Answered: The answer is between 690.5 and 693.5…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Read all: The answer is between 690.5 and 693.5 but I don’t see how to get this. Please answer the question in the method that the textbook shows (images attached) **NOT** using the equation F= Cs x W. Question: Consider a 14-story buildings with a total plan area BxL. In which B=125ft and L=60ft. The average weight of each floor and roof are 95 Ib/ft and 75 Ib/ft, respectively. The 168 ft-tall building is located in a seismic region with Sds=1.6g and Sd1=0.5g for a building supported on rock, where g is the gravitational acceleration. The structure of the . building consists of steel moment frames (all joints are rigid) that have an R value of 8. The importance factor |=1.0 Determine the total design seismic force (kips) acting at the base of the building ( seismic base shear in kips)

Transcribed Image Text:EXAMPLE 2.8 5 @ 12'60' FR=70.8 kips F6=57.4 kips-> F5 = 44.6 kips- F432.3 kips- F3=20.8 kips-> F2=10.1 kips-> (a) V=236 kips (b) roof Determine the design seismic forces acting at each floor of the six-story office building in Figure 2.22. The structure of the building consists of steel moment frames (all joints are rigid) that have an R value of 8. The 75-ft-tall building is located in a high seismic region with Sp = 0.4g and Spi = 1.0g for a building supported on rock, where g is the gravi- tational acceleration. The deadweight of each floor is 700 kips. = 6th floor 5th floor 4th floor 3rd floor 2nd floor Figure 2.22: (a) Six-story building; (b) equiv- alent lateral load profile. { Solution Compute the fundamental period, using Equation 2.12: T= C,h=0.028(75) 0.8 = 0.89 s Assuming that the floor deadweight contains an allowance for the weight of columns, beams, partitions, ceiling, etc., the total weight W of the building is W = 700(6) = 4200 kips. The occupancy importance factor I, is 1 for office buildings. Com- pute the base shear V using Equations 2.11a and c: V= but not more than and not less than F3rd floor SDI T(R/I) Vmax = w3h's Σwh i=1 -V W= 0.4 0.89(8/1) SDs W = R/I 1.0 8/1 (4200) = 236 kips Vmin=0.044SDIW=0.044x 1.0 x 1 x 4200 = 184.8 kips (2.11c) Therefore, use V = 236 kips. Computations of the lateral seismic force at each floor level are sum- marized in Table 2.12. To illustrate these computations, we compute the load at the third floor. Since T = 0.89 s lies between 0.5 and 2.5 s, we must interpolate using Equation 2.14 to compute the k value (Figure 2.21): k = 1 +T-0.5 = 1+ 0.89-0.5 2 2 36,537 415,262 (4200) 525 kips = 1.2 (2.11a) (236) = 20.8 kips (2.11b)

TABLE 2.12 Computation of Seismic Lateral Forces
Floor
Roof
6th
5th
4th
3rd
2nd
Weight
w; (kips)
700
700
700
700
700
700
6
W = Σw; = 4200
i=1
Floor
Height
h; (ft)
75
63
51
39
27
15
wh
124,501
100,997
78,376
56,804
$36,537
18,047
6
Σwh = 415,262
i=1
w.hx
6
Σwh
i=1
0.300
0.243
0.189
0.137
0.088
0.043
Fx (kips)
70.8
57.4
44.6
32.3
20.8
10.1
6
V = Σ F = 236
i=1

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