Question 15.7: Determine the reactions at the supports for the prismatic be...
Determine the reactions at the supports for the prismatic beam and loading shown in Fig. 15.29. (This is the same beam and loading as in Example 15.5 of Sec. 15.5.)
We consider the reaction at B as redundant and release the beam from the support. The reaction R_{B} is now considered as an unknown load (Fig. 15.30a) and will be determined from the condition that the deflection of the beam at B must be zero. The solution is carried out by considering separately the deflection (y_{B})_{w} caused at B by the uniformly distributed load ω (Fig. 15.30b) and the deflection (y_{B})_{R} produced at the same point by the redundant reaction R_{B} (Fig. 15.30c).
From the table of App. C (cases 2 and 1), we find that
(y_{B})_{w} = – \frac{wL^{4}}{8EI} (y_{B})_{R} = + \frac{R_{B}L^{3}}{3EI}
Writing that the deflection at B is the sum of these two quantities and that it must be zero, we have
y_{B} = (y_{B})_{w }+ (y_{B})_{R} = 0
y_{B} = – \frac{wL^{4}}{8EI} + \frac{R_{B}L^{3}}{3EI}= 0
and, solving for R_{B}, R_{B}= \frac{3}{8} wL R_{B} = \frac{3}{8}wL ↑
Drawing the free-body diagram of the beam (Fig. 15.31) and writing the corresponding equilibrium equations, we have
+↑ \sum{F_{y}} = 0: R_{A} + R_{B} – wL = 0 (15.46)
R_{A} = wL – R_{B} = wL – \frac{3}{8} wL = \frac{5}{8} wL
R_{A} = \frac{5}{8} wL ↑
+\curvearrowleft \sum{M_{A}} = 0: M_{A} + R_{B}L – (wL) (\frac{1}{2} L) = 0 (15.47)
M_{A} = \frac{1}{2} wL^{2} – R_{B}L = \frac{1}{2} wL^{2} – \frac{3}{8} wL^{2} = \frac{1}{8}wL^{2}
M_{A} = \frac{1}{8} wL^{2} \curvearrowleft
Alternative Solution. We may consider the couple exerted at the fixed end A as redundant and replace the fixed end by a pin-and-bracket support. The couple M_{A} is now considered as an unknown load (Fig. 15.32a) and will be determined from the condition that the slope of the beam at A must be zero. The solution is carried out by considering separately the slope (θ_{A})_{w} caused at A by the uniformity distributed load w (Fig. 15.32b) and the slope (θ_{A})_{M} produced at the same point by the unknown couple M_{A} (Fig. 15.32c).
Using the table of App. C (cases 6 and 7), and noting that in case 7, A and B must be interchanged, we find that
(θ_{A})_{w} = – \frac{wL^{3}}{24EI} (θ_{A})_{M} = \frac{M_{A}L}{3EI}
Writing that the slope at A is the sum of these two quantities and that it must be zero, we have
θ_{A} = (θ_{A})_{w} + (θ_{A})_{M} = 0
θ_{A} =- \frac{wL^{3}}{25EI}+ \frac{M_{A}L}{3EI} = 0
and, solving for M_{A},
M_{A} = \frac{1}{8} wL^{2} M_{A} = \frac{1}{8} wL^{2} \curvearrowleft
The values of R_{A} and R_{B} may then be found from the equilibrium equations (15.46) and (15.47).
