Question 15.S-P.6: 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 App. C.
Load P. We note that, for this loading, portion BC of the beam is straight.
(θ_{C})_{P} = (θ_{B})_{P} = – \frac{Pa^{2}}{2EI} (y_{C})_{P} = (y_{B})_{P} + (θ_{B})_{P}b= – \frac{Pa^{3}}{3EI} – \frac{Pa^{2}}{2EI} b = – \frac{Pa^{2}}{6EI} (2a + 3b)
Force R_{C} (θ_{C})_{R} = + \frac{R_{C} L^{2}}{2EI} (y_{C})_{R} = + \frac{R_{C} L^{3}}{3EI}
Couple M_{C} (θ_{C})_{M} = + \frac{M_{C} L}{EI} (y_{C})_{M} = + \frac{M_{C }L^{2}}{2EI}
Boundary Conditions. At end C the slope and deflection must be zero:
[ x = L, θ_{C} = 0 ]: θ_{C} = (θ_{C})_{P} + (θ_{C})_{R} + (θ_{C})_{M}0 = – \frac{Pa^{2}}{2EI} + \frac{R_{C} L^{2}}{2EI} + \frac{M_{C} L}{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_{C} L^{3}}{3EI} + \frac{M_{C} L^{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^{2}b}{L^{2}} M_{C} =\frac{Pa^{2}b}{L^{2}} \curvearrowright
Using the methods of statics, we can now determine the reaction at A.
