5 Lateral Component Design
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Transcript of 5 Lateral Component Design
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PCI 6th EditionPCI 6th Edition
Lateral Component Design
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Presentation OutlinePresentation Outline
• Architectural Components– Earthquake Loading
• Shear Wall Systems– Distribution of lateral loads– Load bearing shear wall analysis– Rigid diaphragm analysis
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Architectural ComponentsArchitectural Components
• Must resist seismic forces and be attached to the SFRS
• Exceptions– Seismic Design Category A– Seismic Design Category B with I=1.0
(other than parapets supported by bearing or shear walls).
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Seismic Design Force, FpSeismic Design Force, Fp
Fp=
0.4apS
DSW
p
Rp
1+2z
h
0.3SDS
Wp
Fp
1.6SDS
Wp
Where:ap = component amplification factorfrom Figure 3.10.10
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Seismic Design Force, FpSeismic Design Force, Fp
Fp=
0.4apS
DSW
p
Rp
1+2z
h
0.3SDS
Wp
Fp
1.6SDS
Wp
Where:Rp = component response modification factor from Figure 3.10.10
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Seismic Design Force, FpSeismic Design Force, Fp
Fp=
0.4apS
DSW
p
Rp
1+2z
h
0.3SDS
Wp
Fp
1.6SDS
Wp
Where:h = average roof height of structureSDS= Design, 5% damped, spectral
response acceleration at short periodsWp = component weight
z= height in structure at attachment point < h
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Cladding Seismic Load ExampleCladding Seismic Load Example
• Given:– A hospital building in Memphis, TN – Cladding panels are 7 ft tall by 28 ft long. A 6 ft
high window is attached to the top of the panel, and an 8 ft high window is attached to the bottom.
– Window weight = 10 psf– Site Class C
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Cladding Seismic Load ExampleCladding Seismic Load Example
Problem:– Determine the seismic forces on the panel
• Assumptions– Connections only resist load in direction assumed
– Vertical load resistance at bearing is 71/2” from exterior face of panel
– Lateral Load (x-direction) resistance is 41/2” from exterior face of the panel
– Element being consider is at top of building, z/h=1.0
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Solution StepsSolution Steps
Step 1 – Determine Component Factors Step 2 – Calculate Design Spectral Response
AccelerationStep 3 – Calculate Seismic Force in terms of
panel weightStep 4 – Check limitsStep 5 – Calculate panel loadingStep 6 – Determine connection forcesStep 7 – Summarize connection forces
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Step 1 – Determine ap and RpStep 1 – Determine ap and Rp
• Figure 3.10.10
aapp R Rpp
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Step 2 – Calculate the 5%-Damped Design Spectral Response Acceleration
Step 2 – Calculate the 5%-Damped Design Spectral Response Acceleration
S
DS=1.0
Where:SMS = FaSS
Ss = 1.5 From maps found in IBC 2003Fa = 1.0 From figure 3.10.7
S
DS=
2
3
S
MS
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Step 3 – Calculate Fp in Terms of WpStep 3 – Calculate Fp in Terms of Wp
0.4 1.0 1.0 Wp
2.51+2 1.0 0.48W
p
0.4 1.0 1.0 Wp
2.51+2 1.0 0.48W
p
0.4 1.25 1.0 Wp
1.01+2 1.0 1.5W
p
Wall Element:
Body of Connections:
Fasteners:
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Step 4 – Check Fp LimitsStep 4 – Check Fp Limits
0.3 1.0 Wp
Fp
1.6 1.0 Wp
0.3W
p0.48W
p1.6W
p
0.3W
p0.48W
p1.6W
p
Wall Element:
Body of Connections:
Fasteners:p p p0.3W 1.5W 1.6W
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Step 5 – Panel LoadingStep 5 – Panel Loading
• Gravity Loading
• Seismic Loading Parallel to Panel Face
• Seismic or Wind Loading Perpendicular to Panel Face
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Step 5 – Panel LoadingStep 5 – Panel Loading
• Panel WeightArea = 465.75 in2
Wp=485(28)=13,580 lb
• Seismic Design ForceFp=0.48(13580)=6518 lb
Panel wt=
465.75
144150 485
lb
ft
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Step 5 – Panel LoadingStep 5 – Panel Loading
• Upper Window WeightHeight =6 ft
Wwindow=6(28)(10)=1680 lb
• Seismic Design Force– Inward or Outward– Consider ½ of Window
Wp=3.0(10)=30 plf
Fp=0.48(30)=14.4 plf
14.4(28)=403 lb– Wp=485(28)=13,580 lb
• Seismic Design Force– Fp=0.48(13580)=6518 lb
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Step 5 – Panel LoadingStep 5 – Panel Loading
• Lower Window Weight– No weight on panel
• Seismic Design Force– Inward or outward– Consider ½ of window
height=8 ft
Wp=4.0(10)=40 plf
Fp=0.48(30)=19.2 plf
19.2(28)=538 lb
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Step 5 Loads to Connections
Step 5 Loads to Connections
Dead Load Summary
Wp
(lb)
z
(in)
Wpz
(lb-in)
Panel 13,580 4.5 61,110
Upper Window
1,680 2.0 2,230
Lower Window
0 22.0 0
Total 15,260 64,470
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Step 6Loads to Connections
Step 6Loads to Connections
• Equivalent Load Eccentricity
z=64,470/15,260=4.2 in• Dead Load to Connections
– Vertical
=15,260/2=7630 lb – Horizontal
= 7630 (7.5-4.2)/32.5
=774.7/2=387 lb
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Step 6 – Loads to ConnectionsStep 6 – Loads to Connections
Seismic Load Summary
Fp
(lb)y
(in)Fpy
(lb-in)
Panel 6,518 34.5 224,871
Upper Window 403 84.0 33,852
Lower Window 538 0.0 0.0
Total 7,459 258,723
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Step 6 – Loads to ConnectionsStep 6 – Loads to Connections
Seismic Load Summary
Fp
(lb)z
(in)Fpz
(lb-in)
Panel 6,518 4.5 29,331
Upper Window 403 2.0 806
Lower Window 538 22.0 11,836
Total 7,459 41,973
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Step 6 – Loads to ConnectionsStep 6 – Loads to Connections
• Center of equivalent seismic load from lower left
y=258,723/7459y=34.7 in
z=41,973/7459
z=5.6 in
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Step 6 – Seismic In-Out LoadsStep 6 – Seismic In-Out Loads
• Equivalent Seismic Load
y=34.7 in
Fp=7459 lb• Moments about Rb
Rt=7459(34.7 -27.5)/32.5
Rt=1652 lb• Force equilibrium
Rb=7459-1652
Rb=5807 lb
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Step 6 – Wind Outward LoadsStep 6 – Wind Outward Loads
Outward Wind Load Summary
Fp
(lb)y
(in)Fpy
(lb-in)
Panel 3,430 42.0 144,060
Upper Window 1,470 84.0 123,480
Lower Window 1,960 0.0 0.0
Total 6,860 267,540
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Step 6 – Wind Outward LoadsStep 6 – Wind Outward Loads
• Center of equivalent wind load from lower left
y=267,540/6860
y=39.0 in• Outward Wind Load
Fp=6,860 lb
Fp
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Step 6 – Wind Outward LoadsStep 6 – Wind Outward Loads
• Moments about Rb
Rt=7459(39.0 -27.5)/32.5
Rt=2427 lb• Force equilibrium
Rb=6860-2427
Rb=4433 lb
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Step 6 – Wind Inward LoadsStep 6 – Wind Inward Loads
• Outward Wind Reactions
Rt=2427 lb
Rb=4433 lb• Inward Wind Loads
– Proportional to pressure
Rt=(11.3/12.9)2427 lb
Rt=2126 lb
Rb=(11.3/12.9)4433 lb
Rb=3883 lb
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Step 6 – Seismic Loads Normal to SurfaceStep 6 – Seismic Loads Normal to Surface
• Load distribution (Based on Continuous Beam Model)– Center connections = .58 (Load)– End connections = 0.21 (Load)
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Step 6 – Seismic Loads Parallel to FaceStep 6 – Seismic Loads Parallel to Face
• Parallel load
=+ 7459 lb
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Step 6 – Seismic Loads Parallel to FaceStep 6 – Seismic Loads Parallel to Face
• Up-down load
7459 27.5+32.5-34.7
2 156 605 lb
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Step 6 – Seismic Loads Parallel to FaceStep 6 – Seismic Loads Parallel to Face
• In-out load
7459 5.6-4.5
2 156 26 lb
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Step 7 – Summary of Factored LoadsStep 7 – Summary of Factored Loads
1. Load Factor of 1.2 Applied
2. Load Factor of 1.0 Applied
3. Load Factor of 1.6 Applied
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Distribution of Lateral Loads Shear Wall Systems
Distribution of Lateral Loads Shear Wall Systems
• For Rigid diaphragms– Lateral Load Distributed based on total
rigidity, r
Where:
r=1/D
D=sum of flexural and shear deflections
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Distribution of Lateral Loads Shear Wall Systems
Distribution of Lateral Loads Shear Wall Systems
• Neglect Flexural Stiffness Provided:– Rectangular walls– Consistent materials– Height to length ratio < 0.3
Distribution based on
Cross-Sectional Area
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Distribution of Lateral Loads Shear Wall Systems
Distribution of Lateral Loads Shear Wall Systems
• Neglect Shear Stiffness Provided:– Rectangular walls– Consistent materials– Height to length ratio > 3.0
Distribution based on
Moment of Inertia
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Distribution of Lateral Loads Shear Wall Systems
Distribution of Lateral Loads Shear Wall Systems
• Symmetrical Shear Walls
F
i
ki
r
Vx
Where:Fi = Force Resisted by individual shear wallki=rigidity of wall ir=sum of all wall rigiditiesVx=total lateral load
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Distribution of Lateral Loads “Polar Moment of Stiffness Method”
Distribution of Lateral Loads “Polar Moment of Stiffness Method”
• Unsymmetrical Shear Walls
Force in the y-direction is distributed to a given wall at a given level due to an applied force in the y-direction at that level
Fy
V
yK
y
Ky
T
Vy
xKy
Kyx2 K
xy2
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• Unsymmetrical Shear Walls
Fy
V
yK
y
Ky
T
Vy
xKy
Kyx2 K
xy2
Where:Vy = lateral force at level being consideredKx,Ky = rigidity in x and y directions of wallKx, Ky = summation of rigidities of all wallsT = Torsional Momentx = wall x-distance from the center of stiffnessy = wall y-distance from the center of stiffness
Distribution of Lateral Loads “Polar Moment of Stiffness Method”
Distribution of Lateral Loads “Polar Moment of Stiffness Method”
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• Unsymmetrical Shear Walls
Force in the x-direction is distributed to a given wall at a given level due to an applied force in the y-direction at that level.
Fx
T
Vy
yKx
Kyx2 K
xy2
Distribution of Lateral Loads “Polar Moment of Stiffness Method”
Distribution of Lateral Loads “Polar Moment of Stiffness Method”
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• Unsymmetrical Shear Walls
Fx
T
Vy
yKx
Kyx2 K
xy2
Where:Vy=lateral force at level being consideredKx,Ky=rigidity in x and y directions of wallKx, Ky=summation of rigidities of all wallsT=Torsional Momentx=wall x-distance from the center of stiffnessy=wall y-distance from the center of stiffness
Distribution of Lateral Loads “Polar Moment of Stiffness Method”
Distribution of Lateral Loads “Polar Moment of Stiffness Method”
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Unsymmetrical Shear Wall ExampleUnsymmetrical Shear Wall Example
Given:
– Walls are 8 ft high and 8 in thick
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Unsymmetrical Shear Wall ExampleUnsymmetrical Shear Wall Example
Problem:– Determine the shear in each wall due to the wind load, w
• Assumptions:– Floors and roofs are rigid diaphragms– Walls D and E are not connected to Wall B
• Solution Method:– Neglect flexural stiffness h/L < 0.3– Distribute load in proportion to wall length
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Solution StepsSolution Steps
Step 1 – Determine lateral diaphragm torsion
Step 2 – Determine shear wall stiffness
Step 3 – Determine wall forces
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Step 1 – Determine Lateral Diaphragm TorsionStep 1 – Determine Lateral Diaphragm Torsion
• Total Lateral Load
Vx=0.20 x 200 = 40 kips
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Step 1 – Determine Lateral Diaphragm TorsionStep 1 – Determine Lateral Diaphragm Torsion
• Center of Rigidity from left
x
40 75 30 140 40 180 40 30 40
130.9 ft
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Step 1 – Determine Lateral Diaphragm TorsionStep 1 – Determine Lateral Diaphragm Torsion
• Center of Rigidityy=center of building
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Step 1 – Determine Lateral Diaphragm TorsionStep 1 – Determine Lateral Diaphragm Torsion
• Center of Lateral Load from left
xload=200/2=100 ft
• Torsional Moment
MT=40(130.9-100)=1236 kip-ft
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Step 2 – Determine Shear Wall StiffnessStep 2 – Determine Shear Wall Stiffness
• Polar Moment of Stiffness
Ip
Ixx
Iyy
Ixx
ly2 of east-west wallsI
xx 15 15 2 15 15 2 6750 ft3
Iyy
lx2 of north-south wallsI
yy 40 130.9 75 2 30 140 130.9 2 ...
40 180 130.9 2 223,909 ft3
Ip
6750 223,909 230,659 ft3
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Step 3 – Determine Wall ForcesStep 3 – Determine Wall Forces
• Shear in North-South Walls
F V
xl
l
MTxl
Ip
Wall A 40 40
40 30 40 1236130.9 75 40
230,659
Wall A 14.512.0 26.5 kips
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Step 3 – Determine Wall ForcesStep 3 – Determine Wall Forces
• Shear in North-South Walls
F V
xl
l
MTxl
Ip
Wall B 40 30
40 30 40 1236130.9 140 30
230,659
Wall B 10.91 1.46 9.45 kips
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Step 3 – Determine Wall ForcesStep 3 – Determine Wall Forces
• Shear in North-South Walls
F V
xl
l
MTxl
Ip
Wall C 40 40
40 30 40 1236130.9 180 40
230,659
Wall C 14.5 10.5 4.0 kips
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Step 3 – Determine Wall ForcesStep 3 – Determine Wall Forces
• Shear in East-West Walls
F M
Tyl
Ip
Wall D andE 123615 15
230,6591.21kips
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Load Bearing Shear Wall ExampleLoad Bearing Shear Wall Example
Given:
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Load Bearing Shear Wall ExampleLoad Bearing Shear Wall Example
Given Continued:– Three level parking structure– Seismic Design Controls– Symmetrically placed shear walls– Corner Stairwells are not part of the SFRS
Seismic Lateral Force Distribution
Level Cvx Fx
3 0.500 471
2 0.333 313
1 0.167 157
Total 941
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Load Bearing Shear Wall ExampleLoad Bearing Shear Wall Example
Problem:– Determine the tension steel requirements for
the load bearing shear walls in the north-south direction required to resist seismic loading
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Load Bearing Shear Wall ExampleLoad Bearing Shear Wall Example
• Solution Method:– Accidental torsion must be included in
the analysis– The torsion is assumed to be resisted
by the walls perpendicular to the direction of the applied lateral force
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Solution StepsSolution Steps
Step 1 – Calculate force on wall
Step 2 – Calculate overturning moment
Step 3 – Calculate dead load
Step 4 – Calculate net tension force
Step 5 – Calculate steel requirements
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Step 1 – Calculate Force in Shear WallStep 1 – Calculate Force in Shear Wall
• Accidental Eccentricity=0.05(264)=13.2 ft• Force in two walls
Seismic Lateral Force Distribution
Level Cvx Fx
3 0.500 471
2 0.333 313
1 0.167 157
Total
F2w
941 180 / 213.2
180F
2w540 kips
F1w
540 / 2 270 kips
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Step 1 – Calculate Force in Shear WallStep 1 – Calculate Force in Shear Wall
• Force at each levelLevel 3 F1W=0.500(270)=135 kips
Level 2 F1W=0.333(270)= 90 kips
Level 1 F1W=0.167(270)= 45 kips
Seismic Lateral Force Distribution
Level Cvx Fx
3 0.500 471
2 0.333 313
1 0.167 157
Total 941
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Step 2 – Calculate Overturning Moment Step 2 – Calculate Overturning Moment
• Force at each levelLevel 3 F1W=0.500(270)=135 kips
Level 2 F1W=0.333(270)= 90 kips
Level 1 F1W=0.167(270)= 45 kips
• Overturning moment, MOT
MOT=135(31.5)+90(21)+45(10.5)
MOT=6615 kip-ft
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Step 3 – Calculate Dead LoadStep 3 – Calculate Dead Load
• Load on each Wall– Dead Load = .110 ksf (all components)– Supported Area = (60)(21)=1260 ft2
Wwall=1260(.110)=138.6 kips
• Total LoadWtotal=3(138.6)=415.8~416 kips
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Step 4 – Calculate Tension ForceStep 4 – Calculate Tension Force
• Governing load CombinationU=[0.9-0.2(0.24)]D+1.0E Eq. 3.2.6.7a
U=0.85D+1.0E
• Tension Force
Tu
6615 0.85 416 10
18T
u171kips
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Step 5 – Reinforcement RequirementsStep 5 – Reinforcement Requirements
• Tension Steel, As
• Reinforcement Details– Use 4 - #8 bars = 3.17 in2
– Locate 2 ft from each end
As
Tu
fy
171
0.9 60 3.17 in2
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Rigid Diaphragm Analysis ExampleRigid Diaphragm Analysis Example
Given:
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Rigid Diaphragm Analysis ExampleRigid Diaphragm Analysis Example
Given Continued:– Three level parking structure (ramp at middle bay)– Seismic Design Controls– Seismic Design Category C– Corner Stairwells are not part of the SFRS
Seismic Lateral Force Distribution
Level Cvx Fx
3 0.500 471
2 0.333 313
1 0.167 157
Total 941
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Rigid Diaphragm Analysis ExampleRigid Diaphragm Analysis Example
Problem:– Part A
Determine diaphragm reinforcement required for moment design
– Part B
Determine the diaphragm reinforcement required for shear design
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Solution StepsSolution Steps
Step 1 – Determine diaphragm forceStep 2 – Determine force distributionStep 3 – Determine statics modelStep 4 – Determine design forcesStep 5 – Diaphragm moment designStep 6 – Diaphragm shear design
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Step 1 – Diaphragm Force, FpStep 1 – Diaphragm Force, Fp
• Fp, Eq. 3.8.3.1
Fp = 0.2·IE·SDS·Wp + Vpx
but not less than any force in the lateral force distribution table
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Step 1 – Diaphragm Force, FpStep 1 – Diaphragm Force, Fp
• Fp, Eq. 3.8.3.1
Fp =(1.0)(0.24)(5227)+0.0=251 kips
•
Fp=471 kips
Seismic Lateral Force Distribution
Level Cvx Fx
3 0.500 471
2 0.333 313
1 0.167 157
Total 941
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Step 2 – Diaphragm Force, Fp, DistributionStep 2 – Diaphragm Force, Fp, Distribution
• Assume the forces are uniformly distributed– Total Uniform Load, w
• Distribute the force equally to the three bays
w
471
2641.784 kips / ft
w
1w
3
1.784
30.59 kips / ft
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Step 3 – Diaphragm ModelStep 3 – Diaphragm Model
• Ramp Model
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Step 3 – Diaphragm ModelStep 3 – Diaphragm Model
• Flat Area Model
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Step 3 – Diaphragm ModelStep 3 – Diaphragm Model
• Flat Area Model– Half of the load of the center bay is assumed to be
taken by each of the north and south bays
w2=0.59+0.59/2=0.89 kip/ft
– Stress reduction due to cantilevers is neglected.– Positive Moment design is based on ramp moment
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Step 4 – Design ForcesStep 4 – Design Forces
• Ultimate Positive Moment, +Mu
• Ultimate Negative Moment
• Ultimate Shear
M
u
w1
180 28
0.59 180 2
82390 kip ft
M
u
w2
42 22
0.89 42 2
2 785 kip ft
V
u
w1
180 2
0.59 180
253kips
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Step 5 – Diaphragm Moment DesignStep 5 – Diaphragm Moment Design
• Assuming a 58 ft moment armTu=2390/58=41 kips
• Required Reinforcement, As
– Tensile force may be resisted by:• Field placed reinforcing bars• Welding erection material to embedded plates
As
Tu
fy
41
0.7 60 0.98 in2
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Step 6 – Diaphragm Shear DesignStep 6 – Diaphragm Shear Design
• Force to be transferred to each wall
– Each wall is connected to the diaphragm, 10 ft
Shear/ft=Vwall/10=66.625/10=6.625 klf
– Providing connections at 5 ft centers
Vconnection=6.625(5)=33.125 kips/connection
V
wall
o
Vu
22.5
53
2
66.25 kips
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Step 6 – Diaphragm Shear DesignStep 6 – Diaphragm Shear Design
• Force to be transferred between Tees– For the first interior Tee
Vtransfer=Vu-(10)0.59=47.1 kips
Shear/ft=Vtransfer/60=47.1/60=0.79 klf
– Providing Connections at 5 ft centers
Vconnection=0.79(5)=4 kips
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Questions?Questions?