Question 16.2: Determine the member end moments for the three-span continuo...
Determine the member end moments for the three-span continuous beam shown in Fig. 16.8(a) by the momentdistribution method.
This beam was analyzed previously in Example 15.2 by using the slope-deflection method.
Distribution Factors. From Fig. 16.8(a), we can see that joints B and C of the beam are free to rotate. The distribution factors at joint B are
DF _{B A}=\frac{K_{B A}}{K_{B A}+K_{B C}}=\frac{I / 18}{(I / 18)+(I / 18)}=0.5DF _{B C}=\frac{K_{B C}}{K_{B A}+K_{B C}}=\frac{I / 18}{(I / 18)+(I / 18)}=0.5
Similarly, at joint C,
DF _{C B}=\frac{K_{C B}}{K_{C B}+K_{C D}}=\frac{I / 18}{(I / 18)+(I / 18)}=0.5DF _{C D}=\frac{K_{C D}}{K_{C B}+K_{C D}}=\frac{I / 18}{(I / 18)+(I / 18)}=0.5
Fixed-End Moments.
FEM _{A B}=+\frac{3(18)^{2}}{30}=+32.4 k – ftFEM _{B A}=-\frac{3(18)^{2}}{20}=-48.6 k – ft
FEM _{B C}=+\frac{3(18)^{2}}{12}=+81 k – ft
FEM _{C B}=-81 k – ft
FEM _{C D}=+\frac{3(18)^{2}}{20}=+48.6 k – ft
FEM _{D C}=-\frac{3(18)^{2}}{30}=-32.4 k – ft
Moment Distribution. After recording the distribution factors and the fixed-end moments in the moment-distribution table shown in Fig. 16.8(a), we begin the moment-distribution process by balancing joints B and C. The unbalanced moment at joint B is equal to – 48.6 + 81 = +32.4 k-ft. Thus, the distributed moments at the ends B of members AB and BC are
DM _{B A}= DF _{B A}\left(- UM _{B}\right)=0.5(-32.4)=-16.2 k – ftDM _{B C}= DF _{B C}\left(- UM _{B}\right)=0.5(-32.4)=-16.2 k – ft
Similarly, noting that the unbalanced moment at joint C equals – 81 + 48.6 = – 32.4 k-ft, we determine the distributed moments at the ends C of members BC and CD to be
DM _{C B}= DF _{C B}\left(- UM _{C}\right)=0.5(+32.4)=+16.2 k – ftDM _{C D}= DF _{C D}\left(- UM _{C}\right)=0.5(+32.4)=+16.2 k – ft
One-half of these distributed moments are then carried over to the far ends of the members, as shown on the third line of the moment-distribution table in Fig. 16.8(a). This process is repeated, as shown in the figure, until the unbalanced moments are negligibly small.
Final Moments. The final member end moments, obtained by summing the moments in each column of the momentdistribution table, are recorded on the last line of the table in Fig. 16.8(a). These moments are depicted in Fig. 16.8(b).
The member end shears, support reactions, and shear and bending moment diagrams of the beam were determined in Example 15.2.