**Finding the Reactions of Continuous Beams**

Isolate each span of the beam and consider each as simply supported carrying the original span loading and the computed end moments. Resolve further the simple span into simple beams, one carrying the given loads plus another beam carrying the end moments and couple reactions. With this method, the interior reaction was divided into parts which can be summed up find the total reaction. See example below.

**General instruction**

In the following problems, determine the reactions and sketch the shear diagrams. Then compute the values of maximum vertical shear V and maximum positive bending moment M. In solving the problems, use the moments determined in the reference problems unless otherwise instructed.

## Problem 828 - Reactions of Continuous Beam

A continuous beam carries a uniform load over two equal spans as shown in Fig. P-828.

**Solution 828**

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

$M_1 = M_3 = 0$

$L_1 = L_2 = L$

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

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*