Question 16.17: Determine the support reactions in the continuous beam ABCD ...
Determine the support reactions in the continuous beam ABCD shown in Fig. 16.39; its flexural rigidity EI is constant throughout.
Initially we calculate the FEMs for each of the three spans using the results presented in Table 16.6. Thus
\begin{aligned} & M_{\mathrm{AB}}^{\mathrm{F}}=-M_{\mathrm{BA}}^{\mathrm{F}}=-\frac{8 \times 3^2}{12}=-6.0 \mathrm{kNm} \\ & M_{\mathrm{BC}}^{\mathrm{F}}=-M_{\mathrm{CB}}^{\mathrm{F}}=-\frac{8 \times 2^2}{12}-\frac{20 \times 2}{8}=-7.67 \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{CD}}^{\mathrm{F}}=-M_{\mathrm{DC}}^{\mathrm{F}}=-\frac{8 \times 2^2}{12}=-2.67 \mathrm{kNm} \end{aligned}
In this particular example certain features should be noted. Firstly, the support at A is a fixed support so that it will not be released and clamped in turn. In other words, the moment at A will always be balanced (by the fixed support) but will be continually modified as the beam at B is released and clamped. Secondly, the support at D is an outside pinned support so that the final moment at D must be zero. We can therefore reduce the amount of computation by balancing the beam at D initially and then leaving the support at D pinned so that there will be no carry over of moment from C to D in the subsequent moment distribution. However, the stiffness coefficient of CD must be modified to allow for this since the span CD will then correspond to Case 2 as the beam is released at C and is free to rotate at D. Thus K_{\mathrm{CD}}=K_{\mathrm{DC}}=3E I/L. All other spans correspond to Case 1 where, as we release the beam at a support, that support is a pinned support while the beam at the adjacent support is fixed. Therefore, for the spans AB and BC, the stiffness coefficients are 4EI/L and the COFs are equal to 1/2.
The DFs are obtained from Eq. (16.36).
M_{\mathrm{BA}}^{\prime}={\frac{K_{\mathrm{BA}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}}}(-M_{\mathrm{B}})\qquad M_{\mathrm{BC}}^{\prime}={\frac{K_{\mathrm{BC}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}}}(-M_{\mathrm{B}}) (16.36)
Thus
\begin{aligned} & \mathrm{DF}_{\mathrm{BA}}=\frac{K_{\mathrm{BA}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}}=\frac{4 E I / 3}{4 E I / 3+4 E I / 2}=0.4 \\ & \mathrm{DF}_{\mathrm{BC}}=\frac{K_{\mathrm{BC}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}}=\frac{4 E I / 2}{4 E I / 3+4 E I / 2}=0.6 \\ & \mathrm{DF}_{\mathrm{CB}}=\frac{K_{\mathrm{CB}}}{K_{\mathrm{CB}}+K_{\mathrm{CD}}}=\frac{4 E I / 2}{4 E I / 2+3 E I / 2}=0.57 \\ & \mathrm{DF}_{\mathrm{CD}}=\frac{K_{\mathrm{CD}}}{K_{\mathrm{CB}}+K_{\mathrm{CD}}}=\frac{3 E I / 2}{4 E I / 2+3 E I / 2}=0.43 \end{aligned}
Note that the sum of the DFs at a support must always be equal to unity since they represent the fraction of the out of balance moment which is distributed into the spans meeting at that support. The solution is now completed as shown in Table 16.7.
Note that there is a rapid convergence in the moment distribution. As a general rule it is sufficient to stop the procedure when the distributed moments are of the order of 2% of the original FEMs. In the table the last moment at C in CD is −0.02 which is 0.75% of the original FEM, while the last moment at B in BC is +0.05 which is 0.65% of the original FEM. We could, therefore, have stopped the procedure at least one step earlier and still have retained sufficient accuracy.
The final reactions at the supports are now calculated from the final support moments and the reactions corresponding to the actual loads, i.e. the free reactions; these are calculated as though each span were simply supported. The procedure is identical to that in Ex. 16.15.
For example, in Table 16.8 the final moment reactions in AB form a couple to balance the clockwise moment of 7.19 − 5.42 = 1.77 kN m acting on AB. Thus at A the reaction is 1.77/3.0 = 0.6 kN acting downwards while at B in AB the reaction is 0.6 kN acting upwards. The remaining final moment reactions are calculated in the same way.
Finally the complete reactions at each of the supports are
\begin{array}{l l}{{R_{\mathrm{A}}=11.4\,\mathrm{kN}}} \qquad {{R_{\mathrm{B}}=12.6+18.6=31.2\,\mathrm{kN}}}\\ {{R_{\mathrm{C}}=17.4+10.98=28.38\,\mathrm{kN}}} \qquad {{R_{\mathrm{D}}=5.02\,\mathrm{kN}}}\end{array}
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}} |
TABLE 16.8
| A | B | C | D | |||||||||
| Free reactions | ↑12.0 | 12.0↑ | ↑18.0 | 18.0↑ | ↑8.0 | 8.0↑ | ||||||
| Final moment reactions | ↓0.6 | 0.6↑ | ↑0.6 | 0.6↓ | ↑2.98 | 2.98↓ | ||||||
| Total reactions (kN) | ↑11.4 | 12.6↑ | ↑18.6 | 17.4↑ | ↑10.98 | 5.02↑ | ||||||
