1. CAMPBELL HALL
UNIVERSITY OF CALIFORNIA, BERKELEY
CEE 123 L: Structural Concrete Design Project
Professor J. Moehle
ALEXANDER BILL
MELISSA MEIKLE
ANDREW RICHARD
QUDSIA WAHAB
MAY 9, 2014
9. TABLE OF CONTENTS
Transmittal Letter
Cost Estimation
Design Calculations
General Project Description
Structural system
Building code and references
Loading criteria
Gravity loads and assumed unit weights
Wind loading criteria
Seismic loading criteria
Materials
Load combinations
Analysis and design approach summary
Geotechnical Information
Gravity system design
Slab design
Beam design
Girder design
Column design
Lateral system design
Seismic load calculations
Shear wall design
Wind load calculations
Retaining wall design
Foundation design
Spread footing design
Strip footing design
Structural Drawings
Layout and General Notes
First Floor Framing Plan
Building Elevation
Foundation Plan
Footing Detail
Slab, Beam, and Girder Detail
Beam and Girder Detail
Column Detail
Shear Wall and Retaining Wall Detail
Roof Framing Plan
Basement Framing Plan
Diaphragm Plan
Appendix A
10. CalStructures
Davis Hall
University of California
Berkeley, CA 94720-1712
Jack P Moehle
775 Davis Hall
University of California
Berkeley, CA 94720-1712
Dear Professor Moehle,
Please find attached our submission of our proposal for the Campbell Hall Replacement Project.
The purpose of this proposal is to present you with our unique design solution, along with initial
cost estimate, for this prestigious project. The report was assigned on January 24th 2014. The
fee we are proposing for this project is $2,627,489.33, which we believe is great value for the
design we present in this proposal.
The report includes a brief overview of the project and the specifications provided by the client
and how we have met and exceeded them. We then move into the structural design for this
reinforced concrete solution. We have analyzed our solution for wind, gravity, earth and
earthquake loading ensuring that every scenario is covered and a perfectly functioning building
is achieved. We have included detailed design of the elevated slabs, ground bearing slab,
foundations, retaining walls, shear walls, columns and the building diaphragm. We have also
produced and included detailed construction documents with the detailing of each member
displayed clearly in the attached drawings. A cost estimate has been formulated and is
displayed in this document along with a cost breakdown into each member of the building. All
calculations have been attached also.
We would greatly appreciate to hear your feedback on our findings and we look forward to
hearing the result of who has the honor of delivering this great project.
Yours sincerely,
Alexander Bill, CalStructures Summer Intern
Melissa Meikle, CalStructures Senior Structural Engineer
Andrew Richard, CalStructures Senior Structural Engineer
Qudsia Wahab, CalStructures Senior Structural Engineer
11. Campbell Hall
“COST ESTIMATION”
Date: 03/28/2014
Calculated by: A.B, M.M, A.R, Q.W
CE123L Group Project - Cal Structures (Group 6) - Cost Estimation - 04/27/14
Alexander Bill - Melissa Meikle - Andrew Richard - Qudsia Wahab
Grade beams and spread footings Unit Quantity Cost ($/unit) Total Cost Per Member
Formwork
Spread Footing C1 8'x8'x2.5' SF 320 7.5 $2,400.00
Spread Footing C2 12'x12'x2.5' SF 480 7.5 $3,600.00
Spread Footing C3 15'x15'x3' SF 1620 7.5 $12,150.00
Strip Footing SF 360 7.5 $2,700.00
Total 2780 $20,850.00
Reinforcement
Spread Footing C1 lb 12544 1 $12,544.00
Spread Footing C2 lb 28224 1 $28,224.00
Spread Footing C3 lb 119070 1 $119,070.00
Strip Footing lb 14700 1 $14,700.00
Total 174538 $174,538.00
Concrete & placing - 4000 psi
Spread Footing C1 CY 23.7037 200 $4,740.74
Spread Footing C2 CY 53.33333 200 $10,666.67
Spread Footing C3 CY 225 200 $45,000.00
Strip Footing CY 27.77778 200 $5,555.56
Total 329.8148 $65,962.96
Grade Beams and Spread Footings Total Cost $261,350.96
Slab-on-grade
Edge forms
Basement Slab LF 400 7 $2,800.00
Total 400 $2,800.00
Aggregate base over vapor retarder
Basement Slab SF 10000 1.5 $15,000.00
Total 10000 $15,000.00
Control joints
Basement Slab SF 10000 0.4 $4,000.00
Total 10000 $4,000.00
Reinforcement
Basement Slab lb 130666.7 1 $130,666.67
Total 130666.7 $130,666.67
Concrete & placing - 3500 psi
Basement Slab CY 246.9136 195 $48,148.15
Total 246.9136 $48,148.15
Finish & cure
Basement Slab SF 10000 1 $10,000.00
Total 10000 $10,000.00
Slab-on-grade cost $210,614.81
Concrete Walls
Formwork
Shear Walls SF 9540 15 $143,100.00
Retaining Wall SF 4500 15 $67,500.00
Total 9540 $210,600.00
Reinforcement –LFRS
Shear Walls lb 132300 1.25 $165,375.00
12. Campbell Hall
“COST ESTIMATION”
Date: 03/28/2014
Calculated by: A.B, M.M, A.R, Q.W
Retaining Wall lb 102900 1.25 $128,625.00
Total 132300 $294,000.00
Concrete & placing - 4000 psi
Shear Walls CY 250 210 $52,500.00
Retaining Wall CY 194.4444 210 $40,833.33
Total 250 $93,333.33
Finish & cure - Architectural
Shear Walls SF 9540 2.5 $23,850.00
Retaining Wall SF 4500 2.5 $11,250.00
Total 9540 $35,100.00
Concrete Walls - Building Total $633,033.33
Columns
Formwork
C1 4 x (16"x16") SF 960 10 $9,600.00
C2 4 x (18"x18") SF 1080 10 $10,800.00
C3 9 x (24"x24") SF 3240 10 $32,400.00
Total 5280 $52,800.00
Reinforcement
C1 lb 25088 1 $25,088.00
C2 lb 31752 1 $31,752.00
C3 lb 127008 1 $127,008.00
Total 183848 $183,848.00
Concrete & placing – 4000 psi
C1 CY 47.40741 210 $9,955.56
C2 CY 60 210 $12,600.00
C3 CY 240 210 $50,400.00
Total 347.4074 $72,955.56
Finish & cure - Architectural
C1 SF 960 2.5 $2,400.00
C2 SF 1080 2.5 $2,700.00
C3 SF 3240 2.5 $8,100.00
Total 5280 $13,200.00
Columns Total Cost $322,803.56
Suspended slabs , with or without drop panels
Form & shore slab soffit
S1 12 bays SF 16875 12.5 $210,937.50
S2 4 bays SF 6525 12.5 $81,562.50
Total 23400 $292,500.00
Form & shore beams/girders
B1 SF 1650 12.5 $20,625.00
B2 SF 4400 12.5 $55,000.00
B3 SF 3300 12.5 $41,250.00
G1 SF 1650 12.5 $20,625.00
G2 SF 3300 12.5 $41,250.00
Total 14300 $178,750.00
Reinforcement
S1 lb 294000 1 $294,000.00
S2 lb 98000 1 $98,000.00
B1 lb 11760 1 $11,760.00
13. Campbell Hall
“COST ESTIMATION”
Date: 03/28/2014
Calculated by: A.B, M.M, A.R, Q.W
B2 lb 31360 1 $31,360.00
B3 lb 23520 1 $23,520.00
G1 lb 11760 1 $11,760.00
G2 lb 23520 1 $23,520.00
Total 493920 $493,920.00
Concrete & placing – 4000 psi
S1 CY 555.5556 215 $119,444.44
S2 CY 185.1852 215 $39,814.81
B1 CY 22.22222 215 $4,777.78
B2 CY 59.25926 215 $12,740.74
B3 CY 44.44444 215 $9,555.56
G1 CY 22 215 $4,777.78
G2 CY 44 215 $9,555.56
Total 933 $200,666.67
Finish & cure top
S1 SF 22500 0.5 $11,250.00
S2 SF 7500 0.5 $3,750.00
Total 30000 $15,000.00
Finish & cure exposed soffit
S1 SF 16875 0.5 $8,437.50
S2 SF 6525 0.5 $3,262.50
B1 SF 1650 0.5 $825.00
B2 SF 4400 0.5 $2,200.00
B3 SF 3300 0.5 $1,650.00
G1 SF 1650 0.5 $825.00
G2 SF 3300 0.5 $1,650.00
Total 37700 $18,850.00
Suspended Slabs , With or Without Drop Panels Total Cost $1,199,686.67
Total Project Cost $2,627,489.33
14. GENERAL PROJECT DESCRIPTION
University of California intends to construct a new 30,000 square foot building that will house
classrooms, offices, and library stacks on the U.C. Berkeley Campus. It will also have a roof garden.
Campbell Hall will be constructed of concrete and have a basement and two stories above ground. The
basement is intended for library, the first floor is intended for classrooms, and the second floor is
intended for office spaces. The seismic loading is considered in the design of the building because the
building is located 1,500 feet away from Hayward fault. The project is registered to receive a LEED
Platinum rating. The expected completion date is December 4th
2014.
PROJECT DESIGN AND ENGINEERING
Project Manager: Professor Jack Moehle
Structural Engineer: CalStructures
Geotechnical Engineer: Kutt and Filit
Architect: Mahyar Mostafavy
General Contractor: DPR Construction
STRUCTURAL SYSTEM
The building will have a system consisting of frames to resist gravity loads and structural walls to resist
wind and earthquake loads. The frames consist of beams, girders, and columns, whereas the lateral
system consists of four shear walls located in the perimeter of the building.
BUILDING CODE AND REFERENCES
The building is designed in compliance to ACI 318-11 and ASCE 07-10.
Other references:
USGS Detailed Report (http://earthquake.usgs.gov/designmaps)
NEHRP Seismic Design Technical Brief
LOADING CRITERIA
Unit weight of concrete is assumed to be 150 pcf and is used for the calculation of structural members’
self-weight such as slabs, beams, girders, and columns.
The dead and live load breakdown for the building is tabulated below:
Table 1: Load Breakdown
ROOF DEAD LOAD LIVE LOAD
Self Weight 150 pcf Garden 100 psf
HVAC 15 psf
Lighting 5 psf
15. LEVEL2 DEAD LOAD LIVE LOAD
Self Weight 150 pcf Office 50 psf
HVAC 15 psf
Lighting 5 psf
Cladding 15 psf
LEVEL1 DEAD LOAD LIVE LOAD
Self Weight 150 pcf Classes 100 psf
HVAC 15 psf
Lighting 5 psf
Cladding 15 psf
BASEMENT DEAD LOAD LIVE LOAD
Self Weight 150 pcf Shelf 300 psf
Cladding 15 psf
The loading criteria for seismic loading are as follows:
Site Class: B
Risk Category: II R = 5, Importance factor = 1
Seismic Design Category: D
Deflection Amplification Factor, Cd = 5
Over-strength factor, Ωo = 2.5
Seismic weight =
Base shear, V=
Wind loads are also considered, and the loading criteria are as follows:
Exposure: B
Design wind speed = 85 mph
Gust factor = 0.85
Importance factor = 1
Topographic effect = 1
Wind directionality factor = 0.85
MATERIALS
Table 2: Material Properties
Soils Density = 110 pcf
Concrete Density = 150 pcf, f’c = 4 ksi
Reinforcement grade ASTM A775, fy = 60 ksi
16. LOAD COMBINATIONS
1.4 * Dead Load
1.2 * Dead Load + 1.6 * Live Load
1.0 * Dead Load + 0.25 * Live Load
ANALYSIS AND DESIGN APPROACH SUMMARY
We use LRFD method for the most part of our analysis.
SAP2000 is used to analyze and obtain design moments for slabs, beams, girders and columns.
XTRACT is used to obtain P-M interaction diagrams for columns.
After obtaining the design moment, we generally proceed by selecting the required reinforcement and
making sure that demand is less than capacity. We then perform the necessary checks as required by
ACI 318-11.
GEOTECHNICAL INFORMATION
According to geotechnical firm of Kutt and Filit, the soil bearing capacities and geotechnical design
recommendations are as follows:
Dead load only 3500 psf
Combined dead plus live load 4500 psf
Combined dead plus live plus wind or seismic load 6500 psf
Soil active pressure = 30 pcf per foot of retaining wall height, applied as a triangular load to the
wall, with 0 pressure at grade
Soil passive pressure = 50 pcf per foot of soil applied as a triangular load to the wall with 0
pressure at 18” below grade or basement slab
Soil seismic surcharge = 7 h where h is the height of retained soil, applied as a uniform load to
the wall.
Slab on grade should be a minimum of 6” thick and be built over a vapor retarder placed over a
minimum of 4” of coarse aggregate. This construction can support a live load of up to 150 psf.
Heavier loads should use a slab at least 8” thick
17. Campbell Hall
“SLAB”
Pg. 1 out of 6 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
SLAB DESIGN
The slab is designed as a one-way sab system with a span of 12’ 6”. SAP2000 is used to find the
design moment, Mu. In SAP2000, the slab is modeled using a fixed support at the left or right
end when there is a shear wall. This resulted in a smaller design moment. Using the design
moments, the slab is analyzed a T-beam. If the analysis shows that the slab must be designed as a
T-beam, then the effective flange width, beffective, must be used in the calculations. Otherwise, the
slab can be designed using a rectangular cross-section. When the moment is positive, steel is
placed along the bottom, and when the moment is negative, steel is placed along the top of the
slab. Cracking control spacing and temperature and shrinkage reinforcement are determined for
the maximum design moments. The below calculations adhere to ACI 318-11, ASCE 7, and
ASTM reinforcement.
Assumptions
Design live load for Level 1 is 100 psf (from ASCE 7)
Maximum aggregate size = 1”
f’c = 4000 psi at 28 days
Reinforcing steel shall conform to ASTM A615 Gr. 60 or A616 Gr. 60 or A 617 Gr. 60
moment
shear
Design for Positive Moment Resisting Reinforcement on Bottom (EW Direction):
Choosing reinforcement for positive moment (bottom steel):
From SAP2000 analysis, Mu = 5.68 k-ft = 68.16 k-in
o where
fy = 60 ksi, f’c = 4 ksi
d = 8 – 0.75 – 0.5 = 6.75”
b = 12”
Selecting Bars:
18. Campbell Hall
“SLAB”
Pg. 2 out of 6 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
Table 3.1: Bar Number and Spacing To Resist Moment on Bottom
No. Abar (in2
) Spacing (in)
3 0.11 6.95
4 0.2 12.63
5 0.31 19.6
o Where spacing
o At this stage, pick No. 4 at 12” spacing
Check:
Crack control: s ≤ 15 ( ) -2.5 Cc = 13.13”
≤ 12 ( ) = 12” governs
o where Cc = ¾”, = = = 40,000
Temperature and Shrinkage requirement: = = 0.002 ≥ 0.0018 (OK)
for positive moment resisting reinforcement on bottom (EW direction)
Choosing Temperature and Shrinkage reinforcement:
≥ 0.0018 = and s ≤ 5h ≤ 18”
Selecting Bars:
Table 3.2: Bar Number and Spacing For Temperature and Shrinkage
No. Abar (in2
) Spacing (in)
3 0.11 7.64
4 0.2 13.89
5 0.31 21.5 18 (not OK)
o Where
for Temperature and Shrinkage reinforcement (NS direction)
Design for negative moment resisting reinforcement on top (EW Direction):
Choosing reinforcement for negative moment (top steel) along the beam:
From SAP2000 analysis, Mu = - 7.34 k-ft = -88.1 k-in
Check this moment against
19. Campbell Hall
“SLAB”
Pg. 3 out of 6 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
o b effective = = = 6.25’ governs
≤ bw +16 tf = 12” + 16 (8”) = 140” = 11.67’
≤ lslab = 12.5’
= = - 1.03 k-ft < - 7.34 k-ft (from SAP)
o Hence, use Mu = -7.34 k-ft
o
o where
fy = 60 ksi, f’c = 4 ksi
d = 8 – 0.75 – 0.5 = 6.75”
b = 12”
Selecting Bars:
Table 3.3: Bar Number and Spacing To Resist Moment on Top
No. Abar (in2
) Spacing (in)
3 0.11 5.28
4 0.2 9.6
5 0.31 14.88
b effective
2.625’
BEAM
20. Campbell Hall
“SLAB”
Pg. 3 out of 6 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
o Where
o At this stage, pick No.5 at 12” spacing
Check:
Crack control: s ≤ 15 ( ) -2.5 Cc = 13.13”
≤ 12 ( ) = 12” governs
o where Cc = ¾”, = = = 40,000
Temperature and Shrinkage requirement: = = 0.003 ≥ 0.0018 (OK)
for negative moment resisting reinforcement on top (EW direction)
Design for reinforcement of top steel along girder to prevent crack (NS Direction):
Choosing reinforcement for top steel along the girder to prevent crack:
o b effective = = = 6.25’ governs
≤ bw +16 tf = 12” + 16 (8”) = 140” = 11.67’
≤ lslab = 12.5’
b effective
2.625’
GIRDER
21. Campbell Hall
“SLAB”
Pg. 5 out of 6 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
= = - 1.03 k-ft
o Hence, use Mu = -7.34 k-ft
o
o where
fy = 60 ksi, f’c = 4 ksi
d = 8 – 0.75 – 0.5 = 6.75”
b = 12”
Selecting Bars:
Table 3.4: Bar Number and Spacing To Prevent Crack
No. Abar (in2
) Spacing (in)
3 0.11 39
for reinforcement on top along girder to prevent crack (NS direction)
Sample SAP2000 analysis (for slab):
Dead Load
Combination of Live Load producing maximum moment in slab
Moment diagram for the corresponding dead and live load using load combo 1.2D + 1.6L
22. Campbell Hall
“SLAB”
Pg. 6 out of 6 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
SLAB SCHEDULE
Table 3.5: Slab Schedule
MARK DEPTH REINFORCEMENT
BOTTOM TOP T & S
S1 8” No. 4 @ 12”
(both continuous
and non-
continuous end)
No. 5 @ 12”
(both continuous
and non-
continuous end)
No. 4 @ 12”
Notes:
1. Place main reinforcing steel so that the bottom of steel is 2” above forms in beams and
girders, and ¾” in slabs.
2. For slab S1, the top reinforcement in the non-continuous end must be hooked into the
support, while all the bottom reinforcement in the non-continuous end must be extended
6” into the support.
3. Place #3 x 6.25’ @ 12” reinforcing steels on the top of slab (NS direction) along the
girder.
4. For both slab S1 and S2, one in every four bottom reinforcement shall be continuous all
the way.
23. Campbell Hall
“BEAMS”
Pg. 1 out of 7 Date: 02/14/2014
Calculated by: A.B, M.M, A.R, Q.W
BEAM 1 DESIGN
The beam spans the NS direction with a span of 25’. SAP2000 is used to find the design
moment, Mu. In SAP2000, the beam is modeled using a fixed support where there are columns
and shear walls. Using the design moments, the beam is analyzed as a T-beam. If the analysis
shows that the slab must be designed as a T-beam, then the effective flange width, beffective, must
be used in the calculations. Otherwise, the beam can be designed using a rectangular cross-
section. When the moment is positive, steel is placed along the bottom, and when the moment is
negative, steel is place along the top of the slab. The moment capacity, phi factor, area of steel,
spacing for cracking, inner spacing, and stirrups are determined for the maximum design
moments. The below calculations adhere to ACI 318-11, ASCE 7, and ASTM reinforcement.
Design at Support (Rectangular Cross-Section):
Given From SAP2000:
From load combinations, the maximum negative moment is 189.53 kip-ft.
Assumptions:
No. 8 bar reinforcement
No. 4 stirrup
moment
shear
Parameters:
From ACI 318-11, the coefficient β1 can be determined from the compressive strength of the
concrete.
1.5
24. Campbell Hall
“BEAMS”
Pg. 2 out of 7 Date: 02/14/2014
Calculated by: A.B, M.M, A.R, Q.W
Rectangular Design:
o
Selecting Bars:
Table 3.6: Number of Required Bars for Design a the Support
No. Abar (in2
) No. Req’d As, provided (in2
)
5 0.31 7 2.17
6 0.44 5 2.20
7 0.60 4 2.40
8 0.79 3 2.37
9 1.00 3 3.00
10 1.27 2 2.54
o Choose 3 No. 8 bars
Check:
Moment:
o
o
Phi Factor:
o
o
25. Campbell Hall
“BEAMS”
Pg. 3 out of 7 Date: 02/14/2014
Calculated by: A.B, M.M, A.R, Q.W
o
o
o
As:
o As max:
o As min:
o
Spacing:
Cracking:
o
o
o
contro s
- 2 1.5
Inner Spacing:
o , or ma agg. si e 1
- 2 1.5
Design at Midspan:
Given From SAP2000:
From load combinations, the maximum positive moment is 123.36 kip-ft.
Check for T Beam design:
26. Campbell Hall
“BEAMS”
Pg. 4 out of 7 Date: 02/14/2014
Calculated by: A.B, M.M, A.R, Q.W
If β1c> tf then design as a T-beam
As c is unknown the current design will be carried out assuming it’s a T-beam
Effective width calculation
o Parameters:
Effective width is equal to the lowest of the following:
o
=
Assumptions:
o εs > εy
o φ = 0.9
o Assuming 3 No.8 bars
o Assuming No. 4 stirrups
Depth to neutral axis, c, calculation:
Parameters:
o
o
o
o
b effective
BEAM
27. Campbell Hall
“BEAMS”
Pg. 5 out of 7 Date: 02/14/2014
Calculated by: A.B, M.M, A.R, Q.W
o
Check to see if neutral axis falls below tf:
Check assumptions:
Design for moment steel:
o
Selecting Bars:
Table 3.7: Number of Required Bars for Design at Midspan
No. Abar (in2
) No. Req’d As, provided (in2
)
6 0.44 5 2.20
7 0.60 3 1.80
8 0.79 2 1.58
9 1.00 2 2.00
o Choose 3 No. 7 bars
Stirrup Design:
Shear:
Are stirrups req’d?
28. Campbell Hall
“BEAMS”
Pg. 6 out of 7 Date: 02/14/2014
Calculated by: A.B, M.M, A.R, Q.W
o
o
o
o = 38.0 K
Is beam sufficient for applied shear?
o
Check:
Check Smax
o
o 10 2 contro s
Check AV,min
o
o contro s
Check req’d Vs
o
Select stirrup:
Table 3.8: Stirrup Size and Spacing
Type Av (in2
) S (in)
No. 3 0.22 16
No. 4 0.40 29
Max spacing (in) 10
Choose No. Stirrup at 10” spacing
1st
stirrup is ocated at 5” from the face of support
30. Campbell Hall
“GIRDERS”
Pg. 1 out of 8 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
GIRDER 1 DESIGN
The girder spans the EW direction with a span of 25’. SAP2000 is used to find the design
moment, Mu. In SAP2000, the girder is modeled using a fixed support where there are columns
and shear walls. Using the design moments, the beam is analyzed as a T-beam. If the analysis
shows that the slab must be designed as a T-beam, then the effective flange width, beffective, must
be used in the calculations. Otherwise, the beam can be designed using a rectangular cross-
section. When the moment is positive, steel is placed along the bottom, and when the moment is
negative, steel is place along the top of the slab. The moment capacity, phi factor, area of steel,
spacing for cracking, inner spacing, and stirrups are determined for the maximum design
moments. The below calculations adhere to ACI 318-11, ASCE 7, and ASTM reinforcement.
Design at Support (Rectangular Cross-Section):
Given From SAP2000:
From load combinations, the maximum negative moment is 191.091 kip-ft.
Assumptions:
No. 8 bar reinforcement
No. 4 stirrup
o ent
shear
Parameters:
.5
Rectangular Design:
31. Campbell Hall
“GIRDERS”
Pg. 2 out of 8 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
o
Selecting Bars:
Table 3.10: Number of Required Bars for Design at Support
No. Abar (in2
) No. Req’d As, provided (in2
)
5 0.31 7 2.17
6 0.44 5 2.20
7 0.60 4 2.40
8 0.79 3 2.37
9 1.00 3 3.00
10 1.27 2 2.54
o Choose 3 No. 8 bars
Check:
Moment:
o
o
Phi Factor:
o
o
o
o
o
As:
o As max:
o As min:
32. Campbell Hall
“GIRDERS”
Pg. 3 out of 8 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
o
Spacing:
Cracking:
o
o
o
contro s
Inner Spacing:
o or a agg. si e
contro s
Design at Midspan:
Given From SAP2000:
From load combinations, the maximum positive moment is 183.68 kip-ft.
Check for T Beam design:
If β1c> tf then design as a T-beam
As c is unknown the current design will be carried out assu ing it’s a T-beam
Effective width calculation
o Parameters:
b effective
GIRDER
33. Campbell Hall
“GIRDERS”
Pg. 4 out of 8 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
Effective width is equal to the lowest of the following:
o
=
Assumptions:
o εs > εy
o φ = 0.9
o Assuming 3 No.8 bars
o Assuming No. 4 stirrups
Depth to neutral axis, c, calculation:
Parameters:
o
o
o
o
o
Check to see if neutral axis falls below tf:
Check assumptions:
34. Campbell Hall
“GIRDERS”
Pg. 5 out of 8 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
Design for moment steel:
o
Selecting Bars:
Table 3.11: Number of Required Bars for Design at Midspan
No. Abar (in2
) No. Req’d As, provided (in2
)
7 0.60 4 2.40
8 0.79 3 2.37
9 1.00 2 2.00
o Choose 3 No. 8 bars
Check:
Moment:
o
o
Phi Factor:
o
o
o
o
o
As:
o As max:
o As min:
o
35. Campbell Hall
“GIRDERS”
Pg. 6 out of 8 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
Spacing:
Cracking:
o
o
o
contro s
Inner Spacing:
o or a agg. si e
Stirrup Design:
Shear:
From SAP:
Are stirrups req’d?
o
o
o
o Since and
stirrups are required and required for strength
Is girder sufficient for applied shear?
o
o Okay, girder is sufficient.
Check:
Check Smax
o
o
o 0 2 contro s
Check AV,min
o
o contro s
36. Campbell Hall
“GIRDERS”
Pg. 7 out of 8 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
Check req’d Vs
o
o
Select stirrup:
Table 3.12: Stirrup Size and Spacing
Type Av (in2
) S (in)
No. 3 0.22 22
No. 4 0.40 40
Max spacing (in) 10
o Choose No. 3 Stirrup at 0” spacing
o 1st
stirrup is ocated at 5” fro the face of support
o Stirrups can be discontinued when
SAMPLE SAP2000 ANALYSIS
Dead load is placed along the entire girder. Live load is placed along the first, third, fourth span
to produce the maximum moments.
Moment diagram for the corresponding dead and live load using load combination 1.2D + 1.6L
This generated a positive maximum moment of 183.678 k-ft and a negative maximum moment
of 191.091 k-ft.
Shear diagram for the corresponding dead and live load using load combination 1.2D+1.6L
37. Campbell Hall
“GIRDERS”
Pg. 8 out of 8 Date: 02/21/2014
Calculated by: A.B, M.M, A.R, Q.W
GIRDER SCHEDULE
Table 3.13: Girder Schedule
GIRDER SCHEDULE
MARK SIZE BOTTOM TOP STIRRUPS
W H “A”
BAR
“B” BAR SIZE
NO.
SPACING
G1 12” 24” 2 No. 8 1 No. 8 3 No. 8
CONT.
30 No. 3 @ 5” @ 0”
G2 12” 24” 3 No. 8 2 No. 8 6 No. 8
(2 LAYERS)
30 No. 3 @ ” @ 6”
38. Campbell Hall
“COLUMNS”
Pg. 1 out of 5 Date: 03/07/2014
Calculated by: A.B, M.M, A.R, Q.W
COLUMN DESIGN
The columns for the basement and first levels are designed. The column has a height of 15 feet. The
axial force is calculated from the given loads. SAP2000 is used to find the design moment, Mu. The
dimensions of the column and the amount of steel required are calculated. XTRACT is used to obtain a P-
M diagram for the column. For the required transverse steel, the probable moment strength (Mpr) is
used. The below calculations adhere to ACI 318-11, ASCE 7, and ASTM reinforcement.
Column 1:
To calculate the axial force the on the column, the loads from the slab and the beams/girders are
considered. Calculations are done in Excel and are attached in Appendix A.
In order to calculate the maximum moment on the column, two frames are analysed. Distributed dead
load is on all spans, whereas distributed live load is only on the first and third spans. The figure shows
the models from SAP2000 that are used to determine the maximum moment on the column. Another
figure shows the moment diagram used to obtain the maximum moment.
(a) (b)
a) The location of the distributed dead load on the frame. b) The location of the distributed live load on
the frame.
The moment diagram of the first level
40. Campbell Hall
“COLUMNS”
Pg. 3 out of 5 Date: 03/07/2014
Calculated by: A.B, M.M, A.R, Q.W
The P-M interaction Diagram for column 1, shown in Figure 1 is obtained from XTRACT by using an initial
design for the column. A rectangular column with 8 No.8 bars is input. The type of steel is specified as A
615 Gr 60. A single No 4 hoop with two ties is used. Concrete cover is 1.5”, f’c = 4 ksi and the strain is
0.003.
Figure 1: P-M Interaction Diagram for Column1
As it can be seen in Figure 1, the point Pu, Mu is inside of the curve, which means that the
design of the column is adequate.
Column Shear
(This value is obtained from XTRACT)
Effective Shear Force, Ve, and Shear Resistance, Vn, Calculation
-500
0
500
1000
1500
2000
0 500 1000 1500 2000 2500 3000
AxialLoad(kip)
Moment (kip-in)
COL 1: P-M Interaction Diagram
Mu,Pu
Mn, Pn
ΦMn, ΦPn
41. Campbell Hall
“COLUMNS”
Pg. 4 out of 5 Date: 03/07/2014
Calculated by: A.B, M.M, A.R, Q.W
o
o
o
(This is positive because the column is in compression)
Minimum Spacing for Low Demand
Spacing for column shear (ACI 318-11 §7.10.5.2)
Need to check Pu ≥ 0.35Ag f’c
Is 296 ≥ 0.35 ?
296 K < 358 K
No r qu r
Stirrup and Spacing Selection
42. Campbell Hall
“COLUMNS”
Pg. 5 out of 5 Date: 03/07/2014
Calculated by: A.B, M.M, A.R, Q.W
The calculations for other columns in the system (C2 & C3) can be found in Appendix A.
Table 3.14: Column Schedule for two levels
COLUMN SCHEDULE
MARK DETAIL SIZE VERTICAL
REINFORCEMENT
TIES
FIRST LEVEL
C1 D1 16” x 16” 8 No. 6 No 4 @ 6”
C2 D1 18” x 18” 8 No. 6 No 4 @ 4”
C3 D1 24” x 24” 8 No. 8 No 4 @ 5”
BASEMENT
C1 D1 16” x 16” 8 No. 8 No 4 @ 4.5”
C2 D1 18” x 18” 8 No. 8 No 4 @ 4.5”
C3 D1 24” x 24” 8 No. 10 No 4 @ 6”
43. Campbell Hall
“SHEAR WALLS”
Pg. 1 out of 6 Date: 04/18/2014
Calculated by: A.B, M.M, A.R, Q.W
SHEAR WALL DESIGN
First of all, we need to determine the lateral forces that need to be resisted by the shear wall during an
earthquake. We perform this analysis per ASCE 7-10 Seismic Design Criteria.
Seismic ground motion analysis
From USGS Detailed Report (http://earthquake.usgs.gov/designmaps):
The site is 37.82990
N, 122.257030
W
Per ASCE 7-10 Seismic Design Criteria section 11.4:
o Fundamental vibration period:
o
o
o
o
Calculation of building seismic weight, w
Loads at Roof Level
o
o
o
o
44. Campbell Hall
“SHEAR WALLS”
Pg. 2 out of 6 Date: 04/18/2014
Calculated by: A.B, M.M, A.R, Q.W
o
o
Loads at the Office Level
o
o
o
o
o
o
Calculation of design base shear, V
o
But has to be less than:
And has to be greater than:
o
o
o
45. Campbell Hall
“SHEAR WALLS”
Pg. 3 out of 6 Date: 04/18/2014
Calculated by: A.B, M.M, A.R, Q.W
Calculation of lateral forces at roof and second level on each shear wall
o
o (at second level)
o (at roof level)
Select shear wall thickness, , such that the average shear stress on the wall does not exceed
o
Unfactored Dead Load PD and Live Load calculation on Shear Wall
Calculation of the amount of boundary element longitudinal reinforcement required for moment
strength at the wall base
46. Campbell Hall
“SHEAR WALLS”
Pg. 4 out of 6 Date: 04/18/2014
Calculated by: A.B, M.M, A.R, Q.W
Neutral axis, c, depth calculation
Depth of Boundary Element
o
o -
Boundary Element transverse steel calculation (consider Special Boundary Element)
o
o
49. Campbell Hall
“RETAINING WALL”
Pg. 1 out of 5 Date: 02/14/2014
Calculated by: A.B, M.M, A.R, Q.W
RETAINING WALL DESIGN
The retaining wall will surround the basement level in regions where there is not a shear wall.
The pressure from active Earth produces bending moment on the retaining wall. Using this
design moment, the wall is design as a one-way slab system to find the reinforcement required to
resist moment. The seismic forces generate a moment within the retaining wall. Using these
forces, this moment is calculated and reinforcement is selected to resist this moment. Finally,
temperature and shrinkage reinforcement is found for the retaining wall. The below calculations
adhere to ACI 318-11, ASCE 7, and ASTM reinforcement.
Givens (provided from criteria document and geotechnical report):
Allowable Moment, Mallow, Calculation:
Horizontal Earth Pressure Component
o
Seismic Pressure Component
o
o
30’
15’
16.5’M
C
T
F3
F1
F2
M
T
C
50. Campbell Hall
“RETAINING WALL”
Pg. 2 out of 5 Date: 02/14/2014
Calculated by: A.B, M.M, A.R, Q.W
o
o
o
o
Design as a one-way slab for reinforcement to resist earth pressure:
Initial Values:
Required Steel Area Calculation, As:
o
o
o
o
Table 4.2: Bar Size and Spacing
Bar Size Area (in2
) Spacing (in)
5 0.31 8”
6 0.44 2”
Design like a beam for reinforcement o resist seismic lateral load:
Initial Value:
Required Steel Area Calculation, Aa:
51. Campbell Hall
“RETAINING WALL”
Pg. 3 out of 5 Date: 02/14/2014
Calculated by: A.B, M.M, A.R, Q.W
o
Temperature and Shrinkage Steel:
o
o
o
Table 4.3: Bar Size and Spacing for Temperature and Shrinkage
Bar Size Area (in2
) Spacing (in)
3 0.11 6”
4 0.2 ”
5 0.31 7”
Table 4.4: Retaining Wall Schedule
DESCRIPTION REINFORCEMENT
MAIN STEEL TO RESIST EARTH
PRESSURE
No. 6 @ 2”
MOMENT SEEL TO RESIST SEISMIC
LATERAL LOAD
9 No. 11 (IN 3 LAYERS)
TEMPERATURE AND SHRINKAGE STEEL No. 4 @ ”
3 layers of 9 No. 11
bars
52. Campbell Hall
“RETAINING WALL”
Pg. 4 out of 5 Date: 02/14/2014
Calculated by: A.B, M.M, A.R, Q.W
2F
F
1.5F
.5F1
1.5F
T
TC
C
C
3F
97.5 F
1.5F
48.5F
1.5F
48.5F
1.5F
1.5F
T T C
53. Campbell Hall
“RETAINING WALL”
Pg. 5 out of 5 Date: 02/14/2014
Calculated by: A.B, M.M, A.R, Q.W
In order to calculate reinforcement required for the retaining wall, the following steps are taken.
The wall has an axial force and a bending moment, therefore, it has to be designed like a column.
The axial force on the wall comes from the lateral forces. Since there are two retaining walls, the
axial force on each wall is equal to half of the sum of the two forces. P = 0.5 (2F + F). The
moment on the wall is also caused by the two lateral forces. M = 2F (37.5) + F (22.5). The wall
on the right has a compressive axial force, whereas the wall on the left has a tensile axial force.
54. Campbell Hall
“DIAPHRAGM”
Pg. 2 out of 2 Date: 04/30/2014
Calculated by: A.B, M.M, A.R, Q.W
DIAPHRAGM DESIGN (ROOF LEVEL)
Loading calculation
Chord reinforcement
o
o
Collector
VU
VU
’
11’
39’
25’
Section
cut
25’
VU
’
55. Campbell Hall
“DIAPHRAGM”
Pg. 2 out of 2 Date: 04/30/2014
Calculated by: A.B, M.M, A.R, Q.W
If Earthquake hits in the opposite direction
o
o
Compression check
o
(let s = 4 inches)
o
Note: We have to put the collectors and chord reinforcement all around the building to
anticipate earthquake coming in the other direction. Since collectors can also serve as chord
reinforcement if the earthquake hits in the other direction, and since the amount of collectors
needed is greater than the chord, we will place just the collectors all around the building.
Shear friction
We put all the collectors within the diaphragm region, so all the applied lateral load will be the shear
force in the diaphragm.
o
Pick 50 No. 6
Steel ratio check:
56. Campbell Hall
“DIAPHRAGM DESIGN”
Pg. 1 out of 3 Date: 03/30/2014
Calculated by: A.B, M.M, A.R, Q.W
DIAPHRAGM DESIGN (1ST
FLOOR)
Loading calculation
Chord reinforcement
o
o
Collector
If Earthquake hits in the opposite direction
VU
VU
’
11’
12’
39’
25’VU
”
13’
Section
cut
57. Campbell Hall
“DIAPHRAGM DESIGN”
Pg. 2 out of 3 Date: 03/30/2014
Calculated by: A.B, M.M, A.R, Q.W
o
o
Compression check
o
(let s = 3.5 inches)
o
Note: We have to put the collectors and chord reinforcement all around the building to
anticipate earthquake coming in the other direction. Since collectors can also serve as chord
reinforcement if the earthquake hits in the other direction, and since the amount of collectors
needed is greater than the chord, we will place just the collectors all around the building.
Shear friction
Since we put 6 out of 15 collector bars within the diaphragm, we have the shear force =
there.
Similarly, since we put 9 collectors in the retaining (basement) wall, we have the shear force =
there.
o
For the retaining wall region that has shear wall on top of it, the distributed vertical steel from shear
wall can serve as shear friction as well. In our case, we have 44 No. 5 at 12” as the distributed vertical
reinforcement. Hence, As = 44 * 0.31 = 13.64 inches2
≥ 13.55 inches2
(OK).
For the retaining wall region that has no shear wall on top of it, the vertical moment steel from retaining
wall can serve as shear friction as well. In our case, we have No. 6 at 12” as the vertical moment steel
from retaining wall. This should be good for the shear friction because before we have No. 5 at 12” and
it is sufficient
58. Campbell Hall
“DIAPHRAGM DESIGN”
Pg. 3 out of 3 Date: 03/30/2014
Calculated by: A.B, M.M, A.R, Q.W
For diaphragm shear friction design:
o
o =2.5
Pick 33 No. 6
Steel ratio check:
59. Campbell Hall
“FOUNDATIONS”
Pg. 1 out of 3 Date: 04/27/2014
Calculated by: A.B, M.M, A.R, Q.W
STRIP FOOTING DESIGN
First, the width of the footing is determined based on dead and live loads. Then the footing is checked
for one way shear. The moment is used to determine the reinforcement required for the footing. The
development length of the bars and dowels is also checked. Finally, the length of the lap splice is
determined.
Analysis for loads
Dead Load
Load over 3 levels
Footing width calculation
One way shear check
o
o
60. Campbell Hall
“FOUNDATIONS”
Pg. 2 out of 3 Date: 04/27/2014
Calculated by: A.B, M.M, A.R, Q.W
o
o
o
∴ The footing width is adequate
Moment design
o
o
o
o
o
∴
Moment steel development length check
Provided Development Length Calculation
o
Required Development Length
61. Campbell Hall
“FOUNDATIONS”
Pg. 3 out of 3 Date: 04/27/2014
Calculated by: A.B, M.M, A.R, Q.W
o
o
Dowel placement
Using No. 5 bars, and check if the development length is sufficient
-
o
o
o
o
∴
STRIP FOOTING SCHEDULE
Table 4.2: Strip Footing Schedule
MARK WIDTH DEPTH REINFORCEMENT DOWELS
F4 5’ 1’ 6” No. 4 @ 12” 8 No. 5
62. Campbell Hall
“FOUNDATIONS”
Pg. 1 out of 6 Date: 04/05/2014
Calculated by: A.B, M.M, A.R, Q.W
SPREAD FOOTING DESIGN
Spread Footing for Column 1:
First, the dimensions of the footing are determined based on dead and live loads that were calculated
earlier and are included in the appendix. Then the footing is checked for one way shear and two way
shear. The moment is used to determine the reinforcement required for the footing. Bearing resistance,
the development length of the bars and dowels are also checked. Finally, the length of the lap splice is
determined.
C1 = 8” (square column)
D = 142.5 K
L= 78.13 K
Assumptions
Two cases are examined. In the first case, only dead load (D)
is considered and in the second case, dead and live loads
(D+L) are considered.
Case 1:
(recommended by Kutt and Filit)
63. Campbell Hall
“FOUNDATIONS”
Pg. 2 out of 6 Date: 04/05/2014
Calculated by: A.B, M.M, A.R, Q.W
Case 2:
(recommended by Kutt and Filit)
∴ A footing plan dimension of 8’x8’ is used.
Shear Check
One-way Shear:
First, one way shear is considered. In this case, the critical section is a distance d away from the column.
o
o
λ=1.00
-
o
64. Campbell Hall
“FOUNDATIONS”
Pg. 3 out of 6 Date: 04/05/2014
Calculated by: A.B, M.M, A.R, Q.W
o
∴These plan dimensions are adequate
Two-Way Shear:
o =0.75
o
As the column is positioned central to the footing =40
This will not dominate because it is a square footing & βc =1
Or
65. Campbell Hall
“FOUNDATIONS”
Pg. 4 out of 6 Date: 04/05/2014
Calculated by: A.B, M.M, A.R, Q.W
Or
∴
o
o
o
∴The current dimensions are sufficient
Moment Design
o
o
o
o
o
o
o
∴Use 6 No. 5 bars in both directions
66. Campbell Hall
“FOUNDATIONS”
Pg. 5 out of 6 Date: 04/05/2014
Calculated by: A.B, M.M, A.R, Q.W
Development Length
o
o
o
o
o λ=1
o
37” > 23.7”
∴ The length provided is sufficient
Dowel Placement
Using #8 dowel
b – the straight part of the dowel.
o
o
67. Campbell Hall
“FOUNDATIONS”
Pg. 6 out of 6 Date: 04/05/2014
Calculated by: A.B, M.M, A.R, Q.W
”
∴The provided is currently satisfactory and b=21.75”
Distance from Foundation Level to Lap Splice, a, Calculation
a = max:
∴ a=30” – this is the length of the lap splice
Bearing Resistance
o
o
∴
∴This footing size is adequate
Table 4.1: Spread Footing Schedule
69. Campbell Hall
“APPENDIX A”
Calculated by: A.B, M.M, A.R, Q.W
The P-M interaction Diagram for column 1, shown in Figure 1 is obtained from XTRACT by using an initial
design for the column. A rectangular column with 8 No.8 bars is inputted. The type of steel is specified
as A 615 Gr 60. A single No 4 hoop with two ties is used. Concrete cover is 1.5”, f’c = 4 ksi and the strain
is 0.003.
Figure A.1: P-M Interaction Diagram
As it can be seen in Figure A.1, the point Pu, Mu is inside of the curve, which means that the
design of the column is adequate.
Column Shear
(This value is obtained from XTRACT)
Effective Shear Force, Ve, and Shear Resistance, Vn, Calculation
-500
0
500
1000
1500
2000
2500
0 500 1000 1500 2000 2500 3000 3500 4000
AxialLoad(kip)
Moment (kip-in)
COL 2: P-M Interaction Diagram
Mu, Pu
Mn, Pn
ΦMn, ΦPn
70. Campbell Hall
“APPENDIX A”
Calculated by: A.B, M.M, A.R, Q.W
o
o
o
(This is positive because the column is in compression)
Minimum Spacing for Low Demand
Spacing for column shear (ACI 318-11 §7.10.5.2)
Need to check Pu ≥ 0.35Ag f’c
Is 574 ≥ 0.35 ?
574 K > 454K
Special requirement
71. Campbell Hall
“APPENDIX A”
Calculated by: A.B, M.M, A.R, Q.W
o
o
Stirrup and Spacing Selection
Column 3:
Ultimate Moment, Mu, and Ultimate Axial Load, Pu
Mu = 1445.9 K-in
Initial Values
Gross Cross Sectional Area, Ag, Calculation
72. Campbell Hall
“APPENDIX A”
Calculated by: A.B, M.M, A.R, Q.W
Longitudinal Reinforcement Area, Ast, Calculation
Spacing Requirements
Allowable center-to-center spacing
The P-M interaction Diagram for column 1, shown in Figure 1 is obtained from XTRACT by using an initial
design for the column. A rectangular column with 8 No.8 bars is inputted. The type of steel is specified
as A 615 Gr 60. A single No 4 hoop with two ties is used. Concrete cover is 1.5”, f’c = 4 ksi and the strain
is 0.003.
73. Campbell Hall
“APPENDIX A”
Calculated by: A.B, M.M, A.R, Q.W
Figure A.2: P-M Interaction Diagram
As it can be seen in Figure A.2, the point Pu, Mu is inside of the curve, which means that the
design of the column is adequate.
Column Shear
(This value is obtained from XTRACT)
Effective Shear Force, Ve, and Shear Resistance, Vn, Calculation
o
o
o
-1000
-500
0
500
1000
1500
2000
2500
3000
3500
4000
0 2000 4000 6000 8000 10000
AxialLoad(kip)
Moment (kip-in)
COL 3: P-M Interaction Diagram
Mu, Pu
Mn, Pn
ΦMn, ΦPn
74. Campbell Hall
“APPENDIX A”
Calculated by: A.B, M.M, A.R, Q.W
(This is positive because the column is in compression)
Minimum Spacing for Low Demand
Spacing for column shear (ACI 318-11 §7.10.5.2)
Need to check Pu ≥ 0.35Ag f’c
Is 1112 ≥ 0.35 ?
1112K >806k
S r qu r
o
84. FOOTINGS Choosen OKAY IF BELOW:
Changes demand < capacity
D+L - unfactored Checks - demand (ex: Vu)
A_trib_wall 625 ft2 Checks - capacity (ex: ΦVn)
SLAB (psf) trib (ft)
D_slab
(plf)
factor for
B + G
D_slab per
level (plf) No. slabs
D_slab
total
Wall sw
(plf)
D total
(klf)
DEAD 120 12.5 1500 1.2 1800 3 5400 10125 15.525
3 levels (psf) trib (ft) L total (klf)
LIVE 250 12.5 3.125
COL 1
wall thickness 18 inches 1.5 ft
w D 15.525 ksf
w D + L 18.65 ksf
q_allow (D) 3.5 ksf
q_allow (D + L) 4.5 ksf
ρ RC 0.15 kcf
ρ soil 0.11 kcf
H 24 inches 2 ft
h 18 inches 1.5 ft
H + h = amount below grade
Dead + Live (1' strip)
q = (D + L) / ab + Hρ_soil + hρ_RC ≤ q_allow A = 4.60 ft
q D (klf) D + L (klf)
Hρ_soil
(ksf)
hρ_RC
(ksf)
q_allow
(D) (ksf)
q_allow (D
+ L) (ksf)
15.525 18.65 0.22 0.225 3.5 4.5 » A = 5.08 ft
a = 5.00 ft
60 in
Dead (1' strip)
a
b
85. Bearing Pressure -> Shear
One way shear
f'c 4000 psi 4 ksi Vn = Vc + Vs, Vs = 0
Φ 0.75 Vc = 2*sqrt(f'c)*bw*d*λ
λ 1 Vu = q_u,net * l_b*a
concrete cover 3 inches kips
dia - No. 8 bar 1 inches Vc 22.01 kips
d 14.5 inches ΦVc 16.51 kips
c + d 32.5 inches Vu 2.56 kips
q_u,net = Pu/ab 4.726 ksf 0.032819 ksi
l_b 0.54166667 ft 6.5 inches
Moment
Φ 0.9 M_u2 86.84 k-in 7.236688 k-ft
q_u,net 4.726 ksf 0.0328 ksi Mn 96.49 k-in
l_b 21 inches 1.75 ft Mn = As fy d (1 - 0.59*(As/bd)*(fy/f'c))
fy 60 ksi As = (-b +/- sqrt(b^2-4ac))/(2a)
f'c 4 ksi a = 0.59*fy^2/ (bf'c)
b = -fyd
c = Mn
a 8.85
b -870
c 96.4892 98.194 in^2
As 0.111
just As -> No bars
No Area
(in^2)
No 3 0.11
No 4 0.2
No 5 0.31
No 6 0.44
No 7 0.6
No 8 0.79
86. No 9 1
No 10 1.27
select No. 4 @ 12"
Check l_d
l_provided 18 inches
fy 60000 psi 60 ksi
f'c 4000 psi 4 ksi
≤ No 6 l_d = fy Ψt Ψe db / (25λ sqrt(f'c)) = 18.97367 inches
≥ No 7 l_d = fy Ψt Ψe db / (20 λ sqrt(f'c)) = 23.71708 inches
Ψt 1 bottom bars
Ψe 1
λ 1
db No. 4 0.5
Ψs 0.8 < No. 6
(cb + ktr)/db 2.5
long formula l_d = 3 fy Ψt Ψe Ψs db/ (40 λ sqrt(f'c) (cb+ktr)/db)) = 11.3842 Okay if l_provided > l_d
Placement of dowel steel
db, dowel 0.625 inches assume No 5
db, c wall (No 4) 0.5 inches
fy 60000 psi 60 ksi
f'c 4000 psi 4 ksi
cover 3 inches
compression lap splice
lsc = 0.0005 fy db,dowel = 18.75 inches
ldc = 0.02 fy db, col / sqrt(f'c) = 9.486833 inches
ldc = 0.0003 fy db,col = 9 inches
a = 19 inches
hook end
db No. 4 0.5
ldc = 0.0003 fy db,col = 11.25 inches
l_dc = 0.02 fy db / sqrt(f'c) = 11.85854 inches
b = l_dcprov= h-cover-2db-4db, dow = 12 inches Okay if l_dc provided > l_dc
87. Shear Walls max axial loading - when all spans loaded
Vu = Φ 10 √(f'c) Acw when wall has no reinforcement, use factor 10
Vu = Φ 5 √(f'c) Acw when wall has no reinforcement, use factor 5
Acw = b l, b = height, l = width
length, lw 25 ft
height, hw 30 ft
S_DS 1.643 g
S_D1 0.683 g
To 0.0831406 s
Ts 0.415703 s
hn 30 ft
Ct 0.02 period parameter from Table 12.8-2
x 0.75
T = Ta 0.2563722 s > To < T < Ts < TL
Sa = S_DS 1.643 g
Effective Seismic Weight
Roof
SLAB G + B C 1 C 2 C 3 Wall ∑ D (k)
No. 1 52 4 4 9 4
120 5 2 2.53125 4.5 28.125
D (k) 1200 260 8 10.125 40.5 112.5 1631.125
L (k) 1000
w2 (k) 1881.125
Level 2
SLAB G + B C 1 C 2 C 3 Wall ∑ D (k)
No. 1 52 4 4 9 4
120 5 4 5.0625 9 56.25
D (k) 1200 260 16 20.25 81 225 1802.25
L (k) 500
w1 (k) 1927.25
w (k) 3808.375
Seismic Base Shear
V = CsW
le 1 Risk II
R 5 special reinforced shear wall
Cs 0.3286 g
1.2817302 g
0.072292 g
V base 1251.432 k
88. redundancy factor ρ 1
torsion amp factor 1.3
V design 1626.8616 K
V wall = 1/2V 813.43082 k per wall
Acw = Vu / (0.6*5*sqrt(f'c))
Acw 6595.6259 in2
Acw = 25' x b in
b 21.98542 in
Choose 18 in
Check required wall length - don't need this!!
φvn by NIST 151.78933 psi 0.151789 ksi
A req on wall 5358.946 in2
L req 297.71922 in 24.80994 ft
Vu 9835.9484
Fx = Cvx V
k 1
h 15 ft
∑ wihi 85342.5 k-ft
Cvx2 0.661262
Cvx1 0.338738
Fx2 537.89087 k At roof
Fx1 275.53995 k At level 2
∑ F 813.43082 k
Pu - unfactored
A_trib_wall 625 ft2
Roof + Level 2 + Level 1
SLAB (psf) G + B (k) Wall (k) ∑ D (k)
No. 3 24 1
120 2.5 168.75
D (k) 225 60 168.75 453.75
L (k) 156.25
HOW MUCH STEEL DO YOU NEED IN BOUNDARY?
∑ M = 0
M base = Mu 20269.825 k-ft
M n 22522.028 k-ft 270264.3 k-in
lw 25 ft 300 in
0.1 lw 2.5 ft 30 in
0.2 lw 5 ft 60 in
0.4 lw 10 ft 120 in
0.6 lw 15 ft 180 in
89. 0.8 lw 20 ft 240 in
As'' 8.1 in2
Ts'' = T l 486 k
Pu small 408.375 k
Ts 678.9139 k
As 11.315232 in2
No Area
(in^2)
No. of
bars
No. of
bars
No 3 0.11 102.8657 103
No 4 0.2 56.57616 57
No 5 0.31 36.50075 37
No 6 0.44 25.71644 26
No 7 0.6 18.85872 19
No 8 0.79 14.32308 15
No 9 1 11.31523 12 Choose 12 No. 9 bars
No 10 1.27 8.909631 9
No 11 1.56 7.253354 8
No 14 2.25 5.028992 6
HOW DEEP IS NEUTRAL AXIS?
Pu max 622.625 k
Pu / (Ag f'c) 0.0288252
from NIST graph c/lw = 0.08 -> c = 0.08 lw
c 2 ft 24 in
DEPTH OF B.E. ?
min depth of BE, l_be = max of below conditions
c/2 12 in
c - 0.1 lw -6 in
cover 1.5 in
max depth of BE 21 in
Choose l_be 20 in
CONFINEMENT REINFORCEMENT?
bc1, horiz 17 in
bc2, vert 15 in
spacing
bw 18 in
s ≤ 6 db, longit (no 9) 6.768 in
s ≤ bw/3 6 in <- controls
Choose 4 in
Choose 3 vert legs use bc1
90. Ash ≥ 0.09 s bc f'c / fy 0.408 in2
A per leg 0.136 in2 <- w/ 3 legs at No 4 bar
No Area
(in^2)
No. of
bars
No. of
bars
No 3 0.11 3.709091 4
No 4 0.2 2.04 3
No 5 0.31 1.316129 2
Choose 3 horiz leg use bc2
Ash ≥ 0.09 s bc f'c / fy 0.36 in2
A per leg 0.12 in2 <- w/ 3 legs at No 4 bar
No Area
(in^2)
No. of
bars
No. of
bars
No 3 0.11 1.090909 2
No 4 0.2 0.6 1
No 5 0.31 0.387097 1
WALL SHEAR - MIDDLE OF SHEAR WALL
Vu ≤ phi Vn
hw/ lw 1.2
α 3
pick No 3 for At middle horiz steel
Vn 1355.718 k
ρt 0.001022 < 0.0025
thus, ρt 0.0025
ρt = A t /(bw s) -> s 4.8888889 in Choose 5 in
REQUIRED DISTRIBUTED VERTICAL STEEL - MIDDLE OF SHEAR WALL
ρt 0.0025
ρl = 0.0025 + 0.5*(2.5 - hw/lw) (ρ t - 0.0025)
ρl 0.0025
91. Retaining Wall
σ a 30 pcf > w 792 plf
σ s 7h > w' 161.7 plf
length 16.5 ft
Mmax = M1 13.79981 k-ft soil
M2 5.502853 k-ft EQ
M1 + M2 (allow) 19.30266 k-ft 231.6319 k-in
Mn = Mu/ 0.9 21.4474 k-ft 257.3688 k-in
Design as one-way slab
thickness (l/20) 9.9 in use 14 in
d = 0.8h, h=1' 9.6 in
Mn = As fy d (1 - 0.59(Asfy/(bd*f'c)))
As = (-b +/- sqrt(b^2-4ac))/(2a)
a = 0.59*fy^2/ (bf'c)
b = -fyd
c = Mn
a 37.92857
b -576
c 257.3688
As 0.461 14.726 in^2
just As -> No bars
No Area
(in^2) s = 12*Ab/As
No 3 0.11 2.864564
No 4 0.2 5.208299
No 5 0.31 8.072864
No 6 0.44 11.45826
No 7 0.6 15.6249
No 8 0.79 20.57278
No 9 1 26.0415
No 10 1.27 33.0727
No 11 1.56 40.62473
No 14 2.25 58.59336
Choose No. 6 @ 12"
∑ M Q = - (15+16.5/2)*Fx1 - (30+16.5/2) *(Fx2) + M + F3*16.5/2= 0
F x2 537.8909 k
F x1 275.5399 k
F x3 813.4308 k
M 20269.83 k-ft 243237.9 k-in
0.5 M (b/c 2 side) 10134.91 k-ft 121619 k-in
Design like a beam
b 14 in
h 16.5 ft
92. d = 0.8*h 13.2 ft 158.4 in
Mn = As fy d (1 - 0.59(Asfy/(bd*f'c)))
As = (-b +/- sqrt(b^2-4ac))/(2a)
a = 0.59*fy^2/ (bf'c)
b = -fyd
c = Mn
a 37.92857
b -9504
c 121619
As 13.527 237.049 in^2
just As -> No bars
No Area
(in^2) No bars
No 3 0.11 122.9711 123
No 4 0.2 67.63412 68
No 5 0.31 43.63491 44
No 6 0.44 30.74278 31
No 7 0.6 22.54471 23
No 8 0.79 17.12256 18
No 9 1 13.52682 14
No 10 1.27 10.65104 11
No 11 1.56 8.671041 9
No 14 2.25 6.011922 7
Choose 14 No. 9
T & S
ρ TS 0.0025
b 14 in
A_TS = sh* ρ TS
A TS = 0.035 s
s ≤ A TS/ 0.035
No Area
(in^2) s (in)
No 3 0.11 6.285714 6
No 4 0.2 11.42857 11
No 5 0.31 17.71429 17
Choose 2 No 4 at 10 inches
93. FOOTINGS OKAY IF BELOW: Choosen
demand < capacity Changes
COL 1 Checks - demand (ex: Vu)
C1 16 inches 1.333333 ft Checks - capacity (ex: ΦVn)
C2 16 inches 1.333333 ft
Pu 296 kips
D 142.5 kips
D + L 220.625 kips
q_allow (D) 3.5 ksf
q_allow (D + L) 4.5 ksf
ρ RC 0.15 kcf
ρ soil 0.11 kcf
H 24 inches 2 ft
h 30 inches 2.5 ft
H + h = amount below grade Dead + Live
A = 56.50 ft2
q = (D + L) / ab + Hρ_soil + hρ_RC ≤ q_allow a = b = 7.52 ft
q D (k) D + L (k)
Hρ_soil
(ksf)
hρ_RC
(ksf)
q_allow
(D) (ksf)
q_allow (D
+ L) (ksf)
Dead
142.5 220.625 0.22 0.375 3.5 4.5 » A = 49.05 ft2
a = b = 7.00 ft
a = b= 8 ft
96 in
Bearing Pressure -> Shear
One way shear
f'c 4000 psi 4 ksi Vn = Vc + Vs, Vs = 0
Φ 0.75 Vc = 2*sqrt(f'c)*bw*d*λ
λ 1 Vu = q_u,net * l_b*a
concrete cover 3 inches
dia - No. 8 bar 1 inches Vc 315.72 kips
d 26 inches ΦVc 236.79 kips
c + d 42 inches Vu 43.17 kips
q_u,net 4.625 ksf 0.0321 ksi
a
b
c1
c2
94. l_b 14 inches
Two way shear Vu 239.34 kips
bo 168 inches Vc ≤ 1105.03 kips
Φ 0.75 βc 1657.54 kips
α_s 40 center α_s 2262.67 kips
βc 1 ΦVc 828.77 kips
Moment
Φ 0.9 M_u2 2466.67 k-in
q_u,net 4.625 ksf 0.0321 ksi Mn 2740.74 k-in
l_b 40 inches Mn = As fy d (1 - 0.59*(As/bd)*(fy/f'c))
fy 60 ksi As = (-b +/- sqrt(b^2-4ac))/(2a)
f'c 4 ksi a = 0.59*fy^2/ (bf'c)
b = -fyd
c = Mn
a 5.53125
b -1560
c 2740.741
As 1.768 280.266 in^2
just As -> No bars
No Area
(in^2)
No. of
bars
No 3 0.11 16.07243
No 4 0.2 8.839839
No 5 0.31 5.703122
No 6 0.44 4.018109
No 7 0.6 2.946613
No 8 0.79 2.237934
No 9 1 1.767968
No 10 1.27 1.392101
select 6 No. 5
Check l_d
l_provided 37 inches
fy 60000 psi 60 ksi
95. f'c 4000 psi 4 ksi
≤ No 6 l_d = fy Ψt Ψe db / (25λ sqrt(f'c)) = 23.71708 inches Okay if l_provided > l_d
≥ No 7 l_d = fy Ψt Ψe db / (20 λ sqrt(f'c)) = 29.64635 inches
Ψt 1 bottom bars
Ψe 1
λ 1
db No. 5 0.625
Placement of dowel steel
db, dowel 1 inches assume No 8
db, col (No 8) 1 inches
fy 60000 psi 60 ksi
f'c 4000 psi 4 ksi
cover 3 inches
compression lap splice
lsc = 0.0005 fy db,dowel = 30 inches
ldc = 0.02 fy db, col / sqrt(f'c) = 18.97367 inches
ldc = 0.0003 fy db,col = 18 inches
a = 30 inches
hook end
db No. 5 0.625
l_dc = 0.02 fy db / sqrt(f'c) = 18.97367 inches
l_dcprov= h-cover-2db-4db, dow = 21.75 inches Okay if l_dc provided > l_dc
Check Bearing
Φ 0.65 Pn = 0.85f'cAc*sqrt(A2/A1) + Asfy
y = -2x+2h 60 when x = 0 Pn 2120 kips
fy 60000 psi 60 ksi ΦPn 1378 kips > 296 Pu
f'c 4000 psi 4 ksi Okay
Ac = A col 256 in^2
A1 (col) 256 in^2
A2 (bearing) 18496 in^2
sqrt(A2/A1) ≤ 2 8.5 ≤ 2
Adowel = As 8 No 8 (8 bars b/c its that many in col)
6.32 in^2
