Question 16.10: Determine the member end moments and reactions for the frame...
Determine the member end moments and reactions for the frame shown in Fig. 16.17(a) by using the momentdistribution method.
Distribution Factors. At joint C,
DF _{C A}= DF _{C D}=\frac{I / 20}{2(I / 20)}=0.5At joint D,
DF _{D C}=\frac{I / 20}{(I / 20)+(3 / 4)(I / 14.42)}=0.49DF _{D B}=\frac{(3 / 4)(I / 14.42)}{(I / 20)+(3 / 4)(I / 14.42)}=0.51
Member End Moments Due to an Arbitrary Sidesway Δ′. Since no external loads are applied to the members of the frame, the member end moments M_O in the frame restrained against sidesway will be zero. To determine the member end moments M due to the 30-k lateral load, we subject the frame to an arbitrary known horizontal translation Δ′ at joint C. Figure 16.17(b) shows a qualitative deflected shape of the frame with all joints clamped against rotation and subjected to the horizontal displacement Δ′ at joint C. The procedure for constructing such deflected shapes was discussed in Section 15.5. Note that, since the frame members are assumed to be inextensible and deformations are assumed to be small, an end of a member can translate only in the direction perpendicular to the member. From this figure, we can see that the relative translation \Delta_{A C} between the ends of member AC in the direction perpendicular to the member can be expressed in terms of the joint translation Δ′ as
\Delta_{A C}=C C^{\prime}=\frac{5}{4} \Delta^{\prime}=1.25 \Delta^{\prime}Similarly, the relative translations for members CD and BD are given by
\Delta_{C D}=D_{1} D^{\prime}=\frac{2}{3} \Delta^{\prime}+\frac{3}{4} \Delta^{\prime}=1.417 \Delta^{\prime}\Delta_{B D}=D D^{\prime}=\frac{\sqrt{13}}{3} \Delta^{\prime}=1.202 \Delta^{\prime}
The fixed-end moments due to the relative translations are
FEM _{A C}= FEM _{C A}=\frac{6 E I\left(1.25 \Delta^{\prime}\right)}{(20)^{2}}FEM _{C D}= FEM _{D C}=-\frac{6 E I\left(1.417 \Delta^{\prime}\right)}{(20)^{2}}
FEM _{B D}= FEM _{D B}=\frac{6 E I\left(1.202 \Delta^{\prime}\right)}{(14.42)^{2}}
in which, as shown in Fig. 16.17(b), the fixed-end moments for members AC and BD are counterclockwise (positive), whereas those for member CD are clockwise (negative). If we arbitrarily assume that
FEM _{B D}= FEM _{D B}=\frac{6 E I\left(1.202 \Delta^{\prime}\right)}{(14.42)^{2}}=100 k – ftthen
E I \Delta^{\prime}=2,883.2and, therefore,
FEM _{A C}= FEM _{C A}=54.1 k – ftFEM _{C D}= FEM _{D C}=-61.3 k – ft
These fixed-end moments are distributed by the moment-distribution process, as shown in Fig. 16.17(c), to determine
the member end moments M_Q.
To determine the magnitude of the load Q that corresponds to the member end moments computed in Fig. 16.17(c), we first calculate the shears at the ends of the girder CD by considering the moment equilibrium of the free body of the girder shown in Fig. 16.17(d). The girder shears (5.58 k) thus obtained are then applied to the free bodies of the inclined members AC and BD, as shown in the figure. Next, we apply the equations of moment equilibrium to members AC and BD to calculate the horizontal forces at the lower ends of these members. The magnitude of Q can now be determined by considering the equilibrium of horizontal forces acting on the entire frame as (see Fig. 16.17(d))
+\rightarrow \sum F_{x}=0Q-11.17-8.32=0
Q=19.49 k \rightarrow
Actual Member End Moments. The actual member end moments, M, due to the 30-k lateral load can now be evaluated by multiplying the moments M_Q computed in Fig. 16.17(c) by the ratio 30/Q =30/19.49.
M_{A C}=\frac{30}{19.49}(55.3)=85.1 k – ftM_{C A}=\frac{30}{19.49}(56.5)=87 k – ft
M_{C D}=\frac{30}{19.49}(-56.4)=-86.8 k – ft
M_{D C}=\frac{30}{19.49}(-55.2)=-85 k – ft
M_{D B}=\frac{30}{19.49}(55.2)=85 k – ft
M_{B D}=0
Member End Forces. See Fig. 16.17(e).
Support Reactions. See Fig. 16.17(f ).
Equilibrium Check. The equilibrium equations check.