Question p.6.4: The symmetrical plane rigid jointed frame 1234567, shown in ...
The symmetrical plane rigid jointed frame 1234567, shown in Fig. P.6.4, is fixed to rigid supports at 1 and 5 and supported by rollers inclined at 45° to the horizontal at nodes 3 and 7. It carries a vertical point load P at node 4 and a uniformly distributed load w per unit length on the span 26. Assuming the same flexural rigidity EI for all members, set up the stiffness equations which, when solved, give the nodal displacements of the frame. Explain how the member forces can be obtained.
The uniformly distributed load on the member 26 is equivalent to concentrated loads of wl/4 at nodes 2 and 6 together with a concentrated load of wl/2 at node 4. Thus, referring to Fig. P.6.4 and Fig. 6.3
\begin{array}{lcccccc}\text { Member } & 12 & 23 & 24 & 46 & 56 & 67 \\\text { Length } & l & l & l / 2 & l / 2 & l & l \\\lambda(\cos \theta) & 0 & -1 / \sqrt{2} & 1 & 1 & 0 & 1 / \sqrt{2} \\\mu(\sin \theta) & 1 & 1 / \sqrt{2} & 0 & 0 & 1 & 1 / \sqrt{2}\end{array}From Eq. (6.47) and using the alternative form of Eq. (6.44)
\left[K_{i j}\right]=E I\left[\begin{array}{cccccc}12 \mu^2 / L^3 & & & & &\text { SYM } \\-12 \lambda \mu / L^3 & 12 \lambda^2 / L^3 & & & \\6 \mu / L^2 & -6 \lambda / L^2 & 4 / L & & & \\-12 \mu^2 / L^3& 12 \lambda \mu / L^3 & -6 \mu / L^2 & 12 \mu^2 / L^3 & & \\12 \lambda \mu / L^3 & -12 \lambda^2 / L^3 & 6 \lambda / L^2 & -12\lambda \mu / L^3 & 12 \lambda^2 / L^3 & \\6 \mu / L^2 & -6 \lambda / L^2 & 2 / L & 6 \mu / L^2 & 6 \lambda / L^2 & 4 \lambda / L\end{array}\right] (6.47)
\left\{\begin{array}{l}F_{y, i} \\M_i \\F_{y, j} \\M_j\end{array}\right\}=E I\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]\left\{\begin{array}{c}v_i \\\theta_i \\v_j \\\theta_j\end{array}\right\} (6.44)
\begin{aligned}&\left[K_{12}\right]=\frac{E I}{l^3}\left[\begin{array}{cccccc}12 & & & & & \text { SYM } \\0 & 0 & & & & \\6 & 0 & 4 & & & \\-12 & 0 & -6 & 12 & & \\0 & 0 & 0 & 0 & 0 & \\6 & 0 & 2 & 6 & 0 & 0\end{array}\right]\\&\left[K_{23}\right]=\frac{E I}{l^3}\left[\begin{array}{cccccc}6 & & & & & \text { SYM } \\6 & 6 & & & & \\6 / \sqrt{2} & 6 / \sqrt{2} & 4 & & & \\6 & 6 & -6 \sqrt{2} & 6 & & \\-6 & -6 & -6 / \sqrt{2} & 6 & 6 & \\6 / \sqrt{2} & 6 / \sqrt{2} & 2 & 6 / \sqrt{2} & -6 / \sqrt{2} & -4 / \sqrt{2}\end{array}\right]\\&\left[K_{24}\right]=\left[K_{46}\right]=\frac{E I}{l^3}\left[\begin{array}{cccccc}0 & & & & & \text { SYM } \\0 & 96 & & & & \\0 & -24 & 8 & & & \\0 & 0 & 0 & 0 & & \\0 & -96 & 24 & 0 & 96 & \\0 & -24 & 4 & 0 & 24 & 8\end{array}\right]\end{aligned}\begin{aligned}&\left[K_{56}\right]=\frac{E I}{l^3}\left[\begin{array}{cccccc}12 & & & & & &\text { SYM } \\0 & 0 & & & & \\6 & 0 & 4 & & & \\-12 & 0 & -6 & 12 & & \\0 & 0 & 0 & 0 & 0 & \\6 & 0 & 2 & 6 & 0 & 0\end{array}\right]\\&\left[K_{67}\right]=\frac{E I}{l^3}\left[\begin{array}{cccccc}6 & & & & &\text { SYM } \\-6 & 6 & & & & \\6 / \sqrt{2} & -6 / \sqrt{2} & 4 & & & \\-6 & 6 & -6 / \sqrt{2} & 6 & & \\6 & -6 & 6 / \sqrt{2} & -6 & 6 & \\6 / \sqrt{2} & -6 / \sqrt{2} & 2 & 6 / \sqrt{2} & 6 / \sqrt{2} & 4 / \sqrt{2}\end{array}\right]\end{aligned}
The member stiffness matrices are then assembled into a 21 × 21 symmetrical matrix using the method described in Example 6.1. The known nodal displacements are u_1=v_1=\theta_1=u_5=v_5=\theta_5=u_2=u_4=u_6=\theta_3=\theta_7=0 and the support reactions are obtained from {F} = [K]{δ}. Having obtained the support reactions the internal shear force and bending moment distributions in each member follow (see Example 6.2).
