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

Question

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.

Accepted Answer

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

[latex]T=G J \frac{\mathrm{d} heta}{\mathrm{d} z}[/latex] (3.12)

[latex]M_2=-2 G J \frac{ heta_2}{1.6 l}=-E I \frac{ heta_2}{l}[/latex] (i)

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

[latex]\left\{\begin{array}{l}F_{y, i} \M_i \F_{y, j} \M_j\end{array} ight\}=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} ight]\left\{\begin{array}{l}v_i \ heta_i \v_j \ heta_j\end{array} ight\}[/latex] (6.44)

[latex]\left\{\begin{array}{c}F_{y, 1} \M_1 / l \F_{y, 2} \M_2 / l\end{array} ight\}=\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} ight]\left\{\begin{array}{c}v_1 \ heta_1 L \v_2 \ heta_2 L\end{array} ight\}[/latex] (ii)

and for the member 23

[latex]\left\{\begin{array}{c}F_{y, 2} \M_2 / l \F_{y, 3} \M_3 / l\end{array} ight\}=\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} ight]\left\{\begin{array}{c}v_2 \ heta_2 L \v_3 \ heta_3 L\end{array} ight\}[/latex] (iii)

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

[latex]\left\{\begin{array}{c}F_{y, 1} \M_1 / l \F_{y, 2} \M_2 / l \F_{y, 3} \M_3 / l\end{array} ight\}=\frac{E I}{l^3}\left[\begin{array}{rrrrrr}12 & -6 & -12 & -6 & 0 & 0 \-6 & 4 & 6 & 2 & 0 & 0 \-12 & 6 & 108 & -18 & -96 & -24 \-6 & 2 & -18 & 12 & 21 & 4 \0 & 0 & -96 & 24 & 96 & 24 \0 & 0 & -24 & 4 & 24 & 8\end{array} ight]\left\{\begin{array}{c}v_1 \ heta_1 l \v_2 \ heta_2 l \v_3 \ heta_3 l\end{array} ight\}[/latex] (iv)

In Eq. (iv) [latex]v_1=v_2=0 ext { and } heta_3=0 ext {. Also } M_1=0 ext { and } F_{y, 3}=-P / 2 ext {. }[/latex] Then from Eq. (iv)

[latex]\frac{M_1}{l}=0=\frac{E I}{l^3}\left(4 heta_1 l+2 heta_2 l ight)[/latex]

from which

[latex] heta_1=-\frac{ heta_2}{2}[/latex] (v)

Also, from Eqs (i) and (iv)

[latex]\frac{M_2}{l}=-\frac{E I}{l^2} heta_2=\frac{E I}{l^3}\left(2 heta_1 l+12 heta_2 l+24 v_3 ight)[/latex]

so that

[latex]13 heta_2 l+2 heta_1 l+24 v_3=0[/latex] (vi)

Finally from Eq. (iv)

[latex]F_{y, 3}=-\frac{P}{2}=\frac{E I}{l^3}\left(24 heta_2 l+96 v_3 ight)[/latex]

which gives

[latex]v_3=-\frac{P l^3}{192 E I}-\frac{ heta_2 l}{4}[/latex] (vii)

Substituting in Eq. (vi) for [latex]θ_1[/latex] from Eq. (v) and [latex]v_3[/latex] from Eq. (vii) gives

[latex] heta_2=\frac{P l^2}{48 E I}[/latex]

Then, from Eq. (v)

[latex] heta_1=-\frac{P l^2}{96 E I}[/latex]

and from Eq. (vii)

[latex]v_3=-\frac{P l^3}{96 E I}[/latex]

Now substituting for [latex] heta_1, heta_2 ext { and } v_3 ext { in Eq. (iv) gives } F_{y, 1}=-P / 16, F_{y, 2}=9 P / 16 ext {, }[/latex] [latex]M_2=-P l / 48 ext { (from Eq. (i)) and } M_3=-P l / 6[/latex]. Then the bending moment at 2 in 12 is [latex]F_{y,1}[/latex]l = −Pl/12 and the bending moment at 2 in 32 is −(P/2) (l/2) + [latex]M_3[/latex] = −Pl/12.Also [latex]M_3[/latex] = −Pl/6 so that the bending moment diagram for the member 123 is that shown in Fig. S.6.7(b).

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

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