Question 16.3: Determine the reactions and draw the shear and bending momen...
Determine the reactions and draw the shear and bending moment diagrams for the two-span continuous beam shown in Fig. 16.9(a) by using the moment-distribution method.
Distribution Factors. From Fig. 16.9(a), we can see that joints B and C of the continuous 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{1.5 I / 10}{(1.5 I / 10)+(I / 10)}=0.6DF _{B C}=\frac{K_{B C}}{K_{B A}+K_{B C}}=\frac{I / 10}{(1.5 I / 10)+(I / 10)}=0.4
Similarly, at joint C,
DF _{C B}=\frac{K_{C B}}{K_{C B}}=\frac{0.1 I}{0.1 I}=1Fixed-End Moments.
FEM _{A B}=+\frac{80(10)}{8}=+100 kN \cdot mFEM _{B A}=-100 kN \cdot m
FEM _{B C}=+\frac{40(10)}{8}=+50 kN \cdot m
FEM _{C B}=-50 kN \cdot m
Moment Distribution. After recording the distribution factors and the fixed-end moments in the moment-distribution table shown in Fig. 16.9(b), we begin the moment-distribution process by balancing joints B and C. The unbalanced moment at joint B is equal to – 100 + 50 = – 50 kN m. Thus the distributed moments at the ends B of members AB and BC are
DM _{B A}= DF _{B A}\left(- UM _{B}\right)=0.6(+50)=+30 kN \cdot mDM _{B C}= DF _{B C}\left(- UM _{B}\right)=0.4(+50)=+20 kN \cdot m
Similarly, noting that the unbalanced moment at joint C is 50 kN m, we determine the distributed moment at end C of member BC to be
DM _{C B}= DF _{C B}\left(- UM _{C}\right)=1(+50)=+50 kN \cdot mOne-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. 16.9(b). 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 moment-distribution table, are recorded on the last line of the table in Fig. 16.9(b).
Alternative Method. Because the end support C of the continuous beam is a simple support, the analysis can be simplified by using the reduced relative bending sti¤ness for member BC, which is adjacent to the simple support C:
K_{B C}=\frac{3}{4}\left(\frac{I}{10}\right)Note that the relative bending sti¤ness of member AB remains the same as before. The distribution factors at joint B are now given by
DF _{B A}=\frac{K_{B A}}{K_{B A}+K_{B C}}=\frac{1.5 I / 10}{(1.5 I / 10)+(3 I / 40)}=\frac{2}{3}DF _{B C}=\frac{K_{B C}}{K_{B A}+K_{B C}}=\frac{3 I / 40}{(1.5 I / 10)+(3 I / 40)}=\frac{1}{3}
At joint C, DF _{C B}=K_{C B} / K_{C B}=1 These distribution factors, and the fixed-end moments that remain the same as before, are recorded in the moment-distribution table, as shown in Fig. 16.9(c).
Since we are using the reduced relative bending sti¤ness for member BC, joint C needs to be balanced only once in the moment-distribution process. Thus joints B and C are balanced and the distributed moments are computed in the usual manner, as indicated on the second line of the moment-distribution table (Fig. 16.9(c)). However, as shown on the third line of the table in Fig. 16.9(c), no moment is carried over to end C of member BC. Joint B is balanced once more, and the moment is carried over to the end A of member AB (lines 4 and 5). Because both joints B and C are now balanced, we can end the moment-distribution process and determine the final moments by summing the moments in each
column of the moment-distribution table.
Member End Shears. The member end shears, obtained by considering the equilibrium of each member, are shown in
Fig. 16.9(d).
Support Reactions. See Fig. 16.9(e).
Shear and Bending Moment Diagrams. See Fig. 16.9(f ) and (g).