Given that the force–displacement (stiffness) relationship for the beam element shown in Fig. P.6.6(a) may be expressed in the following form:

Question

[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]

Obtain the force–displacement (stiffness) relationship for the variable section beam (Fig. P.6.6(b)), composed of elements 12, 23 and 34.
Such a beam is loaded and supported symmetrically as shown in Fig. P.6.6(c). Both ends are rigidly fixed and the ties FB, CH have a cross-section area [latex]a_1[/latex] and the ties EB, CG a cross-section area [latex]a_2[/latex]. Calculate the deflections under the loads, the forces in the ties and all other information necessary for sketching the bending moment and shear force diagrams for the beam. Neglect axial effects in the beam. The ties are made from the same material as the beam.

Accepted Answer

The stiffness matrix for each element of the beam is obtained using the given force–displacement relationship, the complete stiffness matrix for the beam is then obtained using the method described in Example 6.1. This gives

[latex]\left\{\begin{array}{c}F_{y, 1} \M_1 / L \F_{y, 2} \M_2 / L \F_{y, 3} \M_3 / L \F_{y, 4} \M_4 / L\end{array} ight\}=\frac{E I}{L^3}\left[\begin{array}{rrrrrrrr}24 & -12 & -24 & -12 & & & & \-12 & 8 & 12 & 4 & & & & \-24 & 12 & 36 & 6 & -12 & -6 & & \-12 & 4 & 6 & 12 & 6 & 2 & & \& & -12 & 6 & 36 & -24 & -24 & -12 \& & -6 & 2 & -6 & 12 & 12 & 4 \& & & & -24 & 12 & 24 & 12 \& & & & -12 & 4 & 12 & 8\end{array} ight]\left\{\begin{array}{c}v_1 \ heta_1 L \v_2 \ heta_2 L \v_3 \ heta_3 L \v_4 \ heta_4 L\end{array} ight\}[/latex] (i)

The ties FB, CH, EB and CG produce vertically upward forces [latex]F_2[/latex] and [latex]F_3 [/latex] at B and C, respectively. These may be found using the method described in S.6.5. Thus

[latex]F_2=-\left(\frac{a_1 E \cos ^2 60^{\circ}}{2 L / \sqrt{3}}+\frac{a_2 E \cos ^2 45^{\circ}}{\sqrt{2} L} ight) v_2[/latex]

But [latex]a_1=384 I / 5 \sqrt{3} L^2 ext { and } a_2=192 I / 5 \sqrt{2} L^2[/latex] so that

[latex]F_2=-\frac{96 E I}{5 L^3} v_2[/latex]

Similarly

[latex]F_3=-\frac{96 E I}{5 L^3} v_3[/latex]

Then

[latex]F_{y, 2}=-P-\frac{96 E I}{5 L^3} v_2 \quad ext { and } \quad F_{y, 3}=-P-\frac{96 E I}{5 L^3} v_3[/latex]

In Eq. (i), [latex]v_1= heta_1=v_4= heta_4=0 ext { and } M_2=M_3=0 ext {. }[/latex] Also, from symmetry, [latex]v_2=v_3[/latex], and [latex] heta_2=- heta_3[/latex]. Then, from Eq. (i)

[latex]M_2=0=6 v_2+12 heta_2 L+6 v_3+2 heta_3 L[/latex]

i.e.

[latex]12 v_2+10 heta_2 L=0[/latex]

which gives

[latex] heta_2=-\frac{6}{5 L} v_2[/latex]

Also from Eq. (i)

[latex]F_{y, 2}=-P-\frac{96 E I}{5 L^3} v_2=\frac{E I}{L^3}\left(36 v_2+6 heta_2 L-12 v_3-6 heta_3 L ight)[/latex]

i.e.

[latex]-P-\frac{96 E I}{5 L^3} v_2=\frac{48 E I}{5 L^3} v_2[/latex]

whence

[latex]v_2=-\frac{5 P L^3}{144 E I}=v_3[/latex]

and

[latex] heta_2=\frac{P L^2}{24 E I}=- heta_3[/latex]

The reactions at the ends of the beam now follow from the above values and Eq. (i).
Thus

[latex]\begin{aligned}F_{y, 1} &=\frac{E I}{L^3}\left(-24 v_2-12 heta_2 L ight)=\frac{P}{3}=F_{y, 4} \M_1 &=\frac{E I}{L^2}\left(12 v_2+4 heta_2 L ight)=-\frac{P L}{4}=-M_4\end{aligned}[/latex]

Also

[latex]F_2=F_3=\frac{96 E I}{5 L^3} \frac{5 P L^3}{144 E I}=\frac{2 P}{3}[/latex]

The forces on the beam are then as shown in Fig. S.6.6(a). The shear force and bending moment diagrams for the beam follow and are shown in Figs S.6.6(b) and (c),respectively.

The forces in the ties are obtained using Eq. (6.32). Thus

[latex]S_{i j}=\frac{A E}{L}\left[\begin{array}{ll}\lambda & \mu \& i j\end{array} ight]\left\{\begin{array}{l}u_j-u_i \v_j-v_i\end{array} ight\}[/latex] (6.32)

[latex]S_{\mathrm{BF}}=S_{\mathrm{CH}}=\frac{a_1 E}{2 L / \sqrt{3}}\left[-\frac{1}{2} \frac{\sqrt{3}}{2} ight]\left\{\begin{array}{c}0-0 \v_2-0\end{array} ight\}[/latex]

i.e.

[latex]S_{\mathrm{BF}}=S_{\mathrm{CH}}=\frac{384 E I \sqrt{3}}{5 \sqrt{3} imes 2 L^3} \frac{1}{2} \frac{5 P L^3}{144 E I}=\frac{2}{3} P[/latex]

and

[latex]S_{\mathrm{BE}}=S_{\mathrm{CG}}=\frac{a_2 E}{\sqrt{2} L}\left[-\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} ight]\left\{\begin{array}{c}0-0 \v_2-0\end{array} ight\}[/latex]

i.e.

[latex]S_{\mathrm{BE}}=S_{\mathrm{CG}}=\frac{192 E I}{5 \sqrt{2} imes \sqrt{2} L^3} \frac{1}{\sqrt{2}} \frac{5 P L^3}{144 E I}=\frac{\sqrt{2} P}{3}[/latex]

Question p.6.6: Given that the force–displacement (stiffness) relationship f...

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