# Problem 735 | Fixed-ended beam with one end not fully restrained

**Problem 735**

The beam shown in Fig. P-735 is perfectly restrained at A but only partially restrained at B, where the slope is w_{o}L^{3}/48EI directed up to the right. Solve for the end moments.

**Solution 735**

$EI \left( \dfrac{w_oL^3}{48EI} \right) = \frac{1}{2}(R_AL)L + M_AL - \frac{1}{3}(\frac{1}{2}w_oL^2)L$

$\frac{1}{48}w_oL^3 = \frac{1}{2}R_AL^2 + M_AL - \frac{1}{6}w_oL^3$

$M_AL = \frac{3}{16}w_oL^3 - \frac{1}{2}R_AL^2$

$M_A = \frac{3}{16}w_oL^2 - \frac{1}{2}R_AL$ → Equation (1)

$EI \, t_{B/A} = (Area_{AB}) \cdot \bar{X}_B = 0$

$\frac{1}{2}(R_AL)L(\frac{1}{3}L) + M_AL(\frac{1}{2}L) - \frac{1}{3}(\frac{1}{2}w_oL^2)L(\frac{1}{4}L) = 0$

$\frac{1}{6}R_AL^3 + \frac{1}{2}M_AL^2 - \frac{1}{24}w_oL^4 = 0$

$\frac{1}{6}R_AL + \frac{1}{2}M_A - \frac{1}{24}w_oL^2 = 0$

Substitute M_{A} defined in equation (1)

$\frac{1}{6}R_AL + \frac{1}{2}(\frac{3}{16}w_oL^2 - \frac{1}{2}R_AL) - \frac{1}{24}w_oL^2 = 0$

$\frac{1}{6}R_AL + \frac{3}{32}w_oL^2 - \frac{1}{4}R_AL - \frac{1}{24}w_oL^2 = 0$

$\frac{1}{12}R_AL = \frac{5}{96}w_oL^2$

$R_A = \frac{5}{8}w_oL$

From equation (1)

$M_A = \frac{3}{16}w_oL^2 - \frac{1}{2}(\frac{5}{8}w_oL)L$

$M_A = \frac{3}{16}w_oL^2 - \frac{5}{16}w_oL^2$

$M_A = -\frac{1}{8}w_oL^2$ *answer*

From moment diagram by parts

$M_B = R_AL + M_A - \frac{1}{2}w_oL^2$

$M_B = (\frac{5}{8}w_oL)L - \frac{1}{8}w_oL^2 - \frac{1}{2}w_oL^2$

$M_B = \frac{5}{8}w_oL^2 - \frac{1}{8}w_oL^2 - \frac{1}{2}w_oL^2$

$M_B = 0$ *answer*