PCI 6th Edition

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PCI 6th Edition

• Lateral Component Design

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Presentation Outline

• Architectural Components
• Earthquake Loading
• Shear Wall Systems
• Distribution of lateral loads
• Load bearing shear wall analysis
• Rigid diaphragm analysis

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Architectural Components

• Must resist seismic forces and be attached to the
  SFRS
• Exceptions
• Seismic Design Category A
• Seismic Design Category B with I1.0 (other than
  parapets supported by bearing or shear walls).

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Seismic Design Force, Fp
Where ap component amplification factorfrom
Figure 3.10.10
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Seismic Design Force, Fp
Where Rp component response modification
factor from Figure 3.10.10
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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 lt h
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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

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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/h1.0

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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

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Step 1 Determine ap and Rp

• Figure 3.10.10

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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
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Step 3 Calculate Fp in Terms of Wp
Wall Element Body of Connections Fasteners

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Step 4 Check Fp Limits
Wall Element Body of Connections Fasteners
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Step 5 Panel Loading

• Gravity Loading
• Seismic Loading Parallel to Panel Face
• Seismic or Wind Loading Perpendicular to Panel
  Face

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Step 5 Panel Loading

• Panel Weight
• Area 465.75 in2
• Wp485(28)13,580 lb
• Seismic Design Force
• Fp0.48(13580)6518 lb

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Step 5 Panel Loading

• Upper Window Weight
• Height 6 ft
• Wwindow6(28)(10)1680 lb
• Seismic Design Force
• Inward or Outward
• Consider ½ of Window
• Wp3.0(10)30 plf
• Fp0.48(30)14.4 plf
• 14.4(28)403 lb
• Wp485(28)13,580 lb
• Seismic Design Force
• Fp0.48(13580)6518 lb

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Step 5 Panel Loading

• Lower Window Weight
• No weight on panel
• Seismic Design Force
• Inward or outward
• Consider ½ of window
• height8 ft
• Wp4.0(10)40 plf
• Fp0.48(30)19.2 plf
• 19.2(28)538 lb

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Step 5 Loads to Connections
Dead Load Summary Dead Load Summary Dead Load Summary Dead Load Summary
Wp (lb) z (in) Wpz (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
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Step 6Loads to Connections

• Equivalent Load Eccentricity
• z64,470/15,2604.2 in
• Dead Load to Connections
• Vertical
• 15,260/27630 lb
• Horizontal
• 7630 (7.5-4.2)/32.5
• 774.7/2387 lb

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Step 6 Loads to Connections
Seismic Load Summary Seismic Load Summary Seismic Load Summary 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
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Step 6 Loads to Connections
Seismic Load Summary Seismic Load Summary Seismic Load Summary 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
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Step 6 Loads to Connections

• Center of equivalent seismic load from lower left
• y258,723/7459 y34.7 in
• z41,973/7459
• z5.6 in

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Step 6 Seismic In-Out Loads

• Equivalent Seismic Load
• y34.7 in
• Fp7459 lb
• Moments about Rb
• Rt7459(34.7 -27.5)/32.5
• Rt1652 lb
• Force equilibrium
• Rb7459-1652
• Rb5807 lb

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Step 6 Wind Outward Loads
Outward Wind Load Summary Outward Wind Load Summary Outward Wind Load Summary Outward Wind Load Summary
Fp (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
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Step 6 Wind Outward Loads

• Center of equivalent wind load from lower left
• y267,540/6860
• y39.0 in
• Outward Wind Load
• Fp6,860 lb

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Step 6 Wind Outward Loads

• Moments about Rb
• Rt7459(39.0 -27.5)/32.5
• Rt2427 lb
• Force equilibrium
• Rb6860-2427
• Rb4433 lb

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Step 6 Wind Inward Loads

• Outward Wind Reactions
• Rt2427 lb
• Rb4433 lb
• Inward Wind Loads
• Proportional to pressure
• Rt(11.3/12.9)2427 lb
• Rt2126 lb
• Rb(11.3/12.9)4433 lb
• Rb3883 lb

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Step 6 Seismic Loads Normal to Surface

• Load distribution (Based on Continuous Beam
  Model)
• Center connections .58 (Load)
• End connections 0.21 (Load)

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Step 6 Seismic Loads Parallel to Face

• Parallel load
• 7459 lb

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Step 6 Seismic Loads Parallel to Face

• Up-down load

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Step 6 Seismic Loads Parallel to Face

• In-out load

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Step 7 Summary of Factored Loads

1. Load Factor of 1.2 Applied
2. Load Factor of 1.0 Applied
3. Load Factor of 1.6 Applied

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Distribution of Lateral Loads Shear Wall Systems

• For Rigid diaphragms
• Lateral Load Distributed based on total rigidity,
  r

Where r1/D Dsum of flexural and shear
deflections
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Distribution of Lateral Loads Shear Wall Systems

• Neglect Flexural Stiffness Provided
• Rectangular walls
• Consistent materials
• Height to length ratio lt 0.3
• Distribution based on
• Cross-Sectional Area

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Distribution of Lateral Loads Shear Wall Systems

• Neglect Shear Stiffness Provided
• Rectangular walls
• Consistent materials
• Height to length ratio gt 3.0
• Distribution based on
• Moment of Inertia

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Distribution of Lateral Loads Shear Wall Systems

• Symmetrical Shear Walls

Where Fi Force Resisted by individual shear
wall kirigidity of wall i Srsum of all wall
rigidities Vxtotal lateral load
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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
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Distribution of Lateral Loads 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
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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.
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Distribution of Lateral Loads Polar Moment of
Stiffness Method

• Unsymmetrical Shear Walls

Where Vylateral force at level being
considered Kx,Kyrigidity in x and y directions
of wall SKx, SKysummation of rigidities of all
walls TTorsional Moment xwall x-distance from
the center of stiffness ywall y-distance from
the center of stiffness
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Unsymmetrical Shear Wall Example

• Given
• Walls are 8 ft high and 8 in thick

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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 lt 0.3
• Distribute load in proportion to wall length

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Solution Steps

• Step 1 Determine lateral diaphragm torsion
• Step 2 Determine shear wall stiffness
• Step 3 Determine wall forces

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Step 1 Determine Lateral Diaphragm Torsion

• Total Lateral Load
• Vx0.20 x 200 40 kips

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Step 1 Determine Lateral Diaphragm Torsion

• Center of Rigidity from left

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Step 1 Determine Lateral Diaphragm Torsion

• Center of Rigidity
• ycenter of building

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Step 1 Determine Lateral Diaphragm Torsion

• Center of Lateral Load from left
• xload200/2100 ft
• Torsional Moment
• MT40(130.9-100)1236 kip-ft

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Step 2 Determine Shear Wall Stiffness

• Polar Moment of Stiffness

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Step 3 Determine Wall Forces

• Shear in North-South Walls

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Step 3 Determine Wall Forces

• Shear in North-South Walls

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Step 3 Determine Wall Forces

• Shear in North-South Walls

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Step 3 Determine Wall Forces

• Shear in East-West Walls

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Load Bearing Shear Wall Example

• Given

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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 Seismic Lateral Force Distribution Seismic Lateral Force Distribution
Level Cvx Fx
3 0.500 471
2 0.333 313
1 0.167 157
Total 941
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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

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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

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Solution Steps

• 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

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Step 1 Calculate Force in Shear Wall

• Accidental Eccentricity0.05(264)13.2 ft
• Force in two walls

Seismic Lateral Force Distribution Seismic Lateral Force Distribution Seismic Lateral Force Distribution
Level Cvx Fx
3 0.500 471
2 0.333 313
1 0.167 157
Total
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Step 1 Calculate Force in Shear Wall

• Force at each level
• Level 3 F1W0.500(270)135 kips
• Level 2 F1W0.333(270) 90 kips
• Level 1 F1W0.167(270) 45 kips

Seismic Lateral Force Distribution Seismic Lateral Force Distribution Seismic Lateral Force Distribution
Level Cvx Fx
3 0.500 471
2 0.333 313
1 0.167 157
Total 941
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Step 2 Calculate Overturning Moment

• Force at each level
• Level 3 F1W0.500(270)135 kips
• Level 2 F1W0.333(270) 90 kips
• Level 1 F1W0.167(270) 45 kips
• Overturning moment, MOT
• MOT135(31.5)90(21)45(10.5)
• MOT6615 kip-ft

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Step 3 Calculate Dead Load

• Load on each Wall
• Dead Load .110 ksf (all components)
• Supported Area (60)(21)1260 ft2
• Wwall1260(.110)138.6 kips
• Total Load
• Wtotal3(138.6)415.8416 kips

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Step 4 Calculate Tension Force

• Governing load Combination
• U0.9-0.2(0.24)D1.0E Eq. 3.2.6.7a
• U0.85D1.0E
• Tension Force

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Step 5 Reinforcement Requirements

• Tension Steel, As
• Reinforcement Details
• Use 4 - 8 bars 3.17 in2
• Locate 2 ft from each end

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Rigid Diaphragm Analysis Example

• Given

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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 Seismic Lateral Force Distribution Seismic Lateral Force Distribution
Level Cvx Fx
3 0.500 471
2 0.333 313
1 0.167 157
Total 941
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Rigid Diaphragm Analysis Example

• Problem
• Part A
• Determine diaphragm reinforcement required for
  moment design
• Part B
• Determine the diaphragm reinforcement required
  for shear design

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Solution Steps

• 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

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Step 1 Diaphragm Force, Fp

• Fp, Eq. 3.8.3.1
• Fp 0.2IESDSWp Vpx
• but not less than any force in the lateral force
  distribution table

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Step 1 Diaphragm Force, Fp

• Fp, Eq. 3.8.3.1
• Fp (1.0)(0.24)(5227)0.0251 kips
• Fp471 kips

Seismic Lateral Force Distribution Seismic Lateral Force Distribution Seismic Lateral Force Distribution
Level Cvx Fx
3 0.500 471
2 0.333 313
1 0.167 157
Total 941
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Step 2 Diaphragm Force, Fp, Distribution

• Assume the forces are uniformly distributed
• Total Uniform Load, w
• Distribute the force equally to the three bays

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Step 3 Diaphragm Model

• Ramp Model

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Step 3 Diaphragm Model

• Flat Area Model

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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 bays
• w20.590.59/20.89 kip/ft
• Stress reduction due to cantilevers is neglected.
• Positive Moment design is based on ramp moment

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Step 4 Design Forces

• Ultimate Positive Moment, Mu
• Ultimate Negative Moment
• Ultimate Shear

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Step 5 Diaphragm Moment Design

• Assuming a 58 ft moment arm
• Tu2390/5841 kips
• Required Reinforcement, As
• Tensile force may be resisted by
• Field placed reinforcing bars
• Welding erection material to embedded plates

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Step 6 Diaphragm Shear Design

• Force to be transferred to each wall
• Each wall is connected to the diaphragm, 10 ft
• Shear/ftVwall/1066.625/106.625 klf
• Providing connections at 5 ft centers
• Vconnection6.625(5)33.125 kips/connection

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Step 6 Diaphragm Shear Design

• Force to be transferred between Tees
• For the first interior Tee
• VtransferVu-(10)0.5947.1 kips
• Shear/ftVtransfer/6047.1/600.79 klf
• Providing Connections at 5 ft centers
• Vconnection0.79(5)4 kips

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Questions?