Question p.6.7: The symmetrical rigid jointed grillage shown in Fig. P.6.7 i...
The symmetrical rigid jointed grillage shown in Fig. P.6.7 is encastré at 6, 7, 8 and 9 and rests on simple supports at 1, 2, 4 and 5. It is loaded with a vertical point load P at 3. Use the stiffness method to find the displacements of the structure and hence calculate the support reactions and the forces in all the members. Plot the bending moment diagram for 123. All members have the same section properties and GJ = 0.8EI.
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)
\frac{M_1}{l}=0=\frac{E I}{l^3}\left(4 \theta_1 l+2 \theta_2 l\right)from which
\theta_1=-\frac{\theta_2}{2} (v)
Also, from Eqs (i) and (iv)
\frac{M_2}{l}=-\frac{E I}{l^2} \theta_2=\frac{E I}{l^3}\left(2 \theta_1 l+12 \theta_2 l+24 v_3\right)so that
13 \theta_2 l+2 \theta_1 l+24 v_3=0 (vi)
Finally from Eq. (iv)
F_{y, 3}=-\frac{P}{2}=\frac{E I}{l^3}\left(24 \theta_2 l+96 v_3\right)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
\theta_2=\frac{P l^2}{48 E I}Then, from Eq. (v)
\theta_1=-\frac{P l^2}{96 E I}and from Eq. (vii)
v_3=-\frac{P l^3}{96 E I}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).
