Solution:-
Design seismic base shear = Ah W
Fundamental natural period = 0.075 (h)0.75
Ta = 0.075 x (11.25)0.75
= 0.46 second
Where h is in meter
Design horizontal seismic coefficient
Ah = Z/2 *I/R *Sa/g
= 0.36/2 *1.5/3 *2.17 = 0.1953
Where Z = zone factor = 0.36 (for South Carolina )
For medical building( I) = 1.5
R =Response reduction factor =3
Sa/g = 1/ Ta = 1/ (0.46) = 2.17
VB =Ah* W
= 0.1953 * 44 *105 =8.59 *105 lb
W = seismic weight of building
Design lateral seismic force at first floor
Q1 = VB * W1 * h12/(W1h12 +W2h22 + W3h32 )
= 8.59 *105 * 17 *105*122/(17*105* 122 + 15 * 105 * 242 + 12 * 105* 362)
= 7.81 * 104 lb Answer
Design lateral force at second floor
Q2 =8.59 *105 *15 *105 * 242/ ( 17 *105 *122 + 15 *105 * 242 + 12 * 105 *362 )
Q2 =2.78 * 105 lb Answer
lateral seismic force At third floor
Q3 = 8.59 * 105 * 12 * 105 * 362 / (17 *105 *122 + 15 * 105 *242 + 12 * 105 * 362)
Q3 = 5.01 * 105 lb Answer
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