Question 8.12: Calculate the response of the tower in Example 8.9 to the lo...
Calculate the response of the tower in Example 8.9 to the loading shown there for the first 1.0 \mathrm{~s} using a numerical integration technique. Assume that damping in the system is 10 \% of critical and use h=0.1 \mathrm{~s}.
We will obtain the response using (a) the constant-acceleration method, (b) the average acceleration method, and (c) the linear acceleration method. For each case,
\begin{aligned}k &=100 \mathrm{kips} / \mathrm{in} . \\m &=2.533 \mathrm{kip} \cdot \mathrm{s}^{2} / \mathrm{in} . \\ \omega &=6.283 \mathrm{rad} / \mathrm{s} \\c &=2 \xi \omega m=3.183 \mathrm{kip} \cdot \mathrm{s} / \mathrm{in} . \\u_{0} &=\dot{u}_{0}=\ddot{u}_{0}=0\end{aligned}
(a) Constant-acceleration method. Using Equations 8.91, 8.89, and 8.104,
we get
Table E8.12a Constant-acceleration method
\begin{matrix}\hline &&&&&u_{n+1} & \dot{u}_{n+1} & \ddot{u}_{n+1}& u_{n}\\ \text{Time }& {u_{n}} & {\dot{u}_{n}} & {\ddot{u}_{n}} & {p_{n+1}} & (Eq. a) & { ( Eq. b) } & { (Eq. c) } & { (\text{theoretical}) } \\\hline 0.0 & 0.0000 & 0.000 & 0.00 & 50.0 & 0.0000 & 0.000 & 19.74 & 0.0000 \\0.1 & 0.0000 & 0.000 & 19.74 & 86.6 & 0.0987 & 1.974 & 27.81 & 0.0323 \\0.2 & 0.0987 & 1.974 & 27.81 & 100.0 & 0.4351 & 4.755 & 16.32 & 0.2254 \\0.3 & 0.4351 & 4.755 & 16.32 & 86.6 & 0.9922 & 6.387 & -13.01 & 0.6204 \\0.4 & 0.9923 & 6.387 & -13.01 & 50.0 & 1.5659 & 5.086 & -48.47 & 1.0961 \\0.5 & 1.5660 & 5.086 & -48.47 & 0.0 & 1.8322 & 0.239 & -72.63 & 1.4251 \\0.6 & 1.8320 & 0.239 & -72.63 & 0.0 & 1.4927 & -7.024 & -50.11 & 1.3772 \\0.7 & 1.4930 & -7.024 & -50.11 & 0.0 & 0.5401 & -12.036 & -6.19 & 0.8683 \\0.8 & 0.5400 & -12.036 & -6.19 & 0.0 & -0.6949 & -12.655 & 43.33 & 0.1105 \\0.9 & -0.6950 & -12.655 & 43.33 & 0.0 & -1.7430 & -8.322 & 79.29 & -0.5974 \\1.0 & -1.7430 & -8.322 & 79.29 & & & & & -1.0073 \\\hline\end{matrix}
\begin{aligned}&\dot{u}_{n+1}=\dot{u}_{n}+0.1 \ddot{u}_{n} \qquad \qquad \qquad (b)\\ &\ddot{u}_{n+1}=\frac{1}{2.533}\left(p_{n+1}-100 u_{n}-13.183 \dot{u}_{n}-0.8183 \ddot{u}_{n}\right) \quad (c) \end{aligned}
Step-by-step time integration is now carried out using Equations a, b, and c. Response calculations for the first 1.0 \mathrm{~s} are shown in Table E8.12a.
(b) Average acceleration method. Substituting the values of m, c, k, and h in Equation 8.124,
\left(\frac{4 m}{h^2}+\frac{2 c}{h}+k\right) u_{n+1}=p_{n+1}+m\left(\frac{4}{h^2} u_n+\frac{4}{h} \dot{u}_n+\ddot{u}_n\right)+c\left(\frac{2}{h} u_n+\dot{u}_n\right) \qquad (8.124)
we get
1176.9 u_{n+1}=p_{n+1}+1076.9 u_{n}+104.5 \dot{u}_{n}+2.533 \ddot{u}_{n} \qquad (d)
Equations 8.123 and 8.122
\begin{aligned}&\dot{u}_{n+1}=-\dot{u}_n+\frac{2}{h}\left(u_{n+1}-u_n\right) \qquad(8.123) \\&\ddot{u}_{n+1}=\frac{4}{h^2}\left(u_{n+1}-u_n-h \dot{u}_n\right)-\ddot{u}_n \qquad \text { (8.122) }\end{aligned}
give
\begin{aligned}&\dot{u}_{n+1}=20\left(u_{n+1}-u_{n}\right)-\dot{u}_{n} \qquad \qquad (e)\\&\ddot{u}_{n+1}=400\left(u_{n+1}-u_{n}\right)-40 \dot{u}_{n}-\ddot{u}_{n} \qquad (f)\end{aligned}
Time integration is carried out using Equations \mathrm{d}, e, and f. Response calculations for the first 1.0 \mathrm{~s} are shown in Table E8.12b. (c) Linear acceleration method. Substitution for m, c, k, and b in Equation 8.133
\left(\frac{6 m}{h^2}+\frac{3 c}{h}+k\right) u_{n+1}=p_{n+1}+m\left(\frac{6}{h^2} u_n+\frac{6}{h} \dot{u}_n+2 \ddot{u}_n\right)+c\left(\frac{3}{h} u_n+2 \dot{u}_n+\frac{h}{2} \ddot{u}_n\right) \qquad (8.133)
gives
1715.3 u_{n+1}=p_{n+1}+1615.3 u_{n}+158.35 \dot{u}_{n}+5.225 \ddot{u}_{n} \qquad (g)
Equations 8.132 and 8.131
\begin{gathered}\dot{u}_{n+1}=\frac{3}{h}\left(u_{n+1}-u_n\right)-2 \dot{u}_n-\frac{h}{2} \ddot{u}_n(8.132) \\\ddot{u}_{n+1}=\frac{6}{h^2}\left(u_{n+1}-u_n-h \dot{u}_n-\frac{h^2}{3} \ddot{u}_n\right)(8.131)\end{gathered}
give
\begin{aligned}&\dot{u}_{n+1}=30\left(u_{n+1}-u_{n}\right)-2 \dot{u}_{n}-0.05 \ddot{u}_{n} \qquad (h) \\&\ddot{u}_{n+1}=600\left(u_{n+1}-u_{n}\right)-60 \dot{u}_{n}-2 \ddot{u}_{n} \qquad (i)\end{aligned}
Time integration is carried out using Equations \mathrm{g}, \mathrm{h}, and i. Response calculations for the first 1.0 \mathrm{~s} are shown in Table E8.12c.
The theoretical response for a sine-wave loading is given by Equation b of Example 8.10. The equation is used to obtain the theoretical response values for the first 0.6 \mathrm{~s} shown in Tables E8.12a, E8.12b, and E8.12c. Theoretical displacement values for t \geq 0.6 \mathrm{~s} are calculated
Table E8.12b Average acceleration method.
Table E8.12c Linear acceleration method.
from Equation \mathrm{j} of Example 8.11 and are entered in Tables E8.12a, E8.12b, and E8.12c. An examination of the response values in these tables shows that the average acceleration and the linear acceleration methods give reasonable results. The time step b of 0.1 \mathrm{~s}, which is one-tenth the natural period of the system and one-twelfth the period of exciting force is probably the maximum one can use to ensure a reasonable accuracy; a smaller step size must be used if better accuracy is desired. The results obtained from the constant-acceleration method are not satisfactory. A considerably smaller step size must be used in this case to obtain acceptable accuracy. In fact, the method is not very effective, for several reasons that will be discussed later.