Beam Handbook

It can be shown that the relationship between L and s that minimizes bending moments, i.e., -M_E = M_G , is:

s =f(L) : -M_E = M_G

s = \Large \frac{2}{2+\sqrt{2}} \normalsize L

The distance e that marks the separation between zero moments at points F and H of the central span is:

e = \sqrt{2Ls-L^2}

Where:

M_F = M_H = 0

x_F = \Large \frac{L-e}{2}

x_H = \Large \frac{L+e}{2}

Characterization of a continuous beam under a distributed load with n sections, 2 symmetric overhangs, n-2 interior spans, and n-1 reactions.

L = d_n = \sum_{i=0}^{n} s_i

d_k = \sum_{i=0}^{k} s_i

Since this is a statically indeterminate problem, as an approximation, it is assumed that each reaction supports half the load of each adjacent span, except for the reactions next to the overhangs, which will support the total overhang load plus half of the adjacent span load.

R_i(x) = \begin{cases} q(s_0+ \Large \frac{s_1}{2}\normalsize) & \text{for}   i = 0 \\ q(\Large \frac{s_i+s_{i+1}}{2}\normalsize)  & \text{for}   i \in [1, n-2] \in \mathbb{N} \\ q(\Large \frac{s_{n-1}}{2}\normalsize +s_n) & \text{for}   i = n-1 \end{cases}

The subscript k denotes the calculation section.

k = \begin{cases} 0 & \text{for left overhang}   \\ \in [1, n-1] \in \mathbb{N} & \text{for interior spans}   \\ n & \text{for right overhang}   \end{cases}

V_k(x) = \begin{cases} -qx & \text{for}   k = 0 \\ -qx+\sum_{i=0}^{k-1} R_i & \text{for}   k \in [1, n-1] \in \mathbb{N} \\ -q(x+L) & \text{for}   k = n \end{cases}

M_k(x) = \begin{cases} -\large \frac{1}{2} \normalsize qx^2 & \text{for}   k = 0 \\ -\large \frac{1}{2} \normalsize qx^2+x\sum_{i=0}^{k-1} R_i - \sum_{i=0}^{k-1} R_i d_i  & \text{for}   k \in [1, n-1] \in \mathbb{N} \\ -\large \frac{1}{2} \normalsize q(x^2-2Lx+L^2) & \text{for}   k = n \end{cases}

The maximum moment occurring at the center of each span can be calculated as:

M_k(x_k) = -\large \frac{1}{2} \normalsize qx_k^2+x_k\sum_{i=0}^{k-1} R_i - \sum_{i=0}^{k-1} R_i d_i 

Where:

x_k = d_k - \Large \frac{s_k}{2}