Question p.6.5: The frame shown in Fig. P.6.5 has the planes xz and yz as pl...

Referring to Fig. P.6.5, u_2 = 0 from symmetry. Consider the members 23 and 29. The forces acting on the member 23 are shown in Fig. S.6.5(a) in which F_{29} is the force applied at 2 in the member 23 due to the axial force in the member 29. Suppose that the node 2 suffers a vertical displacement v_2. The shortening in the member 29 is then v_2 cos θ and the corresponding strain is −(v_2 cos θ)/l. Thus the compressive stress in 29 is −(Ev_2 cos θ)/l and the corresponding compressive force is −(AEv_2 cos θ)/l. Thus

Now A E=6 \sqrt{2} E I / L^2 \cdot \theta=45^{\circ} \text { and } l=\sqrt{2} L Hence

and

F_{y, 2}=-\frac{P}{2}-\frac{3 E I}{L^3} v_2  (i)

Further, from Eq. (3.12)

T=G J \frac{\mathrm{d} \theta}{\mathrm{d} z}  (3.12)

M_3=G J \frac{\mathrm{d} \theta}{\mathrm{d} z}=-2 \times 0.8 E I \frac{\theta_3}{0.8 L}=-\frac{2 E I}{L} \theta_3  (ii)

From the alternative form of Eq. (6.44), for the member 23

\left\{\begin{array}{c}F_{y, 2} \\M_2 / L \\F_{y, 3} \\M_3 / L\end{array}\right\}=\frac{E I}{L^3}\left[\begin{array}{rrrr}12 & -6 &-12 & -6 \\-6 & 4 & 6 & 2 \\-12 & 6 & 12 & 6 \\-6 & 2 & 6 & 4\end{array}\right]\left\{\begin{array}{c}v_2 \\\theta_2 L=0 \\v_3=0 \\\theta_3 L\end{array}\right\}  (iii)

Then, from Eqs (i) and (iii)

Hence

15 v_2-6 \theta_3 L=-\frac{P L^3}{2 E I}  (iv)

From Eqs (ii) and (iii)

which gives θ_3 = v_2/L.
Substituting for θ_3 in Eq. (iv) gives

Then

From Eq. (i)

and from Eq. (ii)

Now, from Eq. (iii)

The force in the member 29 is F_{29} / \cos \theta=\sqrt{2} F_{29}. Thus

S_{29}=S_{28}=\sqrt{2} \frac{3 E I}{L^3} \frac{P L^3}{18 E I}=\frac{\sqrt{2} P}{6}  (tension)

The torques in the members 36 and 37 are given by M_3/2, i.e.

The shear force and bending moment diagrams for the member 123 follow and are shown in Figs S.6.5(b) and (c), respectively.