Reactions of Continuous Beams | Shear Diagrams

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

Where,
$M_1 = M_3 = 0$

$L_1 = L_2 = L$

$\dfrac{6A_1\bar{a}_1}{L_1} = \dfrac{6A_2\bar{b}_2}{L_2} = \frac{1}{4}w_o L^3$

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

$4L \, M_2 + \frac{1}{2}w_o L^3$

$M_2 = -\frac{1}{8}w_o L^2$           answer
 

From the first span
Simple reactions due to loadings
$V_{1L} = V_{1R} = \frac{1}{2}w_o L$
 

Couple reaction due to end moment
${R_1}' = \dfrac{M_2}{L} = \dfrac{\frac{1}{8}w_o L^2}{L}$

${R_1}' = \frac{1}{8}w_o L$
 

Thus,
$R_{1L} = V_{1L} - {R_1}' = \frac{1}{2}w_o L - \frac{1}{8}w_o L$

$R_{1L} = \frac{3}{8}w_o L$
 

$R_{1R} = V_{1R} + {R_1}' = \frac{1}{2}w_o L + \frac{1}{8}w_o L$

$R_{1R} = \frac{5}{8}w_o L$
 

 

From the second span
Simple reactions due to loadings
$V_{2L} = V_{2R} = \frac{1}{2}w_o L$
 

Couple reaction due to end moment
${R_2}' = \dfrac{M_2}{L} = \dfrac{\frac{1}{8}w_o L^2}{L}$

${R_2}' = \frac{1}{8}w_o L$
 

Thus,
$R_{2L} = V_{2L} + {R_2}' = \frac{1}{2}w_o L + \frac{1}{8}w_o L$

$R_{2L} = \frac{5}{8}w_o L$
 

$R_{2R} = V_{2R} - {R_2}' = \frac{1}{2}w_o L - \frac{1}{8}w_o L$

$R_{2R} = \frac{3}{8}w_o L$
 

Note: You can actually use the 'symmetry' to solve for R_2L and R_2R. It is easy easy to see that R_2L = R_1R and R_2R = R_1L. With this, you can shorten the solution by not doing all computations related to the second span.
 

 

From the load diagram
$R_1 = \frac{3}{8}w_o L$           answer

$R_2 = \frac{5}{8}w_o L + \frac{5}{8}w_o L = \frac{5}{4}w_o L$           answer

$R_3 = \frac{3}{8}w_o L$           answer
 

From the shear diagram
By ratio and proportion
$\dfrac{x}{\frac{3}{8}w_o L} = \dfrac{L - x}{\frac{5}{8}w_o L}$

$5x = 3L - 3x$

$x = \frac{3}{8}L$
 

$M_{max\,(+)} = \frac{1}{2}x (\frac{3}{8}w_o L) = \frac{1}{2}( \frac{3}{8}L)(\frac{3}{8}w_o L)$

$M_{max\,(+)} = \frac{9}{128}w_o L^2$           answer