Question 8.13: Solve Example 8.12 by the Wilson θ method using θ = 1.5 and ...
Solve Example 8.12 by the Wilson θ method using θ = 1.5 and h = 0.1 s.
Table E8.13 Wilson-θ method.
\begin{matrix}\hline &&&&&u_{n+\theta} &\ddot{u}_{n+\theta}&\ddot{u}_{n+1}&\dot{u}_{n+1} & {u}_{n+1} & u_n \\ \text{Time}& {{u}_{n}} & {\dot{u}_{n}} & {\ddot{u}_{n}} & p_{n+\theta} & { (Eq. a) } & { (Eq. b) } & { (Eq. c) } & { (Eq. d) }& { (Eq. e) } & { (\text{theoretical}) } \\ \hline 0.0 & 0.0000 & 0.0000 & 0.00 & 75.0 & 0.0894 & 23.837 & 15.89 & 0.7946 & 0.0265 & 0.0000 \\0.1 & 0.0265 & 0.7946 & 15.89 & 104.9 & 0.3508 & 22.923 & 20.58 & 2.6188 & 0.1932 & 0.0323 \\0.2 & 0.1932 & 2.6181 & 20.58 & 106.7 & 0.7634 & 6.197 & 10.99 & 4.1966 & 0.5419 & 0.2254 \\0.3 & 0.5419 & 4.1966 & 10.99 & 79.9 & 1.1806 & -19.507 & -9.34 & 4.2791 & 0.9827 & 0.6204 \\0.4 & 0.9827 & 4.2791 & -9.34 & 31.7 & 1.3934 & -42.927 & -31.73 & 2.2254 & 1.3265 & 1.0961 \\0.5 & 1.3265 & 2.2254 & -31.73 & -25.0 & 1.2237 & -52.993 & -45.91 & -1.6564 & 1.3668 & 1.4251 \\0.6 & 1.3668 & -1.6564 & -45.91 & 0.0 & 0.7011 & -19.467 & -28.28 & -5.3658 & 1.0010 & 1.3772 \\0.7 & 1.0010 & -5.3658 & -28.28 & 0.0 & 0.0143 & 8.039 & -4.07 & -6.9832 & 0.3634 & 0.8683 \\0.8 & 0.3634 & -6.9832 & -4.07 & 0.0 & -0.6018 & 30.041 & 18.67 & -6.2529 & -0.3174 & 0.1105 \\0.9 & -0.3174 & -6.2529 & 18.67 & 0.0 & -0.9640 & 40.323 & 33.11 & -3.6641 & -0.8252 & -0.5974 \\1.0 & -0.8252 & -3.6641 & 33.11 & & & & & & & -1.0073 \\ \hline \end{matrix}
Substitution of m, c, k, h, and \theta in Equation 8.136
\left\{\frac{6 m}{(\theta h)^2}+\frac{3 c}{\theta h}+k\right\} u_{n+\theta}=p_{n+\theta}+m\left\{\frac{6 u_n}{(\theta h)^2}+\frac{6}{\theta h} \dot{u}_n+2 \ddot{u}_n\right\}+c\left\{\frac{3}{\theta h} u_n+2 \dot{u}_n+\frac{\theta h}{2} \ddot{u}_n\right\} \qquad (8.136)
gives
839.13 u_{n+\theta}=p_{n+\theta}+739.13 u_{n}+107.69 \dot{u}_{n}+5.305 \ddot{u}_{n} \qquad (a)
where p_{n+\theta} is obtained from Equation 8.137.
\quad=p_n(1-\theta)+p_{n+1} \theta \quad \quad(8.137)
Equation 8.134
\ddot{u}_{n+\theta}=\frac{6}{(\theta h)^2}\left\{u_{n+\theta}-u_n-\theta h \dot{u}_n-\frac{(\theta h)^2}{3} \ddot{u}_n\right\} \quad(8.134)
now leads to
\ddot{u}_{n+\theta}=266.7\left(u_{n+\theta}-u_{n}\right)-40 \dot{u}_{n}-2 \ddot{u}_{n} \qquad (b)
From Equation 8.138
\ddot{u}_{n+1}=\ddot{u}_n+\frac{\ddot{u}_{n+\theta}-\ddot{u}_n}{\theta} \quad(8.138)
we get
\ddot{u}_{n+1}=0.6667 \ddot{u}_{n+\theta}+0.3333 \ddot{u}_{n} \qquad (c)
Finally, substitution of Equation c in Equations 8.129
\dot{u}_{n+1}=\dot{u}_n+\frac{h}{2}\left(\ddot{u}_n+\ddot{u}_{n+1}\right) \qquad (8.129)
and 8.130
u_{n+1}=u_n+b \dot{u}_n+\frac{h^2}{3} \ddot{u}_n+\frac{h^2}{6} \ddot{u}_{n+1} \quad(8.130)
gives
\begin{aligned}&\dot{u}_{n+1}=\dot{u}_{n}+0.05\left(\ddot{u}_{n}+\ddot{u}_{n+1}\right) \qquad \qquad \qquad (d)\\&u_{n+1}=u_{n}+0.1 \dot{u}_{n}+0.00333 \ddot{u}_{n}+0.00167 \ddot{u}_{n+1} \qquad (e)\end{aligned}
Step-by-step time integration is carried out using Equations a through e. Response calculations for the first 1 \mathrm{~s} are shown in Table E8.13.