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.)

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Accepted Answer

We consider the reaction at B as redundant and release the beam from the support. The reaction [latex]R_{B}[/latex] 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 [latex](y_{B})_{w}[/latex] caused at B by the uniformly distributed load ω (Fig. 15.30b) and the deflection [latex](y_{B})_{R}[/latex] produced at the same point by the redundant reaction [latex]R_{B}[/latex] (Fig. 15.30c).

From the table of App. C (cases 2 and 1), we find that

[latex](y_{B})_{w} = - \frac{wL^{4}}{8EI} (y_{B})_{R} = + \frac{R_{B}L^{3}}{3EI}[/latex]

Writing that the deflection at B is the sum of these two quantities and that it must be zero, we have

[latex]y_{B} = (y_{B})_{w }+ (y_{B})_{R} = 0[/latex]

[latex]y_{B} = - \frac{wL^{4}}{8EI} + \frac{R_{B}L^{3}}{3EI}= 0[/latex]

and, solving for [latex]R_{B}[/latex], [latex]R_{B}= \frac{3}{8} wL R_{B} = \frac{3}{8}wL ↑[/latex]

Drawing the free-body diagram of the beam (Fig. 15.31) and writing the corresponding equilibrium equations, we have

[latex]+↑ \sum{F_{y}} = 0: R_{A} + R_{B} - wL = 0[/latex] (15.46)

[latex]R_{A} = wL - R_{B} = wL - \frac{3}{8} wL = \frac{5}{8} wL[/latex]

[latex]R_{A} = \frac{5}{8} wL ↑[/latex]

[latex]+\curvearrowleft \sum{M_{A}} = 0: M_{A} + R_{B}L - (wL) (\frac{1}{2} L) = 0[/latex] (15.47)

[latex]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}[/latex]

[latex]M_{A} = \frac{1}{8} wL^{2} \curvearrowleft[/latex]

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 [latex]M_{A}[/latex] 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 [latex](θ_{A})_{w}[/latex] caused at A by the uniformity distributed load w (Fig. 15.32b) and the slope [latex](θ_{A})_{M}[/latex] produced at the same point by the unknown couple [latex]M_{A}[/latex] (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

[latex](θ_{A})_{w} = - \frac{wL^{3}}{24EI} (θ_{A})_{M} = \frac{M_{A}L}{3EI}[/latex]

Writing that the slope at A is the sum of these two quantities and that it must be zero, we have

[latex]θ_{A} = (θ_{A})_{w} + (θ_{A})_{M} = 0[/latex]

[latex]θ_{A} =- \frac{wL^{3}}{25EI}+ \frac{M_{A}L}{3EI} = 0[/latex]

and, solving for [latex]M_{A}[/latex],

[latex]M_{A} = \frac{1}{8} wL^{2} M_{A} = \frac{1}{8} wL^{2} \curvearrowleft[/latex]

The values of [latex]R_{A} and R_{B}[/latex] may then be found from the equilibrium equations (15.46) and (15.47).

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