Question 16.9: Determine the reactions for the nonprismatic beam shown in F...
Determine the reactions for the nonprismatic beam shown in Fig. 16.16(a) by using the moment-distribution method.
Since the sti¤ness and carryover relationships derived in Section 16.1 as well as the expressions of fixed-end moments given inside the back cover of the book are valid only for prismatic members, we will analyze the given nonprismatic beam as if it were composed of two prismatic members, AB and BC, rigidly connected at joint B. Note that joint B is free to rotate as well as translate in the vertical direction, as shown in Fig. 16.16(a).
Distribution Factors. The distribution factors at joint B are
DF _{B A}=\frac{I / 30}{(I / 30)+(2 I / 18)}=0.231DF _{B C}=\frac{2 I / 18}{(I / 30)+(2 I / 18)}=0.769
Part I: Joint Translation Prevented. In this part of the analysis, the translation of joint B is prevented by an imaginary roller, as shown in Fig. 16.16(b). The fixed-end moments due to the external load are
FEM _{A B}=+150 k – ft \quad FEM _{B A}=-150 k – ftFEM _{B C}=+54 k – ft \quad FEM _{C B}=-54 k – ft
The moment distribution of these fixed-end moments is performed, as shown in Fig. 16.16(b), to determine the member end moments MO. The restraining force R at the imaginary roller support is then evaluated by considering the equilibrium of members AB and BC and of joint B as shown in Fig. 16.16(c). The restraining force is found to be
R=53.04 k \uparrowPart II: Joint Translation Permitted. Since the actual beam is not supported by a roller at joint B, we neutralize its restraining
e¤ect by applying a downward load R = 53.04 k to the beam, as shown in Fig. 16.16(d). To determine the member end moments M_R due to R, we subject the beam to an arbitrary known translation Δ′, as shown in Fig. 16.16(e). The fixed-end moments due to Δ′ are given by (see Fig. 16.16(f ))
FEM _{B C}= FEM _{C B}=-\frac{6 E(2 I) \Delta^{\prime}}{(18)^{2}}=-\frac{E I \Delta^{\prime}}{27}
If we arbitrarily assume that
FEM _{B C}= FEM _{C B}=-\frac{E I \Delta^{\prime}}{27}=-100 k – ftthen
E I \Delta^{\prime}=2,700and, therefore,
FEM _{A B}= FEM _{B A}=\frac{2,700}{150}=18 k – ftThese fixed-end moments are distributed by the moment-distribution process, as shown in Fig. 16.16(g), to determine the member end moments M_Q. The load Q at the location and in the direction of R that corresponds to these moments
can now be evaluated by considering equilibrium of members AB and BC and of joint B, as shown in Fig. 16.16(h). The magnitude of Q is found to be
Thus, the desired moments MR due to the vertical load R = 53.04 k (Fig. 16.16(d)) must be equal to the moments M_Q
(Fig. 16.16(g)) multiplied by the ratio R/Q = 53.04/8 = 6.63.
Actual Member End Moments. The actual member end moments, M, can now be determined by algebraically summing the member end moments M_O computed in Fig. 16.16(b) and 6.63 times the member end moments M_Q computed in
Fig. 16.16(g).
M_{B A}=-127.8+6.63(36.9)=116.8 k – ft
M_{B C}=127.8+6.63(-36.9)=-116.8 k – ft
M_{C B}=-17.1+6.63(-68.5)=-471.2 k – ft
The member end shears obtained by applying equations of equilibrium are shown in Fig. 16.16(i).
Support Reactions. See Fig. 16.16( j).
Equilibrium Check. The equilibrium equations check.