PCI 6 th Edition Lateral Component Design. Presentation Outline Architectural Components –Earthquake…

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


Fp=0.4apSDSWp

Rp

1+2zh


Where:Rp = component response modification factor from Figure 3.10.10


Fp=0.4apSDSWp

Rp

1+2zh


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


• 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


• 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


• 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


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


• 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


• Center of equivalent wind load from lower left

y=267,540/6860y=39.0 in

• Outward Wind LoadFp=6,860 lb

Fp


• 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


• Up-down load

7459 27.5+32.5-34.7 2 156 605 lb


• 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


• Neglect Flexural Stiffness Provided:– Rectangular walls– Consistent materials– Height to length ratio < 0.3

Distribution based onCross-Sectional Area


• Neglect Shear Stiffness Provided:– Rectangular walls– Consistent materials– Height to length ratio > 3.0

Distribution based onMoment of Inertia


• 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


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



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



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


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


• Center of Rigidity from left

x

40 75 30 140 40 180 40 30 40

130.9 ft


• Center of Rigidityy=center of building


• 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



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



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


• Shear in East-West Walls

F MT yl

Ip

Wall D and E 123615 15

230,6591.21kips

Load Bearing Shear Wall Example

Given:


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


Problem:– Determine the tension steel requirements for

the load bearing shear walls in the north-south direction required to resist seismic loading


• 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


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:


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


Level Cvx Fx

3 0.500 4712 0.333 3131 0.167 157

Total 941


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


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


• Flat Area 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?

PCI 6 th Edition Lateral Component Design. Presentation Outline Architectural Components –Earthquake…

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