Solution.

The direction cosines and their products for each bar are listed in table [tab16.1].

The direction cosines from table [tab16.1] are inserted into eqs. ([eq16.12]) and ([eq16.13]), to get the 4X4 stiffness matrices and the 4X1 fixed-end actions for the truss member. The stiffness matrices are expanded to 6X6 by adding two rows and two columns of zeros, and the column vectors are expanded to 6X1 by adding two rows of zeros. Refer to the discussion in article [sec15.3] on page . The 6X1 vector of forces for bar 1-2 is

$$\begin{aligned} \label{eq16.1a}\tag{a} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{3} \\Q_{4} \\Q_{5} \\Q_{6}\end{array}\right]_{1-2}=\left(\frac{EA}{L}\right)\left[\begin{array}{@{}cccccc@{}}1 & 0 & -1 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 \\-1 & 0 & 1 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4} \\q_{5} \\q_{6}\end{array}\right]-\left[\begin{array}{@{}c@{}}-1 \\0 \\1 \\0 \\0 \\0\end{array}\right]\left(N_{T}\right)_{1-2}.\end{aligned}$$ Rows five and six, and columns five and six, of the stiffness matrix in eq. ([eq16.1a]) contain zeros entries since degrees of freedom five and six do not influence the response of truss bar 1-2. The 6X1 vector of forces for bar 1-3 is $$\begin{aligned} \label{eq16.1b}\tag{b} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{3} \\Q_{4} \\Q_{5} \\Q_{6}\end{array}\right]_{1-3} = \left(\frac{EA}{L}\right)\left[\begin{array}{@{}cccccc@{}}0 & 0 & 0 & 0 & 0 & 0 \\0 & 1 & 0 & 0 & 0 & -1 \\0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 \\0 & -1 & 0 & 0 & 0 & 1\end{array}\right]\left[\begin{array}{@{}c@{}}q_{1} \\q_{2} \\q_{3} \\q_{4} \\q_{5} \\q_{6}\end{array}\right]-\left[\begin{array}{@{}c@{}}0 \\-1 \\0 \\0 \\0 \\1\end{array}\right]\left(N_{T}\right)_{1-3}.\end{aligned}$$ Rows three and four, and columns three and four, of the stiffness matrix in eq. ([eq16.1b]) contain zeros entries since degrees of freedom three and four do not influence the response of truss bar 1-3. The 6X1 vector offorces for bar 2-3 is $$\begin{aligned} \label{eq16.1c}\tag{c} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{3} \\Q_{4} \\Q_{5} \\Q_{6}\end{array}\right]_{2-3} = \left(\frac{EA}{RL}\right)\left[\begin{array}{@{}cccccc@{}}0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 1 / 2 & -1 / 2 & -1 / 2 & 1 / 2 \\0 & 0 & -1 / 2 & 1 / 2 & 1 / 2 & -1 / 2 \\0 & 0 & -1 / 2 & 1 / 2 & 1 / 2 & -1 / 2 \\0 & 0 & 1 / 2 & -1 / 2 & -1 / 2 & 1 / 2\end{array}\right]\left[\begin{array}{@{}c@{}}q_{1} \\q_{2} \\q_{3} \\q_{4} \\q_{5} \\q_{6}\end{array}\right]\left[\begin{array}{@{}c@{}}0 \\0 \\1 / \sqrt{2} \\-1 / \sqrt{2} \\-1 / \sqrt{2} \\1 / \sqrt{2}\end{array}\right]\left(N_{T}\right)_{2-3}.\end{aligned}$$ Rows one and two, and columns one and two, of the stiffness matrix in eq. ([eq16.1c]) contain zeros entries since degrees of freedom one and two do not influence the response of truss bar 1-3. Let $a=\frac{1}{2 \sqrt{2}}$, so that $$\begin{aligned} \label{eq16.1d}\tag{d} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{3} \\Q_{4} \\Q_{5} \\Q_{6}\end{array}\right]_{2-3} = \left(\frac{EA}{L}\right)\left[\begin{array}{@{}cccccc@{}}0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & a & -a & -a & a \\0 & 0 & -a & a & a & -a \\0 & 0 & -a & a & a & -a \\0 & 0 & a & -a & -a & a\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4} \\q_{5} \\q_{6}\end{array}\right]-\left[\begin{array}{@{}c@{}}0 \\0 \\1 / \sqrt{2} \\-1 / \sqrt{2} \\-1 / \sqrt{2} \\1 / \sqrt{2}\end{array}\right]\left(N_{T}\right)_{2-3}.\end{aligned}$$ Addition of the 6X1 force vectors for each truss member equals the external joint force vector acting on the truss. This addition of force vectors satisfies equilibrium at the joints assuming the procedure to expand each truss element to six degrees of freedom to four degrees of freedom is done correctly. Hence, the condition of equilibrium is $$\begin{aligned} \label{eq16.1e}\tag{e} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{3} \\Q_{4} \\Q_{5} \\Q_{6}\end{array}\right]=\left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{3} \\Q_{4} \\Q_{5} \\Q_{6}\end{array}\right]_{1-2}+\left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{3} \\Q_{4} \\Q_{5} \\Q_{6}\end{array}\right]_{1-3}+\left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{3} \\Q_{4} \\Q_{5} \\Q_{6}\end{array}\right]_{2-3}.\end{aligned}$$ Equations ([eq16.1a]), ([eq16.1b]), and ([eq16.1d]) for the force vectors are substituted into eq. ([eq16.1e]) to get the unrestrained stiffness matrix of the truss as $$\begin{aligned} \label{eq16.1f}\tag{f} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{3} \\Q_{4} \\Q_{5} \\Q_{6}\end{array}\right]=\frac{E A}{L}\left[\begin{array}{@{}cccccc@{}}1 & 0 & -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 & 0 & -1 \\-1 & 0 & 1+a & -a & -a & a \\0 & 0 & -a & a & a & -a \\0 & 0 & -a & a & a & -a \\0 & -1 & a & -a & -a & 1+a\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4} \\q_{5} \\q_{6}\end{array}\right]+\left[\begin{array}{@{}c@{}}\left(N_{T}\right)_{1-2} \\\left(N_{T}\right)_{1-3} \\-\left(N_{T}\right)_{1-2}-\left(N_{T}\right)_{2-3} / \sqrt{2} \\\left(N_{T}\right)_{2-3} / \sqrt{2} \\\left(N_{T}\right)_{2-3} / \sqrt{2} \\-\left(N_{T}\right)_{1-3}-\left(N_{T}\right)_{2-3} / \sqrt{2}\end{array}\right].\end{aligned}$$ In compact notation eq. ([eq16.1f]) is $$\begin{aligned} \label{eq16.1g}\tag{g} \{Q\}=[K]\{q\}+\left\{Q^{0}\right\},\end{aligned}$$ where the 6X6 unrestrained structural stiffness matrix is $$\begin{aligned} \label{eq16.1h}\tag{h} [K]=\left(\frac{E A}{L}\right) \begin{array}{@{}c@{}} {\arraycolsep=7.5pt\begin{array}{@{}cccccc@{}} q_{1} & q_{2} & q_{3} & q_{4} & q_{5} & q_{6}\end{array}}\\[6pt] \left[\begin{array}{@{}cccccc@{}}1 & 0 & -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 & 0 & -1 \\-1 & 0 & 1+a & -a & -a & a \\0 & 0 & -a & a & a & -a \\0 & 0 & -a & a & a & -a \\0 & -1 & a & -a & -a & 1+a\end{array}\right]\end{array},\end{aligned}$$ and the 6X1 fixed-end action vector is $$\begin{aligned} \label{eq16.1i}\tag{i} \left\{Q^{0}\right\}=\left[\begin{array}{@{}c@{}}\left(N_{T}\right)_{1-2} \\\left(N_{T}\right)_{1-3} \\-\left(N_{T}\right)_{1-2}-\left(N_{T}\right)_{2-3} / \sqrt{2} \\\left(N_{T}\right)_{2-3} / \sqrt{2} \\\left(N_{T}\right)_{2-3} / \sqrt{2} \\-\left(N_{T}\right)_{1-3}-\left(N_{T}\right)_{2-3} / \sqrt{2}\end{array}\right].\end{aligned}$$ Note that the stiffness matrix in eq. ([eq16.1h]) is symmetric; the sum of column elements equals zero; diagonal elements are positive; and its determinate vanishes.

Assembly algorithm

Consider again the three-bar truss in example [ex16.1] on page . For computer implementation an algorithm is presented to assemble the 6X6 unrestrained structural stiffness matrix from the three 4X4 truss stiffness matrices, and to assemble the 6X1 fixed-end vector from the three 4X1 fixed-end action vectors. Let a truss member be denoted by $m$, where $m = 1$ for bar 1-2, $m = 2$ for bar 1-3, and $m = 3$ for bar 2-3. A description of symbols used in the of assembly algorithm is given in table [tab16.2].

Define the 3X1 “spring” stiffness vector $[K_{t}]$ and the 3X1 thermal force vector $[N T]$ by

$$\begin{aligned} \label{eq16.16} [K t]=\left[\frac{E A}{L} \frac{E A}{L} \frac{E A}{\sqrt{2} L}\right]^{T} \quad[N T]=\left[\left(N_{T}\right)_{1-2}\left(N_{T}\right)_{1-3}\left(N_{T}\right)_{2-3}\right]^{T}.\end{aligned}$$ Direction cosines for each truss bar are specified in the 4X1 matrices $\left[b_{i}\right]$, $i=1,2,3$, in eq. ([eq16.17]) below: $$\begin{aligned} \label{eq16.17} \left[b_{1}\right]=\left[\begin{array}{@{}llll@{}}-1 & 0 & 1 & 0\end{array}\right]^{T} \quad\left[b_{2}\right]=\left[\begin{array}{@{}llll@{}}0 & -1 & 0 & 1\end{array}\right]^{T} \quad\left[b_{3}\right]=\left[\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}\right]^{T}.\end{aligned}$$ Defined a 3X4 connectivity matrix $[C]$ by $$\begin{aligned} \label{eq16.18} &\hspace*{2.5pc}q_{2 i-1} \hspace*{1.2pc} q_{2 i} \hspace*{1.2pc} q_{2 j-1} \hspace*{1.2pc} q_{2 j} \nonumber\\ &[C]= %\begin{array}{@{}c@{}} %{%\arraycolsep=5pt %\hskip-65pt\begin{array}{@{\hspace*{-9pt}}ccc@{\hspace*{14pt}}c@{}} %q_{2 i-1} & q_{2 i} & q_{2 j-1} & q_{2 j}\end{array}}\\[6pt] {\arraycolsep=13.5pt\left[\begin{array}{@{\hspace*{6pt}}cccc@{\hspace*{6pt}}} %q_{2 i-1} & q_{2 i} & q_{2 j-1} & q_{2 j} \\ 1 & 2 & 3 & 4 \\ 1 & 2 & 5 & 6 \\ 3 & 4 & 5 & 6\end{array}\right]} \quad \begin{array}{l}\text {member 1 } \\\text {member 2}\ . \\\text {member 3 }\end{array} %\end{array}\end{aligned}$$

Row one of matrix $[C]$ is assigned to member 1 (bar 1-2), row two to member 2 (bar 1-3), and row three to member 3 (bar 2-3).

Column one contains the DOF for horizontal displacement $q_{2 i-1}$ at the beginning joint $i$ of the member,

column two contains the DOF for the vertical displacement $q_{2 i}$ of the beginning joint i of the member,

column three contains the DOF of the horizontal displacement $q_{2 j-1}$ at the end joint $j$ of the member,

and column four contains the DOF for the vertical displacement $q_{2 j}$ at the end joint $j$ of the member.

Refer to the nomenclature in table [tab16.2], and to matrices defined in eqs. ([eq16.16]), ([eq16.17]), and ([eq16.18]), to understand the flow chart for the assembly algorithm in figure [fig16.3].

[ex16.2]Consider the truss of example [ex16.1] supported in such a manner that joint displacements $q_{1}=q_{2}=q_{4}=q_{5}=0$ as is shown in figure [fig16.4]. The unknown displacements are $q_3$ and $q_6$, and take the corresponding joint forces $Q_{3}=Q_{6}=0$. The thermal forces in bars 1-2, 1-3, and 2-3 are specified as $\left(N_{T}\right)_{1-2}=0$, $\left(N_{T}\right)_{1-3} \neq 0$, and $\left(N_{T}\right)_{2-3}=0$, respectively.

1. Determine the restrained structural stiffness matrix $\left[K_{\alpha \alpha}\right]$, and submatrices $\left[K_{\alpha \beta}\right]$, $\left[K_{\beta \alpha}\right]$, and $\left[K_{\beta \beta}\right]$.

2. Determine the unknown joint displacements $q_{3}$, and $q_{6}$.

3. Determine the unknown support reactions $Q_{1}, Q_{2}, Q_{4}$, and $Q_{5}$.

4. Determine the bar forces $N_{1-2}, N_{1-3}$, and $N_{2-3}$.

Self-strained truss

Strain of the bars in a truss can occur due to temperature changes and also due to the lack of fit during assembly, even in the absence of applied nodal forces. The analysis for lack of fit of bar 1-3 in example [ex16.2] is achieved by replacing the thermal force by

$$\begin{aligned} \left(N_{T}\right)_{1-3} \rightarrow E A(\overline{\Delta} / L)_{1-3},\end{aligned}$$ where $\overline{\Delta}$ is the specified displacement of the bar to connect it to joints 1 and 3. For a gap between joints $\overline{\Delta}>0$ and for an overlap $\overline{\Delta}<0$. Hence, the solution for the bar forces in example [ex16.2] can be interpreted for the problem of lack of fit of bar 1-3 by replacing $\left(N_{T}\right)_{1-3}$ with $E A(\overline{\Delta}/ L)_{1-3}$.

[ex16.3]Now consider a statically determinate configuration of the truss in figure [fig16.2], which is shown in figure [fig16.6]. Support conditions impose displacements $q_{2}=q_{4}=q_{5}=0$. The applied external forces are specified as $Q_{1}=Q_{3}=Q_{6}=0$, and only bar 1-3 is subject to a thermal force $\left(N_{T}\right)_{1-3}=E A(\alpha \Delta T)$.

1. Determine the unknown joint displacements.

2. Determine the unknown joint forces.

3. Determine the elongation of each bar.

Boundary value problem ([eq16.28]). Generalized displacements at the boundaries

The general solution for $v_{1}(z)$ satisfying the differential equation in boundary value problem ([eq16.28]) is a cubic polynomial in the longitudinal coordinate, which is written as

$$\begin{aligned} \label{eq16.33} v_{1}(z)=c_{3} \frac{z^{3}}{6}+c_{2} \frac{z^{2}}{2}+c_{1} z+c_{0},\end{aligned}$$ where the constants $c_{3}, c_{2}, c_{1}$, and $c_{0}$ are to be determined by the four boundary conditions specified in eq. ([eq16.28]). Substitute the general solution ([eq16.33]) into these four boundary conditions and write result as $$\begin{aligned} \label{eq16.34} \left[\begin{array}{@{}cccc@{}}0 & 0 & 0 & 1 \\0 & 0 & -1 & 0 \\ L^{3} / 6 & L^{2} / 2 & L & 1 \\ -L^{2} / 2 & -L & -1 & 0\end{array}\right]\left[\begin{array}{@{}l@{}}c_{3} \\ c_{2} \\ c_{1} \\ c_{0}\end{array}\right]=\left[\begin{array}{@{}l@{}}q_{1} \\ q_{2} \\ q_{3} \\q_{4}\end{array}\right].\end{aligned}$$ Solve eq. ([eq16.34]) for the constants $c_{3}, c_{2}, c_{1}$, and $c_{0}$ to get $$\begin{aligned} \label{eq16.35} \left[\begin{array}{@{}l@{}}c_{3} \\c_{2} \\c_{1} \\c_{0}\end{array}\right]=\left[\begin{array}{@{}cccc@{}}12 / L^{3} & -6 / L^{2} & -12 / L^{3} & -6 / L^{2} \\-6 / L^{2} & 4 / L & 6 / L^{2} & 2 / L \\0 & -1 & 0 & 0 \\1 & 0 & 0 & 0\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right].\end{aligned}$$ Substituting eq. ([eq16.35]) for the constants $c_{3}, c_{2}, c_{1}$, and $c_{0}$ into eq. ([eq16.33]) leads to $$\begin{aligned} \label{eq16.36} v_{1}(z)=\frac{1}{6}\left(\frac{12}{L^{3}} q_{1}-\frac{6}{L^{2}} q_{2}-\frac{12}{L^{3}} q_{3}-\frac{6}{L^{2}} q_{4}\right) z^{3}+\frac{1}{2}\left(-\frac{6}{L^{2}} q_{1}+\frac{4}{L} q_{2}+\frac{6}{L^{2}} q_{3}+\frac{2}{L} q_{4}\right) z^{2}+\left(-q_{2}\right) z+q_{1}.\end{aligned}$$ Rearrange eq. ([eq16.36]) to the form $$\begin{aligned} \label{eq16.37} v_{1}(z)=\left(2 \frac{z^{3}}{L^{3}}-3 \frac{z^{2}}{L^{2}}+1\right) q_{1}+\left(-\frac{z^{3}}{L^{2}}+2 \frac{z^{2}}{L}-z\right) q_{2}+\left(-2 \frac{z^{3}}{L^{3}}+3 \frac{z^{2}}{L^{2}}\right) q_{3}+\left(-\frac{z^{3}}{L^{2}}+\frac{z^{2}}{L}\right) q_{4}.\end{aligned}$$ Equation ([eq16.37]) is further written in the matrix form $$\begin{aligned} \label{eq16.38} v_{1}(z)=\left[\eta_{1}(z)\ \eta_{2}(z)\ \eta_{3}(z)\ \eta_{4}(z)\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right]=[\eta(z)]\{q\}.\end{aligned}$$ The shape functions, or interpolation functions, are defined as $$\begin{aligned} \label{eq16.39} \eta_{1}(z) \equiv 2 \frac{z^{3}}{L^{3}}-3 \frac{z^{2}}{L^{2}}+1 \quad \eta_{2}(z) \equiv-\frac{z^{3}}{L^{2}}+2 \frac{z^{2}}{L}-z \quad \eta_{3}(z) \equiv-2 \frac{z^{3}}{L^{3}}+3 \frac{z^{2}}{L^{2}} \quad \eta_{4}(z) \equiv-\frac{z^{3}}{L^{2}}+\frac{z^{2}}{L}.\end{aligned}$$ From eq. ([eq16.19]) the rotation associated with the lateral displacement function $v_{1}(z)$ is given by $$\begin{aligned} \label{eq16.40} \phi_{x}(z)=-v_{1}^{\prime}(z)=\left[-\eta_{1}^{\prime}(z)-\eta_{2}^{\prime}(z)-\eta_{3}^{\prime}(z)-\eta_{4}{ }^{\prime}(z)\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right],\end{aligned}$$ where $$\begin{aligned} \label{eq16.41} \eta_{1}^{\prime}(z)=\frac{6}{L^{3}} z^{2}-6 \frac{z}{L^{2}} \quad \eta_{2}^{\prime}(z)=-3 \frac{z^{2}}{L^{2}}+4 \frac{z}{L}-1 \quad \eta_{3}{ }^{\prime}(z)=-6 \frac{z^{2}}{L^{3}}+6 \frac{z}{L^{2}} \quad \eta_{4}^{\prime}(z)=-3 \frac{z^{2}}{L^{2}}+2 \frac{z}{L}.\end{aligned}$$ These interpolation functions have the following properties at the end points, or joints, of the beam member: $$\begin{aligned} \label{eq16.42} \begin{array}{@{}llll@{}}\eta_{1}(0)=1 & \eta_{2}(0)=0 & \eta_{3}(0)=0 & \eta_{4}(0)=0 \\\eta_{1}^{\prime}(0)=0 & \eta_{2}^{\prime}(0)=-1 & \eta_{3}^{\prime}(0)=0 & \eta_{4}^{\prime}(0)=0 \\\eta_{1}(L)=0 & \eta_{2}(L)=0 & \eta_{3}(L)=1 & \eta_{4}(L)=0 \\\eta_{1}^{\prime}(L)=0 & \eta_{2}^{\prime}(L)=0 & \eta_{3}^{\prime}(L)=0 & \eta_{4}^{\prime}(L)=-1\end{array}.\end{aligned}$$

The distributions of the shear force ([eq16.32]) and the bending moment ([eq16.30]) for the boundary value problem ([eq16.28]) are

$$\begin{aligned} \label{eq16.43} \left[\begin{array}{@{}l@{}}V_{y}^{1}(z) \\[6pt] M_{x}^{1}(z)\end{array}\right]=E I\left[\begin{array}{@{}c@{}}-v_{1}^{\prime \prime \prime} \\[6pt]-v_{1}^{\prime \prime}\end{array}\right]=E I\left[\begin{array}{@{}cccc@{}}-\eta_{1}{ }^{\prime \prime \prime} & -\eta_{2}^{\prime \prime \prime} & -\eta_{3}^{\prime \prime \prime} & -\eta_{4}^{\prime \prime \prime} \\-\eta_{1}{ }^{\prime \prime} & -\eta_{2}^{\prime \prime} & -\eta_{3}^{\prime \prime} & -\eta_{4}^{\prime \prime}\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right].\end{aligned}$$ Substitute for the shape functions from eq. ([eq16.39]) into eq. ([eq16.43]) to find $$\begin{aligned} \label{eq16.44} \left[\begin{array}{@{}c@{}}V_{y}^{1}(z) \\[6pt] M_{x}^{1}(z)\end{array}\right]=E I\left[\begin{array}{c:c:c:c}-\frac{12}{L^{3}} & \frac{6}{L^{2}} & \frac{12}{L^{3}} & \frac{6}{L^{2}} \\[12pt]\frac{6}{L^{2}}-\frac{12 z}{L^{3}} & -\frac{4}{L}+\frac{6 z}{L^{2}} & -\frac{6}{L^{2}}+\frac{12 z}{L^{3}} & -\frac{2}{L}+\frac{6 z}{L^{2}}\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right].\end{aligned}$$ Since eq. ([eq16.44]) relates the internal actions consisting of the shear force and the bending moment to the joint displacement vector, it defines the 2X4 stress matrix as $$\begin{aligned} \label{eq16.45} \left[S^{1}(z)\right] \equiv E I\left[\begin{array}{c:c:cc}-\frac{12}{L^{3}} & \frac{6}{L^{2}} & \frac{12}{L^{3}} & \frac{6}{L^{2}} \\[12pt]\frac{6}{L^{2}}-\frac{12 z}{L^{3}} & -\frac{4}{L}+\frac{6 z}{L^{2}} & -\frac{6}{L^{2}}+\frac{12 z}{L^{3}} & -\frac{2}{L}+\frac{6 z}{L^{2}}\end{array}\right],\end{aligned}$$ such that $$\begin{aligned} \label{eq16.46} \left[\begin{array}{@{}c@{}}V_{y}^{1}(z) \\[6pt]M_{x}^{1}(z)\end{array}\right]=\left[S^{1}(z)\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right] \quad 0 \leq z \leq L.\end{aligned}$$ Equilibrium at the joints $z = 0$ and $z = \textit{L}$ in ([eq16.25]) leads to $$\begin{aligned} \label{eq16.47} \left[\begin{array}{@{}l@{}}Q_{1}^{1} \\[6pt]Q_{2}^{1}\end{array}\right]=\left[\begin{array}{@{}c@{}}-V_{y}^{1}(0) \\[6pt] -M_{x}^{1}(0)\end{array}\right]=\left[-S^{1}(0)\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right]\left[\begin{array}{@{}l@{}}Q_{3}^{1} \\[6pt]Q_{4}^{1}\end{array}\right]=\left[\begin{array}{@{}c@{}}V_{y}^{1}(L) \\[6pt] M_{x}^{1}(L)\end{array}\right]=\left[S^{1}(L)\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right].\end{aligned}$$ Combine these results into one matrix equation to get $$\begin{aligned} \label{eq16.48} \left[\begin{array}{@{}c@{}}Q_{1}^{1} \\[6pt] Q_{2}^{1} \\[6pt] Q_{3}^{1} \\[6pt] Q_{4}^{1}\end{array}\right]=\left[\begin{array}{@{}c@{}}\left[-S^{1}(0)\right] \\{\left[S^{1}(L)\right]}\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2} \\q_{3} \\q_{4}\end{array}\right] \quad \text { or } \quad\begin{array}{@{}c@{}}\left\{Q^{1}\right\}\\4X1\end{array}=\begin{array}{@{}c@{}}[K]\\4X4\end{array} \begin{array}{@{}c@{}}\{q\}\\ 4X1\end{array}.\end{aligned}$$ The beam element stiffness matrix is defined by $$\begin{aligned} \label{eq16.49} [K]=\left[\begin{array}{@{}c@{}}{\left[-S^{1}(0)\right]} \\[6pt]\left.\left[S^{1}(L)\right]\right]\end{array}\right], \quad \text{which evaluates to } \boxed{[K]=E I\left[\begin{array}{@{}cccc@{}}\frac{12}{L^{3}} & \frac{-6}{L^{2}} & \frac{-12}{L^{3}} & \frac{-6}{L^{2}} \\[6pt]\frac{-6}{L^{2}} & \frac{4}{L} & \frac{6}{L^{2}} & \frac{2}{L} \\[6pt]\frac{-12}{L^{3}} & \frac{6}{L^{2}} & \frac{12}{L^{3}} & \frac{6}{L^{2}} \\[6pt]\frac{-6}{L^{2}} & \frac{2}{L} & \frac{6}{L^{2}} & \frac{4}{L}\end{array}\right]}.\end{aligned}$$ The stiffness matrix of the beam member ([eq16.49]) has the following properties:

• It is symmetric, because the material is linear elastic and the displacements and rotations of the beam are assumed small.

• The column elements satisfy equilibrium for each unit displacement state.

  For example consider unit displacement state one with $\{q\}=\left[\begin{array}{@{}l@{\;}l@{\;}l@{\;}l@{}}1 & 0 & 0 & 0\end{array}\right]^{T}$.

  The corresponding generalized joint forces are

  $$\left\{Q^{1}\right\}=\left[\begin{array}{@{}llll@{}} Q_{1}^{1} & Q_{2}^{1} & Q_{3}^{1} & Q_{4}^{1}\end{array}\right]^{T}=E I\left[\frac{12}{L^{3}} \frac{-6}{L^{2}} \frac{-12}{L^{3}} \frac{-6}{L^{2}}\right]^{T}(1).$$

  The sum of the vertical forces is $Q_{1}^{1}+Q_{3}^{1}=E I\left[\frac{12}{L^{3}}+\left(\frac{-12}{L^{3}}\right)\right](1)=0$.

  The sum of moments about the center of the beam clockwise positive are:

  $$\frac{L}{2} Q_{1}^{1}+Q_{2}^{1}-\frac{L}{2} Q_{3}^{1}+Q_{4}^{1}=E I\left[\frac{L}{2}\left(\frac{12}{L^{3}}\right)+\left(\frac{-6}{L^{2}}\right)-\frac{L}{2}\left(\frac{-12}{L^{3}}\right)+\frac{-6}{L^{2}}\right](1)=0.$$

  As result of the four unit displacement states the elements of the beam stiffness matrix satisfy the following relationships.

  The sum of rows one and three equals zero. $\sum\limits_{j=1}^4 k_{1 j}+k_{3 j}=0$.

  The sum of L/2 times row one plus row two minus L/2 times row three plus row four is equal to zero. $\sum\limits_{j=1}^4(L / 2) k_{1 j}+k_{2 j}-(L / 2) k_{3 j}+k_{4 j}=0$.

  Since the stiffness matrix is symmetric, the column elements satisfy the same relationships as do the row elements

• ${Det}[K]=0$, since the beam member is not restrained against rigid body displacement.

• Its diagonal elements are positive.

Boundary value problem ([eq16.27]). Fixed-end actions

The fixed-end action vector is computed from the boundary value problem ([eq16.27]) to account for the distributed load and the temperature distribution in the direct stiffness method. Many practical problems can be analyzed with a linear distribution of the load intensity and a linear distribution of the cross-sectional temperature gradient. These linear distributions are specified as

$$\begin{aligned} \label{eq16.50} f_{y}(z)=f_{y 1}(1-z / L)+f_{f y 2}(z / L)\mbox{ and }\tau_{y}(z)=\tau_{y 1}(1-z / L)+\tau_{y 2}(z / L).\end{aligned}$$ The values of the distributed load and temperature gradient at $z = 0$ are $f_{y 1}$ and $\tau_{y 1}$, respectively. At $z = \textit{L}$, the distributed load intensity is $f_{y 2}$ and the temperature gradient is $\tau_{y 2}$. The boundary value problem ([eq16.27]) reduces to $$\begin{aligned} \label{eq16.51} \begin{array}{clc}E I v_{0}{}^{\prime \prime \prime\prime} & =f_{y 1}(1-z / L)+f_{f y 2}(z / L) & 0<z<L \\v_{0}(0)=0 & -v_{0}^{\prime}(0)=0 \quad v_{0}(L)=0 & -v_{0}^{\prime}(L)=0\end{array}.\end{aligned}$$ The solution for displacement $v_{0}(z)$ is $$\begin{aligned} \label{eq16.52} v_{0}(z)=\left[\frac{(L-z)^{2}(3 L-z) z^{2}}{120 E I L}\right] f_{y 1}+\left[\frac{(L-z)^{2}(2 L+z) z^{2}}{120 E I L}\right] f_{y 2}.\end{aligned}$$ The distribution of the transverse shear force ([eq16.32]) and the bending moment ([eq16.30]) are $$\begin{aligned} \left[\begin{array}{@{}c@{}}V_{y}^{0} \\[6pt]M_{x}^{0}\end{array}\right]=\frac{1}{60 L}\left[\begin{array}{@{}cc@{}}21 L^{2}-60 L z+30 z^{2} & 9 L^{2}-30 z^{2} \\-3 L^{3}+21 L^{2} z-30 L z^{2}+10 z^{3} & -2 L^{3}+9 L^{2} z-10 z^{3}\end{array}\right]\left[\begin{array}{@{}l@{}}f_{y 1} \\f_{y 2}\end{array}\right]-\frac{E I \alpha}{L}\left[\begin{array}{@{}cc@{}}-1 & 1 \\(L-z) & z\end{array}\right]\left[\begin{array}{@{}l@{}}\tau_{y 1} \\\tau_{y 2}\end{array}\right].\label{eq16.53}\end{aligned}$$ Substitute the results in ([eq16.53]) into joint equilibrium ([eq16.25]) to find the fixed-end actions $$\begin{aligned} \label{eq16.54} \left\{Q^{0}\right\}=\left[\begin{array}{@{}l@{}}Q_{1}^{0} \\[6pt]Q_{2}^{0} \\[6pt]Q_{3}^{0} \\[6pt]Q_{4}^{0}\end{array}\right]=\left[\begin{array}{@{}cc@{}}-7 L / 20 & -3 L / 20 \\L^{2} / 20 & L^{2} / 30 \\-3 L / 20 & -7 L / 20 \\-L^{2} / 30 & -L^{2} / 20\end{array}\right]\left[\begin{array}{@{}l@{}}f_{y 1} \\[3pt]f_{y 2}\end{array}\right]-\frac{E I \alpha}{L}\left[\begin{array}{@{}cc@{}}-1 & 1 \\L & 0 \\1 & -1 \\0 & L\end{array}\right]\left[\begin{array}{@{}c@{}}\tau_{y 1} \\\tau_{y 2}\end{array}\right].\end{aligned}$$

In the case of uniform distributions where $f_{y 1}=f_{y 2}=f_{y 0}$ and $\tau_{y 1}=\tau_{y 2}=\tau_{y 0}$, the bending moment and shear force simplify to

$$\begin{aligned} \label{eq16.55} \left[\begin{array}{@{}c@{}}V_{y}^{0} \\[6pt] M_{x}^{0}\end{array}\right]=\left[\begin{array}{@{}c@{}}(L-2 z) / 2 \\\left(-L^{2}+6 L z-6 z^{2}\right) / 12\end{array}\right] f_{y 0}-E I \alpha\left[\begin{array}{@{}l@{}}0 \\1\end{array}\right] \tau_{y 0},\end{aligned}$$ and the fixed-end actions are $$\begin{aligned} \label{eq16.56} \left\{Q^{0}\right\}=\left[\begin{array}{@{}c@{}}Q_{1}^{0} \\[6pt]Q_{2}^{0} \\[6pt]Q_{3}^{0} \\[6pt]Q_{4}^{0}\end{array}\right]=\left[\begin{array}{@{}c@{}}-L / 2 \\L^{2} / 12 \\-L / 2 \\-L^{2} / 12\end{array}\right] f_{y 0}-E I \alpha\left[\begin{array}{@{}c@{}}0 \\-1 \\0 \\1\end{array}\right] \tau_{y 0}.\end{aligned}$$

Results of the combined superposition solutions for the beam

Joint equilibrium ([eq16.25]) leads to the sum

$$\begin{aligned} \label{eq16.57} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2} \\Q_{3} \\Q_{4}\end{array}\right]=\left[\begin{array}{@{}c@{}}-V_{y}^{0}(0) \\[6pt]-M_{x}^{0}(0) \\[6pt]V_{y}^{0}(L) \\[6pt]M_{x}^{0}(L)\end{array}\right]+\left[\begin{array}{@{}c@{}}-V_{y}^{1}(0) \\[6pt]-M_{x}^{1}(0) \\[6pt]V_{y}^{1}(L) \\[6pt]M_{x}^{1}(L)\end{array}\right]=\left[\begin{array}{@{}l@{}}Q_{1}^{0} \\[6pt]Q_{2}^{0} \\[6pt]Q_{3}^{0} \\[6pt]Q_{4}^{0}\end{array}\right]+\left[\begin{array}{@{}l@{}}Q_{1}^{1} \\[6pt]Q_{2}^{1} \\[6pt]Q_{3}^{1} \\[6pt]Q_{4}^{1}\end{array}\right],\end{aligned}$$ where $\left\{Q^{0}\right\}$ is the 4X1 joint force vector from the fixed-end action boundary value problem ([eq16.27]), and $\left\{Q^{1}\right\}$ is the 4X1 joint force vector from the boundary value problem ([eq16.28]). That is, the total joint force vector is $\{Q\}=\left\{Q^{0}\right\}+\left\{Q^{1}\right\}$. From eq. ([eq16.48]) we have $$\begin{aligned} \label{eq16.58} \left\{Q^{1}\right\}=[K]\{q\},\end{aligned}$$ where the 4X4 beam stiffness matrix is given by eq. ([eq16.49]) and $\{q\}$ is the 4X1 joint displacement vector. Hence, the total joint force vector is given by $$\begin{aligned} \label{eq16.59} \{Q\}=\left\{Q^{0}\right\}+[K]\{q\}.\end{aligned}$$ Equation ([eq16.59]) is written in the form $$\begin{aligned} \label{eq16.60} \{Q\}+(-\{Q^{0}\})=[K]\{q\},\end{aligned}$$ where the vector $-\left\{Q^{0}\right\}$ is called the equivalent joint force vector. It is the negative of the fixed-end action vector.

To summarize, the analysis of a structure composed of beam members, with some members subject to distributed loads and temperature gradients, is as follows:

1. Lock every joint of the structure against translation and rotation, and calculate the fixed-end actions.

2. Apply the fixed-end actions with the opposite sign.

3. Analyze the structure with the specified joint forces and the negative of the fixed-end actions; $\{Q\}+\left(-\left\{Q^{0}\right\}\right)$. Note that the joint displacements computed in this step are the actual joint displacements.

4. Obtain the internal actions consisting of the shear force and bending moment by superposition.

   $$\begin{aligned} \label{eq16.61} \left[\begin{array}{@{}l@{}}V_{y}(z) \\[6pt]M_{x}(z)\end{array}\right]=\left[\begin{array}{@{}l@{}}V_{y}^{0}(z) \\[6pt]M_{x}^{0}(z)\end{array}\right]+\left[\begin{array}{@{}c@{}}V_{y}^{1}(z) \\[6pt]M_{x}^{1}(z)\end{array}\right]=\left[\begin{array}{@{}c@{}}V_{y}^{0}(z) \\[6pt]M_{x}^{0}(z)\end{array}\right]+\left[S^{1}(z)\right]\{q\}\end{aligned}$$ For the linear distributions of the specified external loads, the shear force $V_{y}^{0}(z)$ and bending moment $M_{x}^{0}(z)$ are given by ([eq16.53]). The 2X4 stress matrix $\left[S^{1}(z)\right]$ is given by ([eq16.46]), and $\{q\}$ is the 4X1 joint displacement vector of the beam member obtained from the solution of the assembly of the structural members.

[ex16.6]Consider the multispan uniform beam in figure [fig16.12]. It is subject to equal and opposite couples in the y-z plane at $z = 0$ and $z = \textit{L}$. The magnitude of the moment of these couples is denoted by $M_{a}$. The bending stiffness $E I$ is the same constant in each span.

1. Determine the unknown joint displacements using symmetry to reduce problem size.

2. Draw the shear force and bending moment diagrams.

3. Determine the support reactions.

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Transformation of Cartesian coordinates

Let a coplanar frame assembly be defined with respect to global Cartesian coordinate directions $(X,Y,Z)$. The local Cartesian coordinates of a frame member are $(z, y, x)$ with the $z$-coordinate along the reference axis of the member. The $z$-axis lies in the X-Y plane at an angle $\theta$ with respect to the positive X-direction as shown in figure [fig16.22]. To effect the assembly of member stiffness matrices it is necessary to transform the stiffness matrix ([eq16.64]) of a member from local coordinate directions $(z, y, x)$ to the global coordinate directions $(X,Y,Z)$. The transformation from one Cartesian system $(X,Y,Z)$ to another Cartesian system ($z,y,x$) at joint $i$ is effected by the direction cosines of the latter with respect to the former. For example, denote the cosine of the angle of the $z$-direction with respect to the Z-direction as $\cos (z, Z)$ , the cosine of the angle of the $z$-direction with respect to the $Y$-direction as $\cos (z, Y)$, etc. Then the Cartesian coordinate transformation from global to local directions in terms of the direction cosines is

$$\begin{aligned} \label{eq16.67} \begin{split} z&=\cos (z, X) X+\cos (z, Y) Y+\cos (z, Z) Z \\ y&=\cos (y, X) X+\cos (y, Y) Y+\cos (y, Z) Z \\ x&=\cos (x, X) X+\cos (x, Y) Y+\cos (x, Z) Z \end{split}.\end{aligned}$$ From figure [fig16.22] the directions cosines in terms of angle $\theta$ are as follows. $$\begin{aligned} \cos (z, X)&=\cos \theta, \cos (z, Y)=\cos \left(-\theta+90^{\circ}\right)=\sin \theta, \cos (z, Z)=\cos 90^{\circ}=0\label{eq16.68}\\ \cos (y, X)&=\cos \left(\theta+90^{\circ}\right)=-\sin \theta, \cos (y, Y)=\cos \theta, \cos (y, Z)=\cos 90^{\circ}=0,\label{eq16.69}\\ \cos (x, X)&=\cos 90^{\circ}=0, \cos (x, Y)=\cos 90^{\circ}=0, \cos (x, Z)=\cos 180^{\circ}=-1.\label{eq16.70}\end{aligned}$$ Combine the direction cosines from eqs. ([eq16.68]) to ([eq16.70]) into ([eq16.67]) to get the matrix transformation $$\begin{aligned} \label{eq16.71} \left[\begin{array}{@{}l@{}}z \\y \\x\end{array}\right]=\left[\begin{array}{@{}ccc@{}}\cos \theta & \sin \theta & 0 \\-\sin \theta & \cos \theta & 0 \\0 & 0 & -1\end{array}\right]\left[\begin{array}{@{}l@{}}X \\Y \\Z\end{array}\right].\end{aligned}$$ The inverse Cartesian coordinate transformation from local to global directions is determined from the reverse direction cosines as $$\begin{aligned} \label{eq16.72} \begin{split} X&=\cos (X, z) \cdot z+\cos (X, y) \cdot y+\cos (X, x) \cdot x \\ Y&=\cos (Y, z) \cdot z+\cos (Y, y) \cdot y+\cos (Y, x) \cdot x \\ Z&=\cos (Z, z) \cdot z+\cos (Z, y) \cdot y+\cos (Z, x) \cdot x \end{split}.\end{aligned}$$ The direction cosines of the global directions with respect to the local directions as functions of $\theta$ are $$\begin{aligned} \cos (X, z)&=\cos \theta, \cos (X, y)=\cos \left(\theta+90^{\circ}\right)=-\sin \theta, \cos (X, x)=\cos 90^{\circ}=0,\label{eq16.73}\\ \cos (Y, z)&=\cos \left(-\theta+90^{\circ}\right)=\sin \theta, \cos (Y, y)=\cos \theta, \cos (Y, x)=\cos 90^{\circ}=0\mbox{, and}\label{eq16.74}\\ \cos (Z, z)&=\cos 90^{\circ}=0, \cos (Z, y)=\cos 90^{\circ}=0, \cos (Z, x)=\cos 180^{\circ}=-1.\label{eq16.75}\end{aligned}$$ Substitute the direction cosines from eqs. ([eq16.73]) to ([eq16.75]) into eq. ([eq16.72]) to get the inverse matrix transformation as $$\begin{aligned} \label{eq16.76} \left[\begin{array}{@{}l@{}}X \\Y \\Z\end{array}\right]=\left[\begin{array}{@{}ccc@{}}\cos \theta & -\sin \theta & 0 \\\sin \theta & \cos \theta & 0 \\0 & 0 & -1\end{array}\right]\left[\begin{array}{@{}l@{}}z \\y \\x\end{array}\right].\end{aligned}$$

The generalized displacements corresponding to local coordinates ($z,y,x$) at joint $i$ are $\left\{\bar{q}_{i}\right\}=\left[\begin{array}{@{}l@{\;}l@{\;}l@{}}\bar{q}_{3 i-2} & \bar{q}_{3 i-1} & \bar{q}_{3 i}\end{array}\right]^{T}$, and the generalized displacements corresponding to global coordinates ($X,Y,Z$) at joint $i$ are $\left\{q_{i}\right\}=\left[\begin{array}{@{}l@{\;}l@{\;}l@{}}q_{3 i-2} & q_{3 i-1} & q_{3 i}\end{array}\right]^{T}$. The directions of the generalized displacements coincide with the coordinate directions as is shown in figure [fig16.22]. It follows from the coordinate transformation eq. ([eq16.71]) that the transformation of the generalized displacements from local to global directions is

$$\begin{aligned} \label{eq16.77} \left[\begin{array}{@{}c@{}}\bar{q}_{3i-2} \\\bar{q}_{3i-1} \\\bar{q}_{3 i}\end{array}\right]=\left[\begin{array}{@{}ccc@{}}\cos \theta & \sin \theta & 0 \\-\sin \theta & \cos \theta & 0 \\0 & 0 & -1\end{array}\right]\left[\begin{array}{@{}c@{}}q_{3 i-2} \\q_{3 i-1} \\-q_{3 i}\end{array}\right]=\left[\begin{array}{@{}ccc@{}}\cos \theta & \sin \theta & 0 \\-\sin \theta & \cos \theta & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{@{}c@{}}q_{3 i-2} \\q_{3 i-1} \\q_{3 i}\end{array}\right].\end{aligned}$$ In matrix notation transformation eq. ([eq16.77]) is written as $$\begin{aligned} \label{eq16.78} \left\{\bar{q}_{i}\right\}=[\tau]\left\{q_{i}\right\},\end{aligned}$$ where $$\begin{aligned} \label{eq16.79} [\tau]=\left[\begin{array}{@{}ccc@{}}\cos \theta & \sin \theta & 0 \\-\sin \theta & \cos \theta & 0 \\0 & 0 & 1\end{array}\right].\end{aligned}$$ It follows from the coordinate transformation in eq. ([eq16.76]) that the transformation of the generalized displacements from global to local directions is $$\begin{aligned} \label{eq16.80} \left[\begin{array}{@{}c@{}}q_{3 i-2} \\q_{3 i-1} \\-q_{3 i}\end{array}\right]=\left[\begin{array}{@{}ccc@{}}\cos \theta & -\sin \theta & 0 \\\sin \theta & \cos \theta & 0 \\0 & 0 & -1\end{array}\right]\left[\begin{array}{@{}c@{}}\bar{q}_{3i-2} \\\bar{q}_{3 i-1} \\\bar{q}_{3 i}\end{array}\right].\end{aligned}$$ In matrix notation the transformation in eq. ([eq16.80]) is written as $$\begin{aligned} \label{eq16.81} \left\{q_{i}\right\}=[\tau]^{T}\left\{\bar{q}_{i}\right\}.\end{aligned}$$ The matrix of direction cosines has the following properties: $$\begin{aligned} \label{eq16.82} [\tau]^{T}[\tau]=\left[\begin{array}{@{}ccc@{}}\cos \theta & -\sin \theta & 0 \\\sin \theta & \cos \theta & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{@{}ccc@{}}\cos \theta & \sin \theta & 0 \\-\sin \theta & \cos \theta & 0 \\0 & 0 & 1\end{array}\right]=\left[\begin{array}{@{}ccc@{}}\cos ^{2} \theta+\sin ^{2} \theta & 0 & 0 \\0 & \sin ^{2} \theta+\cos ^{2} \theta & 0 \\0 & 0 & 1\end{array}\right]=\left[\begin{array}{@{}lll@{}}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]=[I],\end{aligned}$$ and ${det}[\tau]=1$. Hence, the inverse of matrix $[\tau]$ is equal to its transpose. Matrix $[\tau]$ is said to be an orthogonal matrix.

Let $c=\cos \theta$ and $s=\sin \theta$. The transformation of the displacements at joint $j$ is the same matrix equation as at joint $i$ except that the components of the vectors are those corresponding to joint $j$. Hence, the transformation of the generalized displacement vector from global to local coordinate directions for frame member i-j can be written in matrix form as

$$\begin{aligned} \label{eq16.83} \left[\begin{array}{@{}c@{}}\bar{q}_{3 i-2} \\\bar{q}_{3 i-1} \\\bar{q}_{3 i} \\\hdashline \bar{q}_{3 j-2} \\\bar{q}_{3 j-1} \\\bar{q}_{3 j}\end{array}\right]=\left[\begin{array}{@{}ccc:ccc}c & s & 0 & 0 & 0 & 0 \\-s & c & 0 & 0 & 0 & 0 \\0 & 0 & 1 & 0 & 0 & 0 \\\hdashline 0 & 0 & 0 & c & s & 0 \\0 & 0 & 0 & -s & c & 0 \\0 & 0 & 0 & 0 & 0 & 1\end{array}\right]\left[\begin{array}{@{}c@{}}q_{3 i-2} \\q_{3 i-1} \\q_{3 i} \\\hdashline q_{3 j-2} \\q_{3 j-1} \\q_{3 j}\end{array}\right].\end{aligned}$$ Equation ([eq16.83]) is written in compact form as $$\begin{aligned} \label{eq16.84} \begin{aligned} &\{\bar{q}\}=[T]\{q\} \\ &6 X 1\quad 6 X 6\ 6 X 1,\end{aligned}\end{aligned}$$ where the 6X6 transformation matrix is $$\begin{aligned} \label{eq16.85} [T]=\left[\begin{array}{@{}c:c@{}}{[\tau]} & 0_{3 X 3} \\\hdashline 0_{3 X 3} & {[\tau]}\end{array}\right]=\left[\begin{array}{@{}ccc:ccc}c & s & 0 & 0 & 0 & 0 \\-s & c & 0 & 0 & 0 & 0 \\0 & 0 & 1 & 0 & 0 & 0 \\\hdashline 0 & 0 & 0 & c & s & 0 \\0 & 0 & 0 & -s & c & 0 \\0 & 0 & 0 & 0 & 0 & 1\end{array}\right].\end{aligned}$$ Transformation matrix $[T]$ is also an orthogonal matrix. That is, its determinate is equal to one and its inverse is equal to its transpose.

Frame stiffness matrix in global coordinate directions

The generalized joint force vector for frame member i-j transforms from global coordinate directions to local coordinate directions in the same manner as the generalized displacement vector does for the member. Hence, from eq. ([eq16.84]) the transformation of the 6X1 generalized force vector for element i-j is

$$\begin{aligned} \label{eq16.86} \{\bar{Q}\}=[T]\{Q\},\end{aligned}$$ where the 6X1 generalized force vector in global directions is $\{Q\}=\left[\begin{array}{@{}lll@{}}Q_{3 i-2} & Q_{3 i-1} & Q_{3 i}\end{array}\right.\left.\begin{array}{@{}lll@{}}Q_{3 j-2} & Q_{3 j-1} & Q_{3 j}\end{array}\right]^{T}$. Since the 6X6 transformation matrix is orthogonal, the inverse transformation from local to global directions is $$\begin{aligned} \label{eq16.87} \{Q\}=[T]^{T}\{\bar{Q}\}.\end{aligned}$$

To obtain the 6X6 frame element stiffness matrix in global coordinate directions, substitute ([eq16.84]) for the generalized displacement vector, and substitute ([eq16.86]) for the generalized force vector, into eq. ([eq16.66]) to get

$$\begin{aligned} \label{eq16.88} [T]\{Q\}=[\bar{K}][T]\{q\}.\end{aligned}$$ Pre-multiply this equation by $[T]^{T}$, recognizing that $[T]^{T}[T]=[I]$, to get $$\begin{aligned} \label{eq16.89} \{Q\}=[T]^{T}[\bar{K}][T]\{q\}=[K]\{q\}.\end{aligned}$$ The 6X6 matrix $[K]=[T]^{T}[\bar{K}][T]$ is the frame member stiffness matrix in global coordinate directions, and is given by $$\begin{aligned} \label{eq16.90} [K]=\left[\begin{array}{@{}ccc@{}}\frac{E A}{L} c^{2}+\frac{12 E I}{L^{3}} s^{2} & \left(\frac{E A}{L}-\frac{12 E I}{L^{3}}\right) c s & \frac{6 E I}{L^{2}} s \\\left(\frac{E A}{L}-\frac{12 E I}{L^{3}}\right) c s & \frac{E A}{L} s^{2}+\frac{12 E I}{L^{3}} c^{2} & \frac{-6 E I}{L^{2}} c \\\frac{6 E I}{L^{2}} s & \frac{-6 E I}{L^{2}} c & \frac{4 E I}{L} \\-\left(\frac{E A}{L} c^{2}+\frac{12 E I}{L^{3}} s^{2}\right) & \left(-\frac{E A}{L}+\frac{12 E I}{L^{3}}\right) c s & \frac{-6 E I}{L^{2}} s \\-\left(\frac{E A}{L}-\frac{12 E I}{L^{3}}\right) c s & -\left(\frac{E A}{L} s^{2}+\frac{12 E I}{L^{3}} c^{2}\right) & \frac{6 E I}{L^{2}} c \\\frac{6 E I}{L^{2}} s & \frac{-6 E I}{L^{2}} c & \frac{2 E I}{L}\end{array}\right.\left.\begin{array}{@{}ccc@{}}-\left(\frac{E A}{L} c^{2}+\frac{12 E I}{L^{3}} s^{2}\right) & -\left(\frac{E A}{L}-\frac{12 E I}{L^{3}}\right) c s & \frac{6 E I}{L^{2}} s \\\left(-\frac{E A}{L}+\frac{12 E I}{L^{3}}\right) c s & -\left(\frac{E A}{L} s^{2}+\frac{12 E I}{L^{3}} c^{2}\right) & \frac{-6 E I}{L^{2}} c \\\frac{-6 E I}{L^{2}} s & \frac{6 E I}{L^{2}} c & \frac{2 E I}{L} \\\frac{E A}{L} c^{2}+\frac{12 E I}{L^{3}} s^{2} & \left(\frac{E A}{L}-\frac{12 E I}{L^{3}}\right) c s & \frac{-6 E I}{L^{2}} s \\\left(\frac{E A}{L}-\frac{12 E I}{L^{3}}\right) c s & \frac{E A}{L} s^{2}+\frac{12 E I}{L^{3}} c^{2} & \frac{6 E I}{L^{2}} c \\\frac{-6 E I}{L^{2}} s & \frac{6 E I}{L^{2}} c & \frac{4 E I}{L}\end{array}\right].\end{aligned}$$ The frame member i-j referenced to global coordinate directions is shown in figure 1.6.

The frame stiffness matrix ([eq16.90]) is symmetric and singular, and the diagonal elements are positive. Equilibrium of the frame member shown in figure 1.6 for each of the six unit displacement states leads to the following relations for the elements of the stiffness matrix.

• Horizontal equilibrium: $k_{1 j}+k_{4 j}=0$, $j=1,2, \ldots, 6$, which implies row 1 plus row 4 = 0.

• Vertical equilibrium: $k_{2 j}+k_{5 j}=0$, $j=1,2, \ldots, 6$, which leads to row 2 plus row 5 = 0.

• Moment equilibrium about joint $i$: $k_{3 j}+(L \sin \theta) k_{4 j}-(L \cos \theta) k_{5 j}+k_{6 j}=0$, $j=1,2, \ldots, 6$, which leads to row 3 plus (L sine($\theta$)) times row 4 minus (L cosine($\theta$)) times row 5 plus row $6 = 0$.

Frame stress matrix

The stress matrix for the frame member i-j relates the internal axial force $N$, the transverse shear force $V$, and the bending moment $M$ to the generalized joint displacement vector. We can combine the stress matrix for the truss member, eq. ([eq16.14]), and the stress matrix for the beam member, eq. ([eq16.44]), if local coordinate direction displacements are employed. With due regard for the joint numbering convention for the frame member relative to the numbering convention of the truss and beam members, the following relationship can be obtained from the stress matrices of the truss and beam members:

$$\begin{aligned} \label{eq16.91} \left[\begin{array}{@{}c@{}}N \\V \\M\end{array}\right]_{i-j}={\underbrace{\left[\begin{array}{@{}cccccc@{}}-E A / L & 0 & 0 & E A / L & 0 & 0 \\0 & -12 E I / L^{3} & 6 E I / L^{2} & 0 & 12 E I / L^{3} & 6 E I / L^{2} \\0 & E I\left(\frac{6}{L^{2}}-\frac{12 z}{L^{3}}\right) & E I\left(-\frac{4}{L}+\frac{6 z}{L^{2}}\right) & 0 & E I\left(-\frac{6}{L^{2}}+\frac{12 z}{L^{3}}\right) & E I\left(-\frac{2}{L}+\frac{6 z}{L^{2}}\right)\end{array}\right]}_{[\bar{S}(z)]_{i-j}}}_{\raisebox{1.4pc}{\small $i{-}j$}}\left[\begin{array}{@{}c@{}}\bar{q}_{3 i-2} \\\bar{q}_{3 i-1} \\\bar{q}_{3 i} \\\bar{q}_{3 j-2} \\\bar{q}_{3 j-1} \\\bar{q}_{3 j}\end{array}\right].\end{aligned}$$ Recall that the axial coordinate $z$ is a local coordinate in the frame element, which is zero at the beginning joint $i$ and equal to the length L of the frame element at end joint $j$. The 3X6 stress matrix $[\bar{S}(z)]_{i-j}$ ([eq16.91]) is referenced to the generalized displacement vector in local coordinate directions. The stress matrix in terms of the generalized displacement vector in global coordinate directions is obtained by substituting ([eq16.84]) for the displacement vector in eq. ([eq16.91]) to get $$\begin{aligned} \label{eq16.92} \left[\begin{array}{@{}c@{}}N \\V \\M\end{array}\right]_{i-j}=[\bar{S}(z)]_{i-j}[T]_{i-j}\left[\begin{array}{@{}c@{}}q_{3 i-2} \\q_{3 i-1} \\q_{3 i} \\q_{3 j-2} \\q_{3 j-1} \\q_{3 j}\end{array}\right]=[S(z)]_{i-j}\left[\begin{array}{@{}c@{}}q_{3 i-2} \\q_{3 i-1} \\q_{3 i} \\q_{3 j-2} \\q_{3 j-1} \\q_{3 j}\end{array}\right] \quad 0 \leq z \leq L,\end{aligned}$$ where $$\begin{aligned} _{i-j}&=\left[\begin{array}{@{}cccc@{}}-E A / L & 0 & 0 & E A / L \\0 & -12 E I / L^{3} & 6 E I / L^{2} & 0 \\0 & E I\left(\frac{6}{L^{2}}-\frac{12 z}{L^{3}}\right) & E I\left(-\frac{4}{L}+\frac{6 z}{L^{2}}\right) & 0\end{array}\right.\left.\begin{array}{@{}ccc@{}}0 & 0 \\12 E I / L^{3} & 6 E I / L^{2} \\E I\left(-\frac{6}{L^{2}}+\frac{12 z}{L^{3}}\right) & E I\left(-\frac{2}{L}+\frac{6 z}{L^{2}}\right)\end{array}\right] \\ &\quad\hspace*{-1.4pt}\left[\begin{array}{@{}cccccc@{}}c & s & 0 & 0 & 0 & 0 \\-s & c & 0 & 0 & 0 & 0 \\0 & 0 & 1 & 0 & 0 & 0 \\0 & 0 & 0 & c & s & 0 \\0 & 0 & 0 & -s & c & 0 \\0 & 0 & 0 & 0 & 0 & 1\end{array}\right].\end{aligned}$$ Perform the matrix multiplication in the last equation to find $$\begin{aligned} _{i-j} &=\left[\begin{array}{@{}ccc@{}}-c E A / L & -s E A / L & 0 \\ s 12 E I / L^{3} & -c 12 E I / L^{3} & 6 E I / L^{2} \\ -E I s\left(\frac{6}{L^{2}}-\frac{12 z}{L^{3}}\right) & E I c\left(\frac{6}{L^{2}}-\frac{12 z}{L^{3}}\right) & E I\left(-\frac{4}{L}+\frac{6 z}{L^{2}}\right)\end{array}\right. \nonumber \\ &\qquad\left.\begin{array}{@{}ccc@{}}c E A / L & s E A / L & 0 \\-s 12 E I / L^{3} & c 12 E I / L^{3} & 6 E I / L^{2} \\-E I s\left(-\frac{6}{L^{2}}+\frac{12 z}{L^{3}}\right) & E I c\left(-\frac{6}{L^{2}}+\frac{12 z}{L^{3}}\right) & E I\left(-\frac{2}{L}+\frac{6 z}{L^{2}}\right)\end{array}\right]_{i-j}. \label{eq16.93}\end{aligned}$$ The 3X6 stress matrix $[S(z)]_{i-j}$, $0 \leq z \leq L$, for the frame member relates the internal actions in local coordinate directions $z$ and $y$ to the generalized displacement vector in global coordinate directions.

[ex16.8]The coplanar rectangular frame shown in figure 1.7 consists of three members: 1-2, 2-3, and 3-4. Joints1 and 4 are restrained against displacement and rotation. At joint 2 there is a rigid connection between members 1-2 and 2-3, and at joint 3 there is a rigid connection between members 2-3 and 3-4. Joints 2 and 3 are moveable, and the generalized displacement vector for these joints is $\left\{q_{\alpha}\right\}=\left[\begin{array}{@{}l@{\;}l@{\;}l@{\;}l@{\;}l@{\;}l@{}}q_{4} & q_{5} & q_{6} & q_{7} & q_{8} & q_{9}\end{array}\right]^{T}$. Each member has a cross-sectional area $A=1,500 \mathrm{~mm}^{2}$, second area moment $I=2.8 \times 10^{6} \mathrm{~mm}^{4}$, and the same modulus of elasticity $E=70 \times 10^{3} \mathrm{~N}/ \mathrm{mm}^{2}$. The direction cosines for member 1-2 are $(c, s)=(0,1)$, for member 2-3 $(c, s)=(1,0)$, and for member 3-4 $(c, s)=(0,-1)$. Determine the generalized displacements of the movable joints 2 and 3, and the bending moment in each member.

The stiffness matrix ([eq16.90]) for each member including only the generalized displacements of joints 2 and 3 are as follows:

$$\begin{aligned} \left[K_{1-2}\right]&= \begin{array}{@{}c@{}} {\arraycolsep=16pt \begin{array}{@{}ccc@{}}\hspace*{-1pc}q_{4} & q_{5} & q_{6}\end{array}} \\[3pt]\left[\begin{array}{@{}ccc@{}}(12 E I) / h^{3} & 0 & (-6 E I) / h^{2} \\0 & (E A) / h & 0 \\(-6 E I) / h^{2} & 0 & (4 E I) / h\end{array}\right]\end{array}=\begin{array}{@{}c@{}} {\arraycolsep=16pt\begin{array}{@{}ccc@{}}q_{4} & q & q_{6}\end{array}} \\[3pt]\left[\begin{array}{@{}ccc@{}}294 & 0 & -294{,}000 \\0 & 52{,}500 & 0 \\-294{,}000 & 0 & 3.92 \times 10^{8}\end{array}\right]\end{array},\nonumber\\ \left[K_{2-3}\right]&=\begin{array}{@{}c@{}} {\arraycolsep=22pt\begin{array}{@{}cccccc@{}} \hspace*{-1pc}q_4 & q_5 & q_6 & q_7 & q_8 & q_9 \end{array}}\\[3pt] \left[\begin{array}{@{}cccccc@{}}(E A) / L & 0 & 0 &(-E A) / L & 0 & 0 \\0 & (12 E I) / L^{3} & (-6 E I) / L^{2} &0 & (-12 E I) / L^{3} & (-6 E I) / L^{2} \\0 & (-6 E I) / L^{2} & (4 E I) / L &0 & (6 E I) / L^{2} & (2 E I) / L \\ (-E A) / L & 0 & 0 &(E A) / L & 0 & 0\\ 0 & (-12 E I) / L^{3} & (6 E I) / L^{2} &0 & (12 E I) / L^{3} & (6 E I) / L^{2} \\ 0 & (-6 E I) / L^{2} & (2 E I) / L &0 & (6 E I) / L^{2} & (4 E I) / L \end{array}\right]\end{array},\label{eq16.8a}\tag{a} \\ \left[K_{2-3}\right]&= \begin{array}{@{}c@{}} {\arraycolsep=20pt\begin{array}{@{}cccccc@{}} q_4 & q_5 & q_6 & q_7 & q_8 & q_9 \end{array}}\\[6pt] \left[\begin{array}{@{}cccccc@{}}65{,}625 & 0 &0&-65{,}625 & 0 & 0 \\ 0 & 574.219 & -459{,}375 &0 & -574.219 & -459{,}375\\ 0& -459{,}375 &4.9 \times 10^{8} &0 & 459{,}375 & 2.45 \times 10^{8} \\ -65{,}625 & 0 &0 & 65{,}625 & 0 & 0 \\ 0 & -574.219 &459{,}375 & 0 & 574.219 & 459,375 \\ 0&-459{,}375 & 2.45 \times 10^{8} & 0 & 459{,}375 & 4.9 \times 10^{8}\end{array}\right]\end{array}\mbox{, and}\label{eq16.8b}\tag{b} \displaybreak\\ \hspace*{2pc}\left[K_{3-4}\right]&= \begin{array}{@{}c@{}} {\arraycolsep=16pt \begin{array}{@{}ccc@{}}q_{7} & q_{8} & q_{9}\end{array}} \\[6pt] \left[\begin{array}{@{}ccc@{}}(12 E I) / h^{3} & 0 & (6 E I) / h^{2} \\0 & (E A) / h & 0 \\(6 E I) / h^{2} & 0 & (4 E I) / h\end{array}\right]\end{array}= \begin{array}{@{}c@{}} {\arraycolsep=16pt \begin{array}{@{}ccc@{}}q_{7} & q_{8} & q_{9}\end{array}} \\[6pt] \left[\begin{array}{@{}ccc@{}}294 & 0 & 294,000 \\0 & 52,500 & 0 \\294,000 & 0 & 3.92 \times 10^{8}\end{array}\right]\end{array}.\label{eq16.8c}\tag{c}\end{aligned}$$ The restrained structural stiffness matrix is obtained by the sum of the member stiffness matrices in eqs. ([eq16.8a]), ([eq16.8b]), and ([eq16.8c]) with due regard to the location of the matrix elements from the individual members to their place in the restrained stiffness matrix. The result is $$\begin{aligned} \label{eq16.8d}\tag{d} \left[K_{\alpha \alpha}\right]=\begin{array}{@{}c@{}} {\arraycolsep=20pt\begin{array}{@{}cccccc@{}} q_4 & q_5 & q_6 & q_7 & q_8 & q_9 \end{array}}\\[6pt] \left[\begin{array}{@{}cccccc@{}}65,919 & 0 & -294,000 & -65,625 & 0 & 0 \\0 & 53,074.2 & -459,375 & 0 & -574.219 & -459,375 \\-294,000 & -459,375 & 8.82 \times 10^{8} & 0 & 459,375 & 2.45 \times 10^{8} \\-65,625 & 0 & 0 & 65,919 & 0 & -294,000 \\0 & -574.219 & 459,375 & 0 & 53,074.2 & 459,375 \\0 & -459,375 & 2.45 \times 10^{8} & -294,000 & 459,375 & 8.82 \times 10^{8}\end{array}\right]\end{array}.\end{aligned}$$ The prescribed external load vector is $\left\{Q_{\alpha}\right\}=\left[\begin{array}{@{}llllll@{}} Q_{4} & Q_{5} & Q_{6} & Q_{7} & Q_{8} & Q_{9}\end{array}\right]^{T}=\left[\begin{array}{@{}l@{\;\;\,\,\,}l@{\;\;\,\,\,}l@{\;\;\,\,\,}l@{\;\;\,\,\,}l@{\;\;\,\,\,}l@{}}18,000 \mathrm{~N} & 0 & 0 & 0 & 0 & 0\end{array}\right]^{T}$,and the matrix equation to determine the generalized displacements is $\left\{Q_{\alpha}\right\}=\left[K_{\alpha \alpha}\right]\left\{q_{\alpha}\right\}$. The solution for the generalized displacements from the latter equation is $$\begin{aligned} \label{eq16.8e}\tag{e} \begin{split} q_{4}=41.6931 \mathrm{~mm} \quad q_{5}=0.18859 \mathrm{~mm} \quad q_{6}=0.0110439\, \mathrm{rad} \text {. }\\ q_{7}=41.5561 \mathrm{~mm} \quad q_{8}=-0.18859 \mathrm{~mm} \quad q_{9}=0.0109807\, \mathrm{rad} \end{split}.\end{aligned}$$ The bending moment in member 1-2 is determined from its stress matrix ([eq16.93]) and the generalized displacement vector for the member. Reading the third row of the stress matrix determines the bending moment as $$\begin{aligned} \label{eq16.8f}\tag{f} (M_{x})_{1-2}=\left[-E I(6 / h^{2}-12 z / h^{3}) 0\ E I(-4 / h+6 z / h^{2})-E I(-6 / h^{2}+12 z / h^{3}) 0\ E I(-2 / h+6 z / h^{2})\right]\left\{q_{1-2}\right\},\end{aligned}$$ where $\left\{q_{1-2}\right\}=\left[\begin{array}{@{}llllll@{}}0 & 0 & 0 & q_{4} & q_{5} & q_{6}\end{array}\right]^{T}$. Numerical evaluation results in $$\begin{aligned} \left(M_{x}\right)_{1-2}=(294,000-294 . z) q_{4}+(-1.96 \times 10^{8}+294,000 z) q_{6} &=1.00931 \times 10^{7}-9,010.84 z \nonumber\\ &\quad 0 \leq z \leq 2,000 \mathrm{~mm}.\label{eq16.8g}\tag{g}\end{aligned}$$ Following the same procedure for members 2-3 and 3-4 we find $$\begin{aligned} \left(M_{x}\right)_{2-3} &=(459,375-574.219 {z}) q_{5}+(-4.9 \times 10^{8}+459,375 {z}) q_{6}+(-459,375+574.219 {z}) q_{8} \nonumber\\ &\quad +(-2.45 \times 10^{8}+459,375 {z}) q_{9}=-7.92854 \times 10^{6}+9,900.99 {z} \quad 0 \leq {z} \leq 1,600 \mathrm{~mm}\mbox{, and}\label{eq16.8h}\tag{h}\\ \hspace*{-4pt}\left(M_{x}\right)_{3-4} &=(294,000-294 {z}) q_{7}+(-3.92 \times 10^{8}+294,000 {z}) q_{9}=7.91305 \times 10^{6}-8,989.16 {z}, 0 \leq {z} \leq 2,000 \mathrm{~mm}.\label{eq16.8i}\tag{i}\end{aligned}$$ The bending moment in each member is plotted with respect to the axial coordinate in figure [fig16.25]. The bending moment distribution is linear in each member and it passes through zero as it changes sign.