Question 19.6: The beam in Example 19.5 is subjected to a vertical upward f...
The beam in Example 19.5 is subjected to a vertical upward force P along d.o.f. 5 of component 2. What are the resulting vertical displacements at each node of the finite element mesh.
Referring to the d.o.f. of component 2 the force vector is given by
\left(\mathbf{f}^{T}\right)^{(2)}=\left[\begin{array}{llllllllll}0 & 0 & 0 & 0 & 100 & 0 & 0 & 0 & 0 & 0\end{array}\right] \qquad (a)
Transforming to the generalized coordinates as per Equation 19.85
\begin{gathered}\mathbf{u}^{(2)}=\mathbf{T}^{(2)} \mathbf{p}^{(2)} \\\left\{\begin{array}{l}\mathbf{u}_i \\\mathbf{u}_a \\\mathbf{u}_c \\\mathbf{u}_r\end{array}\right\}^{(2)}=\left[\begin{array}{cccc}\boldsymbol{\Phi}_{k k} & \boldsymbol{\Psi}_{k a} & \boldsymbol{\Psi}_{k c} & \boldsymbol{\Psi}_{k r} \\\boldsymbol{\Phi}_{a k} & \boldsymbol{\Psi}_{a a} & \boldsymbol{\Psi}_{a c} & \boldsymbol{\Psi}_{a r} \\0 & 0 & \mathbf{I}_{c c} & 0 \\0 & 0 & 0 & \mathbf{I}_{r r}\end{array}\right]^{(2)}\left\{\begin{array}{l}\mathbf{p}_k \\\mathbf{p}_a \\\mathbf{p}_c \\\mathbf{p}_r\end{array}\right\}\end{gathered} (19.85)
On carrying out the operations in Equation (b) we get
\left(\tilde{\mathbf{f}}^{T}\right)^{(2)}=\left[\begin{array}{llllll}50.0260 & -71.2736 & -52.0292 & -0.3064 & 25.0000 & 75.0000\end{array}\right]\qquad (c)
The transformed force vector for component 1, \mathbf{f}^{(1)}, is a null vector of size 5 by 1 .
Assembling the two force vectors and applying the coupling transformation we get
\tilde{\mathbf{f}}=\mathbf{S}^{T}\left\{\begin{array}{c}\tilde{\mathbf{f}}^{(1)} \\\tilde{\mathbf{f}}^{(2)}\end{array}\right\}\qquad (d)
Equation (d) leads to
\tilde{\mathrm{f}}^T = [−0.9686 1.7117 −2.5071 −0.2873 24.1530 50.7790 −69.7566 −54.3539 75.3388] (e)
The generalized displacements are given by
\mathbf{q}=\tilde{\mathbf{K}}^{-1} \tilde{\mathbf{f}} \qquad (f)
Equation (f) yields
\mathrm{q}^T = \frac{l^3}{EI} = [230.61 −38.374 12.503 299.92 4533.3 −306.4 −66.337 14.474 138883.0] (g)
Transforming back to the generalized coordinates for the components
\left\{\begin{array}{l}p^{(1)} \\p^{(2)}\end{array}\right\}=S q (h)
Equation (h) yields
Finally transforming to the physical coordinates of the components we have
\mathbf{u}^{(1)}=T^{(1)} \mathbf{p}^{(1)} \quad \text { and } \quad \mathbf{u}^{(2)}=T^{(2)} p^{(2)} (k)
Equation (\mathrm{k}) yields
(u^T)^{(1)} = \frac{l^3}{EI} [333.3 650.0 1266.7 1200.0 2700.0 1650.0 2000.0 4533.3] (l)
(u^T)^{(2)} = \frac{l^3}{EI} [6667.0 2250.0 9000.0 2400.0 11433.0 2450.0 2000.0 2449.0 4533.0 13883.0] (m)
In Equations (l) and (\mathrm{m}) the slopes \theta have again been given in terms of l \theta.
The vertical displacements in component 1 are obtained from
\left[\begin{array}{llll}u(1) & u(3) & u(5) & u(8)\end{array}\right]^{(1)}=\frac{l^{3}}{E I}\left[\begin{array}{llll}333.3 & 1266.7 & 2700.0 & 4533.3\end{array}\right] (n)
The displacements in component 2 are
\left[\begin{array}{lllll}u(9) & u(1) & u(3) & u(5) & u(10)\end{array}\right]^{(2)}=\frac{l^{3}}{E I}\left[\begin{array}{lllll}4533.0 & 6667.0 & 9000.0 & 11433.0 & 13883.0\end{array}\right] (o)
These displacement values match with the exact results to four significant digits.