Question 16.21: Obtain the bending moment diagram for the frame shown in Fig...
Obtain the bending moment diagram for the frame shown in Fig. 16.44; the flexural rigidity E I is the same for all members.
In this example the frame is unsymmetrical but sway is prevented by the member \mathrm{BC} which is fixed at \mathrm{C}. Also, the member DA is fixed at D while the member EB is pinned at \mathrm{E}.
The FEMs are calculated using the results of Table 16.6 and are
\begin{aligned} & M_{\mathrm{AD}}^{\mathrm{F}}=M_{\mathrm{DA}}^{\mathrm{F}}=0 \quad M_{\mathrm{BE}}^{\mathrm{F}}=M_{\mathrm{EB}}^{\mathrm{F}}=0 \\ & M_{\mathrm{AB}}^{\mathrm{F}}=-M_{\mathrm{BA}}^{\mathrm{F}}=-\frac{12 \times 4 \times 8^{2}}{12^{2}}-\frac{12 \times 8 \times 4^{2}}{12^{2}}=-32 \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{BC}}^{\mathrm{F}}=-M_{\mathrm{CB}}^{\mathrm{F}}=-\frac{1 \times 16^{2}}{12}=-21.3 \mathrm{kN} \mathrm{m} \end{aligned}
Since the vertical member \mathrm{EB} is pinned at \mathrm{E}, the final moment at \mathrm{E} is zero. We may therefore treat \mathrm{E} as an outside pinned support, balance \mathrm{E} initially and reduce the stiffness coefficient, K_{\mathrm{BE}}, as before. However, there is no FEM at \mathrm{E} so that the question of balancing E initially does not arise. The DFs are now calculated
\mathrm{DF}_{\mathrm{AD}}=\frac{K_{\mathrm{AD}}}{K_{\mathrm{AD}}+K_{\mathrm{AB}}}=\frac{4 E I / 12}{4 E I / 12+4 E I / 12}=0.5
Hence
\begin{aligned} & \mathrm{DF}_{\mathrm{AB}}=1-0.5=0.5 \\ & \mathrm{DF}_{\mathrm{BA}}=\frac{K_{\mathrm{BA}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}+K_{\mathrm{BE}}}=\frac{4 E I / 12}{4 E I / 12+4 E I / 16+3 E I / 12}=0.4 \\ & \mathrm{DF}_{\mathrm{BC}}=\frac{K_{\mathrm{BC}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}+K_{\mathrm{BE}}}=\frac{4 E I / 16}{4 E I / 12+4 E I / 16+3 E I / 12}=0.3 \end{aligned}
Hence
\mathrm{DF}_{\mathrm{BE}}=1-0.4-0.3=0.3
The solution is now completed below.
The bending moment diagram is shown in Fig. 16.45 and is drawn on the tension side of each member. The bending moment distributions in the members \mathrm{AB} and \mathrm{BC} are determined by superimposing the fixing moment diagram on the free bending moment diagram, i.e. the bending moment diagram obtained by supposing that \mathrm{AB} and \mathrm{BC} are simply supported.
| E | C | B | A | D | Joint | |||
| EB | CB | BC | BE | BA | AB | AD | DA | Member |
| 1.0 | – | 0.3 | 0.3 | 0.4 | 0.5 | 0.5 | – | DFs |
| 0 | +21.3 | -21.3 | 0 | +32.0 | −32.0 | 0 | 0 | FEMs |
| -3.2 | -3.2 | -4.3 | +16.0 | +16.0 | Balance A and B | |||
| -1.6 | +8.0 | -2.15 | +8.0 | Carry over | ||||
| -2.4 | -2.4 | -3.2 | +1.08 | +1.08 | Balance | |||
| -1.2 | +0.54 | -1.6 | +0.54 | Carry over | ||||
| -0.16 | -0.16 | -0.22 | +0.8 | +0.8 | Balance | |||
| -0.08 | +0.4 | -0.11 | +0.4 | Carry over | ||||
| -0.12 | -0.12 | -0.16 | +0.06 | +0.05 | Balance | |||
| 0 | +18.42 | -27.18 | -5.88 | +33.08 | -17.93 | +17.93 | +8.94 | Final moments |
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}} |
