114

Compute for Base Shear, V:

I= 1.0 (T-208-1)Na=1.0 (T-208-4)Nv=1.0(T-208-5)Soil Type= SD (&-208-2)Ct=0.0731Z= 4.0(T-208-3)R= 8.5(T-208-11A)Ca=0.44Na(T-208-7)Cv=0.63Nv(T-208-8)hn=17.50m

Solve for Period, T:T=Ct(hn)^(3/4) (eq.208-8)T=(0.0731)x(17.50)^(34)= 0.625

Condition:T0.70,since 0.6250.70, thusFt=0use T=0.625

Solve for Base Shear, V: CvIW (0.64)(1)(25,373)V= = RT (8.5)(0.625)V=3,056.70kNCheck for condition: 2.5CaIW (2.5)(1)(1)(25,373)V = R (8.5)V7,462.65kN, thus it's safe!

V>0.11CaIW=(0.11)(0.44)(1)(25,373)V>1,228.05kN, thus it's safe!

Level,i 5 4 3 2 1

Column (kN) Wall (kN) Slabs(kN) Beams(kN) Wi(kN) m(tons)330594594594594

1,9261,9701,9701,9701,970

2,0001,7801,7801,7801,780

1,014926926926926

5,2705,2705,2705,2705,270

25,373

538538538538538

JETRO G. SARAJENASTE 224- STRUCTURAL DYNAMICS