PCI 6th EditionPCI 6th Edition
Lateral Component Design
Presentation OutlinePresentation Outline
• Architectural Components– Earthquake Loading
• Shear Wall Systems– Distribution of lateral loads– Load bearing shear wall analysis– Rigid diaphragm analysis
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).
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
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
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
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
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
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
Step 1 – Determine ap and RpStep 1 – Determine ap and Rp
• Figure 3.10.10
aapp R Rpp
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
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:
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
Step 5 – Panel LoadingStep 5 – Panel Loading
• Gravity Loading
• Seismic Loading Parallel to Panel Face
• Seismic or Wind Loading Perpendicular to Panel Face
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
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
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
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
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
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
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
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
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
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
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
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
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
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)
Step 6 – Seismic Loads Parallel to FaceStep 6 – Seismic Loads Parallel to Face
• Parallel load
=+ 7459 lb
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
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
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
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
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
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
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
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
• 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”
• 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”
• 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”
Unsymmetrical Shear Wall ExampleUnsymmetrical Shear Wall Example
Given:
– Walls are 8 ft high and 8 in thick
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
Solution StepsSolution Steps
Step 1 – Determine lateral diaphragm torsion
Step 2 – Determine shear wall stiffness
Step 3 – Determine wall forces
Step 1 – Determine Lateral Diaphragm TorsionStep 1 – Determine Lateral Diaphragm Torsion
• Total Lateral Load
Vx=0.20 x 200 = 40 kips
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
Step 1 – Determine Lateral Diaphragm TorsionStep 1 – Determine Lateral Diaphragm Torsion
• Center of Rigidityy=center of building
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
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
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
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
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
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
Load Bearing Shear Wall ExampleLoad Bearing Shear Wall Example
Given:
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
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
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
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
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
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
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
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
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
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
Rigid Diaphragm Analysis ExampleRigid Diaphragm Analysis Example
Given:
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
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
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
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
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
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
Step 3 – Diaphragm ModelStep 3 – Diaphragm Model
• Ramp Model
Step 3 – Diaphragm ModelStep 3 – Diaphragm Model
• Flat Area Model
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
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
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
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
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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