Question 16.19: Calculate the reactions at the supports in the beam ABCD sho...
Calculate the reactions at the supports in the beam ABCD shown in Fig. 16.41. The flexural rigidity of the beam is constant throughout.
The beam in Fig. 16.41 is symmetrically supported and loaded about its centre line; we may therefore use this symmetry to reduce the amount of computation.
In the centre span, \mathrm{BC}, M_{\mathrm{BC}}^{\mathrm{F}}=-M_{\mathrm{CB}}^{\mathrm{F}} and will remain so during the distribution. This situation corresponds to Case 3 , so that if we reduce the stiffness \left(K_{\mathrm{BC}}\right) of \mathrm{BC} to 2 E I / L there will be no carry over of moment from B to C (or C to B) and we can consider just half the beam. The outside pinned support at A is treated in exactly the same way as the outside pinned supports in Exs 16.17 and 16.18.
The FEMs are
\begin{aligned} & M_{\mathrm{AB}}^{\mathrm{F}}=-M_{\mathrm{BA}}^{\mathrm{F}}=-\frac{5 \times 6^{2}}{12}=-15 \mathrm{kNm} \\ & M_{\mathrm{BC}}^{\mathrm{F}}=-M_{\mathrm{CB}}^{\mathrm{F}}=-\frac{40 \times 5}{8}=-25 \mathrm{kN} \mathrm{m} \end{aligned}
The DFs are
\mathrm{DF}_{\mathrm{AB}}=\frac{K_{\mathrm{BA}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}}=\frac{3 E I / 6}{3 E I / 6+2 E I / 10}=0.71
Hence
\mathrm{DF}_{\mathrm{BC}}=1-0.71=0.29
The solution is completed as follows:
\begin{array}{lcrcc} \hline & \underline{\quad A\quad} & &&\underline{\quad B\quad} \\ \text{ DFs} & \mathbf{1} & & \mathbf{0 . 7 1} && \mathbf{0 . 2 9} \\ \hline \text{FEMs} & -15.0 & & +15.0 && -25.0 \\ \text{Balance }A & \underline{+15.0}_{\searrow} & & \\ \text{Carry over} & & & +7.5 \\ \text{Balance }B & & & \underline{+1.78} && \underline{+0.72} \\ \text{Final moments} & 0 & & +24.28 && -24.28 \\ \hline \end{array}
Note that we only need to balance the beam at B once. The use of symmetry therefore leads to a significant reduction in the amount of computation.