Question 10.3.2: Figure 10.3-4 shows a portal frame ABCD; it may be assumed t...
Figure 10.3-4 shows a portal frame ABCD; it may be assumed that the horizontal member is infinitely stiff and that the vertical members have negligible mass compared with that of the horizontal member. If there is no damping, determine
(a) the natural frequency ƒ and the natural period T;
(b) the displacement at time t seconds if the member BC is displaced a distance l millimeters and then suddenly released at t = l s .

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(1) From Section 5.2, Eqn 5.2-1, the stiffnesses of the members AB and CD are :
\hspace{5em}k_{AB} =\frac{12EI}{L^3} N/mm;\qquad k_{CD}=\frac{12E(3I)}{(2L)^2}=4.5\frac{EI}{L^3} \ N/mmHence
\hspace{5em}\begin{matrix}k=k_{AB} + k_{CD} & = & 16.5\frac{EI}{L^3} N/mm \\ \\ & = & 16 500 \frac{EI}{L^3} N/m\end{matrix}From Eqn 10.3-17,
\hspace{5em} \omega =\surd \frac{K}{M}(where k is in newtons per metre and M in kilograms), then
\hspace{5em} \omega =\surd{ \left(\frac{16500 EI}{ML^3}\right)}= 128 \surd{\left( \frac{EI}{ML^3}\right)} radians Persecond
From Eqn 10.3-20,
\hspace{5em} T=\frac{2\pi }{\omega }= \frac{2\pi }{128 }\surd \left(\frac{ML^3}{EI} \right) seconds
\hspace{5em} ƒ =\frac{1}{T} = \frac{128}{2\pi }\surd \left(\frac{EI}{ML^3} \right) cycles per second
(2) From Eqn 10.3-25, if the initial conditions are x=x_{\tau } \ m \ and \ \dot{x}=\dot{x}_{\tau} \ m/s at t = τ seconds, then
\hspace{5em} x=\frac{\dot{x}_{\tau }}{\omega } \sin \omega \left(t- \tau \right) + x_\tau \cos \omega \left(t- \tau \right) \ m (10.3-26)
In this example, x_{\tau }=l \ mm = 0.001 \ l \ m, \dot{x}_{\tau }=0, \tau =1s, \ and \ \omega =128\surd \left[\left(EI\right)/\left(ML^3\right) \right], hence
\hspace{5em} x=0.001/\cos \left[128 \surd\left(\frac{EI}{ML^3} \right)\left(t-1\right) \right] mThe reader should pay particular attention to units in structural dynamics problems. The stiffness k should be in absolute units, e.g. N/m; the mass M should be in mass units, e.g. kg. Similarly, in using Eqn 10.3-26, x should be expressed in metres, \dot{x} in m/s, and t and τ in seconds.
\hspace{5em} \begin{matrix} m_{1} \\ m_{2} \end{matrix} =\pm i\surd{\frac{k}{M}=\pm i\omega } (10.3-17)
\hspace{5em} x=\frac{\dot{x}_{\tau }}{\omega } \sin \omega \left(t- \tau \right) + x_\tau \cos \omega \left(t- \tau \right) (10.3-25)
(for x=x_\tau \ and \ dx/dt= \dot{x}_{\tau } at t=τ).