Question 17.5: The frame shown in Fig. 17.18 is for use in a building to be......
The frame shown in Fig. 17.18 is for use in a building to be located on sloping ground. The beams are made much stiffer than columns and can be assumed to be rigid. The cross-section of the columns is 250 mm square but their lengths are 4 m and 8 m respectively. Determine the base shear in the two columns, at the instant of peak response due to El Centro ground motion. Assume damping as 5% of critical damping.
Since the beam is rigid, the stiffness of columns can be taken as
K=\sum\limits_{i=1}^2 \frac{12~ E I_{ci}}{h_i^3}
=\frac{12 ~\times~ 30 ~\times ~10^6~ \times~ 3.254~ \times ~10^{-4}}{4^3}
+\frac{12 ~\times ~30 \times ~10^6~ \times ~3.254 ~\times ~10^{-4}}{8^3}
\begin{array}{l l}(shorter)~~~(longer)\\=1830.9+229\end{array}
= 2059.8 kN m
Mass = 5096 kg
\omega_n=\sqrt{\frac{k}{m}}
=\sqrt{\frac{20 59 800}{5096}}= 20~ rad/s
T_{n}=\frac{\omega_n}{2~\pi} = 0.31~s
Sd = 17 mm
A = 0.76g
Lateral force = 50 × 0.76 = 38 kN
Force shared by two columns
Force shared by short column ={\frac{38}{2059.8}}\times1830.9=33.79\,\mathrm{kN}
Force shared by long column = 38 – 33.79 = 4.21 kN
The shear in columns as well as bending moment diagram are shown in Fig. 17.19.
Observe that both columns go through equal deflection. The stiffer column carries greater force than the flexible column. Sometimes this basic principle is not recognized in building design, leading to unanticipated damage to the stiffer structure.
