (𝑐 +

𝑑

2)2

= 150.06𝑓𝑡^2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 46.58𝑘 b=1 , as=20

𝑏1 = 𝑐 +𝑑

2= 29.843𝑖𝑛 , 𝑏2 = 𝑐 +

𝑑

2= 29.843𝑖𝑛

𝑏0 = 𝑏1 + 𝑏2 = 59.686𝑖𝑛

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=155.2k



Case D Corner Two Way Shear At 150.0652 ft^2


Munb = 147.94k-ft

𝛾𝐹 =1

1 +23 ∗√𝑏1𝑏2

= .6


𝐴𝑐 = (𝑏1 + 𝑏2) ∗ 𝑑 = 697.5𝑖𝑛2

𝐽

𝑐= 𝑏12 ∗ 𝑑(𝑏1 + 4𝑏2) + 𝑑

3(𝑏1 + 𝑏2)

6𝑏1= 22946.5𝑖𝑛3

𝛾𝑣 = 1 − 𝛾𝐹 = .4



𝐽/𝑐= 97.73𝑘/𝑖𝑛2

(𝜙)𝑉𝑐 = .75 ∗ $ ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘






2nd

Floor Two-Way Slab:

Initial slab thickness: h=13in

Per ACI Table 9.5c

Superimposed dead load: DLsup=30psf

Concrete density: γ=150pcf Initial deal load: DL= h* γ=13in*150pcf=162.5psf

Load Combinations:



Dead (D) 192.50 psf 1) 1.4D




Rain ( R) 0.00 psf 5) 1.2D + E + .5L + .2S


Negative Wind (W) 0.00 psf 7) 0.9D + 1E


Results:


1) 269.50 psf - psf Positive: 391.00

2) 391.00 psf - psf Negative: 173.25 3) 281.00 psf 281.00 psf

4) 281.00 psf -

psf

5) 281.00 psf psf

6) 173.25 psf 173.25 psf

7) 173.25 psf - psf



Column cross section dimensions (square): Column Dimensions (CD)= 24in x 24in

L1=L2= 25ft

Ln= L1-CD=22.917ft






8=391.40𝑝𝑠𝑓 ∗ 25𝑓𝑡 ∗ 22.917𝑓𝑡2

8= 642.37𝑘𝑖𝑝 ∗ 𝑓𝑡

2nd Floor Slab Design tables Location along Span Coefficient Moment




End span M+ 0.5 323.1859






End span M+ 0.5 174.5204


Area C Area C

1 19005.44 1 12310.52

2 1781.19 2 4024.057

Sum 20786.63 Sum 16334.58

Variable Value

Ib 18647.85

Is 17968.5

Alpha F1=Ib/Is 1.03780783

Bt 0.578418621






End Span M+ 193.9115625 129.274375


Interior Span M- 315.1062891 105.0354297

Interior Span M+ 135.7380938 90.4920625




End Span M+ 130.8903047 43.63010156


Interior Span M- 170.1573961 30.54107109

Interior Span M+ 91.62321328 30.54107109

Exterior slab panel column strip beam and slab factored

moments k-ft


Slab moment


End Span M+ 120.4190803 54.10132594


Interior Span M- 156.5448044 70.33172372

Interior Span M+ 84.29335622 37.87092816





End Span M+ 120.4190803 44.24363636 164.6627167


Interior Span M- 156.5448044 44.24363636 200.7884408

Interior Span M+ 84.29335622 30.4175 114.7108562

Interior slab panel column strip beam and slab factored



moments k-ft


Slab moment


End Span M+ 222.9982969 100.1876406


Interior Span M- 289.8977859 130.2439328

Interior Span M+ 156.0988078 70.13134844





End Span M+ 222.9982969 44.24363636 267.2419332


Interior Span M- 289.8977859 44.24363636 334.1414223

Interior Span M+ 156.0988078 30.4175 186.5163078


B=12in

HIb=24in




ℎ2 ∗ 𝑥 ∗ ℎ

𝐵 ∗ 𝐻𝑙𝑏 + 𝑥 ∗ ℎ=

242 ∗ 12 ∗ 24 +

112 ∗ 11 ∗ 11

12 ∗ 24 + 11 ∗ 11= 10.18𝑖𝑛

𝐼𝑏 =1


3) + (𝑥(𝐻𝑙𝑏 − 𝐶𝑒𝑛𝑡𝑟𝑜𝑖𝑑)3)]

= 18647.85𝑖𝑛4 Check: Centroid < h 10.18in < 12 in Good!

𝐼𝑠 =1

12[(𝐶𝐷

2+𝐿12) ℎ3 =

1

12[(24

2+25

2) 123 = 17968.5𝑖𝑛4


18647.85

17968.5= 1.038

Check: αF1 > 0.80 1.038 > 0.80 Good!

Reinforcement:


Φ1=0.90




fc’= 5.5ksi

fy=60ksi




=186.35𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 87𝑖𝑛 ∗ 11.8125𝑖𝑛 2

= 204.675𝑝𝑠𝑖

ρ1 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .204675ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))

= 3.50(10−3)



Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 6 (

𝜋

4) (. 875𝑏1

2 ) = 3.61𝑖𝑛2 > 3.594𝑖𝑛2 → 𝐺𝑜𝑜𝑑!

Check: As= 6(0.60in2) = 3.60in

2 > 3.594in

2 Good!


Φ1=0.90

db2= 0.75in

fc’= 5.5ksi

fy=60ksi

ECSW= 87in


𝑀𝑛2 =𝐸𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝐶𝑜𝑙𝑢𝑚𝑛 𝑃𝑎𝑛𝑒𝑙 𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑀𝑜𝑚𝑒𝑛𝑡𝑚𝑎𝑥Φ1 ∗ Exterior Column Panel Width ∗ 𝑑 22

= 193.91 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 87𝑖𝑛 ∗ 11.875𝑖𝑛 2

= 210.74 𝑝𝑠𝑖

ρ2 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .21074ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))

= 3.60(10−3) As = ρ2 ∗ Exterior Column Panel Width ∗ d2 = 3.60(10−3)(87in)(11.875in) = 3.723in2




Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 9 (

𝜋

4) (. 75𝑏1

2 ) = 3.974𝑖𝑛2 > 3.723𝑖𝑛2 → 𝐺𝑜𝑜𝑑!

Check: As= 9(0.44in2) = 3.96in

2 > 3.723 in

2 Good!


Φ1=0.90

db1= 0.875in

fc’= 5.5ksi

fy=60ksi

ICSW= 2*min (.25*L1, .25*L2) = 150in



=339.35 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 150𝑖𝑛 ∗ 11.8125𝑖𝑛 2

= 216.178𝑝𝑠𝑖

ρ1 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .216178ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))

= 3.70(10−3)



Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 12 (

𝜋

4) (. 875𝑏1

2 ) = 7.21 > 6.56 → 𝐺𝑜𝑜𝑑!

Check: As= 12(0.6in2) = 7.20 in

2 > 6.56 in

2 Good!


Φ1=0.90

db2= 0.75in

fc’= 5.5ksi

fy=60ksi

CSW= 150in





=135.74 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 150𝑖𝑛 ∗ 11.875𝑖𝑛 2

= 85.563𝑝𝑠𝑖

ρ2 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .085563ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))

= 1.44(10−3)



Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 6 (

𝜋

4) (. 75𝑏1

2 ) = 2.65𝑖𝑛2 > 2.57𝑖𝑛2 → 𝐺𝑜𝑜𝑑!

Check: As= 6(0.44in2) = 2.64in

2 > 2.57in

2 Good!


Φ1=0.90

db1= 0.75in

fc’= 5.5ksi

fy=60ksi




=113.12 𝑘𝑖𝑝 ∗ 𝑓𝑡 ∗ 12 ∗ 1000

0.90 ∗ 150𝑖𝑛 ∗ 11.875𝑖𝑛 2

= 71.305𝑝𝑠𝑖

ρ1 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .071305ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))



Check: 𝑛𝜋

4(𝑑𝑏12 ) > 𝐴𝑠 → 6 (

𝜋

4) (. 75𝑏1

2 ) = 2.64 𝑖𝑛2 > 2.135in2 → 𝐺𝑜𝑜𝑑!

Check: As= 6(.44in2) = 2.64 in

2 > 2.135 in

2 Good!




Φ1=0.90

db2= 0.875in

fc’= 5.5ksi

fy=60ksi

MSW= 150 in




0.90 ∗ 150𝑖𝑛 ∗ 11.8125𝑖𝑛 2= 82.350𝑝𝑠𝑖

ρ2 = .85 ∗𝑓𝑐′

𝑓𝑦(1 − (1 −

2 ∗ 𝑀𝑛1. 85 ∗ 𝑓𝑐′

) 12) = .85 ∗

5.5𝑘𝑠𝑖

60𝑘𝑠𝑖(1 − ((1 −

2 ∗ .08235ksi

. 85 ∗ 5.5𝑘𝑠𝑖) 12))

= 1.39(10−3)



Check: 𝑛𝜋

4(𝑑𝑏12 ) < 𝐴𝑠 → 6 (

𝜋

4) (. 875𝑏1

2 ) = 3.61𝑖𝑛2 > 2.456𝑖𝑛2 → 𝐺𝑜𝑜𝑑!

Check: As= 6(0.6in2) = 3.60 in

2 > 2.456in

2 Good!

Two-Way Slab Shear 2

nd Floor Slab Shear:

Case A Edge

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 391 psf


db 1.128 inch

qdu 192.5 lbs

qlu 100 lbs

One way Shear Check


𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 263.1542 𝑓𝑡

2



𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 102.9k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 390𝑘

Bw=l2

Case A Edge d 11.686 in


Two-way Slab Shear

𝐴𝑡 = (𝑙12+𝑐

2) ∗ 𝑙2 − 𝑏1 ∗ 𝑏2 = 330.1𝑓𝑡^2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 129.07𝑘 b=1 , as=30

𝑏1 = 𝑐 +𝑑

2= 29.843𝑖𝑛 , 𝑏2 = 𝑐 + 𝑑 = 35.686𝑖𝑛

𝑏0 = 2 ∗ 𝑏1 + 𝑏2 = 95.372𝑖𝑛

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=247.9644k

Case A Edge At 330.1043 ft^2


𝛾𝐹 =1

1+2

3∗√𝑏1𝑏2

= .621

Effective Width = 3h+c = 5.25ft

𝛾𝐹 ∗ 𝑀𝑢𝑛𝑏 = 115.77𝑘 ∗ 𝑓𝑡 𝐴𝑐 = (2𝑏1 + 𝑏2) ∗ 𝑑 = 1199.34𝑖𝑛^2

𝑗

𝑐=2 ∗ 𝑏1

2 ∗ 𝑑(𝑏1 ∗ 2𝑏2) + 𝑑3(2𝑏1 + 𝑏2)

6𝑏1

𝐽

𝑐= 𝑏1 ∗ 𝑑(𝑏1 + 6𝑏2) + 𝑑

3

6= 14686.92𝑖𝑛^3

𝛾𝑣 = 1 − 𝛾𝐹 = .379



𝐽/𝑐= 165.25𝑘/𝑖𝑛2



(𝜙)𝑉𝑐 = .75 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘

Case A Edge Munb 186.35 k-ft



Case B:

Case B Interior

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 391 psf


db 1.128 inch

qdu 192.5 lbs

qlu 100 lbs

One way Shear Check


𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 263.1542 𝑓𝑡

2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 102.9k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 390𝑘

Bw=l2

Case B Interior d 11.686 in


Two-way Slab Shear

𝐴𝑡 = 𝐴𝑡 = (𝑙1) ∗ 𝑙2 − 𝑏1 ∗ 𝑏2 = 616.16𝑓𝑡^2



𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 240.9𝑘 b=1 , as=40

𝑏1 = 𝑐 + 𝑑 = 35.686𝑖𝑛 , 𝑏2 = 𝑐 + 𝑑 = 35.686𝑖𝑛

𝑏0 = 2(𝑏1 + 𝑏2) = 142.744𝑖𝑛

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=371.13k

Case B Interior At 616.1563 ft^2


Munb = 24.23k-ft

𝛾𝐹 =1

1 +23 ∗√𝑏1𝑏2

= .6


𝐴𝑐 = 2(𝑏1 + 𝑏2) ∗ 𝑑 = 1668.11𝑖𝑛^2

𝐽

𝑐= 𝑏1 ∗ 𝑑(𝑏1 + 3𝑏2) + 𝑑

3

3= 20374.64𝑖𝑛^3

𝛾𝑣 = 1 − 𝛾𝐹 = .40



𝐽/𝑐= 155.33𝑘𝑖𝑛2

(𝜙)𝑉𝑐 = .75 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘



Case B Interior Munb 24.23 k-ft



Case C:

Case C Edge

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 391 psf


db 1.128 inch

qdu 192.5 lbs

qlu 100 lbs

One way Shear Check


𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 263.1542 𝑓𝑡

2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 102.9k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 390𝑘

Bw=l2



Two-way Slab Shear



𝐴𝑡 = (𝑙12+𝑐

2) ∗ 𝑙2 − 𝑏1 ∗ 𝑏2 = 330.1𝑓𝑡^2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 129.07𝑘 b=1 , as=30

𝑏1 = 𝑐 +𝑑

2= 29.84𝑖𝑛 , 𝑏2 = 𝑐 + 𝑑 = 35.686

𝑏0 = 2 ∗ 𝑏1 + 𝑏2 = 95.372

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=247.644k

Case C Edge At 330.1043 ft^2


Munb = 186.35k-ft

𝛾𝐹 =1

1 +23 ∗√𝑏1𝑏2

= .62


𝐴𝑐 = (𝑏1 + 2𝑏2) ∗ 𝑑 = 1114.517𝑖𝑛^2

𝐽

𝑐= 2 ∗ 𝑏1

2 ∗ 𝑑(𝑏1 ∗ 2𝑏2) + 𝑑3(2𝑏1 + 𝑏2)

6𝑏1= 12616.1𝑖𝑛^3

𝛾𝑣 = 1 − 𝛾𝐹 = .378



𝐽/𝑐= 182.94𝑘/𝑖𝑛2

(𝜙)𝑉𝑐 = .75 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘





Case D:

Case D Corner

Column Size 24 inch

f'c 5500 psi

fy 60000 psi

L1 25 ft

L2 25 ft

qu 391 psf


db 1.128 inch

qdu 192.5 lbs

qlu 100 lbs

One way Shear Check


𝐴𝑡 = (𝑙12−𝑐

2− 𝑑) ∗ 𝑙2 = 263.1542 𝑓𝑡

2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 102.9k

FVc = (phi) ∗ 2 ∗ √𝑓′𝑐 ∗ 𝑏𝑤 ∗ 𝑑 = 390𝑘

Bw=l2

Case D Corner d 11.686 in




Two-way Slab Shear

𝐴𝑡 = (𝑙12) ∗𝑙22− (𝑐 +

𝑑

2)2

= 150.07𝑓𝑡^2

𝑉𝑢 = 𝐴𝑡 ∗ 𝑞𝑢 = 58.67𝑘 b=1 , as=20

𝑏1 = 𝑐 +𝑑

2= 29.843𝑖𝑛 , 𝑏2 = 𝑐 +

𝑑

2= 29.843𝑖𝑛

𝑏0 = 𝑏1 + 𝑏2 = 59.686𝑖𝑛

(𝜙)𝑉𝑐 = (𝜙)min (2 +4

𝛽, 2 +

𝛼𝑠∗𝑑

𝑏0, 4) 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑=155.1819k

Case D Corner At 150.0652 ft^2


Munb=full slab moment (col+mid) for and edge column=End span exterior M- = 186.35k-ft

𝛾𝐹 =1

1 +23 ∗√𝑏1𝑏2

= .6

𝛾𝐹 ∗ 𝑀𝑢𝑛𝑏 = 111.81𝑘 ∗ 𝑓𝑡

Effective Width = 3h+c = 5ft

𝐴𝑐 = (2𝑏1 + 𝑏2) ∗ 𝑑 = 697.5𝑖𝑛^2

𝐽

𝑐= 𝑏12 ∗ 𝑑(𝑏1 + 4𝑏2) + 𝑑

3(𝑏1 + 𝑏2)

6𝑏1= 22946.5𝑖𝑛^3

𝛾𝑣 = 1 − 𝛾𝐹 = .38



𝐽/𝑐= 123.1𝑘/𝑖𝑛2

(𝜙)𝑉𝑐 = .75 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 = 222.486𝑘






CONTINUOUS L-BEAM DESIGN Pg 143


CONCRETE CONTINUOUS L-BEAM DESIGN:

SPANNING GREEN ROOF AND THIRD FLOOR:

Design the continuous beam for the following loads by taking worst case for span:

𝑤𝑢 = 10.3 𝑘/𝑓𝑡


𝐹𝑜𝑟 𝑙𝑜𝑛𝑔 𝑡𝑒𝑟𝑚(𝑑𝑒𝑎𝑑) + 𝑖𝑚𝑚𝑒𝑑𝑖𝑎𝑡𝑒(𝑑𝑒𝑎𝑑 + 𝑙𝑖𝑣𝑒 + 𝑤𝑖𝑛𝑑) =𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛

𝐹𝑜𝑟 𝑖𝑚𝑚𝑒𝑑𝑖𝑎𝑡𝑒 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 (𝑙𝑖𝑣𝑒 + 𝑤𝑖𝑛𝑑 𝑜𝑛𝑙𝑦) =𝐿

360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .833 𝑖𝑛

Use ASTM A615 Gr 60 rebar and 5,500 psi concrete.

GRAVITY MOMENT DESIGN:

From two-way slab design/analysis: 𝑀𝑢 = 230 𝑘 ∗ 𝑓𝑡

Estimate the beam sizeℎ = 24 𝑖𝑛, 𝑏𝑤𝑒𝑏 = 16 𝑖𝑛, 𝑑 = 21.5 𝑖𝑛, 𝑡 = 12 𝑖𝑛.

Effective Flange Width Conditions:

𝑏𝑒 =𝐿

12= 25 𝑖𝑛, 𝑏𝑒 = 6𝑡 + 𝑏𝑤 = 88 𝑖𝑛, 𝑏𝑒 = 𝑏 = 25 𝑖𝑛.

Select smallest value of 25in.

Solve for steel percentage using the following quadratic equation:

𝑀𝑛 = 𝜌𝑓𝑦𝑏𝑑2 (1 −

𝜌𝑓𝑦

1.7𝑓′𝑐) → 𝜌 = .0065

Check against the minimum percentage:

𝜌𝑚𝑖𝑛 = max(200

𝑓𝑦,3√𝑓′𝑐

𝑓𝑦) = 0.00371 < .0065 → 𝑂𝐾

Assume Lever Arm (z):

𝑧 = 0.9𝑑 = 19.35 𝑜𝑟 𝑧 = 𝑑 −𝑡

2= 15.5

Select Largest Value: 19.35 in.

Find the required area of steel:

𝐴𝑠 = 𝑀𝑛/𝑓𝑦𝑧 = (230 ∗ 12)/(60,000 ∗ 19.35) = 2.38 𝑖𝑛2

Compute 𝐴𝑐:

0.85𝑓′𝑐𝐴𝑐 = 𝐴𝑠𝑓𝑦 → 𝐴𝑐 =

𝐴𝑠𝑓𝑦

0.85𝑓′𝑐

= 30.51 𝑖𝑛2



𝐴𝑐 < (𝑡 ∗ 𝑏𝑒) < 108 𝑖𝑛2 𝑂𝐾

Compute a:

a=𝐴𝑐/𝑏𝑒 = 1.22 𝑖𝑛

Compute z:

𝑧 = 𝑑 −𝑎

2= 20.89 𝑖𝑛

Calculate 𝐴𝑠 with revised z:

𝐴𝑠 = 𝑀𝑛/𝑓𝑦𝑧 = (230 ∗ 12)/(60,000 ∗ 20.89) = 2.20 𝑖𝑛2

Compute 𝐴𝑐 𝑤𝑖𝑡ℎ 𝑟𝑒𝑣𝑖𝑠𝑒𝑑 𝐴𝑠 :


𝐴𝑠𝑓𝑦

0.85𝑓′𝑐

= 28.26 𝑖𝑛2

Compute a:


Compute z:

𝑧 = 𝑑 −𝑎

2= 20.93 𝑖𝑛


𝐴𝑠 = 𝑀𝑛/𝑓𝑦𝑧 = (230 ∗ 12)/(60,000 ∗ 20.93) = 2.20 𝑖𝑛2

OK (Close to previous Value)

Checking Minimum Reinforcing:

𝐴𝑠 𝑚𝑖𝑛 =3√𝑓′𝑐𝑏𝑤𝑑

𝑓𝑦=3 ∗ √5500 ∗ 16 ∗ 21.5

60000 = 1.28 𝑖𝑛2

But not less than:

𝐴𝑠 𝑚𝑖𝑛 =200𝑏𝑤𝑑

𝑓𝑦=200 ∗ 16 ∗ 21.5

60000 = 1.15 𝑖𝑛2 < 2.2 𝑖𝑛2 → 𝑂𝐾

Or:

𝐴𝑠 𝑚𝑖𝑛 = 𝜌𝑚𝑖𝑛𝑏𝑤𝑑 = 0.0033 ∗ 16 ∗ 21.5 = 1.14 𝑖𝑛2 < 2.2 𝑖𝑛2 → 𝑂𝐾

Check steel strain:

Compute c:

𝛽1 = .775 @ 5500 𝑝𝑠𝑖, 𝑐 =𝑎

𝛽1= 1.57𝑖𝑛

Compute Steel Strain 𝜀𝑡:



𝜀𝑡 = 0.003𝑑 − 𝑐

𝑐= 0.003

21.5 − 1.57

1.57= .038 > .005 → 𝑂𝐾

Check Φ:

Φ = 0.9 𝑓𝑜𝑟 𝜀𝑡 > .005

Select Reinforcing:

𝐴𝑠 𝑟𝑒𝑞′𝑑: 2.20 𝑖𝑛2

Bar Design:

4#9 bars with 𝐴𝑠 : 4.00 𝑖𝑛2 with diameter 1.128 in

Stirrup Design:

#3 with diameter 0.375 in

Check fit of bars within beam width:

𝑏𝑚𝑖𝑛 = 1.5 + .375 + 2(. 375) + 4(1.128) + 3(. 375) + .375 + 1.5 = 15.76 𝑖𝑛 < 16 𝑖𝑛 → 𝑂𝐾

Verify capacity:

Φ𝑀𝑛 = Φ𝐴𝑠𝑓𝑦 (𝑑 −𝑎

2) = .9(4)(60) (21.5 −

1.13

2) = 376.83 𝑘 ∗ 𝑓𝑡 > 𝑀𝑢 → 𝑂𝐾

Check shear:

Find shear load at support:

𝑉𝑢 = 70.45 𝑘 𝑓𝑟𝑜𝑚 𝑉𝑖𝑠𝑢𝑎𝑙 𝐴𝑛𝑎𝑙𝑦𝑠𝑖𝑠 At a distance d from the support:

𝑉𝑢𝑑 = 70.45 −21.5

12∗ 10.3 = 52.00 𝑘

Required Vn:

𝑉𝑛 =𝑉𝑢𝑑Φ=52.00

0.75= 69.33 𝑘

Finding shear strength of concrete:

𝑉𝑐 = 2𝜆√𝑓′𝑐𝑏𝑑 = 2(1)(√5500)(16)(21.5) = 51.02 𝑘

Vn exceeds Vc, so reinforcement is required.

Find Vs:

𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 69.33 − 51.00 = 18.30 𝑘

Use #3 stirrups (𝐴𝑣 = 2 ∗ .11 = .22 𝑖𝑛2)

Maximum stirrup spacing:

𝑠 =𝐴𝑣𝑓𝑦𝑑

𝑉𝑠=. 22 ∗ 60 ∗ 21.5

18.30= 15.5 𝑖𝑛



Use 7 inches.

Check max spacing limits:

𝑠𝑚𝑎𝑥 = min(𝐴𝑣𝑓𝑦

. 75𝑏√𝑓′𝑐,𝐴𝑣𝑓𝑦

50𝑏) = min(14.83,16.5) = 14.83 𝑖𝑛

Based on beam geometry:

4√𝑓′𝑐𝑏𝑑 = 102.05𝑘

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑠𝑚𝑎𝑥 = min (𝑑

2, 24) = 11 𝑖𝑛 𝑆𝑖𝑛𝑐𝑒 𝑖𝑡 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑡ℎ𝑎𝑛 15.5𝑖𝑛 𝑢𝑠𝑒 11𝑖𝑛 𝑠𝑝𝑎𝑐𝑖𝑛𝑔𝑠

Check max Vs:

𝑉𝑠,𝑚𝑎𝑥 = 8√𝑓′𝑐𝑏𝑑 = 204.09𝑘 > 51.02𝑘 → 𝑂𝐾

Verify capacity:

𝑉𝑠 =𝐴𝑣𝑓𝑦𝑑

𝑠= 18.92 𝑘 > 18.30𝑘 → 𝑂𝐾

Φ𝑉𝑛 = .75(51.02 + 18.30) = 52.00 𝑘𝑖𝑝 ≥ 𝑉𝑢𝑑 → 𝑂𝐾

Check deflection:

Due to uniform load,

∆=𝑤𝑙3

192𝐸𝑐𝐼𝑒

Finding Ec:

𝐸𝑐 = 57000√𝑓′𝑐 = 4227 𝑘𝑠𝑖

Finding modulus of rupture:

𝑓𝑟 = 7.5𝜆√𝑓′𝑐 = 556 𝑝𝑠𝑖=.556ksi

Gross moment of inertia:

𝐼𝑔 = 22809.6 𝑖𝑛4

Distance from neutral axis of gross (uncracked section to tension face):

𝑦𝑡 =ℎ

2=24

2= 12 𝑖𝑛

Cracking moment:

𝑀𝑐𝑟 =𝑓𝑟𝐼𝑔

𝑦𝑡= 1057.25 𝑘 ∗ 𝑖𝑛

Service moment:

𝑀𝑎 =𝑤𝑢𝑙

2

8=7.23 ∗ 252

8= 4781.25 𝑘 ∗ 𝑖𝑛, 7.23 = 𝐴𝑆𝐷 𝑤𝑢



Transform steel to concrete:

𝑛 =𝐸𝑠𝐸𝑐=29000

4227= 6.86

Gravity: 𝐴𝑠 = 4.00 𝑖𝑛2, 𝑑 = 21.5 𝑖𝑛

Solve for x:

𝑏𝑥𝑥

2= 𝑛𝐴𝑠(𝑑 − 𝑥) → 𝑥 = 7.0422 𝑖𝑛

Find Icr:

𝐼𝑐𝑟 =𝑏𝑥3

3+ 𝑛𝐴𝑠(𝑑 − 𝑥)

2 = 7598.58 𝑖𝑛4

Find Ie:

𝐼𝑒 = [(𝑀𝑐𝑟𝑀𝑎)3

] 𝐼𝑔 + [1 − (𝑀𝑐𝑟𝑀𝑎)3

] 𝐼𝑐𝑟 = 7649.88 𝑖𝑛4

Checking deflection:

∆=𝑤𝑙3

192𝐸𝑐𝐼𝑒= 0.03 𝑖𝑛

𝐹𝑜𝑟 𝑙𝑖𝑣𝑒 𝑎𝑛𝑑 𝑤𝑖𝑛𝑑 𝑜𝑛𝑙𝑦, ∆< .833 𝑖𝑛 → 𝑂𝐾

For long term deflection:

Gravity dead only: Δ𝐷 = 1.2 𝑖𝑛

𝜆Δ =𝜉

1 + 50𝜌′= 2.0 𝑓𝑜𝑟 𝑛𝑜 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙

Δ𝐿𝑇 = 𝜆ΔΔ𝐷 + Δ𝐼𝑀𝐺 = 0.12 𝑖𝑛

Δ𝑇 = (1 + 𝜆𝐷)Δ𝐷 + Δ𝐼𝑀𝐺 = 0.17 𝑖𝑛 < 1.2 𝑖𝑛 → 𝑂𝐾

MOMENT DIAGRAM:

Note: Moment values in analysis were taken from slab analysis.

SHEAR DIAGRAM:



Note: Highest shear taken for use in analysis for worst case.

DEFLECTION DIAGRAM:

Note: Deflection in analysis and Visual Analysis both pass. Visual Analysis doesn’t account for slab.

LOADING CASES:

L-Beam 3rd Floor Dead Load: Note: Loading values are calculated by taking dead load in psf from LRFD load combination Excel and multiplying by the span. This value is the peak of the triangular loadings (5.015 k/ft). The uniform load is the beam stem weight distributed across the beam (0.2k/ft).

L-Beam 3rd Floor Live Load Note: Loading values are calculated by taking live load in psf from LRFD load combination Excel and multiplying by the span. This value is the peak of the triangular loadings (2.5 k/ft).



L-Beam 3rd Floor Wind Load Note: Loading values are calculated by taking live load in psf from LRFD load combination Excel and multiplying by the span. This value is the peak of the uniform load (0.59 k/ft).

L-Beam 3rd Floor Wind Uplift Load Note: Note: Loading values are calculated by taking live load in psf from LRFD load combination Excel and multiplying by the span. This value is the peak of the uniform load (0.86 k/ft).

SPANNING SECOND FLOOR:

Design the continuous beam for the following loads by taking worst case for span:

𝑤𝑢 = 9.78 𝑘/𝑓𝑡


𝐹𝑜𝑟 𝑙𝑜𝑛𝑔 𝑡𝑒𝑟𝑚(𝑑𝑒𝑎𝑑) + 𝑖𝑚𝑚𝑒𝑑𝑖𝑎𝑡𝑒(𝑑𝑒𝑎𝑑 + 𝑙𝑖𝑣𝑒 + 𝑤𝑖𝑛𝑑) =𝐿

240=25 𝑓𝑡 ∗ 12 𝑖𝑛

240= 1.25 𝑖𝑛

𝐹𝑜𝑟 𝑖𝑚𝑚𝑒𝑑𝑖𝑎𝑡𝑒 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 (𝑙𝑖𝑣𝑒 + 𝑤𝑖𝑛𝑑 𝑜𝑛𝑙𝑦) =𝐿

360=25 𝑓𝑡 ∗ 12 𝑖𝑛

360= .833 𝑖𝑛

Use ASTM A615 Gr 60 rebar and 5,500 psi concrete.

GRAVITY MOMENT DESIGN:

From two-way slab design/analysis: 𝑀𝑢 = 217 𝑘 ∗ 𝑓𝑡

Estimate the beam size ℎ = 24 𝑖𝑛, 𝑏𝑤𝑒𝑏 = 16 𝑖𝑛, 𝑑 = 21.5 𝑖𝑛, 𝑡 = 12 𝑖𝑛.

Effective Flange Width Conditions:

𝑏𝑒 =𝐿

12= 25 𝑖𝑛, 𝑏𝑒 = 6𝑡 + 𝑏𝑤 = 88 𝑖𝑛, 𝑏𝑒 = 𝑏 = 25 𝑖𝑛.

Select smallest value of 25in.

Solve for steel percentage using the following quadratic equation:

𝑀𝑛 = 𝜌𝑓𝑦𝑏𝑑2 (1 −

𝜌𝑓𝑦

1.7𝑓′𝑐) → 𝜌 = .0061

Check against the minimum percentage:



𝜌𝑚𝑖𝑛 = max(200

𝑓𝑦,3√𝑓′𝑐

𝑓𝑦) = 0.00371 < .0061 → 𝑂𝐾

Assume Lever Arm (z):

𝑧 = 0.9𝑑 = 19.35 𝑜𝑟 𝑧 = 𝑑 −𝑡

2= 15.5

Select Largest Value: 19.35 in.

Find the required area of steel:

𝐴𝑠 = 𝑀𝑛/𝑓𝑦𝑧 = (217 ∗ 12)/(60,000 ∗ 19.35) = 2.24 𝑖𝑛2

Compute 𝐴𝑐:


𝐴𝑠𝑓𝑦

0.85𝑓′𝑐

= 28.79 𝑖𝑛2

𝐴𝑐 < (𝑡 ∗ 𝑏𝑒) < 108 𝑖𝑛2 𝑂𝐾

Compute a:


Compute z:

𝑧 = 𝑑 −𝑎

2= 20.92 𝑖𝑛


𝐴𝑠 = 𝑀𝑛/𝑓𝑦𝑧 = (217 ∗ 12)/(60,000 ∗ 20.92) = 2.24 𝑖𝑛2

Compute 𝐴𝑐 𝑤𝑖𝑡ℎ 𝑟𝑒𝑣𝑖𝑠𝑒𝑑 𝐴𝑠 :


𝐴𝑠𝑓𝑦

0.85𝑓′𝑐

= 26.62 𝑖𝑛2

Compute a:


Compute z:

𝑧 = 𝑑 −𝑎

2= 20.97 𝑖𝑛


𝐴𝑠 = 𝑀𝑛/𝑓𝑦𝑧 = (230 ∗ 12)/(60,000 ∗ 20.93) = 2.07 𝑖𝑛2

OK (Close to previous Value)

Checking Minimum Reinforcing:

𝐴𝑠 𝑚𝑖𝑛 =3√𝑓′𝑐𝑏𝑤𝑑

𝑓𝑦=3 ∗ √5500 ∗ 16 ∗ 21.5

60000 = 1.28 𝑖𝑛2



But not less than:

𝐴𝑠 𝑚𝑖𝑛 =200𝑏𝑤𝑑

𝑓𝑦=200 ∗ 16 ∗ 21.5

60000 = 1.15 𝑖𝑛2 < 2.07 𝑖𝑛2 → 𝑂𝐾

Or:

𝐴𝑠 𝑚𝑖𝑛 = 𝜌𝑚𝑖𝑛𝑏𝑤𝑑 = 0.0033 ∗ 16 ∗ 21.5 = 1.14 𝑖𝑛2 < 2. .07 𝑖𝑛2 → 𝑂𝐾

Check steel strain:

Compute c:

𝛽1 = .775 @ 5500 𝑝𝑠𝑖, 𝑐 =𝑎

𝛽1= 1.49𝑖𝑛

Compute Steel Strain 𝜀𝑡:

𝜀𝑡 = 0.003𝑑 − 𝑐

𝑐= 0.003

21.5 − 1.57

1.57= .04 > .005 → 𝑂𝐾

Check Φ:

Φ = 0.9 𝑓𝑜𝑟 𝜀𝑡 > .005

Select Reinforcing:

𝐴𝑠 𝑟𝑒𝑞′𝑑: 2.07 𝑖𝑛2

Bar Design:

3#9 bars with 𝐴𝑠 : 3.00 𝑖𝑛2 with diameter 1.128 in

Stirrup Design:

#3 with diameter 0.375 in

Check fit of bars within beam width:

𝑏𝑚𝑖𝑛 = 1.5 + .375 + 2(. 375) + 3(1.128) + 2(. 375) + .375 + 1.5 = 12.63 𝑖𝑛 < 16 𝑖𝑛 → 𝑂𝐾

Verify capacity:

Φ𝑀𝑛 = Φ𝐴𝑠𝑓𝑦 (𝑑 −𝑎

2) = .9(3)(60) (21.5 −

1.13

2) = 283.06 𝑘 ∗ 𝑓𝑡 > 𝑀𝑢 → 𝑂𝐾

Check shear:

Find shear load at support:

𝑉𝑢 = 60 𝑘 𝑓𝑟𝑜𝑚 𝑉𝑖𝑠𝑢𝑎𝑙 𝐴𝑛𝑎𝑙𝑦𝑠𝑖𝑠 At a distance d from the support:

𝑉𝑢𝑑 = 60 −21.5

12∗ 9.78 = 42.48 𝑘

Required Vn:



𝑉𝑛 =𝑉𝑢𝑑Φ=42.48

0.75= 56.64 𝑘

Finding shear strength of concrete:

𝑉𝑐 = 2𝜆√𝑓′𝑐𝑏𝑑 = 2(1)(√5500)(16)(21.5) = 51.02 𝑘

Vn exceeds Vc, so reinforcement is required.

Find Vs:

𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 56.64 − 51.02 = 5.61 𝑘

Use #3 stirrups (𝐴𝑣 = 2 ∗ .11 = .22 𝑖𝑛2)

Maximum stirrup spacing:

𝑠 =𝐴𝑣𝑓𝑦𝑑

𝑉𝑠=. 22 ∗ 60 ∗ 21.5

5.61= 50.56 𝑖𝑛

Use 7 inches.

Check max spacing limits:

𝑠𝑚𝑎𝑥 = min(𝐴𝑣𝑓𝑦

. 75𝑏√𝑓′𝑐,𝐴𝑣𝑓𝑦

50𝑏) = min(14.83,16.5) = 14.83 𝑖𝑛

Based on beam geometry:

4√𝑓′𝑐𝑏𝑑 = 102.05𝑘

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑠𝑚𝑎𝑥 = min (𝑑

2, 24) = 11 𝑖𝑛 𝑆𝑖𝑛𝑐𝑒 𝑖𝑡 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑡ℎ𝑎𝑛 50.6𝑖𝑛 𝑢𝑠𝑒 11𝑖𝑛 𝑠𝑝𝑎𝑐𝑖𝑛𝑔𝑠

Check max Vs:

𝑉𝑠,𝑚𝑎𝑥 = 8√𝑓′𝑐𝑏𝑑 = 204.09𝑘 > 51.02𝑘 → 𝑂𝐾

Verify capacity:

𝑉𝑠 =𝐴𝑣𝑓𝑦𝑑

𝑠= 5.68𝑘 > 5.61𝑘 → 𝑂𝐾

Φ𝑉𝑛 = .75(51.02 + 5.61) = 42.48 𝑘𝑖𝑝 ≥ 𝑉𝑢𝑑 → 𝑂𝐾

Check deflection:

Due to uniform load,

∆=𝑤𝑙3

192𝐸𝑐𝐼𝑒

Finding Ec:

𝐸𝑐 = 57000√𝑓′𝑐 = 4227 𝑘𝑠𝑖

Finding modulus of rupture:

𝑓𝑟 = 7.5𝜆√𝑓′𝑐 = 556 𝑝𝑠𝑖=.556ksi



Gross moment of inertia:

𝐼𝑔 = 22809.6 𝑖𝑛4

Distance from neutral axis of gross (uncracked section to tension face):

𝑦𝑡 =ℎ

2=24

2= 12 𝑖𝑛

Cracking moment:

𝑀𝑐𝑟 =𝑓𝑟𝐼𝑔

𝑦𝑡= 1057.25 𝑘 ∗ 𝑖𝑛

Service moment:

𝑀𝑎 =𝑤𝑢𝑙

2

8=7.31 ∗ 252

8= 6853.13 𝑘 ∗ 𝑖𝑛, 7.31 = 𝐴𝑆𝐷 𝑤𝑢

Transform steel to concrete:

𝑛 =𝐸𝑠𝐸𝑐=29000

4227= 6.86

Gravity: 𝐴𝑠 = 3.00 𝑖𝑛2, 𝑑 = 21.5 𝑖𝑛

Solve for x:

𝑏𝑥𝑥

2= 𝑛𝐴𝑠(𝑑 − 𝑥) → 𝑥 = 6.2613 𝑖𝑛

Find Icr:

𝐼𝑐𝑟 =𝑏𝑥3

3+ 𝑛𝐴𝑠(𝑑 − 𝑥)

2 = 6088.4 𝑖𝑛4

Find Ie:

𝐼𝑒 = [(𝑀𝑐𝑟𝑀𝑎)3

] 𝐼𝑔 + [1 − (𝑀𝑐𝑟𝑀𝑎)3

] 𝐼𝑐𝑟 = 6149.79 𝑖𝑛4

Checking deflection:

∆=𝑤𝑙3

192𝐸𝑐𝐼𝑒= 0.04 𝑖𝑛

𝐹𝑜𝑟 𝑙𝑖𝑣𝑒 𝑎𝑛𝑑 𝑤𝑖𝑛𝑑 𝑜𝑛𝑙𝑦, ∆< .833 𝑖𝑛 → 𝑂𝐾

For long term deflection:

Gravity dead only: Δ𝐷 = 1.2 𝑖𝑛

𝜆Δ =𝜉

1 + 50𝜌′= 2.0 𝑓𝑜𝑟 𝑛𝑜 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙

Δ𝐿𝑇 = 𝜆ΔΔ𝐷 + Δ𝐼𝑀𝐺 = 0.15 𝑖𝑛

Δ𝑇 = (1 + 𝜆𝐷)Δ𝐷 + Δ𝐼𝑀𝐺 = 0.21 𝑖𝑛 < 1.2 𝑖𝑛 → 𝑂𝐾



MOMENT DIAGRAM:

Note: Moment values in analysis were taken from slab analysis.

SHEAR DIAGRAM:

DEFLECTION DIAGRAM:

Note: Deflection in analysis and Visual Analysis both pass. Visual Analysis doesn’t account for slab.

LOADING CASES: L-Beam 2nd Floor Dead Load

Note: Loading values are calculated by taking dead load in psf from LRFD load combination Excel and multiplying by the span. This value is the peak of the triangular loadings (4.89 k/ft). The uniform load is the beam stem weight distributed across the beam (0.2k/ft).



L-Beam 2nd Floor Live Load

Note: Loading values are calculated by taking live load in psf from LRFD load combination Excel and multiplying by the span. This value is the peak of the triangular loadings (2.5 k/ft).

CONCRETE COLUMN DESIGNS Pg 156


CONCRETE COLUMN DESIGN:

SPANNING GREEN ROOF AND THIRD FLOOR:

Design the continuous column with f’c = 4000 psi and fy = 60000 psi.

COMPUTING DESIGN VALUES:

Assuming F=0.65

Obtaining Pu from Slab Analysis Axial Loading:

Pu = 885.82k

Solving for Pn:

𝑃𝑛 =𝑃𝑛0.65

=885.82𝑘

0.65= 1362.8𝑘

From Slab Analysis Unbalanced Moment:

Mux = 123.6 k-ft

Muy = 123.6 k-ft

Mnx = 190.2 k-ft

Mnx = 190.2 k-ft

As a result of biaxial bending, the design moment about the x- or y- axis is assumed to equal:

𝑀𝑛𝑥 +𝑀𝑛𝑦 = 190.2𝑘 − 𝑓𝑡 + 190.2 𝑘 − 𝑓𝑡 = 380.3𝑘 − 𝑓𝑡

DETERMINING STEEL REQUIRED:

𝑒𝑥 = 𝑒𝑦 =(12𝑖𝑛𝑓𝑡) (380.3𝑘 − 𝑓𝑡)

1362.8𝑘= 3.35 𝑖𝑛

𝛾 =20𝑖𝑛

24𝑖𝑛= 0.83

Obtaining rg from Graph 8 Appendix A:

CONCRETE COLUMN DESIGNS Pg 157


𝑅𝑛 =𝑃𝑛𝑒

𝑓′𝑐𝐴𝑔ℎ

=(1362.8𝑘)(3.35𝑖𝑛)

(4000𝑝𝑠𝑖)(24𝑖𝑛 ∗ 24𝑖𝑛)(24𝑖𝑛)= 0.08

𝐾𝑛 =𝑃𝑛𝑓′𝑐𝐴𝑔=

(1362.8𝑘)

(4000𝑝𝑠𝑖)(24𝑖𝑛 ∗ 24𝑖𝑛)= 0.59

Thus by interpolation from Graph 8 Appendix A with bars on all four faces: rg = 0.01

Determining As:

As = rg bd = (0.01)(24in)(24in)= 5.76in2

Use 8#8 Bars with As = 6.32 in2

DETERMINING TIES:

From ACI 7.10.5.2 and ACI 7.10.5.3 select minimum of 3 spacing conditions:

Assume #3 Ties

1) 48in x Tie Dia.

2) 16in x Vert. Bar Dia.

3) Least Dimension.

Computing Conditions:

1) 48in x 3/8in = 18in

2) 16in x 1in = 16in

3) Least Dim. = 24in

Select 16in for tie spacing.

Use #3 ties @ 16in

SIMPLE SHEAR CONNECTIONS DESIGNS Pg 158


SIMPLE SHEAR CONNECTIONS

The simple shear connections were designed using the recommended maximum bolting from the

AISC manual (Tables 10-1 and 10-2) for double angles. The design used Group A bolts, threads

included in the shear plane and standard sized holes. The following table summarizes the

maximum amount of bolts per connection based on the depth of each wide flange in our design.

It also shows the angle thickness for each of these connections.

Beam Depth No. Bolts Bolt Plate

Thickness

10 2 3/4" 1/4"

12 3 3/4" 1/4"

14 3 3/4" 1/4"

16 4 3/4" 5/16"

18 5 3/4" 3/8"

Table 10-1 in the manual was used to determine the capacity of each bolted connection, which

was then compared to the shear values found from the beam designs. An Excel “IF” function

was used to check whether the capacities meet these shear loadings. The results are shown

below:

Beam No. Bolts Bolt Dia. Angle

Thickness Shear (k) Capacity

(k) Check

12x16 3 3/4" 1/4" 4.99 50.9 OK

12x40 3 3/4" 1/4" 2.2 50.9 OK

12x87 3 3/4" 1/4" 21.6 50.9 OK

14x26 3 3/4" 1/4" 7.85 50.9 OK

14x30 3 3/4" 1/4" 14.7 50.9 OK

14x30 3 3/4" 1/4" 4.513 50.9 OK

16x31 4 3/4" 5/16" 16.6 83.9 OK

16x31 4 3/4" 5/16" 15.4 83.9 OK

16x31 4 3/4" 5/16" 15.91 83.9 OK

18x35 5 3/4" 3/8" 18.92 119 OK

18x55 5 3/4" 3/8" 7.62 119 OK

18x55 5 3/4" 3/8" 28.07 119 OK

18x55 5 3/4" 3/8" 19.2 119 OK

18x97 5 3/4" 3/8" 63.5 119 OK

18x97 5 3/4" 3/8" 68.8 119 OK

18x97 5 3/4" 3/8" 55.1 119 OK

As the results show, all of the simple shear bolted connections passed the design checks.

SIMPLE SHEAR CONNECTIONS DESIGNS Pg 159


However, the design calls for bolts in the beam flanges and welds to the column webs. Per AISC

Table 10-2, this calls for Weld type B. Using this table, the following capacities and design

checks are summarized below:

Beam

Col. Flange

Thickness Col. Web Thickness

Angle Thickness

Weld Length

(in)

Weld Size

(D/16) Capacity

(k)

Min Support

Thickness (in.)

Shear (k) Check

12x16 .86" .525" 1/4" 8.5 0.25 32.1 0.19 4.99 OK

12x40 .86" .525" 1/4" 8.5 0.25 32.1 0.19 2.2 OK

12x87 .86" .525" 1/4" 8.5 0.25 32 0.19 21.6 OK

14x26 .86" .525" 1/4" 8.5 0.25 32.1 0.19 7.85 OK

14x30 .86" .525" 1/4" 8.5 0.25 32.1 0.19 14.7 OK

14x30 .86" .525" 1/4" 8.5 0.25 32.1 0.19 4.513 OK

16x31 .86" .525" 5/16" 11.5 0.25 53.3 0.19 16.6 OK

16x31 .86" .525" 5/16" 11.5 0.25 53.3 0.19 15.4 OK

16x31 .86" .525" 5/16" 11.5 0.25 53.3 0.19 15.91 OK

18x35 .86" .525" 3/8" 14.5 0.25 76.4 0.19 18.92 OK

18x55 .86" .525" 3/8" 14.5 0.25 76.4 0.19 7.62 OK

18x55 .86" .525" 3/8" 14.5 0.25 76.4 0.19 28.07 OK

18x55 .86" .525" 3/8" 14.5 0.25 76.4 0.19 19.2 OK

18x97 .86" .525" 3/8" 14.5 0.25 76.4 0.19 63.5 OK

18x97 .86" .525" 3/8" 14.5 0.25 76.4 0.19 68.8 OK

18x97 .86" .525" 3/8" 14.5 0.25 76.4 0.19 55.1 OK

ALL DESIGN CHECKS ARE SATISFIED. THEREFORE CONSTRUCT ALL SIMPLE

SHEAR CONNECTIONS ACCORDING TO AUTOCAD DRAWINGS.

ROOF CONNECTIONS DESIGN Pg 160


ROOF CONNECTIONS

FOR WELDED CONNECTIONS FROM JOIST TO JOIST GIRDER AND JOIST TO

PERIMETER BEAM (W12x87):

According to the Vulcraft Manual, one should use the following minimum welds: two 1/8 inch

fillet welds that are 1 inch long.

From the roof joist design, the largest ASD load combination amounts to 282.55 lb/ft. Using this

information, one can find the end reactions of the joist:

𝑤𝑙

2=(282.55

𝑙𝑏𝑓𝑡∗ 50 𝑓𝑡)

2= 7.06 𝑘

For .81-in. flange thickness, 𝐷𝑚𝑖𝑛 = 5 from AISC Specification Table J2.4.

Try 5/16-in fillet welds, 𝐷 = 5. From AISC equation 8-2:

𝐿𝑟𝑒𝑞 =𝑃

2𝐷(0.928)=

7.06 𝑘𝑖𝑝𝑠

2(5)(0.928)= .761 𝑖𝑛

Weld the minimum length required by Vulcraft:

𝐿 = 1 𝑖𝑛 > .761 𝑖𝑛 → 𝑂𝐾

Check the maximum size of fillet weld:

𝑀𝑎𝑥 𝑠𝑖𝑧𝑒 = .375 − .0625 = .3125 𝑖𝑛 = .3125 𝑖𝑛 → 𝑂𝐾

Check the minimum length:

𝐿𝑚𝑖𝑛 = 4(. 1875 𝑖𝑛) = .75 𝑖𝑛 < 1 𝑖𝑛 → 𝑂𝐾

Check the weld length to weld size ratio: 𝐿

𝑤=

1

. 3125= 3.2 < 100 → 𝐶ℎ𝑒𝑐𝑘 𝛽 𝑓𝑎𝑐𝑡𝑜𝑟 𝑛𝑜𝑡 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦

Check Base Metals by the largest effective weld size. (𝑎 =𝐷

16)

For the Perimeter Beam:

𝑎 =Ω𝐹𝑢

. 707𝐹𝐸𝑋𝑋𝑡𝐵𝑀 =

2 ∗ 65 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 81 𝑖𝑛 = 2.63 𝑖𝑛 > .3125 𝑖𝑛 → 𝑂𝐾

THEREFORE USE 2 1 INCH 5/16 FILLET WELDS.

FOR UNSTIFFENED SEATED CONNECTIONS FROM JOIST GIRDERS TO COLUMNS:

Solving for joist girder reactions gives a design load of 𝑅 = 61.2 𝑘𝑖𝑝𝑠.



Use ¾-in-diamter Group A bolts in standard holes to connect the supported joist girder to the

seat. Use 70 ksi electrode welds to connect the seat to the column flange (W14x109) and ASTM

A36 angles.

Shear Yielding and Flexural Yielding of Angle

According to Vulcraft, assume 𝑙𝑏 = .5 𝑖𝑛. Use AISC Table 10-6 with an angle length of

8 inches and thickness of 5/8-in.

𝐴𝑛𝑔𝑙𝑒 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ: 𝑅

Ω= 72 𝑘𝑖𝑝𝑠 > 61.2 𝑘𝑖𝑝𝑠 → 𝑂𝐾

𝑊𝑒𝑙𝑑 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ, 8𝑥4 𝑠𝑒𝑎𝑡 𝑎𝑛𝑔𝑙𝑒 𝑠𝑖𝑧𝑒, .5 𝑖𝑛 𝑓𝑖𝑙𝑙𝑒𝑡 𝑤𝑒𝑙𝑑𝑠: 𝑅

Ω= 71.2 𝑘𝑖𝑝𝑠 > 61.2 𝑘𝑖𝑝𝑠

→ 𝑂𝐾

𝐵𝑜𝑙𝑡 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ, 𝐶𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛 𝑇𝑦𝑝𝑒 𝐸:𝑅

Ω= 71.6 𝑘𝑖𝑝𝑠 > 61.2 𝑘𝑖𝑝𝑠 → 𝑂𝐾

THEREFORE USE AN L8X4X5/8 WITH TWO ¾-IN DIAMETER GROUP A BOLTS

FROM GRIDER TO SEAT AND 2 ¼-INCH FILLET WELDS FROM SEAT TO

COLUMN.

.



DIAPHRAGM DESIGN:

Using interpolated MWRFS wind loads, find the diaphragm load, shear, and chord forces for

each steel floor. Use a diaphragm depth of 150 feet and length of 100 feet. Use the following

wind pressures:

MWRFS wind pressures (psf): Height (ft) Windward wall Leeward wall Side wall

38 7.8 13.9 -10.4 -4.2 -13.3 -7.1

53 8.9 15.0 -10.4 -4.2 -13.3 -7.1

68 9.7 15.9 -10.4 -4.2 -13.3 -7.1

83 10.5 16.7 -10.4 -4.2 -13.3 -7.1

98 11.2 17.4 -10.4 -4.2 -13.3 -7.1

100 11.4 17.6 -10.4 -4.2 -13.3 -7.1

Finding the loads for each diaphragm on each floor using half of the floor height above and

below the level being considered gives the below table.

w in lb/ft for diaphragms Height Windward p Leeward p Side p

38 58.18 104.52 -77.89 -31.54 -99.78 -53.43

53 124.59 217.29 -155.79 -63.09 -199.56 -106.86

68 139.16 231.86 -155.79 -63.09 -199.56 -106.86

83 151.32 244.01 -155.79 -63.09 -199.56 -106.86

98 78.57 124.92 -77.89 -31.54 -99.78 -53.43

100 84.28 130.63 -77.89 -31.54 -99.78 -53.43

An example of the calculations used in this table is as follows:

𝑊𝑖𝑛𝑑𝑤𝑎𝑟𝑑 𝑑𝑖𝑎𝑝ℎ𝑟𝑎𝑚 𝑙𝑜𝑎𝑑 𝑎𝑡 53 𝑓𝑡 = (7.5 𝑓𝑡)(7.8 𝑝𝑠𝑓) + (7.5 𝑓𝑡)(8.9 𝑝𝑠𝑓) = 125 𝑝𝑙𝑓

Use superposition to add the windward and leeward maximum positive and negative values to

obtain the worst case scenario on each diaphragm. Assume side pressures cancel each other and

therefore do not contribute to the diaphragm loading. Use the equations 𝑉𝑢 =𝑤𝑙

2 and 𝑀𝑢 =

𝑤𝑙2

8 to

find the maximum shear and moment on each diaphragm.

Maximum Shear and Moment on diaphragms

Height (ft) Vu (kip) Mu (k*ft)

38 13.68 228.02

53 27.98 466.34

68 29.07 484.56



83 29.99 499.75

98 15.21 253.51

100 15.64 260.65

Find diaphragm shear and chord force by dividing by the diaphragm depth of 150 ft. Also,

multiply the diaphragm shear by an ASD load factor of 0.6.

Diaphragm Shear and Chord Force Height (ft) Vu (plf) T (kip) Vu with ASD load factor (plf)

38 91.21 1.52 54.73

53 186.54 3.11 111.92

68 193.82 3.23 116.29

83 199.90 3.33 119.94

98 101.40 1.69 60.84

100 104.26 1.74 62.56

From the Vulcraft manual using a 1.5B22 deck with a span of 5 ft, choose 5/8 puddle welds for

support fasteners and #10 TEK screws for sidelap fasteners. Use 5 sidelap fasteners to give a

strength of 537 plf, which exceeds the values found above. Therefore, use this type of

diaphragm for all of the steel floors.

LATERAL STEEL ANALYSIS Pg 164


LATERAL STEEL ANALYSIS:

Visual Analysis was used to analyze the steel lateral force resisting system. A 3D model of the

steel was created and the diaphragm reactions at each of the floors were applied to this model.

These reactions were found using the same distributed loads found in the previous diaphragm

design. The reactions were calculated using the equation 𝑉𝑢 =𝑤𝑙

2.

Diaphragm reactions (k) Height Windward Leeward Side

38 4.36 7.84 -5.84 -2.37 -7.48 -4.01

53 9.34 16.30 -11.68 -4.73 -14.97 -8.01

68 10.44 17.39 -11.68 -4.73 -14.97 -8.01

83 11.35 18.30 -11.68 -4.73 -14.97 -8.01

98 5.89 9.37 -5.84 -2.37 -7.48 -4.01

100 6.32 9.80 -5.84 -2.37 -7.48 -4.01

A series of pictures showing the inputs of the model follows on the next few pages. All of the

material designations and section properties were made to the match previous designs and

AutoCAD drawings. An assumed 2L6x6x1/2 was used for each brace. Fixed supports were

used in order to make the program run; Visual Analysis will not run if pinned supports are used

in a 3D model. Although fixed supports are not the best approximation of the connections, they

at least allow one to obtain results and run the program. There are no moment reactions found

by the program, and any errors these supports may cause will be factored into load combinations

and factors of safety. Therefore, the fixed supports should be an okay in the first order analysis

completed by the program. Outputs vary on which member is being analyzed, therefore results

will be shown in a case by case basis in the subsequent analyses and design.

[PICUTRES APPEAR ON NEXT PAGE]



Fixed Supports:



Member Releases:

Note: Moment releases are shown as circles on the ends of each member.



Member Materials:

Note: The part of the model not seen in this picture follows the same settings.



Member Sections:

Note: The members not show in this picture match the sections designated in the AutoCAD

drawings and previous designs.



Rigid Diaphragms:



Node Locations:

Note: Nodes not found in this picture follow the locations designated in the AutoCAD drawings.



Picture View:



Windward Wall Loads:



Leeward Wall Loads:



Sidewall Loads



More sidewall Loads:

Note: Loads not seen in this picture match the loads in the previous photograph.



Dead Loads:

Note: These loads are the same in the other corner of the building.



Live loads:

Note: Loads are the same in the other corner of the building.



Live roof load:

Note: Loads are same in the other corner of building.



Roof Wind Load:

Note: Loads are the same in the other corner of the building.

APPROXIMATE SECOND ORDER ANALYSIS Pg 180


APPROXIMATE SECOND ORDER ANALYSIS:

Perform an approximate second order analysis for the column circled below using ASD. Use a

2D model so that lateral translation can be restrained properly and pin supports can be used.

Apply the lateral loads found from the diaphragm design that add both the windward and

leeward loads. A similar analysis should be conducted for every member in the LRFS, but for

the scope of this project, only one member is analyzed. Place artificial rollers at each floor to

restrain lateral translation (AJR). From iterations not shown in this paper, try W14x109

columns.

Building Floor dimensions: 𝐿 = 100 𝑓𝑡, 𝐵 = 150 𝑓𝑡 Story height: ℎ = 15 𝑓𝑡 Floor and Roof Loads:



𝑤𝑑𝑒𝑎𝑑 𝑓𝑙𝑜𝑜𝑟 = 𝑤𝑑𝑓 = 32.5 𝑝𝑠𝑓

𝑤𝑙𝑖𝑣𝑒 𝑓𝑙𝑜𝑜𝑟 = 𝑤𝑙 = 65 𝑝𝑠𝑓

𝑤𝑤𝑖𝑛𝑑 = 𝑤𝑤 = 16 𝑝𝑠𝑓

𝑤𝑙𝑖𝑣𝑒 𝑟𝑜𝑜𝑓 = 𝑤𝑙𝑟 = 20 𝑝𝑠𝑓

𝑤𝑑𝑒𝑎𝑑 𝑟𝑜𝑜𝑓 = 𝑤𝑑𝑟 = 30 𝑝𝑠𝑓

Story loads:

𝑃𝐷𝑠𝑡𝑜𝑟𝑦 = 𝐿𝐵(3𝑤𝑑𝑓 + 𝑤𝑑𝑟) = (100)(150)(3 ∗ 32.5 + 30) = 1913 𝑘𝑖𝑝

𝑃𝐿𝑟𝑠𝑡𝑜𝑟𝑦 = 𝐿𝐵𝑤𝑙𝑟 = (100)(150)(20) = 300 𝑘𝑖𝑝

𝑃𝐿𝑠𝑡𝑜𝑟𝑦 = 𝐿𝐵3𝑤𝑙 = (100)(150)(3)(65) = 2925 𝑘𝑖𝑝

𝑃𝑤𝑠𝑡𝑜𝑟𝑦 = 𝐿𝐵𝑤𝑤 = (100)(150)(16) = 240 𝑘𝑖𝑝

Bottom column axial loads when translation is restrained (results obtained from visual analysis):

𝑃𝐷𝑛𝑡 = 231.3 𝑘𝑖𝑝

𝑃𝐿𝑛𝑡 = 341 𝑘𝑖𝑝

𝑃𝐿𝑟𝑛𝑡 = 33.1 𝑘𝑖𝑝

𝑃𝑊𝑛𝑡 = 9.85 𝑘𝑖𝑝 Bottom column bending moment when translation is restrained:

𝑀𝐷𝑛𝑡 = 0 𝑘𝑖𝑝 ∗ 𝑖𝑛

𝑀𝐿𝑛𝑡 = 0 𝑘𝑖𝑝 ∗ 𝑖𝑛

𝑀𝐿𝑟𝑛𝑡 = 0 𝑘𝑖𝑝 ∗ 𝑖𝑛

𝑀𝑊𝑛𝑡 = 0 𝑘𝑖𝑝 ∗ 𝑖𝑛 Artificial Joint Reactions due to the restraints (rollers in x) provided at each floor to restrain

translation (nt):

𝐴𝐽𝑅𝐷 = (

1.333.133.9510.48

)𝑘𝑖𝑝

𝐴𝐽𝑅𝐿𝑟 = (

1.57−.230.201.17

)𝑘𝑖𝑝

𝐴𝐽𝑅𝐿 = (

−2.306.716.9916.53

)𝑘𝑖𝑝

𝐴𝐽𝑅𝑊 = (

21.1928.4224.1319.92

)𝑘𝑖𝑝

Bottom column axial loads when translation is restrained (results obtained from visual analysis):

𝑃𝐷𝑙𝑡 = 19.96 𝑘𝑖𝑝

𝑃𝐿𝑙𝑡 = 24.67 𝑘𝑖𝑝



𝑃𝐿𝑟𝑙𝑡 = 4.75 𝑘𝑖𝑝

𝑃𝑊𝑙𝑡 = 144.7 𝑘𝑖𝑝 Bottom column bending moment when translation is restrained:

𝑀𝐷𝑙𝑡 = 3.15 𝑘𝑖𝑝 ∗ 𝑖𝑛

𝑀𝐿𝑙𝑡 = 1.25 𝑘𝑖𝑝 ∗ 𝑖𝑛

𝑀𝐿𝑟𝑙𝑡 = 1.02 𝑘𝑖𝑝 ∗ 𝑖𝑛

𝑀𝑊𝑙𝑡 = 98.52 𝑘𝑖𝑝 ∗ 𝑖𝑛 Load factor matrix (LRFD load conditions ASCE 7-10 2.3):

𝛾𝑠𝑡𝑟 =

(

1 0 0 01 1 0 01 0 1 01 . 75 . 75 01 0 0 . 61 . 75 . 75 . 451 . 75 0 0. 6 0 0 . 6. 6 0 0 0 )

Factored Loads:

𝑃𝑢𝑠𝑡𝑜𝑟𝑦 = 𝛾𝑠𝑡𝑟 (

19132925300240

) =

(

191348382213433220574440410712921148)

𝑘𝑖𝑝

𝑃𝑛𝑡 = 𝛾𝑠𝑡𝑟 (

231.334133.19.85

) =

(

231.3572.3264.4511.88237.21516.31487.05144,69138.78)

𝑘𝑖𝑝



𝑃𝑙𝑡 = 𝛾𝑠𝑡𝑟 (

19.9624.674.75144.7

) =

(

19.9644.6324.7142.03106.78107.1438.4698.8011.98 )

𝑘𝑖𝑝

𝑀𝑛𝑡 = 𝛾𝑠𝑡𝑟 (

0000

) =

(

000000)


𝑀𝑙𝑡 = 𝛾𝑠𝑡𝑟 (

3.151.251.0298.52

) =

(

3.154.404.174.8562.2649.194.0961.001.89 )


Factored lateral loads on the “lt” frame. Note these are the cumulative loads on the bottom

column since we are examining the bottom column:

𝑝1 = 𝛾𝑠𝑡𝑟 (

1.33−2.31.5721.19

) =

(

1.33−.972.90. 7814.0410.32−.4013.51. 80 )

𝑘𝑖𝑝


3.136.71−.2328.42

) =

(

3.139.842.907.9920.1820.788.1618.931.88 )

𝑘𝑖𝑝




3.956.99. 2024.13

) =

(

3.9510.944.159.3418.4320.209.1916.852.37 )

𝑘𝑖𝑝


10.4816.531.1719.92

) =

(

10.4827.0111.6523.7622.4332.7222.8818.246.29 )

𝑘𝑖𝑝

𝐻𝑢∑𝑝𝑖 =

(

18.8946.8221.641.8775.0984.0239.8467.5311.33)

𝑘𝑖𝑝

Deflection of the joint at the top of the bottom column due to each load case (found in Visual

Analysis):

∆= (

. 0290

. 0422

. 0044

. 1508

) 𝑖𝑛

Factored deflections corresponding to loads Hu:

∆𝐻= 𝛾𝑠𝑡𝑟∆=

(

. 029. 0712. 0334. 0640. 1195. 1318. 0607. 1079. 0174)

𝑖𝑛

ASD load factor: 𝛼 = 1.6

Factored axial load in moment frames. (This structure has no moment frames):



𝑃𝑢𝑚𝑓 =

(

000000)

𝑘𝑖𝑝

Rm factor:

𝑅𝑚 = 1 − .15𝑃𝑢𝑚𝑓

𝑃𝑢𝑠𝑡𝑜𝑟𝑦=

(

111111)

Buckling load per story:

𝑃𝑒𝑠𝑡𝑜𝑟𝑦 = (𝑅𝑚𝐻𝑢ℎ

∆𝐻) =

(

117248.3118365.2116407.2117759.4113106.3114746.6118141.7112654.3117206.9)

𝑘𝑖𝑝

P-Big Delta Amplification factor B2:

𝐵2 =

(

𝑏𝑖 = max (1,1

1 −𝛼𝑃𝑢𝑠𝑡𝑜𝑟𝑦𝑖𝑃𝑒𝑠𝑡𝑜𝑟𝑦𝑖 )

=

(

1.031.071.031.061.031.071.061.021.02)

Find P-Little Delta factor B1:

Note: Bottom column is pinned at base. M1 is always zero. CM therefore is always 0.6.

𝐶𝑀 = 0.6 Required axial design strength:

𝑃𝑟 = (𝑃𝑛𝑡 + 𝐵2𝑃𝑙𝑡) =

(

250619286555341627527241140)

𝑘𝑖𝑝



Steel properties and column section properties (W14X109):

𝐸𝑠 = 29000 𝑘𝑠𝑖 𝐼 = 1240 𝑖𝑛4

Effective length factor: 𝑘 = 1.0 (pin-pin connection)

Column buckling load:

𝑃𝑒1 =𝜋𝐸𝑠𝐼

(𝑘ℎ)2= 3486.80 𝑘𝑖𝑝

P-small delta Amplification factor B1:

𝐵1 = (𝑏𝑖 = max (1,𝐶𝑀

1 −𝛼𝑃𝑟𝑖𝑃𝑒1

) =

(

111111)

Required bending moment design strength:

𝑀𝑟 = (𝐵1𝑀𝑛𝑡 + 𝐵2𝑀𝑙𝑡) =

(

3.244.714.305.1464.1352.634.3462.221.93 )


Checking Member:

𝑃𝑐 = 808 𝑘𝑖𝑝

𝐿𝑏 = 15 𝑓𝑡, 𝐿𝑟 = 48.5 𝑓𝑡, 𝐿𝑝 = 13.2 𝑓𝑡 → 𝑍𝑜𝑛𝑒 𝐼𝐼 𝑀𝑝

Ω= 479 𝑘 ∗ 𝑓𝑡,

𝐵𝐹

Ω= 5.01 𝑘𝑖𝑝

𝑀𝑛Ω= 𝐶𝑏 [

𝑀𝑝

Ω−𝐵𝐹

Ω(𝐿𝑏 − 𝐿𝑝)] = 469.98 𝑘 ∗ 𝑓𝑡

𝑃𝑟

𝑃𝑐=627

808= 0.78 > 0.2

𝑃𝑟𝑃𝑐+8

9(𝑀𝑟𝑀𝑐) =

627

808+8

9(64.13/12

469.98) = 0.79 < 1.0 → 𝑂𝐾!

BRACED FRAME DESIGN Pg 187


BRACED FRAME DESIGN:

Design the braced frame connections for the braces and maximum factored loads shown below.

Use ASD.

Brace 1: Maximum factored load of 92.55 kip.

Brace 2: Maximum factored load of 101.0 kip.



DESIGN OF BRACE 1:

Draw the brace in AutoCAD. Use the following properties for all members involved:

Beam:

W18x55 𝑑 = 18.1 𝑖𝑛

ASTM A992 𝑡𝑓 = .63 𝑖𝑛

𝐹𝑦 = 50 𝑘𝑠𝑖

𝐹𝑢 = 65 𝑘𝑠𝑖

Column:

W14X109 𝑑 = 14.3 𝑖𝑛


𝐹𝑦 = 50 𝑘𝑠𝑖 𝑏𝑓 = 14.6 𝑖𝑛

𝐹𝑢 = 65 𝑘𝑠𝑖 𝑡𝑤 = .525 𝑖𝑛

Diagonal Brace:

2L6x6x1/2x3/8 𝑡 = .5 𝑖𝑛

ASTM A36 𝐴 = 11.5 𝑖𝑛2

𝐹𝑦 = 36 𝑘𝑠𝑖 �� = 1.67 𝑖𝑛


Clip Angle:

2L6X6X3/8

ASTM A36



Gusset Plate:

𝑡 = .375 𝑖𝑛 ASTM A36



Solve for the dimensions of the gusset plate/center of gravity for beam to gusset and column to

gusset connections:

𝑒𝑏 =18.1

2= 9.05 𝑖𝑛 (𝑓𝑜𝑟 𝑊18𝑋55)

𝑒𝑐 =14.3

2= 7.15 𝑖𝑛 (𝑓𝑜𝑟 𝑊14𝑋109)

𝛽 = 6 𝑖𝑛

𝜃 = 59∘ Solving for Alpha:

𝛼 = 𝑒𝑏 tan 𝜃 − 𝑒𝑐 + 𝛽 tan𝜃



𝛼 = 17.9 𝑖𝑛

Finding loads per Uniform Force Method (AISC Part 13)

𝑟 = √(𝛼 + 𝑒𝑐)2 + (𝛽 + 𝑒𝑏)2 = 29.22 𝑖𝑛

𝑉𝑐 =𝛽

𝑟𝑃 =

6 𝑖𝑛

29.22 𝑖𝑛92.55 𝑘𝑖𝑝 = 19 𝑘𝑖𝑝

𝐻𝑐 =𝑒𝑐𝑟𝑃 =

7.15 𝑖𝑛

29.22 𝑖𝑛92.55 𝑘𝑖𝑝 = 22.65 𝑘𝑖𝑝

𝑉𝑏 =𝑒𝑏𝑟𝑃 =

9.05 𝑖𝑛

29.22 𝑖𝑛92.55 𝑘𝑖𝑝 = 28.66 𝑘𝑖𝑝

𝐻𝑏 =𝛼

𝑟𝑃 =

17.9 𝑖𝑛

29.22 𝑖𝑛92.55 𝑘𝑖𝑝 = 56.70 𝑘𝑖𝑝

Design of welds connecting Diagonal brace to the Gusset Plate:

For ½-in. angles, 𝐷𝑚𝑖𝑛 = 3 from AISC Specification Table J2.4.



4𝐷(0.928)=

92.55 𝑘𝑖𝑝

4(3)(0.928)= 8.31 𝑖𝑛

Weld the entire length of the angle minus some clearance.

𝐿 = 19 𝑖𝑛 > 8.31 𝑖𝑛 → 𝑂𝐾


𝑀𝑎𝑥 𝑠𝑖𝑧𝑒 = .375 − .0625 = .3125 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾


𝐿𝑚𝑖𝑛 = 4(. 1875 𝑖𝑛) = .75 𝑖𝑛 < 19 𝑖𝑛 → 𝑂𝐾

𝐿𝑚𝑖𝑛 = 6 𝑖𝑛 < 19 𝑖𝑛 → 𝑂𝐾


𝑤=

19

. 1875= 101.33 > 100 → 𝐶ℎ𝑒𝑐𝑘 𝛽 𝑓𝑎𝑐𝑡𝑜𝑟

𝛽 = 1.2 − .002 ∗𝑙

𝑤= 1.2 − .002 ∗ 101.33 = .997

Recheck strength: 𝑅𝑛Ω= (. 997)(4)(3)(. 928)(19) = 211 𝑘𝑖𝑝 > 92.55 𝑘𝑖𝑝 → 𝑂𝐾

Check Base Metals by the largest effective weld size (𝑎 =𝐷

16):

For the Diagonal brace:

𝑎 =Ω𝐹𝑢


2 ∗ 58 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 5 𝑖𝑛 = 1.17 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾



For the Gusset plate:

𝑎 =Ω𝐹𝑢


2 ∗ 58 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 375 𝑖𝑛 = .879 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾

Tensile Yielding of Diagonal Brace:

𝑅𝑛Ω=𝐹𝑦𝐴𝑔

Ω=36 𝑘𝑠𝑖 ∗ 11.5 𝑖𝑛2

1.67= 247.9 𝑘𝑖𝑝 > 92.55 𝑘𝑖𝑝 → 𝑂𝐾

Tensile Rupture of Diagonal Brace:

𝐴𝑛 = 𝐴𝑔 = 11.5 𝑖𝑛2

𝑈 = 1 −��

𝑙= 1 −

1.67 𝑖𝑛

19 𝑖𝑛= 0.912

𝐴𝑒 = 𝐴𝑛𝑈 = 11.5 𝑖𝑛2 ∗ .912 = 10.48 𝑖𝑛2

𝑅𝑛Ω=𝐹𝑢𝐴𝑒Ω

=58 𝑘𝑠𝑖 ∗ 10.48 𝑖𝑛2

2= 303.9 𝑘𝑖𝑝𝑠 > 92.55 𝑘𝑖𝑝𝑠 → 𝑂𝐾

Welds connecting Clip Angle to Column

Design weld for 𝑉𝑐 + 𝐻𝑐 = 19 𝑘𝑖𝑝 + 22.65 𝑘𝑖𝑝 = 41.65 𝑘𝑖𝑝𝑠 For 3/8-in. angles, 𝐷𝑚𝑖𝑛 = 3 from AISC Specification Table J2.4.


𝐿𝑟𝑒𝑞 =𝑉𝑐 + 𝐻𝑐2𝐷(0.928)

=41.65 𝑘𝑖𝑝

2(5)(0.928)= 4.49 𝑖𝑛

Fillet weld for the entire length of gusset for full length minus some clearance:

𝐿 = 9.5 𝑖𝑛 > 4.49 𝑖𝑛 → 𝑂𝐾


𝑀𝑎𝑥 𝑠𝑖𝑧𝑒 = .375 − .0625 = .3125 𝑖𝑛 = .3125 𝑖𝑛 → 𝑂𝐾


𝐿𝑚𝑖𝑛 = 4(. 3125 𝑖𝑛) = 1.25 𝑖𝑛 < 9.5 𝑖𝑛 → 𝑂𝐾

𝐿𝑚𝑖𝑛 = 7.375 𝑖𝑛 < 9.5 𝑖𝑛 → 𝑂𝐾


𝑤=

9.5


Check Base Metals by the largest effective weld size.

For the Clip Angle:



𝑎 =Ω𝐹𝑢


2 ∗ 58 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 375 𝑖𝑛 = .879 𝑖𝑛 > .3125 𝑖𝑛 → 𝑂𝐾

For the Column Flange:

𝑎 =Ω𝐹𝑢


2 ∗ 65 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 595 𝑖𝑛 = 1.56 𝑖𝑛 > .3125 𝑖𝑛 → 𝑂𝐾

Bolts connecting Clip Angle from Column to Gusset Plate

Use the following load for design: 𝑅 = √𝑉𝑐2 + 𝐻𝑐

2 = 29.56 𝑘𝑖𝑝𝑠

From AISC Manual Table 7-1, find the minimum number of 1-in. Group A, threads

included, bolts.

𝑛𝑚𝑖𝑛 =𝑅𝑟𝑛Ω

=29.56 𝑘𝑖𝑝𝑠

21.2 𝑘𝑖𝑝𝑠= 1.4 𝑏𝑜𝑙𝑡𝑠

Try 4 standard sized bolts at 2.5 inches on center.

Shear Yielding of Clip Angle from column to Gusset Plate

Follow AISC specification J4-2. 𝑅𝑛Ω=. 6𝐹𝑦𝐴𝑔𝑣

Ω=. 6 ∗ 36 𝑘𝑠𝑖 ∗ 12 𝑖𝑛 ∗ .375 𝑖𝑛 ∗ 2

1.5= 129.6 𝑘𝑖𝑝𝑠 > 29.56 𝑘𝑖𝑝𝑠

→ 𝑂𝐾

Shear Rupture of Clip Angle from column to Gusset Plate

Following AISC specification J4-2:

𝐴𝑛𝑣 = 2(. 375 𝑖𝑛)[12 − 4(1 𝑖𝑛 + .125 𝑖𝑛)] = 5.625 𝑖𝑛2

𝑅𝑛Ω=. 6𝐹𝑢𝐴𝑛𝑣Ω

=. 6 ∗ 58 𝑘𝑠𝑖 ∗ 5.625 𝑖𝑛2

2= 97.9 𝑘𝑖𝑝𝑠 > 29.56 𝑘𝑖𝑝𝑠

Block Shear Rupture of Clip Angle from column to Gusset Plate

Assume uniform tension stress, so use 𝑈𝑏𝑠 = 1.

𝐴𝑔𝑣 = 2(12 𝑖𝑛 − 2.25 𝑖𝑛). 375 𝑖𝑛 = 7.31 𝑖𝑛2

𝐴𝑛𝑣 = 7.31 𝑖𝑛2 − .375 𝑖𝑛(3.5)(1.125 𝑖𝑛)2 = 4.36 𝑖𝑛2

𝐴𝑛𝑡 = [(3 𝑖𝑛 ∗ .375 𝑖𝑛) − (.5 ∗ 1.125 𝑖𝑛 ∗ .375 𝑖𝑛)]2 = 1.83 𝑖𝑛2

From AISC Specification J4-3:

𝑅𝑛 = .6𝐹𝑢𝐴𝑛𝑣 + 𝑈𝑏𝑠𝐹𝑢𝐴𝑛𝑡 ≤ .5𝐹𝑦𝐴𝑔𝑣 + 𝑈𝑏𝑠𝐹𝑢𝐴𝑛𝑡

𝑅𝑛 = .6(58)(4.36) + 1(58)(1.83) ≤ .5(36)(7.31) + 1(58)(1.83) 𝑅𝑛 = 257.9 𝑘𝑖𝑝𝑠 ≤ 237.7 𝑘𝑖𝑝𝑠 𝑅𝑛Ω=237.7

2= 118.9 𝑘𝑖𝑝𝑠 > 29.56 𝑘𝑖𝑝𝑠 → 𝑂𝐾



Shear Bearing Strength at Bolt Holes

𝑙𝑐 = 2.25 −1.125

2= 1.6875 𝑖𝑛

Consider deformation at the bolthole at service load a design consideration.

𝑅𝑛 = 1.2𝑙𝑐𝑡𝐹𝑢 ≤ 2.4𝑑𝑡𝐹𝑢

𝑅𝑛 = 1.2(1.6875 𝑖𝑛)(. 375 𝑖𝑛)(58 𝑘𝑠𝑖)4 ≤ 2.4(1 𝑖𝑛)(. 375 𝑖𝑛)(58 𝑘𝑠𝑖)4

𝑅𝑛 = 176.18 𝑘𝑖𝑝𝑠 ≤ 208.8 𝑘𝑖𝑝𝑠 𝑅𝑛Ω=176.18

2= 88.09 𝑘𝑖𝑝𝑠 > 29.56 𝑘𝑖𝑝𝑠 → 𝑂𝐾

Design of welds connecting Beam to the Gusset Plate:

Design connection for 𝑉𝑏 +𝐻𝑏 = 28.66 𝑘𝑖𝑝 + 56.7 𝑘𝑖𝑝 = 85.36 𝑘𝑖𝑝𝑠 For 3/8-in. plate thickness, 𝐷𝑚𝑖𝑛 = 3 from AISC Specification Table J2.4.



2𝐷(0.928)=85.36 𝑘𝑖𝑝𝑠

2(3)(0.928)= 15.33 𝑖𝑛

Weld the entire length of the minus some clearance.

𝐿 = 33 𝑖𝑛 > 15.33 𝑖𝑛 → 𝑂𝐾


𝑀𝑎𝑥 𝑠𝑖𝑧𝑒 = .375 − .0625 = .3125 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾


𝐿𝑚𝑖𝑛 = 4(. 1875 𝑖𝑛) = .75 𝑖𝑛 < 33 𝑖𝑛 → 𝑂𝐾

𝐿𝑚𝑖𝑛 = .375 𝑖𝑛 < 33 𝑖𝑛 → 𝑂𝐾


𝑤=

33

. 1875= 176 > 100 → 𝐶ℎ𝑒𝑐𝑘 𝛽 𝑓𝑎𝑐𝑡𝑜𝑟

𝛽 = 1.2 − .002 ∗𝑙

𝑤= 1.2 − .002 ∗ 176 = .848

Recheck strength: 𝑅𝑛Ω= (. 848)(2)(3)(. 928)(33) = 155.8 𝑘𝑖𝑝 > 85.36 𝑘𝑖𝑝 → 𝑂𝐾


16)

For the Beam:

𝑎 =Ω𝐹𝑢


2 ∗ 65 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 63 𝑖𝑛 = 1.65 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾




𝑎 =Ω𝐹𝑢


2 ∗ 58 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 375 𝑖𝑛 = .88 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾

Gusset Plate Compression

Whitmore section effective width, 𝑤 = 29.42 𝑖𝑛

According to AISC Part 9, “The Whitmore section may spread across the joint between

connecting elements, but cannot spread beyond an unconnected edge.” Thus, this section

is ok.

𝐿1 = 6.18 𝑖𝑛

𝐿2 = 6.7 𝑖𝑛

𝐿3 = −16 𝑖𝑛

Unbraced length, 𝑙 =𝐿1+𝐿2+𝐿3

3= −1.04 𝑖𝑛

Because this value is negative, use the alternate method discussed by William A.

Thornton, P.E., PH.D. and Carlo Lini, P.E. in The Whitmore Section, Steel Wise, Modern

Steel Construction, July 2011. In this method, take the unbraced length as:

𝐿2 = 𝑙 = 6.7 𝑖𝑛



𝑟 =𝑡

√12=. 375 𝑖𝑛

√12= .108 𝑖𝑛

By AISC Manual specification J4: 𝑘𝑙

𝑟=1(6.7 𝑖𝑛)

.108 𝑖𝑛= 62.04

From Table 4-22, 𝐹𝑐𝑟

Ω= 17.5 𝑘𝑠𝑖

𝐴𝑔 = 𝑤𝑡 = 29.42 𝑖𝑛 ∗ .375 𝑖𝑛 = 11.03 𝑖𝑛2

𝑃𝑛Ω=𝐹𝑐𝑟𝐴𝑔

Ω=17.5 𝑘𝑠𝑖 ∗ 11.03 𝑖𝑛2

1.67= 115.6 𝑘𝑖𝑝𝑠 > 92.55 𝑘𝑖𝑝𝑠 → 𝑂𝐾

Tensile Yielding of Gusset Plate


Ω=36 𝑘𝑠𝑖 ∗ 11.03 𝑖𝑛2

1.67= 237.8 𝑘𝑖𝑝 > 92.55 𝑘𝑖𝑝 → 𝑂𝐾

Tensile Rupture of Gusset Plate:

𝐴𝑛 = 𝐴𝑔 = 11.03 𝑖𝑛2

Case 4 in AISC table D3.1: 𝑈 = 1.0

𝐴𝑒 = 𝐴𝑛𝑈 = 11.03 𝑖𝑛2 ∗ 1 = 11.03 𝑖𝑛2




=58 𝑘𝑠𝑖 ∗ 11.03 𝑖𝑛2

2= 319.9 𝑘𝑖𝑝𝑠 > 92.55 𝑘𝑖𝑝𝑠 → 𝑂𝐾

Connection Geometry

𝛼 = 17.9 𝑖𝑛

𝛽 = 6 𝑖𝑛

𝜃 = 59°

𝑒𝑏 =18.1

2= 9.05 𝑖𝑛 (𝑓𝑜𝑟 𝑊18𝑋55)

𝑒𝑐 =14.3

2= 7.15 𝑖𝑛 (𝑓𝑜𝑟 𝑊14𝑋109)


17.9 = 9.05tan 59° − 7.15 + 6 tan 59° → 𝑇𝑟𝑢𝑒 → 𝑂𝐾

ALL CHECKS ARE SATISFIED; THEREFORE CONSTRUCT BRACE ACCORDING

TO AUTOCAD DRAWING.

DESIGN OF BRACE 2:

Draw the brace in AutoCAD. Use the following properties for all members involved:

Beam:

W14x30 𝑑 = 13.8 𝑖𝑛






Column:

W14X109 𝑑 = 14.3 𝑖𝑛


𝐹𝑦 = 50 𝑘𝑠𝑖 𝑏𝑓 = 14.6 𝑖𝑛

𝐹𝑢 = 65 𝑘𝑠𝑖 𝑡𝑤 = .525 𝑖𝑛

Diagonal Brace:

2L6x6x1/2x3/8 𝑡 = .5 𝑖𝑛

ASTM A36 𝐴 = 11.5 𝑖𝑛2

𝐹𝑦 = 36 𝑘𝑠𝑖 �� = 1.67 𝑖𝑛


Clip Angle:

2L6X6X3/8

ASTM A36



Gusset Plate:

𝑡 = .375 𝑖𝑛 ASTM A36



Solve for the dimensions of the gusset plate/center of gravity for beam to gusset and column to

gusset connections:

𝑒𝑏 =13.8

2= 6.90 𝑖𝑛 (𝑓𝑜𝑟 𝑊14𝑥30)

𝑒𝑐 =14.3

2= 7.15 𝑖𝑛 (𝑓𝑜𝑟 𝑊14𝑋109)

𝛽 = 6 𝑖𝑛

𝜃 = 59° Solving for Alpha:


𝛼 = 14.32 𝑖𝑛

Finding loads per Uniform Force Method (AISC Part 13)

𝑟 = √(𝛼 + 𝑒𝑐)2 + (𝛽 + 𝑒𝑏)2 = 25.05 𝑖𝑛

𝑉𝑐 =𝛽

𝑟𝑃 =

6 𝑖𝑛

25.05 𝑖𝑛101 𝑘𝑖𝑝 = 24.2 𝑘𝑖𝑝

𝐻𝑐 =𝑒𝑐𝑟𝑃 =

7.15 𝑖𝑛

25.05 𝑖𝑛101 𝑘𝑖𝑝 = 28.8 𝑘𝑖𝑝



𝑉𝑏 =𝑒𝑏𝑟𝑃 =

6.90 𝑖𝑛

25.05 𝑖𝑛101 𝑘𝑖𝑝 = 27.8 𝑘𝑖𝑝

𝐻𝑏 =𝛼

𝑟𝑃 =

14.32 𝑖𝑛

25.05 𝑖𝑛101 𝑘𝑖𝑝 = 57.74 𝑘𝑖𝑝

Design of welds connecting Diagonal brace to the Gusset Plate:

For ½-in. angles, 𝐷𝑚𝑖𝑛 = 3 from AISC Specification Table J2.4.



4𝐷(0.928)=

101 𝑘𝑖𝑝

4(3)(0.928)= 9.07 𝑖𝑛

Weld the entire length of the angle minus some clearance.

𝐿 = 16 𝑖𝑛 > 9.07 𝑖𝑛 → 𝑂𝐾


𝑀𝑎𝑥 𝑠𝑖𝑧𝑒 = .375 − .0625 = .3125 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾


𝐿𝑚𝑖𝑛 = 4(. 1875 𝑖𝑛) = .75 𝑖𝑛 < 16 𝑖𝑛 → 𝑂𝐾

𝐿𝑚𝑖𝑛 = 6 𝑖𝑛 < 16 𝑖𝑛 → 𝑂𝐾


𝑤=

16

. 1875= 85.33 < 100 → 𝛽 𝑓𝑎𝑐𝑡𝑜𝑟 𝐶ℎ𝑒𝑐𝑘 𝑛𝑜𝑡 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦


16)

For the Diagonal brace:

𝑎 =Ω𝐹𝑢


2 ∗ 58 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 5 𝑖𝑛 = 1.17 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾


𝑎 =Ω𝐹𝑢


2 ∗ 58 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 375 𝑖𝑛 = .88 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾

Tensile Yielding of Diagonal Brace:


Ω=36 𝑘𝑠𝑖 ∗ 11.5 𝑖𝑛2

1.67= 247.9 𝑘𝑖𝑝 > 101 𝑘𝑖𝑝 → 𝑂𝐾

Tensile Rupture of Diagonal Brace:

𝐴𝑛 = 𝐴𝑔 = 11.5 𝑖𝑛2

𝑈 = 1 −��

𝑙= 1 −

1.67 𝑖𝑛

16 𝑖𝑛= 0.899



𝐴𝑒 = 𝐴𝑛𝑈 = 11.5 𝑖𝑛2 ∗ .899 = 10.34 𝑖𝑛2


=58 𝑘𝑠𝑖 ∗ 10.34 𝑖𝑛2

2= 299.9 𝑘𝑖𝑝𝑠 > 101 𝑘𝑖𝑝𝑠 → 𝑂𝐾

Welds connecting Clip Angle to Column

Design weld for 𝑉𝑐 + 𝐻𝑐 = 24.2 𝑘𝑖𝑝 + 28.8 𝑘𝑖𝑝 = 53 𝑘𝑖𝑝𝑠 For 3/8-in. angles, 𝐷𝑚𝑖𝑛 = 3 from AISC Specification Table J2.4.


𝐿𝑟𝑒𝑞 =𝑉𝑐 + 𝐻𝑐2𝐷(0.928)

=53 𝑘𝑖𝑝

2(5)(0.928)= 5.71 𝑖𝑛

Fillet weld for the entire length of gusset for full length minus some clearance:

𝐿 = 9.5 𝑖𝑛 > 5.71 𝑖𝑛 → 𝑂𝐾


𝑀𝑎𝑥 𝑠𝑖𝑧𝑒 = .375 − .0625 = .3125 𝑖𝑛 = .3125 𝑖𝑛 → 𝑂𝐾


𝐿𝑚𝑖𝑛 = 4(. 3125 𝑖𝑛) = 1.25 𝑖𝑛 < 9.5 𝑖𝑛 → 𝑂𝐾

𝐿𝑚𝑖𝑛 = 7.375 𝑖𝑛 < 9.5 𝑖𝑛 → 𝑂𝐾


𝑤=

9.5


Check Base Metals by the largest effective weld size.

For the Clip Angle:

𝑎 =Ω𝐹𝑢


2 ∗ 58 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 375 𝑖𝑛 = .879 𝑖𝑛 > .3125 𝑖𝑛 → 𝑂𝐾

For the Column Flange:

𝑎 =Ω𝐹𝑢


2 ∗ 65 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 595 𝑖𝑛 = 1.56 𝑖𝑛 > .3125 𝑖𝑛 → 𝑂𝐾

Bolts connecting Clip Angle from Column to Gusset Plate

Use the following load for design: 𝑅 = √𝑉𝑐2 + 𝐻𝑐

2 = 37.62 𝑘𝑖𝑝𝑠

From AISC Manual Table 7-1, find the minimum number of 1-in. Group A, threads

included, bolts.

𝑛𝑚𝑖𝑛 =𝑅𝑟𝑛Ω

=37.62 𝑘𝑖𝑝𝑠

21.2 𝑘𝑖𝑝𝑠= 1.77 𝑏𝑜𝑙𝑡𝑠

Try 4 standard sized bolts at 2.5 inches on center.



Shear Yielding of Clip Angle from column to Gusset Plate

Follow AISC specification J4-2. 𝑅𝑛Ω=. 6𝐹𝑦𝐴𝑔𝑣

Ω=. 6 ∗ 36 𝑘𝑠𝑖 ∗ 12 𝑖𝑛 ∗ .375 𝑖𝑛 ∗ 2

1.5= 129.6 𝑘𝑖𝑝𝑠 > 37.62 𝑘𝑖𝑝𝑠

→ 𝑂𝐾

Shear Rupture of Clip Angle from column to Gusset Plate

Following AISC specification J4-2:

𝐴𝑛𝑣 = 2(. 375 𝑖𝑛)[12 − 4(1 𝑖𝑛 + .125 𝑖𝑛)] = 5.625 𝑖𝑛2

𝑅𝑛Ω=. 6𝐹𝑢𝐴𝑛𝑣Ω

=. 6 ∗ 58 𝑘𝑠𝑖 ∗ 5.625 𝑖𝑛2

2= 97.9 𝑘𝑖𝑝𝑠 > 37.62 𝑘𝑖𝑝𝑠

Block Shear Rupture of Clip Angle from column to Gusset Plate

Assume uniform tension stress, so use 𝑈𝑏𝑠 = 1.

𝐴𝑔𝑣 = 2(12 𝑖𝑛 − 2.25 𝑖𝑛). 375 𝑖𝑛 = 7.31 𝑖𝑛2

𝐴𝑛𝑣 = 7.31 𝑖𝑛2 − .375 𝑖𝑛(3.5)(1.125 𝑖𝑛)2 = 4.36 𝑖𝑛2

𝐴𝑛𝑡 = [(3 𝑖𝑛 ∗ .375 𝑖𝑛) − (.5 ∗ 1.125 𝑖𝑛 ∗ .375 𝑖𝑛)]2 = 1.83 𝑖𝑛2

From AISC Specification J4-3:

𝑅𝑛 = .6𝐹𝑢𝐴𝑛𝑣 + 𝑈𝑏𝑠𝐹𝑢𝐴𝑛𝑡 ≤ .5𝐹𝑦𝐴𝑔𝑣 + 𝑈𝑏𝑠𝐹𝑢𝐴𝑛𝑡

𝑅𝑛 = .6(58)(4.36) + 1(58)(1.83) ≤ .5(36)(7.31) + 1(58)(1.83) 𝑅𝑛 = 257.9 𝑘𝑖𝑝𝑠 ≤ 237.7 𝑘𝑖𝑝𝑠 𝑅𝑛Ω=237.7

2= 118.9 𝑘𝑖𝑝𝑠 > 37.62 𝑘𝑖𝑝𝑠 → 𝑂𝐾

Shear Bearing Strength at Bolt Holes

𝑙𝑐 = 2.25 −1.125

2= 1.6875 𝑖𝑛

Consider deformation at the bolthole at service load a design consideration.

𝑅𝑛 = 1.2𝑙𝑐𝑡𝐹𝑢 ≤ 2.4𝑑𝑡𝐹𝑢

𝑅𝑛 = 1.2(1.6875 𝑖𝑛)(. 375 𝑖𝑛)(58 𝑘𝑠𝑖)4 ≤ 2.4(1 𝑖𝑛)(. 375 𝑖𝑛)(58 𝑘𝑠𝑖)4

𝑅𝑛 = 176.18 𝑘𝑖𝑝𝑠 ≤ 208.8 𝑘𝑖𝑝𝑠 𝑅𝑛Ω=176.18

2= 88.09 𝑘𝑖𝑝𝑠 > 37.62 𝑘𝑖𝑝𝑠 → 𝑂𝐾

Design of welds connecting Beam to the Gusset Plate:

Design connection for 𝑉𝑏 +𝐻𝑏 = 27.8 𝑘𝑖𝑝 + 57.74 𝑘𝑖𝑝 = 85.54 𝑘𝑖𝑝𝑠 For 3/8-in. plate thickness, 𝐷𝑚𝑖𝑛 = 3 from AISC Specification Table J2.4.





2𝐷(0.928)=85.54 𝑘𝑖𝑝𝑠

2(3)(0.928)= 15.36 𝑖𝑛

Weld the entire length of the minus some clearance.

𝐿 = 25 𝑖𝑛 > 15.36 𝑖𝑛 → 𝑂𝐾


𝑀𝑎𝑥 𝑠𝑖𝑧𝑒 = .375 − .0625 = .3125 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾


𝐿𝑚𝑖𝑛 = 4(. 1875 𝑖𝑛) = .75 𝑖𝑛 < 25 𝑖𝑛 → 𝑂𝐾

𝐿𝑚𝑖𝑛 = .375 𝑖𝑛 < 25 𝑖𝑛 → 𝑂𝐾


𝑤=

25

. 1875= 133.3 > 100 → 𝐶ℎ𝑒𝑐𝑘 𝛽 𝑓𝑎𝑐𝑡𝑜𝑟

𝛽 = 1.2 − .002 ∗𝑙

𝑤= 1.2 − .002 ∗ 133.3 = .933

Recheck strength: 𝑅𝑛Ω= (. 933)(2)(3)(. 928)(25) = 129.9 𝑘𝑖𝑝 > 85.54 𝑘𝑖𝑝 → 𝑂𝐾


16)

For the Beam:

𝑎 =Ω𝐹𝑢


2 ∗ 65 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 63 𝑖𝑛 = 1.65 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾


𝑎 =Ω𝐹𝑢


2 ∗ 58 𝑘𝑠𝑖

. 707 ∗ 70 𝑘𝑠𝑖. 375 𝑖𝑛 = .88 𝑖𝑛 > .1875 𝑖𝑛 → 𝑂𝐾

Gusset Plate Compression

Whitmore section effective width, 𝑤 = 25.48 𝑖𝑛

According to AISC Part 9, “The Whitmore section may spread across the joint between

connecting elements, but cannot spread beyond an unconnected edge.” Thus, this section

is ok.



𝐿1 = 3.87 𝑖𝑛

𝐿2 = 7.68 𝑖𝑛

𝐿3 = −11.34𝑖𝑛

Unbraced length, 𝑙 =𝐿1+𝐿2+𝐿3

3= 0.21 𝑖𝑛



𝑟 =𝑡

√12=. 375 𝑖𝑛

√12= .108 𝑖𝑛

By AISC Manual specification J4: 𝑘𝑙

𝑟=1(0.21 𝑖𝑛)

. 108 𝑖𝑛= 1.94

From Table 4-22, 𝐹𝑐𝑟

Ω= 21.6 𝑘𝑠𝑖

𝐴𝑔 = 𝑤𝑡 = 25.48 𝑖𝑛 ∗ .375 𝑖𝑛 = 9.56 𝑖𝑛2

𝑃𝑛Ω=𝐹𝑐𝑟𝐴𝑔

Ω=21.6 𝑘𝑠𝑖 ∗ 9.56 𝑖𝑛2

1.67= 123.7 𝑘𝑖𝑝𝑠 > 101 𝑘𝑖𝑝𝑠 → 𝑂𝐾

Tensile Yielding of Gusset Plate


Ω=36 𝑘𝑠𝑖 ∗ 9.56 𝑖𝑛2

1.67= 206.1 𝑘𝑖𝑝 > 101 𝑘𝑖𝑝 → 𝑂𝐾

Tensile Rupture of Gusset Plate:

𝐴𝑛 = 𝐴𝑔 = 9.56 𝑖𝑛2

Case 4 in AISC table D3.1: 𝑈 = 1.0

𝐴𝑒 = 𝐴𝑛𝑈 = 9.56 𝑖𝑛2 ∗ 1 = 9.56 𝑖𝑛2




=58 𝑘𝑠𝑖 ∗ 9.56 𝑖𝑛2

2= 277.2 𝑘𝑖𝑝𝑠 > 101 𝑘𝑖𝑝𝑠 → 𝑂𝐾

Connection Geometry

𝛼 = 14.32 𝑖𝑛

𝛽 = 6 𝑖𝑛

𝜃 = 59°

𝑒𝑏 =13.8

2= 6.90 𝑖𝑛 (𝑓𝑜𝑟 𝑊14𝑥30)

𝑒𝑐 =14.3

2= 7.15 𝑖𝑛 (𝑓𝑜𝑟 𝑊14𝑋109)


14.32 = 6.9tan 59° − 7.15 + 6 tan 59° → 𝑇𝑟𝑢𝑒 → 𝑂𝐾

ALL CHECKS ARE SATISFIED; THEREFORE CONSTRUCT BRACE ACCORDING

TO AUTOCAD DRAWING.

SHEAR WALL DESIGN Pg 204


SHEAR WALL DESIGN

Concrete self-weight = 150 pcf (0.8333ft) + 15psf for finishes = 140psf

Loading from the lateral framing system:

Fx Fy

D (kip) L (kip) D (kip) L (kip)

0.024 0.025 165.61 223.698

0.04 0.057 155.296 210.07

-0.39 -1.502 208.637 297.482

Wind Load:

Wind Pressure p = qGCp – qi(GCpi) (ksi):

Height Windward (in) p Leeward (out) p Side (out) p

15 11.4 -10.4 -13.3

30 13.2 -10.4 -13.3

40 14.1 -10.4 -13.3

48 14.8 -10.4 -13.3

60 15.4 -10.4 -13.3

70 16 -10.4 -13.3

90 17 -10.4 -13.3

98 17.4 -10.4 -13.3

Wind Pressure p = qGCp – qi(GCpi) (kip*ft):

Height Windward (in) p Leeward (out) p Side (out) p

15 41.04 -37.44 -47.88

30 47.52 -37.44 -47.88

40 50.76 -37.44 -47.88

48 53.28 -37.44 -47.88

60 55.44 -37.44 -47.88

70 57.6 -37.44 -47.88

90 61.2 -37.44 -47.88

98 62.64 -37.44 -47.88

Roof Wind Pressure p = qGCp – qi(GCpi) (ksi):

Distance From

Edge

Case A

(positive)

Case B

(negative)

0 to h/2 -16.2 -5.7

h/2 to h -16.2 -5.7

h to 2h -10.4 -5.7

> 2h -7.5 -5.7

Roof Wind Pressure p = qGCp – qi(GCpi) (kip*ft):

Distance From Case A Case B



Edge (positive) (negative)

0 to h/2 -58.32 -20.52

h/2 to h -58.32 -20.52

h to 2h -37.44 -20.52

> 2h -27 -20.52

Moment about the slab:

+↺ΣM=0;

Mu= 1678.351 kip*ft

Summing the forces in the x-direction:

+ ΣFx=0;

Vu= 50.76

Summing the moments in the y-direction:

↑+ ΣFy=0;

Nu= -239.184 kip

Loadings

Mu (kip): Moment 1678.351

Nu (kip): Horizontal -239.182

Vu (kip): Vertical 50.76

Wall Classification: Short ACI 318 App

A

h(w): 10 in

l(w): 25 ft ℎ𝑤

𝑙𝑤< 2

38

25= 1.52 < 2 We may now use the strut-and-tie method ACI 318 Appendix A for

the shear walls.

Moment Capacity:

Moment Capacity

As (in^2): 7.2

f'c (psi): 8000

fy (psi): 60000

h (in): 10

lw (ft): 25

ϕ: 0.65

λ: 1

d=0.90*lw=22.5 ft



𝑇 = 𝐴𝑠 ∗ 𝑓𝑦 = 432 𝑘𝑖𝑝

𝑎 =𝑇

0.85 ∗ 𝑓 ′𝑐 ∗ ℎ= 6.35 𝑖𝑛

𝑎1 =| 𝐴𝑠 ∗ 𝑓𝑦 + 𝑁𝑢|

0.85 ∗ 𝑓 ′𝑐 ∗ ℎ= 2.85𝑖𝑛

𝐶 = 0.85 ∗ 𝑓 ′𝑐 ∗ ℎ ∗ 𝑎1 = 192.816 𝑘𝑖𝑝

𝑀𝑛 = 𝑇 (𝑑 −𝑎

2) + 𝑁𝑢 (

𝑙𝑤 − 𝑎

2) = 6679.16 𝑘𝑖𝑝 ∗ 𝑓𝑡

𝜙𝑀𝑛 = 0.65 ∗ 6679.43 = 4341.45 𝑘𝑖𝑝 ∗ 𝑓𝑡 𝜙𝑀𝑛 > 𝑀𝑢 4341.63 kip*ft > 1678.351 kip*ft Ok!

Shear Capacity

𝑉(𝑐1) = 3.3λ(𝑓′𝑐)12 ∗ ℎ ∗ 𝑑 +

|𝑁𝑢| ∗ 𝑑

4 ∗ 𝑙𝑤= 796.88 𝑘𝑖𝑝 ∗ 𝑓𝑡

𝑉(𝑐2) = (0.60λ(𝑓 ′𝑐)12 +

lw(1.25 ∗ λ ∗ (𝑓 ′𝑐)12 +

0.20|𝑁𝑢|𝑙𝑤 ∗ ℎ

)

|𝑀𝑢||𝑉𝑢|

−𝑙𝑤2

) ∗ ℎ ∗ 𝑑 = 629.66 𝑘𝑖𝑝 ∗ 𝑓𝑡

Choose the smaller of the two Vc values.

Vcmin = 629.66 kip

𝜙𝑉𝑐 = 472.25 𝑘𝑖𝑝

|𝜙𝑉𝑐| > |𝑁𝑢| 472.25 kip > 239.162 kip Ok!

Check Steel: |𝜙𝑉𝑐|

2= 236.12 < 𝑉𝑢 < 𝜙𝑉𝑐 236.12 < 239.162 < 472.25 Provide minimum shear

wall steel per ACI 318 11.9.9.2 and 11.9.9.4 hw

lw=38

25= 1.5 < 2 ρ(min ) = 0.0025 → ρ(t) = ρ(l)

Horizontal Steel:

As(h)= 3.6in2 #7 Bars @ 16” O.C.

ρ(t) = ρ(l)

ρ(𝑡) =𝐴𝑠(ℎ)

ℎ ∗ 𝑡= ρ(l) = 0.0225

ρ(𝑚𝑖𝑛) = 0.0025 + .5(2.5 −ℎ

𝑙𝑤)(ρ(t) − .0025) = 0.0235

As(v)= ρ(min) ∗ h ∗ t = 3.76in2

Vertical Steel:

As(v)= 3.8in2 #7 Bars @ 16” O.C.

Moment resisting at the end of shear walls:

Trial value of As: 8 #7 Bars As= 4.8in2

𝑇 = 𝐴𝑠 ∗ 𝑓𝑦 = 288 𝑘𝑖𝑝

𝑎 =𝑇 + 𝑁𝑛

0.85 ∗ 𝑓 ′𝑐 ∗ ℎ= 7.75 𝑖𝑛



𝑀𝑛 = 𝑇 (𝑑 −𝑎

2) + 𝑁𝑢 (

𝑙𝑤 − 𝑎

2) = 111594.762 𝑘𝑖𝑝 ∗ 𝑖𝑛

𝜙𝑀𝑛 = 9299.56 𝑘𝑖𝑝 ∗ 𝑓𝑡 > 𝑀𝑢Ok!

Provide 8 #7 Bars at each end.

Steel Strain

𝑐 = 11.12 𝑖𝑛

ε(t)= 0.0537 >.004 Ok!

ε(t)= 0.0537 >.005 Ok! 𝜙 = 0.90

ρ(g) =n ∗ cross sectional area

boundary width ∗ boundary length= 0.0121

ρ(g) = 0.01875 > 0.01 Although ties are not ideal ties are required.

Add #4 closed ties at 16” O.C. vertical maximum at each horizontal bar.

ALL CHECKS ARE SATISFIED; THEREFORE CONSTRUCT SHEAR WALLS

ACCORDING TO AUTOCAD DRAWING.

FOUNDATION DESIGN Pg 208


FOUNDATION DESIGN GREEN ROOF EDGE COLUMN FOOTING

Given Data and assumptions

Variable Value Unit

qa 8000 psf

Soil Depth weight γw 110 pcf

Slab weight with Thickness (4")

50 psf

Soil Depth 24 in

LL Surcharge 100 psf

Point Dead Load 118.9 k

Point Live Load 31.25 k

Soil Weight gw 220 psf

Column Size (c) 24 in^2

Ka 0.33

Kp 3

Theta 30 degrees

f'c 4000 psi

Reinforcement fy 60000 psi

Cover Cc 3 in

Calculations

𝑞𝑛𝑒𝑡 = 8000𝑝𝑠𝑓– (𝑁𝑜𝑛 𝑐𝑜𝑙𝑢𝑚𝑛 𝐿𝑜𝑎𝑑𝑠)𝑝𝑠𝑓 = 8000𝑝𝑠𝑓 − 100𝑝𝑠𝑓 − 50𝑝𝑠𝑓 − 220𝑝𝑠𝑓= 7630𝑝𝑠𝑓 𝑆𝑢𝑚 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 118.9𝑘 + 31.25𝑘 = 150.15𝑘

𝐵 = 𝐿 = √150.15𝑘/7.63𝑘𝑠𝑓 = 4.44𝑓𝑡

Say 4.5ft 𝑄𝑑 = (𝐶𝑜𝑙𝑢𝑚𝑛 𝐷𝑒𝑎𝑑 𝐿𝑜𝑎𝑑) + .27 ∗ 𝐵2 = 118.9𝑘 + .27𝑘𝑠𝑓 ∗ (4.5𝑓𝑡)2 = 124.3675𝑘

𝑄𝐿 = (𝐶𝑜𝑙𝑢𝑚𝑛 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑) + .1 ∗ 𝐵2 = 31.25𝑘 + .1𝑘𝑠𝑓 ∗ (4.5𝑓𝑡)2 = 33.275𝑘 𝑄𝑢 = 1.2 ∗ (𝑄𝑑) + 1.6(𝑄𝐿) = 202.481𝑘

𝑞𝑢 = 𝑄𝑢

𝐵2=202.481𝑘

(4.5𝑓𝑡)2= 10𝑘𝑠𝑓

𝐴𝑠𝑠𝑢𝑚𝑒 𝑑 = 10𝑖𝑛 Calculate One-way and Two-way Shear

One way

𝑉𝑢 = (𝐵

2−𝑐

2− 𝑑) ∗ 𝐿 ∗ 𝑞𝑢 = (

4.5𝑓𝑡

2−2𝑓𝑡

2− (10

12)𝑓𝑡) ∗ 4.5𝑓𝑡 ∗ 10𝑘𝑠𝑓 = 18.75𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 2 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝐿 ∗ 𝑑 =. 75 ∗ 2 ∗ 1 ∗ √4000𝑝𝑠𝑖 ∗ 4.5𝑓𝑡 ∗ 12 ∗ 10𝑖𝑛

1000𝑙𝑏= 51.23𝑘

𝐶ℎ𝑒𝑐𝑘 𝜙𝑉𝑐 > 𝑉𝑢

51.23𝑘 > 18.75𝑘 𝑂𝐾



Two-Way

𝑏0 = 4 ∗ (𝑐 + 𝑑) = 4(24 + 10) = 136𝑖𝑛

𝑉𝑢 = (𝐵 ∗ 𝐿 − (𝑏04)2

) ∗ 𝑞𝑢 = 122.22𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑 = 258.01𝑘


258.01𝑘 > 122.22𝑘 𝑂𝐾 Add 3 to 4 inches to initial d value

𝑑 = 13𝑖𝑛 Check Moment

𝑀𝑢 = 𝑞𝑢 ∗ 𝐿 ∗(𝐵2 −

𝑐2)2

2= 35.15𝑘 ∗ 𝑓𝑡

𝑀𝑛 =𝑀𝑢𝜙=35.15𝑘 ∗ 𝑓𝑡

. 9= 39.06𝑘 ∗ 𝑓𝑡

𝑀𝑛 = 𝜌𝑓𝑦𝐿𝑑2(1 −

𝜌𝑓𝑦

1.7 ∗ 𝑓′𝑐)

𝑆𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝜌 𝑢𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑓𝑜𝑟𝑚𝑢𝑙𝑎

𝜌 = .0008

𝐴𝑠 = 𝜌𝐵𝑑 = (. 0008) ∗ 4.5𝑓𝑡 ∗12𝑖𝑛

𝑓𝑡∗ 13𝑖𝑛 = .54454𝑖𝑛2

USE 3 #4 BARS IN EACH DIRECTON

𝐴𝑠 = .6𝑖𝑛2

𝑎 =(𝐴𝑆𝑓𝑦)

. 85𝑓′𝑐𝐿= .196

𝜙𝑀𝑛 = .9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 −𝑎

2) = 34.84𝑘 ∗ 𝑓𝑡

𝜙𝑀𝑛314 = .9 ∗ .85 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ 𝑑 = 29.835𝑘 ∗ 𝑓𝑡

𝑅𝑎𝑡𝑖𝑜 =𝑀𝑛314𝑀𝑛

= .856

Summary: Design for a 4.5ft long by 4.5ft wide by 13in deep foundation imbedded 2ft below the

foundation slab with 3 #4 bars spanning both directions.



SHEAR WALL FOUNDATION DESIGN Given Data and assumptions

Variable Value Unit

qa 8000 psf

Concrete Weight wc 150 pcf


Slab weight with Thickness (4") 50 psf

Soil Depth 24 in


Ka 0.33

Kp 3

Theta 30 degrees

f'c 4000 psi


Cover Cc 3 in

Assumed foundation dimension (parallel) B

30 ft

Assumed foundation dimension (perpendicular) L

10 ft

Foundation Thickness 24 in

Wall Thickness twall 10 in

Wall Length lw 25 ft

Calculations

𝐹𝑜𝑜𝑡𝑖𝑛𝑔 𝑤𝑒𝑖𝑔ℎ𝑡 𝑃𝑓 = 𝐵 ∗ 𝐿 ∗ ℎ ∗ 𝑤𝑐 = 30𝑓𝑡 ∗ 10𝑓𝑡 ∗ 24𝑖𝑛 ∗𝑓𝑡

12𝑖𝑛∗ 150𝑝𝑐𝑓 = 90𝑘

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑓𝑜𝑜𝑡𝑖𝑛𝑔 𝑉𝑓 = 𝐵 ∗ 𝐿 ∗ ℎ =30𝑓𝑡 ∗ 10𝑓𝑡 ∗ 24𝑖𝑛 ∗

𝑓𝑡12𝑖𝑛

27𝑓𝑡3= 22.23𝑦𝑑3

𝑆𝑜𝑖𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑃𝑠𝑜𝑖𝑙 = 𝐵 ∗ 𝐿 ∗ ℎ𝑠𝑜𝑖𝑙 ∗ 𝛾𝑠𝑜𝑖𝑙 = 30𝑓𝑡 ∗ 10𝑓𝑡 ∗ 24𝑖𝑛 ∗1𝑓𝑡

12𝑖𝑛∗ 110𝑝𝑐𝑓 = 66𝑘

𝑆𝑢𝑟𝑐ℎ𝑎𝑟𝑔𝑒 𝑊𝑒𝑖𝑔ℎ𝑡 𝑃𝑠𝑢𝑟𝐿 = 𝐵 ∗ 𝐿 ∗ 𝑤𝑠𝑢𝑟𝐿 = 30𝑘 Total Loads from Shear Wall Design

𝑃𝑤 = 239𝑘 𝑉𝑤 = 50.76𝑘

𝑀𝑤 = 1678.351𝑘 ∗ 𝑓𝑡 Factored Service Loads 𝑄𝑠 = 1(𝑃𝑓 + 𝑃𝑤) = 329𝑘

𝑉𝑠 = .6 ∗ 𝑉𝑤 = 30.46𝑘



𝑀𝑠 = 𝑀𝑤 ∗ .6 + (𝑉𝑠 ∗ℎ𝑠𝑜𝑖𝑙12) = 1068𝑘 ∗ 𝑓𝑡

Factored Strength Loads 𝑄𝑢 = 1.4 ∗ (𝑃𝑓 + 𝑃𝑤) = 460.6𝑘

𝑉𝑢 = .5 ∗ (𝑉𝑤) = 25.38𝑘

𝑀𝑢 = 𝑀𝑤 + (𝑉𝑢 ∗ℎ𝑠𝑜𝑖𝑙12) = 1729.111𝑘 ∗ 𝑓𝑡

𝑆𝑜𝑖𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑒𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦 𝑞𝑐𝑜𝑛 =𝑄𝑠𝐵𝐿= 1096.67𝑝𝑠𝑓

𝐷𝑒𝑚𝑎𝑛𝑑 ∶ 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜𝑠 𝑞𝑐𝑜𝑛𝑞𝑎

= .137 < 1 𝑂𝐾

𝐸𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦 =𝑀𝑠𝑄𝑠= 3.25𝑓𝑡

𝐵

6= 5𝑓𝑡

𝐸𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑦 <𝐵

6

Max soil pressure with eccentricity

𝑞𝑚𝑎𝑥 = 𝑞𝑐𝑜𝑛 ∗ (1 + 6 ∗𝐸𝑐𝑐

𝐵) = 1808.62𝑝𝑠𝑓

𝑞𝑚𝑎𝑥𝑞𝑎

= .226 < 1 𝑂𝐾

Minimum soil pressure with eccentricity

𝑞𝑚𝑖𝑛 = 𝑞𝑐𝑜𝑛 ∗ (1 − 6 ∗𝐸𝑐𝑐

𝐵) = 384.72𝑝𝑠𝑓

Check Sliding Resistance

Coefficient of friction 𝜇 = tan (2

3∗ 𝜙) = .36

𝐷𝑒𝑝𝑡ℎ 𝑓𝑟𝑜𝑚 𝑡𝑜𝑝 𝑜𝑓 𝑠𝑜𝑖𝑙 𝑡𝑜 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑓𝑜𝑜𝑡𝑖𝑛𝑔 𝐻 = ℎ + ℎ𝑠𝑜𝑖𝑙 = 4𝑓𝑡

𝑁𝑒𝑡 𝑒𝑎𝑟𝑡ℎ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑜 𝑠𝑙𝑖𝑑𝑖𝑛𝑔 𝑃𝑛𝑒𝑡 =1

2(𝐾𝑝 − 𝐾𝑎)𝛾𝑠𝑜𝑖𝑙 ∗ 𝐻

2 ∗ 𝐿 = 58.74𝑘

𝑇𝑜𝑡𝑎𝑙 𝑓𝑜𝑢𝑛𝑑𝑎𝑡𝑖𝑜𝑛 𝑠𝑙𝑖𝑑𝑖𝑛𝑔 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑉𝑓𝑡𝑔 = 𝑄𝑠 ∗ 𝜇 + 𝑃𝑛𝑒𝑡 = 177.18𝑘

𝑆𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠𝑙𝑖𝑑𝑖𝑛𝑔 𝐹𝑆𝑠𝑙𝑑 =𝑉𝑓𝑡𝑔

𝑉𝑠= 5.82 > 1.67 𝑂𝐾

Design reinforced concrete foundation for bending moment and shear

𝑆𝑜𝑖𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑒𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦 𝑞𝑢𝑐𝑜𝑛 =𝑄𝑢𝐵 ∗ 𝐿

= 1535.33𝑝𝑠𝑓

Max LRFD soil pressure

𝑞𝑢𝑚𝑎𝑥 = 𝑞𝑢𝑐𝑜𝑛 ∗ (1 + 6 ∗𝐸𝑐𝑐

𝐵) = 2532.1𝑝𝑠𝑓

Min LRFD soil pressure

𝑞𝑢𝑚𝑖𝑛 = 𝑞𝑢𝑐𝑜𝑛 ∗ (1 − 6 ∗𝐸𝑐𝑐

𝐵) = 538.61𝑝𝑠𝑓

Design for shear Parallel to Wall Length



Choose Size and # of tension steel bars Bar size=9 Bar num=20

𝐴𝑠 = 20𝑖𝑛2

𝑆𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑎𝑙 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑓𝑜𝑜𝑡𝑖𝑛𝑔 𝑑 = ℎ − 𝑐𝑐 − 1.5𝑑𝑏 = 24 − 3 − 1.5 ∗ 1.128 = 1.61𝑓𝑡

𝐶𝑎𝑛𝑡𝑖𝑙𝑒𝑣𝑒𝑟 𝑓𝑜𝑜𝑡𝑖𝑛𝑔 𝑙𝑒𝑛𝑔𝑡ℎ 𝑙𝑣 =𝐵

2−𝑙𝑤2− 𝑑 = .891𝑓𝑡

Interpolating using lv to find the critical section soil pressure 𝑞𝑢2 = 2472.86𝑝𝑠𝑓

𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒 𝐹𝑢 =𝑞𝑢𝑚𝑎𝑥 + 𝑞𝑢2

2∗ 𝑙𝑣 ∗ 𝐿 = 22.3𝑘

𝜙𝑉𝑐 = .75 ∗ 2 ∗ √4000 ∗ 𝐿 ∗ 𝑑 = 219.81𝑘 𝐹𝑢𝜙𝑉𝑐

= .102 < 1 𝑂𝐾

Check two-way shear 𝑏1 = 𝑙𝑤 + 𝑑 = 26.61𝑓𝑡 𝑏2 = 𝑡𝑤𝑎𝑙𝑙 + 𝑑 = 2.443𝑓𝑡

𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎 𝐴𝑐 = 2(𝑏1 + 𝑏2) ∗ 𝑑 = 93.49𝑓𝑡2

𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑑𝑢𝑙𝑜𝑢𝑠 𝐽𝑐 =𝑏1 ∗ 𝑑 ∗ (𝑏1 + 3 ∗ 𝑏2) + 𝑑

3

3= 485.7𝑓𝑡3

𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑒𝑠𝑠 𝑜𝑛 𝑤𝑎𝑙𝑙 𝑒𝑛𝑑 𝑣𝑢 =𝐹𝑢𝐴𝑐+𝑀𝑢𝐽𝑐= 26.38𝑝𝑠𝑖

𝜙𝑣𝑐 = .75 ∗ 4 ∗ √4000 = 189.74𝑝𝑠𝑖 𝑣𝑢𝜙𝑣𝑐

= .14 < 1 𝑂𝐾

Bending Moment Parallel to wall

𝑙𝑚 =𝐵

2−𝑙𝑤2= 2.5𝑓𝑡

Interpolating using lm to find the critical section soil pressure 𝑞𝑢2 =2365.939psf

𝐹𝑢 =𝑞𝑢𝑚𝑎𝑥 + 𝑞𝑢2

2∗ 𝑙𝑚 ∗ 𝐿 = 61.225𝑘

𝐹𝑜𝑟𝑐𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑟𝑚 𝑎𝑠 = 𝑙𝑚 −𝑙𝑚∗(2∗𝑞𝑢2+𝑞𝑢𝑚𝑎𝑥)

3∗(𝑞𝑢2+𝑞𝑢𝑚𝑎𝑥)=1.264ft

𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑀𝑢𝑝𝑟𝑙 = (𝐹𝑢 ∗ 𝑎𝑠) = 77.4𝑘 ∗ 𝑓𝑡

𝜙𝑀𝑛 = .9 ∗ .85 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ 𝑑 =1477.1k*ft 𝑀𝑢𝑝𝑟𝑙

𝜙𝑀𝑛= .053 < 1 𝑂𝐾

Design for Shear force Perpendicular to wall length Choose Bar size and Spacing Bar size= 9, Spacing = 10in

𝐴𝑠 =1.2𝑖𝑛2

𝑓𝑡

𝑙𝑣 =𝐿

2−𝑡𝑤𝑎𝑙𝑙2− 𝑑 = 2.975𝑓𝑡



𝐹𝑢 = 𝑞𝑚𝑎𝑥 ∗ 𝑙𝑣 = 5.4𝑘

𝑓𝑡

𝜙𝑉𝑐 = .75 ∗ 2 ∗ √4000 ∗ 𝑑 = 21.98𝑘

𝑓𝑡

𝐹𝑢𝜙𝑉𝑐

= .245 < 1 𝑂𝐾

Design for bending moment perpendicular to wall

𝑙𝑚 =𝐿

2−𝑡𝑤𝑎𝑙𝑙2

= 2.584𝑓𝑡

𝑀𝑢𝑝𝑟𝑝 = 𝑞𝑚𝑎𝑥 ∗𝑙𝑚2

2= 18 𝑘 ∗

𝑓𝑡

𝑓𝑡

𝜙𝑀𝑛 = .9 ∗ .85 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ 𝑑 = 88.63 𝑘 ∗𝑓𝑡

𝑓𝑡

𝑀𝑢𝑝𝑟𝑝

𝜙𝑀𝑛= .215 < 1 𝑂𝐾

Summary Provide foundation B=30ft long, L=10ft wide, h=2ft thick with 20 # 9 bars top and bottom in the long direction and #9 bars at 10in spacing OC top and bottom in the short direction. Provide footing of this specification at all shear wall locations and overlap as necessary.

APPENDIX Pg 214


APPENDIX

AUTOCAD Pg 215


FOUNDATION DESIGN GREEN ROOF CORNER COLUMN FOOTING


Variable Value Unit

qa 8000 psf



50 psf

Soil Depth 24 in






Ka 0.33

Kp 3

Theta 30 degrees

f'c 4000 psi


Cover Cc 3 in

Calculations

𝑞𝑛𝑒𝑡 = 8000𝑝𝑠𝑓– (𝑁𝑜𝑛 𝑐𝑜𝑙𝑢𝑚𝑛 𝐿𝑜𝑎𝑑𝑠)𝑝𝑠𝑓 = 8000𝑝𝑠𝑓 − 100𝑝𝑠𝑓 − 50𝑝𝑠𝑓 − 220𝑝𝑠𝑓= 7630𝑝𝑠𝑓 𝑆𝑢𝑚 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 59.5𝐾 + 15.63𝐾 = 75.13𝑘

𝐵 = 𝐿 = √75.13𝑘/7.63𝑘𝑠𝑓3.14𝑓𝑡

Say 3.5ft 𝑄𝑑 = (𝐶𝑜𝑙𝑢𝑚𝑛 𝐷𝑒𝑎𝑑 𝐿𝑜𝑎𝑑) + .27 ∗ 𝐵2 = 59.5𝑘 + .27𝑘𝑠𝑓 ∗ (3.5𝑓𝑡)2 = 62.81𝐾 𝑄𝐿 = (𝐶𝑜𝑙𝑢𝑚𝑛 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑) + .1 ∗ 𝐵2 = 15.63𝑘 + .1𝑘𝑠𝑓 ∗ (3.5𝑓𝑡)2 = 16.86𝑘

𝑄𝑢 = 1.2 ∗ (𝑄𝑑) + 1.6(𝑄𝐿) = 102.34𝑘

𝑞𝑢 = 𝑄𝑢

𝐵2= 8.35𝑘𝑠𝑓

𝐴𝑠𝑠𝑢𝑚𝑒 𝑑 = 9𝑖𝑛 Calculate One-way and Two-way Shear

One way

𝑉𝑢 = (𝐵

2−𝑐

2− 𝑑) ∗ 𝐿 ∗ 𝑞𝑢 = 0𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 2 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝐿 ∗ 𝑑 =. 75 ∗ 2 ∗ 1 ∗ √4000𝑝𝑠𝑖 ∗ 3.5𝑓𝑡 ∗ 12 ∗ 9𝑖𝑛

1000𝑙𝑏= 35.86𝑘


AUTOCAD Pg 216


35.86𝑘 > 0𝑘 𝑂𝐾

Two-Way

𝑏0 = 4 ∗ (𝑐 + 𝑑) = 4(24 + 9) = 132𝑖𝑛

𝑉𝑢 = (𝐵 ∗ 𝐿 − (𝑏04)2

) ∗ 𝑞𝑢 = 39.16𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑 = 225.41𝑘




𝑀𝑢 = 𝑞𝑢 ∗ 𝐿 ∗(𝐵2 −

𝑐2)2

2= 8.224𝑘 ∗ 𝑓𝑡


. 9= 9.14𝑘 ∗ 𝑓𝑡


𝜌𝑓𝑦

1.7 ∗ 𝑓′𝑐)


𝜌 = .0003

𝐴𝑠 = 𝜌𝐵𝑑 = (. 0003) ∗ 3.5𝑓𝑡 ∗12𝑖𝑛

𝑓𝑡∗ 12𝑖𝑛 = .134𝑖𝑛2


𝐴𝑠 = .22𝑖𝑛2


. 85𝑓′𝑐𝐿= .092

𝜙𝑀𝑛 = .9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 −𝑎

2) = 11.83𝑘 ∗ 𝑓𝑡

𝜙𝑀𝑛314 = .9 ∗ .85 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ 𝑑 = 10.1𝑘 ∗ 𝑓𝑡


= .853



AUTOCAD Pg 217


FOUNDATION DESIGN GREEN ROOF INTERIOR COLUMN FOOTING


Variable Value Unit

qa 8000 psf



50 psf

Soil Depth 24 in






Ka 0.33

Kp 3

Theta 30 degrees

f'c 4000 psi


Cover Cc 3 in

Calculations

𝑞𝑛𝑒𝑡 = 8000𝑝𝑠𝑓– (𝑁𝑜𝑛 𝑐𝑜𝑙𝑢𝑚𝑛 𝐿𝑜𝑎𝑑𝑠)𝑝𝑠𝑓 = 8000𝑝𝑠𝑓 − 100𝑝𝑠𝑓 − 50𝑝𝑠𝑓 − 220𝑝𝑠𝑓= 7630𝑝𝑠𝑓 𝑆𝑢𝑚 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 237.9𝐾 + 62.5𝐾 = 300.4𝑘

𝐵 = 𝐿 = √300.4𝑘/7.63𝑘𝑠𝑓 = 6.27𝑓𝑡

Say 6.5ft 𝑄𝑑 = (𝐶𝑜𝑙𝑢𝑚𝑛 𝐷𝑒𝑎𝑑 𝐿𝑜𝑎𝑑) + .27 ∗ 𝐵2 = 249.31𝐾 𝑄𝐿 = (𝐶𝑜𝑙𝑢𝑚𝑛 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑) + .1 ∗ 𝐵2 = 66.73𝑘

𝑄𝑢 = 1.2 ∗ (𝑄𝑑) + 1.6(𝑄𝐿) = 405.93𝑘

𝑞𝑢 = 𝑄𝑢

𝐵2= 9.61𝑘𝑠𝑓

𝐴𝑠𝑠𝑢𝑚𝑒 𝑑 = 16𝑖𝑛

Calculate One-way and Two-way Shear One way

𝑉𝑢 = (𝐵

2−𝑐

2− 𝑑) ∗ 𝐿 ∗ 𝑞𝑢 = 57.25𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 2 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝐿 ∗ 𝑑 = 118.4𝑘


AUTOCAD Pg 218


118.4𝑘 > 57.25𝑘 𝑂𝐾

Two-Way

𝑏0 = 4 ∗ (𝑐 + 𝑑) = 4(24 + 16) = 160𝑖𝑛

𝑉𝑢 = (𝐵 ∗ 𝐿 − (𝑏04)2

) ∗ 𝑞𝑢 = 300𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑 = 485.73𝑘


485.73𝑘 > 300𝑘 𝑂𝐾 Add 3 to 4 inches to initial d value


𝑀𝑢 = 𝑞𝑢 ∗ 𝐿 ∗(𝐵2 −

𝑐2)2

2= 158.1𝑘 ∗ 𝑓𝑡


. 9= 175.64𝑘 ∗ 𝑓𝑡


𝜌𝑓𝑦

1.7 ∗ 𝑓′𝑐)


𝜌 = .0011

𝐴𝑠 = 𝜌𝐵𝑑 = 1.681𝑖𝑛2


𝐴𝑠 = 1.76𝑖𝑛2


. 85𝑓′𝑐𝐿= .398

𝜙𝑀𝑛 = .9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 −𝑎

2) = 148.9𝑘 ∗ 𝑓𝑡

𝜙𝑀𝑛314 = .9 ∗ .85 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ 𝑑 = 127.91𝑘 ∗ 𝑓𝑡


= .859



AUTOCAD Pg 219


FOUNDATION DESIGN GREEN ROOF/INTERIOR EDGE COLUMN FOOTING


Variable Value Unit

qa 8000 psf



50 psf

Soil Depth 24 in



Point Live Load 164 k



Ka 0.33

Kp 3

Theta 30 degrees

f'c 4000 psi


Cover Cc 3 in

Calculations

𝑞𝑛𝑒𝑡 = 8000𝑝𝑠𝑓– (𝑁𝑜𝑛 𝑐𝑜𝑙𝑢𝑚𝑛 𝐿𝑜𝑎𝑑𝑠)𝑝𝑠𝑓 = 8000𝑝𝑠𝑓 − 100𝑝𝑠𝑓 − 50𝑝𝑠𝑓 − 220𝑝𝑠𝑓= 7630𝑝𝑠𝑓 𝑆𝑢𝑚 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 158.4𝑘 + 61.7𝑘 = 220.1𝑘

𝐵 = 𝐿 = √220.1𝑘/7.63𝑘𝑠𝑓 = 5.37𝑓𝑡

Say 5.5ft 𝑄𝑑 = (𝐶𝑜𝑙𝑢𝑚𝑛 𝐷𝑒𝑎𝑑 𝐿𝑜𝑎𝑑) + .27 ∗ 𝐵

2 = 166.5675𝐾 𝑄𝐿 = (𝐶𝑜𝑙𝑢𝑚𝑛 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑) + .1 ∗ 𝐵

2 = 64.73𝑘 𝑄𝑢 = 1.2 ∗ (𝑄𝑑) + 1.6(𝑄𝐿) = 303.441𝑘

𝑞𝑢 = 𝑄𝑢

𝐵2= 10.03𝑘𝑠𝑓



𝑉𝑢 = (𝐵

2−𝑐

2− 𝑑) ∗ 𝐿 ∗ 𝑞𝑢 = 32.2𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 2 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝐿 ∗ 𝑑 = 87.66𝑘


AUTOCAD Pg 220


87.66𝑘 > 32.2𝑘 𝑂𝐾

Two-Way

𝑏0 = 4 ∗ (𝑐 + 𝑑) = 4(24 + 14) = 152𝑖𝑛

𝑉𝑢 = (𝐵 ∗ 𝐿 − (𝑏04)2

) ∗ 𝑞𝑢 = 202.9𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑 = 403.8𝑘




𝑀𝑢 = 𝑞𝑢 ∗ 𝐿 ∗(𝐵2 −

𝑐2)2

2= 84.5 𝑘 ∗ 𝑓𝑡

𝑀𝑛 =𝑀𝑢𝜙= 93.867 𝑘 ∗ 𝑓𝑡


𝜌𝑓𝑦

1.7 ∗ 𝑓′𝑐)


𝜌 = .0009

𝐴𝑠 = 𝜌𝐵𝑑 = 1.002𝑖𝑛2


𝐴𝑠 = 1.32𝑖𝑛2


. 85𝑓′𝑐𝐿= .353

𝜙𝑀𝑛 = .9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 −𝑎

2) = 99.93𝑘 ∗ 𝑓𝑡

𝜙𝑀𝑛314 = .9 ∗ .85 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ 𝑑 = 85.833𝑘 ∗ 𝑓𝑡


= .859



AUTOCAD Pg 221


FOUNDATION DESIGN GREEN ROOF/INTERIOR INTERIOR COLUMN FOOTING


Variable Value Unit

qa 8000 psf



50 psf

Soil Depth 24 in






Ka 0.33

Kp 3

Theta 30 degrees

f'c 4000 psi


Cover Cc 3 in

Calculations

𝑞𝑛𝑒𝑡 = 8000𝑝𝑠𝑓– (𝑁𝑜𝑛 𝑐𝑜𝑙𝑢𝑚𝑛 𝐿𝑜𝑎𝑑𝑠)𝑝𝑠𝑓 = 8000𝑝𝑠𝑓 − 100𝑝𝑠𝑓 − 50𝑝𝑠𝑓 − 220𝑝𝑠𝑓= 7630𝑝𝑠𝑓

𝑆𝑢𝑚 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 547.9𝑘

𝐵 = 𝐿 = √220.1𝑘/7.63𝑘𝑠𝑓 = 8.47𝑓𝑡


2 = 403.4075 𝑄𝐿 = (𝐶𝑜𝑙𝑢𝑚𝑛 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑) + .1 ∗ 𝐵

2 = 171.225𝑘 𝑄𝑢 = 1.2 ∗ (𝑄𝑑) + 1.6(𝑄𝐿) = 758.05𝑘

𝑞𝑢 = 𝑄𝑢

𝐵2= 10.5𝑘𝑠𝑓



𝑉𝑢 = (𝐵

2−𝑐

2− 𝑑) ∗ 𝐿 ∗ 𝑞𝑢 = 104.05𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 2 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝐿 ∗ 𝑑 = 241.9𝑘

AUTOCAD Pg 222



241.9𝑘 > 104.05𝑘 𝑂𝐾

Two-Way

𝑏0 = 4 ∗ (𝑐 + 𝑑) = 4(24 + 25) = 196𝑖𝑛

𝑉𝑢 = (𝐵 ∗ 𝐿 − (𝑏04)2

) ∗ 𝑞𝑢 = 583.11𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑 = 929.71𝑘




𝑀𝑢 = 𝑞𝑢 ∗ 𝐿 ∗(𝐵2 −

𝑐2)2

2= 470 𝑘 ∗ 𝑓𝑡

𝑀𝑛 =𝑀𝑢𝜙= 523.32 𝑘 ∗ 𝑓𝑡


𝜌𝑓𝑦

1.7 ∗ 𝑓′𝑐)


𝜌 = .0012

𝐴𝑠 = 𝜌𝐵𝑑 = 3.4𝑖𝑛2


𝐴𝑠 = 3.52𝑖𝑛2


. 85𝑓′𝑐𝐿= .609

𝜙𝑀𝑛 = .9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 −𝑎

2) = 438.7𝑘 ∗ 𝑓𝑡

𝜙𝑀𝑛314 = .9 ∗ .85 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ 𝑑 = 377𝑘 ∗ 𝑓𝑡


= .859



AUTOCAD Pg 223


FOUNDATION DESIGN INTERIOR ROOF EDGE COLUMN FOOTING


Variable Value Unit

qa 8000 psf



50 psf

Soil Depth 24 in






Ka 0.33

Kp 3

Theta 30 degrees

f'c 4000 psi


Cover Cc 3 in

Calculations



𝐵 = 𝐿 = √397.7𝑘/7.63𝑘𝑠𝑓 = 7.2𝑓𝑡


2 = 280.1𝑘 𝑄𝐿 = (𝐶𝑜𝑙𝑢𝑚𝑛 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑) + .1 ∗ 𝐵

2 = 138.43𝑘 𝑄𝑢 = 1.2 ∗ (𝑄𝑑) + 1.6(𝑄𝐿) = 557.6𝑘

𝑞𝑢 = 𝑄𝑢

𝐵2= 9.91𝑘𝑠𝑓



𝑉𝑢 = (𝐵

2−𝑐

2− 𝑑) ∗ 𝐿 ∗ 𝑞𝑢 = 62𝑘

AUTOCAD Pg 224


𝜙𝑉𝑐 = 𝜙 ∗ 2 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝐿 ∗ 𝑑 = 196.4𝑘


196.4𝑘 > 62𝑘 𝑂𝐾

Two-Way

𝑏0 = 4 ∗ (𝑐 + 𝑑) = 4(24 + 23) = 188𝑖𝑛

𝑉𝑢 = (𝐵 ∗ 𝐿 − (𝑏04)2

) ∗ 𝑞𝑢 = 405.5226𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑 = 820.42𝑘




𝑀𝑢 = 𝑞𝑢 ∗ 𝐿 ∗(𝐵2 −

𝑐2)2

2= 281.12 𝑘 ∗ 𝑓𝑡

𝑀𝑛 =𝑀𝑢𝜙= 312.25𝑘 ∗ 𝑓𝑡


𝜌𝑓𝑦

1.7 ∗ 𝑓′𝑐)


𝜌 = .0009

𝐴𝑠 = 𝜌𝐵𝑑 = 2.18𝑖𝑛2


𝐴𝑠 = 2.2𝑖𝑛2


. 85𝑓′𝑐𝐿= .431

𝜙𝑀𝑛 = .9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 −𝑎

2) = 255.26𝑘 ∗ 𝑓𝑡

𝜙𝑀𝑛314 = .9 ∗ .85 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ 𝑑 = 218.8𝑘 ∗ 𝑓𝑡


= .857



AUTOCAD Pg 225


FOUNDATION DESIGN INTERIOR ROOF INTERIOR COLUMN FOOTING


Variable Value Unit

qa 8000 psf



50 psf

Soil Depth 24 in






Ka 0.33

Kp 3

Theta 30 degrees

f'c 4000 psi


Cover Cc 3 in

Calculations



𝐵 = 𝐿 = √397.7𝑘/7.63𝑘𝑠𝑓 = 9.38𝑓𝑡



2 = 193.33𝑘 𝑄𝑢 = 1.2 ∗ (𝑄𝑑) + 1.6(𝑄𝐿) = 923.2𝑘

𝑞𝑢 = 𝑄𝑢

𝐵2= 10.23𝑘𝑠𝑓



𝑉𝑢 = (𝐵

2−𝑐

2− 𝑑) ∗ 𝐿 ∗ 𝑞𝑢 = 145.8𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 2 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝐿 ∗ 𝑑 = 292𝑘


AUTOCAD Pg 226


292𝑘 > 145.8𝑘 𝑂𝐾

Two-Way

𝑏0 = 4 ∗ (𝑐 + 𝑑) = 4(24 + 27) = 204𝑖𝑛

𝑉𝑢 = (𝐵 ∗ 𝐿 − (𝑏04)2

) ∗ 𝑞𝑢 = 738.43𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑 = 1045.1𝑘




𝑀𝑢 = 𝑞𝑢 ∗ 𝐿 ∗(𝐵2 −

𝑐2)2

2= 683.3 𝑘 ∗ 𝑓𝑡

𝑀𝑛 =𝑀𝑢𝜙= 759.21𝑘 ∗ 𝑓𝑡


𝜌𝑓𝑦

1.7 ∗ 𝑓′𝑐)


𝜌 = .0013

𝐴𝑠 = 𝜌𝐵𝑑 = 4.61𝑖𝑛2


𝐴𝑠 = 4.8𝑖𝑛2


. 85𝑓′𝑐𝐿= .743

𝜙𝑀𝑛 = .9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 −𝑎

2) = 639.98𝐾 ∗ 𝑓𝑡

𝜙𝑀𝑛314 = .9 ∗ .85 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ 𝑑 = 550.8𝑘 ∗ 𝑓𝑡


= .86



AUTOCAD Pg 227


FOUNDATION DESIGN INTERIOR ROOF CORNER COLUMN FOOTING


Variable Value Unit

qa 8000 psf



50 psf

Soil Depth 24 in


Point Dead Load 265 k




Ka 0.33

Kp 3

Theta 30 degrees

f'c 4000 psi


Cover Cc 3 in

Calculations



𝐵 = 𝐿 = √397.7𝑘/7.63𝑘𝑠𝑓 = 7.22𝑓𝑡



2 = 138.43𝑘 𝑄𝑢 = 1.2 ∗ (𝑄𝑑) + 1.6(𝑄𝐿) = 557.71𝑘

𝑞𝑢 = 𝑄𝑢

𝐵2= 9.915𝑘𝑠𝑓



𝑉𝑢 = (𝐵

2−𝑐

2− 𝑑) ∗ 𝐿 ∗ 𝑞𝑢 = 62𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 2 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝐿 ∗ 𝑑 = 196.4𝑘


AUTOCAD Pg 228


196.4𝑘 > 62𝑘 𝑂𝐾

Two-Way

𝑏0 = 4 ∗ (𝑐 + 𝑑) = 4(24 + 23) = 188𝑖𝑛

𝑉𝑢 = (𝐵 ∗ 𝐿 − (𝑏04)2

) ∗ 𝑞𝑢 = 405.61𝑘

𝜙𝑉𝑐 = 𝜙 ∗ 4 ∗ 𝜆 ∗ √𝑓′𝑐 ∗ 𝑏0 ∗ 𝑑 = 820.42𝑘




𝑀𝑢 = 𝑞𝑢 ∗ 𝐿 ∗(𝐵2 −

𝑐2)2

2= 281.2𝑘 ∗ 𝑓𝑡

𝑀𝑛 =𝑀𝑢𝜙= 312.42𝑘 ∗ 𝑓𝑡


𝜌𝑓𝑦

1.7 ∗ 𝑓′𝑐)


𝜌 = .0009

𝐴𝑠 = 𝜌𝐵𝑑 = 2.18𝑖𝑛2


𝐴𝑠 = 2.2𝑖𝑛2


. 85𝑓′𝑐𝐿= .43

𝜙𝑀𝑛 = .9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 −𝑎

2) = 255.3𝐾 ∗ 𝑓𝑡

𝜙𝑀𝑛314 = .9 ∗ .85 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ 𝑑 = 218.8𝑘 ∗ 𝑓𝑡


= .857



AUTOCAD Pg 229


AUTOCAD DRAWINGS

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REFERENCES Pg 230


REFERENCES:

Florida Building Code 2010, Building. Tallahassee, FL: Florida Building Commission, 2011.

International Code Council. International Code Council, Inc., 2011. Web. 24 Sept.

2014. <www.ecodes.biz>.

Microsoft Excel. Redmond, WA: Microsoft Corp., 1991. Computer software.

Minimum Design Loads for Buildings and Other Structures (7-10). Reston, VA: American

Society of Civil Engineers, 2010. Print.

Steel Construction Manual. 14th ed. Chicago, IL: American Institute of Steel Construction,

2011. Print.

Visual Analysis Edu. Vers. 10.0. Bozeman, MT: IES, Inc., 2012. Computer software.

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