Question 10.6.1: Suppose the shear building of Example 10.5-1 is acted on by ...

The uncoupled equations are as given by Eqns 10.6-3,

$\frac{d^2q_1}{dt^2}+\omega ^{2}_{1}q_1= \ _1z_1P_1+ \ _1z_2P_2+\cdot \cdot \cdot + \ _1z_nP_n$
$\frac{d^2q_2}{dt^2}+\omega ^{2}_{2}q_2= \ _2z_1P_1+ \ _2z_2P_2+\cdot \cdot \cdot + \ _2z_nP_n$
$\vdots$
$\frac{d^2q_n}{dt^2}+\omega ^{2}_{n}q_n= \ _nz_1P_1+ \ _nz_2P_2+\cdot \cdot \cdot + \ _nz_nP_n$               (10.6-3)

in which the force components should be expressed in newtons and not kilonewtons (in imperial units, they should be expressed in the pound-force unit lbf). Using the value of $\omega^2$ and Z in Example 10.5-1, Eqns 10.5-36 and 10.5-38, we have

$\omega^2=\left[\begin{matrix} \omega^{2}_{1} & 0 & 0 \\ 0 & \omega^{2}_{2} & 0 \\ 0 & 0 & \omega^{2}_{3} \end{matrix} \right]=\left[\begin{matrix} 167 & 0 & 0 \\ 0 & 1000 & 0 \\ 0 & 0 & 1710 \end{matrix} \right]s^{-2}$            (10.5-36)

$Z=\left[\begin{matrix} \frac{_1e_1}{L_1} & \frac{_2e_1}{L_2} & \frac{_3e_1}{L_3} \\\\ \frac{_1e_2}{L_1} & \frac{_2e_2}{L_2} & \frac{_3e_2}{L_3} \\\\ \frac{_1e_3}{L_1} & \frac{_2e_3}{L_2} & \frac{_3e_3}{L_3} \end{matrix} \right]=\left[\begin{matrix} 5.1 & 10.3 & 10.9 \\\\ 11.1 & 5.1 & -10.0 \\\\ 14.2 & -15.4 & 7.9 \end{matrix} \right]\times 10^{-3} \ kg^{-1/2}$      (10.5-38)

$\ddot{q}_1+167q_1=1110(1-t)$    (10.6-4(a))

$\ddot{q}_2+1000q_2=510(1-t)$    (10.6-4(b))

$\ddot{q}_3+1710q_3=-1000(1-t)$   (10.6-4(c))

The solution of each of these three equations consists of a complementary function (of the standard form in Eqns 10.5-30) and a particular integral, namely

$q_1=A_1\sin \omega _1t+B_1\cos \omega _1t,$  or  $C_1\sin (\omega _1t+\alpha _1),$  0r  $D_1\cos (\omega _1t+\beta _1)$
$q_2=A_2\sin \omega _2t+B_2\cos \omega _2t,$  or  $C_2\sin (\omega _2t+\alpha _2),$  0r  $D_2\cos (\omega _2t+\beta _2)$
$\vdots$
$q_n=A_n\sin \omega _nt+B_n\cos \omega _nt,$  or  $C_n\sin (\omega _nt+\alpha _n),$  0r  $D_n\cos (\omega _nt+\beta _n)$  (10.5-30)

$q_1=A_1\sin √(167)t+B_1\cos √(167)t+\frac{1110(1-t)}{167}$

$q_2=A_2\sin √(1000)t+B_2\cos √(1000)t+\frac{510(1-t)}{1000}$

$q_3=A_3\sin √(1710)t+B_3\cos √(1710)t-\frac{1000(1-t)}{1710}$             (10.6-5)

Differentiating with respect to t,

$\dot{q}_1=A_1 √(167)\cos√(167)t-B_1 √(167)\sin √(167)t-6.65$

$\dot{q}_2=A_2 √(1000)\cos√(1000)t-B_2 √(1000)\sin √(1000)t-0.51$

$\dot{q}_3=A_3 √(1710)\cos√(1710)t-B_3 √(1710)\sin √(1710)t+0.59$                     (10.6-6)

The initial conditions on q and $\dot{q}$ are determined from those prescribed on x and $\dot{x}$, using Eqns 10.5-40 of Example 10.5-1.

In this particular case, x and $\dot{x}$ are both zero at t = 0; hence by inspection, at t = 0:

$\left[\begin{matrix} q_1 \\ q_2 \\ q_3 \end{matrix} \right]=0$;               $\left[\begin{matrix} \dot{q}_1 \\ \dot{q}_2 \\ \dot{q}_3 \end{matrix} \right]=0$           (10.6-7)

Substituting $q_1$ = 0 and t = 0 into the first equation in Eqns 10.6-5,

$B_1=-6.65$

Substituting $\dot{q}_1 = 0$, and t = 0 into the first equation in Eqns 10.6-6,

$A_1=6.65/√(167)=0.515$

The other four constants of integrations can be determined in a similar way, resulting in:

whence, from Eqns 10.5-32 and 10.5-38

x=Zq                 (10.5-32)

where q is as given in Eqn 10.6-8.