Question 16.8: Determine the member end moments for the frame shown in Fig....
Determine the member end moments for the frame shown in Fig. 16.15(a) by using the moment-distribution method.
This frame was analyzed in Example 15.10 by the slope-deflection method.
Distribution Factors. At joint C,
DF _{C A}= DF _{C D}=\frac{I / 7}{2(I / 7)}=0.5At joint D,
DF _{D C}=\frac{I / 7}{(I / 7)+(I / 5)}=0.417DF _{D B}=\frac{I / 5}{(I / 7)+(I / 5)}=0.583
DF _{D C}+ DF _{D B}=0.417+0.583=1 Checks
Part I: Sidesway Prevented. In the first part of the analysis, the sidesway of the frame is prevented by adding an imaginary roller at joint C, as shown in Fig. 16.15(b). Assuming that joints C and D of this frame are clamped against rotation, we calculate the fixed-end moments due to the external load to be
FEM _{C D}=+39.2 kN \cdot m \quad FEM _{D C}=-29.4 kN \cdot mFEM _{A C}= FEM _{C A}= FEM _{B D}= FEM _{D B}=0
The moment-distribution of these fixed-end moments is then performed, as shown on the moment-distribution table in Fig. 16.15(c), to determine the member end moments M_O in the frame with sidesway prevented.
To evaluate the restraining force R that develops at the imaginary roller support, we first calculate the shears at the lower ends of the columns AC and BD by considering the moment equilibrium of the free bodies of the columns shown in Fig. 16.15(d). Next, by considering the equilibrium of the horizontal forces acting on the entire frame (Fig. 16.15(e)), we determine the restraining force R to be
+\rightarrow \sum F_{X}=0 \quad R+5.14-7.2=0R=2.06 kN \rightarrow
Note that the restraining force acts to the right, indicating that if the roller would not have been in place, the frame would have swayed to the left.
Part II: Sidesway Permitted. Since the actual frame is not supported by a roller at joint C, we neutralize the e¤ect of the restraining force by applying a lateral load R = 2.06 kN in the opposite direction (i.e., to the left) to the frame, as shown in Fig. 16.15(f ). As discussed previously, since the moment-distribution method cannot be used directly to compute member end moments MR due to the lateral load R = 2.06 kN, we use an indirect approach in which the frame is subjected to an arbitrary known joint translation D0 caused by an unknown load Q acting at the location and in the direction of R, as shown in Fig. 16.15(g). Assuming that the joints C and D of the frame are clamped against rotation, as shown in Fig. 16.15(h), the fixed-end moments due to the translation Δ′ are given by
FEM _{A C}= FEM _{C A}=-\frac{6 E I \Delta^{\prime}}{(7)^{2}}=-\frac{6 E I \Delta^{\prime}}{49}FEM _{B D}= FEM _{D B}=-\frac{6 E I \Delta^{\prime}}{(5)^{2}}=-\frac{6 E I \Delta^{\prime}}{25}
FEM _{C D}= FEM _{D C}=0
in which negative signs have been assigned to the fixed-end moments for the columns, because these moments must act in the clockwise direction, as shown in Fig. 16.15(h).
Instead of arbitrarily assuming a numerical value for Δ′ to compute the fixed-end moments, it is usually more convenient to assume a numerical value for one of the fixed-end moments, evaluate Δ′ from the expression of that fixed-end moment, and use the value of Δ′ thus obtained to compute the remaining fixed-end moments. Thus, we arbitrarily assume the fixed-end moment FEM _{A C} to be – 50 kN m; that is,
FEM _{A C}= FEM _{C A}=-\frac{6 E I \Delta^{\prime}}{49}=-50 kN \cdot mBy solving for Δ′ , we obtain
\Delta^{\prime}=\frac{408.33}{E I}By substituting this value of Δ′ into the expressions for FEM_{BD} and FEM _{D B}, we determine the consistent values of these moments to be
FEM _{B D}= FEM _{D B}=-\frac{6(408.33)}{25}=-98 kN \cdot mThe foregoing fixed-end moments are then distributed by the usual moment-distribution process, as shown in Fig. 16.15(i), to determine the member end momentsM_{Q}caused by the yet-unknown load Q.
To evaluate the magnitude of Q that corresponds to these member end moments, we first calculate shears at the lower ends of the columns by considering their moment equilibrium (Fig. 16.15( j)) and then apply the equation of equilibrium in the horizontal direction to the entire frame:
+\rightarrow \sum F_{X}=0-Q+10.97+23.44=0
Q=34.41 kN \leftarrow
which indicates that the moments M_Q computed in Fig. 16.15(i) are caused by a lateral load Q = 34.41 kN. Since the
moments are linearly proportional to the magnitude of the load, the desired moments M_R due to the lateral load R = 2.06 kN must be equal to the moments M_Q (Fig. 16.15(i)) multiplied by the ratio R/Q = 2.06/34.41.
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. 16.15(c) and 2.06/34.41 times the member end moments M_Q computed in Fig. 16.15(i). Thus
M_{A C}=-12+\left(\frac{2.06}{34.41}\right)(-42.3)=-14.5 kN \cdot mM_{C A}=-24+\left(\frac{2.06}{34.41}\right)(-34.5)=-26.1 kN \cdot m
M_{C D}=23.9+\left(\frac{2.06}{34.41}\right)(34.3)=26 kN \cdot m
M_{D C}=-24+\left(\frac{2.06}{34.41}\right)(45.4)=-21.3 kN \cdot m
M_{D B}=24+\left(\frac{2.06}{34.41}\right)(-45.4)=21.3 kN \cdot m
M_{B D}=12+\left(\frac{2.06}{34.41}\right)(-71.8)=7.7 kN \cdot m
These moments are depicted in Fig. 16.15(k).