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# PCI 6 th Edition - PowerPoint PPT Presentation

PCI 6 th 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.

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Presentation Transcript

### PCI 6th Edition

Lateral Component Design

• Architectural Components

• Shear Wall Systems

• Load bearing shear wall analysis

• Rigid diaphragm analysis

• 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).

Where:

ap = component amplification factorfrom Figure 3.10.10

Where:

Rp = component response modification factor from Figure 3.10.10

Where:

h = average roof height of structure

SDS= Design, 5% damped, spectral response acceleration at short periods

Wp = component weight

z= height in structure at attachment point < h

• 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

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

Step 1 – Determine Component Factors

Step 2 – Calculate Design Spectral Response Acceleration

Step 3 – Calculate Seismic Force in terms of panel weight

Step 4 – Check limits

Step 6 – Determine connection forces

Step 7 – Summarize connection forces

Step 1 – Determine ap and Rp

ap Rp

Where:

SMS=FaSS

Ss =1.5 From maps found in IBC 2003

Fa= 1.0 From figure 3.10.7

Step 3 – Calculate F Accelerationp in Terms of Wp

Wall Element:

Body of Connections:

Fasteners:

Step 4 – Check F Accelerationp Limits

Wall Element:

Body of Connections:

Fasteners:

• Panel Weight

• Area = 465.75 in2

• Wp=485(28)=13,580 lb

• Seismic Design Force

• Fp=0.48(13580)=6518 lb

• Upper Window Weight

• Height =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

z=64,470/15,260=4.2 in

• Vertical

=15,260/2=7630 lb

• Horizontal

= 7630 (7.5-4.2)/32.5

=774.7/2=387 lb

Step 6 – Loads to Connections Acceleration

• Center of equivalent seismic load from lower left

• y=258,723/7459 y=34.7 in

• z=41,973/7459

• z=5.6 in

Step 6 – Seismic In-Out Loads Acceleration

• y=34.7 in

• Fp=7459 lb

• Rt=7459(34.7 -27.5)/32.5

• Rt=1652 lb

• Force equilibrium

• Rb=7459-1652

• Rb=5807 lb

Step 6 – Wind Outward Loads Acceleration

Fp

• Center of equivalent wind load from lower left

• y=267,540/6860

• y=39.0 in

• Fp=6,860 lb

Step 6 – Wind Outward Loads Acceleration

• Rt=7459(39.0 -27.5)/32.5

• Rt=2427 lb

• Force equilibrium

• Rb=6860-2427

• Rb=4433 lb

Step 6 – Wind Inward Loads Acceleration

• Outward Wind Reactions

• Rt=2427 lb

• Rb=4433 lb

• 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 Surface Acceleration

• Load distribution (Based on Continuous Beam Model)

• Center connections = .58 (Load)

• End connections = 0.21 (Load)

Step 6 – Seismic Loads Parallel to Face Acceleration

• =+ 7459 lb

Step 7 – Summary of Factored Loads Acceleration

• Load Factor of 1.2 Applied

• Load Factor of 1.0 Applied

• Load Factor of 1.6 Applied

Distribution of Lateral Loads AccelerationShear 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 AccelerationShear 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 AccelerationShear 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 AccelerationShear Wall Systems

• Symmetrical Shear Walls

Where:

Fi = Force Resisted by individual shear wall

ki=rigidity of wall i

Sr=sum of all wall rigidities

Distribution of Lateral Loads Acceleration“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

Distribution of Lateral Loads Acceleration“Polar Moment of Stiffness Method”

• Unsymmetrical Shear Walls

Where:

Vy = lateral force at level being considered

Kx,Ky = rigidity in x and y directions of wall

SKx, SKy = summation of rigidities of all walls

T = Torsional Moment

x = wall x-distance from the center of stiffness

y = wall y-distance from the center of stiffness

Distribution of Lateral Loads Acceleration“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.

Distribution of Lateral Loads Acceleration“Polar Moment of Stiffness Method”

• Unsymmetrical Shear Walls

Where:

Vy=lateral force at level being considered

Kx,Ky=rigidity in x and y directions of wall

SKx, SKy=summation of rigidities of all walls

T=Torsional Moment

x=wall x-distance from the center of stiffness

y=wall y-distance from the center of stiffness

Unsymmetrical Shear Wall Example Acceleration

Given:

• Walls are 8 ft high and 8 in thick

Unsymmetrical Shear Wall Example Acceleration

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 Acceleration

Step 1 – Determine lateral diaphragm torsion

Step 2 – Determine shear wall stiffness

Step 3 – Determine wall forces

Step 1 – Determine Lateral Diaphragm Torsion Acceleration

Vx=0.20 x 200 = 40 kips

Step 1 – Determine Lateral Diaphragm Torsion Acceleration

• Center of Rigidity from left

Step 1 – Determine Lateral Diaphragm Torsion Acceleration

• Center of Rigidity

y=center of building

Step 1 – Determine Lateral Diaphragm Torsion Acceleration

• Center of Lateral Load from left

• Torsional Moment

MT=40(130.9-100)=1236 kip-ft

Step 2 – Determine Shear Wall Stiffness Acceleration

• Polar Moment of Stiffness

Step 3 – Determine Wall Forces Acceleration

• Shear in North-South Walls

Step 3 – Determine Wall Forces Acceleration

• Shear in North-South Walls

Step 3 – Determine Wall Forces Acceleration

• Shear in North-South Walls

Step 3 – Determine Wall Forces Acceleration

• Shear in East-West Walls

Load Bearing Shear Wall Example Acceleration

Given:

Load Bearing Shear Wall Example Acceleration

Given Continued:

• Three level parking structure

• Seismic Design Controls

• Symmetrically placed shear walls

• Corner Stairwells are not part of the SFRS

Load Bearing Shear Wall Example Acceleration

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 Acceleration

• 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 Acceleration

Step 1 – Calculate force on wall

Step 2 – Calculate overturning moment

Step 4 – Calculate net tension force

Step 5 – Calculate steel requirements

Step 1 – Calculate Force in Shear Wall Acceleration

• Accidental Eccentricity=0.05(264)=13.2 ft

• Force in two walls

Step 1 – Calculate Force in Shear Wall Acceleration

• Force at each level

Level 3 F1W=0.500(270)=135 kips

Level 2 F1W=0.333(270)= 90 kips

Level 1 F1W=0.167(270)= 45 kips

Step 2 – Calculate Overturning Moment Acceleration

• Force at each level

Level 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

• Supported Area = (60)(21)=1260 ft2

Wwall=1260(.110)=138.6 kips

Wtotal=3(138.6)=415.8~416 kips

Step 4 – Calculate Tension Force Acceleration

U=[0.9-0.2(0.24)]D+1.0E Eq. 3.2.6.7a

U=0.85D+1.0E

• Tension Force

Step 5 – Reinforcement Requirements Acceleration

• Tension Steel, As

• Reinforcement Details

• Use 4 - #8 bars = 3.17 in2

• Locate 2 ft from each end

Rigid Diaphragm Analysis Example Acceleration

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

Rigid Diaphragm Analysis Example Acceleration

Problem:

• Part A

Determine diaphragm reinforcement required for moment design

• Part B

Determine the diaphragm reinforcement required for shear design

Solution Steps Acceleration

Step 1 – Determine diaphragm force

Step 2 – Determine force distribution

Step 3 – Determine statics model

Step 4 – Determine design forces

Step 5 – Diaphragm moment design

Step 6 – Diaphragm shear design

Step 1 – Diaphragm Force, F Accelerationp

• 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, F Accelerationp

• Fp, Eq. 3.8.3.1

Fp =(1.0)(0.24)(5227)+0.0=251 kips

Fp=471 kips

Step 2 – Diaphragm Force, F Accelerationp, Distribution

• Assume the forces are uniformly distributed

• Distribute the force equally to the three bays

Step 3 – Diaphragm Model Acceleration

• Ramp Model

Step 3 – Diaphragm Model Acceleration

• Flat Area Model

Step 3 – Diaphragm Model Acceleration

• 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 Forces Acceleration

• Ultimate Positive Moment, +Mu

• Ultimate Negative Moment

• Ultimate Shear

Step 5 – Diaphragm Moment Design Acceleration

• Assuming a 58 ft moment arm

Tu=2390/58=41 kips

• Required Reinforcement, As

• Tensile force may be resisted by:

• Field placed reinforcing bars

• Welding erection material to embedded plates

Step 6 – Diaphragm Shear Design Acceleration

• 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

Step 6 – Diaphragm Shear Design Acceleration

• 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