Question 16.20: Calculate the end moments at the supports in the beam shown ...
Calculate the end moments at the supports in the beam shown in Fig. 16.42 if the support at B is subjected to a settlement of 12 \mathrm{~mm}. Furthermore, the second moment of area of the cross section of the beam is 9 \times 10^{6} \mathrm{~mm}^{4} in the span \mathrm{AB} and 12 \times 10^{6} \mathrm{~mm}^{4} in the span \mathrm{BC}; Young’s modulus, E, is 200000 \mathrm{~N} / \mathrm{mm}^{2}.
In this example the FEMs produced by the applied loads are modified by additional moments produced by the sinking support. Thus, using Table 16.6
\begin{aligned} & M_{\mathrm{AB}}^{\mathrm{F}}=-\frac{6 \times 5^{2}}{12}-\frac{6 \times 200000 \times 9 \times 10^{6} \times 12}{\left(5 \times 10^{3}\right)^{2} \times 10^{6}}=-17.7 \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{BA}}^{\mathrm{F}}=+\frac{6 \times 5^{2}}{12}-\frac{6 \times 200000 \times 9 \times 10^{6} \times 12}{\left(5 \times 10^{3}\right)^{2} \times 10^{6}}=+7.3 \mathrm{kN} \mathrm{m} \end{aligned}
Since the support at \mathrm{C} is an outside pinned support, the effect on the FEMs in BC of the settlement of B is reduced (see the last case in Table 16.6). Thus
\begin{aligned} & M_{\mathrm{BC}}^{\mathrm{F}}=-\frac{40 \times 6}{8}+\frac{3 \times 200000 \times 12 \times 10^{6} \times 12}{\left(6 \times 10^{3}\right)^{2} \times 10^{6}}=-27.6 \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{CB}}^{\mathrm{F}}=+\frac{40 \times 6}{8}=+30.0 \mathrm{kN} \mathrm{m} \end{aligned}
The DFs are
\mathrm{DF}_{\mathrm{BA}}=\frac{K_{\mathrm{BA}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}}=\frac{\left(4 E \times 9 \times 10^{6}\right) / 5}{\left(4 E \times 9 \times 10^{6}\right) / 5+\left(3 E \times 12 \times 10^{6}\right) / 6}=0.55
Hence
\mathrm{DF}_{\mathrm{BC}}=1-0.55=0.45
\begin{array}{lcccc} \hline & A & &B &&C \\ \hline { DFs } & – & 0.55 && 0.45 & 1.0\\ \hline \text { FEMs } & -17.7 & +7.3 && -27.6 & +30.0\\ \text { Balance C }&&&&& _{\swarrow} \underline{-30.0} \\ \text{Carrv over} & & && -15.0 & \\ \text{Balance B} & & _{\swarrow} \underline{+19.41} && \underline{+15.89} & \\ \text{Carry over} & +9.71 & & & \\ \text{Final moments} & -7.99 & +26.71& & -26.71 & 0 \\ \hline \end{array}
Note that in this example balancing the beam at \mathrm{B} has a significant effect on the fixing moment at \mathrm{A}; we therefore complete the distribution after a carry over to A.
TABLE 16.6
FEMs  |
||
| Load case | \textstyle{M}_{\mathrm{AB}}^{\mathrm{F}} |
\textstyle{M}_{\mathrm{BA}}^{\mathrm{F}} |
 |
-{\frac{W L}{\mathrm{8}}} |
+{\frac{W L}{\mathrm{8}}} |
 |
-{\frac{W a b^{2}}{L^2}} |
+{\frac{W a^{2} b}{L^2}} |
 |
-{\frac{w L^{2}}{12}} |
+{\frac{w L^{2}}{12}} |
 |
-\frac{w}{L^{2}}\left[\frac{L^{2}}{2}(b^{2}-a^{2})-\frac{2}{3}L(b^{3}-a^{3})+\frac{1}{4}(b^{4}-a^{4})\right] |
+\frac{w b^{3}}{L^{2}}\left(\frac{L}{3}-\frac{b}{4}\right) |
 |
+\frac{M_{0}b}{L^{2}}(2a-b) |
+\frac{M_{0}a}{L^{2}}(2b-a) |
 |
-{\frac{6EIδ}{L^2}} |
-{\frac{6EIδ}{L^2}} |
 |
0 |
-{\frac{3EIδ}{L^2}} |