Question 16.2: The uniform shear beam shown in Figure E16.2a has a mass m p...
The uniform shear beam shown in Figure E16.2a has a mass m per unit length and shear stiffness k. It is subjected to a rectangular ground acceleration pulse as shown in Figure E16.2b. Obtain expressions for the maximum displacement and shear in each mode of vibration. By taking the absolute sum of the modal values, obtain an upper bound estimate for the base shear.
The equation of motion for a shear beam subjected to a support excitation u_{g}(t) is given by
-k \frac{\partial^{2} u}{\partial x^{2}}+m \frac{\partial^{2} u}{\partial t^{2}}=-m \ddot{u}_{g}(t) \qquad (a)
where u is the displacement relative to the support.
The uncoupled modal equations are
\ddot{y}_{n}(t)+\omega_{n}^{2} y_{n}(t)=p_{n} \qquad (b)
where
p_{n}=\int_{0}^{L}-m \ddot{u}_{g}(t) \phi_{n}(x) d x \qquad (c)
and
\begin{aligned}\phi_{n}(x) &=\sqrt{\frac{2}{m L}} \sin \frac{(2 n-1) \pi x}{2 L} \qquad (d)\\\omega_{n} &=\frac{(2 n-1) \pi}{2} \sqrt{\frac{k}{m L^{2}}} \qquad (e)\end{aligned}
On substituting for \ddot{u}_{g} and \phi_{n}(x) in Equation c and ignoring the negative sign, we get
\begin{aligned}p_{n} &=m a \sqrt{\frac{2}{m L}} \int_{0}^{L} \sin \frac{(2 n-1) \pi x}{2 L} d x \\&=m a \sqrt{\frac{2}{m L}} \frac{2 L}{(2 n-1) \pi} \qquad \qquad (f)\end{aligned}
Equation \mathrm{b} has the solution
y_{n}=\frac{p_{n}}{\omega_{n}^{2}}\left(1-\cos \omega_{n} t\right) \qquad (g)
Transformation to physical coordinates gives
u(x, t)=\sum_{n=1}^{\infty} u_{n} \qquad (h)
where
The modal shears are given by
\begin{aligned}V_{n} &=k \frac{\partial u_{n}}{\partial x} \\&=\frac{2 a k}{L \omega_{n}^{2}} \cos \frac{(2 n-1) \pi x}{2 L}\left(1-\cos \omega_{n} t\right) \qquad (j)\end{aligned}
Substitution for \omega_{n} gives
V_{n}=\frac{8 a m L}{\pi^{2}(2 n-1)^{2}} \cos \frac{(2 n-1) \pi x}{2 L}\left(1-\cos \omega_{n} t\right) \qquad (k)
The maximum modal displacement is obtained from Equation i with \cos \omega_{n} t=-1, so that
\left(u_{n}\right)_{\max }=\frac{8 a}{(2 n-1) \pi \omega_{n}^{2}} \sin \frac{(2 n-1) \pi x}{2 L} \qquad (l)
On substituting for \omega_{n} and setting x=L in Equation 1 to obtain the maximum displacement at the top, we get
\left\{u_{n}(L)\right\}_{\max }=\pm \frac{32 a m L^{2}}{\pi^{3} k} \frac{1}{(2 n-1)^{3}} \qquad (m)
The maximum value of V_{n} is obtained from Equation \mathrm{j} with \cos \omega_{n} t=-1
\left(V_{n}\right)_{\max }=\frac{16 a m L}{\pi^{2}(2 n-1)^{2}} \cos \frac{(2 n-1) \pi x}{2 L} \qquad (n)
The corresponding base shear is
\left(V_{0 n}\right)_{\max }=\frac{16 a m L}{\pi^{2}(2 n-1)^{2}} \qquad (o)
The absolute sum of the maximum base shears in all modes is given by
\left(V_{0}\right)_{\max }=\frac{16 a m L}{\pi^{2}} \sum_{n=1}^{\infty} \frac{1}{(2 n-1)^{2}} \qquad (p)
The series in Equation \mathrm{p} converges to \pi^{2} / 8. Hence
\left(V_{0}\right)_{\max }=2 a m L \qquad (q)