Transcript Slide 1

PCI
th
6
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
• 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
0.4ap SDS Wp 
z
Fp =
1+2 
Rp
h

0.3SDS Wp  Fp  1.6SDS Wp
Where:
ap = component
amplification factor
from Figure 3.10.10
Seismic Design Force, Fp
0.4ap SDS Wp 
z
Fp =
1+2 
Rp
h

0.3SDS Wp  Fp  1.6SDS Wp
Where:
Rp = component response
modification factor from
Figure 3.10.10
Seismic Design Force, Fp
0.4ap SDS Wp 
z
Fp =
1+2 
Rp
h

0.3SDS Wp  Fp  1.6SDS Wp
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
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/h=1.0
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 ap and Rp
• Figure 3.10.10
ap
Rp
Step 2 – Calculate the 5%-Damped Design
Spectral Response Acceleration
 2
SDS =   SMS
 3
Where:
SMS = FaSS
Ss = 1.5 From maps found in IBC 2003
Fa = 1.0 From figure 3.10.7
SDS =1.0
Step 3 – Calculate Fp in Terms of Wp
Wall Element:
  
0.4 1.0 1.0 Wp

2.5
1+2 1.0 0.48W
p
Body of Connections:

Fasteners:

  
0.4 1.0 1.0 Wp
2.5
  
1+2 1.0 0.48W
0.4 1.25 1.0 Wp
1.0
p
1+2 1.0 1.5W
p
Step 4 – Check Fp Limits
 
 
0.3 1.0 Wp  Fp  1.6 1.0 Wp
Wall Element:
0.3Wp  0.48Wp  1.6Wp
Body of Connections:
0.3Wp  0.48Wp  1.6Wp
Fasteners:
0.3Wp  1.5Wp  1.6Wp
Step 5 – Panel Loading
• Gravity Loading
• Seismic Loading Parallel to
Panel Face
• Seismic or Wind Loading
Perpendicular to Panel Face
Step 5 – Panel Loading
• Panel Weight
Area = 465.75 in2
Panel wt=
465.75
144
 
150  485
Wp=485(28)=13,580 lb
• Seismic Design Force
Fp=0.48(13580)=6518 lb
lb
ft
Step 5 – Panel Loading
• 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
• 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
Panel
Wp
z
(lb)
(in)
13,580 4.5
Wp z
(lb-in)
61,110
Upper
1,680 2.0 2,230
Window
Lower
0
22.0
0
Window
Total
15,260
64,470
Step 6
Loads 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 Connections
Seismic Load Summary
Fp
(lb)
y
(in)
Fpy
(lb-in)
6,518
34.5
224,871
Upper
Window
403
84.0
33,852
Lower
Window
538
0.0
0.0
Panel
Total
7,459
258,723
Step 6 – Loads to Connections
Seismic Load Summary
Fp
(lb)
z
(in)
Fpz
(lb-in)
6,518
4.5
29,331
Upper
Window
403
2.0
806
Lower
Window
538
22.0
11,836
Panel
Total
7,459
41,973
Step 6 – Loads to Connections
• 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
• 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
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
Step 6 – Wind Outward Loads
• Center of equivalent wind
load from lower left
y=267,540/6860
y=39.0 in
• Outward Wind Load
Fp=6,860 lb
Fp
Step 6 – Wind Outward Loads
• 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
• 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
• Load distribution (Based on Continuous Beam Model)
– Center connections = .58 (Load)
– End connections = 0.21 (Load)
Step 6 – Seismic Loads Parallel to Face
• Parallel load
=+ 7459 lb
Step 6 – Seismic Loads Parallel to Face
• Up-down load


7459 27.5+32.5-34.7
 
2 156
  605 lb
Step 6 – Seismic Loads Parallel to Face
• In-out load


  26 lb
2 156 
7459 5.6-4.5
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
Distribution of Lateral Loads
Shear 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 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 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 Systems
• Symmetrical Shear Walls
Fi 
 k 
i


r
 
Vx
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”
• Unsymmetrical Shear Walls
Fy 
Vy  K y
K
y

TV  x  K y
K
y
y
 x2   K x  y2
Force in the y-direction is distributed to a given wall
at a given level due to an applied force in the ydirection at that level
Distribution of Lateral Loads
“Polar Moment of Stiffness Method”
• Unsymmetrical Shear Walls
Fy 
Vy  K y
K
y

TV  x  K y
K
y
y
 x2   K x  y2
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 Method”
• Unsymmetrical Shear Walls
Fx 
TV  y  K x
K
y
y
 x2   K x  y2
Force in the x-direction is distributed to a given wall
at a given level due to an applied force in the ydirection at that level.
Distribution of Lateral Loads
“Polar Moment of Stiffness Method”
• Unsymmetrical Shear Walls
Fx 
TV  y  K x
K
y
y
 x2   K x  y2
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
Given:
– Walls are 8 ft high and 8 in thick
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 < 0.3
– Distribute load in proportion to wall length
Solution Steps
Step 1 – Determine lateral diaphragm torsion
Step 2 – Determine shear wall stiffness
Step 3 – Determine wall forces
Step 1 – Determine Lateral Diaphragm Torsion
• Total Lateral Load
Vx=0.20 x 200 = 40 kips
Step 1 – Determine Lateral Diaphragm Torsion
• Center of Rigidity from left
x
 
 
   130.9 ft
40 75  30 140  40 180
40  30  40
Step 1 – Determine Lateral Diaphragm Torsion
• Center of Rigidity
y=center of building
Step 1 – Determine Lateral Diaphragm Torsion
• 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
• Polar Moment of Stiffness
Ip  I xx  I yy
I xx 
 l y
2
of east-west walls
       6750 ft
2
I xx  15 15  15 15
I yy 
 l x
2
2
3
of north-south walls
 
  
40180  130.9  223, 909 ft
2
I yy  40 130.9  75  30 140  130.9
2
Ip  6750  223, 909  230, 659 ft 3
3
  ...
2
Step 3 – Determine Wall Forces
• Shear in North-South Walls
F
Vx l
l

Wall A 
MT  x  l
Ip
4040

40  30  40
 

1236  130.9  75  40
Wall A  14.5  12.0  26.5 kips
230, 659
Step 3 – Determine Wall Forces
• Shear in North-South Walls
F
Vx l
l

Wall B 
MT  x  l
Ip
4030

40  30  40
 

1236  130.9  140  30
Wall B  10.91  1.46  9.45 kips
230, 659
Step 3 – Determine Wall Forces
• Shear in North-South Walls
F
Vx l
l

Wall C 
MT  x  l
Ip
4040

40  30  40
 

1236  130.9  180  40
Wall C  14.5  10.5  4.0 kips
230, 659
Step 3 – Determine Wall Forces
• Shear in East-West Walls
F
MT  y  l
Ip
Wall D and E 
     1.21 kips
1236  15  15
230, 659
Load Bearing Shear Wall Example
Given:
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
Level
Cvx
Fx
3
0.500
471
2
0.333
313
1
0.167
157
Total
941
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
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
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
Step 1 – Calculate Force in Shear Wall
• Accidental Eccentricity=0.05(264)=13.2 ft
• Force in two walls
F2w
941180 / 2  13.2


180
F2w  540 kips
F1w  540 / 2  270 kips
Seismic Lateral Force
Distribution
Level
Cvx
Fx
3
0.500
471
2
0.333
313
1
0.167
157
Total
Step 1 – Calculate Force in Shear Wall
• 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
Seismic Lateral Force Distribution
Level
Cvx
Fx
3
0.500
471
2
0.333
313
1
0.167
157
Total
941
Step 2 – Calculate Overturning Moment
• 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
• 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
• Governing load Combination
U=[0.9-0.2(0.24)]D+1.0E
U=0.85D+1.0E
• Tension Force
Tu 
  
6615  0.85 416 10
Tu  171 kips
18
Eq. 3.2.6.7a
Step 5 – Reinforcement Requirements
• Tension Steel, As
As 
Tu
 fy

171
 
0.9 60
 3.17 in2
• Reinforcement Details
– Use 4 - #8 bars = 3.17 in2
– Locate 2 ft from each end
Rigid Diaphragm Analysis Example
Given:
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
Level
Cvx
Fx
3
0.500
471
2
0.333
313
1
0.167
157
Total
941
Rigid Diaphragm Analysis Example
Problem:
– Part A
Determine diaphragm reinforcement required
for moment design
– Part B
Determine the diaphragm reinforcement
required for shear design
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
Step 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, Fp
• Fp, Eq. 3.8.3.1
Fp =(1.0)(0.24)(5227)+0.0=251 kips
•
Seismic Lateral Force
Distribution
Level
Cvx
Fx
3
0.500
471
2
0.333
313
1
0.167
157
Total
941
Fp=471 kips
Step 2 – Diaphragm Force, Fp, Distribution
• Assume the forces are uniformly distributed
– Total Uniform Load, w
w
471
264
 1.784 kips / ft
• Distribute the force equally to the three bays
w1  w3 
1.784
3
 0.59 kips / ft
Step 3 – Diaphragm Model
• Ramp Model
Step 3 – Diaphragm Model
• Flat Area
Model
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
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
• Ultimate Positive Moment, +Mu
Mu 
   0.59 180  2390 kip  ft
w1 180
2
8
2
8
• Ultimate Negative Moment
Mu 
 
w2 42
2
2
   785 kip  ft
0.89 42
2
• Ultimate Shear
Vu 
2
   0.59 180  53 kips
w1 180
2
2
Step 5 – Diaphragm Moment Design
• Assuming a 58 ft moment arm
Tu=2390/58=41 kips
• Required Reinforcement, As
As 
Tu
 fy

41
 
0.7 60
 0.98 in2
– Tensile force may be resisted by:
• Field placed reinforcing bars
• Welding erection material to embedded plates
Step 6 – Diaphragm Shear Design
• Force to be transferred to each wall
Vwall
 53 
 o
 2.5    66.25 kips
2
 2
Vu
– 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
• 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?