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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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Pci 6 th edition

PCI 6th Edition

Lateral Component Design


Presentation outline
Presentation Outline

  • Architectural Components

    • Earthquake Loading

  • Shear Wall Systems

    • Distribution of lateral loads

    • Load bearing shear wall analysis

    • Rigid diaphragm analysis


Architectural components
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 f p
Seismic Design Force, Fp

Where:

ap = component amplification factorfrom Figure 3.10.10


Seismic design force f p1
Seismic Design Force, Fp

Where:

Rp = component response modification factor from Figure 3.10.10


Seismic design force f p2
Seismic Design Force, Fp

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


Cladding seismic load example
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 example1

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

Cladding Seismic Load Example


Solution steps
Solution Steps

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 5 – Calculate panel loading

Step 6 – Determine connection forces

Step 7 – Summarize connection forces


Step 1 determine a p and r p

Figure 3.10.10

Step 1 – Determine ap and Rp

ap Rp


Step 2 calculate the 5 damped design spectral response acceleration
Step 2 – Calculate the 5%-Damped Design Spectral Response Acceleration

Where:

SMS=FaSS

Ss =1.5 From maps found in IBC 2003

Fa= 1.0 From figure 3.10.7


Step 3 calculate f p in terms of w p
Step 3 – Calculate F Accelerationp in Terms of Wp

Wall Element:

Body of Connections:

Fasteners:


Step 4 check f p limits
Step 4 – Check F Accelerationp Limits

Wall Element:

Body of Connections:

Fasteners:


Step 5 panel loading
Step 5 – Panel Loading Acceleration

  • Gravity Loading

  • Seismic Loading Parallel to Panel Face

  • Seismic or Wind Loading Perpendicular to Panel Face


Step 5 panel loading1
Step 5 – Panel Loading Acceleration

  • Panel Weight

    • Area = 465.75 in2

    • Wp=485(28)=13,580 lb

  • Seismic Design Force

    • Fp=0.48(13580)=6518 lb


Step 5 panel loading2
Step 5 – Panel Loading Acceleration

  • 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


Step 5 panel loading3
Step 5 – Panel Loading Acceleration

  • 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 AccelerationLoads to Connections


Step 6 loads to connections
Step 6 AccelerationLoads 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 connections3
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
Step 6 – Seismic In-Out Loads Acceleration

  • 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 loads1
Step 6 – Wind Outward Loads Acceleration

Fp

  • Center of equivalent wind load from lower left

  • y=267,540/6860

  • y=39.0 in

  • Outward Wind Load

  • Fp=6,860 lb


Step 6 wind outward loads2
Step 6 – Wind Outward Loads Acceleration

  • 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 loads
Step 6 – Wind Inward Loads Acceleration

  • 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 surface
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
Step 6 – Seismic Loads Parallel to Face Acceleration

  • Parallel load

  • =+ 7459 lb




Step 7 summary of factored loads
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 shear wall systems
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 shear wall systems1
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 shear wall systems2
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 shear wall systems3
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

Vx=total lateral load


Distribution of lateral loads polar moment of stiffness method
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 polar moment of stiffness method1
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 polar moment of stiffness method2
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 polar moment of stiffness method3
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
Unsymmetrical Shear Wall Example Acceleration

Given:

  • Walls are 8 ft high and 8 in thick


Unsymmetrical shear wall example1
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 steps1
    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
    Step 1 – Determine Lateral Diaphragm Torsion Acceleration

    • Total Lateral Load

      Vx=0.20 x 200 = 40 kips


    Step 1 determine lateral diaphragm torsion1
    Step 1 – Determine Lateral Diaphragm Torsion Acceleration

    • Center of Rigidity from left


    Step 1 determine lateral diaphragm torsion2
    Step 1 – Determine Lateral Diaphragm Torsion Acceleration

    • Center of Rigidity

      y=center of building


    Step 1 determine lateral diaphragm torsion3
    Step 1 – Determine Lateral Diaphragm Torsion Acceleration

    • 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 stiffness
    Step 2 – Determine Shear Wall Stiffness Acceleration

    • Polar Moment of Stiffness


    Step 3 determine wall forces
    Step 3 – Determine Wall Forces Acceleration

    • Shear in North-South Walls


    Step 3 determine wall forces1
    Step 3 – Determine Wall Forces Acceleration

    • Shear in North-South Walls


    Step 3 determine wall forces2
    Step 3 – Determine Wall Forces Acceleration

    • Shear in North-South Walls


    Step 3 determine wall forces3
    Step 3 – Determine Wall Forces Acceleration

    • Shear in East-West Walls


    Load bearing shear wall example
    Load Bearing Shear Wall Example Acceleration

    Given:


    Load bearing shear wall example1
    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 example2
    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 example3
    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 steps2
    Solution Steps Acceleration

    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 wall
    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 wall1
    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
    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


    Step 3 calculate dead load
    Step 3 – Calculate Dead Load Acceleration

    • Load on each Wall

      • Dead Load = .110 ksf (all components)

      • Supported Area = (60)(21)=1260 ft2

        Wwall=1260(.110)=138.6 kips

    • Total Load

      Wtotal=3(138.6)=415.8~416 kips


    Step 4 calculate tension force
    Step 4 – Calculate Tension Force Acceleration

    • Governing load Combination

      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
    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 example1
    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 example2
    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 steps3
    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 p
    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 p1
    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 p distribution
    Step 2 – Diaphragm Force, F Accelerationp, Distribution

    • Assume the forces are uniformly distributed

      • Total Uniform Load, w

    • Distribute the force equally to the three bays


    Step 3 diaphragm model
    Step 3 – Diaphragm Model Acceleration

    • Ramp Model


    Step 3 diaphragm model1
    Step 3 – Diaphragm Model Acceleration

    • Flat Area Model


    Step 3 diaphragm model2
    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
    Step 4 – Design Forces Acceleration

    • Ultimate Positive Moment, +Mu

    • Ultimate Negative Moment

    • Ultimate Shear


    Step 5 diaphragm moment design
    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
    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 design1
    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


    Questions

    Questions? Acceleration