Problem 851 | Continuous Beams with Fixed Ends

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

$0 + 2M_1(0 + L) + M_2(L) + 0 + \frac{3}{8}PL^2 = 0$

$2L\,M_1 + L\,M_2 + \frac{3}{8}PL^2 = 0$

$2M_1 + M_2 + \frac{3}{8}PL = 0$

$2M_1 = -M_2 - \frac{3}{8}PL$

$M_1 = -\frac{1}{2}M_2 - \frac{3}{16}PL$
 

 

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

$M_1(L) + 2M_2(L + \alpha L) + M_3(\alpha L) + \frac{3}{8}PL^2 + 0 = 0$

$L\,M_1 + 2(1 + \alpha)L\,M_2 + \alpha L\,M_3 + \frac{3}{8}PL^2 = 0$

$M_1 + 2(1 + \alpha)M_2 + \alpha M_3 + \frac{3}{8}PL = 0$

$(-\frac{1}{2}M_2 - \frac{3}{16}PL) + 2(1 + \alpha)M_2 + \alpha M_3 + \frac{3}{8}PL = 0$

$-\frac{1}{2}M_2 - \frac{3}{16}PL + 2M_2 + 2\alpha M_2 + \alpha M_3 + \frac{3}{8}PL = 0$

$\frac{3}{2}M_2 + 2\alpha M_2 + \alpha M_3 + \frac{3}{16}PL = 0$

$(\frac{3}{2} + 2\alpha)M_2 + (\alpha M_3 + \frac{3}{16}PL) = 0$

$(\frac{3}{2} + 2\alpha)M_2 = -(\alpha M_3 + \frac{3}{16}PL)$

$M_2 = -\dfrac{\alpha M_3 + \frac{3}{16}PL}{\frac{3}{2} + 2\alpha}$
 

Apply three-moment equation between spans (2) and (3)
$M_2L_2 + 2M_3(L_2 + L_3) + M_4L_3 + \dfrac{6A_2\bar{a}_2}{L_2} + \dfrac{6A_3\bar{b}_3}{L_3} = 0$

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

$\alpha L\,M_2 + 2\alpha L\,M_3 = 0$

$M_2 + 2M_3 = 0$

$-\dfrac{\alpha M_3 + \frac{3}{16}PL}{\frac{3}{2} + 2\alpha} + 2M_3 = 0$

$-(\alpha M_3 + \frac{3}{16}PL) + 2(\frac{3}{2} + 2\alpha)M_3 = 0$

$-\alpha M_3 - \frac{3}{16}PL + 3M_3 + 4\alpha M_3 = 0$

$3M_3 + 3\alpha M_3 = \frac{3}{16}PL$

$(3 + 3\alpha)M_3 = \frac{3}{16}PL$

$M_3 = \dfrac{3PL}{16} \cdot \dfrac{1}{3 + 3\alpha}$           answer
 

$M_2 = -\dfrac{\alpha M_3 + \frac{3}{16}PL}{\frac{3}{2} + 2\alpha} = -\dfrac{\alpha \left( \dfrac{3PL}{16} \cdot \dfrac{1}{3 + 3\alpha} \right) + \dfrac{3PL}{16}}{\frac{3}{2} + 2\alpha}$

$M_2 = -\dfrac{3PL}{16} \cdot \dfrac{\dfrac{\alpha}{3 + 3\alpha} + 1}{\frac{3}{2} + 2\alpha} = -\dfrac{3PL}{16} \cdot \dfrac{\dfrac{\alpha + (3 + 3\alpha)}{3 + 3\alpha}}{\dfrac{3 + 4\alpha}{2}}$

$M_2 = -\dfrac{3PL}{16} \cdot \dfrac{\dfrac{3 + 4\alpha}{3 + 3\alpha}}{\dfrac{3 + 4\alpha}{2}} = -\dfrac{3PL}{16} \cdot \dfrac{\dfrac{1}{3 + 3\alpha}}{\dfrac{1}{2}}$

$M_2 = -\dfrac{3PL}{16} \cdot \dfrac{2}{3 + 3\alpha}$           answer
 

$M_1 = -\frac{1}{2}M_2 - \frac{3}{16}PL = -\dfrac{1}{2} \left( -\dfrac{3PL}{16} \cdot \dfrac{2}{3 + 3\alpha} \right) - \dfrac{3PL}{16}$

$M_1 = \dfrac{3PL}{16} \cdot \dfrac{1}{3 + 3\alpha} - \dfrac{3PL}{16} = -\dfrac{3PL}{16} \left( -\dfrac{1}{3 + 3\alpha} + 1 \right)$

$M_1 = -\dfrac{3PL}{16} \cdot \dfrac{-1 + (3 + 3\alpha)}{3 + 3\alpha}$

$M_1 = -\dfrac{3PL}{16} \cdot \dfrac{2 + 3\alpha}{3 + 3\alpha}$           answer