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

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

y = – \frac{w}{24EI} (x^{4} – 2L x^{3} + L^{3}x)

At point B, x = \frac{2}{3} L:

(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}

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

(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}

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

y_{B} = (y_{B})_{w} +(y_{B})_{R}

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

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

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

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

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

Finally, θ_{A}= (θ_{A})_{w} + (θ_{A})_{R}