Question 15.SP.5: For the uniform beam and loading shown, determine (a) the re...
For the uniform beam and loading shown, determine (a) the reaction at each support, (b) the slope at end A.
STRATEGY: The beam is statically indeterminate to the first degree. Strategically selecting the reaction at B as the redundant, you can use the method of superposition to model the given problem by using a summation of load cases for which deflection formulae are readily available.
MODELING: The reaction \pmb{R}_B is selected as redundant and considered as an unknown load. Applying the principle of superposition, the deflections due to the distributed load and to the reaction \pmb{R}_B are considered separately as shown in Fig. 1.
ANALYSIS: For each loading case, the deflection at point B is found by using the table of Beam Deflections and Slopes in Appendix E.
Distributed Loading. Use case 6, Appendix E:
y=-\frac{w}{24 E I}\left(x^4-2 L x^3+L^3 x\right)At point B, x=\frac{2}{3} L:
\left(y_B\right)_w=-\frac{w}{24 E I}\left[\left(\frac{2}{3} L\right)^4-2 L\left(\frac{2}{3} L\right)^3+L^3\left(\frac{2}{3} L\right)\right]=-0.01132 \frac{w L^4}{E I}Redundant Reaction Loading. From case 5, Appendix E, with a=\frac{2}{3} L and b=\frac{1}{3} L,
\left(y_B\right)_R=-\frac{P a^2 b^2}{3 E I L}=+\frac{R_B}{3 E I L}\left(\frac{2}{3} L\right)^2\left(\frac{L}{3}\right)^2=0.01646 \frac{R_B L^3}{E I}a. Reactions at Supports. Recalling that y_B=0,
\begin{aligned}y_B &=\left(y_B\right)_w+\left(y_B\right)_R \\0 &=-0.01132 \frac{w L^4}{E I}+0.01646 \frac{R_B L^3}{E I} \quad \quad \pmb{R}_B=0.688 w L \uparrow\end{aligned}Because the reaction R_B is now known, use the methods of statics to determine the other reactions (Fig. 2):
\pmb{R}_A=0.271 w L \uparrow \quad \pmb{R}_C=0.0413 w L \uparrowb. Slope at End A. Referring again to Appendix E,
Distributed Loading. \left(\theta_A\right)_w=-\frac{w L^3}{24 E I}=-0.04167 \frac{w L^3}{E I}
Redundant Reaction Loading. For P=-R_B=-0.688 w L and b=\frac{1}{3} L,
\left(\theta_A\right)_R=-\frac{P b\left(L^2-b^2\right)}{6 E I L}=+\frac{0.688 w L}{6 E I L}\left(\frac{L}{3}\right)\left[L^2-\left(\frac{L}{3}\right)^2\right]=0.03398 \frac{w L^3}{E I}Finally, \theta_A=\left(\theta_A\right)_w+\left(\theta_A\right)_R
=-0.04167 \frac{w L^3}{E I}+0.03398 \frac{w L^3}{E I}=-0.00769 \frac{w L^3}{E I}\theta_A=0.00769 \frac{w L^3}{E I} ⦪
