5_Lateral_Component_Design.ppt

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

0.4apS

0.3SDS

1.6SDS

Where:ap = component amplification factorfrom Figure 3.10.10

0.4apS

0.3SDS

1.6SDS

Where:Rp = component response modification factor from Figure 3.10.10

0.4apS

0.3SDS

1.6SDS

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

DS=1.0

Where:SMS = FaSS

Ss = 1.5 From maps found in IBC 2003Fa = 1.0 From figure 3.10.7

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

0.4 1.0 1.0 Wp

2.51+2 1.0 0.48W

0.4 1.25 1.0 Wp

1.01+2 1.0 1.5W

Wall Element:

Body of Connections:

Fasteners:

Step 4 – Check Fp LimitsStep 4 – Check Fp Limits

0.3 1.0 Wp

1.6 1.0 Wp

p0.48W

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

• 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

• 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

• 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

Dead Load Summary

(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

• 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

(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

(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=41,973/7459

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

(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=267,540/6860

y=39.0 in• Outward Wind Load

Fp=6,860 lb

• 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

• 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 LoadsStep 7 – Summary of Factored Loads

1. Load Factor of 1.2 Applied

Distribution of Lateral Loads Shear Wall Systems

• For Rigid diaphragms– Lateral Load Distributed based on total

rigidity, r

Where:

D=sum of flexural and shear deflections

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

Distribution based on

Cross-Sectional Area

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

Distribution based on

Moment of Inertia

• Symmetrical Shear Walls

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

Kyx2 K

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.

Kyx2 K

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

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

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

• Center of Rigidity from left

40 75 30 140 40 180 40 30 40

130.9 ft

• Center of Rigidityy=center of building

• 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

ly2 of east-west wallsI

xx 15 15 2 15 15 2 6750 ft3

lx2 of north-south wallsI

yy 40 130.9 75 2 30 140 130.9 2 ...

40 180 130.9 2 223,909 ft3

6750 223,909 230,659 ft3

Step 3 – Determine Wall ForcesStep 3 – Determine Wall Forces

• Shear in North-South Walls

Wall A 40 40

40 30 40 1236130.9 75 40

230,659

Wall A 14.512.0 26.5 kips

Wall B 40 30

40 30 40 1236130.9 140 30

230,659

Wall B 10.91 1.46 9.45 kips

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

Wall D andE 123615 15

230,6591.21kips

Load Bearing Shear Wall ExampleLoad 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 0.500 471

2 0.333 313

1 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

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

Level Cvx Fx

3 0.500 471

2 0.333 313

1 0.167 157

941 180 / 213.2

2w540 kips

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

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

6615 0.85 416 10

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

0.9 60 3.17 in2

Rigid Diaphragm Analysis ExampleRigid 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 0.500 471

2 0.333 313

1 0.167 157

Total 941

Problem:– Part A

Determine diaphragm reinforcement required for moment design

– Part B

Determine the diaphragm reinforcement required for shear design

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

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

2641.784 kips / ft

30.59 kips / ft

Step 3 – Diaphragm ModelStep 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 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

180 28

0.59 180 2

82390 kip ft

0.89 42 2

2 785 kip ft

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

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

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

Questions?Questions?

5_Lateral_Component_Design.ppt

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