# Problem 734 | Restrained beam with uniform load over half the span

**Problem 734**

Determine the end moments for the restrained beams shown in Fig. P-734.

**Solution 734**

$EI\,\theta_{AB} = (\text{Area}_{AB}) = 0$

$\frac{1}{2}L(R_AL) + MAL - \frac{1}{3}(\frac{1}{2}L)(\frac{1}{8}w_oL^2) = 0$

$\frac{1}{2}R_AL^2 + M_AL - \frac{1}{48}w_oL^3 = 0$

$M_A = \frac{1}{48}w_oL^2 - \frac{1}{2}R_AL$ → equation (1)

The deviation of B from a tangent line through A is zero

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

$\frac{1}{2}L(R_A L)(\frac{1}{3}L) + M_A L(\frac{1}{2}L) - \frac{1}{3}(\frac{1}{2}L)(\frac{1}{8}w_o L^2)[ \, \frac{1}{4}(\frac{1}{2}L) \, ] = 0$

$\frac{1}{6}R_A L^3 + \frac{1}{2} M_A L^2 - \frac{1}{384}w_oL^4 = 0$

$64R_AL + 192M_A = w_oL^2$ → equation (2)

Substitute M_{A} of equation (1) to equation (2) above

$64R_AL + 192(\frac{1}{48}w_oL^2 - \frac{1}{2}R_AL) = w_oL^2$

$64R_AL + 4w_oL^2 - 96R_AL = w_oL^2$

$32R_AL = 3w_oL^2$

$R_A = \frac{3}{32}w_oL$

Substitute R_{A} to equation (1)

$M_A = \frac{1}{48}w_oL^2 - \frac{1}{2}(\frac{3}{32}w_oL)L$

$M_A = \frac{1}{48}w_oL^2 - \frac{3}{64}w_oL^2$

$M_A = -\frac{5}{192}w_oL^2$ *answer*

Sum up moments at right support B

$M_B = M_A + R_AL - \frac{1}{8}w_oL^2$ → see the right end of moment diagram by parts

$M_B = -\frac{5}{192}w_oL^2 + (\frac{3}{32}w_oL)L - \frac{1}{8}w_oL^2$

$M_B = -\frac{5}{192}w_oL^2 + \frac{3}{32}w_oL^2 - \frac{1}{8}w_oL^2$

$M_B = -\frac{11}{192}w_oL^2$ *answer*