Question p.6.6: Given that the force–displacement (stiffness) relationship f...
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:
\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\}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 a_1 and the ties EB, CG a cross-section area a_2. 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.
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
The ties FB, CH, EB and CG produce vertically upward forces F_2 and F_3 at B and C, respectively. These may be found using the method described in S.6.5. Thus
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}\right) v_2But a_1=384 I / 5 \sqrt{3} L^2 \text { and } a_2=192 I / 5 \sqrt{2} L^2 so that
F_2=-\frac{96 E I}{5 L^3} v_2Similarly
F_3=-\frac{96 E I}{5 L^3} v_3Then
F_{y, 2}=-P-\frac{96 E I}{5 L^3} v_2 \quad \text { and } \quad F_{y, 3}=-P-\frac{96 E I}{5 L^3} v_3In Eq. (i), v_1=\theta_1=v_4=\theta_4=0 \text { and } M_2=M_3=0 \text {. } Also, from symmetry, v_2=v_3, and \theta_2=-\theta_3. Then, from Eq. (i)
M_2=0=6 v_2+12 \theta_2 L+6 v_3+2 \theta_3 Li.e.
12 v_2+10 \theta_2 L=0which gives
\theta_2=-\frac{6}{5 L} v_2Also from Eq. (i)
F_{y, 2}=-P-\frac{96 E I}{5 L^3} v_2=\frac{E I}{L^3}\left(36 v_2+6 \theta_2 L-12 v_3-6 \theta_3 L\right)i.e.
-P-\frac{96 E I}{5 L^3} v_2=\frac{48 E I}{5 L^3} v_2whence
v_2=-\frac{5 P L^3}{144 E I}=v_3and
\theta_2=\frac{P L^2}{24 E I}=-\theta_3The reactions at the ends of the beam now follow from the above values and Eq. (i).
Thus
Also
F_2=F_3=\frac{96 E I}{5 L^3} \frac{5 P L^3}{144 E I}=\frac{2 P}{3}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
S_{i j}=\frac{A E}{L}\left[\begin{array}{ll}\lambda & \mu \\& i j\end{array}\right]\left\{\begin{array}{l}u_j-u_i \\v_j-v_i\end{array}\right\} (6.32)
S_{\mathrm{BF}}=S_{\mathrm{CH}}=\frac{a_1 E}{2 L / \sqrt{3}}\left[-\frac{1}{2} \frac{\sqrt{3}}{2}\right]\left\{\begin{array}{c}0-0 \\v_2-0\end{array}\right\}i.e.
S_{\mathrm{BF}}=S_{\mathrm{CH}}=\frac{384 E I \sqrt{3}}{5 \sqrt{3} \times 2 L^3} \frac{1}{2} \frac{5 P L^3}{144 E I}=\frac{2}{3} Pand
S_{\mathrm{BE}}=S_{\mathrm{CG}}=\frac{a_2 E}{\sqrt{2} L}\left[-\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}\right]\left\{\begin{array}{c}0-0 \\v_2-0\end{array}\right\}i.e.
S_{\mathrm{BE}}=S_{\mathrm{CG}}=\frac{192 E I}{5 \sqrt{2} \times \sqrt{2} L^3} \frac{1}{\sqrt{2}} \frac{5 P L^3}{144 E I}=\frac{\sqrt{2} P}{3}