Question 9.S-P.9: For the beam and loading shown, determine the reaction at th...
For the beam and loading shown, determine the reaction at the fixed support C.
Principle of Superposition. Assuming the axial force in the beam to be zero, the beam ABC is indeterminate to the second degree and we choose two reaction components as redundant, namely, the vertical force R_C and the couple M_C. The deformations caused by the given load P, the force R_C, and the couple M_C will be considered separately as shown.
For each load, the slope and deflection at point C will be 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) |
Load P. We note that, for this loading, portion BC of the beam is straight.
\left( u _{C}\right)_{P}=\left( u _{B}\right)_{P}=-\frac{P a^{2}}{2 E I}\begin{aligned}\left(y_{C}\right)_{P}&=\left(y_{B}\right)_{P}+\left( u _{B}\right)_{p} b \\&=-\frac{P a^{3}}{3 E I}-\frac{P a^{2}}{2 E I} b=-\frac{P a^{2}}{6 E I}(2 a+3 b)\end{aligned}
Force R_C \left( u_{C}\right)_{R}=+\frac{R_{C} L^{2}}{2 E I} \left(y_{C}\right)_{R}=+\frac{R_{C} L^{3}}{3 E I}
Couple M_{C} \left( u _{C}\right)_{M}=+\frac{M_{C} L}{E I} \left(y_{C}\right)_{M}=+\frac{M_{C} L^{2}}{2 E I}
Boundary Conditions. At end C the slope and deflection must be zero:
[x = L, u_C = 0] : u_C = (u_C)_P + (u_C)_R + (u_C)_M
0 = -\frac{Pa^2}{2EI} + \frac{R_CL^2}{2EI} + \frac{M_CL}{EI} (1)
[x = L, y_C = 0] : y_C = (y_C)_P + (y_C)_R + (y_C)_M
0 = -\frac{Pa^2}{6EI} (2a + 3b) + \frac{R_CL^3}{3EI} + \frac{M_CL^2}{2EI} (2)
Reaction Components at C. Solving simultaneously Eqs. (1) and (2), we find after reductions
R_C = +\frac{Pa^2}{L^3}(a + 3b) R_C = \frac{Pa^2}{L^3}(a + 3b)↑
M_C = – \frac{Pa^2b}{L^2} M_C = \frac{Pa^2b}{L^2} \curvearrowright
Using the methods of statics, we can now determine the reaction at A.
