Question 16.7: Determine the member end moments for the frame shown in Fig....
Determine the member end moments for the frame shown in Fig. 16.13(a) by using the moment-distribution method.
This frame was analyzed in Example 15.8 by the slope-deflection method.
Distribution Factors. At joint C,
DF _{C A}=\frac{\left(\frac{800}{20}\right)}{\left(\frac{800}{20}\right)+\left(\frac{1600}{30}\right)}=0.429 \quad DF _{C D}=\frac{\left(\frac{1600}{30}\right)}{\left(\frac{800}{20}\right)+\left(\frac{1600}{30}\right)}=0.571
DF _{C A}+ DF _{C D}=0.429+0.571=1 Checks
At joint D,
DF _{D B}=\frac{\left(\frac{800}{20}\right)}{\left(\frac{800}{20}\right)+\left(\frac{1600}{30}\right)+\left(\frac{3}{4}\right)\left(\frac{1600}{30}\right)}=0.3DF _{D C}=\frac{\left(\frac{1600}{30}\right)}{\left(\frac{800}{20}\right)+\left(\frac{1600}{30}\right)+\left(\frac{3}{4}\right)\left(\frac{1600}{30}\right)}=0.4
DF _{D E}=\frac{\left(\frac{3}{4}\right)\left(\frac{1600}{30}\right)}{\left(\frac{800}{20}\right)+\left(\frac{1600}{30}\right)+\left(\frac{3}{4}\right)\left(\frac{1600}{30}\right)}=0.3
DF _{D B}+ DF _{D E}+ DF _{D C}=2(0.3)+0.4=1 Checks
At joint E,
DF _{E D}=1Fixed-End Moments. By using the fixed-end moment expressions, we obtain
FEM _{A C}=+100 k – ft FEM _{C A}=-100 k – ft
FEM _{B D}= FEM _{D B}=0
FEM _{C D}= FEM _{D E}=+150 k – ft \quad FEM _{D C}= FEM _{E D}=-150 k – ft
Moment Distribution. The moment-distribution process is carried out in tabular form, as shown in Fig. 16.13(b). The table,which is similar in form to those used previously for the analysis of continuous beams, contains one column for each member end of the structure. Note that the columns for all member ends, which are connected to the same joint, are grouped together, so that any unbalanced moment at the joint can be conveniently distributed among the members connected to it. Also, when the columns for two ends of a member cannot be located adjacent to each other, then an overhead arrow connecting the columns for the member ends may serve as a reminder to carry over moments from one end of the member to the other. In Fig. 16.13(b), such an arrow is used between the columns for the ends of member BD. This arrow indicates that a distributed moment at end D of member BD induces a carryover moment at the far end B. Note, however, that no moment can be carried over from end B to end D of member BD, because joint B, which is at a fixed support, will not be released during the moment-distribution process.
The moment distribution is carried out in the same manner as discussed previously for continuous beams. Note that any unbalanced moment at joint D must be distributed to the ends D of the three members connected to it in accordance with their distribution factors.
Final Moments. The final member end moments are obtained by summing all the moments in each column of the moment-distribution table. Note that the final moments, which are recorded on the last line of the moment-distribution table and are depicted in Fig. 16.13(c), satisfy the equations of moment equilibrium at joints C and D of the frame.