PCI 6 th Edition Lateral Component Design. Presentation Outline Architectural Components...
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Transcript of PCI 6 th Edition Lateral Component Design. Presentation Outline Architectural Components...
PCI 6th Edition
Lateral Component Design
Presentation Outline
• Architectural Components– Earthquake Loading
• Shear Wall Systems– Distribution of lateral loads– Load bearing shear wall analysis– Rigid diaphragm analysis
Architectural 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, Fp
Fp=0.4apSDSWp
Rp
1+2zh
0.3SDSWp Fp 1.6SDSWp
Where:ap = component amplification factorfrom Figure 3.10.10
Seismic Design Force, Fp
Fp=0.4apSDSWp
Rp
1+2zh
0.3SDSWp Fp 1.6SDSWp
Where:Rp = component response modification factor from Figure 3.10.10
Seismic Design Force, Fp
Fp=0.4apSDSWp
Rp
1+2zh
0.3SDSWp Fp 1.6SDSWp
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 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 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 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 Rp
• Figure 3.10.10
aapp R Rpp
Step 2 – Calculate the 5%-Damped Design Spectral Response Acceleration
SDS=1.0
Where:SMS = FaSS
Ss = 1.5 From maps found in IBC 2003Fa = 1.0 From figure 3.10.7
SDS=
23
SMS
Step 3 – Calculate Fp in Terms of Wp
0.4 1.0 1.0 Wp
2.51+2 1.0 0.48Wp
0.4 1.0 1.0 Wp
2.51+2 1.0 0.48Wp
0.4 1.25 1.0 Wp
1.01+2 1.0 1.5Wp
Wall Element:
Body of Connections:
Fasteners:
Step 4 – Check Fp Limits
0.3 1.0 Wp Fp 1.6 1.0 Wp
0.3Wp 0.48Wp 1.6Wp
0.3Wp 0.48Wp 1.6Wp
Wall Element:
Body of Connections:
Fasteners:p p p0.3W 1.5W 1.6W
Step 5 – Panel Loading
• Gravity Loading
• Seismic Loading Parallel to Panel Face
• Seismic or Wind Loading Perpendicular to Panel Face
Step 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 485lb
ft
Step 5 – Panel Loading
• Upper Window WeightHeight =6 ftWwindow=6(28)(10)=1680 lb
• Seismic Design Force– Inward or Outward– Consider ½ of Window
Wp=3.0(10)=30 plfFp=0.48(30)=14.4 plf14.4(28)=403 lb
– Wp=485(28)=13,580 lb
• Seismic Design Force– Fp=0.48(13580)=6518 lb
Step 5 – Panel Loading
• Lower Window Weight– No weight on panel
• Seismic Design Force– Inward or outward– Consider ½ of window
height=8 ftWp=4.0(10)=40 plfFp=0.48(30)=19.2 plf19.2(28)=538 lb
Step 5 Loads to Connections
Dead Load SummaryWp
(lb)
z(in)
Wpz
(lb-in)Panel 13,580 4.5 61,110Upper Window 1,680 2.0 2,230
Lower Window 0 22.0 0
Total 15,260 64,470
Step 6Loads to Connections
• Equivalent Load Eccentricityz=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 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 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 Connections
• Center of equivalent seismic load from lower left
y=258,723/7459y=34.7 in
z=41,973/7459z=5.6 in
Step 6 – Seismic In-Out Loads
• Equivalent Seismic Loady=34.7 inFp=7459 lb
• Moments about RbRt=7459(34.7
-27.5)/32.5Rt=1652 lb
• Force equilibriumRb=7459-1652Rb=5807 lb
Step 6 – Wind Outward Loads
Outward Wind Load SummaryFp
(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 Loads
• Center of equivalent wind load from lower left
y=267,540/6860y=39.0 in
• Outward Wind LoadFp=6,860 lb
Fp
Step 6 – Wind Outward Loads
• Moments about RbRt=7459(39.0
-27.5)/32.5Rt=2427 lb
• Force equilibriumRb=6860-2427Rb=4433 lb
Step 6 – Wind Inward Loads
• Outward Wind ReactionsRt=2427 lbRb=4433 lb
• Inward Wind Loads– Proportional to pressure
Rt=(11.3/12.9)2427 lbRt=2126 lbRb=(11.3/12.9)4433 lbRb=3883 lb
Step 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 Face
• Parallel load=+ 7459 lb
Step 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 Face
• In-out load
7459 5.6-4.5 2 156 26 lb
Step 7 – Summary of Factored Loads
1. Load Factor of 1.2 Applied2. Load Factor of 1.0 Applied3. Load Factor of 1.6 Applied
Distribution of Lateral Loads Shear Wall Systems
• For Rigid diaphragms– Lateral Load Distributed based on total
rigidity, r
Where:r=1/DD=sum of flexural and shear deflections
Distribution of Lateral Loads Shear Wall Systems
• Neglect Flexural Stiffness Provided:– Rectangular walls– Consistent materials– Height to length ratio < 0.3
Distribution based onCross-Sectional Area
Distribution of Lateral Loads Shear Wall Systems
• Neglect Shear Stiffness Provided:– Rectangular walls– Consistent materials– Height to length ratio > 3.0
Distribution based onMoment of Inertia
Distribution of Lateral Loads Shear Wall Systems
• Symmetrical Shear Walls
Fi
kir
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”
• 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
Vy Ky
Ky
TVyxKy
Ky x2 Kx y2
• Unsymmetrical Shear Walls
Fy
Vy Ky
Ky
TVyxKy
Ky x2 Kx y2
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”
• 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
TVyyKx
Ky x2 Kx y2
Distribution of Lateral Loads “Polar Moment of Stiffness Method”
• Unsymmetrical Shear Walls
Fx
TVyyKx
Ky x2 Kx y2Where:
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”
Unsymmetrical Shear Wall Example
Given:
– Walls are 8 ft high and 8 in thick
Unsymmetrical 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 Steps
Step 1 – Determine lateral diaphragm torsionStep 2 – Determine shear wall stiffnessStep 3 – Determine wall forces
Step 1 – Determine Lateral Diaphragm Torsion
• Total Lateral LoadVx=0.20 x 200 = 40 kips
Step 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 Torsion
• Center of Rigidityy=center of building
Step 1 – Determine Lateral Diaphragm Torsion
• Center of Lateral Load from leftxload=200/2=100 ft
• Torsional MomentMT=40(130.9-100)=1236 kip-ft
Step 2 – Determine Shear Wall Stiffness
• Polar Moment of Stiffness
Ip I xx I yy
I xx ly2 of east-west wallsI xx 15 15 2 15 15 2 6750 ft3
I yy 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 Forces
• Shear in North-South Walls
F Vxl
l
MT xlIp
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 Forces
• Shear in North-South Walls
F Vxl
l
MT xlIp
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 Forces
• Shear in North-South Walls
F Vxl
l
MT xlIp
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 Forces
• Shear in East-West Walls
F MT yl
Ip
Wall D and E 123615 15
230,6591.21kips
Load Bearing Shear Wall Example
Given:
Load 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 4712 0.333 3131 0.167 157
Total 941
Load 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 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 Steps
Step 1 – Calculate force on wallStep 2 – Calculate overturning momentStep 3 – Calculate dead loadStep 4 – Calculate net tension forceStep 5 – Calculate steel requirements
Step 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 4712 0.333 3131 0.167 157
Total
F2w 941 180 / 213.2
180F2w 540 kipsF1w 540 / 2 270 kips
Step 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 DistributionLevel Cvx Fx
3 0.500 4712 0.333 3131 0.167 157
Total 941
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 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 Force
• Governing load CombinationU=[0.9-0.2(0.24)]D+1.0E Eq. 3.2.6.7aU=0.85D+1.0E
• Tension Force
Tu 6615 0.85 416 10
18Tu 171kips
Step 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 Example
Given:
Rigid 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 4712 0.333 3131 0.167 157
Total 941
Rigid Diaphragm Analysis Example
Problem:– Part A
Determine diaphragm reinforcement required for moment design
– Part BDetermine the diaphragm reinforcement required for shear design
Solution 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, 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, 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 4712 0.333 3131 0.167 157
Total 941
Step 2 – Diaphragm Force, Fp, Distribution
• Assume the forces are uniformly distributed– Total Uniform Load, w
• Distribute the force equally to the three bays
w
471264
1.784 kips / ft
w1 w3
1.7843
0.59 kips / ft
Step 3 – Diaphragm Model
• Ramp Model
Step 3 – Diaphragm Model
• Flat Area Model
Step 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 baysw2=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 Forces
• Ultimate Positive Moment, +Mu
• Ultimate Negative Moment
• Ultimate Shear
Mu
w1 180 28
0.59 180 2
82390 kip ft
Mu
w2 42 22
0.89 42 2
2 785 kip ft
Vu
w1 180 2
0.59 180
253kips
Step 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
410.7 60 0.98 in2
Step 6 – Diaphragm Shear Design
• Force to be transferred to each wall
– Each wall is connected to the diaphragm, 10 ftShear/ft=Vwall/10=66.625/10=6.625 klf
– Providing connections at 5 ft centersVconnection=6.625(5)=33.125 kips/connection
Vwall o
Vu
22.5 53
2
66.25 kips
Step 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 centersVconnection=0.79(5)=4 kips
Questions?