Question 16.5: Determine the member end moments for the continuous beam sho...
Determine the member end moments for the continuous beam shown in Fig. 16.11(a) due to a settlement of 20 mm at support B. Use the moment-distribution method.
This beam was analyzed previously in Example 15.5 by using the slope-deflection method.
Distribution Factors. At joint B,
DF_{BA} = \frac{I/8}{(I/8)+(I/8)} = 0.5DF_{BC} = \frac{I/8}{(I/8)+(I/8)} = 0.5
At joint C,
DF_{CB} = \frac{I/8}{(I/8)+(I/8)} = 0.5DF_{CD} = \frac{I/8}{(I/8)+(I/8)} = 0.5
Fixed-End Moments. A qualitative deflected shape of the continuous beam with all joints clamped against rotation and subjected to the specified support settlement is depicted in Fig. 16.11(b) using an exaggerated scale. It can be seen from this figure that the relative settlements for the three members are \Delta_{A B}=\Delta_{B C}=0.02 m, and \Delta_{C D}=0
By using the fixed-end moment expressions, we determine the fixed-end moments due to the support settlement to be
FEM _{A B}= FEM _{B A}=+\frac{6 E I \Delta}{L^{2}}=+\frac{6(70)(800)(0.02)}{(8)^{2}}=+105 kN \cdot mFEM _{B C}= FEM _{C B}=-\frac{6 E I \Delta}{L^{2}}=-\frac{6(70)(800)(0.02)}{(8)^{2}}=-105 kN \cdot m
FEM _{C D}= FEM _{D C}=0
Moment Distribution. The moment distribution is carried out in the usual manner, as shown on the moment-distribution table in Fig. 16.11(c).
Final Moments. See the moment-distribution table and Fig. 16.11(d).