Question p.6.7: The symmetrical rigid jointed grillage shown in Fig. P.6.7 i...

The forces acting on the member 123 are shown in Fig. S.6.7(a). The moment $M_2$ arises from the torsion of the members 26 and 28 and, from Eq. (3.12), is given by

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

$M_2=-2 G J \frac{\theta_2}{1.6 l}=-E I \frac{\theta_2}{l}$  (i)

Now using the alternative form of Eq. (6.44) for the member 12

$\left\{\begin{array}{c}F_{y, 1} \\M_1 / l \\F_{y, 2} \\M_2 / l\end{array}\right\}=\frac{E I}{l^3}\left[\begin{array}{rrrr}12 & -6 & -12 & -6 \\-6 & 4 & 6 & 2 \\-12 & 6 & 12 & 6 \\-6 & 2 & 6 & 4\end{array}\right]\left\{\begin{array}{c}v_1 \\\theta_1 L \\v_2 \\\theta_2 L\end{array}\right\}$  (ii)

and for the member 23

$\left\{\begin{array}{c}F_{y, 2} \\M_2 / l \\F_{y, 3} \\M_3 / l\end{array}\right\}=\frac{E I}{l^3}\left[\begin{array}{rrrr}96 & -24 & -96 & -24 \\-24 & 8 & 24 & 4 \\-96 & 24 & 96 & 24 \\-24 & 4 & 24 & 8\end{array}\right]\left\{\begin{array}{c}v_2 \\\theta_2 L \\v_3 \\\theta_3 L\end{array}\right\}$  (iii)

Combining Eqs (ii) and (iii) using the method described in Example 6.1

In Eq. (iv) $v_1=v_2=0 \text { and } \theta_3=0 \text {. Also } M_1=0 \text { and } F_{y, 3}=-P / 2 \text {. }$ Then from Eq. (iv)

from which

$\theta_1=-\frac{\theta_2}{2}$  (v)

Also, from Eqs (i) and (iv)

so that

$13 \theta_2 l+2 \theta_1 l+24 v_3=0$  (vi)

Finally from Eq. (iv)

which gives

$v_3=-\frac{P l^3}{192 E I}-\frac{\theta_2 l}{4}$  (vii)

Substituting in Eq. (vi) for $θ_1$ from Eq. (v) and $v_3$ from Eq. (vii) gives

Then, from Eq. (v)

and from Eq. (vii)

Now substituting for $\theta_1, \theta_2 \text { and } v_3 \text { in Eq. (iv) gives } F_{y, 1}=-P / 16, F_{y, 2}=9 P / 16 \text {, }$ $M_2=-P l / 48 \text { (from Eq. (i)) and } M_3=-P l / 6$. Then the bending moment at 2 in 12 is $F_{y,1}$l = −Pl/12 and the bending moment at 2 in 32 is −(P/2) (l/2) + $M_3$ = −Pl/12.Also $M_3$ = −Pl/6 so that the bending moment diagram for the member 123 is that shown in Fig. S.6.7(b).