Question 9.S-P.8: 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.
Principle of Superposition. The reaction R_B is designated as redundant and considered as an unknown load. The deflections due to the distributed load and to the reaction R_B are considered separately as shown below.
For each loading the deflection at point B is found by using the table of Beam Deflections and Slopes in Appendix D.
| Beam Deflections and Slopes | ||||
| Beam and Loading | Elastic Curve | Maximum Deflection | Slope at End | Equation of Elastic Curve |
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-\frac{P L^{3}}{3 E I} | -\frac{P L^{2}}{2 E I} | y=\frac{P}{6 EI}\left(x^{3}-3 L x^{2}\right) |
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-\frac{w L^{4}}{8 E I} | -\frac{w L^{3}}{6 E I} | y=-\frac{w}{24 E I}\left(x^{4}-4 L x^{3}+6 L^{2} x^{2}\right) |
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-\frac{M L^{2}}{2 E I} | -\frac{M L}{E I} | y=-\frac{M}{2 E I} x^{2} |
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-\frac{P L^{3}}{48 E I} | \pm \frac{P L^{2}}{16 E I} | For x \leq \frac{1}{2} L :
y=\frac{P}{48 E I}\left(4 x^{3}-3 L^{2} x\right) |
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For a > b:
-\frac{P b\left(L^{2}-b^{2}\right)^{3 / 2}}{9 \sqrt{3} E I L} at x_{m}=\sqrt{\frac{L^{2}-b^{2}}{3}} |
u _{A}=-\frac{P b\left(L^{2}-b^{2}\right)}{6 E I L}
u_{B}=+\frac{P a\left(L^{2}-a^{2}\right)}{6 E I L} |
For x < a:
y=\frac{P b}{6 E I L}\left[x^{3}-\left(L^{2}-b^{2}\right) x\right] For x = a: y=-\frac{P a^{2} b^{2}}{3 E I L} |
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-\frac{5 w L^{4}}{384 E I} | \pm \frac{w L^{3}}{24 E I} | y=-\frac{w}{24 E I}\left(x^{4}-2 L x^{3}+L^{3} x\right) |
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\frac{M L^{2}}{9 \sqrt{3} E I} | u _{A}=+\frac{M L}{6 E I}
u _{B}=-\frac{M L}{3 E I} |
y=-\frac{M}{6 E I L}\left(x^{3}-L^{2} x\right) |
Distributed Loading. We use case 6, Appendix D
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 D, with a=\frac{2}{3} L and b=\frac{1}{3} L, we have
\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, we write
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} R _{B}=0.688 w L \uparrow
Since the reaction R_B is now known, we may use the methods of statics to determine the other reactions:
R _{A}=0.271 w L \uparrow R _{C}=0.0413 w L \uparrow
b. Slope at End A. Referring again to Appendix D, we have
Distributed Loading. \left( u _{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( u _{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] \left( u _{A}\right)_{R}=0.03398 \frac{w L^{3}}{E I}
Finally, u _{A}=\left( u _{A}\right)_{w}+\left( u _{A}\right)_{R}
u _{A}=-0.04167 \frac{w L^{3}}{E I}+0.03398 \frac{w L^{3}}{E I}=-0.00769 \frac{w L^{3}}{E I} u_{A}=0.00769 \frac{w L^{3}}{E I} c
