# Problem 820 | Continuous Beam by Three-Moment Equation

**Problem 820**

Solve Prob. 819 if the concentrated load is replaced by a uniformly distributed load of intensity w_{o} over the first span.

**Solution 820**

$M_1 L_1 + 2M_2(L_1 + L_2) + M_3 L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

Where,

$M_1 = 0$

$L_1 = L$

$L_2 = \alpha L$

$\dfrac{6A_1\bar{a}_1}{L_1} = \frac{1}{4}w_o L^3$

$\dfrac{6A_2\bar{b}_2}{L_2} = 0$

$0 + 2M_2(L + \alpha L) + M_3(\alpha L) + \frac{1}{4}w_o L^3 + 0 = 0$

$2(1 + \alpha)L \, M_2 + \alpha L \, M_3 = -\frac{1}{4}w_o L^3$

$2(1 + \alpha)M_2 + \alpha M_3 = -\frac{1}{4}w_o L^2$ ← equation (1)

Three-moment equation between spans (2) and (3)

$M_2 L_2 + 2M_3(L_2 + L_3) + M_4 L_3 + \dfrac{6A_2\bar{a}_2}{L_2} + \dfrac{6A_3\bar{b}_3}{L_3} = 0$

Where,

$L_2 = \alpha L$

$L_3 = \beta L$

$M_4 = 0$

$\dfrac{6A_2\bar{a}_2}{L_2} = \dfrac{6A_3\bar{b}_3}{L_3} = 0$

Thus,

$M_2(\alpha L) + 2M_3(\alpha L + \beta L) + 0 + 0 + 0 = 0$

$\alpha L \, M_2 + 2(\alpha + \beta)L \, M_3 = 0$

$\alpha M_2 + 2(\alpha + \beta)M_3 = 0$

$M_3 = -\dfrac{\alpha M_2}{2(\alpha + \beta)}$ ← equation (2)

Substitute M_{3} in equation (2) to equation (1)

$2(1 + \alpha) \, M_2 + \alpha \left[ -\dfrac{\alpha M_2}{2(\alpha + \beta)} \right] = -\frac{1}{4}w_o L^2$

$2(1 + \alpha) \, M_2 - \dfrac{\alpha^2 M_2}{2(\alpha + \beta)} = -\frac{1}{4}w_o L^2$

$\left[ 2(1 + \alpha) - \dfrac{\alpha^2}{2(\alpha + \beta)} \right] M_2 = -\frac{1}{4}w_o L^2$

$\left[ \dfrac{4(1 + \alpha)(\alpha + \beta) - \alpha^2}{2(\alpha + \beta)} \right] M_2 = -\frac{1}{4}w_o L^2$

$M_2 = -\dfrac{w_o L^2}{4} \cdot \dfrac{2(\alpha + \beta)}{4(1 + \alpha)(\alpha + \beta) - \alpha^2}$ *answer*

From equation (2)

$M_3 = -\dfrac{\alpha}{2(\alpha + \beta)} \left[ -\dfrac{w_o L^2}{4} \cdot \dfrac{2(\alpha + \beta)}{4(1 + \alpha)(\alpha + \beta) - \alpha^2} \right]$

$M_3 = \dfrac{w_o L^2}{4} \cdot \dfrac{\alpha}{4(1 + \alpha)(\alpha + \beta) - \alpha^2}$ *answer*