Problem 819 | Continuous Beam by Three-Moment Equation

Three-moment equation between spans (1) and (2)
$M_1 L_1 + 2M_2(L_1 + L_2) + M_3 L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$
 

Where,
$M_1 = 0$

$L_1 = L$

$L_2 = \alpha L$

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

$\dfrac{6A_2\bar{b}_2}{L_2} = 0$
 

Thus,
$0 + 2M_2(L + \alpha L) + M_3(\alpha L) + \dfrac{Pa}{L}(L^2 - a^2) + 0 = 0$

$2(1 + \alpha)L \, M_2 + \alpha L \, M_3 = -\dfrac{Pa}{L}(L^2 - a^2)$

$2(1 + \alpha) \, M_2 + \alpha \, M_3 = -\dfrac{Pa}{L^2}(L^2 - a^2)$   ←   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 M₃ in equation (2) to equation (1)
$2(1 + \alpha) \, M_2 + \alpha \left[ -\dfrac{\alpha M_2}{2(\alpha + \beta)} \right] = -\dfrac{Pa}{L^2}(L^2 - a^2)$

$2(1 + \alpha) \, M_2 - \dfrac{\alpha^2 M_2}{2(\alpha + \beta)} = -\dfrac{Pa}{L^2}(L^2 - a^2)$

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

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

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

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

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