The frame shown in Fig. P.6.5 has the planes xz and yz as planes of symmetry. The nodal coordinates of one quarter of the frame are given in Table P.6.5(i). In this structure the deformation of each member is due to a single effect, this being axial, bending or torsional. The mode of deformation

Question

The frame shown in Fig. P.6.5 has the planes xz and yz as planes of symmetry. The nodal coordinates of one quarter of the frame are given in Table P.6.5(i). In this structure the deformation of each member is due to a single effect, this being axial, bending or torsional. The mode of deformation of each member is given in Table P.6.5(ii), together with the relevant rigidity.

[latex]\begin{aligned}& ext { Table P.6.5(i) }\&\begin{array}{llll}\hline ext { Node } & x & y & z \\hline 2 & 0 & 0 & 0 \3 & L & 0 & 0 \7 & L & 0.8 L & 0 \9 & L & 0 & L \\hline\end{array}\end{aligned}[/latex]

Table P.6.5(ii)

[latex]\begin{array}{llcc}\hline & { ext { Effect }} \ ext { Member } & ext { Axial } & ext { Bending } & ext { Torsional } \\hline 23 & - & E I & - \37 & - & - & G J=0.8 E I \29 & E A=6 \sqrt{2} \frac{E I}{L^2} & - & -\end{array}[/latex]

Use the direct stiffness method to find all the displacements and hence calculate the forces in all the members. For member 123 plot the shear force and bending moment diagrams.
Briefly outline the sequence of operations in a typical computer program suitable for linear frame analysis.

Accepted Answer

Referring to Fig. P.6.5, [latex]u_2[/latex] = 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 [latex]F_{29}[/latex] 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 [latex]v_2[/latex]. The shortening in the member 29 is then [latex]v_2[/latex] cos θ and the corresponding strain is −([latex]v_2[/latex] cos θ)/l. Thus the compressive stress in 29 is −(E[latex]v_2[/latex] cos θ)/l and the corresponding compressive force is −(AE[latex]v_2[/latex] cos θ)/l. Thus

[latex]F_{29}=-\left(A E v_2 \cos ^2 heta ight) / l[/latex]

Now [latex]A E=6 \sqrt{2} E I / L^2 \cdot heta=45^{\circ} ext { and } l=\sqrt{2} L[/latex] Hence

[latex]F_{29}=-\frac{3 E T}{L^3} v_2[/latex]

and

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

Further, from Eq. (3.12)

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

[latex]M_3=G J \frac{\mathrm{d} heta}{\mathrm{d} z}=-2 imes 0.8 E I \frac{ heta_3}{0.8 L}=-\frac{2 E I}{L} heta_3[/latex] (ii)

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

[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, 2} \M_2 / L \F_{y, 3} \M_3 / 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_2 \ heta_2 L=0 \v_3=0 \ heta_3 L\end{array} ight\}[/latex] (iii)

Then, from Eqs (i) and (iii)

[latex]F_{y, 2}=-\frac{P}{2}-\frac{3 E I}{L^3} v_2=\frac{12 E I}{L^3} v_2-\frac{6 E I}{L^2} 0_3[/latex]

Hence

[latex]15 v_2-6 heta_3 L=-\frac{P L^3}{2 E I}[/latex] (iv)

From Eqs (ii) and (iii)

[latex]\frac{M_3}{L}=-\frac{2 E I}{L^2} heta_3=-\frac{6 E I}{L^3} v_2+\frac{4 E I}{L^2} heta_3[/latex]

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

[latex]v_2=-\frac{P L^3}{18 E I}[/latex]

Then

[latex] heta_3=-\frac{P L^2}{18 E I}[/latex]

From Eq. (i)

[latex]F_{y, 2}=-\frac{P}{2}+\frac{3 E I}{L^3} \frac{P L^3}{18 E I}=-\frac{P}{3}[/latex]

and from Eq. (ii)

[latex]M_3=\frac{2 E I}{I} \frac{P L^2}{18 E I}=\frac{P L}{9}=-M_1[/latex]

Now, from Eq. (iii)

[latex]\begin{gathered}\frac{M_2}{L}=-\frac{E I}{L^3} 6 v_2+\frac{2 E I}{L^3} heta_3 L=\frac{2 P L}{9} \F_{y, 3}=-\frac{12 E I}{L^3} v_2+\frac{6 E I}{L^3} heta_3 L=\frac{P}{3}\end{gathered}[/latex]

The force in the member 29 is [latex]F_{29} / \cos heta=\sqrt{2} F_{29}[/latex]. Thus

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

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

[latex]M_{36}=M_{37}=P L / 18[/latex]

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

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

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