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ESE Civil 2018 Official Paper

Option 1 : 1350 kN

CT 3: Building Materials

3235

10 Questions
20 Marks
12 Mins

As per IS 1893 (part 1), CL 6.4.2:

Horizontal seismic coefficient,

\({A_h} = \frac{Z}{2} \times \frac{I}{R} \times \frac{S}{g}\)

Where

Z = Zone factor; Z = 0.36 (For Zone V) (Table 2 of IS : 1893)

I = Importance factor; for **important building**, I = 1.5 (Table 6 of IS: 1893)

S/g = 2.5 (acceleration factor)

R= Response Reduction factor; R = 5 (for ductility consideration or Moment resisting frame)

Now,

\({A_h} = \frac{{0.36}}{2} \times \frac{{1.5}}{5} \times 2.5\) = 0.135

Now, Base Shear, \(V = \;{A_h}\;W\)

Seismic Weight, W = 10000 kN

\({\bf{V}} = \;0.135 \times 10000 = 1350\;{\bf{kN}}\)