Question 16.6: Determine the member end moments for the three-span continuo...

Determine the member end moments for the three-span continuous beam shown in Fig. 16.12(a) due to the uniformly distributed load and due to the support settlements of \frac{5}{8} in. at B, 1 \frac{1}{2} in. at C, and \frac{3}{4} in. at D. Use the moment-distribution method.

 

16.12
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This beam was previously analyzed in Example 15.6 by the slope-deflection method.

Distribution Factors. At joint A,

DF _{A B}=1

At joint B,

DF _{B A}=\frac{3 I / 80}{(3 I / 80)+(I / 20)}=0.429

 

DF _{B C}=\frac{I / 20}{(3 I / 80)+(I / 20)}=0.571

At joint C,

DF _{C B}=\frac{I / 20}{(3 I / 80)+(I / 20)}=0.571

 

DF _{C D}=\frac{3I / 80}{(3 I / 80)+(I / 20)}=0.429

At joint D,

DF _{D C}=1

Fixed-End Moments. A qualitative deflected shape of the continuous beam with all joints clamped against rotation and subjected to the specified support settlements is depicted in Fig. 16.12(b) using an exaggerated scale. It can be seen from this figure that the relative settlements for the three members are \Delta_{A B}=\frac{5}{8} \text { in., } \Delta_{B C}=1 \frac{1}{2}-\frac{5}{8}=\frac{7}{8} \text { in. }, and \Delta_{C D}=1 \frac{1}{2}-\frac{3}{4}=\frac{3}{4} in. By using the fixed-end-moment expressions, we determine the fixed-end moments due to the support settlements to be

FEM _{A B}= FEM _{B A}=+\frac{6 E I \Delta}{L^{2}}=+\frac{6(29,000)(7,800)\left(\frac{5}{8}\right)}{(20)^{2}(12)^{3}}

= +1,227.2 k-ft

FEM _{B C}= FEM _{C B}=+\frac{6(29,000)(7,800)\left(\frac{7}{8}\right)}{(20)^{2}(12)^{3}}=+1,718.1 k – ft

 

FEM _{C D}= FEM _{D C}=-\frac{6(29,000)(7,800)\left(\frac{3}{4}\right)}{(20)^{2}(12)^{3}}=-1,472.7 k – ft

The fixed-end moments due to the 2-k/ft external load are

FEM _{A B}= FEM _{B C}= FEM _{C D}=+\frac{2(20)^{2}}{12}=+66.7 k – ft

 

FEM _{B A}= FEM _{C B}= FEM _{D C}=-66.7 k – ft

Thus, the total fixed-end moments due to the combined e¤ect of the external load and the support settlements are

FEM _{A B}=+1,293.9 k – ft    \quad FEM _{B A}=+1,160.5 k – ft

 

FEM _{B C}=+1,784.8 k – ft    \quad FEM _{C B}=+1,651.4 k – ft

 

FEM _{C D}=-1,406 k – ft \quad FEM _{D C}=-1,539.4 k – ft

Moment Distribution. The moment distribution is carried out in the usual manner, as shown on the momentdistribution table in Fig. 16.12(c). Note that the joints A and D at the simple end supports are balanced only once and that no moments are carried over to these joints.

Final Moments. See the moment-distribution table and Fig. 16.12(d).

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