For the uniform beam and loading shown, determine (a) the reaction at each support, (b) the slope at end A.

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Principle of Superposition. The reaction [latex]R_{B}[/latex] is designated as redundant and considered as an unknown load. The deflections due to the distributed load and to the reaction [latex]R_{B}[/latex] 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 App. C.

Distributed Loading. We use case 6, App. C.

[latex]y = - \frac{w}{24EI} (x^{4} - 2L x^{3} + L^{3}x)[/latex]

At point B,[latex] x = \frac{2}{3} L[/latex]:

[latex](y_{B})_{w} = -\frac{w}{24EI} \left[ \left(\frac{2}{3} L\right)^{4} - 2L \left( \frac{2}{3} L\right)^{3} + L^{3}\left( \frac{2}{3} L\right) \right] = -0.01132 \frac{wL^{4}}{EI}[/latex]

Redundant Reaction Loading. From case 5, App. C, with [latex]a = \frac{2}{3} L and b = \frac{1}{3} L[/latex], we have

[latex](y_{B})_{R} = -\frac{Pa^{2}b^{2}}{3EIL} = + \frac{R_{B}}{3EIL} \left( \frac{2}{3} L\right)^{2} \left(\frac{L}{3} \right)^{2} = 0.01646 \frac{R_{B}L^{3}}{EI}[/latex]

a. Reactions at Supports. Recalling that [latex]y_{B} = 0[/latex], we write

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

[latex]0 = -0.01132 \frac{wL^{4}}{EI} + 0.01646 \frac{R_{B}L^{3}}{EI} R_{B} = 0.688wL↑[/latex]

Since the reaction [latex]R_{B}[/latex] is now known, we may use the methods of statics to determine the other reactions: [latex]R_{A} = 0.271wL ↑ R_{C} = 0.0413wL↑[/latex]

b. Slope at End A. Referring again to App. C, we have

Distributed Loading. [latex](θ_{A})_{w} = -\frac{wL^{3}}{24EI} = -0.04167 \frac{wL^{3}}{EI}[/latex]

dundant Reaction Loading. For [latex]P = -R_{B} = -0.688wL and b = \frac{1}{3} L[/latex]

[latex](θ_{A})_{R} = - \frac{Pb(L^{2} - b^{2})}{6EIL} = + \frac{0.688wL}{6EIL} \left(\frac{L}{3} \right) \left[ L^{2} - \left(\frac{L}{3} \right)^{2} \right] (θ_{A})_{R} = 0.03398 \frac{wL^{3}}{EI}[/latex]

Finally, [latex]θ_{A}= (θ_{A})_{w} + (θ_{A})_{R}[/latex]

[latex]θ_{A}= -0.04167 \frac{wL^{3}}{EI} + 0.03398 \frac{wL^{3}}{EI} = -0.00769 \frac{wL^{3}}{EI} θ_{A}= 0.00769 \frac{wL^{3}}{EI} ⦪[/latex]

Question 15.S-P.5: For the uniform beam and loading shown, determine (a) the re...

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