The building is a 45 ft tall, 3-story hotel building with a footprint of 70 time
ID: 1709860 • Letter: T
Question
The building is a 45 ft tall, 3-story hotel building with a footprint of 70 times 350 ft. The building is located in Boise, ID, and will be located at 43degree 32' 9.7" N, 116degree 10' 7.82" W. The seismic weights are 490 kips per story including the roof. The soil conditions for this site were determined by a geotechnical engineer as having a shear wave velocity of 2000 ft/sec. Define the Seismic Design Category. Then determine the seismic forces on the building following the Equivalent Linear Force Method (ELF).Explanation / Answer
The Site class in Class B with shear velocity ranges between 2500-5000 ft/s.
Solution :
Building: 45 Ft tall
3 storey with base plan of 70ft X 350 ft
Located at 43degree 32Minutes 9.7” north. And 116 degree 10 Minutes 7.82 ” west.
Seismic Weight per floor: 490 kips
V = Cs x W
Cs = seismic base shear coefficient
W = structure’s seismic weight
CS = 0.44SDSI (any value lesser than this acceptable)
Lateral Earthquake force at each story is
0, the superscript k has a value of unity for structures with a fundamental period (T) less than or equal to 0.5 second, has a value of 2 for structures with a fundamental period greater than or equal to 2.5 seconds, and has a value that is linearly interpolated from these values for structures with a fundamental period that falls between these values.
So ,
CS = SDS /(R/I)
R= 3
I=1
Adjusted maximum considered earthquake response acceleration:
SDS= 2/3 SMS
SDS= 2/3 SMS
Design spectral response acceleration
SMS = FaSs
SM1=F vS1
and from Table 9.4.1.2.4a and 9.4.1.2.4b: Fa=1.0 Fv=1.5
Maximum considered earthquake ground motion , from Figure 9.4.1.1: Ss=2.05g S1=0.81g
CS = SDS /(R/I)
= 1.36/3 = 0.453
Total base shear force (Eq. 9.5.5.2.1): V=CSW=0.453*490=221.97 kips
Approximate fundamental period: Ta=Cthnx =0.028*(39)0.8=0.525
T=CuTa=1.4*0.525=0.735 sec
T = 0.735 sec, so: k=1+(2-1)*(0.735-0.5)/(2-1)=1.235 sec
Fx=Cvx x V
Level
Weight kips
Heights
WiHi^k
Cvx
Force
Story Shear
R
35.45
39
3270
0.56
250
0
3
32.76
26
1832
0.31
150
250
2
32.76
13
778
0.13
90
400
1
0
0
0
0
490
The Site class in Class B with shear velocity ranges between 2500-5000 ft/s.
Solution :
Building: 45 Ft tall
3 storey with base plan of 70ft X 350 ft
Located at 43degree 32Minutes 9.7” north. And 116 degree 10 Minutes 7.82 ” west.
Seismic Weight per floor: 490 kips
V = Cs x W
Cs = seismic base shear coefficient
W = structure’s seismic weight
CS = 0.44SDSI (any value lesser than this acceptable)
Lateral Earthquake force at each story is
0, the superscript k has a value of unity for structures with a fundamental period (T) less than or equal to 0.5 second, has a value of 2 for structures with a fundamental period greater than or equal to 2.5 seconds, and has a value that is linearly interpolated from these values for structures with a fundamental period that falls between these values.
So ,
CS = SDS /(R/I)
R= 3
I=1
Adjusted maximum considered earthquake response acceleration:
SDS= 2/3 SMS
SDS= 2/3 SMS
Design spectral response acceleration
SMS = FaSs
SM1=F vS1
and from Table 9.4.1.2.4a and 9.4.1.2.4b: Fa=1.0 Fv=1.5
Maximum considered earthquake ground motion , from Figure 9.4.1.1: Ss=2.05g S1=0.81g
CS = SDS /(R/I)
= 1.36/3 = 0.453
Total base shear force (Eq. 9.5.5.2.1): V=CSW=0.453*490=221.97 kips
Approximate fundamental period: Ta=Cthnx =0.028*(39)0.8=0.525
T=CuTa=1.4*0.525=0.735 sec
T = 0.735 sec, so: k=1+(2-1)*(0.735-0.5)/(2-1)=1.235 sec
Fx=Cvx x V
Level
Weight kips
Heights
WiHi^k
Cvx
Force
Story Shear
R
35.45
39
3270
0.56
250
0
3
32.76
26
1832
0.31
150
250
2
32.76
13
778
0.13
90
400
1
0
0
0
0
490
Level
Weight kips
Heights
WiHi^k
Cvx
Force
Story Shear
R
35.45
39
3270
0.56
250
0
3
32.76
26
1832
0.31
150
250
2
32.76
13
778
0.13
90
400
1
0
0
0
0
490
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