Question 18.2: Determine the peak displacement, storey shears, floor overtu......
Determine the peak displacement, storey shears, floor overturning moment for the above frame with m = 45 412 kg, k = 5530 kN/m; h = 3.657 m due to ground motion characterized by design spectrum scaled to peak ground acceleration of 0.25g. The spectrum is given in Fig. 18.5.
1. Compute natural period
\omega_n=0.302~ \sqrt{\frac{k}{m}}
=0.302 ~\sqrt{\frac{5530 ~\times ~10^3}{45412}}=3.335 ~rad / s
T=\frac{2~ \pi}{\omega_n}=1.89~ s
\frac{A}{g}=(1.89)^{-1} 1.80 \times 0.25=0.238
A = 0.238 × 9.81 = 2.334 m/s²
D=\frac{A}{\omega_n^2}=\frac{2.334}{3.335^2}=0.212 ~m
Z=\bar{\Gamma}~ D
\bar{\Gamma}=\tilde{L} / \tilde{m}
=\frac{3 ~m}{11~ m} \times 5=\frac{15}{11}
Z=\frac{15}{11} \times 0.212=0.286~ m
The displacement diagram is shown in Fig. 18.6.
\text { Storey shear }=\omega^2 ~m_j u_j
=\omega^2~ m_j \psi_j~ Z
=m_j \psi_j A_j ~\bar{\Gamma}
=\frac{15}{11} \times 2.334 \times 45412 \times \psi_j
= 144.5 ψ kN
The base shear is 430 kN. The show force diagram is shown in Fig. 18.7.
\text { Total weight }=45412 \times \frac{9.81}{1000} \times 5
= 2227 kN
V_b=430
\frac{430}{2227}
= 19.3% of total weight of the building.
