Question 16.1: Determine the member end moments for the two-span continuous...
Determine the member end moments for the two-span continuous beam shown in Fig. 16.7(a) by using the momentdistribution method.
This beam was previously analyzed in Example 15.1 by the slope-deflection method.
Distribution Factors. Only joint B is free to rotate. The distribution factors at this joint are
DF _{B A}=\frac{K_{B A}}{K_{B A}+K_{B C}}=\frac{I / 25}{(I / 25)+(I / 30)}=0.545DF _{B C}=\frac{K_{B C}}{K_{B A}+K_{B C}}=\frac{I / 30}{(I / 25)+(I / 30)}=0.455
Note that the sum of the distribution factors at joint B is equal to 1; that is,
DF _{B A}+ DF _{B C}=0.545+0.455=1 Checks
The distribution factors are recorded in boxes beneath the corresponding member ends on top of the momentdistribution table, as shown in Fig. 16.7(a).
Fixed-End Moments. Assuming that joint B is clamped against rotation, we calculate the fixed-end moments due to the external loads by using the fixed-end moment expressions given inside the back cover of the book:
FEM _{A B}=\frac{18(10)(15)^{2}}{(25)^{2}}=64.8 k – ft \circlearrowleft \quad \text { or } \quad+64.8 k – ftFEM _{B A}=\frac{18(10)^{2}(15)}{(25)^{2}}=43.2 k – ft \circlearrowright \quad \text { or } \quad-43.2 k – ft
FEM _{B C}=\frac{2(30)^{2}}{12}=150 k – ft \circlearrowleft \quad \text { or } \quad+150 k – ft
FEM _{C B}=150 k – ft \circlearrowright \quad \text { or } \quad-150 k – ft
These fixed-end moments are recorded on the first line of the moment-distribution table, as shown in Fig. 16.7(a).
Moment Distribution. Since joint B is actually not clamped, we release the joint and determine the unbalanced moment acting on it by summing the moments at ends B of members AB and BC:
UM _{B}=-43.2+150=+106.8 k – ftThis unbalanced moment at joint B induces distributed moments at the ends B of members AB and BC, which can be determined by multiplying the negative of the unbalanced moment by the distribution factors:
DM _{B A}= DF _{B A}\left(- UM _{B}\right)=0.545(-106.8)=-58.2 k – ftDM _{B C}= DF _{B C}\left(- UM _{B}\right)=0.455(-106.8)=-48.6 k – ft
These distributed moments are recorded on line 2 of the moment-distribution table, and a line is drawn beneath them to indicate that joint B is now balanced. The carryover moments at the far ends A and C of members AB and BC, respectively, are then computed as
COM _{A B}=\frac{1}{2}\left( DM _{B A}\right)=\frac{1}{2}(-58.2)=-29.1 k – ft\operatorname{COM}_{C B}=\frac{1}{2}\left( DM _{B C}\right)=\frac{1}{2}(-48.6)=-24.3 k – ft
The carryover moments are recorded on the next line (line 3) of the moment-distribution table, with an inclined arrow pointing from each distributed moment to its carryover moment, as shown in Fig. 16.7(a).
Joint B is the only joint of the structure that is free to rotate, and because it has been balanced, we end the momentdistribution process.
Final Moments. The final member end moments are obtained by algebraically summing all the moments in each column of the moment-distribution table. The final moments thus obtained are recorded on the last line of the table in Fig. 16.7(a). Note that these final moments satisfy the equation of moment equilibrium at joint B. A positive answer for an end moment indicates that its sense is counterclockwise, whereas a negative answer for an end moment implies a clockwise sense. The final member end moments are depicted in Fig. 16.7(b).
The member end shears and support reactions can now be determined by considering the equilibrium of the members and joints of the continuous beam, as discussed in Example 15.1. The shear and bending moment diagrams of the beam were also constructed in Example 15.1.