**Problem 656**

Find the value of EIδ at the point of application of the 200 N·m couple in Fig. P-656.

**Solution 656**

$\Sigma M_D = 0$

$4R_1 = 3(500) + 200$

$R_1 = 425 \, \text{ N}$

$\Sigma M_A = 0$

$4R_2 + 200 = 1(500)$

$R_2 = 75 \, \text{ N}$

From the conjugate beam

$\Sigma M_A = 0$

$4F_2 + \frac{1}{2}(2)(1000)[1 + \frac{2}{3}(2)] = \frac{1}{2}(3)(1275)[\frac{2}{3}(3)] + \frac{1}{2}(1)(75)[3 + \frac{1}{3}(1)]$

$F_2 = 404.17 \, \text{ N}\cdot\text{m}^3$

$M_C = \frac{1}{2}(1)(75)[\frac{1}{3}(1)] - 1(F_2)$

$M_C = 12.5 - 1(404.17)$

$M_C = -391.67 \, \text{ N}\cdot\text{m}^3$

Therefore, the deflection at C is

$EI \, \delta_C = M_C$

$EI \, \delta_C = -391.67 \, \text{ N}\cdot\text{m}^3$

$EI \, \delta_C = 391.67 \, \text{ N}\cdot\text{m}^3$ downward *answer*