# Problem 712 | Propped beam with initial clearance at the roller support

**Problem 712**

There is a small initial clearance D between the left end of the beam shown in Fig. P-712 and the roller support. Determine the reaction at the roller support after the uniformly distributed load is applied.

**Solution**

$\delta_R + \Delta = \delta_w + \frac{1}{2}L\theta$

See Case 1 and Case 3 of Superposition Method for formulas:

$\dfrac{RL^3}{3EI} + \Delta = \dfrac{w_o(\frac{1}{2}L)^4}{8EI} + \dfrac{L}{2} \left[ \dfrac{w_o(\frac{1}{2}L)^3}{6EI} \right]$

$\dfrac{RL^3}{3EI} + \Delta = \dfrac{w_oL^4}{128EI} + \dfrac{w_oL^4}{96EI}$

$\dfrac{RL^3}{3EI} + \Delta = \dfrac{7w_oL^4}{384EI}$

$\dfrac{RL^3}{3EI} = \dfrac{7w_oL^4}{384EI} - \Delta$

$R = \dfrac{7w_oL}{128} - \dfrac{3EI \Delta}{L^3}$

$R = \dfrac{1}{128L^3}(7w_oL^4 - 384EI \Delta)$ *answer*

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