Question 9.SP.14: For the uniform beam and loading shown, determine the reacti......
For the uniform beam and loading shown, determine the reaction at B.
STRATEGY: Applying the superposition concept, you can model this statically indeterminate problem as a summation of the displacements for the given load and the redundant load cases. The redundant reaction can then be found by noting that a displacement associated with the two cases must be consistent with the geometry of the original beam.
MODELING: The beam is indeterminate to the first degree. Referring to Fig. 1, the reaction R _B is chosen as redundant, and the distributed load and redundant reaction load are considered separately. Next the tangent at A is selected as the reference tangent. From the similar triangles ABB′ and ACC′,
\frac{t_{C / A}}{L}=\frac{t_{B / A}}{\frac{2}{3} L} \quad t_{C / A}=\frac{3}{2} t_{B / A} (1)
For each loading, we draw the M/EI diagram and then determine the tangential deviations of B and C with respect to A.
ANALYSIS:
Distributed Loading (Fig. 2). Considering the M/EI diagram from end A to an arbitrary point X,
\left(t_{X / A}\right)_w=A_1 \frac{x}{3}+A_2 \frac{x}{4}=\left(\frac{1}{2} \frac{w L x}{2 E I} x\right) \frac{x}{3}+\left(-\frac{1}{3} \frac{w x^2}{2 E I} x\right) \frac{x}{4}=\frac{w x^3}{24 E I}(2 L-x)
Letting x = L and x=\frac{2}{3} L ,
\left(t_{C / A}\right)_w=\frac{w L^4}{24 E I} \quad\left(t_{B / A}\right)_w=\frac{4}{243} \frac{w L^4}{E I}
Redundant Reaction Loading (Fig. 3).
\begin{gathered} \left(t_{C / A}\right)_R=A_3 \frac{L}{9}+A_4 \frac{L}{3}=\left(\frac{1}{2} \frac{R_B L}{3 E I} \frac{L}{3}\right) \frac{L}{9}+\left(-\frac{1}{2} \frac{R_B L}{3 E I} L\right) \frac{L}{3}=-\frac{4}{81} \frac{R_B L^3}{E I} \\ \left(t_{B / A}\right)_R=A_5 \frac{2 L}{9}=\left[-\frac{1}{2} \frac{2 R_B L}{9 E I}\left(\frac{2 L}{3}\right)\right] \frac{2 L}{9}=-\frac{4}{243} \frac{R_B L^3}{E I} \end{gathered}
Combined Loading. Adding the results gives
t_{C / A}=\frac{w L^4}{24 E I}-\frac{4}{81} \frac{R_B L^3}{E I} \quad t_{B / A}=\frac{4}{243} \frac{\left(w L^4-R_B L^3\right)}{E I}
Reaction at B. Substituting for t_{C / A} \text { and } t_{B / A} into Eq. (1),
\begin{aligned} \left(\frac{w L^4}{24 E I}-\frac{4}{81} \frac{R_B L^3}{E I}\right) & =\frac{3}{2}\left[\frac{4}{243} \frac{\left(w L^4-R_B L^3\right)}{E I}\right] \\ R_B & =0.6875 w L \quad R_B=0.688 w L \uparrow \end{aligned}
REFLECT and THINK: Note that an alternate strategy would be to determine the deflections at B for the given load and the redundant reaction and to set the sum equal to zero.
