# Problem 704 | Propped beam with some uniform load by moment-area method

**Problem 704**

Find the reaction at the simple support of the propped beam shown in Figure PB-001 by using moment-area method.

**Solution**

The moment at C due to reaction R

_{A}is R_{A}L and the moment at C due to uniform load w_{o}is –w_{o}b(0.5b) = -½w_{o}b^{2}.

The deviation at A from tangent line through C is zero. Thus,

$EI \, t_{A/C} = (Area_{AC}) \bar{X}_A = 0$

$\frac{1}{2}L(R_AL)(\frac{2}{3}L) - \frac{1}{3}b(\frac{1}{2}w_ob^2)(L - \frac{1}{4}b) = 0$

$\frac{1}{3}R_AL^3 - \frac{1}{6}w_ob^3(L - \frac{1}{4}b) = 0$

$\frac{1}{3}R_AL^3 - \frac{1}{24}w_ob^3(4L - b) = 0$

$\frac{1}{3}R_AL^3 = \frac{1}{24}w_ob^3(4L - b)$

$R_A = \dfrac{w_ob^3}{8L^3}(4L - b)$ *answer*

**Learn More with Our Video**

Subscribe to MATHalino.com on