Problem 855 | Continuous Beams with Fixed Ends

By symmetry, M₄ = M₁ and M₃ = M₂
 

 

Apply three-moment equation between spans (0) and (1)
$M_0L_0 + 2M_1(L_0 + L_1) + M_2L_1 + \dfrac{6A_0\bar{a}_0}{L_0} + \dfrac{6A_1\bar{b}_1}{L_1} = 0$

$0 + 2M_1(0 + \alpha L) + M_2(\alpha L) + 0 + 0 = 0$

$2\alpha L \, M_1 + \alpha L \, M_2 = 0$

$M_1 = -\frac{1}{2}M_2$
 

Apply three-moment equation between spans (1) and (2)
$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

$-\frac{1}{2}M_2(\alpha L) + 2M_2(\alpha L + L) + M_2(L) + 0 + \frac{3}{8}PL^2 = 0$

$-\frac{1}{2}\alpha L \, M_2 + 2\alpha L \, M_2 + 2L \, M_2 + L \, M_2 = -\frac{3}{8}PL^2$

$\frac{3}{2}\alpha L \, M_2 + 3L \, M_2 = -\frac{3}{8}PL^2$

$\frac{3}{2}\alpha \, M_2 + 3 \, M_2 = -\frac{3}{8}PL$

$\frac{3}{2}(\alpha + 2)M_2 = -\frac{3}{8}PL$

$M_2 = -\dfrac{3PL}{8} \cdot \dfrac{2}{3(\alpha + 2)}$

$M_2 = -\dfrac{PL}{8} \cdot \dfrac{2}{\alpha + 2} = M_3$           answer
 

$M_1 = -\dfrac{1}{2}\left( -\dfrac{PL}{8} \cdot \dfrac{2}{\alpha + 2} \right)$

$M_1 = \dfrac{PL}{8} \cdot \dfrac{1}{\alpha + 2} = M_4$           answer