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### PCI 6th Edition

### Questions? Acceleration

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

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

- 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

Cladding Seismic Load ExampleSolution 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 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 5 – Panel Loading Acceleration

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

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 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 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 AccelerationLoads 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

- Vertical

Step 6 – Loads to Connections Acceleration

Step 6 – Loads to Connections Acceleration

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

- 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 Loads Acceleration

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 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 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 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

- Parallel load
- =+ 7459 lb

Step 6 – Seismic Loads Parallel to Face Acceleration

- Up-down load

Step 6 – Seismic Loads Parallel to Face Acceleration

- In-out load

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

Vx=total lateral load

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 Assumptions: Solution Method:

Problem:

- Determine the shear in each wall due to the wind load, w

- Floors and roofs are rigid diaphragms
- Walls D and E are not connected to Wall B

- 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

- Total Lateral Load
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
xload=200/2=100 ft

- 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 3 – Calculate dead load

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

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 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 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:

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
- Total Uniform Load, w

- 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

- Half of the load of the center bay is assumed to be taken by each of the north and south bays

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

- Tensile force may be resisted by:

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

- Each wall is connected to the diaphragm, 10 ft

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

- For the first interior Tee

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