# Problem 707 | Propped Beam with Moment Load

**Problem 707**

A couple M is applied at the propped end of the beam shown in Fig. P-707. Compute R at the propped end and also the wall restraining moment.

**Solution 04**

## Click here to expand or collapse this section

The moment at any point point on the beam which is at distance x from the left support is

$M_x = M - Rx$

$M_x = M - Rx$

By double integration method

$EI \, y'' = M_x$

$EI \, y'' = M - Rx$

$EI \, y' = Mx - \frac{1}{2}Rx^2 + C_1$

$EI \, y = \frac{1}{2}Mx^2 - \frac{1}{6}Rx^3 + C_1x + C_2$

**Boundary conditions**

At x = 0, y = 0; C_{2} = 0

At x = L, y = 0;

$0 = \frac{1}{2}ML^2 - \frac{1}{6}RL^3 + C_1L$

$C_1 = \frac{1}{6}RL^2 - \frac{1}{2}ML$

At x = L, y' = 0;

$0 = ML - \frac{1}{2}RL^2 + (\frac{1}{6}RL^2 - \frac{1}{2}ML)$

$0 = M - \frac{1}{2}RL + \frac{1}{6}RL - \frac{1}{2}M$

$\frac{1}{3}RL = \frac{1}{2}M$

$R = \dfrac{3M}{2L}$ *answer*

- Log in to post comments