Question 16.18: Calculate the support reactions in the beam shown in Fig. 16...
Calculate the support reactions in the beam shown in Fig. 16.40; the flexural rigidity, E I, of the beam is constant throughout.
This example differs slightly from Ex. 16.17 in that there is no fixed support and there is a cantilever overhang at the right-hand end of the beam. We therefore treat the support at \mathrm{A} in exactly the same way as the support at \mathrm{D} in the previous example. The effect of the cantilever overhang may be treated in a similar manner since we know that the final value of moment at \mathrm{D} is -5 \times 4=-20 \mathrm{kN} \mathrm{m}. We therefore calculate the FEMs M_{\mathrm{DE}}^{\mathrm{F}}(=-20 \mathrm{kNm}) and M_{\mathrm{DC}}^{\mathrm{F}}, balance the beam at \mathrm{D}, carry over to \mathrm{C} and then leave the beam at \mathrm{D} balanced and pinned; again the stiffness coefficient, K_{\mathrm{DC}}, is modified to allow for this (Case 2).
The FEMs are again calculated using the appropriate results from Table 16.6. Thus
\begin{aligned} & M_{\mathrm{AB}}^{\mathrm{F}}=-M_{\mathrm{BA}}^{\mathrm{F}}=-\frac{12 \times 14}{8}=-21 \mathrm{kNm} \\ & M_{\mathrm{BC}}^{\mathrm{F}}=-M_{\mathrm{CB}}^{\mathrm{F}}=-\frac{7 \times 4 \times 8^{2}}{12^{2}}-\frac{7 \times 8 \times 4^{2}}{12^{2}}=-18.67 \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{CD}}^{\mathrm{F}}=-M_{\mathrm{DC}}^{\mathrm{F}}=-\frac{22 \times 12}{12}=-22 \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{DE}}^{\mathrm{F}}=-5 \times 4=-20 \mathrm{kNm} \end{aligned}
The DFs are calculated as follows
\mathrm{DF}_{\mathrm{BA}}=\frac{K_{\mathrm{BA}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}}=\frac{3 E I / 14}{3 E I / 14+4 E I / 12}=0.39
Hence
\begin{aligned}& \mathrm{DF}_{\mathrm{BC}}=1-0.39=0.61 \\& \mathrm{DF}_{\mathrm{CB}}=\frac{K_{\mathrm{CB}}}K_{\mathrm{CB}}+K_{\mathrm{CD}}=\frac{4 E I / 12}{4 E I / 12+3 E I / 12}=0.57\end{aligned}
Hence
\mathrm{DF}_{\mathrm{CD}}=1-0.57=0.43
The solution is completed as follows:
The support reactions are now calculated in an identical manner to that in Ex. 16.17 and are
R_{\mathrm{A}}=4.18 \mathrm{kN} \quad R_{\mathrm{B}}=15.35 \mathrm{kN} \quad R_{\mathrm{C}}=17.4 \mathrm{kN} \quad R_{D}=16.07 \mathrm{kN}
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}} |
