CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT...
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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
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TABLE OF CONTENTS Pg 1
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
TABLE OF CONTENTS:
TABLE OF CONTENTSβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦... 1
EXECUTIVE SUMMARYβ¦β¦β¦β¦β¦β¦β¦β¦.β¦..β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 3
SCHEDULEβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦... 4
TABULATION OF SPECIFIED LOADSβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 7
MWFRS CALCULATIONSβ¦β¦β¦.β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 9
COMPONENTS & CLADDING CALCULATIONSβ¦β¦β¦β¦β¦β¦β¦β¦β¦..β¦β¦β¦β¦β¦. 12
ROOF DECK DESIGNβ¦β¦β¦β¦β¦β¦...β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. 17
ROOF JOIST DESIGNβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦......20
ROOF JOIST GIRDER DESIGNβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..23
ROOF PERIMETER BEAM DESIGNβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..... 25
COMPOSITE FLOOR DECK DESIGNβ¦β¦β¦β¦...β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 35
FLOOR BEAM DESIGNβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦...β¦... 40
STEEL COLUMN DESIGNβ¦β¦β¦β¦β¦..β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦83
TWO-WAY SLAB DESIGNβ¦β¦β¦.β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 95
CONCRETE CONTINUOUS L-BEAM DESIGNβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..β¦β¦β¦β¦β¦. 143
CONCRETE COLUMN DESIGNβ¦β¦..β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. 156
SIMPLE SHEAR CONNECTIONSβ¦β¦.β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦......158
ROOF CONNECTIONS...β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..160
DIAPHRAGM DESIGNβ¦...β¦β¦β¦.β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ .162
LATERAL STEEL ANALYSISβ¦β¦β¦β¦β¦β¦β¦β¦.β¦β¦β¦β¦β¦β¦β¦β¦β¦..β¦β¦β¦β¦β¦..164
APPROXIMATE SECOND ORDER ANALYSISβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦180
BRACED FRAME DESIGNβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦......187
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TABLE OF CONTENTS Pg 2
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
SHEAR WALL DESIGNβ¦β¦β¦.β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..204
FOUNDATION DESIGNβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..β¦β¦β¦β¦β¦β¦β¦β¦β¦........................ 208
APPENDIXβ¦β¦β¦β¦β¦..β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦...β¦... 214
AUTOCADβ¦β¦β¦β¦β¦β¦..β¦β¦β¦β¦β¦..β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 229
REFERENCESβ¦β¦β¦β¦β¦.β¦β¦β¦.β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 230
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EXECUTIVE SUMMARY Pg 3
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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 projectβs 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 frameβs 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 engineerβs involved had limited experience with ASD. ASD
also makes the design more conservative because it is based on a memberβs 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 Constructionβs Specification for Structural Steel Buildings June 22, 2010, the
14th
edition of the AISC Steel Construction Manual, the American Society of Civil Engineersβ
Minimum Design Loads for Buildings and Other Structures (7-10), the 2012 International
Building Code, and the American Concrete Instituteβs 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
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SCHEDULE Pg 4
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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.
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SCHEDULE Pg 5
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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SCHEDULE Pg 6
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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TABULATION OF SPECIFIED LOADS Pg 7
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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
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TABULATION OF SPECIFIED LOADS Pg 8
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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
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MWRFS CALCULATIONS Pg 9
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
MWRFS CALCULATIONS
The Main Wind Resisting Force System calculations were completed in Excel. The equations
used by this excel program are as follows:
ππ§ = 0.00256πΎπ§πΎπ§π‘πΎππ2
π = ππΊπΆπ β πππΊπΆππ
Note: πΎπ§ 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
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MWRFS CALCULATIONS Pg 10
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Wind Pressure p = qGCp β 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 β 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
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MWRFS CALCULATIONS Pg 11
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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C&C CALCULATIONS Pg 12
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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 πΎπ§ 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,
π΄πππ,ππππ π‘ = 50 ππ‘ β50
3 ππ‘ = 833.33 ππ‘2
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:
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C&C CALCULATIONS Pg 13
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
πΏππ£ππ 1: 20 ππ‘ β 20 ππ‘
3 = 133.33 ππ‘2
πΏππ£ππ 2: 18 ππ‘ β18 ππ‘
3= 108 ππ‘2
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 ππ§ 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, π = 140 ππβ
3) Determining directionality factor from section 26.6 gives: πΎπ = 0.85
From project specifications, use exposure category B.
From project specifications, the topographic factor is πΎπ§π‘ = 1.0
From table 26.11-1, the internal pressure coefficient is (πΊπΆππ) = {0.18β0.18
4) From table 26.9-1, πΌ = 7.0 and π§π = 1200 ππ‘.
From table 30.3-1:
πΎπ§ = 2.01 (π§
π§π)
2πΌ
= 2.01 (98 ππ‘
1200 ππ‘)
27.0= 0.98
Using Excel to calculate kz values for values of z at every floor height of the building
gives:
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C&C CALCULATIONS Pg 14
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
ππ§ = 0.00256πΎπ§πΎπ§π‘πΎππ2 = 0.00256(0.98)(1)(. 85)(140 ππβ)2 = 41.91 ππ π
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 ππ§(πΊπΆπ) β ππ§(πΊπΆππ) and taking the highest negative value:
For the joist, use the height of 98 ft:
Zone 1: π = {41.91ππ π(β0.9) β 41.91ππ π(. 18) = βππ. ππππ
41.91ππ π(β0.9) β 41.91ππ π(β.18) = β30.2ππ π
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:
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C&C CALCULATIONS Pg 15
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Zone Load with GCP+ (psf)
Load with GCP- (psf) Design loads (psf)
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 Load with GCP+ (psf)
Load with GCP- (psf) Design loads (psf)
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
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C&C CALCULATIONS Pg 16
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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
Tributary area (sq. ft) Zone 4 Zone 5
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
Tributary area (sq. ft) Zone 4 Zone 5
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
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ROOF DECK DESIGN Pg 17
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
πΏπ = 20 ππ π
π = β141.6 ππ π, 16 ππ π
π΄π = 5 ππ‘ β 50 ππ‘ = 250 ππ‘2
π· = 30 ππ π (πππ‘ πππππ’ππππ π πππ π€πππβπ‘)
LIVE LOAD REDUCTIONS:
Use ASCE 7-10 Section 4.8
πΏπ = πΏ0π 1π 2
12 β€ πΏπ = 20 ππ π β€ 20 ππ π β ππΎ
π 1 = 1.2 β 0.001π΄π = 1.2 β 0.001(250 ππ‘2) = .95 πππππ’π π 200 ππ‘2 < π΄π < 400 ππ‘
2
πΉ = .25 πππβππ ππ πππ π/ππππ‘ π 2 = 1
Therefore, πΏπ = (20 ππ π)(. 95)(1) = 19 ππ π
DESIGN:
Assume a dead self-weight of 2.5 psf.
Therefore, π· = 30 ππ π + 2.5 ππ π = 32.5 ππ π
Finding ASD load combinations from an Excel File:
[LOAD COMBINATIONS FOUND ON NEXT PAGE]
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ROOF DECK DESIGN Pg 18
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Using Vulcraft manual, try a 1.5B22. This meets the max SDI Construction Span of 6β-11β 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 deckβs capacity still meets the required load using the actual self-weight.
CHECK UPLIFT:
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ROOF DECK DESIGN Pg 19
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πΉπ = 0.6πΉπ¦ = 0.6(33 ππ π) = 19.8 ππ π
π =π€π2
8=65.89 ππ π β (5 ππ‘)2
8= 2.47
πππ β ππ
ππ‘
ππππ’ππππ π πππ‘πππ ππππ’ππ’π =π
πΉπ=2.47
πππ β ππππ‘
19.8 ππ π= 0.125
ππ3
ππ‘
This meets the section modulus of a 1.5B22 (ππ = .192ππ3
ππ‘).
CHECK UPLIFT DEFLECTION:
βπππ₯=πΏ
240=5 ππ‘ β 12 ππ
240= 0.25 ππ
β= 0.0052π€π4
πΈπΌ
πΌπππ = 0.0052π€π4
πΈβπππ₯= 0.0052
65.89 ππππ‘2
β1 πππ1000 ππ
β (ππ‘12 ππ)
2
(5 ππ‘ β 12 ππ)4
29000 ππ π β .25 ππ= 0.004
ππ4
ππ‘
This minimum I meets the I of the deck, which is 0.184 ππ4
ππ‘β ππΎ
THEREFORE USE A 1.5B22 ROOF DECK.
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ROOF JOIST DESIGN Pg 20
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
πΏπ = 20 ππ π β 5 ππ‘ = 100 ππ/ππ‘
π = β103.9 ππ π β 5 ππ‘ = β519.5ππ
ππ‘, 16 ππ π β 5 ππ‘ = 80 ππ/ππ‘
π΄π = 5 ππ‘ β 50 ππ‘ = 250 ππ‘2
π· = 30 ππ π + 1.78 ππ π ππππ ππππ ππππ = 31.78 ππ π β 5ππ‘= 158.9 ππ/ππ‘ (πππ‘ πππππ’ππππ π πππ π€πππβπ‘)
LIVE LOAD REDUCTIONS:
Use ASCE 7-10 Section 4.8
πΏπ = πΏ0π 1π 2
12 β€ πΏπ = 20 ππ π β€ 20 ππ π β ππΎ
π 1 = 1.2 β 0.001π΄π = 1.2 β 0.001(250 ππ‘2) = .95 πππππ’π π 200 ππ‘2 < π΄π < 400 ππ‘
2
πΉ = .25 πππβππ ππ πππ π/ππππ‘ π 2 = 1
Therefore, πΏπ = (20 ππ π)(. 95)(1) = 19 ππ π β 5ππ‘ = 95 ππ/ππ‘
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]
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ROOF JOIST DESIGN Pg 21
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
[LOAD COMBINATIONS FOUND ON NEX PAGE]
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ROOF JOIST DESIGN Pg 22
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
The 30K11βs 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.
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ROOF JOIST GIRDER DESIGN Pg 23
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
π΄π =(50 ππ‘ + 50 ππ‘)
2β50
10= 250 ππ‘2
πΏπ = 20 ππ π β 250 ππ‘2 = 5 πππ
π = β103.9 ππ π β 250 ππ‘2 = β26 πππ, 16 ππ π β 250 ππ‘2 = 4 πππ
π· = 30 ππ π + 1.78 ππ π ππππ ππππ ππππ
= ((31.78 ππ π β 50ππ‘) + 16.4ππ
ππ‘ ππππ π‘ π€πππβπ‘) β 5ππ‘ = 8.03 πππ
LIVE LOAD REDUCTIONS:
Use ASCE 7-10 Section 4.8
πΏπ = πΏ0π 1π 2
12 β€ πΏπ = 20 ππ π β€ 20 ππ π β ππΎ
π 1 = 1.2 β 0.001π΄π = 1.2 β 0.001(250 ππ‘2) = .95 πππππ’π π 200 ππ‘2 < π΄π < 400 ππ‘
2
πΉ = .25 πππβππ ππ πππ π/ππππ‘ π 2 = 1
Therefore, πΏπ = (20 ππ π)(. 95)(1) = 19 ππ π β 250ππ‘2 = 4.75 πππ
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:
[LOAD COMBINATIONS FOUND ON NEX PAGE]
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ROOF JOIST GIRDER DESIGN Pg 24
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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 girderβs deflection:
πΌ = 0.027πππΏπ = 0.027(10)(13.6)(50)(50) = 9180 ππ4
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
Ξ£(PiNi)= 10.54
π· = 1.15β(ππππ)π
3
6πΈπΌ= 1.15
10.54(50 ππ‘ β 12 ππ)3
6(29000 ππ π)(9180 ππ4)= 1.43 ππ
π·πππ₯ =πΏ
360=50 ππ‘ β 12 ππ
360= 1.667 ππ > 1.43 ππ β ππΎ
THEREFORE USE 50G10N13.6K. THIS JOIST GIRDER HAS AN APPROXIMATE
SELF-WEIGHT OF 78 LB/FT.
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ROOF PERIMETER BEAM DESIGN Pg 25
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
GIVEN/KNOWN VALUES:
For the east to west beams:
π΄π =(50 ππ‘)
2β50
10= 125 ππ‘2
πΏπ = 20 ππ π β 125ππ‘2 = 2.5 πππ
π = β103.9 ππ π β 125 ππ‘2 = β13 πππ, 16 ππ π β 125 ππ‘2 = 2 πππ
π· = [(30 ππ π + 1.78 ππ π ππππ ππππ ππππ) (50 ππ‘
2) + (16.4
ππ
ππ‘ ππππ ππππ π‘π )] (5 ππ‘)
= 4.05 π (ππ₯πππ’ππππ π πππ π€πππβπ‘)
For the north to south beams:
π΄π =(5 ππ‘)
2β100
4= 62.5 ππ‘2
πΏπ = 20 ππ π β5
2ππ‘ = .05 π/ππ‘
π = β103.9 ππ π β 2.5 ππ‘ = β.26π
ππ‘, 16 ππ π β 2.5 ππ‘2 = .04 π/ππ‘
π· = [(30 ππ π + 1.78 ππ π ππππ ππππ ππππ) (5 ππ‘
2)] = 0.08
π
ππ‘(ππ₯πππ’ππππ π πππ π€πππβπ‘)
LIVE LOAD REDUCTIONS:
Use ASCE 7-10 Section 4.8
πΏπ = πΏ0π 1π 2
12 β€ πΏπ = 20 ππ π β€ 20 ππ π β ππΎ
π 1 = 1 πππππ’π π π΄π < 200 ππ‘2
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ROOF PERIMETER BEAM DESIGN Pg 26
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πΉ = .25 πππβππ ππ πππ π/ππππ‘ π 2 = 1
Therefore, πΏπ = (20 ππ π)(1)(1) = 20 ππ π β 2.5 ππ‘2 = .05 π/ππ‘
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 π· + .75(πΏ + .6π + πΏπ) gives the maximum positive moment:
![Page 28: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/28.jpg)
ROOF PERIMETER BEAM DESIGN Pg 27
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ASD load combination . 6π· + .6π 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:
πΉππ =πΆππ
2πΈ
(πΏπππ‘π )2β1 + .078
π½π
ππ₯β0(πΏπππ‘π )2
=1.14π2(29000)
(25 β 122.21 )
2β1 + .078
. 906 β 1
51.5 β 11.4(25 β 12
2.21)2
= 31.8 ππ π ππΞ©=πΉππππ₯Ξ©
=31.8 β 51.5
1.67= 81.7 π β ππ‘ > 3.75 π β ππ‘ β ππΎ
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:
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ROOF PERIMETER BEAM DESIGN Pg 28
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ57ππ3
1.67= 142.2 π β ππ‘ > 13.7 π β ππ‘ β ππΎ
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 beamβs Zone III capacity meets the demand:
ππΞ©=πΉππππ₯Ξ©
=31.8 β 51.5
1.67= 81.7 β ππ‘ > 6.6 π β ππ‘ β ππΎ
Running visual analysis to find the maximum shear values gives:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-26 of 70.2k:
ππΞ©= 70.2π > 2.2π β ππΎ
ππΞ©= 70.2π > 1.05π β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏπ +π:
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ROOF PERIMETER BEAM DESIGN Pg 29
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
These deflections meet the requirements:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > 0.09 ππ β ππΎ
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > 0.17 ππ β ππΎ
For πΏπ +π:
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ROOF PERIMETER BEAM DESIGN Pg 30
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
These deflections meet the requirements: πΏ
360=25 ππ‘ β 12 ππ
360= 0.833 ππ > 0.21 ππ β ππΎ
πΏ
360=25 ππ‘ β 12 ππ
360= 0.833 ππ > 0.09 ππ β ππΎ
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:
Dead load with an assumed self-weight of 100lb/ft.:
Live roof load:
Wind in the positive direction:
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ROOF PERIMETER BEAM DESIGN Pg 31
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Wind in the negative direction (uplift):
ASD load combination π· + .75(πΏ + .6π + πΏπ) gives the maximum positive moment:
ASD load combination . 6π· + .6π 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:
πΏπ = 10.8 ππ‘ < πΏπ = 25 ππ‘ < πΏπ = 43.1 ππ‘ β ππππ πΌπΌ Checking LTB:
32.9 ππ π ππΞ©= πΆπ [
ππ
Ξ©βπ΅πΉ
Ξ©(πΏπ β πΏπ)] = 1.0[329 β 3.81(25 β 10.8)] = 275 π β ππ‘ > 76.1 π β ππ‘
β ππΎ
Finding the max positive moment using its actual self-weight of 87 lb/ft and the controlling ASD
load combinations mentioned previously in visual analysis gives:
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ROOF PERIMETER BEAM DESIGN Pg 32
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
The beamβs Zone I capacity:
ππ
Ξ©=πΉπ¦ππ₯
Ξ©=50 ππ πβ132ππ3
1.67= 329 π β ππ‘ > 110 π β ππ‘ β πΊπππ
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 beamβs Zone II capacity meets the demand:
ππΞ©= πΆπ [
ππ
Ξ©βπ΅πΉ
Ξ©(πΏπ β πΏπ)] = 1.0[329 β 3.81(25 β 10.8)] = 275 π β ππ‘ > 76.1 π β ππ‘
Checking Deflection:
For Lr+W:
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ROOF PERIMETER BEAM DESIGN Pg 33
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Check the selected beam for deflection:
πΏ
360=25 ππ‘ β 12 ππ
360= .833 ππ > .832 ππ β πΊπππ
πΏ
360=25 ππ‘ β 12 ππ
360= .833 ππ > .36 ππ β πΊπππ
For D+Lr+W:
Check the selected beam for deflection:
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ROOF PERIMETER BEAM DESIGN Pg 34
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > .48 ππ β πΊπππ
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > .71 ππ β πΊπππ
Check the selected beam for maximum shear:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-24 of 129k: ππΞ©= 129π > 21.6π β ππΎ
ππΞ©= 129π > 15.5π β ππΎ
THEREFORE USE A W12X87 FOR THE EAST TO WEST PERIMETER BEAMS.
FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 21.6 K
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COMPOSITE FLOOR DECK DESIGN Pg 35
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
π· = 9.5 ππ π (πππ‘ πππππ’ππππ π πππ β π€πππβπ‘) πΏ = 65 ππ π
π΄π = 12.5 ππ‘ β 25 ππ‘ = 312.5 ππ‘2
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 1
Therefore, π΄ππΎπΏπΏ = 312.5ππ‘2 β 1 < 400 ππ‘2 β π πππ’ππ‘πππ πππ‘ ππππππ‘π‘ππ
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:
πΆππππππ‘π¦: 74.50 ππ π < 90 ππ π β ππΎ
πππ₯ ππ·πΌ ππππ: 12.5 ππ‘ < 13 ππ‘ 8 ππ β ππΎ
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COMPOSITE FLOOR DECK DESIGN Pg 36
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
For future reference, this deckβs 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:
π· = 9.5 ππ π (πππ‘ πππππ’ππππ π πππ β π€πππβπ‘) πΏ = 65 ππ π
π΄π = 7.75 ππ‘ β 25ππ‘ = 193.75 ππ‘2
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 1
Therefore, π΄ππΎπΏπΏ = 193.75 ππ‘2 β 1 < 400 ππ‘2 β π πππ’ππ‘πππ πππ‘ ππππππ‘π‘ππ
DESIGN:
Factoring the loads using ASD and finding the superimposed load:
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,
![Page 38: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/38.jpg)
COMPOSITE FLOOR DECK DESIGN Pg 37
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
πΆππππππ‘π¦ = 270 β (270 β 244)7.75 β 7.5
8 β 7.5= 257 ππ π
πΆππππππ‘π¦: 74.50 ππ π < 257 ππ π β ππΎ
πππ₯ ππ·πΌ ππππ: 5 ππ‘ < 11 ππ‘ 4 ππ β ππΎ For future reference, this deckβs 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 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:
π· = 9.5 ππ π (πππ‘ πππππ’ππππ π πππ β π€πππβπ‘) πΏ = 65 ππ π
π΄π = 4.75 ππ‘ β 16ππ‘ = 76 ππ‘2
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 1
Therefore, π΄ππΎπΏπΏ = 76 ππ‘2 β 1 < 400 ππ‘2 β π πππ’ππ‘πππ πππ‘ ππππππ‘π‘ππ
DESIGN:
Factoring the loads using ASD and finding the superimposed load:
[LOAD COMBINATIONS FOUND ON NEX PAGE]
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COMPOSITE FLOOR DECK DESIGN Pg 38
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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 that 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 4.75 ft and by extrapolation:
πΆππππππ‘π¦ = 302 β (302 β 0)7 β 4.75
7 β 0= 204.9 ππ π
πΆππππππ‘π¦: 74.50 ππ π < 204.9 ππ π β ππΎ
πππ₯ ππ·πΌ ππππ: 5 ππ‘ < 13 ππ‘ 8 ππ β ππΎ For future reference, this deckβs 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 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:
π· = 9.5 ππ π (πππ‘ πππππ’ππππ π πππ β π€πππβπ‘) πΏ = 65 ππ π
π΄π = 8.85 ππ‘ β 25ππ‘ = 221.25 ππ‘2
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COMPOSITE FLOOR DECK DESIGN Pg 39
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 1
Therefore, π΄ππΎπΏπΏ = 221.15 ππ‘2 β 1 < 400 ππ‘2 β π πππ’ππ‘πππ πππ‘ ππππππ‘π‘ππ
DESIGN:
Factoring the loads using ASD and finding the superimposed load:
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 that 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 8.85 ft and by interpolation:
πΆππππππ‘π¦ = 222 β (222 β 203)8.85 β 8.5
9 β 8.5= 208.7 ππ π
πΆππππππ‘π¦: 74.50 ππ π < 208.7 ππ π β ππΎ
πππ₯ ππ·πΌ ππππ: 5 ππ‘ < 11 ππ‘ 4 ππ β ππΎ For future reference, this deckβs 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.
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FLOOR BEAM DESIGN Pg 40
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
GIVEN/KNOWN VALUES:
π΄π =(8.85 ππ‘)
2β 25 ππ‘ = 110.625 ππ‘2
πΏ = 65 ππ π β8.85 ππ‘
2= .287625 π/ππ‘
π· = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ + 8 ππ π ππππ π€πππ )]8.85 ππ‘
2= .314175 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 2
Therefore, π΄ππΎπΏπΏ = 110.625 ππ‘2 β 2 = 221.25 < 400 ππ‘2 β π πππ’ππ‘πππ πππ‘ ππππππ‘π‘ππ
DESIGN OF BEAM:
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
Live load:
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FLOOR BEAM DESIGN Pg 41
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ASD load combination π· + πΏ gives the maximum positive moment:
The top flange is fully braced, so use the plastic moment to find the required section modulus:
ππππ =ππ’Ξ©
πΉπ¦=(54.828 π β ππ‘)(1.67) β 12ππ
50 ππ π= 21.98 ππ3
From the AISC manual, table 3-2, try a W12X19.
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ24.7ππ3
1.67= 61.63 π β ππ‘ > 54.9 π β ππ‘ β ππΎ
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 beamβs Zone I capacity meets the demand:
ππΞ©==
πΉπ¦ππ₯
Ξ©=50 ππ π β 24.7ππ3
1.67= 61.63 π β ππ‘ > 48.5 π β ππ‘ β ππΎ
Running visual analysis to find the maximum shear values gives:
![Page 43: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/43.jpg)
FLOOR BEAM DESIGN Pg 42
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
These values meet the beamβs shear capacity found in the AISC manual pg. 3-26 of 57.3k: ππΞ©= 57.3π > 7.76π β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
These deflections do not meet the requirements:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ < 1.4472 ππ β πππ‘ ππππ
Retry with a W14X26.
Finding the max positive moment using its actual self-weight of 26 lb/ft and the controlling ASD
load combinations mentioned previously in visual analysis gives:
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ40.2ππ3
1.67= 100.3 π β ππ‘ > 49.06 π β ππ‘ β ππΎ
Running visual analysis to find the maximum shear values gives:
![Page 44: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/44.jpg)
FLOOR BEAM DESIGN Pg 43
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
These values meet the beamβs shear capacity found in the AISC manual pg. 3-26 of 70.9k: ππΞ©= 70.9π > 7.85π β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
These deflections meet the requirements:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > 0.777 ππ β ππΎ
For πΏ:
These deflections meet the requirements:
πΏ
360=25 ππ‘ β 12 ππ
360= .883 ππ > 0.356 ππ β ππΎ
THEREFORE USE A W14X26 FOR THE OUTSIDE STAIRWELL BEAM.
FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 7.85K.
![Page 45: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/45.jpg)
FLOOR BEAM DESIGN Pg 44
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
FOR THE INSIDE STAIRWELL 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:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
GIVEN/KNOWN VALUES:
π΄π =(8.85 + 12.5ππ‘)
2β 25 ππ‘ = 266.875 ππ‘2
πΏ = 65 ππ π β(8.85 ππ‘ + 12.5 ππ‘)
2= .694 π/ππ‘
π· = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](8.85 ππ‘ + 12.5ππ‘)
2= .673 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 2
Therefore, π΄ππΎπΏπΏ = 266.875 ππ‘2 β 2 = 533.75 > 400 ππ‘2 β π πππ’ππ‘πππ ππππππ‘π‘ππ
πΏ = πΏ0 (0.25 +15
βπΎπΏπΏπ΄π)
πΏ = .694 (0.25 +15
β2 β 266.875) = .624
π
ππ‘
. 624π
ππ‘> .5 β .694 = .347 β ππΎ
DESIGN OF BEAM:
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
Live load:
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FLOOR BEAM DESIGN Pg 45
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ASD load combination π· + πΏ gives the maximum positive moment:
The top flange is fully braced, so use the plastic moment to find the required section modulus:
ππππ =ππ’Ξ©
πΉπ¦=(109.14 π β ππ‘)(1.67) β 12ππ
50 ππ π= 43.74 ππ3
From the AISC manual, table 3-2, try a W14X30.
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ47.3ππ3
1.67= 118.01 π β ππ‘ > 109.14 π β ππ‘ β
ππΎ Test the beam using its self-weight of 30lb/ft instead of the assumed weight as before. The max
moment found in visual analysis is:
The beamβs Zone I capacity meets the demand:
ππΞ©==
πΉπ¦ππ₯
Ξ©=50 ππ π β 47.3ππ3
1.67= 118.01 π β ππ‘ > 103.68 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
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FLOOR BEAM DESIGN Pg 46
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
These deflections do not meet the requirements:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ < 1.3822 ππ β πππ‘ ππππ
Retry with a W16X31.
Finding the max positive moment using its actual self-weight of 31 lb/ft and the controlling ASD
load combinations mentioned previously in visual analysis gives:
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ54 ππ3
1.67= 134.7 π β ππ‘ > 103.8 π β ππ‘ β ππΎ
Running visual analysis to find the maximum shear values gives:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-26 of 87.5k: ππΞ©= 87.5π > 16.6π β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
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FLOOR BEAM DESIGN Pg 47
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
These deflections meet the requirements:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > 1.07 ππ β ππΎ
For πΏ:
These deflections meet the requirements:
πΏ
360=25 ππ‘ β 12 ππ
360= .883 ππ > 0.504 ππ β ππΎ
THEREFORE USE A W16X31 FOR THE INSIDE STAIRWELL BEAM.
FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 16.6K.
FOR THE INSIDE MOST ELEVATOR 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 16 feet.
Use the following maximum deflections:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
GIVEN/KNOWN VALUES:
π΄π =(4.75 + 4.75ππ‘)
2β 16 ππ‘ = 76 ππ‘2
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FLOOR BEAM DESIGN Pg 48
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πΏ = 65 ππ π β(4.75 ππ‘ + 4.75 ππ‘)
2= .30875 π/ππ‘
π· = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](4.75 ππ‘ + 4.75 ππ‘)
2= .29925 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 2
Therefore, π΄ππΎπΏπΏ = 76 ππ‘2 β 2 = 152 ππ‘2 < 400 ππ‘2 β π πππ’ππ‘πππ πππ‘ ππππππ‘π‘ππ
DESIGN OF BEAM:
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
Live load:
ASD load combination π· + πΏ gives the maximum positive moment:
The top flange is fully braced, so use the plastic moment to find the required section modulus:
ππππ =ππ’Ξ©
πΉπ¦=(22.656 π β ππ‘)(1.67) β 12ππ
50 ππ π= 8.92 ππ3
From the AISC manual, table 3-2, try a W12X16.
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ20.1ππ3
1.67= 50.15 π β ππ‘ > 22.66 π β ππ‘ β ππΎ
Test the beam using its self-weight of 16 lb/ft instead of the assumed weight as before. The max
moment found in visual analysis is:
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FLOOR BEAM DESIGN Pg 49
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
The beamβs Zone I capacity meets the demand:
ππΞ©==
πΉπ¦ππ₯
Ξ©=50 ππ π β 20.1ππ3
1.67= 50.15 π β ππ‘ > 19.97 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
This deflections meets the requirement:
πΏ
240=16 ππ‘ β 12 ππ
240= .8 ππ > .308 ππ β ππΎ
For L:
This deflections meets the requirement:
πΏ
360=16 ππ‘ β 12 ππ
360= .533 ππ > .152 ππ β ππΎ
Running visual analysis to find the maximum shear values gives:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-2 of 52.8k:
ππΞ©= 52.8π > 4.99π β ππΎ
THEREFORE USE A W12X16 FOR THE MOST INSIDE ELEVATOR BEAM.
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FLOOR BEAM DESIGN Pg 50
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 4.99K.
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.
Use the following maximum deflections:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
KNOWN/GIVEN VALUES:
πΏ = 65 ππ π β 8 ππ‘ = .52 π/ππ‘
π΄π =(16 ππ‘)
2β 9.5 ππ‘ = 76 ππ‘2
π· = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](8 ππ‘) = .504 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
ππ· = ((9.5 ππ π + 53.5 ππ π) β 8 ππ‘ β 4.75 ππ‘) + 16ππ
ππ‘β 8ππ‘ = 2.52 πππ
ππΏ = 65 ππ π β 8 ππ‘ β 4.75 ππ‘ = 2.47 πππ
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 2
Therefore, π΄ππΎπΏπΏ = 76ππ‘2 β 2 = 152 ππ‘2 < 400 ππ‘2 β π πππ’ππ‘πππ πππ‘ ππππππ‘π‘ππ
DESIGN OF BEAM:
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100 lb/ft.:
Live load:
![Page 52: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/52.jpg)
FLOOR BEAM DESIGN Pg 51
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ASD load combination π· + πΏ gives the maximum positive moment:
The top flange is fully braced, so use the plastic moment to find the required section modulus:
ππππ =ππ’Ξ©
πΉπ¦=(24.5 π β ππ‘)(1.67) β 12ππ
50 ππ π= 9.82 ππ3
From the AISC manual, table 3-2, try a W18X55. Using the actual self-weight gives:
The beamβs Zone I capacity of ππ
Ξ©=πΉπ¦ππ₯
Ξ©=50 ππ πβ112 ππ3
1.67= 279.4 π β ππ‘ > 24 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
![Page 53: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/53.jpg)
FLOOR BEAM DESIGN Pg 52
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
This deflection meets the requirements:
πΏ
240=9.5 ππ‘ β 12 ππ
240= .475 ππ > .01 ππ β ππΎ
For πΏ:
This deflection meets the requirement: πΏ
360=9.5 ππ‘ β 12 ππ
360= .32 ππ > .007 ππ β πΊπππ
Running visual analysis to find the maximum shear values gives:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-25 of 141k: ππΞ©= 141π > 7.62π β ππΎ
![Page 54: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/54.jpg)
FLOOR BEAM DESIGN Pg 53
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
THEREFORE USE A W18X55 FOR THE SMALL ELEVATOR GIRDER. FOR THE
CONNECTION FORCE, USE GRAVITY SHEAR AT 7.62K.
FOR THE MIDDLE ELEVATOR BEAMS:
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:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
GIVEN/KNOWN VALUES:
π΄π =(4.75 + 7.75ππ‘)
2β 25 ππ‘ = 156.25 ππ‘2
πΏ = 65 ππ π β(4.75 ππ‘ + 7.75 ππ‘)
2= .41 π/ππ‘
π· = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](4.75 ππ‘ + 7.75 ππ‘)
2= .40 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
ππ· = 3.92 π ππππ ππππππ πππππ‘πππ
ππΏ = 3.71 π ππππ ππππππ πππππ‘πππ
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 2
Therefore, π΄ππΎπΏπΏ = 156.25 ππ‘2 β 2 = 312.5 ππ‘2 < 400 ππ‘2 β π πππ’ππ‘πππ πππ‘ ππππππ‘π‘ππ
DESIGN OF BEAM:
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
Live load:
![Page 55: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/55.jpg)
FLOOR BEAM DESIGN Pg 54
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ASD load combination π· + πΏ gives the maximum positive moment:
The top flange is fully braced, so use the plastic moment to find the required section modulus:
ππππ =ππ’Ξ©
πΉπ¦=(109.6 π β ππ‘)(1.67) β 12ππ
50 ππ π= 43.9 ππ3
From the AISC manual, table 3-2, try a W16X31.
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ54ππ3
1.67= 134.7 π β ππ‘ > 109.6 π β ππ‘ β ππΎ
Test the beam using its self-weight of 31 lb/ft instead of the assumed weight as before. The max
moment found in visual analysis is:
The beamβs Zone I capacity meets the demand:
ππΞ©=πΉπ¦ππ₯
Ξ©=50 ππ π β 54ππ3
1.67= 134.7 π β ππ‘ > 104.5 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
![Page 56: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/56.jpg)
FLOOR BEAM DESIGN Pg 55
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
For π· + πΏ:
This deflection meets the requirement:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > 1.03 ππ β ππΎ
For L:
This deflection meets the requirement:
πΏ
360=25 ππ‘ β 12 ππ
360= .833 ππ > .503 ππ β ππΎ
Running visual analysis to find the maximum shear values gives:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-26 of 87.5k:
ππΞ©= 87.5π > 15.4π β ππΎ
THEREFORE USE A W16X31 FOR THE MIDDLE ELEVATOR BEAMS.
![Page 57: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/57.jpg)
FLOOR BEAM DESIGN Pg 56
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 15.4K.
FOR THE OUTSIDE ELEVATOR BEAMS:
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:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
GIVEN/KNOWN VALUES:
π΄π =(7.75 + 12.5ππ‘)
2β 25 ππ‘ = 253.125 ππ‘2
πΏ = 65 ππ π β(7.75 ππ‘ + 12.5 ππ‘)
2= .658125 π/ππ‘
π· = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](7.75 ππ‘ + 12.5 ππ‘)
2= .637875 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 2
Therefore, π΄ππΎπΏπΏ = 253.125 ππ‘2 β 2 = 506.25 ππ‘2 > 400 ππ‘2 β π πππ’ππ‘πππ ππππππ‘π‘ππ
πΏ = πΏ0 (0.25 +15
βπΎπΏπΏπ΄π)
πΏ = .658125 (0.25 +15
β2 β 253.125) = .6034
π
ππ‘
. 6034π
ππ‘> .5 β .658125 = .329 β ππΎ
DESIGN OF BEAM:
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
![Page 58: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/58.jpg)
FLOOR BEAM DESIGN Pg 57
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Live load:
ASD load combination π· + πΏ gives the maximum positive moment:
The top flange is fully braced, so use the plastic moment to find the required section modulus:
ππππ =ππ’Ξ©
πΉπ¦=(104.79 π β ππ‘)(1.67) β 12ππ
50 ππ π= 42 ππ3
From the AISC manual, table 3-2, try a W14X30.
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ47.3ππ3
1.67= 118 π β ππ‘ > 104.79 π β ππ‘ β ππΎ
Test the beam using its self-weight of 30 lb/ft instead of the assumed weight as before. The max
moment found in visual analysis is:
The beamβs Zone I capacity meets the demand:
ππΞ©==
πΉπ¦ππ₯
Ξ©=50 ππ π β 47.3ππ3
1.67= 118 π β ππ‘ > 99.3 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
This deflection does not meet the requirement:
![Page 59: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/59.jpg)
FLOOR BEAM DESIGN Pg 58
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ < 1.3242 ππ β πππ‘ ππππ
Retry with a W16X31. Visual Analysis gives the following maximum moment considering a
self-weight of 31 lb/ft:
The beamβs Zone I capacity meets the demand:
ππΞ©==
πΉπ¦ππ₯
Ξ©=50 ππ π β 54ππ3
1.67= 134.7 π β ππ‘ > 99.41 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For D+L:
This deflection meets the requirement:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > 1.03 ππ β ππΎ
For L:
This deflection meets the requirement:
πΏ
360=25 ππ‘ β 12 ππ
360= .833 ππ > .488 ππ β ππΎ
Running visual analysis to find the maximum shear values gives:
![Page 60: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/60.jpg)
FLOOR BEAM DESIGN Pg 59
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
These values meet the beamβs shear capacity found in the AISC manual pg. 3-26 of 87.5k:
ππΞ©= 87.5π > 15.91π β ππΎ
THEREFORE USE A W16X31 FOR THE OUTSIDE ELEVATOR BEAMS.
FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 15.91K.
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.
Use the following maximum deflections:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
KNOWN/GIVEN VALUES:
π΄π = (12.5ππ‘)(25 ππ‘) + (12.5ππ‘)(7.75ππ‘) = 409.3ππ‘2
πΏ1 = 65 ππ π(25 ππ‘) = 1.63 π/ππ‘ πΏ2 = 65 ππ π(12.5 ππ‘) = .81 π/ππ‘
π·1 = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](25 ππ‘)= 1.58 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
π·2 = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](12.5 ππ‘)= .79 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
ππ· = 7.9 π ππππ ππππ£πππ’π πππ πππ
ππΏ = 7.5 π ππππ ππππ£πππ’π πππ πππ
ππ·2 = (9.5 + 53.5)(10.125 β 25) + 35(12.5) + 31(12.5) = 16.8 π
ππΏ2 = 65(10.125 β 25) = 16.5 π
LIVE LOAD REDUCTIONS:
Do not consider live load reductions because of the change in areas/loading.
DESIGN OF BEAM:
![Page 61: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/61.jpg)
FLOOR BEAM DESIGN Pg 60
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
Live load:
ASD load combination π· + πΏ gives the maximum positive moment:
From the AISC manual, table 3-2, try a W18X97.
Checking the maximum moment using the actual self-weight of 97 lb/ft:
![Page 62: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/62.jpg)
FLOOR BEAM DESIGN Pg 61
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
The beamβs Zone I capacity of ππ
Ξ©=πΉπ¦ππ₯
Ξ©=50 ππ πβ211 ππ3
1.67= 526 π β ππ‘ > 494 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
This deflection meets the requirements:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > 1.01 ππ β ππΎ
For πΏ:
![Page 63: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/63.jpg)
FLOOR BEAM DESIGN Pg 62
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
This deflection meets the requirement: πΏ
360=25 ππ‘ β 12 ππ
360= .833 ππ > .50 ππ β πΊπππ
Running visual analysis to find the maximum shear values gives:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-24 of 199k: ππΞ©= 199π > 63.5π β ππΎ
THEREFORE USE A W18X97 FOR THE GIRDERS NORTH OF THE ELEVATOR.
FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 63.5K.
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.
Use the following maximum deflections:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
KNOWN/GIVEN VALUES:
π΄π = (25 ππ‘)(25 ππ‘) = 625 ππ‘2
πΏ = 65 ππ π(25 ππ‘) = 1.63 π/ππ‘ π· = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](25 ππ‘) = 1.58 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘) ππ· = 6.8 π ππππ ππππ£πππ’π πππ πππ
ππΏ = 6.5 π ππππ ππππ£πππ’π πππ πππ
ππ·2 = (9.5 + 53.5)(10.125 β 25) + 35(12.5) + 31(12.5) = 16.8 π
ππΏ2 = 65(10.125 β 25) = 16.5 π
LIVE LOAD REDUCTIONS:
![Page 64: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/64.jpg)
FLOOR BEAM DESIGN Pg 63
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Do not consider live load reductions because of the point loads β they may complicate the
design.
DESIGN OF BEAM:
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
Live load:
ASD load combination π· + πΏ gives the maximum positive moment:
From the AISC manual, table 3-2, try a W18X97.
Checking the maximum moment using the actual self-weight of 97 lb/ft:
![Page 65: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/65.jpg)
FLOOR BEAM DESIGN Pg 64
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
The beamβs Zone I capacity of ππ
Ξ©=πΉπ¦ππ₯
Ξ©=50 ππ πβ211 ππ3
1.67= 526 π β ππ‘ > 499 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
This deflection meets the requirements:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > 1.02 ππ β ππΎ
For πΏ:
This deflection meets the requirement: πΏ
360=25 ππ‘ β 12 ππ
360= .833 ππ > .50 ππ β πΊπππ
![Page 66: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/66.jpg)
FLOOR BEAM DESIGN Pg 65
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Running visual analysis to find the maximum shear values gives:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-24 of 199k: ππΞ©= 199π > 68.8π β ππΎ
THEREFORE USE A W18X97 FOR THE GIRDERS SOUTH OF THE ELEVATOR.
FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 68.8K.
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.
Use the following maximum deflections:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
GIVEN/KNOWN VALUES:
π΄π = 25ππ‘ β 25ππ‘ = 625ππ‘2
πΏ = 65 ππ π β 25ππ‘2 = 1.625 π/ππ‘ π· = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](25 ππ‘)
= 1.575 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
ππ· = (12.5 ππ‘ β (9.5 ππ π + 53.5 ππ π) + 35ππ
ππ‘) 25 ππ‘ = 20.6 π
ππΏ = (12.5 ππ‘)(25 ππ‘)(65 ππ π) = 20.3 π
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 2
Therefore, π΄ππΎπΏπΏ = 625 ππ‘2 β 2 = 1250 ππ‘2 > 400 ππ‘2 β π πππ’ππ‘πππ ππππππ‘π‘ππ
πΏ = πΏ0 (0.25 +15
βπΎπΏπΏπ΄π)
![Page 67: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/67.jpg)
FLOOR BEAM DESIGN Pg 66
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πΏ = 1.625 (0.25 +15
β2 β 625) = 1.1
π
ππ‘
1.1 > .5 β 1.625 = .8125 β ππΎ
Visual Analysis was used to design the 25 ft. beams. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
Live load:
ASD load combination π· + πΏ gives the maximum positive moment:
The top flange is fully braced, so use the plastic moment to find the required section modulus:
ππππ =ππ’Ξ©
πΉπ¦=(472.42 π β ππ‘)(1.67) β 12ππ
50 ππ π= 189 ππ3
From the AISC manual, table 3-2, try a W18X97.
Finding the max positive moment using its actual self-weight of 97 lb/ft and the controlling ASD
load combinations mentioned previously in visual analysis gives:
![Page 68: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/68.jpg)
FLOOR BEAM DESIGN Pg 67
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ211 ππ3
1.67= 526 π β ππ‘ > 472 π β ππ‘ β ππΎ
Running visual analysis to find the maximum shear values gives:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-24 of 199k: ππΞ©= 199π > 55.1π β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
![Page 69: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/69.jpg)
FLOOR BEAM DESIGN Pg 68
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
These deflections meet the requirements:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > .93 ππ β ππΎ
Finding the deflections from Visual Analysis:
For πΏ:
These deflections meet the requirements: πΏ
360=25 ππ‘ β 12 ππ
360= .883 ππ > 0.42 ππ β ππΎ
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.
Use the following maximum deflections:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
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FLOOR BEAM DESIGN Pg 69
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
GIVEN/KNOWN VALUES:
π΄π = 25ππ‘ β 12.5ππ‘ = 312.5 ππ‘2
πΏ = 65 ππ π β 12.5ππ‘2 = .8125 π/ππ‘ π· = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](12.5 ππ‘)
= .7875 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 2
Therefore, π΄ππΎπΏπΏ = 312.5 ππ‘2 β 2 = 625 ππ‘2 > 400 ππ‘2 β π πππ’ππ‘πππ ππππππ‘π‘ππ
πΏ = πΏ0 (0.25 +15
βπΎπΏπΏπ΄π)
πΏ = .8125 (0.25 +15
β2 β 312.5) = .6907
π
ππ‘
. 6907 > .5 β .8125 = .406 β ππΎ
Visual Analysis was used to design the 25 ft. beams. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
Live load:
ASD load combination π· + πΏ gives the maximum positive moment:
The top flange is fully braced, so use the plastic moment to find the required section modulus:
![Page 71: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/71.jpg)
FLOOR BEAM DESIGN Pg 70
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππππ =ππ’Ξ©
πΉπ¦=(123.3 π β ππ‘)(1.67) β 12ππ
50 ππ π= 49.42 ππ3
From the AISC manual, table 3-2, try a W18X35.
Finding the max positive moment using its actual self-weight of 35 lb/ft and the controlling ASD
load combinations mentioned previously in visual analysis gives:
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ66.5ππ3
1.67= 166π β ππ‘ > 118.3 π β ππ‘ β ππΎ
Running visual analysis to find the maximum shear values gives:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-25 of 81.1k: ππΞ©= 81.1π > 18.92π β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
These deflections meet the requirements:
![Page 72: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/72.jpg)
FLOOR BEAM DESIGN Pg 71
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > .8995 ππ β ππΎ
Finding the deflections from Visual Analysis:
For πΏ:
These deflections meet the requirements: πΏ
360=25 ππ‘ β 12 ππ
360= .883 ππ > 0.4106 ππ β ππΎ
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.
Use the following maximum deflections:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
KNOWN/GIVEN VALUES:
π΄1 = (25 ππ‘ β 12.5 ππ‘) = 312.5 ππ‘2
π΄2 = (12.5 ππ‘ β 8.85 ππ‘) = 110.625 ππ‘2
πΏ1 = 65 ππ π(12.5 ππ‘) = .8125 π/ππ‘ πΏ2 = .8125 β 2 = 1.625 π/ππ‘ π·1 = (9.5 ππ π + 53.5 ππ π)(12.5 ππ‘) = .7875 π/ππ‘ π·2 = 2(. 7875) = 1.575 π/ππ‘
LIVE LOAD REDUCTIONS:
![Page 73: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/73.jpg)
FLOOR BEAM DESIGN Pg 72
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
None of the tributary areas are greater than 400 sq. ft., therefore reductions are not permitted.
DESIGN OF BEAM:
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
Live load:
ASD load combination π· + πΏ gives the maximum positive moment:
From the AISC manual, table 3-2, try a W18X55.
Checking the maximum moment using the actual self-weight of 55 lb/ft:
![Page 74: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/74.jpg)
FLOOR BEAM DESIGN Pg 73
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ112 ππ3
1.67= 279.4 π β ππ‘ > 161.3 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
This deflection meets the requirements:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > .70 ππ β ππΎ
For πΏ:
![Page 75: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/75.jpg)
FLOOR BEAM DESIGN Pg 74
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
This deflection meets the requirement: πΏ
360=25 ππ‘ β 12 ππ
360= .833 ππ > .28 ππ β πΊπππ
Running visual analysis to find the maximum shear values gives:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-25 of 141k: ππΞ©= 141π > 28.07π β ππΎ
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.
Use the following maximum deflections:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
GIVEN/KNOWN VALUES:
π΄π =(12.5ππ‘)
2β 25 ππ‘ = 156.25 ππ‘2
πΏ = 65 ππ π β(12.5 ππ‘)
2= .40625 π/ππ‘
π· = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](12.5 ππ‘)
2+ (23 ππ π ππππ π€πππ)(15 ππ‘)
= .73875 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
![Page 76: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/76.jpg)
FLOOR BEAM DESIGN Pg 75
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 2
Therefore, π΄ππΎπΏπΏ = 156.25 ππ‘2 β 2 = 312.5 ππ‘2 < 400 ππ‘2 β π πππ’ππ‘πππ πππ‘ ππππππ‘π‘ππ
DESIGN OF BEAM:
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
Live load:
ASD load combination π· + πΏ gives the maximum positive moment:
The top flange is fully braced, so use the plastic moment to find the required section modulus:
ππππ =ππ’Ξ©
πΉπ¦=(97.266 π β ππ‘)(1.67) β 12ππ
50 ππ π= 38.98 ππ3
From the AISC manual, table 3-2, try a W14X26.
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ40.2ππ3
1.67= 100.3 π β ππ‘ > 97.27 π β ππ‘ β ππΎ
Test the beam using its self-weight of 26 lb/ft instead of the assumed weight as before. The max
moment found in visual analysis is:
![Page 77: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/77.jpg)
FLOOR BEAM DESIGN Pg 76
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
The beamβs Zone I capacity meets the demand:
ππΞ©==
πΉπ¦ππ₯
Ξ©=50 ππ π β 40.2ππ3
1.67= 100.3 π β ππ‘ > 91.5 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
This deflection does not meet the requirement:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ < 1.45 ππ β πππ‘ ππππ
Retry with a W14X30. Visual Analysis gives the following maximum moment considering a
self-weight of 30 lb/ft:
The beamβs Zone I capacity meets the demand:
ππΞ©==
πΉπ¦ππ₯
Ξ©=50 ππ π β 47.3ππ3
1.67= 118 π β ππ‘ > 91.8 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For D+L:
This deflection meets the requirement:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > 1.23 ππ β ππΎ
For L:
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FLOOR BEAM DESIGN Pg 77
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
This deflection meets the requirement:
πΏ
360=25 ππ‘ β 12 ππ
360= .833 ππ > .423 ππ β ππΎ
Running visual analysis to find the maximum shear values gives:
These values meet the beamβs shear capacity found in the AISC manual pg. 3-26 of 74.5k:
ππΞ©= 74.5π > 14.7π β ππΎ
THEREFORE USE A W14X30 FOR THE NORTH TO SOUTH PERIMETER BEAMS.
FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 14.7K.
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.
Use the following maximum deflections:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
GIVEN/KNOWN VALUES:
π΄π = 12.5 ππ‘ β 25 ππ‘ = 300 ππ‘2
πΏ = 65 ππ π β 12.5 ππ‘ = .813 π/ππ‘ π· = [(9.5 ππ π + 53.5 ππ π ππππ πππππ ππππ)](12.5 ππ‘)
= .788 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
LIVE LOAD REDUCTIONS:
From ASCE 7-10 Table 4-2, πΎπΏπΏ = 2
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FLOOR BEAM DESIGN Pg 78
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Therefore, π΄ππΎπΏπΏ = 156.25 ππ‘2 β 2 = 300 ππ‘2 > 600 ππ‘2 β π πππ’ππ‘πππ ππππππ‘π‘ππ
πΏ = πΏ0 (0.25 +15
βπΎπΏπΏπ΄π)
πΏ = .813 (0.25 +15
β2 β 300) = .7
π
ππ‘
. 7 > .5 β .813 = .41 β ππΎ
DESIGN OF BEAM:
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
Live load:
ASD load combination π· + πΏ gives the maximum positive moment:
![Page 80: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/80.jpg)
FLOOR BEAM DESIGN Pg 79
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
From the AISC manual, table 3-10, try a W21X55.
Test the beam using its self-weight of 55 lb/ft instead of the assumed weight as before. The max
moment found in visual analysis is:
The beamβs Zone I capacity meets the demand:
ππΞ©==
πΉπ¦ππ₯
Ξ©=50 ππ π β 112ππ3
1.67= 279 π β ππ‘ > 120 π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For π· + πΏ:
This deflection meets the requirement:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > .53 ππ β ππΎ
For L only:
This deflection does meet the requirement:
πΏ
240=25 ππ‘ β 12 ππ
240= .833 ππ > .24 ππ β πΊπππ
Running visual analysis to find the maximum shear values gives:
![Page 81: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/81.jpg)
FLOOR BEAM DESIGN Pg 80
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
These values meet the beamβs shear capacity found in the AISC manual pg. 3-25 of 141k: ππΞ©= 141π > 19.2π β ππΎ
THEREFORE USE A W18X55 FOR THE NORTH TO SOUTH PERIMETER BEAMS.
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.
Use the following maximum deflections:
π·πππ + πΏππ£π +ππππ =πΏ
240
πΏππ£π +ππππ ππππ¦ =πΏ
360ππ 1.5"
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
GIVEN/KNOWN VALUES:
π΄π =(16.15ππ‘)
2β 25 ππ‘ = 201.872 ππ‘2
π· = (23 ππ π ππππ π€πππ)(15 ππ‘) = .345 π/ππ‘ (ππ₯πππ’ππππ π πππ π€πππβπ‘)
DESIGN OF BEAM:
Visual Analysis was used to analyze the beam. The following load cases were used:
Dead load with an assumed self-weight of 100lb/ft.:
![Page 82: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/82.jpg)
FLOOR BEAM DESIGN Pg 81
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ASD load combination π· gives the maximum positive moment:
The top flange is fully braced, so use the plastic moment to find the required section modulus:
ππππ =ππ’Ξ©
πΉπ¦=(34.77 π β ππ‘)(1.67) β 12ππ
50 ππ π= 13.9ππ3
From the AISC manual, table 3-6, try a W12X16.
The beamβs Zone I capacity of ππ
Ξ©==
πΉπ¦ππ₯
Ξ©=50 ππ πβ20.1ππ3
1.67= 50.15π β ππ‘ > 34.77π β ππ‘ β ππΎ
Test the beam using its self-weight of 16lb/ft instead of the assumed weight as before. The max
moment found in visual analysis is:
The beamβs Zone I capacity meets the demand:
ππΞ©=πΉπ¦ππ₯
Ξ©=50 ππ π β 20.1ππ3
1.67= 50.15 π β ππ‘ > 28.21π β ππ‘ β ππΎ
Finding the deflections from Visual Analysis:
For π·:
This deflection does not meet the requirement:
πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ > 1.062 ππ β πΊπππ
Running visual analysis to find the maximum shear values gives:
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FLOOR BEAM DESIGN Pg 82
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
These values meet the beamβs shear capacity found in the AISC manual pg. 3-26 of 52.8k:
ππΞ©= 52.8π > 4.513π β ππΎ
THEREFORE USE A W14X30 FOR THE NORTH TO SOUTH PERIMETER BEAMS.
FOR THE CONNECTION FORCE, USE GRAVITY SHEAR AT 4.513K.
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STEEL COLUMN DESIGNS Pg 83
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
STEEL COLUMN DESIGN:
FINDING THE LOADS FOR EACH OF THE COLUMNS FROM THE ROOF:
For the columns on the corners of the building:
πΏπ = 20 ππ π(12.5 )(12.5) = 3.1 π
π· = (30 + 1.78 ππ π)(12.5 ππ‘)(12.5 ππ‘) + 40ππ
ππ‘(12.5 ππ‘) + 87
ππ
ππ‘(12.5 ππ‘)
+ 16.4ππ
ππ‘(12.5 ππ‘)(2) = 7.0 π
π = {β103.9 ππ π(12.5 ππ‘)(12.5 ππ‘) = β16.2 π
16 ππ π(12.5 ππ‘)(12.5 ππ‘) = 2.5 π
For the columns in the east to west direction on the perimeter:
πΏπ = 20 ππ π(25 ππ‘)(25 ππ‘) = 12.5 π
π· = (30 + 1.78 ππ π)(25 ππ‘)(25 ππ‘) + 87ππ
ππ‘(25 ππ‘) + 16.4
ππ
ππ‘(25 ππ‘)(5) = 24.1 π
π = {β103.9(25 ππ‘)(25 ππ‘) = β64.9 π
16(25 ππ‘)(25 ππ‘) = 10 π
For the columns supporting the joist girder:
πΏπ = 20 ππ π(50 ππ‘)(50 ππ‘) = 50 π
π· = (30 + 1.78 ππ π)(50 ππ‘)(50 ππ‘) + 16.4ππ
ππ‘(50 ππ‘)(11) + 78
ππ
ππ‘(50 ππ‘) = 92.4 π
π = {β103.9 ππ π(50 ππ‘)(50 ππ‘) = β259.8 π
16 ππ π(50 ππ‘)(50 ππ‘) = 40 π
For the columns in the north to south direction on the perimeter:
πΏπ = 20 ππ π(25 ππ‘)(25 ππ‘) = 12.5 π
π· = (30 + 1.78 ππ π)(25 ππ‘)(25 ππ‘) + 16.4ππ
ππ‘(25 ππ‘)(5) + 78
ππ
ππ‘(25 ππ‘) = 23.9 π
π = {β103.9 ππ π(25 ππ‘)(25 ππ‘) = β64.9 π
16 ππ π(25 ππ‘)(25 ππ‘) = 10 π
FINDING THE LOADS FOR EACH OF THE COLUMNS FROM THE FLOOR:
For columns E3 and A9:
πΏ = 65 ππ π(12.5 ππ‘)(12.5 ππ‘) = 10.2 π
π· = (9.5 + 53.5 ππ π)(12.5 ππ‘)(12.5 ππ‘) + 30ππ
ππ‘(12.5 ππ‘) + 55
ππ
ππ‘ (12.5 ππ‘)
+ 35ππ
ππ‘(12.5 ππ‘) + 23 ππ π(15 ππ‘)(25 ππ‘) = 20π
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STEEL COLUMN DESIGNS Pg 84
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
For the columns E4 to E7 and A5 to A8:
πΏ = 65 ππ π(25 ππ‘)(12.5 ππ‘) = 20.3 π
π· = (9.5 + 53.5 ππ π)(12.5 ππ‘)(25 ππ‘) + 55ππ
ππ‘(25 ππ‘) + 35
ππ
ππ‘(12.5 ππ‘)(3)
+ 23 ππ π(15 ππ‘)(25 ππ‘) = 31 π
For the columns E9 and A3:
π· =30ππ
ππ‘(12.5 ππ‘) + 55
ππ
ππ‘(12.5 ππ‘) + 23 ππ π(15 ππ‘)(25 ππ‘) = 9.69 π
For the columns A4 and E8:
πΏ = 65 ππ π(12.5 ππ‘)(12.5 + 8.85 ππ‘) = 17.35 π
π· = (9.5 + 53.5 ππ π)(12.5 + 8.85 ππ‘) (12.5 ππ‘) +31ππ
ππ‘(12.5 ππ‘) +
26ππ
ππ‘(12.5 ππ‘)
+55ππ
ππ‘(25 ππ‘) +
35ππ
ππ‘(12.5 ππ‘) + 23 ππ π(15 ππ‘)(25 ππ‘) = 30 π
For the columns D4 to D7, C4, C5, C7, C8, B5, B7, and B8:
πΏ = 65 ππ π(25 ππ‘)(25 ππ‘) = 40.6 π
π· = (9.5 + 53.5 ππ π)(25 ππ‘)(25 ππ‘) + 35ππ
ππ‘(3)(25 ππ‘) + 97
ππ
ππ‘(25 ππ‘) = 44.4 π
For the column B6:
πΏ = 65 ππ π(25 ππ‘)(25 ππ‘) = 40.6 π
π· = (9.5 + 53.5 ππ π)(25 ππ‘)(25 ππ‘) + 35ππ
ππ‘(3)(12.5 ππ‘) + 31
ππ
ππ‘(12.5 ππ‘)(4)
+ 16ππ
ππ‘(12.5 ππ‘) + 97
ππ
ππ‘(25 ππ‘) = 44.9π
For the column C6:
πΏ = 65 ππ π[(25 ππ‘)(25 ππ‘) β (9.5 ππ‘)(9 ππ‘)] = 35.1 π
π· = (9.5 + 53.5 ππ π)[(25 ππ‘)(25 ππ‘) β (9.5 ππ‘)(9 ππ‘)] + 35ππ
ππ‘(3)(12.5 ππ‘)
+ 31ππ
ππ‘ (12.5 ππ‘)(4) + 16
ππ
ππ‘(12.5 β 8 ππ‘) + 55
ππ
ππ‘(9.5 ππ‘) + 97
ππ
ππ‘(25 ππ‘)
= 39.9π
For the columns B3 and D9:
πΏ = 65 ππ π(12.5 ππ‘)(12.5 ππ‘) = 10.2 π
π· = (9.5 + 53.5 ππ π)(12.5 ππ‘)(12.5 ππ‘) + 30ππ
ππ‘(25 ππ‘) + 35
ππ
ππ‘ (12.5 ππ‘) + 97
ππ
ππ‘(12.5 ππ‘)
+ 23 ππ π(15 ππ‘)(25 ππ‘) = 20.9π
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STEEL COLUMN DESIGNS Pg 85
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
For the columns D3, C3, C9 and B9:
πΏ = 65 ππ π(12.5 ππ‘)(25 ππ‘) = 20.3 π
π· = (9.5 + 53.5 ππ π)(12.5 ππ‘)(25 ππ‘) + 30ππ
ππ‘(25 ππ‘) + 35
ππ
ππ‘ 25 + 97
ππ
ππ‘(12.5 ππ‘)
+ 23 ππ π(15 ππ‘)(25 ππ‘) = 31.2π
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.
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
GIVEN/KNOWN VALUES:
Use the following maximum loading for internal beams per floor:
π·πππ ππππ ππππππ = 44π
π·πππ ππππ ππππ = 92.4π
πΏππ£π ππππ ππππππ = 41π
πΏππ£π ππππ ππππ = 50π
ππππ ππππ ππππ = 40π 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.
πππ‘ππ π·πππ = π·πππ ππππ ππππ + π·πππ ππππ ππππππ β 3 = 224.4 π
πππ‘ππ πΏππ£π = πΏππ£π ππππ ππππππ β 3 = 123 π
ASD Load combinations are found on the next page.
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STEEL COLUMN DESIGNS Pg 86
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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: ππΞ©= 361π > 347π β ππΎ
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:
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 support.
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
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STEEL COLUMN DESIGNS Pg 87
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
GIVEN/KNOWN VALUES:
Use the following maximum loading for corner beams per floor:
π·πππ ππππ ππππ = 7π
π·πππ ππππ ππππππ = 9.7π
πΏππ£π ππππ ππππππ = 0π
πΏππ£π ππππ = 3.1π
ππππ = 2.5π DESIGN OF COLUMN:
For 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 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.
πππ‘ππ π·πππ = π·πππ ππππ ππππ + π·πππππππ ππππππ β 3 = 36.1π πππ‘ππ πΏππ£π = πΏππ£π β 3 = 0π
ASD load combinations are as follows:
[LOAD COMBINATIONS FOUND ON NEX PAGE]
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STEEL COLUMN DESIGNS Pg 88
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
From Table 4.1 in AISC manual, choose W14X48. This meets the factored loads:
ππΞ©= 221π > 36.1π β ππΎ
THEREFORE USE A W14X48 FOR THE CORNER COLUMNS NEAR THE
STAIRWELLS. THESE COLUMNS ARE SPECIFIED ON THE FLOOR LAYOUT AS
E9 AND A3.
*FROM THE FUTURE SECOND ORDER ANALYSIS, USE A W14X109 INSTEAD.*
PERIMETER 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 support.
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
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STEEL COLUMN DESIGNS Pg 89
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
GIVEN/KNOWN VALUES:
Use the following maximum loading:
π·πππ ππππ ππππ = 24.1π
π·πππ ππππ ππππππ = 31.2π πΏππ£π ππππ ππππππ = 20.3π
πΏππ£π ππππ = 12.5π
ππππ = 10π DESIGN OF COLUMN:
For 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 31.2k and 20.3k by 3 to have the total loading to factor.
Negative wind loading is neglected.
πππ‘ππ π·πππ = π·πππ ππππ ππππ + π·πππππππ ππππππ β 3 = 117.7π πππ‘ππ πΏππ£π = πΏππ£π β 3 = 60.9π
The ASD load combinations are found on the following page:
[LOAD COMBINATIONS FOUND ON NEX PAGE]
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STEEL COLUMN DESIGNS Pg 90
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Checking Table 4.1 in the AISC manual for 178.6k choose a W14X48 for perimeter columns
reaching the roof: ππΞ©= 221π > 178.6π β ππΎ
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.
*FROM THE FUTURE SECOND ORDER ANALYSIS, USE A W14X109 INSTEAD.*
ELEVATOR 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 support.
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
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STEEL COLUMN DESIGNS Pg 91
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
GIVEN/KNOWN VALUES:
Use the following maximum loading:
π·πππ ππππ ππππ = 0π
π·πππ ππππ ππππππ = 44.9π πΏππ£π ππππ ππππππ = 40.6π
πΏππ£π ππππ = 0π
ππππ = 0π 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.
πππ‘ππ π·πππ = π·πππ ππππ ππππ + π·πππππππ ππππππ β 3 = 134.7π πππ‘ππ πΏππ£π = πΏππ£π β 3 = 121.8 π
ASD load combinations in Excel are as follows:
[LOAD COMBINATIONS FOUND ON NEX PAGE]
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STEEL COLUMN DESIGNS Pg 92
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Checking Table 4.1 in the AISC manual for 256.6k choose a W14X61 for elevator columns: ππΞ©= 361π > 256.6π β ππΎ
THEREFORE USE A W14X61 FOR THE ELEVATOR COLUMNS. THESE COLUMNS
ARE SPECIFIED ON THE FLOOR LAYOUT AS C6 AND B6.
*FROM THE FUTURE SECOND ORDER ANALYSIS, USE A W14X109 INSTEAD.*
STAIRWELL 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 support.
Also, use AISC Manual Table 2-4 ASTM A992 for material properties:
πΉπ¦ = 50 ππ π, πΉπ’ = 65 ππ π
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STEEL COLUMN DESIGNS Pg 93
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
GIVEN/KNOWN VALUES:
Use the following maximum loading for beams per floor:
π·πππ ππππ ππππ = 24.1π
π·πππ ππππ ππππππ = 30π πΏππ£π ππππ ππππππ = 17.4π
πΏππ£π ππππ = 12.5π
ππππ = 10π 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.
πππ‘ππ π·πππ = π·πππ ππππ ππππ + π·πππππππ ππππππ β 3 = 114.1π πππ‘ππ πΏππ£π = πΏππ£π β 3 = 52.2π
ASD load combinations from Excel are shown on the next page:
[LOAD COMBINATIONS FOUND ON NEX PAGE]
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STEEL COLUMN DESIGNS Pg 94
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
ππΞ©= 221π > 166.3π β ππΎ
THEREFORE USE A W14X48 FOR THE STAIRWELL COLUMNS. THESE
COLUMNS ARE SPECIFIED ON THE FLOOR LAYOUT AS A3, A4, B3, E8, E9, AND
D9.
*FROM THE FUTURE SECOND ORDER ANALYSIS, USE A W14X109 INSTEAD.*
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TWO-WAY SLAB DESIGN Pg 95
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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: Ξ³=150pcf Initial deal load: DL= h* Ξ³=12in*150pcf=150psf
Load Combinations:
[COMBINATIONS PROVIDED ON NEXT PAGE]
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TWO-WAY SLAB DESIGN Pg 96
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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 =ππ’ β πΏ1 β πΏπ2
8=412.52 β 25 β 22.9172
8= 451.36πππ β ππ‘
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TWO-WAY SLAB DESIGN Pg 97
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Green Roof Slab design Tables Location along Span Coefficient Moment
Interior Span M- 0.65 443.2656
Interior Span M- 0.35 238.6815
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
Interior Span M- 0.65 239.3634
Interior Span M- 0.35 128.888
End span interior M- 0.7 257.776
End span M+ 0.5 184.1257
End Span Exterior M- 0.3 110.4754
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 Exterior M- 196.400772 8.1833655
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
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TWO-WAY SLAB DESIGN Pg 98
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Exterior Slab Panel Factored Moments k-ft
Location Column Strip Middle Strip
End Span Exterior M- 106.0564169 4.41901737
End Span M+ 138.0942928 46.03143094
End Span Interior 193.3320099 64.44400331
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 Exterior M- 76.22804963 34.24738462
End Span M+ 127.0467494 57.07897436
End Span Interior 177.8654491 79.91056411
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
Location Beam Moments from slab
Slab moment
End Span Exterior M- 141.1630549 63.42108263
End Span M+ 235.2717581 105.7018044
End Span Interior 329.3804614 147.9825261
Interior Span M- 305.8532856 137.4123457
Interior Span M+ 164.6902307 73.99126306
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TWO-WAY SLAB DESIGN Pg 99
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Interior 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- 141.1630549 44.24363636 185.4066912
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) =
π»ππ2 β π΅ β π»ππ +
β2 β π₯ β β
π΅ β π»ππ + π₯ β β=
242 β 12 β 24 +
122 β 12 β 12
12 β 24 + 12 β 12= 10ππ
πΌπ =1
3[(π΅ + π₯)(πΆπππ‘ππππ2) + (π΅(π»ππ β πΆπππ‘ππππ)
3) + (π₯(π»ππ β πΆπππ‘ππππ)3)] = 19008ππ4
Check: Centroid < h 10in < 12 in Good!
πΌπ =1
12[(πΆπ·
2+πΏ12) β3 =
1
12[(24
2+25
2) 123 = 23328ππ4
πΌπΉ1 = (πΌππΌπ ) =
19008
23328= 0.815
Check: Ξ±F1 > 0.80 0.815 > 0.80 Good! We may now use direct design method!
Reinforcement:
Exterior Column Panel Negative Steel Reinforcement:
Ξ¦1=0.90
db1 = diameter of steel reinforcement = 0.875in
fcβ= 5.5ksi
fy=60ksi
ECSW= CSW/2 +CD/2 = 87in
Critical Section: π1 = β β 0.75ππ β 0.50 β ππΌπ = 12 β .75 β 0.5(0.875) = 10.8125ππ
ππ1 =πΈπ₯π‘πππππ πΆπππ’ππ πππππ πππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Exterior Column Panel Width β π 21
=199.33πππ β ππ‘ β 12 β 1000
0.90 β 87ππ β 10.8125ππ 2
= 261.30ππ π
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TWO-WAY SLAB DESIGN Pg 100
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Ο1 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .2613ksi
. 85 β 5.5ππ π) 12))
= 4.48(10β3)
As = Ο1 β Exterior Column Panel Width β d1 = 4.48(10β3)(87in)(10.8125in) = 4.218in2
Top Steel: Number of bars 8 #7 bar
Check: ππ
4(ππ12 ) > π΄π β 8 (
π
4) (. 875π1
2 ) = 4.81 > 4.218 β πΊπππ!
Check: As= 8(.60in2) = 4.8in
2 > 4.218 in
2 Good!
Exterior Column Panel Positive Steel Reinforcement:
Ξ¦1=0.90
db2= 0.75in
fcβ= 5.50ksi
fy=60ksi
ECSW= 87in
Critical Section: π2 = β β 0.75ππ β 0.50 β ππ2 = 12 β 0.75 β 0.50(0.75ππ) = 10.875ππ
ππ2 =πΈπ₯π‘πππππ πΆπππ’ππ πππππ ππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Exterior Column Panel Width β π 22
=204.60 πππ β ππ‘ β 12 β 1000
0.90 β 87ππ β 10.875ππ 2
= 265.135 ππ π
Ο2 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .265135ksi
. 85 β 5.5ππ π) 12))
= 4.56(10β3)
As = Ο2 β Exterior Column Panel Width β d2 = 4.56(10β3)(87in)(10.875in) = 4.32in2
Bottom Steel: Number of bars 10 #6 bar
Check: ππ
4(ππ12 ) > π΄π β 10 (
π
4) (. 75π1
2 ) = 4.42ππ2 > 4.32ππ2 β πΊπππ!
Check: As= 10(.44in2) = 4.4in
2 > 4.32 in
2 Good!
Interior Column Panel Negative Steel Reinforcement:
![Page 102: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/102.jpg)
TWO-WAY SLAB DESIGN Pg 101
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Ξ¦1=0.90
db1= 0.875in
fcβ= 5ksi
fy=60ksi
ICSW= 2*min (.25*L1, .25*L2) = 150in
Critical Section: π1 = β β 0.75ππ β 0.50 β ππΌπ = 12 β .75 β 0.5(0.875) = 10.8125ππ
ππ1 =πΌππ‘πππππ πΆπππ’ππ πππππ πππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Interior Column Strip Width β π 21
=358.02 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 10.8125ππ 2
= 272.21ππ π
Ο1 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5ππ π
60ππ π(1 β ((1 β
2 β .27221ksi
. 85 β 5ππ π) 12))
= 4.69(10β3)
As = Ο1 β Interior Column Panel Width β d1 = 4.69(10β3)(150in)(10.8125in) = 7.61in2
Top Steel: Number of bars 13 #7 bar
Check: ππ
4(ππ12 ) > π΄π β 13 (
π
4) (. 875π1
2 ) = 7.813 > 7.61 β πΊπππ!
Check: As= 13(0.60in2) = 7.80 in
2 > 7.61 in
2 Good!
Interior Column Panel Positive Steel Reinforcement:
Ξ¦1=0.90
db2= 0.75in
fcβ= 5ksi
fy=60ksi
CSW= 150in
Critical Section: π2 = β β 0.75ππ β 0.50 β ππ2 = 12 β 0.75 β 0.50(0.75ππ) = 10.875ππ
ππ2 =πΌππππ πΆπππ’ππ πππππ ππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Interior Column Panel Width β π 22
=143.20 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 10.875ππ 2
= 107.63ππ π
Ο2 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5ππ π
60ππ π(1 β ((1 β
2 β .10763ksi
. 85 β 5ππ π) 12))
= 1.823(10β3)
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TWO-WAY SLAB DESIGN Pg 102
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
As = Ο2 β Interior Column Panel Width β d2 = 1.823(10β3)(150in)(10.875in) = 2.964in2
Bottom Steel: Number of bars 7 #6 bar
Check: ππ
4> π΄π β 7 (
π
4) (. 75π1
2 ) = 3.09ππ2 > 2.964ππ2 β πΊπππ!
Check: As= 7(0.44in2) = 3.08in
2 > 2.964in
2 Good!
Middle Column Panel Negative Steel Reinforcement:
Ξ¦1=0.90
db1= 0.75in
fcβ= 5ksi
fy=60ksi
MSW= L2-CSW = (25ft)(12in) -150in = 150in
Critical Section: π1 = β β 0.75ππ β 0.50 β ππΌπ = 12 β .75 β 0.5(0.75) = 10.875 ππ
ππ1 =ππππππ πΆπππ’ππ πππππ πππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Middle Column Panel Width β π 21
=119.30 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 10.875ππ 2
= 89.67ππ π
Ο1 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5ππ π
60ππ π(1 β ((1 β
2 β .08967ksi
. 85 β 5ππ π) 12))
= 1.51(10β3) As = Ο1 β Middle Column Panel Width β d1 = 1.51(10β3)(150in)(10.875in) = 2.464in2
Top Steel: Number of bars 6 #6 bar
Check: ππ
4(ππ12 ) > π΄π β 6 (
π
4) (. 75π1
2 ) = 2.65 > 2.464 β πΊπππ!
Check: As= 6(0.44in2) = 2.64 in
2 > 2.464 in
2 Good!
Middle Column Panel Positive Steel Reinforcement:
Ξ¦1=0.90
db2= 0.875in
fcβ= 5ksi
fy=60ksi
MSW= 150 in
Critical Section: π2 = β β 0.75ππ β 0.50 β ππ2 = 12 β 0.75 β 0.50(0.875ππ) = 10.8125 ππ
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TWO-WAY SLAB DESIGN Pg 103
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππ2 =πΆπππ’ππ πππππ ππππ‘ππ£π ππππππ‘πππ₯
Ξ¦1 β Middle Column Panel Width β π 22=136.40 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 10.8125ππ 2= 103.71ππ π
Ο2 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5ππ π
60ππ π(1 β ((1 β
2 β .10371ksi
. 85 β 5ππ π) 12))
= 1.75(10β3)
As = Ο2 β Middle Column Panel Width β d2 = 1.75(10β3)(150in)(10.8125in) = 2.84in2
Bottom Steel: Number of bars 5 #7 bar
Check: ππ
4(ππ12 ) > π΄π β 5 (
π
4) (. 875π1
2 ) = 3.01ππ2 > 2.85ππ2 β πΊπππ!
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
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 10.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 265.2375 ππ‘
2
ππ’ = π΄π‘ β ππ’ = 109.4158k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 356.6π
Bw=l2
![Page 105: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/105.jpg)
TWO-WAY SLAB DESIGN Pg 104
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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
π΄π‘ = (π12+π
2) β π2 β π1 β π2 = 330.4ππ‘^2
ππ’ = π΄π‘ β ππ’ = 136.3π b=1 , as=30
π1 = π +π
2= 29.41ππ , π2 = π + π = 34.686
π0 = 2 β π1 + π2 = 93.372
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=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
πΎπΉ =1
1 +23 ββπ1π2
= .62
πΎπΉ β ππ’ππ = 123.54π β ππ‘ Effective Width = 3h+c = 5ft
π΄π = (2π1 + π2) β π = 1054.9ππ^2
π½
π= π1 β π(π1 + 6π2) + π
3
6= 12613ππ^3
πΎπ£ = 1 β πΎπΉ = .38
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 201.3π/ππ2
(π)ππ = .75 β 4 β π β βπβ²π = 222.486π
![Page 106: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/106.jpg)
TWO-WAY SLAB DESIGN Pg 105
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 10.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 265.2375 ππ‘
2
ππ’ = π΄π‘ β ππ’ = 109.4158k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 356.6π
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
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TWO-WAY SLAB DESIGN Pg 106
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
π΄π‘ = (π1) β π2 β π1 β π2 = 616.645ππ‘2
ππ’ = π΄π‘ β ππ’ = 254.384π
b=1 , as=40
π1 = π + π = 34.8125ππ , π2 = π + π = 34.686ππ
π0 = π1 + π2 = 138.744ππ
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=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
πΎπΉ =1
1 +23 ββπ1π2
= .6
πΎπΉ β ππ’ππ = 15.3426π β ππ‘ Effective Width = 3h+c = 5ft
π΄π = 2(π1 + π2) β π = 1482.618ππ^2
π½
π= π1 β π(π1 + 3π2) + π
3
3= 17548.8ππ^3
πΎπ£ = 1 β πΎπΉ = .4
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 184.2π/ππ2
(π)ππ = .75 β 4 β π β βπβ²π = 222.486π
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
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TWO-WAY SLAB DESIGN Pg 107
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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
Slab thickness 12 inch
db 1.128 inch
qdu 200.6 lbs
qlu 100 lbs
One way Shear Check
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 10.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 265.2375 ππ‘
2
ππ’ = π΄π‘ β ππ’ = 109.4158k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 356.6π
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
π΄π‘ = (π12+π
2) β π2 β π1 β π2 = 330.4ππ‘^2
ππ’ = π΄π‘ β ππ’ = 136.3π b=1 , as=30
π1 = π +π
2= 29.41ππ , π2 = π + π = 34.78125
π0 = 2 β π1 + π2 = 93.372
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=222k
![Page 109: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/109.jpg)
TWO-WAY SLAB DESIGN Pg 108
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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
πΎπΉ =1
1 +23 ββπ1π2
= .62
πΎπΉ β ππ’ππ = 123.54π β ππ‘ Effective Width = 3h+c = 5ft
π΄π = (π1 + 2π2) β π = 997.77ππ^2
π
π=2 β π1
2 β π(π1 β 2π2) + π3(2π1 + π2)
6π1= 10964.82ππ^3
πΎπ£ = 1 β πΎπΉ = .38
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 219.5π/ππ2
(π)ππ = .75 β $ β π β βπβ²π = 222.486π
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
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TWO-WAY SLAB DESIGN Pg 109
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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
Slab thickness 12 inch
db 1.128 inch
qdu 200.6 lbs
qlu 100 lbs
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 10.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 265.2375 ππ‘
2
ππ’ = π΄π‘ β ππ’ = 109.4158k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 356.6π
Bw=l2
Case D Corner d 10.686 in
At 265.2375 ft^2 Vu 109.4158 kip (F)Vc 356.6227 kip OK
Two-way Slab Shear
π΄π‘ = (π12) βπ22β (π +
π
2)2
= 150.27ππ‘^2
ππ’ = π΄π‘ β ππ’ = 62π b=1 , as=30
π1 = π +π
2= 29.41ππ , π2 = π +
π
2= 29.41ππ
π0 = π1 + π2 = 58.69ππ
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=139.53k
![Page 111: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/111.jpg)
TWO-WAY SLAB DESIGN Pg 110
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Case D Corner At 150.2708 ft^2
Vu 61.98969 kip b 1 as 20 b1 29.343 in b2 29.343 in b0 58.686 in (F)Vc 139.5251 kip OK
Munb = 199.3k-ft
πΎπΉ =1
1 +23 ββπ1π2
= .6
Effective Width = 3h+c = 5ft
πΎπΉ β ππ’ππ = 119.58π β ππ‘ π΄π = (π1 + π2) β π = 627.12ππ
2
π½
π= π12 β π(π1 + 4π2) + π
3(π1 + π2)
6π1= 21664.01ππ3
πΎπ£ = 1 β πΎπΉ = .4
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 143π/ππ2
(π)ππ = .75 β $ β π β βπβ²π = 222.486π
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
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TWO-WAY SLAB DESIGN Pg 111
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Superimposed dead load: DLsup=9.50psf
Concrete density: Ξ³=150pcf Initial deal load: DL= h* Ξ³=13in*150pcf=162.5psf
Load Combinations:
LRFD LOAD COMBINATIONS
Loads: Combinations:
Dead (D) 172.00 psf 1) 1.4D
Live (L) 65.00 psf 2) 1.2D + 1.6L + .5(Lr or S or R)
Live Roof (Lr) 0.00 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) 0.00 psf 6) 0.9D + W
Negative Wind (W) 0.00 psf 7) 0.9D + 1E
Seismic (E) 0.00 psf
Results:
Positive: Negative: DESIGN VALUES (psf):
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
LRFD Max Positive Moment: Mmax positive= 310.40 psf
LRFD Max Negative Moment: Mmax negative= 154.80 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 13in > 8.33in OK!
qu=Mmax positive=310.40psf
Mo =ππ’ β πΏ1 β πΏπ2
8=310.40ππ π β 25ππ‘ β 22.917ππ‘2
8= 509.43πππ β ππ‘
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TWO-WAY SLAB DESIGN Pg 112
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
3rd Floor Slab design tables Location along Span Coefficient Moment
Interior Span M- 0.65 333.5345
Interior Span M- 0.35 179.5955
End span interior M- 0.7 359.191
End span M+ 0.5 256.565
End Span Exterior M- 0.3 153.939
Location along Span Coefficient Moment
Interior Span M- 0.65 180.1086
Interior Span M- 0.35 96.98157
End span interior M- 0.7 193.9631
End span M+ 0.5 138.5451
End Span Exterior M- 0.3 83.12706
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
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TWO-WAY SLAB DESIGN Pg 113
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Interior Slab Panel Factored Moments k-ft
Location Column Strip Middle Strip
End Span Exterior M- 145.010538 8.928462
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
Exterior Slab Panel Factored Moments k-ft
Location Column Strip Middle Strip
End Span Exterior M- 78.30569052 4.82136948
End Span M+ 103.908825 34.636275
End Span Interior M- 145.472355 48.490785
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
Location Beam Moments from slab
Slab moment
End Span Exterior M- 57.3576714 25.7693886
End Span M+ 95.596119 42.948981
End Span Interior 133.8345666 60.1285734
Interior Span M- 124.2749547 55.8336753
Interior Span M+ 66.9172833 30.0642867
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TWO-WAY SLAB DESIGN Pg 114
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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-
57.3576714 44.24363636 101.6013078
End Span M+ 95.596119 44.24363636 139.8397554
End Span Interior 133.8345666 48.668 182.5025666
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
Location Beam Moments from slab
Slab moment
End Span Exterior M- 106.21791 47.72109
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
Interior 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- 106.21791
44.24363636 150.4615464
End Span M+ 177.02985 44.24363636 221.2734864
End Span Interior 247.84179 48.668 296.50979
Interior Span M- 230.138805 44.24363636 274.3824414
Interior Span M+ 123.920895 30.4175 154.338395
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TWO-WAY SLAB DESIGN Pg 115
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Finding the centroid:
B=12in
HIb=24in
hw= HIb-h=24in-13in=11in
Centroid (from the top) =
π»ππ2 β π΅ β π»ππ +
β2 β π₯ β β
π΅ β π»ππ + π₯ β β=
242 β 12 β 24 +
112 β 11 β 11
12 β 24 + 11 β 11= 10.18ππ
πΌπ =1
3[(π΅ + π₯)(πΆπππ‘ππππ2) + (π΅(π»ππ β πΆπππ‘ππππ)
3) + (π₯(π»ππ β πΆπππ‘ππππ)3)]
= 18647.85ππ4 Check: Centroid < h 10.18in < 12 in Good!
πΌπ =1
12[(πΆπ·
2+πΏ12) β3 =
1
12[(24
2+25
2) 123 = 17968.5ππ4
πΌπΉ1 = (πΌππΌπ ) =
18647.85
17968.5= 1.038
Check: Ξ±F1 > 0.80 1.038 > 0.80 Good!
Reinforcement:
Exterior Column Panel Negative Steel Reinforcement:
Ξ¦1=0.90
db1 = diameter of steel reinforcement = 0.875in
fcβ= 5.5ksi
fy=60ksi
ECSW= CSW/2 +CD/2 = 87in
Critical Section: π1 = β β 0.75ππ β 0.50 β ππΌπ = 13 β .75 β 0.5(0.875) = 11.8125ππ
ππ1 =πΈπ₯π‘πππππ πΆπππ’ππ πππππ πππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Exterior Column Panel Width β π 21
=147.94πππ β ππ‘ β 12 β 1000
0.90 β 87ππ β 11.8125ππ 2
= 162.49ππ π
Ο1 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .16249ksi
. 85 β 5.5ππ π) 12))
= 2.76(10β3)
As = Ο1 β Exterior Column Panel Width β d1 = 2.76(10β3)(87in)(11.8125in) = 2.83in2
Top Steel: Number of bars 5 #7 bar
Check: ππ
4(ππ12 ) > π΄π β 5 (
π
4) (. 875π1
2 ) = 3.01ππ2 > 2.83ππ2 β πΊπππ!
Check: As= 6(0.60in2) = 3.0in
2 > 2.83in
2 Good!
![Page 117: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/117.jpg)
TWO-WAY SLAB DESIGN Pg 116
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Exterior Column Panel Positive Steel Reinforcement:
Ξ¦1=0.90
db2= 0.75in
fcβ= 5.5ksi
fy=60ksi
ECSW= 87in
Critical Section: π2 = β β 0.75ππ β 0.50 β ππ2 = 13 β 0.75 β 0.50(0.75ππ) = 11.875ππ
ππ2 =πΈπ₯π‘πππππ πΆπππ’ππ πππππ πππ ππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Exterior Column Panel Width β π 22
= 153.94 πππ β ππ‘ β 12 β 1000
0.90 β 87ππ β 11.875ππ 2
= 167.303 ππ π
Ο2 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .1673ksi
. 85 β 5.5ππ π) 12))
= 2.84(10β3) As = Ο2 β Exterior Column Panel Width β d2 = 2.861(10β3)(87in)(11.875in) = 2.93in2
Bottom Steel: Number of bars 7 #6 bar
Check: ππ
4(ππ12 ) > π΄π β 7 (
π
4) (. 75π1
2 ) = 3.09ππ2 > 2.93ππ2 β πΊπππ!
Check: As= 7(0.44in2) = 3.08in
2 > 2.93 in
2 Good!
Interior Column Panel Negative Steel Reinforcement:
Ξ¦1=0.90
db1= 0.875in
fcβ= 5.5ksi
fy=60ksi
ICSW= 2*min (.25*L1, .25*L2) = 150in
Critical Section: π1 = β β 0.75ππ β 0.50 β ππΌπ = 13 β .75 β 0.5(0.75) = 11.8125ππ
ππ1 =πΌππ‘πππππ πΆπππ’ππ πππππ πππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Interior Column Strip Width β π 21
=269.39 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 11.8125ππ 2
= 171.61ππ π
Ο1 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
4ππ π
60ππ π(1 β ((1 β
2 β .17161ksi
. 85 β 4ππ π) 12))
= 2.91(10β3)
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TWO-WAY SLAB DESIGN Pg 117
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
As = Ο1 β Interior Column Panel Width β d1 = 2.91(10β3)(150in)(11.8125in) = 5.16in2
Top Steel: Number of bars 9 #7 bar
Check: ππ
4(ππ12 ) > π΄π β 9 (
π
4) (. 875π1
2 ) = 5.41 > 5.16 β πΊπππ!
Check: As= 9(0.6in2) = 5.40 in
2 > 5.16 in
2 Good!
Interior Column Panel Positive Steel Reinforcement:
Ξ¦1=0.90
db2= 0.75in
fcβ= 5.5ksi
fy=60ksi
CSW= 150in
Critical Section: π2 = β β 0.75ππ β 0.50 β ππ2 = 13 β 0.75 β 0.50(0.675ππ) = 11.875ππ
ππ2 =πΌππππ πΆπππ’ππ πππππ ππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Interior Column Panel Width β π 22
=107.76 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 11.875ππ 2
= 67.93ππ π
Ο2 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .06793ksi
. 85 β 5.5ππ π) 12))
= 1.14(10β3)
As = Ο2 β Interior Column Panel Width β d2 = 1.14(10β3)(150in)(11.875in) = 2.03in2
Bottom Steel: Number of bars 5#6 bar
Check: ππ
4(ππ12 ) > π΄π β 3 (
π
4) (. 75π1
2 ) = 2.21ππ2 > 2.03ππ2 β πΊπππ!
Check: As= 5(0.44in2) = 2.20in
2 > 2.03in
2 Good!
Middle Column Panel Negative Steel Reinforcement:
Ξ¦1=0.90
db1= 0.75in
fcβ= 5.5ksi
fy=60ksi
MSW= L2-CSW = (25ft)(12in) -150in = 150in
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TWO-WAY SLAB DESIGN Pg 118
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Critical Section: π1 = β β 0.75ππ β 0.50 β ππΌπ = 13 β .75 β 0.5(0.5) = 11.875 ππ
ππ1 =ππππππ πΆπππ’ππ πππππ πππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Middle Column Panel Width β π 21
=89.8 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 11.875ππ 2
= 56.61ππ π
Ο1 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .8967ksi
. 85 β 5.5ππ π) 12))
= 0.95(10β3) As = Ο1 β Middle Column Panel Width β d1 = 0.95(10β3)(150in)(11.875in) = 1.69in2
Top Steel: Number of bars 4 #6 bar
Check: ππ
4(ππ12 ) > π΄π β 4 (
π
4) (. 75π1
2 ) = 1.77 ππ2 > 1.69in2 β πΊπππ!
Check: As= 4(0.44in2) = 1.76 in
2 > 1.69 in
2 Good!
Middle Column Panel Positive Steel Reinforcement:
Ξ¦1=0.90
db2= 0.875in
fcβ= 5.5ksi
fy=60ksi
MSW= 150 in
Critical Section: π2 = β β 0.75ππ β 0.50 β ππ2 = 13 β 0.75 β 0.50(0.875ππ) = 11.8125 ππ
ππ2 =πΆπππ’ππ πππππ ππππ‘ππ£π ππππππ‘πππ₯
Ξ¦1 β Middle Column Panel Width β π 22=102.63 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 11.8125ππ 2= 65.38ππ π
Ο2 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .06538ksi
. 85 β 5.5ππ π) 12))
= 1.10(10β3)
As = Ο2 β Middle Column Panel Width β d2 = 1.10(10β3)(150in)(11.8125in) = 1.94in2
Bottom Steel: Number of bars 4 #7 bar
Check: ππ
4(ππ12 ) < π΄π β 4 (
π
4) (. 875π1
2 ) = 2.41ππ2 > 1.94ππ2 β πΊπππ!
Check: As= 4(0.6in2) = 2.40 in
2 > 1.94 in
2 Good!
![Page 120: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/120.jpg)
TWO-WAY SLAB DESIGN Pg 119
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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
Slab thickness 13 inch
db 1.128 inch
qdu 172 lbs
qlu 65 lbs
One way Shear Check
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 11.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 263.2 ππ‘
2
ππ’ = π΄π‘ β ππ’ = 81.7k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 390π
bw=l2
Case A Edge One Way Shear d 11.686 in
At 263.1542 ft^2 Vu 81.68305 kip (F)Vc 389.9956 kip OK
Two-way Slab Shear
π΄π‘ = (π12+π
2) β π2 β π1 β π2 = 330.1ππ‘^2
ππ’ = π΄π‘ β ππ’ = 102.46π b=1 , as=30
π1 = π +π
2= 29.91ππ , π2 = π + π = 35.686ππ
π0 = 2 β π1 + π2 = 95.372ππ
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=247.96k
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TWO-WAY SLAB DESIGN Pg 120
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Case A Edge Two Way Shear At 330.1043 ft^2
Vu 102.4644 kip b 1 as 30 b1 29.843 in b2 35.686 in b0 95.372 in (F)Vc 247.9644 kip OK
Munb= 147.94k-ft
πΎπΉ =1
1 +23 ββπ1π2
= .62
πΎπΉ β ππ’ππ = 91.9π β ππ‘ Effective Width = 3h+c = 5.25ft
π΄π = (2π1 + π2) β π = 1182.8ππ^2
π½
π= π1 β π(π1 + 6π2) + π
3
6= 14445.92ππ^3
πΎπ£ = 1 β πΎπΉ = .38
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 133.17π/ππ2
(π)ππ = .75 β 4 β π β βπβ²π = 222.486π
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
![Page 122: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/122.jpg)
TWO-WAY SLAB DESIGN Pg 121
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Case B Interior
Column Size 24 inch
f'c 5500 psi
fy 60000 psi
L1 25 ft
L2 25 ft
qu 310.4 psf
Slab thickness 13 inch
db 1.128 inch
qdu 172 lbs
qlu 65 lbs
One way Shear Check
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 11.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 263.2 ππ‘
2
ππ’ = π΄π‘ β ππ’ = 81.7k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 390π
bw=l2
Case B Interior One Way Shear d 11.686 in
At 263.1542 ft^2 Vu 81.68305 kip (F)Vc 389.9956 kip OK
Two-way Slab Shear
π΄π‘ = (π1) β π2 β π1 β π2 = 616.15ππ‘2
ππ’ = π΄π‘ β ππ’ = 191.25π
b=1 , as=40
π1 = π + π = 35.686ππ , π2 = π + π = 35.686ππ
π0 = π1 + π2 = 142.744ππ
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=371.13k
![Page 123: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/123.jpg)
TWO-WAY SLAB DESIGN Pg 122
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Case B Interior Two Way Shear At 616.1563 ft^2
Vu 191.2549 kip b 1 as 40 b1 35.686 in b2 35.686 in b0 142.744 in (F)Vc 371.1302 kip OK
Munb=19.24 k-ft
πΎπΉ =1
1+2
3ββπ1π2
= .6
πΎπΉ β ππ’ππ = 91.9π β ππ‘ Effective Width = 3h+c = 5.25ft
π΄π = 2(π1 + π2) β π = 1668.101ππ^2
π½
π= π1 β π(π1 + 3π2) + π
3
3= 20374.35ππ^3
πΎπ£ = 1 β πΎπΉ = .4
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 121.74π/ππ2
(π)ππ = .75 β 4 β π β βπβ²π = 222.486π
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
![Page 124: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/124.jpg)
TWO-WAY SLAB DESIGN Pg 123
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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
Slab thickness 13 inch
db 1.128 inch
qdu 172 lbs
qlu 65 lbs
One way Shear Check
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 11.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 263.2 ππ‘
2
ππ’ = π΄π‘ β ππ’ = 81.7k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 390π
bw=l2
Case C Edge d 11.686 in
At 263.1542 ft^2 Vu 81.68305 kip (F)Vc 389.9956 kip OK
Two-way Slab Shear
π΄π‘ = (π12+π
2) β π2 β π1 β π2 = 330.06ππ‘^2
ππ’ = π΄π‘ β ππ’ = 102.46π b=1 , as=30
π1 = π +π
2= 29.84ππ , π2 = π + π = 35.686ππ
π0 = 2 β π1 + π2 = 95.372ππ
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=247.96k
![Page 125: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/125.jpg)
TWO-WAY SLAB DESIGN Pg 124
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Case C Edge Two Way Shear At 330.1043 ft^2
Vu 102.4644 kip b 1 as 30 b1 29.843 in b2 35.686 in b0 95.372 in (F)Vc 247.9644 kip OK
Munb = 147.4k-ft
πΎπΉ =1
1 +23 ββπ1π2
= .62
πΎπΉ β ππ’ππ = 91.9π β ππ‘ Effective Width = 3h+c = 5.25ft
π΄π = (π1 + 2π2) β π = 1114.52ππ^2
π
π=2 β π1
2 β π(π1 β 2π2) + π3(2π1 + π2)
6π1= 12616.1ππ^3
πΎπ£ = 1 β πΎπΉ = .38
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 145.1π/ππ2
(π)ππ = .75 β $ β π β βπβ²π = 222.486π
Case C 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 1114.517 in^2 J/c 12616.1 in^3 Yv 0.378747 Vmax 145.2317 Kips/in^2 (F)Vu 222.486 kips OK
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TWO-WAY SLAB DESIGN Pg 125
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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
Slab thickness 13 inch
db 1.128 inch
qdu 172 lbs
qlu 65 lbs
One way Shear Check
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 11.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 263.2 ππ‘
2
ππ’ = π΄π‘ β ππ’ = 81.7k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 390π
bw=l2
Case D Corner One Way Shear d 11.686 in
At 263.1542 ft^2 Vu 81.68305 kip (F)Vc 389.9956 kip OK
Two-way Slab Shear
π΄π‘ = (π12) βπ22β (π +
π
2)2
= 150.06ππ‘^2
ππ’ = π΄π‘ β ππ’ = 46.58π b=1 , as=20
π1 = π +π
2= 29.843ππ , π2 = π +
π
2= 29.843ππ
π0 = π1 + π2 = 59.686ππ
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=155.2k
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TWO-WAY SLAB DESIGN Pg 126
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Case D Corner Two Way Shear At 150.0652 ft^2
Vu 46.58025 kip b 1 as 20 b1 29.843 in b2 29.843 in b0 59.686 in (F)Vc 155.1819 kip OK
Munb = 147.94k-ft
πΎπΉ =1
1 +23 ββπ1π2
= .6
πΎπΉ β ππ’ππ = 88.76π β ππ‘ Effective Width = 3h+c = 5.25ft
π΄π = (π1 + π2) β π = 697.5ππ2
π½
π= π12 β π(π1 + 4π2) + π
3(π1 + π2)
6π1= 22946.5ππ3
πΎπ£ = 1 β πΎπΉ = .4
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 97.73π/ππ2
(π)ππ = .75 β $ β π β βπβ²π = 222.486π
Case D Corner Munb 147.94 k-ft
Yf 0.6 Yf*Munb 88.764 Effective
Width 5.25 ft Min Munb 30.08688 k-ft Ac 697.4906 in^2 J/c 22946.53 in^3 Yv 0.4 Vmax 97.729 Kips/in^2 (F)Vu 222.486 kips OK
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TWO-WAY SLAB DESIGN Pg 127
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
2nd
Floor Two-Way Slab:
Initial slab thickness: h=13in
Per ACI Table 9.5c
Superimposed dead load: DLsup=30psf
Concrete density: Ξ³=150pcf Initial deal load: DL= h* Ξ³=13in*150pcf=162.5psf
Load Combinations:
LRFD LOAD COMBINATIONS
Loads: Combinations:
Dead (D) 192.50 psf 1) 1.4D
Live (L) 100.00 psf 2) 1.2D + 1.6L + .5(Lr or S or R)
Live Roof (Lr) 0.00 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) 0.00 psf 6) 0.9D + W
Negative Wind (W) 0.00 psf 7) 0.9D + 1E
Seismic (E) 0.00 psf
Results:
Positive: Negative: DESIGN VALUES (psf):
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
LRFD Max Positive Moment: Mmax positive= 391.0 psf
LRFD Max Negative Moment: Mmax negative= 173.25 psf
Column cross section dimensions (square): Column Dimensions (CD)= 24in x 24in
L1=L2= 25ft
Ln= L1-CD=22.917ft
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TWO-WAY SLAB DESIGN Pg 128
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
hmin=Ln/33=(22.917*12)/33=8.333in Check: h > hmin 13in > 8.33in OK!
qu=Mmax positive=391.40psf
Mo =ππ’ β πΏ1 β πΏπ2
8=391.40ππ π β 25ππ‘ β 22.917ππ‘2
8= 642.37πππ β ππ‘
2nd Floor Slab Design tables Location along Span Coefficient Moment
Interior Span M- 0.65 420.1417
Interior Span M- 0.35 226.2302
End span interior M- 0.7 452.4603
End span M+ 0.5 323.1859
End Span Exterior M- 0.3 193.9116
Location along Span Coefficient Moment
Interior Span M- 0.65 226.8765
Interior Span M- 0.35 122.1643
End span interior M- 0.7 244.3286
End span M+ 0.5 174.5204
End Span Exterior M- 0.3 104.7122
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
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TWO-WAY SLAB DESIGN Pg 129
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Interior Slab Panel Factored Moments k-ft
Location Column Strip Middle Strip
End Span Exterior M- 182.6646919 11.24687063
End Span M+ 193.9115625 129.274375
End Span Interior M- 339.3452344 113.1150781
Interior Span M- 315.1062891 105.0354297
Interior Span M+ 135.7380938 90.4920625
Exterior Slab Panel Factored Moments k-ft
Location Column Strip Middle Strip
End Span Exterior M- 98.63893361 6.073310138
End Span M+ 130.8903047 43.63010156
End Span Interior 183.2464266 61.08214219
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
Location Beam Moments from slab
Slab moment
End Span Exterior M- 72.25144819 32.46079556
End Span M+ 120.4190803 54.10132594
End Span Interior 168.5867124 75.74185631
Interior Span M- 156.5448044 70.33172372
Interior Span M+ 84.29335622 37.87092816
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- 72.25144819 44.24363636 116.4950846
End Span M+ 120.4190803 44.24363636 164.6627167
End Span Interior 168.5867124 48.668 217.2547124
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
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TWO-WAY SLAB DESIGN Pg 130
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
moments k-ft
Location Beam Moments from slab
Slab moment
End Span Exterior M- 133.7989781 60.11258438
End Span M+ 222.9982969 100.1876406
End Span Interior 312.1976156 140.2626969
Interior Span M- 289.8977859 130.2439328
Interior Span M+ 156.0988078 70.13134844
Interior 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- 133.7989781 44.24363636 178.0426145
End Span M+ 222.9982969 44.24363636 267.2419332
End Span Interior 312.1976156 48.668 360.8656156
Interior Span M- 289.8977859 44.24363636 334.1414223
Interior Span M+ 156.0988078 30.4175 186.5163078
Finding the centroid:
B=12in
HIb=24in
hw= HIb-h=24in-13in=11in
Centroid (from the top) =
π»ππ2 β π΅ β π»ππ +
β2 β π₯ β β
π΅ β π»ππ + π₯ β β=
242 β 12 β 24 +
112 β 11 β 11
12 β 24 + 11 β 11= 10.18ππ
πΌπ =1
3[(π΅ + π₯)(πΆπππ‘ππππ2) + (π΅(π»ππ β πΆπππ‘ππππ)
3) + (π₯(π»ππ β πΆπππ‘ππππ)3)]
= 18647.85ππ4 Check: Centroid < h 10.18in < 12 in Good!
πΌπ =1
12[(πΆπ·
2+πΏ12) β3 =
1
12[(24
2+25
2) 123 = 17968.5ππ4
πΌπΉ1 = (πΌππΌπ ) =
18647.85
17968.5= 1.038
Check: Ξ±F1 > 0.80 1.038 > 0.80 Good!
Reinforcement:
Exterior Column Panel Negative Steel Reinforcement:
Ξ¦1=0.90
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TWO-WAY SLAB DESIGN Pg 131
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
db1 = diameter of steel reinforcement = 0.875in
fcβ= 5.5ksi
fy=60ksi
ECSW= CSW/2 +CD/2 = 87in
Critical Section: π1 = β β 0.75ππ β 0.50 β ππΌπ = 13 β .75 β 0.5(0.875) = 11.8125ππ
ππ1 =πΈπ₯π‘πππππ πΆπππ’ππ πππππ πππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Exterior Column Panel Width β π 21
=186.35πππ β ππ‘ β 12 β 1000
0.90 β 87ππ β 11.8125ππ 2
= 204.675ππ π
Ο1 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .204675ksi
. 85 β 5.5ππ π) 12))
= 3.50(10β3)
As = Ο1 β Exterior Column Panel Width β d1 = 3.50(10β3)(87in)(11.8125in) = 3.594in2
Top Steel: Number of bars 6 #7 bar
Check: ππ
4(ππ12 ) > π΄π β 6 (
π
4) (. 875π1
2 ) = 3.61ππ2 > 3.594ππ2 β πΊπππ!
Check: As= 6(0.60in2) = 3.60in
2 > 3.594in
2 Good!
Exterior Column Panel Positive Steel Reinforcement:
Ξ¦1=0.90
db2= 0.75in
fcβ= 5.5ksi
fy=60ksi
ECSW= 87in
Critical Section: π2 = β β 0.75ππ β 0.50 β ππ2 = 13 β 0.75 β 0.50(0.75ππ) = 11.875ππ
ππ2 =πΈπ₯π‘πππππ πΆπππ’ππ πππππ πππ ππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Exterior Column Panel Width β π 22
= 193.91 πππ β ππ‘ β 12 β 1000
0.90 β 87ππ β 11.875ππ 2
= 210.74 ππ π
Ο2 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .21074ksi
. 85 β 5.5ππ π) 12))
= 3.60(10β3) As = Ο2 β Exterior Column Panel Width β d2 = 3.60(10β3)(87in)(11.875in) = 3.723in2
Bottom Steel: Number of bars 9 #6 bar
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TWO-WAY SLAB DESIGN Pg 132
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Check: ππ
4(ππ12 ) > π΄π β 9 (
π
4) (. 75π1
2 ) = 3.974ππ2 > 3.723ππ2 β πΊπππ!
Check: As= 9(0.44in2) = 3.96in
2 > 3.723 in
2 Good!
Interior Column Panel Negative Steel Reinforcement:
Ξ¦1=0.90
db1= 0.875in
fcβ= 5.5ksi
fy=60ksi
ICSW= 2*min (.25*L1, .25*L2) = 150in
Critical Section: π1 = β β 0.75ππ β 0.50 β ππΌπ = 13 β .75 β 0.5(0.75) = 11.8125ππ
ππ1 =πΌππ‘πππππ πΆπππ’ππ πππππ πππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Interior Column Strip Width β π 21
=339.35 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 11.8125ππ 2
= 216.178ππ π
Ο1 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .216178ksi
. 85 β 5.5ππ π) 12))
= 3.70(10β3)
As = Ο1 β Interior Column Panel Width β d1 = 3.70(10β3)(150in)(11.8125in) = 6.56in2
Top Steel: Number of bars 12 #7 bar
Check: ππ
4(ππ12 ) > π΄π β 12 (
π
4) (. 875π1
2 ) = 7.21 > 6.56 β πΊπππ!
Check: As= 12(0.6in2) = 7.20 in
2 > 6.56 in
2 Good!
Interior Column Panel Positive Steel Reinforcement:
Ξ¦1=0.90
db2= 0.75in
fcβ= 5.5ksi
fy=60ksi
CSW= 150in
Critical Section: π2 = β β 0.75ππ β 0.50 β ππ2 = 13 β 0.75 β 0.50(0.675ππ) = 11.875ππ
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TWO-WAY SLAB DESIGN Pg 133
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππ2 =πΌππππ πΆπππ’ππ πππππ ππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Interior Column Panel Width β π 22
=135.74 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 11.875ππ 2
= 85.563ππ π
Ο2 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .085563ksi
. 85 β 5.5ππ π) 12))
= 1.44(10β3)
As = Ο2 β Interior Column Panel Width β d2 = 1.44(10β3)(150in)(11.875in) = 2.57in2
Bottom Steel: Number of bars 6 #6 bar
Check: ππ
4(ππ12 ) > π΄π β 6 (
π
4) (. 75π1
2 ) = 2.65ππ2 > 2.57ππ2 β πΊπππ!
Check: As= 6(0.44in2) = 2.64in
2 > 2.57in
2 Good!
Middle Column Panel Negative Steel Reinforcement:
Ξ¦1=0.90
db1= 0.75in
fcβ= 5.5ksi
fy=60ksi
MSW= L2-CSW = (25ft)(12in) -150in = 150in
Critical Section: π1 = β β 0.75ππ β 0.50 β ππΌπ = 13 β .75 β 0.5(0.5) = 11.875 ππ
ππ1 =ππππππ πΆπππ’ππ πππππ πππππ‘ππ£π ππππππ‘πππ₯Ξ¦1 β Middle Column Panel Width β π 21
=113.12 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 11.875ππ 2
= 71.305ππ π
Ο1 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .071305ksi
. 85 β 5.5ππ π) 12))
= 1.20(10β3) As = Ο1 β Middle Column Panel Width β d1 = 1.20(10β3)(150in)(11.875in) = 2.135in2
Top Steel: Number of bars 6 #6 bar
Check: ππ
4(ππ12 ) > π΄π β 6 (
π
4) (. 75π1
2 ) = 2.64 ππ2 > 2.135in2 β πΊπππ!
Check: As= 6(.44in2) = 2.64 in
2 > 2.135 in
2 Good!
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TWO-WAY SLAB DESIGN Pg 134
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Middle Column Panel Positive Steel Reinforcement:
Ξ¦1=0.90
db2= 0.875in
fcβ= 5.5ksi
fy=60ksi
MSW= 150 in
Critical Section: π2 = β β 0.75ππ β 0.50 β ππ2 = 13 β 0.75 β 0.50(0.875ππ) = 11.8125 ππ
ππ2 =πΆπππ’ππ πππππ ππππ‘ππ£π ππππππ‘πππ₯
Ξ¦1 β Middle Column Panel Width β π 22=129.27 πππ β ππ‘ β 12 β 1000
0.90 β 150ππ β 11.8125ππ 2= 82.350ππ π
Ο2 = .85 βππβ²
ππ¦(1 β (1 β
2 β ππ1. 85 β ππβ²
) 12) = .85 β
5.5ππ π
60ππ π(1 β ((1 β
2 β .08235ksi
. 85 β 5.5ππ π) 12))
= 1.39(10β3)
As = Ο2 β Middle Column Panel Width β d2 = 1.39(10β3)(150in)(11.8125in) = 2.456in2
Bottom Steel: Number of bars 6 #7 bar
Check: ππ
4(ππ12 ) < π΄π β 6 (
π
4) (. 875π1
2 ) = 3.61ππ2 > 2.456ππ2 β πΊπππ!
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
Slab thickness 13 inch
db 1.128 inch
qdu 192.5 lbs
qlu 100 lbs
One way Shear Check
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 11.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 263.1542 ππ‘
2
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TWO-WAY SLAB DESIGN Pg 135
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππ’ = π΄π‘ β ππ’ = 102.9k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 390π
Bw=l2
Case A Edge d 11.686 in
At 263.1542 ft^2 Vu 102.8933 kip (F)Vc 389.9956 kip OK
Two-way Slab Shear
π΄π‘ = (π12+π
2) β π2 β π1 β π2 = 330.1ππ‘^2
ππ’ = π΄π‘ β ππ’ = 129.07π b=1 , as=30
π1 = π +π
2= 29.843ππ , π2 = π + π = 35.686ππ
π0 = 2 β π1 + π2 = 95.372ππ
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=247.9644k
Case A Edge At 330.1043 ft^2
Vu 129.0708 kip b 1 as 30 b1 29.843 in b2 35.686 in b0 95.372 in (F)Vc 247.9644 kip OK
πΎπΉ =1
1+2
3ββπ1π2
= .621
Effective Width = 3h+c = 5.25ft
πΎπΉ β ππ’ππ = 115.77π β ππ‘ π΄π = (2π1 + π2) β π = 1199.34ππ^2
π
π=2 β π1
2 β π(π1 β 2π2) + π3(2π1 + π2)
6π1
π½
π= π1 β π(π1 + 6π2) + π
3
6= 14686.92ππ^3
πΎπ£ = 1 β πΎπΉ = .379
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 165.25π/ππ2
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TWO-WAY SLAB DESIGN Pg 136
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
(π)ππ = .75 β 4 β π β βπβ²π = 222.486π
Case A Edge Munb 186.35 k-ft
Yf 0.621253 Yf*Munb 115.7705 Effective
Width 5.25 ft Min Munb 46.2875 k-ft Ac 1182.798 in^2 J/c 14445.9 in^3 Yv 0.378747 Vmax 167.7526 Kips/in^2 (F)Vu 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 391 psf
Slab thickness 13 inch
db 1.128 inch
qdu 192.5 lbs
qlu 100 lbs
One way Shear Check
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 11.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 263.1542 ππ‘
2
ππ’ = π΄π‘ β ππ’ = 102.9k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 390π
Bw=l2
Case B Interior d 11.686 in
At 263.1542 ft^2 Vu 102.8933 kip (F)Vc 389.9956 kip OK
Two-way Slab Shear
π΄π‘ = π΄π‘ = (π1) β π2 β π1 β π2 = 616.16ππ‘^2
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TWO-WAY SLAB DESIGN Pg 137
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππ’ = π΄π‘ β ππ’ = 240.9π b=1 , as=40
π1 = π + π = 35.686ππ , π2 = π + π = 35.686ππ
π0 = 2(π1 + π2) = 142.744ππ
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=371.13k
Case B Interior At 616.1563 ft^2
Vu 240.9171 kip b 1 as 40 b1 35.686 in b2 35.686 in b0 142.744 in (F)Vc 371.1302 kip OK
Munb = 24.23k-ft
πΎπΉ =1
1 +23 ββπ1π2
= .6
πΎπΉ β ππ’ππ = 14.54π β ππ‘ Effective Width = 3h+c = 5.25ft
π΄π = 2(π1 + π2) β π = 1668.11ππ^2
π½
π= π1 β π(π1 + 3π2) + π
3
3= 20374.64ππ^3
πΎπ£ = 1 β πΎπΉ = .40
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 155.33πππ2
(π)ππ = .75 β 4 β π β βπβ²π = 222.486π
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TWO-WAY SLAB DESIGN Pg 138
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Case B Interior Munb 24.23 k-ft
Yf 0.6 Yf*Munb 14.538 Effective
Width 5.25 ft Min Munb 46.2875 k-ft Ac 1668.106 in^2 J/c 20374.64 in^3 Yv 0.4 Vmax 155.3302 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 391 psf
Slab thickness 13 inch
db 1.128 inch
qdu 192.5 lbs
qlu 100 lbs
One way Shear Check
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 11.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 263.1542 ππ‘
2
ππ’ = π΄π‘ β ππ’ = 102.9k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 390π
Bw=l2
Case C Edge d 11.686 in
At 263.1542 ft^2 Vu 102.8933 kip (F)Vc 389.9956 kip OK
Two-way Slab Shear
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TWO-WAY SLAB DESIGN Pg 139
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
π΄π‘ = (π12+π
2) β π2 β π1 β π2 = 330.1ππ‘^2
ππ’ = π΄π‘ β ππ’ = 129.07π b=1 , as=30
π1 = π +π
2= 29.84ππ , π2 = π + π = 35.686
π0 = 2 β π1 + π2 = 95.372
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=247.644k
Case C Edge At 330.1043 ft^2
Vu 129.0708 kip b 1 as 30 b1 29.843 in b2 35.686 in b0 95.372 in (F)Vc 247.9644 kip OK
Munb = 186.35k-ft
πΎπΉ =1
1 +23 ββπ1π2
= .62
πΎπΉ β ππ’ππ = 115.77π β ππ‘ Effective Width = 3h+c = 5.25ft
π΄π = (π1 + 2π2) β π = 1114.517ππ^2
π½
π= 2 β π1
2 β π(π1 β 2π2) + π3(2π1 + π2)
6π1= 12616.1ππ^3
πΎπ£ = 1 β πΎπΉ = .378
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 182.94π/ππ2
(π)ππ = .75 β 4 β π β βπβ²π = 222.486π
![Page 141: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/141.jpg)
TWO-WAY SLAB DESIGN Pg 140
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Case C Edge Munb 186.35 k-ft
Yf 0.621253 Yf*Munb 115.7705 Effective Width 5.25 ft Min Munb 46.2875 k-ft Ac 1114.517 in^2 J/c 12616.1 in^3 Yv 0.378747 Vmax 182.9415 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 391 psf
Slab thickness 13 inch
db 1.128 inch
qdu 192.5 lbs
qlu 100 lbs
One way Shear Check
π = ππππ π‘βππππππ π β .75 β .5(ππ) = 11.686 ππ
π΄π‘ = (π12βπ
2β π) β π2 = 263.1542 ππ‘
2
ππ’ = π΄π‘ β ππ’ = 102.9k
FVc = (phi) β 2 β βπβ²π β ππ€ β π = 390π
Bw=l2
Case D Corner d 11.686 in
At 263.1542 ft^2 Vu 102.8933 kip (F)Vc 389.9956 kip OK
![Page 142: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/142.jpg)
TWO-WAY SLAB DESIGN Pg 141
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Two-way Slab Shear
π΄π‘ = (π12) βπ22β (π +
π
2)2
= 150.07ππ‘^2
ππ’ = π΄π‘ β ππ’ = 58.67π b=1 , as=20
π1 = π +π
2= 29.843ππ , π2 = π +
π
2= 29.843ππ
π0 = π1 + π2 = 59.686ππ
(π)ππ = (π)min (2 +4
π½, 2 +
πΌπ βπ
π0, 4) π β βπβ²π β π0 β π=155.1819k
Case D Corner At 150.0652 ft^2
Vu 58.67551 kip b 1 as 20 b1 29.843 in b2 29.843 in b0 59.686 in (F)Vc 155.1819 kip OK
Munb=full slab moment (col+mid) for and edge column=End span exterior M- = 186.35k-ft
πΎπΉ =1
1 +23 ββπ1π2
= .6
πΎπΉ β ππ’ππ = 111.81π β ππ‘
Effective Width = 3h+c = 5ft
π΄π = (2π1 + π2) β π = 697.5ππ^2
π½
π= π12 β π(π1 + 4π2) + π
3(π1 + π2)
6π1= 22946.5ππ^3
πΎπ£ = 1 β πΎπΉ = .38
ππππ₯ =ππ’
π΄π+ πΎπ£ β ππ’ππ β 12
π½/π= 123.1π/ππ2
(π)ππ = .75 β 4 β π β βπβ²π = 222.486π
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TWO-WAY SLAB DESIGN Pg 142
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Case D Corner Munb 186.35 k-ft
Yf 0.6 Yf*Munb 111.81 Effective
Width 5.25 ft Min Munb 46.2875 k-ft Ac 697.4906 in^2 J/c 22946.53 in^3 Yv 0.4 Vmax 123.1048 Kips/in^2 (F)Vu 222.486 kips OK
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CONTINUOUS L-BEAM DESIGN Pg 143
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
π€π’ = 10.3 π/ππ‘
Use the following maximum deflections:
πΉππ ππππ π‘πππ(ππππ) + ππππππππ‘π(ππππ + πππ£π + π€πππ) =πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ
πΉππ ππππππππ‘π πππππππ‘πππ (πππ£π + π€πππ ππππ¦) =πΏ
360=25 ππ‘ β 12 ππ
360= .833 ππ
Use ASTM A615 Gr 60 rebar and 5,500 psi concrete.
GRAVITY MOMENT DESIGN:
From two-way slab design/analysis: ππ’ = 230 π β ππ‘
Estimate the beam sizeβ = 24 ππ, ππ€ππ = 16 ππ, π = 21.5 ππ, π‘ = 12 ππ.
Effective Flange Width Conditions:
ππ =πΏ
12= 25 ππ, ππ = 6π‘ + ππ€ = 88 ππ, ππ = π = 25 ππ.
Select smallest value of 25in.
Solve for steel percentage using the following quadratic equation:
ππ = πππ¦ππ2 (1 β
πππ¦
1.7πβ²π) β π = .0065
Check against the minimum percentage:
ππππ = max(200
ππ¦,3βπβ²π
ππ¦) = 0.00371 < .0065 β ππΎ
Assume Lever Arm (z):
π§ = 0.9π = 19.35 ππ π§ = π βπ‘
2= 15.5
Select Largest Value: 19.35 in.
Find the required area of steel:
π΄π = ππ/ππ¦π§ = (230 β 12)/(60,000 β 19.35) = 2.38 ππ2
Compute π΄π:
0.85πβ²ππ΄π = π΄π ππ¦ β π΄π =
π΄π ππ¦
0.85πβ²π
= 30.51 ππ2
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CONTINUOUS L-BEAM DESIGN Pg 144
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
π΄π < (π‘ β ππ) < 108 ππ2 ππΎ
Compute a:
a=π΄π/ππ = 1.22 ππ
Compute z:
π§ = π βπ
2= 20.89 ππ
Calculate π΄π with revised z:
π΄π = ππ/ππ¦π§ = (230 β 12)/(60,000 β 20.89) = 2.20 ππ2
Compute π΄π π€ππ‘β πππ£ππ ππ π΄π :
0.85πβ²ππ΄π = π΄π ππ¦ β π΄π =
π΄π ππ¦
0.85πβ²π
= 28.26 ππ2
Compute a:
a=π΄π/ππ = 1.13 ππ
Compute z:
π§ = π βπ
2= 20.93 ππ
Calculate π΄π with revised z:
π΄π = ππ/ππ¦π§ = (230 β 12)/(60,000 β 20.93) = 2.20 ππ2
OK (Close to previous Value)
Checking Minimum Reinforcing:
π΄π πππ =3βπβ²πππ€π
ππ¦=3 β β5500 β 16 β 21.5
60000 = 1.28 ππ2
But not less than:
π΄π πππ =200ππ€π
ππ¦=200 β 16 β 21.5
60000 = 1.15 ππ2 < 2.2 ππ2 β ππΎ
Or:
π΄π πππ = ππππππ€π = 0.0033 β 16 β 21.5 = 1.14 ππ2 < 2.2 ππ2 β ππΎ
Check steel strain:
Compute c:
π½1 = .775 @ 5500 ππ π, π =π
π½1= 1.57ππ
Compute Steel Strain ππ‘:
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CONTINUOUS L-BEAM DESIGN Pg 145
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππ‘ = 0.003π β π
π= 0.003
21.5 β 1.57
1.57= .038 > .005 β ππΎ
Check Ξ¦:
Ξ¦ = 0.9 πππ ππ‘ > .005
Select Reinforcing:
π΄π πππβ²π: 2.20 ππ2
Bar Design:
4#9 bars with π΄π : 4.00 ππ2 with diameter 1.128 in
Stirrup Design:
#3 with diameter 0.375 in
Check fit of bars within beam width:
ππππ = 1.5 + .375 + 2(. 375) + 4(1.128) + 3(. 375) + .375 + 1.5 = 15.76 ππ < 16 ππ β ππΎ
Verify capacity:
Ξ¦ππ = Ξ¦π΄π ππ¦ (π βπ
2) = .9(4)(60) (21.5 β
1.13
2) = 376.83 π β ππ‘ > ππ’ β ππΎ
Check shear:
Find shear load at support:
ππ’ = 70.45 π ππππ πππ π’ππ π΄ππππ¦π ππ At a distance d from the support:
ππ’π = 70.45 β21.5
12β 10.3 = 52.00 π
Required Vn:
ππ =ππ’πΞ¦=52.00
0.75= 69.33 π
Finding shear strength of concrete:
ππ = 2πβπβ²πππ = 2(1)(β5500)(16)(21.5) = 51.02 π
Vn exceeds Vc, so reinforcement is required.
Find Vs:
ππ = ππ β ππ = 69.33 β 51.00 = 18.30 π
Use #3 stirrups (π΄π£ = 2 β .11 = .22 ππ2)
Maximum stirrup spacing:
π =π΄π£ππ¦π
ππ =. 22 β 60 β 21.5
18.30= 15.5 ππ
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CONTINUOUS L-BEAM DESIGN Pg 146
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Use 7 inches.
Check max spacing limits:
π πππ₯ = min(π΄π£ππ¦
. 75πβπβ²π,π΄π£ππ¦
50π) = min(14.83,16.5) = 14.83 ππ
Based on beam geometry:
4βπβ²πππ = 102.05π
πβπππππππ, π πππ₯ = min (π
2, 24) = 11 ππ πππππ ππ‘ ππ π ππππππ π‘βππ 15.5ππ π’π π 11ππ π πππππππ
Check max Vs:
ππ ,πππ₯ = 8βπβ²πππ = 204.09π > 51.02π β ππΎ
Verify capacity:
ππ =π΄π£ππ¦π
π = 18.92 π > 18.30π β ππΎ
Ξ¦ππ = .75(51.02 + 18.30) = 52.00 πππ β₯ ππ’π β ππΎ
Check deflection:
Due to uniform load,
β=π€π3
192πΈππΌπ
Finding Ec:
πΈπ = 57000βπβ²π = 4227 ππ π
Finding modulus of rupture:
ππ = 7.5πβπβ²π = 556 ππ π=.556ksi
Gross moment of inertia:
πΌπ = 22809.6 ππ4
Distance from neutral axis of gross (uncracked section to tension face):
π¦π‘ =β
2=24
2= 12 ππ
Cracking moment:
πππ =πππΌπ
π¦π‘= 1057.25 π β ππ
Service moment:
ππ =π€π’π
2
8=7.23 β 252
8= 4781.25 π β ππ, 7.23 = π΄ππ· π€π’
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CONTINUOUS L-BEAM DESIGN Pg 147
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Transform steel to concrete:
π =πΈπ πΈπ=29000
4227= 6.86
Gravity: π΄π = 4.00 ππ2, π = 21.5 ππ
Solve for x:
ππ₯π₯
2= ππ΄π (π β π₯) β π₯ = 7.0422 ππ
Find Icr:
πΌππ =ππ₯3
3+ ππ΄π (π β π₯)
2 = 7598.58 ππ4
Find Ie:
πΌπ = [(πππππ)3
] πΌπ + [1 β (πππππ)3
] πΌππ = 7649.88 ππ4
Checking deflection:
β=π€π3
192πΈππΌπ= 0.03 ππ
πΉππ πππ£π πππ π€πππ ππππ¦, β< .833 ππ β ππΎ
For long term deflection:
Gravity dead only: Ξπ· = 1.2 ππ
πΞ =π
1 + 50πβ²= 2.0 πππ ππ πππππππ π πππ π π‘πππ
ΞπΏπ = πΞΞπ· + ΞπΌππΊ = 0.12 ππ
Ξπ = (1 + ππ·)Ξπ· + ΞπΌππΊ = 0.17 ππ < 1.2 ππ β ππΎ
MOMENT DIAGRAM:
Note: Moment values in analysis were taken from slab analysis.
SHEAR DIAGRAM:
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CONTINUOUS L-BEAM DESIGN Pg 148
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Note: Highest shear taken for use in analysis for worst case.
DEFLECTION DIAGRAM:
Note: Deflection in analysis and Visual Analysis both pass. Visual Analysis doesnβt 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).
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CONTINUOUS L-BEAM DESIGN Pg 149
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
π€π’ = 9.78 π/ππ‘
Use the following maximum deflections:
πΉππ ππππ π‘πππ(ππππ) + ππππππππ‘π(ππππ + πππ£π + π€πππ) =πΏ
240=25 ππ‘ β 12 ππ
240= 1.25 ππ
πΉππ ππππππππ‘π πππππππ‘πππ (πππ£π + π€πππ ππππ¦) =πΏ
360=25 ππ‘ β 12 ππ
360= .833 ππ
Use ASTM A615 Gr 60 rebar and 5,500 psi concrete.
GRAVITY MOMENT DESIGN:
From two-way slab design/analysis: ππ’ = 217 π β ππ‘
Estimate the beam size β = 24 ππ, ππ€ππ = 16 ππ, π = 21.5 ππ, π‘ = 12 ππ.
Effective Flange Width Conditions:
ππ =πΏ
12= 25 ππ, ππ = 6π‘ + ππ€ = 88 ππ, ππ = π = 25 ππ.
Select smallest value of 25in.
Solve for steel percentage using the following quadratic equation:
ππ = πππ¦ππ2 (1 β
πππ¦
1.7πβ²π) β π = .0061
Check against the minimum percentage:
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CONTINUOUS L-BEAM DESIGN Pg 150
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππππ = max(200
ππ¦,3βπβ²π
ππ¦) = 0.00371 < .0061 β ππΎ
Assume Lever Arm (z):
π§ = 0.9π = 19.35 ππ π§ = π βπ‘
2= 15.5
Select Largest Value: 19.35 in.
Find the required area of steel:
π΄π = ππ/ππ¦π§ = (217 β 12)/(60,000 β 19.35) = 2.24 ππ2
Compute π΄π:
0.85πβ²ππ΄π = π΄π ππ¦ β π΄π =
π΄π ππ¦
0.85πβ²π
= 28.79 ππ2
π΄π < (π‘ β ππ) < 108 ππ2 ππΎ
Compute a:
a=π΄π/ππ = 1.15 ππ
Compute z:
π§ = π βπ
2= 20.92 ππ
Calculate π΄π with revised z:
π΄π = ππ/ππ¦π§ = (217 β 12)/(60,000 β 20.92) = 2.24 ππ2
Compute π΄π π€ππ‘β πππ£ππ ππ π΄π :
0.85πβ²ππ΄π = π΄π ππ¦ β π΄π =
π΄π ππ¦
0.85πβ²π
= 26.62 ππ2
Compute a:
a=π΄π/ππ = 1.06 ππ
Compute z:
π§ = π βπ
2= 20.97 ππ
Calculate π΄π with revised z:
π΄π = ππ/ππ¦π§ = (230 β 12)/(60,000 β 20.93) = 2.07 ππ2
OK (Close to previous Value)
Checking Minimum Reinforcing:
π΄π πππ =3βπβ²πππ€π
ππ¦=3 β β5500 β 16 β 21.5
60000 = 1.28 ππ2
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CONTINUOUS L-BEAM DESIGN Pg 151
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
But not less than:
π΄π πππ =200ππ€π
ππ¦=200 β 16 β 21.5
60000 = 1.15 ππ2 < 2.07 ππ2 β ππΎ
Or:
π΄π πππ = ππππππ€π = 0.0033 β 16 β 21.5 = 1.14 ππ2 < 2. .07 ππ2 β ππΎ
Check steel strain:
Compute c:
π½1 = .775 @ 5500 ππ π, π =π
π½1= 1.49ππ
Compute Steel Strain ππ‘:
ππ‘ = 0.003π β π
π= 0.003
21.5 β 1.57
1.57= .04 > .005 β ππΎ
Check Ξ¦:
Ξ¦ = 0.9 πππ ππ‘ > .005
Select Reinforcing:
π΄π πππβ²π: 2.07 ππ2
Bar Design:
3#9 bars with π΄π : 3.00 ππ2 with diameter 1.128 in
Stirrup Design:
#3 with diameter 0.375 in
Check fit of bars within beam width:
ππππ = 1.5 + .375 + 2(. 375) + 3(1.128) + 2(. 375) + .375 + 1.5 = 12.63 ππ < 16 ππ β ππΎ
Verify capacity:
Ξ¦ππ = Ξ¦π΄π ππ¦ (π βπ
2) = .9(3)(60) (21.5 β
1.13
2) = 283.06 π β ππ‘ > ππ’ β ππΎ
Check shear:
Find shear load at support:
ππ’ = 60 π ππππ πππ π’ππ π΄ππππ¦π ππ At a distance d from the support:
ππ’π = 60 β21.5
12β 9.78 = 42.48 π
Required Vn:
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CONTINUOUS L-BEAM DESIGN Pg 152
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππ =ππ’πΞ¦=42.48
0.75= 56.64 π
Finding shear strength of concrete:
ππ = 2πβπβ²πππ = 2(1)(β5500)(16)(21.5) = 51.02 π
Vn exceeds Vc, so reinforcement is required.
Find Vs:
ππ = ππ β ππ = 56.64 β 51.02 = 5.61 π
Use #3 stirrups (π΄π£ = 2 β .11 = .22 ππ2)
Maximum stirrup spacing:
π =π΄π£ππ¦π
ππ =. 22 β 60 β 21.5
5.61= 50.56 ππ
Use 7 inches.
Check max spacing limits:
π πππ₯ = min(π΄π£ππ¦
. 75πβπβ²π,π΄π£ππ¦
50π) = min(14.83,16.5) = 14.83 ππ
Based on beam geometry:
4βπβ²πππ = 102.05π
πβπππππππ, π πππ₯ = min (π
2, 24) = 11 ππ πππππ ππ‘ ππ π ππππππ π‘βππ 50.6ππ π’π π 11ππ π πππππππ
Check max Vs:
ππ ,πππ₯ = 8βπβ²πππ = 204.09π > 51.02π β ππΎ
Verify capacity:
ππ =π΄π£ππ¦π
π = 5.68π > 5.61π β ππΎ
Ξ¦ππ = .75(51.02 + 5.61) = 42.48 πππ β₯ ππ’π β ππΎ
Check deflection:
Due to uniform load,
β=π€π3
192πΈππΌπ
Finding Ec:
πΈπ = 57000βπβ²π = 4227 ππ π
Finding modulus of rupture:
ππ = 7.5πβπβ²π = 556 ππ π=.556ksi
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CONTINUOUS L-BEAM DESIGN Pg 153
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Gross moment of inertia:
πΌπ = 22809.6 ππ4
Distance from neutral axis of gross (uncracked section to tension face):
π¦π‘ =β
2=24
2= 12 ππ
Cracking moment:
πππ =πππΌπ
π¦π‘= 1057.25 π β ππ
Service moment:
ππ =π€π’π
2
8=7.31 β 252
8= 6853.13 π β ππ, 7.31 = π΄ππ· π€π’
Transform steel to concrete:
π =πΈπ πΈπ=29000
4227= 6.86
Gravity: π΄π = 3.00 ππ2, π = 21.5 ππ
Solve for x:
ππ₯π₯
2= ππ΄π (π β π₯) β π₯ = 6.2613 ππ
Find Icr:
πΌππ =ππ₯3
3+ ππ΄π (π β π₯)
2 = 6088.4 ππ4
Find Ie:
πΌπ = [(πππππ)3
] πΌπ + [1 β (πππππ)3
] πΌππ = 6149.79 ππ4
Checking deflection:
β=π€π3
192πΈππΌπ= 0.04 ππ
πΉππ πππ£π πππ π€πππ ππππ¦, β< .833 ππ β ππΎ
For long term deflection:
Gravity dead only: Ξπ· = 1.2 ππ
πΞ =π
1 + 50πβ²= 2.0 πππ ππ πππππππ π πππ π π‘πππ
ΞπΏπ = πΞΞπ· + ΞπΌππΊ = 0.15 ππ
Ξπ = (1 + ππ·)Ξπ· + ΞπΌππΊ = 0.21 ππ < 1.2 ππ β ππΎ
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CONTINUOUS L-BEAM DESIGN Pg 154
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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 doesnβt 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).
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CONTINUOUS L-BEAM DESIGN Pg 155
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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).
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CONCRETE COLUMN DESIGNS Pg 156
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
CONCRETE COLUMN DESIGN:
SPANNING GREEN ROOF AND THIRD FLOOR:
Design the continuous column with fβc = 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:
ππ =ππ0.65
=885.82π
0.65= 1362.8π
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:
πππ₯ +πππ¦ = 190.2π β ππ‘ + 190.2 π β ππ‘ = 380.3π β ππ‘
DETERMINING STEEL REQUIRED:
ππ₯ = ππ¦ =(12ππππ‘) (380.3π β ππ‘)
1362.8π= 3.35 ππ
πΎ =20ππ
24ππ= 0.83
Obtaining rg from Graph 8 Appendix A:
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CONCRETE COLUMN DESIGNS Pg 157
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
π π =πππ
πβ²ππ΄πβ
=(1362.8π)(3.35ππ)
(4000ππ π)(24ππ β 24ππ)(24ππ)= 0.08
πΎπ =πππβ²ππ΄π=
(1362.8π)
(4000ππ π)(24ππ β 24ππ)= 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
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SIMPLE SHEAR CONNECTIONS DESIGNS Pg 158
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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 βIFβ 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.
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SIMPLE SHEAR CONNECTIONS DESIGNS Pg 159
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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.
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ROOF CONNECTIONS DESIGN Pg 160
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
π€π
2=(282.55
ππππ‘β 50 ππ‘)
2= 7.06 π
For .81-in. flange thickness, π·πππ = 5 from AISC Specification Table J2.4.
Try 5/16-in fillet welds, π· = 5. From AISC equation 8-2:
πΏπππ =π
2π·(0.928)=
7.06 ππππ
2(5)(0.928)= .761 ππ
Weld the minimum length required by Vulcraft:
πΏ = 1 ππ > .761 ππ β ππΎ
Check the maximum size of fillet weld:
πππ₯ π ππ§π = .375 β .0625 = .3125 ππ = .3125 ππ β ππΎ
Check the minimum length:
πΏπππ = 4(. 1875 ππ) = .75 ππ < 1 ππ β ππΎ
Check the weld length to weld size ratio: πΏ
π€=
1
. 3125= 3.2 < 100 β πΆβπππ π½ ππππ‘ππ πππ‘ πππππ π πππ¦
Check Base Metals by the largest effective weld size. (π =π·
16)
For the Perimeter Beam:
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 65 ππ π
. 707 β 70 ππ π. 81 ππ = 2.63 ππ > .3125 ππ β ππΎ
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 π = 61.2 ππππ .
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ROOF CONNECTIONS DESIGN Pg 161
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Use ΒΎ-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 ππ = .5 ππ. Use AISC Table 10-6 with an angle length of
8 inches and thickness of 5/8-in.
π΄ππππ ππ‘πππππ‘β: π
Ξ©= 72 ππππ > 61.2 ππππ β ππΎ
ππππ ππ‘πππππ‘β, 8π₯4 π πππ‘ πππππ π ππ§π, .5 ππ ππππππ‘ π€ππππ : π
Ξ©= 71.2 ππππ > 61.2 ππππ
β ππΎ
π΅πππ‘ ππ‘πππππ‘β, πΆππππππ‘πππ ππ¦ππ πΈ:π
Ξ©= 71.6 ππππ > 61.2 ππππ β ππΎ
THEREFORE USE AN L8X4X5/8 WITH TWO ΒΎ-IN DIAMETER GROUP A BOLTS
FROM GRIDER TO SEAT AND 2 ΒΌ-INCH FILLET WELDS FROM SEAT TO
COLUMN.
.
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ROOF CONNECTIONS DESIGN Pg 162
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
πππππ€πππ ππππβπππ ππππ ππ‘ 53 ππ‘ = (7.5 ππ‘)(7.8 ππ π) + (7.5 ππ‘)(8.9 ππ π) = 125 πππ
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 ππ’ =π€π
2 and ππ’ =
π€π2
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
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ROOF CONNECTIONS DESIGN Pg 163
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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.
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LATERAL STEEL ANALYSIS Pg 164
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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 ππ’ =π€π
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]
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LATERAL STEEL ANALYSIS Pg 165
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Fixed Supports:
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LATERAL STEEL ANALYSIS Pg 166
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Member Releases:
Note: Moment releases are shown as circles on the ends of each member.
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LATERAL STEEL ANALYSIS Pg 167
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Member Materials:
Note: The part of the model not seen in this picture follows the same settings.
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LATERAL STEEL ANALYSIS Pg 168
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Member Sections:
Note: The members not show in this picture match the sections designated in the AutoCAD
drawings and previous designs.
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LATERAL STEEL ANALYSIS Pg 169
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Rigid Diaphragms:
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LATERAL STEEL ANALYSIS Pg 170
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Node Locations:
Note: Nodes not found in this picture follow the locations designated in the AutoCAD drawings.
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LATERAL STEEL ANALYSIS Pg 171
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Picture View:
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LATERAL STEEL ANALYSIS Pg 172
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Windward Wall Loads:
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LATERAL STEEL ANALYSIS Pg 173
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Leeward Wall Loads:
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LATERAL STEEL ANALYSIS Pg 174
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Sidewall Loads
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LATERAL STEEL ANALYSIS Pg 175
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
More sidewall Loads:
Note: Loads not seen in this picture match the loads in the previous photograph.
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LATERAL STEEL ANALYSIS Pg 176
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Dead Loads:
Note: These loads are the same in the other corner of the building.
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LATERAL STEEL ANALYSIS Pg 177
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Live loads:
Note: Loads are the same in the other corner of the building.
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LATERAL STEEL ANALYSIS Pg 178
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Live roof load:
Note: Loads are same in the other corner of building.
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LATERAL STEEL ANALYSIS Pg 179
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Roof Wind Load:
Note: Loads are the same in the other corner of the building.
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APPROXIMATE SECOND ORDER ANALYSIS Pg 180
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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: πΏ = 100 ππ‘, π΅ = 150 ππ‘ Story height: β = 15 ππ‘ Floor and Roof Loads:
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APPROXIMATE SECOND ORDER ANALYSIS Pg 181
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
π€ππππ πππππ = π€ππ = 32.5 ππ π
π€πππ£π πππππ = π€π = 65 ππ π
π€π€πππ = π€π€ = 16 ππ π
π€πππ£π ππππ = π€ππ = 20 ππ π
π€ππππ ππππ = π€ππ = 30 ππ π
Story loads:
ππ·π π‘πππ¦ = πΏπ΅(3π€ππ + π€ππ) = (100)(150)(3 β 32.5 + 30) = 1913 πππ
ππΏππ π‘πππ¦ = πΏπ΅π€ππ = (100)(150)(20) = 300 πππ
ππΏπ π‘πππ¦ = πΏπ΅3π€π = (100)(150)(3)(65) = 2925 πππ
ππ€π π‘πππ¦ = πΏπ΅π€π€ = (100)(150)(16) = 240 πππ
Bottom column axial loads when translation is restrained (results obtained from visual analysis):
ππ·ππ‘ = 231.3 πππ
ππΏππ‘ = 341 πππ
ππΏπππ‘ = 33.1 πππ
ππππ‘ = 9.85 πππ Bottom column bending moment when translation is restrained:
ππ·ππ‘ = 0 πππ β ππ
ππΏππ‘ = 0 πππ β ππ
ππΏπππ‘ = 0 πππ β ππ
ππππ‘ = 0 πππ β ππ Artificial Joint Reactions due to the restraints (rollers in x) provided at each floor to restrain
translation (nt):
π΄π½π π· = (
1.333.133.9510.48
)πππ
π΄π½π πΏπ = (
1.57β.230.201.17
)πππ
π΄π½π πΏ = (
β2.306.716.9916.53
)πππ
π΄π½π π = (
21.1928.4224.1319.92
)πππ
Bottom column axial loads when translation is restrained (results obtained from visual analysis):
ππ·ππ‘ = 19.96 πππ
ππΏππ‘ = 24.67 πππ
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APPROXIMATE SECOND ORDER ANALYSIS Pg 182
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππΏπππ‘ = 4.75 πππ
ππππ‘ = 144.7 πππ Bottom column bending moment when translation is restrained:
ππ·ππ‘ = 3.15 πππ β ππ
ππΏππ‘ = 1.25 πππ β ππ
ππΏπππ‘ = 1.02 πππ β ππ
ππππ‘ = 98.52 πππ β ππ Load factor matrix (LRFD load conditions ASCE 7-10 2.3):
πΎπ π‘π =
(
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:
ππ’π π‘πππ¦ = πΎπ π‘π (
19132925300240
) =
(
191348382213433220574440410712921148)
πππ
πππ‘ = πΎπ π‘π (
231.334133.19.85
) =
(
231.3572.3264.4511.88237.21516.31487.05144,69138.78)
πππ
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APPROXIMATE SECOND ORDER ANALYSIS Pg 183
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πππ‘ = πΎπ π‘π (
19.9624.674.75144.7
) =
(
19.9644.6324.7142.03106.78107.1438.4698.8011.98 )
πππ
πππ‘ = πΎπ π‘π (
0000
) =
(
000000)
πππ β ππ
πππ‘ = πΎπ π‘π (
3.151.251.0298.52
) =
(
3.154.404.174.8562.2649.194.0961.001.89 )
πππ β ππ
Factored lateral loads on the βltβ frame. Note these are the cumulative loads on the bottom
column since we are examining the bottom column:
π1 = πΎπ π‘π (
1.33β2.31.5721.19
) =
(
1.33β.972.90. 7814.0410.32β.4013.51. 80 )
πππ
π2 = πΎπ π‘π (
3.136.71β.2328.42
) =
(
3.139.842.907.9920.1820.788.1618.931.88 )
πππ
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APPROXIMATE SECOND ORDER ANALYSIS Pg 184
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
π3 = πΎπ π‘π (
3.956.99. 2024.13
) =
(
3.9510.944.159.3418.4320.209.1916.852.37 )
πππ
π4 = πΎπ π‘π (
10.4816.531.1719.92
) =
(
10.4827.0111.6523.7622.4332.7222.8818.246.29 )
πππ
π»π’βππ =
(
18.8946.8221.641.8775.0984.0239.8467.5311.33)
πππ
Deflection of the joint at the top of the bottom column due to each load case (found in Visual
Analysis):
β= (
. 0290
. 0422
. 0044
. 1508
) ππ
Factored deflections corresponding to loads Hu:
βπ»= πΎπ π‘πβ=
(
. 029. 0712. 0334. 0640. 1195. 1318. 0607. 1079. 0174)
ππ
ASD load factor: πΌ = 1.6
Factored axial load in moment frames. (This structure has no moment frames):
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APPROXIMATE SECOND ORDER ANALYSIS Pg 185
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππ’ππ =
(
000000)
πππ
Rm factor:
π π = 1 β .15ππ’ππ
ππ’π π‘πππ¦=
(
111111)
Buckling load per story:
πππ π‘πππ¦ = (π ππ»π’β
βπ») =
(
117248.3118365.2116407.2117759.4113106.3114746.6118141.7112654.3117206.9)
πππ
P-Big Delta Amplification factor B2:
π΅2 =
(
ππ = max (1,1
1 βπΌππ’π π‘πππ¦ππππ π‘πππ¦π )
=
(
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.
πΆπ = 0.6 Required axial design strength:
ππ = (πππ‘ + π΅2πππ‘) =
(
250619286555341627527241140)
πππ
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APPROXIMATE SECOND ORDER ANALYSIS Pg 186
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Steel properties and column section properties (W14X109):
πΈπ = 29000 ππ π πΌ = 1240 ππ4
Effective length factor: π = 1.0 (pin-pin connection)
Column buckling load:
ππ1 =ππΈπ πΌ
(πβ)2= 3486.80 πππ
P-small delta Amplification factor B1:
π΅1 = (ππ = max (1,πΆπ
1 βπΌπππππ1
) =
(
111111)
Required bending moment design strength:
ππ = (π΅1πππ‘ + π΅2πππ‘) =
(
3.244.714.305.1464.1352.634.3462.221.93 )
πππ β ππ
Checking Member:
ππ = 808 πππ
πΏπ = 15 ππ‘, πΏπ = 48.5 ππ‘, πΏπ = 13.2 ππ‘ β ππππ πΌπΌ ππ
Ξ©= 479 π β ππ‘,
π΅πΉ
Ξ©= 5.01 πππ
ππΞ©= πΆπ [
ππ
Ξ©βπ΅πΉ
Ξ©(πΏπ β πΏπ)] = 469.98 π β ππ‘
ππ
ππ=627
808= 0.78 > 0.2
ππππ+8
9(ππππ) =
627
808+8
9(64.13/12
469.98) = 0.79 < 1.0 β ππΎ!
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BRACED FRAME DESIGN Pg 187
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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.
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BRACED FRAME DESIGN Pg 188
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
DESIGN OF BRACE 1:
Draw the brace in AutoCAD. Use the following properties for all members involved:
Beam:
W18x55 π = 18.1 ππ
ASTM A992 π‘π = .63 ππ
πΉπ¦ = 50 ππ π
πΉπ’ = 65 ππ π
Column:
W14X109 π = 14.3 ππ
ASTM A992 π‘π = .595 ππ
πΉπ¦ = 50 ππ π ππ = 14.6 ππ
πΉπ’ = 65 ππ π π‘π€ = .525 ππ
Diagonal Brace:
2L6x6x1/2x3/8 π‘ = .5 ππ
ASTM A36 π΄ = 11.5 ππ2
πΉπ¦ = 36 ππ π οΏ½οΏ½ = 1.67 ππ
πΉπ’ = 58 ππ π
Clip Angle:
2L6X6X3/8
ASTM A36
πΉπ¦ = 36 ππ π
πΉπ’ = 58 ππ π
Gusset Plate:
π‘ = .375 ππ ASTM A36
πΉπ¦ = 36 ππ π
πΉπ’ = 58 ππ π
Solve for the dimensions of the gusset plate/center of gravity for beam to gusset and column to
gusset connections:
ππ =18.1
2= 9.05 ππ (πππ π18π55)
ππ =14.3
2= 7.15 ππ (πππ π14π109)
π½ = 6 ππ
π = 59β Solving for Alpha:
πΌ = ππ tan π β ππ + π½ tanπ
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BRACED FRAME DESIGN Pg 189
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πΌ = 17.9 ππ
Finding loads per Uniform Force Method (AISC Part 13)
π = β(πΌ + ππ)2 + (π½ + ππ)2 = 29.22 ππ
ππ =π½
ππ =
6 ππ
29.22 ππ92.55 πππ = 19 πππ
π»π =ππππ =
7.15 ππ
29.22 ππ92.55 πππ = 22.65 πππ
ππ =ππππ =
9.05 ππ
29.22 ππ92.55 πππ = 28.66 πππ
π»π =πΌ
ππ =
17.9 ππ
29.22 ππ92.55 πππ = 56.70 πππ
Design of welds connecting Diagonal brace to the Gusset Plate:
For Β½-in. angles, π·πππ = 3 from AISC Specification Table J2.4.
Try 3/16-in fillet welds, π· = 3. From AISC equation 8-2:
πΏπππ =π
4π·(0.928)=
92.55 πππ
4(3)(0.928)= 8.31 ππ
Weld the entire length of the angle minus some clearance.
πΏ = 19 ππ > 8.31 ππ β ππΎ
Check the maximum size of fillet weld:
πππ₯ π ππ§π = .375 β .0625 = .3125 ππ > .1875 ππ β ππΎ
Check the minimum length:
πΏπππ = 4(. 1875 ππ) = .75 ππ < 19 ππ β ππΎ
πΏπππ = 6 ππ < 19 ππ β ππΎ
Check the weld length to weld size ratio: πΏ
π€=
19
. 1875= 101.33 > 100 β πΆβπππ π½ ππππ‘ππ
π½ = 1.2 β .002 βπ
π€= 1.2 β .002 β 101.33 = .997
Recheck strength: π πΞ©= (. 997)(4)(3)(. 928)(19) = 211 πππ > 92.55 πππ β ππΎ
Check Base Metals by the largest effective weld size (π =π·
16):
For the Diagonal brace:
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 58 ππ π
. 707 β 70 ππ π. 5 ππ = 1.17 ππ > .1875 ππ β ππΎ
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BRACED FRAME DESIGN Pg 190
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
For the Gusset plate:
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 58 ππ π
. 707 β 70 ππ π. 375 ππ = .879 ππ > .1875 ππ β ππΎ
Tensile Yielding of Diagonal Brace:
π πΞ©=πΉπ¦π΄π
Ξ©=36 ππ π β 11.5 ππ2
1.67= 247.9 πππ > 92.55 πππ β ππΎ
Tensile Rupture of Diagonal Brace:
π΄π = π΄π = 11.5 ππ2
π = 1 βοΏ½οΏ½
π= 1 β
1.67 ππ
19 ππ= 0.912
π΄π = π΄ππ = 11.5 ππ2 β .912 = 10.48 ππ2
π πΞ©=πΉπ’π΄πΞ©
=58 ππ π β 10.48 ππ2
2= 303.9 ππππ > 92.55 ππππ β ππΎ
Welds connecting Clip Angle to Column
Design weld for ππ + π»π = 19 πππ + 22.65 πππ = 41.65 ππππ For 3/8-in. angles, π·πππ = 3 from AISC Specification Table J2.4.
Try 5/16-in fillet welds, π· = 5. From AISC equation 8-2:
πΏπππ =ππ + π»π2π·(0.928)
=41.65 πππ
2(5)(0.928)= 4.49 ππ
Fillet weld for the entire length of gusset for full length minus some clearance:
πΏ = 9.5 ππ > 4.49 ππ β ππΎ
Check the maximum size of fillet weld:
πππ₯ π ππ§π = .375 β .0625 = .3125 ππ = .3125 ππ β ππΎ
Check the minimum length:
πΏπππ = 4(. 3125 ππ) = 1.25 ππ < 9.5 ππ β ππΎ
πΏπππ = 7.375 ππ < 9.5 ππ β ππΎ
Check the weld length to weld size ratio: πΏ
π€=
9.5
. 3125= 30.4 < 100 β πΆβπππ π½ ππππ‘ππ πππ‘ πππππ π πππ¦
Check Base Metals by the largest effective weld size.
For the Clip Angle:
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BRACED FRAME DESIGN Pg 191
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 58 ππ π
. 707 β 70 ππ π. 375 ππ = .879 ππ > .3125 ππ β ππΎ
For the Column Flange:
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 65 ππ π
. 707 β 70 ππ π. 595 ππ = 1.56 ππ > .3125 ππ β ππΎ
Bolts connecting Clip Angle from Column to Gusset Plate
Use the following load for design: π = βππ2 + π»π
2 = 29.56 ππππ
From AISC Manual Table 7-1, find the minimum number of 1-in. Group A, threads
included, bolts.
ππππ =π ππΞ©
=29.56 ππππ
21.2 ππππ = 1.4 ππππ‘π
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. π πΞ©=. 6πΉπ¦π΄ππ£
Ξ©=. 6 β 36 ππ π β 12 ππ β .375 ππ β 2
1.5= 129.6 ππππ > 29.56 ππππ
β ππΎ
Shear Rupture of Clip Angle from column to Gusset Plate
Following AISC specification J4-2:
π΄ππ£ = 2(. 375 ππ)[12 β 4(1 ππ + .125 ππ)] = 5.625 ππ2
π πΞ©=. 6πΉπ’π΄ππ£Ξ©
=. 6 β 58 ππ π β 5.625 ππ2
2= 97.9 ππππ > 29.56 ππππ
Block Shear Rupture of Clip Angle from column to Gusset Plate
Assume uniform tension stress, so use πππ = 1.
π΄ππ£ = 2(12 ππ β 2.25 ππ). 375 ππ = 7.31 ππ2
π΄ππ£ = 7.31 ππ2 β .375 ππ(3.5)(1.125 ππ)2 = 4.36 ππ2
π΄ππ‘ = [(3 ππ β .375 ππ) β (.5 β 1.125 ππ β .375 ππ)]2 = 1.83 ππ2
From AISC Specification J4-3:
π π = .6πΉπ’π΄ππ£ + πππ πΉπ’π΄ππ‘ β€ .5πΉπ¦π΄ππ£ + πππ πΉπ’π΄ππ‘
π π = .6(58)(4.36) + 1(58)(1.83) β€ .5(36)(7.31) + 1(58)(1.83) π π = 257.9 ππππ β€ 237.7 ππππ π πΞ©=237.7
2= 118.9 ππππ > 29.56 ππππ β ππΎ
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BRACED FRAME DESIGN Pg 192
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Shear Bearing Strength at Bolt Holes
ππ = 2.25 β1.125
2= 1.6875 ππ
Consider deformation at the bolthole at service load a design consideration.
π π = 1.2πππ‘πΉπ’ β€ 2.4ππ‘πΉπ’
π π = 1.2(1.6875 ππ)(. 375 ππ)(58 ππ π)4 β€ 2.4(1 ππ)(. 375 ππ)(58 ππ π)4
π π = 176.18 ππππ β€ 208.8 ππππ π πΞ©=176.18
2= 88.09 ππππ > 29.56 ππππ β ππΎ
Design of welds connecting Beam to the Gusset Plate:
Design connection for ππ +π»π = 28.66 πππ + 56.7 πππ = 85.36 ππππ For 3/8-in. plate thickness, π·πππ = 3 from AISC Specification Table J2.4.
Try 3/16-in fillet welds, π· = 3. From AISC equation 8-2:
πΏπππ =π
2π·(0.928)=85.36 ππππ
2(3)(0.928)= 15.33 ππ
Weld the entire length of the minus some clearance.
πΏ = 33 ππ > 15.33 ππ β ππΎ
Check the maximum size of fillet weld:
πππ₯ π ππ§π = .375 β .0625 = .3125 ππ > .1875 ππ β ππΎ
Check the minimum length:
πΏπππ = 4(. 1875 ππ) = .75 ππ < 33 ππ β ππΎ
πΏπππ = .375 ππ < 33 ππ β ππΎ
Check the weld length to weld size ratio: πΏ
π€=
33
. 1875= 176 > 100 β πΆβπππ π½ ππππ‘ππ
π½ = 1.2 β .002 βπ
π€= 1.2 β .002 β 176 = .848
Recheck strength: π πΞ©= (. 848)(2)(3)(. 928)(33) = 155.8 πππ > 85.36 πππ β ππΎ
Check Base Metals by the largest effective weld size. (π =π·
16)
For the Beam:
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 65 ππ π
. 707 β 70 ππ π. 63 ππ = 1.65 ππ > .1875 ππ β ππΎ
For the Gusset plate:
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BRACED FRAME DESIGN Pg 193
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π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 58 ππ π
. 707 β 70 ππ π. 375 ππ = .88 ππ > .1875 ππ β ππΎ
Gusset Plate Compression
Whitmore section effective width, π€ = 29.42 ππ
According to AISC Part 9, βThe Whitmore section may spread across the joint between
connecting elements, but cannot spread beyond an unconnected edge.β Thus, this section
is ok.
πΏ1 = 6.18 ππ
πΏ2 = 6.7 ππ
πΏ3 = β16 ππ
Unbraced length, π =πΏ1+πΏ2+πΏ3
3= β1.04 ππ
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:
πΏ2 = π = 6.7 ππ
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BRACED FRAME DESIGN Pg 194
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π =π‘
β12=. 375 ππ
β12= .108 ππ
By AISC Manual specification J4: ππ
π=1(6.7 ππ)
.108 ππ= 62.04
From Table 4-22, πΉππ
Ξ©= 17.5 ππ π
π΄π = π€π‘ = 29.42 ππ β .375 ππ = 11.03 ππ2
ππΞ©=πΉπππ΄π
Ξ©=17.5 ππ π β 11.03 ππ2
1.67= 115.6 ππππ > 92.55 ππππ β ππΎ
Tensile Yielding of Gusset Plate
π πΞ©=πΉπ¦π΄π
Ξ©=36 ππ π β 11.03 ππ2
1.67= 237.8 πππ > 92.55 πππ β ππΎ
Tensile Rupture of Gusset Plate:
π΄π = π΄π = 11.03 ππ2
Case 4 in AISC table D3.1: π = 1.0
π΄π = π΄ππ = 11.03 ππ2 β 1 = 11.03 ππ2
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BRACED FRAME DESIGN Pg 195
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π πΞ©=πΉπ’π΄πΞ©
=58 ππ π β 11.03 ππ2
2= 319.9 ππππ > 92.55 ππππ β ππΎ
Connection Geometry
πΌ = 17.9 ππ
π½ = 6 ππ
π = 59Β°
ππ =18.1
2= 9.05 ππ (πππ π18π55)
ππ =14.3
2= 7.15 ππ (πππ π14π109)
πΌ = ππ tan π β ππ + π½ tanπ
17.9 = 9.05tan 59Β° β 7.15 + 6 tan 59Β° β πππ’π β ππΎ
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 π = 13.8 ππ
ASTM A992 π‘π = .385 ππ
πΉπ¦ = 50 ππ π
πΉπ’ = 65 ππ π
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BRACED FRAME DESIGN Pg 196
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Column:
W14X109 π = 14.3 ππ
ASTM A992 π‘π = .595 ππ
πΉπ¦ = 50 ππ π ππ = 14.6 ππ
πΉπ’ = 65 ππ π π‘π€ = .525 ππ
Diagonal Brace:
2L6x6x1/2x3/8 π‘ = .5 ππ
ASTM A36 π΄ = 11.5 ππ2
πΉπ¦ = 36 ππ π οΏ½οΏ½ = 1.67 ππ
πΉπ’ = 58 ππ π
Clip Angle:
2L6X6X3/8
ASTM A36
πΉπ¦ = 36 ππ π
πΉπ’ = 58 ππ π
Gusset Plate:
π‘ = .375 ππ ASTM A36
πΉπ¦ = 36 ππ π
πΉπ’ = 58 ππ π
Solve for the dimensions of the gusset plate/center of gravity for beam to gusset and column to
gusset connections:
ππ =13.8
2= 6.90 ππ (πππ π14π₯30)
ππ =14.3
2= 7.15 ππ (πππ π14π109)
π½ = 6 ππ
π = 59Β° Solving for Alpha:
πΌ = ππ tan π β ππ + π½ tanπ
πΌ = 14.32 ππ
Finding loads per Uniform Force Method (AISC Part 13)
π = β(πΌ + ππ)2 + (π½ + ππ)2 = 25.05 ππ
ππ =π½
ππ =
6 ππ
25.05 ππ101 πππ = 24.2 πππ
π»π =ππππ =
7.15 ππ
25.05 ππ101 πππ = 28.8 πππ
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BRACED FRAME DESIGN Pg 197
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ππ =ππππ =
6.90 ππ
25.05 ππ101 πππ = 27.8 πππ
π»π =πΌ
ππ =
14.32 ππ
25.05 ππ101 πππ = 57.74 πππ
Design of welds connecting Diagonal brace to the Gusset Plate:
For Β½-in. angles, π·πππ = 3 from AISC Specification Table J2.4.
Try 3/16-in fillet welds, π· = 3. From AISC equation 8-2:
πΏπππ =π
4π·(0.928)=
101 πππ
4(3)(0.928)= 9.07 ππ
Weld the entire length of the angle minus some clearance.
πΏ = 16 ππ > 9.07 ππ β ππΎ
Check the maximum size of fillet weld:
πππ₯ π ππ§π = .375 β .0625 = .3125 ππ > .1875 ππ β ππΎ
Check the minimum length:
πΏπππ = 4(. 1875 ππ) = .75 ππ < 16 ππ β ππΎ
πΏπππ = 6 ππ < 16 ππ β ππΎ
Check the weld length to weld size ratio: πΏ
π€=
16
. 1875= 85.33 < 100 β π½ ππππ‘ππ πΆβπππ πππ‘ πππππ π πππ¦
Check Base Metals by the largest effective weld size. (π =π·
16)
For the Diagonal brace:
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 58 ππ π
. 707 β 70 ππ π. 5 ππ = 1.17 ππ > .1875 ππ β ππΎ
For the Gusset plate:
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 58 ππ π
. 707 β 70 ππ π. 375 ππ = .88 ππ > .1875 ππ β ππΎ
Tensile Yielding of Diagonal Brace:
π πΞ©=πΉπ¦π΄π
Ξ©=36 ππ π β 11.5 ππ2
1.67= 247.9 πππ > 101 πππ β ππΎ
Tensile Rupture of Diagonal Brace:
π΄π = π΄π = 11.5 ππ2
π = 1 βοΏ½οΏ½
π= 1 β
1.67 ππ
16 ππ= 0.899
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BRACED FRAME DESIGN Pg 198
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
π΄π = π΄ππ = 11.5 ππ2 β .899 = 10.34 ππ2
π πΞ©=πΉπ’π΄πΞ©
=58 ππ π β 10.34 ππ2
2= 299.9 ππππ > 101 ππππ β ππΎ
Welds connecting Clip Angle to Column
Design weld for ππ + π»π = 24.2 πππ + 28.8 πππ = 53 ππππ For 3/8-in. angles, π·πππ = 3 from AISC Specification Table J2.4.
Try 5/16-in fillet welds, π· = 5. From AISC equation 8-2:
πΏπππ =ππ + π»π2π·(0.928)
=53 πππ
2(5)(0.928)= 5.71 ππ
Fillet weld for the entire length of gusset for full length minus some clearance:
πΏ = 9.5 ππ > 5.71 ππ β ππΎ
Check the maximum size of fillet weld:
πππ₯ π ππ§π = .375 β .0625 = .3125 ππ = .3125 ππ β ππΎ
Check the minimum length:
πΏπππ = 4(. 3125 ππ) = 1.25 ππ < 9.5 ππ β ππΎ
πΏπππ = 7.375 ππ < 9.5 ππ β ππΎ
Check the weld length to weld size ratio: πΏ
π€=
9.5
. 3125= 30.4 < 100 β πΆβπππ π½ ππππ‘ππ πππ‘ πππππ π πππ¦
Check Base Metals by the largest effective weld size.
For the Clip Angle:
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 58 ππ π
. 707 β 70 ππ π. 375 ππ = .879 ππ > .3125 ππ β ππΎ
For the Column Flange:
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 65 ππ π
. 707 β 70 ππ π. 595 ππ = 1.56 ππ > .3125 ππ β ππΎ
Bolts connecting Clip Angle from Column to Gusset Plate
Use the following load for design: π = βππ2 + π»π
2 = 37.62 ππππ
From AISC Manual Table 7-1, find the minimum number of 1-in. Group A, threads
included, bolts.
ππππ =π ππΞ©
=37.62 ππππ
21.2 ππππ = 1.77 ππππ‘π
Try 4 standard sized bolts at 2.5 inches on center.
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BRACED FRAME DESIGN Pg 199
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Shear Yielding of Clip Angle from column to Gusset Plate
Follow AISC specification J4-2. π πΞ©=. 6πΉπ¦π΄ππ£
Ξ©=. 6 β 36 ππ π β 12 ππ β .375 ππ β 2
1.5= 129.6 ππππ > 37.62 ππππ
β ππΎ
Shear Rupture of Clip Angle from column to Gusset Plate
Following AISC specification J4-2:
π΄ππ£ = 2(. 375 ππ)[12 β 4(1 ππ + .125 ππ)] = 5.625 ππ2
π πΞ©=. 6πΉπ’π΄ππ£Ξ©
=. 6 β 58 ππ π β 5.625 ππ2
2= 97.9 ππππ > 37.62 ππππ
Block Shear Rupture of Clip Angle from column to Gusset Plate
Assume uniform tension stress, so use πππ = 1.
π΄ππ£ = 2(12 ππ β 2.25 ππ). 375 ππ = 7.31 ππ2
π΄ππ£ = 7.31 ππ2 β .375 ππ(3.5)(1.125 ππ)2 = 4.36 ππ2
π΄ππ‘ = [(3 ππ β .375 ππ) β (.5 β 1.125 ππ β .375 ππ)]2 = 1.83 ππ2
From AISC Specification J4-3:
π π = .6πΉπ’π΄ππ£ + πππ πΉπ’π΄ππ‘ β€ .5πΉπ¦π΄ππ£ + πππ πΉπ’π΄ππ‘
π π = .6(58)(4.36) + 1(58)(1.83) β€ .5(36)(7.31) + 1(58)(1.83) π π = 257.9 ππππ β€ 237.7 ππππ π πΞ©=237.7
2= 118.9 ππππ > 37.62 ππππ β ππΎ
Shear Bearing Strength at Bolt Holes
ππ = 2.25 β1.125
2= 1.6875 ππ
Consider deformation at the bolthole at service load a design consideration.
π π = 1.2πππ‘πΉπ’ β€ 2.4ππ‘πΉπ’
π π = 1.2(1.6875 ππ)(. 375 ππ)(58 ππ π)4 β€ 2.4(1 ππ)(. 375 ππ)(58 ππ π)4
π π = 176.18 ππππ β€ 208.8 ππππ π πΞ©=176.18
2= 88.09 ππππ > 37.62 ππππ β ππΎ
Design of welds connecting Beam to the Gusset Plate:
Design connection for ππ +π»π = 27.8 πππ + 57.74 πππ = 85.54 ππππ For 3/8-in. plate thickness, π·πππ = 3 from AISC Specification Table J2.4.
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BRACED FRAME DESIGN Pg 200
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Try 3/16-in fillet welds, π· = 3. From AISC equation 8-2:
πΏπππ =π
2π·(0.928)=85.54 ππππ
2(3)(0.928)= 15.36 ππ
Weld the entire length of the minus some clearance.
πΏ = 25 ππ > 15.36 ππ β ππΎ
Check the maximum size of fillet weld:
πππ₯ π ππ§π = .375 β .0625 = .3125 ππ > .1875 ππ β ππΎ
Check the minimum length:
πΏπππ = 4(. 1875 ππ) = .75 ππ < 25 ππ β ππΎ
πΏπππ = .375 ππ < 25 ππ β ππΎ
Check the weld length to weld size ratio: πΏ
π€=
25
. 1875= 133.3 > 100 β πΆβπππ π½ ππππ‘ππ
π½ = 1.2 β .002 βπ
π€= 1.2 β .002 β 133.3 = .933
Recheck strength: π πΞ©= (. 933)(2)(3)(. 928)(25) = 129.9 πππ > 85.54 πππ β ππΎ
Check Base Metals by the largest effective weld size. (π =π·
16)
For the Beam:
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 65 ππ π
. 707 β 70 ππ π. 63 ππ = 1.65 ππ > .1875 ππ β ππΎ
For the Gusset plate:
π =Ξ©πΉπ’
. 707πΉπΈπππ‘π΅π =
2 β 58 ππ π
. 707 β 70 ππ π. 375 ππ = .88 ππ > .1875 ππ β ππΎ
Gusset Plate Compression
Whitmore section effective width, π€ = 25.48 ππ
According to AISC Part 9, βThe Whitmore section may spread across the joint between
connecting elements, but cannot spread beyond an unconnected edge.β Thus, this section
is ok.
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BRACED FRAME DESIGN Pg 201
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πΏ1 = 3.87 ππ
πΏ2 = 7.68 ππ
πΏ3 = β11.34ππ
Unbraced length, π =πΏ1+πΏ2+πΏ3
3= 0.21 ππ
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BRACED FRAME DESIGN Pg 202
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π =π‘
β12=. 375 ππ
β12= .108 ππ
By AISC Manual specification J4: ππ
π=1(0.21 ππ)
. 108 ππ= 1.94
From Table 4-22, πΉππ
Ξ©= 21.6 ππ π
π΄π = π€π‘ = 25.48 ππ β .375 ππ = 9.56 ππ2
ππΞ©=πΉπππ΄π
Ξ©=21.6 ππ π β 9.56 ππ2
1.67= 123.7 ππππ > 101 ππππ β ππΎ
Tensile Yielding of Gusset Plate
π πΞ©=πΉπ¦π΄π
Ξ©=36 ππ π β 9.56 ππ2
1.67= 206.1 πππ > 101 πππ β ππΎ
Tensile Rupture of Gusset Plate:
π΄π = π΄π = 9.56 ππ2
Case 4 in AISC table D3.1: π = 1.0
π΄π = π΄ππ = 9.56 ππ2 β 1 = 9.56 ππ2
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BRACED FRAME DESIGN Pg 203
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π πΞ©=πΉπ’π΄πΞ©
=58 ππ π β 9.56 ππ2
2= 277.2 ππππ > 101 ππππ β ππΎ
Connection Geometry
πΌ = 14.32 ππ
π½ = 6 ππ
π = 59Β°
ππ =13.8
2= 6.90 ππ (πππ π14π₯30)
ππ =14.3
2= 7.15 ππ (πππ π14π109)
πΌ = ππ tan π β ππ + π½ tanπ
14.32 = 6.9tan 59Β° β 7.15 + 6 tan 59Β° β πππ’π β ππΎ
ALL CHECKS ARE SATISFIED; THEREFORE CONSTRUCT BRACE ACCORDING
TO AUTOCAD DRAWING.
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SHEAR WALL DESIGN Pg 204
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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 β 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 β 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 β 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 β qi(GCpi) (kip*ft):
Distance From Case A Case B
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SHEAR WALL DESIGN Pg 205
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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:
+βΊΞ£M=0;
Mu= 1678.351 kip*ft
Summing the forces in the x-direction:
+ Ξ£Fx=0;
Vu= 50.76
Summing the moments in the y-direction:
β+ Ξ£Fy=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 βπ€
ππ€< 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
Ο: 0.65
Ξ»: 1
d=0.90*lw=22.5 ft
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SHEAR WALL DESIGN Pg 206
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
π = π΄π β ππ¦ = 432 πππ
π =π
0.85 β π β²π β β= 6.35 ππ
π1 =| π΄π β ππ¦ + ππ’|
0.85 β π β²π β β= 2.85ππ
πΆ = 0.85 β π β²π β β β π1 = 192.816 πππ
ππ = π (π βπ
2) + ππ’ (
ππ€ β π
2) = 6679.16 πππ β ππ‘
πππ = 0.65 β 6679.43 = 4341.45 πππ β ππ‘ πππ > ππ’ 4341.63 kip*ft > 1678.351 kip*ft Ok!
Shear Capacity
π(π1) = 3.3Ξ»(πβ²π)12 β β β π +
|ππ’| β π
4 β ππ€= 796.88 πππ β ππ‘
π(π2) = (0.60Ξ»(π β²π)12 +
lw(1.25 β Ξ» β (π β²π)12 +
0.20|ππ’|ππ€ β β
)
|ππ’||ππ’|
βππ€2
) β β β π = 629.66 πππ β ππ‘
Choose the smaller of the two Vc values.
Vcmin = 629.66 kip
πππ = 472.25 πππ
|πππ| > |ππ’| 472.25 kip > 239.162 kip Ok!
Check Steel: |πππ|
2= 236.12 < ππ’ < πππ 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 Ο(min ) = 0.0025 β Ο(t) = Ο(l)
Horizontal Steel:
As(h)= 3.6in2 #7 Bars @ 16β O.C.
Ο(t) = Ο(l)
Ο(π‘) =π΄π (β)
β β π‘= Ο(l) = 0.0225
Ο(πππ) = 0.0025 + .5(2.5 ββ
ππ€)(Ο(t) β .0025) = 0.0235
As(v)= Ο(min) β h β t = 3.76in2
Vertical Steel:
As(v)= 3.8in2 #7 Bars @ 16β O.C.
Moment resisting at the end of shear walls:
Trial value of As: 8 #7 Bars As= 4.8in2
π = π΄π β ππ¦ = 288 πππ
π =π + ππ
0.85 β π β²π β β= 7.75 ππ
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SHEAR WALL DESIGN Pg 207
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππ = π (π βπ
2) + ππ’ (
ππ€ β π
2) = 111594.762 πππ β ππ
πππ = 9299.56 πππ β ππ‘ > ππ’Ok!
Provide 8 #7 Bars at each end.
Steel Strain
π = 11.12 ππ
Ξ΅(t)= 0.0537 >.004 Ok!
Ξ΅(t)= 0.0537 >.005 Ok! π = 0.90
Ο(g) =n β cross sectional area
boundary width β boundary length= 0.0121
Ο(g) = 0.01875 > 0.01 Although ties are not ideal ties are required.
Add #4 closed ties at 16β O.C. vertical maximum at each horizontal bar.
ALL CHECKS ARE SATISFIED; THEREFORE CONSTRUCT SHEAR WALLS
ACCORDING TO AUTOCAD DRAWING.
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FOUNDATION DESIGN Pg 208
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
FOUNDATION DESIGN GREEN ROOF EDGE COLUMN FOOTING
Given Data and assumptions
Variable Value Unit
qa 8000 psf
Soil Depth weight Ξ³w 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
ππππ‘ = 8000ππ πβ (πππ ππππ’ππ πΏππππ )ππ π = 8000ππ π β 100ππ π β 50ππ π β 220ππ π= 7630ππ π ππ’π π‘βπ ππππ’ππ πππππ = 118.9π + 31.25π = 150.15π
π΅ = πΏ = β150.15π/7.63ππ π = 4.44ππ‘
Say 4.5ft ππ = (πΆπππ’ππ π·πππ πΏπππ) + .27 β π΅2 = 118.9π + .27ππ π β (4.5ππ‘)2 = 124.3675π
ππΏ = (πΆπππ’ππ πΏππ£π πΏπππ) + .1 β π΅2 = 31.25π + .1ππ π β (4.5ππ‘)2 = 33.275π ππ’ = 1.2 β (ππ) + 1.6(ππΏ) = 202.481π
ππ’ = ππ’
π΅2=202.481π
(4.5ππ‘)2= 10ππ π
π΄π π π’ππ π = 10ππ Calculate One-way and Two-way Shear
One way
ππ’ = (π΅
2βπ
2β π) β πΏ β ππ’ = (
4.5ππ‘
2β2ππ‘
2β (10
12)ππ‘) β 4.5ππ‘ β 10ππ π = 18.75π
πππ = π β 2 β π β βπβ²π β πΏ β π =. 75 β 2 β 1 β β4000ππ π β 4.5ππ‘ β 12 β 10ππ
1000ππ= 51.23π
πΆβπππ πππ > ππ’
51.23π > 18.75π ππΎ
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FOUNDATION DESIGN Pg 209
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Two-Way
π0 = 4 β (π + π) = 4(24 + 10) = 136ππ
ππ’ = (π΅ β πΏ β (π04)2
) β ππ’ = 122.22π
πππ = π β 4 β π β βπβ²π β π0 β π = 258.01π
πΆβπππ πππ > ππ’
258.01π > 122.22π ππΎ Add 3 to 4 inches to initial d value
π = 13ππ Check Moment
ππ’ = ππ’ β πΏ β(π΅2 β
π2)2
2= 35.15π β ππ‘
ππ =ππ’π=35.15π β ππ‘
. 9= 39.06π β ππ‘
ππ = πππ¦πΏπ2(1 β
πππ¦
1.7 β πβ²π)
ππππ£π πππ π π’π πππ π‘βπ ππ’πππππ‘ππ πππππ’ππ
π = .0008
π΄π = ππ΅π = (. 0008) β 4.5ππ‘ β12ππ
ππ‘β 13ππ = .54454ππ2
USE 3 #4 BARS IN EACH DIRECTON
π΄π = .6ππ2
π =(π΄πππ¦)
. 85πβ²ππΏ= .196
πππ = .9 β π΄π β ππ¦ β (π βπ
2) = 34.84π β ππ‘
πππ314 = .9 β .85 β π΄π β ππ¦ β π = 29.835π β ππ‘
π ππ‘ππ =ππ314ππ
= .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.
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FOUNDATION DESIGN Pg 210
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
SHEAR WALL FOUNDATION DESIGN Given Data and assumptions
Variable Value Unit
qa 8000 psf
Concrete Weight wc 150 pcf
Soil Depth weight Ξ³w 110 pcf
Slab weight with Thickness (4") 50 psf
Soil Depth 24 in
LL Surcharge 100 psf
Ka 0.33
Kp 3
Theta 30 degrees
f'c 4000 psi
Reinforcement fy 60000 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
πΉπππ‘πππ π€πππβπ‘ ππ = π΅ β πΏ β β β π€π = 30ππ‘ β 10ππ‘ β 24ππ βππ‘
12ππβ 150πππ = 90π
ππππ’ππ ππ ππππ‘πππ ππ = π΅ β πΏ β β =30ππ‘ β 10ππ‘ β 24ππ β
ππ‘12ππ
27ππ‘3= 22.23π¦π3
ππππ π€πππβπ‘ ππ πππ = π΅ β πΏ β βπ πππ β πΎπ πππ = 30ππ‘ β 10ππ‘ β 24ππ β1ππ‘
12ππβ 110πππ = 66π
ππ’ππβππππ ππππβπ‘ ππ π’ππΏ = π΅ β πΏ β π€π π’ππΏ = 30π Total Loads from Shear Wall Design
ππ€ = 239π ππ€ = 50.76π
ππ€ = 1678.351π β ππ‘ Factored Service Loads ππ = 1(ππ + ππ€) = 329π
ππ = .6 β ππ€ = 30.46π
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FOUNDATION DESIGN Pg 211
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
ππ = ππ€ β .6 + (ππ ββπ πππ12) = 1068π β ππ‘
Factored Strength Loads ππ’ = 1.4 β (ππ + ππ€) = 460.6π
ππ’ = .5 β (ππ€) = 25.38π
ππ’ = ππ€ + (ππ’ ββπ πππ12) = 1729.111π β ππ‘
ππππ ππππ π π’ππ π€ππ‘βππ’π‘ ππππππ‘πππππ‘π¦ ππππ =ππ π΅πΏ= 1096.67ππ π
π·πππππ βΆ πΆππππππ‘π¦ πππ‘πππ ππππππ
= .137 < 1 ππΎ
πΈπππππ‘πππππ‘π¦ =ππ ππ = 3.25ππ‘
π΅
6= 5ππ‘
πΈπππππ‘ππππ¦ <π΅
6
Max soil pressure with eccentricity
ππππ₯ = ππππ β (1 + 6 βπΈππ
π΅) = 1808.62ππ π
ππππ₯ππ
= .226 < 1 ππΎ
Minimum soil pressure with eccentricity
ππππ = ππππ β (1 β 6 βπΈππ
π΅) = 384.72ππ π
Check Sliding Resistance
Coefficient of friction π = tan (2
3β π) = .36
π·πππ‘β ππππ π‘ππ ππ π πππ π‘π πππ‘π‘ππ ππ ππππ‘πππ π» = β + βπ πππ = 4ππ‘
πππ‘ ππππ‘β ππππ π π’ππ πππ ππ π‘ππππ π‘π π ππππππ ππππ‘ =1
2(πΎπ β πΎπ)πΎπ πππ β π»
2 β πΏ = 58.74π
πππ‘ππ πππ’ππππ‘πππ π ππππππ πππ ππ π‘ππππ πππ‘π = ππ β π + ππππ‘ = 177.18π
πππππ‘π¦ ππππ‘ππ ππππππ π‘ π ππππππ πΉππ ππ =πππ‘π
ππ = 5.82 > 1.67 ππΎ
Design reinforced concrete foundation for bending moment and shear
ππππ ππππ π π’ππ π€ππ‘βππ’π‘ ππππππ‘πππππ‘π¦ ππ’πππ =ππ’π΅ β πΏ
= 1535.33ππ π
Max LRFD soil pressure
ππ’πππ₯ = ππ’πππ β (1 + 6 βπΈππ
π΅) = 2532.1ππ π
Min LRFD soil pressure
ππ’πππ = ππ’πππ β (1 β 6 βπΈππ
π΅) = 538.61ππ π
Design for shear Parallel to Wall Length
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FOUNDATION DESIGN Pg 212
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
Choose Size and # of tension steel bars Bar size=9 Bar num=20
π΄π = 20ππ2
ππ‘ππ’ππ‘π’πππ ππππ‘β ππ ππππ‘πππ π = β β ππ β 1.5ππ = 24 β 3 β 1.5 β 1.128 = 1.61ππ‘
πΆπππ‘ππππ£ππ ππππ‘πππ πππππ‘β ππ£ =π΅
2βππ€2β π = .891ππ‘
Interpolating using lv to find the critical section soil pressure ππ’2 = 2472.86ππ π
π ππ π’ππ‘πππ‘ πππππ πΉπ’ =ππ’πππ₯ + ππ’2
2β ππ£ β πΏ = 22.3π
πππ = .75 β 2 β β4000 β πΏ β π = 219.81π πΉπ’πππ
= .102 < 1 ππΎ
Check two-way shear π1 = ππ€ + π = 26.61ππ‘ π2 = π‘π€πππ + π = 2.443ππ‘
πΆπππ‘ππππ π πππ‘πππ ππππ π΄π = 2(π1 + π2) β π = 93.49ππ‘2
πΆπππ‘ππππ π πππ‘πππ ππππ’πππ’π π½π =π1 β π β (π1 + 3 β π2) + π
3
3= 485.7ππ‘3
πβπππ π π‘ππ π ππ π€πππ πππ π£π’ =πΉπ’π΄π+ππ’π½π= 26.38ππ π
ππ£π = .75 β 4 β β4000 = 189.74ππ π π£π’ππ£π
= .14 < 1 ππΎ
Bending Moment Parallel to wall
ππ =π΅
2βππ€2= 2.5ππ‘
Interpolating using lm to find the critical section soil pressure ππ’2 =2365.939psf
πΉπ’ =ππ’πππ₯ + ππ’2
2β ππ β πΏ = 61.225π
πΉππππ ππππππ‘ πππ ππ = ππ βππβ(2βππ’2+ππ’πππ₯)
3β(ππ’2+ππ’πππ₯)=1.264ft
π΅ππππππ ππππππ‘ ππ‘ ππππ‘ππππ π πππ‘πππ ππ’πππ = (πΉπ’ β ππ ) = 77.4π β ππ‘
πππ = .9 β .85 β π΄π β ππ¦ β π =1477.1k*ft ππ’πππ
πππ= .053 < 1 ππΎ
Design for Shear force Perpendicular to wall length Choose Bar size and Spacing Bar size= 9, Spacing = 10in
π΄π =1.2ππ2
ππ‘
ππ£ =πΏ
2βπ‘π€πππ2β π = 2.975ππ‘
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FOUNDATION DESIGN Pg 213
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πΉπ’ = ππππ₯ β ππ£ = 5.4π
ππ‘
πππ = .75 β 2 β β4000 β π = 21.98π
ππ‘
πΉπ’πππ
= .245 < 1 ππΎ
Design for bending moment perpendicular to wall
ππ =πΏ
2βπ‘π€πππ2
= 2.584ππ‘
ππ’πππ = ππππ₯ βππ2
2= 18 π β
ππ‘
ππ‘
πππ = .9 β .85 β π΄π β ππ¦ β π = 88.63 π βππ‘
ππ‘
ππ’πππ
πππ= .215 < 1 ππΎ
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.
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APPENDIX Pg 214
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
APPENDIX
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AUTOCAD Pg 215
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
FOUNDATION DESIGN GREEN ROOF CORNER COLUMN FOOTING
Given Data and assumptions
Variable Value Unit
qa 8000 psf
Soil Depth weight Ξ³w 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
ππππ‘ = 8000ππ πβ (πππ ππππ’ππ πΏππππ )ππ π = 8000ππ π β 100ππ π β 50ππ π β 220ππ π= 7630ππ π ππ’π π‘βπ ππππ’ππ πππππ = 59.5πΎ + 15.63πΎ = 75.13π
π΅ = πΏ = β75.13π/7.63ππ π3.14ππ‘
Say 3.5ft ππ = (πΆπππ’ππ π·πππ πΏπππ) + .27 β π΅2 = 59.5π + .27ππ π β (3.5ππ‘)2 = 62.81πΎ ππΏ = (πΆπππ’ππ πΏππ£π πΏπππ) + .1 β π΅2 = 15.63π + .1ππ π β (3.5ππ‘)2 = 16.86π
ππ’ = 1.2 β (ππ) + 1.6(ππΏ) = 102.34π
ππ’ = ππ’
π΅2= 8.35ππ π
π΄π π π’ππ π = 9ππ Calculate One-way and Two-way Shear
One way
ππ’ = (π΅
2βπ
2β π) β πΏ β ππ’ = 0π
πππ = π β 2 β π β βπβ²π β πΏ β π =. 75 β 2 β 1 β β4000ππ π β 3.5ππ‘ β 12 β 9ππ
1000ππ= 35.86π
πΆβπππ πππ > ππ’
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AUTOCAD Pg 216
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
35.86π > 0π ππΎ
Two-Way
π0 = 4 β (π + π) = 4(24 + 9) = 132ππ
ππ’ = (π΅ β πΏ β (π04)2
) β ππ’ = 39.16π
πππ = π β 4 β π β βπβ²π β π0 β π = 225.41π
πΆβπππ πππ > ππ’
225.41π > 39.16π ππΎ Add 3 to 4 inches to initial d value
π = 12ππ Check Moment
ππ’ = ππ’ β πΏ β(π΅2 β
π2)2
2= 8.224π β ππ‘
ππ =ππ’π=35.15π β ππ‘
. 9= 9.14π β ππ‘
ππ = πππ¦πΏπ2(1 β
πππ¦
1.7 β πβ²π)
ππππ£π πππ π π’π πππ π‘βπ ππ’πππππ‘ππ πππππ’ππ
π = .0003
π΄π = ππ΅π = (. 0003) β 3.5ππ‘ β12ππ
ππ‘β 12ππ = .134ππ2
USE 2 #3 BARS IN EACH DIRECTON
π΄π = .22ππ2
π =(π΄πππ¦)
. 85πβ²ππΏ= .092
πππ = .9 β π΄π β ππ¦ β (π βπ
2) = 11.83π β ππ‘
πππ314 = .9 β .85 β π΄π β ππ¦ β π = 10.1π β ππ‘
π ππ‘ππ =ππ314ππ
= .853
Summary: Design for a 3.5ft long by 3.5ft wide by 12in deep foundation imbedded 2ft below the
foundation slab with 2 #3 bars spanning both directions.
![Page 218: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/218.jpg)
AUTOCAD Pg 217
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
FOUNDATION DESIGN GREEN ROOF INTERIOR COLUMN FOOTING
Given Data and assumptions
Variable Value Unit
qa 8000 psf
Soil Depth weight Ξ³w 110 pcf
Slab weight with Thickness (4")
50 psf
Soil Depth 24 in
LL Surcharge 100 psf
Point Dead Load 158.4 k
Point Live Load 61.7 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
ππππ‘ = 8000ππ πβ (πππ ππππ’ππ πΏππππ )ππ π = 8000ππ π β 100ππ π β 50ππ π β 220ππ π= 7630ππ π ππ’π π‘βπ ππππ’ππ πππππ = 237.9πΎ + 62.5πΎ = 300.4π
π΅ = πΏ = β300.4π/7.63ππ π = 6.27ππ‘
Say 6.5ft ππ = (πΆπππ’ππ π·πππ πΏπππ) + .27 β π΅2 = 249.31πΎ ππΏ = (πΆπππ’ππ πΏππ£π πΏπππ) + .1 β π΅2 = 66.73π
ππ’ = 1.2 β (ππ) + 1.6(ππΏ) = 405.93π
ππ’ = ππ’
π΅2= 9.61ππ π
π΄π π π’ππ π = 16ππ
Calculate One-way and Two-way Shear One way
ππ’ = (π΅
2βπ
2β π) β πΏ β ππ’ = 57.25π
πππ = π β 2 β π β βπβ²π β πΏ β π = 118.4π
πΆβπππ πππ > ππ’
![Page 219: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/219.jpg)
AUTOCAD Pg 218
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
118.4π > 57.25π ππΎ
Two-Way
π0 = 4 β (π + π) = 4(24 + 16) = 160ππ
ππ’ = (π΅ β πΏ β (π04)2
) β ππ’ = 300π
πππ = π β 4 β π β βπβ²π β π0 β π = 485.73π
πΆβπππ πππ > ππ’
485.73π > 300π ππΎ Add 3 to 4 inches to initial d value
π = 19ππ Check Moment
ππ’ = ππ’ β πΏ β(π΅2 β
π2)2
2= 158.1π β ππ‘
ππ =ππ’π=35.15π β ππ‘
. 9= 175.64π β ππ‘
ππ = πππ¦πΏπ2(1 β
πππ¦
1.7 β πβ²π)
ππππ£π πππ π π’π πππ π‘βπ ππ’πππππ‘ππ πππππ’ππ
π = .0011
π΄π = ππ΅π = 1.681ππ2
USE 4 #6 BARS IN EACH DIRECTON
π΄π = 1.76ππ2
π =(π΄πππ¦)
. 85πβ²ππΏ= .398
πππ = .9 β π΄π β ππ¦ β (π βπ
2) = 148.9π β ππ‘
πππ314 = .9 β .85 β π΄π β ππ¦ β π = 127.91π β ππ‘
π ππ‘ππ =ππ314ππ
= .859
Summary: Design for a 6.5ft long by 6.5ft wide by 19in deep foundation imbedded 2ft below the
foundation slab with 4 #6 bars spanning both directions.
![Page 220: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/220.jpg)
AUTOCAD Pg 219
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
FOUNDATION DESIGN GREEN ROOF/INTERIOR EDGE COLUMN FOOTING
Given Data and assumptions
Variable Value Unit
qa 8000 psf
Soil Depth weight Ξ³w 110 pcf
Slab weight with Thickness (4")
50 psf
Soil Depth 24 in
LL Surcharge 100 psf
Point Dead Load 383.9 k
Point Live Load 164 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
ππππ‘ = 8000ππ πβ (πππ ππππ’ππ πΏππππ )ππ π = 8000ππ π β 100ππ π β 50ππ π β 220ππ π= 7630ππ π ππ’π π‘βπ ππππ’ππ πππππ = 158.4π + 61.7π = 220.1π
π΅ = πΏ = β220.1π/7.63ππ π = 5.37ππ‘
Say 5.5ft ππ = (πΆπππ’ππ π·πππ πΏπππ) + .27 β π΅
2 = 166.5675πΎ ππΏ = (πΆπππ’ππ πΏππ£π πΏπππ) + .1 β π΅
2 = 64.73π ππ’ = 1.2 β (ππ) + 1.6(ππΏ) = 303.441π
ππ’ = ππ’
π΅2= 10.03ππ π
π΄π π π’ππ π = 14ππ
Calculate One-way and Two-way Shear One way
ππ’ = (π΅
2βπ
2β π) β πΏ β ππ’ = 32.2π
πππ = π β 2 β π β βπβ²π β πΏ β π = 87.66π
πΆβπππ πππ > ππ’
![Page 221: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/221.jpg)
AUTOCAD Pg 220
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
87.66π > 32.2π ππΎ
Two-Way
π0 = 4 β (π + π) = 4(24 + 14) = 152ππ
ππ’ = (π΅ β πΏ β (π04)2
) β ππ’ = 202.9π
πππ = π β 4 β π β βπβ²π β π0 β π = 403.8π
πΆβπππ πππ > ππ’
403.8π > 202.9π ππΎ Add 3 to 4 inches to initial d value
π = 17ππ Check Moment
ππ’ = ππ’ β πΏ β(π΅2 β
π2)2
2= 84.5 π β ππ‘
ππ =ππ’π= 93.867 π β ππ‘
ππ = πππ¦πΏπ2(1 β
πππ¦
1.7 β πβ²π)
ππππ£π πππ π π’π πππ π‘βπ ππ’πππππ‘ππ πππππ’ππ
π = .0009
π΄π = ππ΅π = 1.002ππ2
USE 3 #6 BARS IN EACH DIRECTON
π΄π = 1.32ππ2
π =(π΄πππ¦)
. 85πβ²ππΏ= .353
πππ = .9 β π΄π β ππ¦ β (π βπ
2) = 99.93π β ππ‘
πππ314 = .9 β .85 β π΄π β ππ¦ β π = 85.833π β ππ‘
π ππ‘ππ =ππ314ππ
= .859
Summary: Design for a 5.5ft long by 5.5ft wide by 17in deep foundation imbedded 2ft below the
foundation slab with 3 #6 bars spanning both directions.
![Page 222: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/222.jpg)
AUTOCAD Pg 221
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
FOUNDATION DESIGN GREEN ROOF/INTERIOR INTERIOR COLUMN FOOTING
Given Data and assumptions
Variable Value Unit
qa 8000 psf
Soil Depth weight Ξ³w 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
ππππ‘ = 8000ππ πβ (πππ ππππ’ππ πΏππππ )ππ π = 8000ππ π β 100ππ π β 50ππ π β 220ππ π= 7630ππ π
ππ’π π‘βπ ππππ’ππ πππππ = 547.9π
π΅ = πΏ = β220.1π/7.63ππ π = 8.47ππ‘
Say 8.5ft ππ = (πΆπππ’ππ π·πππ πΏπππ) + .27 β π΅
2 = 403.4075 ππΏ = (πΆπππ’ππ πΏππ£π πΏπππ) + .1 β π΅
2 = 171.225π ππ’ = 1.2 β (ππ) + 1.6(ππΏ) = 758.05π
ππ’ = ππ’
π΅2= 10.5ππ π
π΄π π π’ππ π = 25ππ
Calculate One-way and Two-way Shear One way
ππ’ = (π΅
2βπ
2β π) β πΏ β ππ’ = 104.05π
πππ = π β 2 β π β βπβ²π β πΏ β π = 241.9π
![Page 223: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/223.jpg)
AUTOCAD Pg 222
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πΆβπππ πππ > ππ’
241.9π > 104.05π ππΎ
Two-Way
π0 = 4 β (π + π) = 4(24 + 25) = 196ππ
ππ’ = (π΅ β πΏ β (π04)2
) β ππ’ = 583.11π
πππ = π β 4 β π β βπβ²π β π0 β π = 929.71π
πΆβπππ πππ > ππ’
929.71π > 583.11π ππΎ Add 3 to 4 inches to initial d value
π = 28ππ Check Moment
ππ’ = ππ’ β πΏ β(π΅2 β
π2)2
2= 470 π β ππ‘
ππ =ππ’π= 523.32 π β ππ‘
ππ = πππ¦πΏπ2(1 β
πππ¦
1.7 β πβ²π)
ππππ£π πππ π π’π πππ π‘βπ ππ’πππππ‘ππ πππππ’ππ
π = .0012
π΄π = ππ΅π = 3.4ππ2
USE 8 #6 BARS IN EACH DIRECTON
π΄π = 3.52ππ2
π =(π΄πππ¦)
. 85πβ²ππΏ= .609
πππ = .9 β π΄π β ππ¦ β (π βπ
2) = 438.7π β ππ‘
πππ314 = .9 β .85 β π΄π β ππ¦ β π = 377π β ππ‘
π ππ‘ππ =ππ314ππ
= .859
Summary: Design for a 8.5ft long by 8.5ft wide by 28in deep foundation imbedded 2ft below the
foundation slab with 8 #6 bars spanning both directions.
![Page 224: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/224.jpg)
AUTOCAD Pg 223
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
FOUNDATION DESIGN INTERIOR ROOF EDGE COLUMN FOOTING
Given Data and assumptions
Variable Value Unit
qa 8000 psf
Soil Depth weight Ξ³w 110 pcf
Slab weight with Thickness (4")
50 psf
Soil Depth 24 in
LL Surcharge 100 psf
Point Dead Load 264.9 k
Point Live Load 132.8 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
ππππ‘ = 8000ππ πβ (πππ ππππ’ππ πΏππππ )ππ π = 8000ππ π β 100ππ π β 50ππ π β 220ππ π= 7630ππ π
ππ’π π‘βπ ππππ’ππ πππππ = 397.7π
π΅ = πΏ = β397.7π/7.63ππ π = 7.2ππ‘
Say 7.5ft ππ = (πΆπππ’ππ π·πππ πΏπππ) + .27 β π΅
2 = 280.1π ππΏ = (πΆπππ’ππ πΏππ£π πΏπππ) + .1 β π΅
2 = 138.43π ππ’ = 1.2 β (ππ) + 1.6(ππΏ) = 557.6π
ππ’ = ππ’
π΅2= 9.91ππ π
π΄π π π’ππ π = 23ππ
Calculate One-way and Two-way Shear One way
ππ’ = (π΅
2βπ
2β π) β πΏ β ππ’ = 62π
![Page 225: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/225.jpg)
AUTOCAD Pg 224
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
πππ = π β 2 β π β βπβ²π β πΏ β π = 196.4π
πΆβπππ πππ > ππ’
196.4π > 62π ππΎ
Two-Way
π0 = 4 β (π + π) = 4(24 + 23) = 188ππ
ππ’ = (π΅ β πΏ β (π04)2
) β ππ’ = 405.5226π
πππ = π β 4 β π β βπβ²π β π0 β π = 820.42π
πΆβπππ πππ > ππ’
820.42π > 405.52π ππΎ Add 3 to 4 inches to initial d value
π = 26ππ Check Moment
ππ’ = ππ’ β πΏ β(π΅2 β
π2)2
2= 281.12 π β ππ‘
ππ =ππ’π= 312.25π β ππ‘
ππ = πππ¦πΏπ2(1 β
πππ¦
1.7 β πβ²π)
ππππ£π πππ π π’π πππ π‘βπ ππ’πππππ‘ππ πππππ’ππ
π = .0009
π΄π = ππ΅π = 2.18ππ2
USE 5 #6 BARS IN EACH DIRECTON
π΄π = 2.2ππ2
π =(π΄πππ¦)
. 85πβ²ππΏ= .431
πππ = .9 β π΄π β ππ¦ β (π βπ
2) = 255.26π β ππ‘
πππ314 = .9 β .85 β π΄π β ππ¦ β π = 218.8π β ππ‘
π ππ‘ππ =ππ314ππ
= .857
Summary: Design for a 7.5ft long by 7.5ft wide by 26in deep foundation imbedded 2ft below the
foundation slab with 5 #6 bars spanning both directions.
![Page 226: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/226.jpg)
AUTOCAD Pg 225
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
FOUNDATION DESIGN INTERIOR ROOF INTERIOR COLUMN FOOTING
Given Data and assumptions
Variable Value Unit
qa 8000 psf
Soil Depth weight Ξ³w 110 pcf
Slab weight with Thickness (4")
50 psf
Soil Depth 24 in
LL Surcharge 100 psf
Point Dead Load 487.2 k
Point Live Load 184.3 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
ππππ‘ = 8000ππ πβ (πππ ππππ’ππ πΏππππ )ππ π = 8000ππ π β 100ππ π β 50ππ π β 220ππ π= 7630ππ π
ππ’π π‘βπ ππππ’ππ πππππ = 671.5π
π΅ = πΏ = β397.7π/7.63ππ π = 9.38ππ‘
Say 9.5ft ππ = (πΆπππ’ππ π·πππ πΏπππ) + .27 β π΅
2 = 511.57π ππΏ = (πΆπππ’ππ πΏππ£π πΏπππ) + .1 β π΅
2 = 193.33π ππ’ = 1.2 β (ππ) + 1.6(ππΏ) = 923.2π
ππ’ = ππ’
π΅2= 10.23ππ π
π΄π π π’ππ π = 27ππ
Calculate One-way and Two-way Shear One way
ππ’ = (π΅
2βπ
2β π) β πΏ β ππ’ = 145.8π
πππ = π β 2 β π β βπβ²π β πΏ β π = 292π
πΆβπππ πππ > ππ’
![Page 227: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/227.jpg)
AUTOCAD Pg 226
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
292π > 145.8π ππΎ
Two-Way
π0 = 4 β (π + π) = 4(24 + 27) = 204ππ
ππ’ = (π΅ β πΏ β (π04)2
) β ππ’ = 738.43π
πππ = π β 4 β π β βπβ²π β π0 β π = 1045.1π
πΆβπππ πππ > ππ’
1045.1π > 738.43π ππΎ Add 3 to 4 inches to initial d value
π = 30ππ Check Moment
ππ’ = ππ’ β πΏ β(π΅2 β
π2)2
2= 683.3 π β ππ‘
ππ =ππ’π= 759.21π β ππ‘
ππ = πππ¦πΏπ2(1 β
πππ¦
1.7 β πβ²π)
ππππ£π πππ π π’π πππ π‘βπ ππ’πππππ‘ππ πππππ’ππ
π = .0013
π΄π = ππ΅π = 4.61ππ2
USE 8 #7 BARS IN EACH DIRECTON
π΄π = 4.8ππ2
π =(π΄πππ¦)
. 85πβ²ππΏ= .743
πππ = .9 β π΄π β ππ¦ β (π βπ
2) = 639.98πΎ β ππ‘
πππ314 = .9 β .85 β π΄π β ππ¦ β π = 550.8π β ππ‘
π ππ‘ππ =ππ314ππ
= .86
Summary: Design for a 9.5ft long by 9.5ft wide by 30in deep foundation imbedded 2ft below the
foundation slab with 8 #7 bars spanning both directions.
![Page 228: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/228.jpg)
AUTOCAD Pg 227
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
FOUNDATION DESIGN INTERIOR ROOF CORNER COLUMN FOOTING
Given Data and assumptions
Variable Value Unit
qa 8000 psf
Soil Depth weight Ξ³w 110 pcf
Slab weight with Thickness (4")
50 psf
Soil Depth 24 in
LL Surcharge 100 psf
Point Dead Load 265 k
Point Live Load 231.8 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
ππππ‘ = 8000ππ πβ (πππ ππππ’ππ πΏππππ )ππ π = 8000ππ π β 100ππ π β 50ππ π β 220ππ π= 7630ππ π
ππ’π π‘βπ ππππ’ππ πππππ = 379.8π
π΅ = πΏ = β397.7π/7.63ππ π = 7.22ππ‘
Say 7.5ft ππ = (πΆπππ’ππ π·πππ πΏπππ) + .27 β π΅
2 = 280.2π ππΏ = (πΆπππ’ππ πΏππ£π πΏπππ) + .1 β π΅
2 = 138.43π ππ’ = 1.2 β (ππ) + 1.6(ππΏ) = 557.71π
ππ’ = ππ’
π΅2= 9.915ππ π
π΄π π π’ππ π = 23ππ
Calculate One-way and Two-way Shear One way
ππ’ = (π΅
2βπ
2β π) β πΏ β ππ’ = 62π
πππ = π β 2 β π β βπβ²π β πΏ β π = 196.4π
πΆβπππ πππ > ππ’
![Page 229: CES 4743C Structural Design - Dominick Tota, E.I....CES 4743C Structural Design FINAL PROJECT SUBMITTAL Team Vandelay: Jonathon Ambar Paul Heagney Dominick Tota David Lutz Due Date:](https://reader034.fdocuments.in/reader034/viewer/2022042212/5eb5954f68f1ac06994a5b24/html5/thumbnails/229.jpg)
AUTOCAD Pg 228
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
196.4π > 62π ππΎ
Two-Way
π0 = 4 β (π + π) = 4(24 + 23) = 188ππ
ππ’ = (π΅ β πΏ β (π04)2
) β ππ’ = 405.61π
πππ = π β 4 β π β βπβ²π β π0 β π = 820.42π
πΆβπππ πππ > ππ’
820.42π > 405.61π ππΎ Add 3 to 4 inches to initial d value
π = 26ππ Check Moment
ππ’ = ππ’ β πΏ β(π΅2 β
π2)2
2= 281.2π β ππ‘
ππ =ππ’π= 312.42π β ππ‘
ππ = πππ¦πΏπ2(1 β
πππ¦
1.7 β πβ²π)
ππππ£π πππ π π’π πππ π‘βπ ππ’πππππ‘ππ πππππ’ππ
π = .0009
π΄π = ππ΅π = 2.18ππ2
USE 5 #6 BARS IN EACH DIRECTON
π΄π = 2.2ππ2
π =(π΄πππ¦)
. 85πβ²ππΏ= .43
πππ = .9 β π΄π β ππ¦ β (π βπ
2) = 255.3πΎ β ππ‘
πππ314 = .9 β .85 β π΄π β ππ¦ β π = 218.8π β ππ‘
π ππ‘ππ =ππ314ππ
= .857
Summary: Design for a 7.5ft long by 7.5ft wide by 26in deep foundation imbedded 2ft below the
foundation slab with 5 #6 bars spanning both directions.
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AUTOCAD Pg 229
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
AUTOCAD DRAWINGS
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REFERENCES Pg 230
Vandelay Industries CES 4743 FINAL PROJECT SUBMITTAL
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.