Solution to part (a).

Rearrange the unrestrained stiffness matrix in eq. ([eq16.1h]) of example [ex16.1] so that the order of the rows and columns correspond to degrees of freedom 3, 6, 1, 2, 4, and 5. \[\begin{aligned} \label{eq16.2a}\tag{a} [K]=\left(\frac{E A}{L}\right) \begin{array}{@{}c@{}} {\arraycolsep=7.5pt\begin{array}{@{}cccccc@{}} q_3 & q_6 & q_1 & q_2 & q_4 & q_5 \end{array}}\\[6pt] \left[\begin{array}{cc;{2pt/2pt}cccc} 1+a & a & -1 & 0 & -a & -a \\ a & 1+a & 0 & -1 & -a & -a \\\hdashline[2pt/2pt] -1 & 0 & 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 & 0 \\ -a & -a & 0 & 0 & a & a \\ -a & -a & 0 & 0 & a & a\end{array}\right]\end{array}.\end{aligned}\] Compare the matrix in eq. ([eq16.2a]) to the general form ([eq15.27]) on page to identify \[\begin{aligned} \label{eq16.2b}\tag{b} \left[K_{\alpha \alpha}\right]=\left(\frac{E A}{L}\right)\left[\begin{array}{@{}cc@{}}1+a & a \\a & 1+a\end{array}\right] \quad\left[K_{\alpha \beta}\right]=\left(\frac{E A}{L}\right)\left[\begin{array}{@{}cccc@{}}-1 & 0 & -a & -a \\0 & -1 & -a & -a\end{array}\right],\end{aligned}\] and \[\begin{aligned} \label{eq16.2c}\tag{c} \left[K_{\beta \alpha}\right]=\left(\frac{E A}{L}\right)\left[\begin{array}{@{}cc@{}}-1 & 0 \\0 & -1 \\-a & -a \\-a & -a\end{array}\right] \quad\left[K_{\beta \beta}\right]=\left(\frac{E A}{L}\right)\left[\begin{array}{@{}llll@{}}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & a & a \\0 & 0 & a & a\end{array}\right].\end{aligned}\] The restrained structural stiffness matrix \([K_{\alpha \alpha}]\) is symmetric, and the sum of its column elements is not zero. Also note that the restrained stiffness structural matrix can be obtained from the unrestrained structural stiffness matrix in eq. ([eq16.2h]) by merely crossing out rows and columns 1, 2, 4, and 5: \[\begin{aligned} \label{eq16.2d}\tag{d} [K]=\left(\frac{E A}{L}\right) \begin{array}{@{}c@{}} {\arraycolsep=7.5pt\begin{array}{@{}cccccc@{}} q_1 & q_2 & q_3 & q_4 & q_5 & q_6 \end{array}}\\[6pt] \left[\begin{array}{@{}cccccc@{}} \multirow{6}{2pt}{\hspace*{3pt}\rule{0.5pt}{76pt}}\hspace*{-1pt}1 & \multirow{6}{2pt}{\hspace*{3pt}\rule{0.5pt}{76pt}}\hspace*{-1pt}0 & -1 & \multirow{6}{2pt}{\hspace*{3pt}\rule{0.5pt}{76pt}}\hspace*{-1pt}0 & \multirow{6}{2pt}{\hspace*{3pt}\rule{0.5pt}{76pt}}\hspace*{-1pt}0 & 0 \\[-7pt] \hline\\[-4pt] 0 & 1 & 0 & 0 & 0 & -1 \\[-7pt] \hline\\[-4pt] -1 & 0 & 1+a & -a & -a & a \\ 0 & 0 & -a & a & a & -a \\[-7pt] \hline\\[-4pt] 0 & 0 & -a & a & a & -a \\[-7pt] \hline\\[-4pt] 0 & -1 & a & -a & -a & 1+a\end{array}\right]\end{array}.\end{aligned}\] The fixed-end action vector in eq. ([eq16.1i]) of example [ex16.1] for the unrestrained truss reduces to \[\begin{aligned} \label{eq16.2e}\tag{e} \left\{Q^{0}\right\}=\left[\begin{array}{@{}llllll@{}}0 & \left(N_{T}\right)_{1-3} & 0 & 0 & 0 & -\left(N_{T}\right)_{1-3}\end{array}\right]^{T}.\end{aligned}\] Elements in rows 3 and 6 constitute \(\left\{Q_{\alpha}^{0}\right\}\) while the remaining rows constitute \(\left\{Q_{\beta}^{0}\right\}\). Thus, \[\begin{aligned} \label{eq16.2f}\tag{f} \left\{Q_{\alpha}^{0}\right\}=\left[\begin{array}{@{}c@{}}0 \\-\left(N_{T}\right)_{1-3}\end{array}\right] \quad\left\{Q_{\beta}^{0}\right\}=\left[\begin{array}{@{}c@{}}0 \\\left(N_{T}\right)_{1-3} \\0 \\0\end{array}\right].\end{aligned}\]

Solution to part (b).

Equation ([eq15.31]) on page with the addition of the fixed-end action vector is \[\begin{aligned} \label{eq16.2g}\tag{g} \left\{Q_{\alpha}\right\}=\left[K_{\alpha \alpha}\right]\left\{q_{\alpha}\right\}+\left[K_{\alpha \beta}\right]\left\{q_{\beta}\right\}+\left\{Q_{\alpha}^{0}\right\}.\end{aligned}\] The fixed-end action vector is subtracted from each side of this equation, since it is a known vector determined from the specified temperature changes in the bars. That is, eq. ([eq16.2g]) is written in the form \[\begin{aligned} \label{eq16.2h}\tag{h} \left\{Q_{\alpha}\right\}+\underbrace{(-\{Q_{\alpha}^{0}\})}_{\hspace*{-11pc}\mbox{equivalent joint force vector---}}=\left[K_{\alpha \alpha}\right]\left\{q_{\alpha}\right\}+\left[K_{\alpha \beta}\right]\left\{q_{\beta}\right\}.\end{aligned}\] The vector \(-\left\{Q^{0}\right\}\) is called the equivalent joint force vector. In this example the prescribed joint displacement vector is \(\left\{q_{\beta}\right\}=\left[\begin{array}{@{}llll@{}}q_{1} & q_{2} & q_{4} & q_{5}\end{array}\right]^{T}=\left[\begin{array}{@{}llll@{}}0 & 0 & 0 & 0\end{array}\right]^{T}\), and the prescribed joint force vector is \(\left\{Q_{\alpha}\right\}^{T}=\left[\begin{array}{@{}ll@{}}Q_{3} & Q_{6}\end{array}\right]=\left[\begin{array}{@{}ll@{}}0 & 0\end{array}\right]\). The solution for the unknown joint displacement vector is \[\begin{aligned} \label{eq16.2i}\tag{i} \left\{q_{\alpha}\right\}=\left[K_{\alpha \alpha}\right]^{-1}\left\{-Q_{\alpha}^{0}\right\}\mbox{, where the inverse matrix is }\left[K_{\alpha \alpha}\right]^{-1}=\left(a d j\left[K_{\alpha \alpha}\right]\right) /\left({det}\left[K_{\alpha \alpha}\right]\right).\end{aligned}\] The adjoint of the restrained structural stiffness matrix and its determinate are¹ \[\begin{aligned} \label{eq16.2j}\tag{j} a d j\left[K_{\alpha \alpha}\right]=\left(\frac{E A}{L}\right)\left[\begin{array}{@{}cc@{}}1+a & -a \\-a & 1+a\end{array}\right] \quad {det}\left[K_{\alpha \alpha}\right]=\left(\frac{E A}{L}\right)^{2}\left((1+a)^{2}-a^{2}\right)=\left(\frac{E A}{L}\right)^{2}(1+2 a).\end{aligned}\] So the inverse of the restrained structural stiffness matrix is \[\begin{aligned} \label{eq16.2k}\tag{k} \left[K_{\alpha \alpha}\right]^{-1}=\left(\frac{L}{E A}\right)\left(\frac{1}{1+2 a}\right)\left[\begin{array}{@{}cc@{}}1+a & -a \\-a & 1+a\end{array}\right].\end{aligned}\] Perform a check of this inverse. Is \(\left[K_{\alpha \alpha}\right]\left[K_{\alpha \alpha}\right]^{-1}=[I]\)? \[\begin{aligned} \left(\frac{E A}{L}\right)&\left[\begin{array}{@{}cc@{}}1+a & a \\a & 1+a\end{array}\right]\left(\frac{L}{E A}\right)\left(\frac{1}{1+2 a}\right)\left[\begin{array}{@{}cc@{}}1+a & -a \\-a & 1+a\end{array}\right]=\left(\frac{1}{1+2 a}\right)\left[\begin{array}{@{}cc@{}}1+a & a \\a & 1+a\end{array}\right]\left[\begin{array}{@{}cc@{}}1+a & -a \\-a & 1+a\end{array}\right] \nonumber\\ &=\frac{1}{1+2 a}\left[\begin{array}{@{}cc@{}}(1+a)^{2}+\left(-a^{2}\right) &(1+a)(-a)+a(1+a) \\ a(1+a)+(1+a)(-a)&-a^{2}+(1+a)^2\end{array}\right]=\frac{1}{1+2 a}\left[\begin{array}{@{}cc@{}}1+2 a & 0 \\0 & 1+2 a\end{array}\right]=\left[\begin{array}{@{}cc@{}}1 & 0 \\0 & 1\end{array}\right].\label{eq16.2l}\tag{l}\end{aligned}\] Hence, the inverse satisfies \(\left[K_{\alpha \alpha}\right]\left[K_{\alpha \alpha}\right]^{-1}=[I]\). The solution for the unknown nodal displacement vector is \[\begin{aligned} \label{eq16.2m}\tag{m} \left[\begin{array}{@{}l@{}}q_{3} \\q_{6}\end{array}\right]=\left(\frac{L}{E A}\right)\left[\begin{array}{@{}l@{}}\frac{-\sqrt{2}}{4+2 \sqrt{2}} \\[6pt]\frac{4+\sqrt{2}}{4+2 \sqrt{2}}\end{array}\right]\left(N_{T}\right)_{1-3}.\end{aligned}\]

Solution to part (c).

The support reactions are determined from eq. ([eq15.32]) on page , which is repeated below as eq. ([eq16.2n]). \[\begin{aligned} \label{eq16.2n}\tag{n} \left\{Q_{\beta}\right\}=\left[K_{\beta \alpha}\right]\left\{q_{\alpha}\right\}+\left[K_{\beta \beta}\right]\left\{q_{\beta}\right\}+\left\{Q_{\beta}^{0}\right\}.\end{aligned}\] The prescribed joint displacement vector \(\left\{q_{\beta}\right\}=0_{4 X 1}\), and submatrix \(\left[K_{\beta \alpha}\right]\) was determined in part ([eq16.2a]). Hence, \[\begin{aligned} \label{eq16.2o}\tag{o} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{4} \\Q_{5}\end{array}\right]=\left(\frac{E A}{L}\right)\left[\begin{array}{@{}cc@{}}-1 & 0 \\0 & -1 \\-a & -a \\-a & -a\end{array}\right]\left[\begin{array}{@{}l@{}}q_{3} \\q_{6}\end{array}\right]+\left[\begin{array}{@{}c@{}}0 \\\left(N_{T}\right)_{1-3} \\0 \\0\end{array}\right].\end{aligned}\] Substitute eq. ([eq16.2m]) for the displacement vector into eq. ([eq16.2o]) to get \[\begin{aligned} \label{eq16.2p}\tag{p} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{4} \\Q_{5}\end{array}\right]=\left(\frac{E A}{L}\right)\left[\begin{array}{@{}cc@{}} -1 & 0 \\0 & -1 \\-a & -a \\-a & -a\end{array}\right]\left(\frac{L}{E A}\right)\left[\begin{array}{@{}c@{}}\frac{-1}{2\,+\,2 \sqrt{2}} \\[6pt]\frac{4\,+\,\sqrt{2}}{4\,+\,2 \sqrt{2}}\end{array}\right]\left(N_{T}\right)_{1-3}+\left[\begin{array}{@{}c@{}}0 \\\left(N_{T}\right)_{1-3} \\0 \\0\end{array}\right].\end{aligned}\] After matrix algebra the reactive joint forces are \[\begin{aligned} \label{eq16.2q}\tag{q} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{4} \\Q_{5}\end{array}\right]=\left[\begin{array}{@{}c@{}}\frac{1}{2\,+\,2 \sqrt{2}} \\[6pt]\frac{1}{2\,+\,2 \sqrt{2}} \\[6pt]\frac{1}{2}-\frac{1}{\sqrt{2}} \\[6pt]\frac{1}{2}-\frac{1}{\sqrt{2}}\end{array}\right]\left(N_{T}\right)_{1-3}.\end{aligned}\] A free body diagram of all the joint forces is shown in figure [fig16.5].

The condition for horizontal equilibrium is \(Q_{1}+Q_{5}=0\). Substitute the results for these reactive forces from eq. ([eq16.2q]) into condition for horizontal equilibrium to get \[\begin{aligned} \label{eq16.2r}\tag{r} Q_{1}+Q_{5}=\left(\frac{1}{2+2 \sqrt{2}}+\frac{1}{2}-\frac{1}{\sqrt{2}}\right)\left(N_{T}\right)_{1-3}.\end{aligned}\]

L132pt

Extract a common denominator in eq. ([eq16.2r]): \[\begin{aligned} \label{eq16.2s}\tag{s} Q_{1}+Q_{5}=\left(\frac{1}{2+2 \sqrt{2}}\right)\left[1+(1+\sqrt{2})-\frac{(2+2 \sqrt{2})}{\sqrt{2}}\right]\left(N_{T}\right)_{1-3}.\end{aligned}\] Combine terms in eq. ([eq16.2s]) to get the final result \[\begin{aligned} Q_{1}+Q_{5}&=\left(\frac{1}{2+2 \sqrt{2}}\right)[2+\sqrt{2}-(\sqrt{2}+2)]\left(N_{T}\right)_{1-3}=\left(\frac{1}{2+2 \sqrt{2}}\right)[0]\left(N_{T}\right)_{1-3} \nonumber\\ &=0. \label{eq16.2t}\tag{t}\end{aligned}\] Hence, the matrix solution for reactive forces \(Q_1\) and \(Q_5\) satisfy horizontal equilibrium. The condition for vertical equilibrium is \(Q_{2}+Q_{4}=0\). Substitute the results for these reactive forces from eq. ([eq16.2q]) into the condition for vertical equilibrium to get \[\begin{aligned} \label{eq16.2u}\tag{u} Q_{2}+Q_{4}=\left[\frac{1}{2+2 \sqrt{2}}+\frac{1}{2}-\frac{1}{\sqrt{2}}\right]\left(N_{T}\right)_{1-3}=0.\end{aligned}\] Note that the algebra in eq. ([eq16.2u]) is the same as the algebra detailed in eq. ([eq16.2r]) to eq. ([eq16.2t]). So the condition for vertical equilibrium is satisfied. Vanishing of the moment about joint 1 requires \(L Q_{4}-L Q_{5}=0\). Substitute the results for these reactive forces from eq. ([eq16.2q]) into the condition for moment equilibrium to get \[\begin{aligned} \label{eq16.2v}\tag{v} L Q_{4}-L Q_{5}=L\left[\frac{1}{2}-\frac{1}{\sqrt{2}}-\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)\right]\left(N_{T}\right)_{1-3}=0.\end{aligned}\] Hence, the matrix solution for the reactive forces plus the applied forces satisfies equilibrium of the free body diagram for the entire truss.

Solution to part (d).

The axial normal force in the bar between joints \(i\) and \(j\) from eq. ([eq16.14]) is \[\begin{aligned} \label{eq16.2w}\tag{w} N_{i-j}=\left[S_{i-j}\right]\{q\}_{i-j}-\left(N_{T}\right)_{i-j},\end{aligned}\] where the stress matrix ([eq16.15]) is \[\begin{aligned} \label{eq16.2x}\tag{x} \left[S_{i-j}\right] \equiv\left(\frac{E A}{L}\right)_{i-j}\left[\begin{array}{@{}l@{\;}l@{\;}l@{\;}l@{}}-c & -s & c & s\end{array}\right]_{i-j}.\end{aligned}\] The direction cosines for each bar are listed in table [tab16.1].

For bar 1-2, the axial normal force is \[\begin{aligned} \label{eq16.2y}\tag{y} N_{1-2}=\left(\frac{E A}{L}\right)\left[\begin{array}{@{}l@{\;}l@{\;}l@{\;}l@{}}-1 & 0 & 1 & 0\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right]-\left(N_{T}\right)_{1-2}=\left(\frac{E A}{L}\right)\left[\begin{array}{@{}l@{\;}l@{\;}l@{\;}l@{}}-1 & 0 & 1 & 0\end{array}\right]\left[\begin{array}{@{}c@{}}0 \\0 \\q_{3} \\0\end{array}\right]-0=\left(\frac{E A}{L}\right) q_{3}.\end{aligned}\] From eq. ([eq16.2m]) the solution for the displacement is \(q_{3}=\left(\frac{L}{E A}\right)\left(\frac{-\sqrt{2}}{4\,{+}\,2 \sqrt{2}}\right)\left(N_{T}\right)_{1-3}\), Substitute the result for \(q_3\) into eq. ([eq16.2y]) to find \[\begin{aligned} N_{1-2}=\left(\frac{E A}{L}\right)\left(\frac{L}{E A}\right)\left(\frac{-\sqrt{2}}{4+2 \sqrt{2}}\right)\left(N_{T}\right)_{1-3}=\left(\frac{-\sqrt{2}}{4+2 \sqrt{2}}\right)\left(N_{T}\right)_{1-3}.\end{aligned}\] For bar 1-3, the axial normal force is \[\begin{aligned} \label{eq16.2z}\tag{z} N_{1-3}=\left(\frac{E A}{L}\right)\left[\begin{array}{@{}l@{\;}l@{\;}l@{\;}l@{}}0 & -1 & 0 & 1\end{array}\right]\left[\begin{array}{@{}c@{}}0 \\0 \\0 \\q_{6}\end{array}\right]-\left(N_{T}\right)_{1-3}=\left(\frac{E A}{L}\right) q_{6}-\left(N_{T}\right)_{1-3}.\end{aligned}\] From eq. ([eq16.2m]) the solution for the displacement is \(q_{6}=\left(\frac{L}{E A}\right) \frac{4+\sqrt{2}}{4\,{+}\,2 \sqrt{2}}\left(N_{T}\right)_{1-3}\). Substitute the result for \(q_6\) into eq. ([eq16.2z]) to find \[\begin{aligned} \label{eq16.2aa}\tag{aa} N_{1-3}=\left(\frac{E A}{L}\right)\left[\left(\frac{L}{E A}\right) \frac{4+\sqrt{2}}{4+2 \sqrt{2}}\left(N_{T}\right)_{1-3}\right]-\left(N_{T}\right)_{1-3}=\left(\frac{-1}{2+2 \sqrt{2}}\right)\left(N_{T}\right)_{1-3}.\end{aligned}\] For bar 2-3, the axial normal force is \[\begin{aligned} \label{eq16.2ab}\tag{ab} N_{2-3}=\left(\frac{E A}{\sqrt{2} L}\right)\left[\begin{array}{@{}lll@{}}1 / \sqrt{2}-1 / \sqrt{2}-1 / \sqrt{2}\ 1\; /\; \sqrt{2}\end{array}\hspace*{-3pt}\right]\left[\begin{array}{@{}l@{}}q_{3} \\ q_{4} \\q_{5} \\q_{6}\end{array}\right]-\left(N_{T}\right)_{2-3}=\left(\frac{E A}{\sqrt{2} L}\right)\left[\begin{array}{@{}lll@{}}1 / \sqrt{2}-1 / \sqrt{2}-1 / \sqrt{2}\ 1 / \sqrt{2}\end{array}\hspace*{-3pt}\right]\left[\begin{array}{@{}c@{}}q_{3} \\0 \\0 \\q_{6}\end{array}\right]-0.\end{aligned}\] Expand eq. ([eq16.2ab]) to get \[\begin{aligned} N_{2-3}&=\left(\frac{E A}{2 L}\right)\left[\begin{array}{@{}l@{\;}l@{}}1 & 1\end{array}\right]\left[\begin{array}{@{}l@{}}q_{3} \\q_{6}\end{array}\right]-\left(N_{T}\right)_{2-3}=\frac{E A}{2 L}\left(q_{3}+q_{6}\right)-0 \nonumber\\ &=\frac{E A}{2 L}\left[\left(\frac{L}{E A}\right)\left(\frac{-\sqrt{2}}{4+2 \sqrt{2}}\right)\left(N_{T}\right)_{1-3}+\left(\frac{L}{E A}\right) \frac{4+\sqrt{2}}{4+2 \sqrt{2}}\left(N_{T}\right)_{1-3}\right] \nonumber\\ &=\left(\frac{1}{2}\right)\left[\frac{-\sqrt{2}}{4+2 \sqrt{2}}+\frac{4+\sqrt{2}}{4+2 \sqrt{2}}\right]\left(N_{T}\right)_{1-3}=\left(\frac{1}{2}\right)\left[\frac{4}{4+2 \sqrt{2}}\right]\left(N_{T}\right)_{1-3}.\label{eq16.2ac}\tag{ac}\end{aligned}\] The final result for the force in bar 2-3 is \[\begin{aligned} \label{eq16.2ad}\tag{ad} N_{2-3}=\frac{\left(N_{T}\right)_{1-3}}{2+\sqrt{2}}.\end{aligned}\] Note that for this statically indeterminate truss all three bar forces are proportional to the change in temperature of bar 1-3.

Solution to part (a).

The matrix equation to determine the unknown joint displacements is \[\begin{aligned} \label{eq16.3a}\tag{a} \left\{Q_{\alpha}\right\}=\left[K_{\alpha \alpha}\right]\left\{q_{\alpha}\right\}+\left[K_{\alpha \beta}\right]\left\{q_{\beta}\right\}+\left\{Q_{\alpha}^{0}\right\}.\end{aligned}\] Refer to the stiffness matrix in eq. ([eq16.3h]) and the fixed-end vector in eq. ([eq16.1i]) of example [ex16.1]. Then the matrices in eq. ([eq16.3a]) are \[\begin{gathered} \left\{Q_{\alpha}\right\}=\left[\begin{array}{@{}c@{}}Q_{1} \\Q_{3} \\Q_{6}\end{array}\right]=\left[\begin{array}{@{}l@{}}0 \\0 \\0\end{array}\right], \left[K_{\alpha \alpha}\right]=\frac{E A}{L}\left[\begin{array}{@{}ccc@{}}1 & -1 & 0 \\-1 & 1+a & a \\0 & a & 1+a\end{array}\right], \left\{q_{\alpha}\right\}=\left[\begin{array}{@{}l@{}}q_{1} \\q_{3} \\q_{6}\end{array}\right], \left[K_{\alpha \beta}\right]=\frac{E A}{L}\left[\begin{array}{@{}ccc@{}}0 & 0 & 0 \\0 & -a & -a \\-1 & -a & -a\end{array}\right],\label{eq16.3b}\tag{b}\\ \left\{q_{\beta}\right\}=\left[\begin{array}{@{}l@{}}q_{2} \\q_{4} \\q_{5}\end{array}\right]=\left[\begin{array}{@{}l@{}}0 \\0 \\0\end{array}\right]\mbox{, and }\left\{Q_{\alpha}^{0}\right\}=\left[\begin{array}{r}Q_{1}^{0} \\[6pt] Q_{3}^{0} \\[6pt] Q_{6}^{0}\end{array}\right]=\left[\begin{array}{@{}c@{}}0 \\0 \\-\left(N_{T}\right)_{1-3}\end{array}\right].\label{eq16.3c}\tag{c}\end{gathered}\] The solution to eq. ([eq16.3a]) for the joint displacements is \[\begin{aligned} \label{eq16.3d}\tag{d} q_{1}=-L(\alpha \Delta T) \quad q_{3}=-L(\alpha \Delta T) \quad q_{6}=L(\alpha \Delta T).\end{aligned}\]

Solution to part (b).

The matrix equation to determine the unknown joint forces is \[\begin{aligned} \label{eq16.3e}\tag{e} \left\{Q_{\beta}\right\}=\left[K_{\beta \alpha}\right]\left\{q_{\alpha}\right\}+\left[K_{\beta \beta}\right]\left\{q_{\beta}\right\}+\left\{Q_{\beta}^{0}\right\}.\end{aligned}\]

The matrices in eq. ([eq16.3e]) are \[\begin{aligned} \label{eq16.3f}\tag{f} \left\{Q_{\beta}\right\}=\left[\begin{array}{@{}c@{}}Q_{2} \\Q_{4} \\Q_{5}\end{array}\right], \left[K_{\beta \alpha}\right]=\frac{E A}{L}\left[\begin{array}{@{}ccc@{}}0 & 0 & -1 \\0 & -a & -a \\0 & -a & -a\end{array}\right], \left[K_{\beta \beta}\right]=\frac{E A}{L}\left[\begin{array}{@{}lll@{}}1 & 0 & 0 \\0 & a & a \\0 & a & a\end{array}\right]\mbox{, and }\left\{Q_{\beta}^{0}\right\}=\left[\begin{array}{@{}l@{}}Q_{2}^{0} \\[6pt] Q_{4}^{0} \\[6pt] Q_{5}^{0}\end{array}\right]=\left[\begin{array}{@{}c@{}}\left(N_{T}\right)_{1-3} \\0 \\0\end{array}\right].\end{aligned}\] The solution of eq. ([eq16.3e]) for the unknown joint forces, or the reactive forces, is \[\begin{aligned} \label{eq16.3g}\tag{g} Q_{2}=Q_{4}=Q_{5}=0.\end{aligned}\] There no external forces acting on the truss since both the applied and reactive forces are zero. Consequently, it is reasonable to surmise that the internal forces in the bars vanish. That is, \[\begin{aligned} \label{eq16.3h}\tag{h} N_{1-2}=N_{1-3}=N_{2-3}=0,\end{aligned}\] which can be verified using eq. ([eq16.14]).

Solution to part (c).

The elongation of a truss is determined from eq. ([eq16.4]). Using the direction cosines listed in table [tab16.1], the elongation of each bar is given by \[\begin{aligned} \label{eq16.3i}\tag{i} \begin{split} \Delta_{1-2}&=(1)\left(q_{3}-q_{1}\right)+(0)\left(q_{4}-q_{2}\right)=0 \\ \Delta_{1-3}&=(0)\left(q_{5}-q_{1}\right)+(1)\left(q_{6}-q_{2}\right)=L(\alpha \Delta T) \\ \Delta_{2-3}&=\left(\frac{-1}{\sqrt{2}}\right)\left(q_{5}-q_{3}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(q_{6}-q_{4}\right)=\left(\frac{1}{\sqrt{2}}\right)\left(q_{6}+q_{3}\right)=0 \end{split}.\end{aligned}\] Hence, bars 1-2 and 2-3 do not change in length, and the new length of bar 1-3 is \(L(1+\alpha \Delta T)\). Assuming \(\alpha \Delta T>0\), the displaced truss is shown with respect to initial configuration in figure [fig16.7].

[ex16.4]The five-bar truss shown in figure [fig16.8] is restrained against rigid body motion, since joints 1 and 4 are fixed. pins All bars have the same extensional stiffness \(E A\). Determine the restrained structural stiffness matrix \(\left[K_{\alpha \alpha}\right]\).

Solution.

The dimensions of the restrained structural stiffness matrix is 4X4 in displacement degrees of freedom \(q_{3}, q_{4}, q_{5}\), and \(q_{6}\). The direction cosines for the truss bars are listed in table [tab16.3].

From eq. ([eq16.12]) the following member stiffness matrices are constructed using the direction cosines in table [tab16.3]. Only elements contributing to rows and columns 3, 4, 5, and 6 of the restrained structural stiffness matrix are extracted from the individual element stiffness matrices. These member stiffness matrices follow: \[\begin{gathered} \label{eq16.4a}\tag{a} \left[K_{\alpha \alpha}\right]_{1-2}=\!\begin{array}{@{}c@{}} \begin{array}{@{}ccc@{}}\phantom{0.}q_{3}& &q_{4}\phantom{0} \end{array}\\[6pt]\left[\begin{array}{@{}cr@{}}E A / L & 0 \\0 & 0\end{array}\right]\end{array} \quad\left[K_{\alpha \alpha}\right]_{1-3}=\frac{E A}{L /(\cos \alpha)}\begin{array}{@{}c@{}} {\arraycolsep=5pt\begin{array}{@{}cc@{}}q_{5}\phantom{0000} & \phantom{0000}q_{6} \end{array}}\\[6pt]\left[\begin{array}{@{}cc@{}}\cos ^{2} \alpha & \cos \alpha \sin \alpha \\\cos \alpha \sin \alpha & \sin ^{2} \alpha\end{array}\right]\end{array}=\frac{E A}{L}\begin{array}{@{}c@{}}{\arraycolsep=5pt\begin{array}{@{}cc@{}} q_{5}\phantom{0000} & \phantom{0000}q_{6} \end{array}}\\[6pt]\left[\begin{array}{@{}ll@{}}\cos ^{3} \alpha & \cos ^{2} \alpha \sin \alpha \\\cos ^{2} \alpha \sin \alpha & \cos ^{2} \sin ^{2} \alpha\end{array}\right]\end{array}\\ \left[K_{\alpha \alpha}\right]_{2-3}=\left(\frac{E A}{L \tan \alpha}\right)\left[\begin{array}{@{}cccc@{}}0 & 0 & 0 & 0 \\0 & 1 & 0 & -1 \\0 & 0 & 0 & 0 \\0 & -1 & 0 & 1\end{array}\right]=\left(\frac{E A}{L}\right)\begin{array}{@{}c@{}}{\arraycolsep=7.5pt\begin{array}{@{}cccc@{}}q_{3}\phantom{.} & q_{4}\phantom{0} & q_{5}\phantom{0} & q_{6}\phantom{00} \end{array}}\\[6pt]\left[\begin{array}{@{}cccc@{}}0 & 0 & 0 & 0 \\0 & \cot \alpha & 0 & -\cot \alpha \\0 & 0 & 0 & 0 \\0 & -\cot \alpha & 0 & \cot \alpha\end{array}\right]\end{array}\quad \left[K_{\alpha \alpha}\right]_{3-4}=\begin{array}{@{}c@{}}\begin{array}{@{}cc@{}}\phantom{0}q_{5} & \phantom{00}q_{6} \end{array}\\[6pt]\left[\begin{array}{@{}cc@{}}E A / L & 0 \\0 & 0\end{array}\right]\end{array}.\label{eq16.4b}\tag{b}\\ \left[K_{\alpha \alpha}\right]_{2-4}=\left(\frac{E A}{L /(\cos \alpha)}\right)\left[\begin{array}{@{}cc@{}}\cos ^{2} \alpha & -\cos \alpha \sin \alpha \\-\cos \alpha \sin \alpha & \sin ^{2} \alpha\end{array}\right]=\frac{E A}{L}\begin{array}{@{}c@{}}\begin{array}{@{}cc@{}}q_{3}\phantom{0000} & \phantom{0000}q_{4} \end{array}\\[6pt]\left[\begin{array}{@{}cc@{}}\cos ^{3} \alpha & -\cos ^{2} \alpha \sin \alpha \\-\cos ^{2} \alpha \sin \alpha & \cos \alpha \sin ^{2} \alpha\end{array}\right]\end{array}.\label{eq16.4c}\tag{c}\end{gathered}\] Assemblage of the restrained structural stiffness matrix is accomplished by adding like row and column elements from the stiffness matrices of each truss bar. The result for the restrained structural stiffness matrix is \[\begin{aligned} \label{eq16.4d}\tag{d} \left[K_{\alpha \alpha}\right]=\left(\frac{E A}{L}\right)\begin{array}{@{}c@{}}{\begin{array}{@{\hspace*{-6pt}}c@{\qquad\qquad\qquad}c@{\qquad\qquad\qquad}c@{\qquad\qquad\qquad\;\;}c@{}} \hspace*{-0.8pc}q_{3}\hspace*{6pt} &\hspace*{6pt} q_{4}&\hspace*{9pt} \!\!q_{5}\hspace*{-6pt} & \phantom{00}q_{6} \end{array}}\\[6pt]\left[\begin{array}{@{}cccc@{}}\left(1+\cos ^{3} \alpha\right) & -\cos ^{2} \alpha \sin \alpha & 0 & 0 \\-\cos ^{2} \alpha \sin \alpha & \left(\cot \alpha+\cos \alpha \sin ^{2} \alpha\right) & 0 & -\cot \alpha \\0 & 0 & \left(\cos ^{3} \alpha+1\right) & \cos ^{2} \alpha \sin \alpha \\0 & -\cot \alpha & \cos ^{2} \alpha \sin \alpha & \left(\cos \alpha \sin ^{2} \alpha+\cot \alpha\right)\end{array}\right]\end{array}.\end{aligned}\] Note that the matrix is symmetric and the sum of the column elements do not add to zero. If we take \(\alpha=30^{\circ}\), then the restrained structural stiffness matrix reduces to \[\begin{aligned} \label{eq16.4e}\tag{e} \left[K_{\alpha \alpha}\right]=\left(\frac{E A}{L}\right)\begin{array}{@{}c@{}}\begin{array}{@{}c@{\qquad\qquad}c@{\qquad\qquad\quad}c@{\qquad\qquad}c@{}}q_{3} & q_{4} & q_{5} & q_{6} \end{array}\\[3pt]\left[\begin{array}{@{}cccc@{}}1+(3 \sqrt{3}) / 8 & -3 / 8 & 0 & 0 \\-3 / 8 & \sqrt{3}(9 / 8) & 0 & -\sqrt{3} \\0 & 0 & 1+(3 \sqrt{3}) / 8 & 3 / 8 \\0 & -\sqrt{3} & 3 / 8 & \sqrt{3}(9 / 8)\end{array}\right]\end{array}. \\[-1.4pc] \nonumber {\hfill\qed\hspace*{-7.4pc}}\end{aligned}\]

[ex16.5]Consider the five-bar truss problem of example [ex16.4] with \(\alpha=30^{\circ}\) that is subject to prescribed nodal forces \(Q_{3}, Q_{4}, Q_{5}, \text { and } Q_{6}\). Use symmetry to reduce the problem size to solve for the unknown joint displacements.

Solution.

We note that the structure and boundary conditions are symmetric about a horizontal axis through the center of the truss. The joint displacements and corresponding forces can be decomposed into a symmetric and antisymmetric sets about this horizontal axis of symmetry as shown in figure [fig16.10]. The joint displacements and the corresponding forces are related to the symmetric and antisymmetric counterparts by \[\begin{aligned} \label{eq16.5a}\tag{a} \left[\begin{array}{@{}l@{}}q_{3} \\q_{4} \\q_{5} \\q_{6}\end{array}\right]=\left[\begin{array}{@{}c@{}}x_{a} \\-y_{a} \\x_{a} \\y_{a}\end{array}\right]+\left[\begin{array}{@{}c@{}}-x_{b} \\y_{b} \\x_{b} \\y_{b}\end{array}\right]=\left[\begin{array}{@{}cccc@{}}1 & 0 & -1 & 0 \\0 & -1 & 0 & 1 \\1 & 0 & 1 & 0 \\0 & 1 & 0 & 1\end{array}\right]\left[\begin{array}{@{}l@{}}x_{a} \\y_{a} \\x_{b} \\y_{b}\end{array}\right]\mbox{, and }\left[\begin{array}{@{}c@{}}Q_{3} \\Q_{4} \\Q_{5} \\Q_{6}\end{array}\right]=\left[\begin{array}{@{}c@{}}X_{a} \\-Y_{a} \\X_{a} \\Y_{a}\end{array}\right]+\left[\begin{array}{@{}c@{}}-X_{b} \\Y_{b} \\X_{b} \\Y_{b}\end{array}\right]=\left[\begin{array}{@{}cccc@{}}1 & 0 & -1 & 0 \\0 & -1 & 0 & 1 \\1 & 0 & 1 & 0 \\0 & 1 & 0 & 1\end{array}\right]\left[\begin{array}{@{}c@{}}X_{a} \\Y_{a} \\X_{b} \\Y_{b}\end{array}\right].\end{aligned}\]

The expressions in eq. ([eq16.5a]) are written in compact form as \[\begin{aligned} \label{eq16.5b}\tag{b} \left\{q_{\alpha}\right\}=[A]\{x\}\mbox{, and }\left\{Q_{\alpha}\right\}=[A]\{X\},\end{aligned}\]

where the elements of the 4X4 matrix \([A]\) are either –1, 0, or 1.The force vector is related to the displacement vector by \(\left\{Q_{\alpha}\right\}=\left[K_{\alpha \alpha}\right]\left\{q_{\alpha}\right\}\), where matrix \(\left[K_{\alpha \alpha}\right]\) is given by eq. ([eq16.5e]) in example [ex16.4]. Substitute eq. ([eq16.5b]) into the matrix equation relating the force vector to the displacement vector to get \[\begin{aligned} \label{eq16.5c}\tag{c} [A]\{X\}=\left[K_{\alpha \alpha}\right][A]\{x\}.\end{aligned}\] Pre-multiply eq. ([eq16.5c]) by the inverse of matrix \([A]\) to find \[\begin{aligned} \label{eq16.5d}\tag{d} \{X\}=[A]^{-1}\left[K_{\alpha \alpha}\right][A]\{x\}.\end{aligned}\] Define stiffness matrix by \(\left[\bar{K}_{\alpha \alpha}\right]=[A]^{-1}\left[K_{\alpha \alpha}\right][A]\). The the matrices to compute \(\left[\bar{K}_{\alpha \alpha}\right]\) are \[\begin{aligned} \label{eq16.5e}\tag{e} \left[\bar{K}_{\alpha \alpha}\right]=\frac{1}{2}\left[\begin{array}{@{}cccc@{}}1 & 0 & 1 & 0 \\0 & -1 & 0 & 1 \\-1 & 0 & 1 & 0 \\0 & 1 & 0 & 1\end{array}\right]\left(\frac{E A}{L}\right)\left[\begin{array}{ccccc}1+(3 \sqrt{3}) / 8 & -3 / 8 & 0 & 0 \\-3 / 8 & \sqrt{3}(9 / 8) & 0 & -\sqrt{3} \\0 & 0 & 1+(3 \sqrt{3}) / 8 & 3 / 8\\0 & -\sqrt{3} &3 / 8 & \sqrt{3}(9 / 8) \end{array}\right] \left[\begin{array}{@{}cccc@{}}1 & 0 & -1 & 0 \\0 & -1 & 0 & 1 \\1 & 0 & 1 & 0 \\0 & 1 & 0 & 1\end{array}\right].\end{aligned}\] The result of the matrix multiplications in eq. ([eq16.5e]) is \[\begin{aligned} \label{eq16.5f}\tag{f} \left[\bar{K}_{\alpha \alpha}\right]=\frac{E A}{L}\left|\begin{array}{@{}cccc@{}}1+(3 \sqrt{3}) / 8 & 3 / 8 & 0 & 0 \\3 / 8 & (17 \sqrt{3}) / 8 & 0 & 0 \\0 & 0 & 1+(3 \sqrt{3}) / 8 & 3 / 8 \\0 & 0 & 3 / 8 & (\sqrt{3}) / 8\end{array}\right|=\left[\begin{array}{@{}c@{}}{\left[\bar{K}_{a}\right]\left[\begin{array}{@{}c@{}}0_{2 X 2}\end{array}\right]} \\[6pt]\left.\left[0_{2 X 2}\right]\left[\bar{K}_{b}\right]\right]\end{array}\right].\end{aligned}\] Note that the partitioned form of \(\left[\bar{K}_{\alpha \alpha}\right]\) is diagonal, and the 2X2 sub-matrices on the diagonal are \[\begin{aligned} \label{eq16.5g}\tag{g} \left[\bar{K}_{a}\right]=\frac{E A}{L}\left[\begin{array}{@{}cc@{}}1+\frac{3 \sqrt{3}}{8} & \frac{3}{8} \\[6pt]\frac{3}{8} & \frac{17 \sqrt{3}}{8}\end{array}\right]\mbox{, and }\left[\bar{K}_{b}\right]=\frac{E A}{L}\left[\begin{array}{@{}cc@{}}1+\frac{3 \sqrt{3}}{8} & \frac{3}{8} \\[6pt]\frac{3}{8} & \frac{\sqrt{3}}{8}\end{array}\right].\end{aligned}\] The inverses of the matrices in eq. ([eq16.5g]) are \[\begin{aligned} \label{eq16.5h}\tag{h} \left[\bar{K}_{a}\right]^{-1}=\frac{L}{E A} \left[\begin{array}{@{}l@{}}\frac{17}{181}(17-6 \sqrt{3}) \frac{1}{181}(18-17 \sqrt{3}) \\[6pt]\frac{1}{181}(18-17 \sqrt{3}) \frac{1}{543}(9+82 \sqrt{3})\end{array}\right]\mbox{, and }\left[\bar{K}_{b}\right]^{-1}=\frac{L}{E A}\left[\begin{array}{@{}cc@{}}1 & -\sqrt{3} \\-\sqrt{3} & (3+8 / \sqrt{3})\end{array}\right].\end{aligned}\] Then the inverse of eq. ([eq16.5f]) is given by \[\begin{aligned} \label{eq16.5i}\tag{i} \left[\bar{K}_{\alpha \alpha}\right]^{-1}=\left[\begin{array}{@{}l@{}}{\left[\bar{K}_{a}\right]^{-1}\left[\begin{array}{@{}l@{}}0_{2 X 2}\end{array}\right]} \\[6pt]{\left[0_{2 X 2}\right]\left[\bar{K}_{b}\right]^{-1}}\end{array}\right].\end{aligned}\] Hence, the solution for the displacement vector \(\{x\}\) in terms of the force vector \(\{X\}\) is \[\begin{aligned} \label{eq16.5j}\tag{j} \{x\}=\left[\bar{K}_{\alpha \alpha}\right]^{-1}\{X\}.\end{aligned}\] From eq. ([eq16.5b]) \(\{x\}=[A]^{-1}\left\{q_{\alpha}\right\}\mbox{ and }\{X\}=[A]^{-1}\left\{Q_{\alpha}\right\}\) Substitute the latter relations into eq. ([eq16.5j]) to get \[\begin{aligned} \label{eq16.5k}\tag{k} [A]^{-1}\left\{q_{\alpha}\right\}=\left[\bar{K}_{\alpha \alpha}\right]^{-1}[A]^{-1}\left\{Q_{\alpha}\right\}.\end{aligned}\] Pre-multiply eq. ([eq16.5k]) by matrix \([A]\) to write the result for the unknown displacements as \[\begin{aligned} \label{eq16.5l}\tag{l} \left\{q_{\alpha}\right\}=\left[C_{\alpha \alpha}\right]\left\{Q_{\alpha}\right\},\end{aligned}\] where the compliance matrix is \[\begin{aligned} \label{eq16.5m}\tag{m} \left[C_{\alpha \alpha}\right]=[A]\left[\bar{K}_{\alpha \alpha}\right]^{-1}[A]^{-1}=\frac{L}{E A}\left[\begin{array}{@{}cccc@{}}0.810306 & 0.897641 & -0.189694 & 0.83441 \\0.897641 & 3.94847 & -0.83441 & 3.67033 \\-0.189694 & -0.83441 & 0.810306 & -0.897641 \\0.83441 & 3.67033 & -0.897641 & 3.94847\end{array}\right].\end{aligned}\] The compliance matrix in eq. ([eq16.5m]) was obtained by inverting two 2X2 sub-matrices, rather than directly inverting the 4X4 stiffness matrix \(\left[K_{\alpha \alpha}\right]\). Exploiting the symmetry conditions as illustrated in figure [fig16.10], reduces the number of computations to find the inverse of matrix \(\left[K_{\alpha \alpha}\right]\).

Solution for the unknown joint displacements.

The joints are taken at the support locations and are numbered one to five from left to right. Hence, there are ten degrees of freedom (DOFs) as is shown in the top sketch in figure 1.1. The support conditions mean the vertical displacements vanish; i.e., \[\begin{aligned} \label{eq16.6a}\tag{a} \left\{q_{\beta}\right\}=\left[\begin{array}{@{}l@{\;}l@{\;}l@{\;}l@{\;}l@{}}q_{1} & q_{3} & q_{5} & q_{7} & q_{9}\end{array}\right]^{T}=0_{5X1}.\end{aligned}\] The geometry, boundary conditions, and material properties of the structure are symmetric about the vertical centerline. If the top sketch of the beam and its DOFs are rotated \(180^{\circ}\) about this vertical centerline, the bottom sketch is obtained. See figure 1.1. The displacements and rotations at the joints in the top and bottom sketch must be the same. Hence, symmetry implies the joint rotations must satisfy

\[\begin{aligned} \label{eq16.6b}\tag{b} q_{10}=-q_{2} \quad q_{8}=-q_{4} \quad q_{6}=-q_{6}.\end{aligned}\] Clearly, the last symmetry condition on the rotations means rotation of the center joint vanishes; \(q_{6}=0\). Then, the analysis for the response of the beam reduces to a two-span beam, clamped at its right end as is shown in figure [fig16.14]. The two active degrees of freedom are rotations \(q_{2}\) and \(q_{4}\). The stiffness matrices ([eq16.49]) for beam members 1-2 and 2-3 are

\[\begin{aligned} \label{eq16.6c}\tag{c} \left[K_{1-2}\right]=E I \begin{array}{@{}c@{}} {\arraycolsep=13pt\begin{array}{@{}cccc@{}} q_1 & q_2 & q_3 & q_4 \end{array}}\\[6pt] \left[\begin{array}{@{}cccc@{}}12 / L^{3} & -6 / L^{2} & -12 / L^{3} & -6 / L^{2} \\-6 / L^{2} & 4 / L & 6 / L^{2} & 2 / L \\-12 / L^{3} & 6 / L^{2} & 12 / L^{3} & 6 / L^{2} \\ -6 / L^{2} & 2 / L & 6 / L^{2} & 4 / L\end{array}\right] \end{array} \left[K_{2-3}\right]=E I \begin{array}{@{}c@{}} {\arraycolsep=13pt\begin{array}{@{}cccc@{}} q_3 & q_4 & q_5 & q_6 \end{array}}\\[6pt] \left[\begin{array}{@{}cccc@{}}12 / L^{3} & -6 / L^{2} & -12 / L^{3} & -6 / L^{2} \\-6 / L^{2} & 4 / L & 6 / L^{2} & 2 / L \\-12 / L^{3} & 6 / L^{2} & 12 / L^{3} & 6 / L^{2} \\-6 / L^{2} & 2 / L & 6 / L^{2} & 4 / L\end{array}\right] \end{array}.\end{aligned}\] The 6X6 unrestrained structural stiffness matrix is the sum \(\left[K_{1-2}\right]+\left[K_{2-3}\right]\) with each member stiffness matrix expanded to 6X6 by adding two rows and two columns of zeros for degrees of freedom not contained in the member. The result is \[\begin{aligned} \label{eq16.6d}\tag{d} [K]=E I \begin{array}{@{}c@{}} {\arraycolsep=15pt\begin{array}{@{}cccccc@{}} q_1 & q_2 & q_3 & q_4 & q_5 & q_6 \end{array}}\\[6pt] \left[\begin{array}{@{}cccccc@{}}12 / L^{3} & -6 / L^{2} & -12 / L^{3} & -6 / L^{2} & 0 & 0 \\-6 / L^{2} & 4 / L & 6 / L^{2} & 2 / L & 0 & 0 \\-12 / L^{3} & 6 / L^{2} & 24 / L^{3} & 0 & -12 / L^{3} & -6 / L^{2} \\-6 / L^{2} & 2 / L & 0 & 8 / L & 6 / L^{2} & 2 / L \\0 & 0 & -12 / L^{3} & 6 / L^{2} & 12 / L^{3} & 6 / L^{2} \\0 & 0 & -6 / L^{2} & 2 / L & 6 / L^{2} & 4 / L\end{array}\right] \end{array}.\end{aligned}\] Partition the unrestrained structural stiffness matrix in terms of unknowns and knowns to get \[\begin{aligned} \label{eq16.6e}\tag{e} [K]=E I \begin{array}{@{}c@{}} {\arraycolsep=15pt\begin{array}{@{}cccccc@{}} q_2 & q_4 & q_1 & q_3 & q_5 & q_6 \end{array}}\\[6pt] \left[\begin{array}{cc:cccc}4 / L & 2 / L & -6 / L^{2} & 6 / L^{2} & 0 & 0 \\2 / L & 8 / L & -6 / L^{2} & 0 & 6 / L^{2} & 2 / L \\\hdashline-6 / L^{2} & -6 / L^{2} & 12 / L^{3} & -12 / L^{3} & 0 & 0 \\6 / L^{3} & 0 & -12 / L^{3} & 24 / L^{3} & -12 / L^{3} & -6 / L^{2} \\0 & 6 / L^{2} & 0 & -12 / L^{3} & 12 / L^{3} & 6 / L^{2} \\0 & 2 / L & 0 & -6 / L^{2} & 6 / L^{2} & 4 / L\end{array}\right] \end{array} =\left[\begin{array}{@{}l@{}}{\left[K_{\alpha \alpha}\right]\left[K_{\alpha \beta}\right]} \\{\left[K_{\beta \alpha}\right]\left[K_{\beta \beta}\right]}\end{array}\right].\end{aligned}\] The restrained structural stiffness matrix is \[\begin{aligned} \label{eq16.6f}\tag{f} \left[K_{\alpha \alpha}\right]=E I \begin{array}{@{}c@{}} {\arraycolsep=12pt\begin{array}{@{}cc@{}}q_{2} &q_{4}\end{array}} \\[3pt]\left[\begin{array}{@{}cc@{}}4 / L & 2 / L \\2 / L & 8 / L\end{array}\right]\end{array}.\end{aligned}\] The unknown rotations are determined from \[\begin{aligned} \label{eq16.6g}\tag{g} \left[\begin{array}{@{}c@{}}Q_{2} \\Q_{4}\end{array}\right]=\left[\begin{array}{@{}c@{}}M_{a} \\0\end{array}\right]=E I\left[\begin{array}{@{}ll@{}}4 / L & 2 / L \\2 / L & 8 / L\end{array}\right]\left[\begin{array}{@{}l@{}}q_{2} \\q_{4}\end{array}\right].\end{aligned}\] Solve eq. ([eq16.6g]) for the nodal rotations to find \[\begin{aligned} \label{eq16.6h}\tag{h} \left[\begin{array}{@{}l@{}}q_{2} \\q_{4}\end{array}\right]=\frac{L}{E I} \frac{1}{(32-4)}\left[\begin{array}{@{}cc@{}}8 & -2 \\-2 & 4\end{array}\right]\left[\begin{array}{@{}c@{}}M_{a} \\0\end{array}\right]=\frac{M_{a} L}{14 E I}\left[\begin{array}{@{}c@{}}4 \\-1\end{array}\right].\end{aligned}\] By symmetry the joint rotations for the entire structure are \[\begin{aligned} \label{eq16.6i}\tag{i} \left[\begin{array}{@{}c@{}}q_{2} \\q_{4} \\q_{6} \\q_{8} \\q_{10}\end{array}\right]=\frac{M_{a} L}{14 E I}\left[\begin{array}{@{}c@{}}4 \\-1 \\0 \\1 \\-4\end{array}\right].\end{aligned}\]

Solution for the shear force and bending moment distributions.

The shear force and bending moment distribution in beam members 1-2 and 2-3 are determined from eq. ([eq16.44]). For member 1-2, we have \[\begin{aligned} \label{eq16.6j}\tag{j} \left[\begin{array}{@{}l@{}}V_{1}(z) \\M_{1}(z)\end{array}\right]_{1-2}=E I\left[\begin{array}{@{}cccc@{}}-\frac{12}{L^{3}} & \frac{6}{L^{2}} & \frac{12}{L^{3}} & \frac{6}{L^{2}} \\\left(\frac{6}{L^{2}}-\frac{12 z}{L^{3}}\right) & \left(-\frac{4}{L}+\frac{6 z}{L^{2}}\right) & \left(-\frac{6}{L^{2}}+\frac{12 z}{L^{3}}\right) & \left(-\frac{2}{L}+\frac{6 z}{L^{2}}\right)\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right].\end{aligned}\] But \(q_{1}=q_{3}=0\), so \[\begin{aligned} \label{eq16.6k}\tag{k} \left[\begin{array}{@{}l@{}}V_{1}(z) \\M_{1}(z)\end{array}\right]_{1-2}=E I\left[\begin{array}{@{}cc@{}}\frac{6}{L^{2}} & \frac{6}{L^{2}} \\[6pt]\left(-\frac{4}{L}+\frac{6 z}{L^{2}}\right)&\left(-\frac{2}{L}+\frac{6 z}{L^{2}}\right)\end{array}\right]\left[\begin{array}{@{}l@{}}q_{2} \\q_{4}\end{array}\right].\end{aligned}\] Substitute the solution for the rotations from eq. ([eq16.6h]) into eq. ([eq16.6k]) to get \[\begin{aligned} \label{eq16.6l}\tag{l} \left[\begin{array}{@{}l@{}}V_{1}(z) \\M_{1}(z)\end{array}\right]_{1-2}=E I\left[\begin{array}{@{}cc@{}}\frac{6}{L^{2}} & \frac{6}{L^{2}} \\\left(-\frac{4}{L}+\frac{6 z}{L^{2}}\right)&\left(-\frac{2}{L}+\frac{6 z}{L^{2}}\right)\end{array}\right] \frac{M_{a} L}{14 E I}\left[\begin{array}{@{}c@{}}4 \\-1\end{array}\right]=\frac{M_{a} L}{14}\left[\begin{array}{@{}c@{}}\frac{18}{L^{2}} \\-\frac{14}{L}+18 \frac{z}{L^{2}}\end{array}\right]=\left[\begin{array}{@{}c@{}}\frac{9 M_{a}}{7 L} \\-M_{a}+\frac{9 M_{a}}{7 L }z\end{array}\right].\end{aligned}\] Note that the coordinate z is local to the member in the formulas for the shear force and bending moment. For member 2-3, we have

\[\begin{aligned} \label{eq16.6m}\tag{m} \left[\begin{array}{@{}l@{}}V_{1}(z) \\M_{1}(z)\end{array}\right]_{2-3}=E I\left[\begin{array}{@{}cccc@{}}-\frac{12}{L^{3}} & \frac{6}{L^{2}} & \frac{12}{L^{3}} & \frac{6}{L^{2}} \\[6pt]\frac{6}{L^{2}}-\frac{12 z}{L^{3}} & -\frac{4}{L}+\frac{6 z}{L^{2}} & -\frac{6}{L^{2}}+\frac{12 z}{L^{3}} & -\frac{2}{L}+\frac{6 z}{L^{2}}\end{array}\right]\left[\begin{array}{@{}l@{}}q_{3} \\q_{4} \\q_{5} \\q_{6}\end{array}\right].\end{aligned}\] But \(q_{3}=q_{5}=q_{6}=0\), so \[\begin{aligned} \label{eq16.6n}\tag{n} \left[\begin{array}{@{}l@{}}V_{1}(z) \\M_{1}(z)\end{array}\right]_{2-3}=E I\left[\begin{array}{@{}c@{}}\frac{6}{L^{2}} \\[6pt]-\frac{4}{L}+\frac{6 z}{L^{2}}\end{array}\right] q_{4}.\end{aligned}\] Substitute the solution for rotation \(q_{4}\) from eq. ([eq16.6h]) into eq. ([eq16.6n]) to get \[\begin{aligned} \label{eq16.6o}\tag{o} \left[\begin{array}{@{}l@{}}V_{1}(z) \\M_{1}(z)\end{array}\right]_{2-3}=E I\left[\begin{array}{@{}c@{}}\frac{6}{L^{2}} \\[6pt]-\frac{4}{L}+\frac{6 z}{L^{2}}\end{array}\right]\left(-\frac{M_{a} L}{14 E I}\right)=\left[\begin{array}{@{}c@{}}-\frac{3 M_{a}}{7 L} \\[6pt]\frac{2 M_{a}}{7}-\frac{3 M_{a}}{7 L} z\end{array}\right].\end{aligned}\] Again, note that the \(z\)-coordinate in these formulas for the shear force and bending moment in member 2-3 is local to the member and runs from zero to L. However, the beginning joint 2 corresponds to the global longitudinal coordinate L, and end joint 3 corresponds to the global longitudinal coordinate 2L. The relationship between the member local coordinate and the global structural coordinate has to be taken into account when drawing the shear force and bending moment diagrams. The shear force and bending moment diagrams are shown in figure [fig16.15]. In this example, the shear force diagram in antisymmetric, and the moment diagram is symmetric, about the center of the multispan beam.

Solution for the support reactions.

The support reactions are \(\left[Q_{\beta}\right]=\left[K_{\beta \alpha}\right]\left\{q_{\alpha}\right\}\), since \(\left\{q_{\beta}\right\}=0_{4 X 1}\). Hence, \[\begin{aligned} \label{eq16.6p}\tag{p} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{3} \\Q_{5} \\O_{6}\end{array}\right]=E I\left[\begin{array}{@{}cc@{}}-6 / L^{2} & -6 / L^{2} \\6 / L^{2} & 0 \\0 & 6 / L^{2} \\0 & 2 / L\end{array}\right]\left[\begin{array}{@{}l@{}}q_{2} \\q_{4}\end{array}\right]=E I\left[\begin{array}{@{}cc@{}}-6 / L^{2} & -6 / L^{2} \\6 / L^{2} & 0 \\0 & 6 / L^{2} \\0 & 2 / L\end{array}\right] \frac{M_{a} L}{14 E I}\left[\begin{array}{@{}c@{}}4 \\-1\end{array}\right]=M_{a}\left[\begin{array}{@{}c@{}}-9 /(7 L) \\12 /(7 L) \\-3 /(7 L) \\-1 / 7\end{array}\right].\end{aligned}\] Note that these are the support reactions for the left half of the beam. By symmetry the support reactions on the right half of the beam are obtained by a rotation of the left half by \(180^{\circ}\) as shown in figure 1.2.

Joining the left half and right half we get the support reactions for the overall free body diagram of the multispan beam as shown in figure 1.3.

[ex16.7]The beam structure shown in figure [fig16.18](a) has a step change in thickness at midspan, and is clamped at each end. The left half of has a uniform flexural stiffness \(2 E I\), and the right half has a uniform flexural stiffness \(E I\). Each half has a length denoted by \(a\). A vertical linear elastic spring of stiffness \(k=6 E I / a^{3}\) is connected at midspan. The structure is subject to a vertical distributed load and a vertical point force \(P\) applied at midspan. The distributed load is uniform on the left half with intensity \(f_{y 0}\), and decreases linearly from \(f_{y 0}\) to zero on the right half. Model the response of the beam with two beam members, one in each half, and a spring member. Determine

1. The restrained structural stiffness matrix.

2. The fixed-end action vector.

3. The unknown joint displacements.

4. The support reactions.

5. The shear force and bending moment in the left half of the beam.

Solution to part (a).

The unrestrained structure has four joints and seven degrees of freedom as shown in figure [fig16.18](b). The size of the unrestrained structural stiffness matrix is 7X7. The support conditions impose the vanishing of the following generalized displacement vector: \(\left\{q_{\beta}\right\}=\left[\begin{array}{@{}l@{\;}l@{\;}l@{\;}l@{\;}l@{}}q_{1} & q_{2} & q_{5} & q_{6} & q_{7}\end{array}\right]^{T}=0_{5X1}\). The active, or unknown, displacement vector is \(\left\{q_{\alpha}\right\}=\left[\begin{array}{@{}l@{\;}l@{}}q_{3} & q_{4}\end{array}\right]^{T}\).

The stiffness matrices for the two beam members are obtained from eq. ([eq16.49]) as \[\begin{aligned} \label{eq16.7a}\tag{a} \left[K_{1-2}\right]=E I \begin{array}{@{}c@{}} {\arraycolsep=13pt\begin{array}{@{}cccc@{}} q_1 & q_2 & q_3 & q_4 \end{array}}\\[6pt] \left[\begin{array}{@{}cccc@{}}24 / a^{3} & -12 / a^{2} & -24 / a^{3} & -12 / a^{2} \\-12 / a^{2} & 8 / a & 12 / a^{2} & 4 / a \\-24 / a^{3} & 12 / a^{2} & 24 / a^{3} & 12 / a^{2} \\-12 / a^{2} & 4 / a & 12 / a^{2} & 8 / a\end{array}\right] \end{array} \left[K_{2-3}\right]=E I \begin{array}{@{}c@{}} {\arraycolsep=13pt\begin{array}{@{}cccc@{}} q_3 & q_4 & q_5 & q_6 \end{array}}\\[6pt] \left[\begin{array}{@{}cccc@{}}12 / a^{3} & -6 / a^{2} & -12 / a^{3} & -6 / a^{2} \\-6 / a^{2} & 4 / a & 6 / a^{2} & 2 / a \\-12 / a^{3} & 6 / a^{2} & 12 / a^{3} & 6 / a^{2} \\-6 / a^{2} & 2 / a & 6 / a^{2} & 4 / a\end{array}\right] \end{array}.\end{aligned}\] The spring stiffness matrix is obtained from eq. ([eq15.8]) on page  as \[\begin{aligned} \label{eq16.7b}\tag{b} \left[K_{4-2}\right]=E I \begin{array}{@{}c@{}} {\arraycolsep=13pt\begin{array}{@{}cc@{}} q_7 & q_3 \end{array}}\\[6pt] \left[\begin{array}{@{}cc@{}}6 / a^{3} & -6 / a^{3} \\-6 / a^{3} & 6 / a^{3}\end{array}\right] \end{array}.\end{aligned}\] Assembly of the element stiffness matrices is by summation of the element stiffness matrices with attention to the location of the matrix elements in the 7X7 unrestrained structural stiffness matrix. The result is \[\begin{aligned} \label{eq16.7c}\tag{c} [K]=E I \begin{array}{@{}c@{}} {\arraycolsep=15pt\begin{array}{@{}ccccccc@{}} q_1 & q_2 & q_3 & q_4 & q_5 & q_6 & q_7 \end{array}}\\[6pt] \left[\begin{array}{@{}ccccccc@{}}24 / a^{3} & -12 / a^{2} & -24 / a^{3} & -12 / a^{2} & 0 & 0 & 0 \\-12 / a^{2} & 8 / a & 12 / a^{2} & 4 / a & 0 & 0 & 0 \\-24 / a^{3} & 12 / a^{2} & 42 / a^{3} & 6 / a^{2} & 12 / a^{3} & -6 / a^{2} & -6 / a^{3} \\-12 / a^{2} & 4 / a & 6 / a^{2} & 12 / a & 6 / a^{2} & 2 / a & 0 \\0 & 0 & -12 / a^{3} & 6 / a^{2} & 12 / a^{3} & 6 / a^{2} & 0 \\0 & 0 & -6 / a^{2} & 2 / a & 6 / a^{2} & 4 / a & 0 \\0 & 0 & -6 / a^{3} & 0 & 0 & 0 & 6 / a^{3}\end{array}\right] \end{array}.\end{aligned}\] Partition the unrestrained structural stiffness matrix in eq. ([eq16.7c]) so that rows and columns are in the order 3, 4, 1, 2, 5, 6, 7: \[\begin{aligned} \label{eq16.7d}\tag{d} [K]=E I \begin{array}{@{}c@{}} {\arraycolsep=16pt\begin{array}{@{}ccccccc@{}} q_3 & q_4 & q_1 & q_2 & q_5 & q_6 & q_7 \end{array}}\\[6pt] \left[\begin{array}{cc;{2pt/2pt}ccccc}42 / a^{3} & 6 / a^{2} & -24 / a^{3} & 12 / a^{2} & -12 / a^{3} & -6 / a^{2} & -6 / a^{3} \\6 / a^{2} & 12 / a & -12 / a^{2} & 4 / a & 6 / a^{2} & 2 / a & 0 \\\hdashline[2pt/2pt] -24/ a^{3} & -12 / a^{2} & 24 / a^{3} & -12 / a^{2} & 0 & 0 & 0 \\12 / a^{2} & 4 / a & -12 / a^{2} & 8 / a & 0 & 0 & 0 \\-12 / a^{3} & 6 / a^{2} & 0 & 0 & 12 / a^{3} & 6 / a^{2} & 0 \\-6 / a^{2} & 2 / a & 0 & 0 & 6 / a^{2} & 4 / a & 0 \\-6 / a^{3} & 0 & 0 & 0 & 0 & 0 & 6 / a^{3}\end{array}\right] \end{array} =\left[\begin{array}{@{}ll@{}}{\left[K_{\alpha \alpha}\right]} & {\left[K_{\alpha \beta}\right]} \\{\left[K_{\beta \alpha}\right]} & {\left[K_{\beta \beta}\right]}\end{array}\right].\end{aligned}\] From the partitioned form of eq. ([eq16.7d]) the restrained structural stiffness matrix is \[\begin{aligned} \label{eq16.7e}\tag{e} \left[K_{\alpha \alpha}\right]=E I\left[\begin{array}{@{}cc@{}}42 / a^{3} & 6 / a^{2} \\6 / a^{2} & 12 / a\end{array}\right].\end{aligned}\]

Solution to part (b).

From eqs. ([eq16.54]) and ([eq16.56]) fixed-end actions of the beam member are \[\begin{aligned} \label{eq16.7f}\tag{f} \left[\begin{array}{@{}l@{}}Q_{1}^{0} \\[6pt]Q_{2}^{0} \\[6pt]Q_{3}^{0} \\[6pt]Q_{4}^{0}\end{array}\right]_{1-2}=\left[\begin{array}{@{}c@{}}-a f_{y 0} / 2 \\[3pt]a^{2} f_{y 0} / 12 \\[3pt]-a f_{y 0} / 2 \\[3pt]-a^{2} f_{y 0} / 12\end{array}\right]\left[\begin{array}{@{}l@{}}Q_{3}^{0} \\[6pt]Q_{4}^{0} \\[6pt]Q_{5}^{0} \\[6pt]Q_{6}^{0}\end{array}\right]_{2-3}=\left[\begin{array}{@{}c@{}}-7 a f_{y 0} / 20 \\[3pt]a^{2} f_{y 0} / 20 \\[3pt]-3 a f_{y 0} / 20 \\[3pt]-a^{2} f_{y 0} / 30\end{array}\right].\end{aligned}\] The assembled 7X1 fixed-end action vector in the natural order 1, 2, 3, 4, 5, 6, 7 is \[\left\{Q^{0}\right\}=\left[\frac{-a}{2} \frac{a^{2}}{12} \frac{-17 a}{20} \frac{-a^{2}}{30} \frac{-3 a}{20} \frac{-a^{2}}{30} 0\right]^{T} f_{y 0}.\] Partitioning the fixed-end action vector in the order 3, 4, 1, 2, 5, 6, 7 we get \[\left\{Q^{0}\right\}=\left[\frac{-17 a}{20} \frac{-a^{2}}{30} \bigg| \frac{-a}{2} \frac{a^{2}}{12} \frac{-3 a}{20} \frac{-a^{2}}{30} 0\right]^{T} f_{y 0}=\left\{Q_{\alpha}^{0}\right\}+\left\{Q_{\beta}^{0}\right\},\] where \[\begin{aligned} \label{eq16.7g}\tag{g} \left\{Q_{\alpha}^{0}\right\}=\left[\frac{-17 a}{20} \frac{-a^{2}}{30}\right]^{T} f_{y 0} \quad\left\{Q_{\beta}^{0}\right\}=\left[\frac{-a}{2} \frac{a^{2}}{12} \frac{-3 a}{20} \frac{-a^{2}}{30} 0\right]^{T} f_{y 0}.\end{aligned}\]

Solution to part (c).

The matrix equation to determine the unknown joint displacement \(\left\{q_{\alpha}\right\}\) is \[\begin{aligned} \label{eq16.7h}\tag{h} \left\{Q_{\alpha}\right\}=\left[K_{\alpha \alpha}\right]\left\{q_{\alpha}\right\}+\left[K_{\alpha \beta}\right]\left\{q_{\beta}\right\}+\left\{Q_{\alpha}^{0}\right\}.\end{aligned}\] The prescribed joint force vector \(\left\{Q_{\alpha}\right\}\) is \[\begin{aligned} \label{eq16.7i}\tag{i} \left[\begin{array}{@{}c@{}}Q_{3} \\Q_{4}\end{array}\right]_{\alpha}=\left[\begin{array}{@{}l@{}}P \\0\end{array}\right].\end{aligned}\] The specified displacement vector \(\left\{q_{\beta}\right\}=0{ }_{5 X 1}\). The matrix equation for the solution of the displacement vector \(\left\{q_{\alpha}\right\}\) reduces to \[\begin{aligned} \label{eq16.7j}\tag{j} \left[K_{\alpha \alpha}\right]\left\{q_{\alpha}\right\}=\left\{Q_{\alpha}\right\}+(-\{Q_{\alpha}^{0}\}),\end{aligned}\]

where \(\left(-\left\{Q_{\alpha}^{0}\right\}\right)\) is the equivalent joint force vector. See figure [fig16.19]. The explicit form of the matrix equation to determine the unknown displacements is \[E I\left[\begin{array}{@{}cc@{}}42 / a^{3} & 6 / a^{2} \\6 / a^{2} & 12 / a\end{array}\right]\left[\begin{array}{@{}l@{}}q_{3} \\q_{4}\end{array}\right]=\left[\begin{array}{@{}c@{}}P \\0\end{array}\right]+\left[\begin{array}{@{}c@{}}17 a / 20 \\a^{2} / 30\end{array}\right] f_{y 0}.\] The solution for the generalized displacements is \[\begin{aligned} \label{eq16.7k}\tag{k} \left[\begin{array}{@{}l@{}}q_{3} \\q_{4}\end{array}\right]=\frac{1}{E I}\left[\begin{array}{@{}cc@{}}a^{3} / 39 & -a^{2} / 78 \\-a^{2} / 78 & 7 a / 78\end{array}\right]\left[\begin{array}{@{}c@{}}P+17 a f_{y 0} / 20 \\a^{2} f_{y 0} / 30\end{array}\right]=\frac{1}{39 E I}\left[\begin{array}{@{}c@{}} a^{3}\left(P+5 a f_{y 0} / 18\right) \\ \frac{-a^{2}}{2}\left(P+37 a f_{y 0} / 60\right) \end{array}\right].\end{aligned}\]

Solution to part (d).

The governing matrix equation for the unknown joint forces, or the support reactions, is \[\begin{aligned} \label{eq16.7l}\tag{l} \left\{Q_{\beta}\right\}=\left[K_{\beta \alpha}\right]\left\{q_{\alpha}\right\}+\left[K_{\beta \beta}\right]\left\{q_{\beta}\right\}+\left\{Q_{\beta}^{0}\right\}.\end{aligned}\] The matrix 5X2 \(\left[K_{\beta \alpha}\right]\) is obtained from the partitioned form eq. ([eq16.7d]). Writing eq. ([eq16.7l]) in detail we have \[\begin{aligned} \label{eq16.7m}\tag{m} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{5} \\Q_{6} \\Q_{7}\end{array}\right]=E I\left[\begin{array}{@{}cc@{}}-24 / a^{3} & -12 / a^{2} \\12 / a^{2} & 4 / a \\-12 / a^{3} & 6 / a^{2} \\-6 / a^{2} & 2 / a \\-6 / a^{3} & 0\end{array}\right] \frac{1}{39 E I}\left[\begin{array}{@{}c@{}}a^{3} P+5 a f_{y 0} / 18 \\\frac{-a^{2}}{2} P+37 a f_{y 0} / 60\end{array}\right]+\left[\begin{array}{@{}c@{}}-a / 2 \\a^{2} / 12 \\-3 a / 20 \\-a^{2} / 30 \\0\end{array}\right] f_{y 0}.\end{aligned}\] After performing the matrix algebra in eq. ([eq16.7m]) the result for the support reactions is \[\begin{aligned} \label{eq16.7n}\tag{n} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{5} \\Q_{6} \\O_{7}\end{array}\right]=\left[\begin{array}{@{}cc@{}}-6 / 13 & -179 / 195 \\10 a / 39 & 721 a / 2,340 \\-5 / 13 & -59 / 130 \\7 a / 39 & -83 a / 468 \\2 / 13 & -5 / 39\end{array}\right]\left[\begin{array}{@{}c@{}}P \\a f_{y 0}\end{array}\right].\end{aligned}\]

Solution to part e.

Referring to eq. ([eq16.61]) on page , the shear force and bending moment in beam member 1-2 is given by the superposition of the fixed-end solution and the displacement solution as \[\begin{aligned} \label{eq16.7o}\tag{o} \left[\begin{array}{@{}c@{}}V_{y}(z) \\M_{x}(z)\end{array}\right]_{1-2}=\left[\begin{array}{@{}c@{}}V_{y}^{0}(z) \\M_{x}^{0}(z)\end{array}\right]_{1-2}+\left[\begin{array}{@{}c@{}}V_{y}^{1}(z) \\M_{x}^{1}(z)\end{array}\right]_{1-2}=\left[\begin{array}{@{}c@{}}V_{y}^{0}(z) \\M_{x}^{0}(z)\end{array}\right]_{1-2}+\left[S^{1}(z)\right]_{1-2}\{q\}_{1-2}.\end{aligned}\] From the fixed-end action solution ([eq16.55]) the vector of the shear force and bending moment is \[\begin{aligned} \label{eq16.7p}\tag{p} \left[\begin{array}{@{}c@{}}V_{y}^{0}(z) \\[3pt]M_{x}^{0}(z)\end{array}\right]_{1-2}=\left[\begin{array}{@{}c@{}}(a-2 z) / 2 \\\left(-a^{2}+6 a z-6 z^{2}\right) / 12\end{array}\right] f_{y 0} \quad 0 \leq z \leq a.\end{aligned}\] Equations ([eq16.45]) and ([eq16.46]) combine to determine the shear and moment from the displacements of the member. That is, \[\begin{aligned} \label{eq16.7q}\tag{q} \left[\begin{array}{@{}c@{}}V_{y}^{1}(z) \\[3pt]M_{x}^{1}(z)\end{array}\right]_{1-2}=\left[S^{1}(z)\right]_{1-2}\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right]=E I\left[\begin{array}{c:c:c:c}-\frac{12}{a^{3}} & \frac{6}{a^{2}} & \frac{12}{a^{3}} & \frac{6}{a^{2}} \\[6pt]\frac{6}{a^{2}}-\frac{12 z}{a^{3}} & -\frac{4}{a}+\frac{6 z}{a^{2}} & -\frac{6}{a^{2}}+\frac{12 z}{a^{3}} & -\frac{2}{a}+\frac{6 z}{a^{2}}\end{array}\right]\left[\begin{array}{@{}c@{}}0 \\0 \\q_{3} \\q_{4}\end{array}\right].\end{aligned}\] Perform the matrix algebra in eq. ([eq16.7q]) to find \[\begin{aligned} \label{eq16.7r}\tag{r} \left[\begin{array}{@{}c@{}}V_{y}^{1}(z) \\[3pt]M_{x}^{1}(z)\end{array}\right]_{1-2}=E I\left[\begin{array}{@{}cc@{}}\frac{12}{a^{3}} & \frac{6}{a^{2}} \\[6pt]-\frac{6}{a^{2}}+\frac{12 z}{a^{3}} &-\frac{2}{a}+\frac{6 z}{a^{2}}\end{array}\right] \frac{1}{39 E I}\left[\begin{array}{@{}c@{}}a^{3} P+5 a f_{y 0} / 18 \\-\frac{a^{2}}{2} P+37 a f_{y 0} / 60\end{array}\right]=\left[\begin{array}{@{}cc@{}}\frac{6}{13} &\frac{163a}{390}\\[6pt] \left(\frac{-10 a}{39}+\frac{6 z}{13}\right)&\left(\frac{-263 a^{2}}{1170}+\frac{163 a z}{390}\right)\end{array}\right]\left[\begin{array}{@{}c@{}}P \\f_{y 0}\end{array}\right].\end{aligned}\] Finally, substitute eqs. ([eq16.7p]) and ([eq16.7r]) into ([eq16.7o]) to get \[\begin{aligned} \left[\begin{array}{@{}l@{}}V_{y}(z) \\[3pt]M_{x}(z)\end{array}\right]_{1-2}=\left[\begin{array}{@{}cc@{}}\frac{6}{13} & \frac{179 a}{195}-z \\[3pt]\left(\frac{-10 a}{39}+\frac{6 z}{13}\right)&\left(\frac{-721 a^{2}}{2,340}+\frac{179 a z}{195}-\frac{z^{2}}{2}\right)\end{array}\right]\left[\begin{array}{@{}c@{}}P \\f_{y 0}\end{array}\right] \quad 0 \leq z \leq a. \\[-2.5pc] \text{\hfill\qed\hspace*{-83pt}}\end{aligned}\]