# Problem 815 | Continuous Beam by Three-Moment Equation

**Problem 815**

Find the moment at R_{2} and R_{3} of the continuous beam shown in Fig. P-815.

**Solution 815**

$y = 60 ~ \text{lb}/\text{ft}$

Moment under R_{3}

$M_3 = \frac{1}{2}(8y)[ \, \frac{1}{3}(8) \, ] = -\frac{32}{3}y = -\frac{32}{3}(60)$

$M_3 = -640 ~ \text{lb}\cdot\text{ft}$ *answer*

Solving for moment under R_{2}

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

Where

$L_1 = 3(4) = 12 ~ \text{ft}$

$L_2 = 10 ~ \text{ft}$

$\dfrac{6A_1\bar{a}_1}{L_1} = \sum\dfrac{Pa}{L}(L^2 - a^2)$

$\dfrac{6A_1\bar{a}_1}{L_1} = \dfrac{300(4)}{12}(12^2 - 4^2) + \dfrac{300(8)}{12}(12^2 - 8^2)$

$\dfrac{6A_1\bar{a}_1}{L_1} = 28\,800 ~ \text{lb}\cdot\text{ft}^2$

$\dfrac{6A_2\bar{b}_2}{L_2} = \dfrac{8w_oL^3}{60} + \dfrac{7yL^3}{60}$

$\dfrac{6A_2\bar{b}_2}{L_2} = \dfrac{8(135)(10^3)}{60} + \dfrac{7(60)(10^3)}{60}$

$\dfrac{6A_2\bar{b}_2}{L_2} = 25\,000 ~ \text{lb}\cdot\text{ft}^2$

Thus,

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

$0 + 2M_2(12 + 10) - 640(10) + 28\,800 + 25\,000 = 0$

$M_2 = -1077.27 ~ \text{lb}\cdot\text{ft}$ *answer*