**Problem 813**

Determine the moment over the support R_{2} of the beam shown in Fig. P-813.

**Solution 813**

$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

Where

$M_1 = M_3 = 0$

$M_1 = M_3 = 0$

$L_1 = L_2 = 4 \, \text{ m}$

$\dfrac{6A_1\bar{a}_1}{L_1} = \dfrac{8w_oL^3}{60} = \dfrac{8(1400)(4^3)}{60}$

$\dfrac{6A_1\bar{a}_1}{L_1} = 11\,946.67 \, \text{ N}\cdot\text{m}^2$

$\dfrac{6A_2\bar{b}_2}{L_2} = \dfrac{Pb}{L}(L^2 - b^2) + \dfrac{w_od^2}{4L}(2L^2 - d^2)$

$\dfrac{6A_2\bar{b}_2}{L_2} = \dfrac{900(3)}{4}(4^2 - 3^2) + \dfrac{800(2^2)}{4(4)}[ \, 2(4^2) - 2^2 \, ]$

$\dfrac{6A_2\bar{b}_2}{L_2} = 10\,325 \, \text{ N}\cdot\text{m}^2$

Thus,

$0 + 2M_2(4 + 4) + 0 + 11\,946.67 + 10\,325$

$M_2 = -1391.98 \, \text{ N}\cdot\text{m}$ *answer*