Beam Handbook

F_1 = F_2 = \Large \frac{qL}{2}

V(x) = \begin{cases} -qx & \text{for}   x \le \large \frac{L-s}{2} \\ -q(x-\Large \frac{L}{2}\normalsize) & \text{for}   \large \frac{L-s}{2} \normalsize < x \le \small L-\large \frac{L-s}{2} \\ -q(x+L) & \text{for}   x > \small L-\large \frac{L-s}{2} \end{cases}

-V_A = V_D = \Large \frac{q (L-s)}{2}

V_B = -V_C = \Large \frac{qs}{2}

M(x) = \begin{cases} -\large \frac{1}{2} \normalsize qx^2 & \text{for}   x \le \large \frac{L-s}{2} \\ -\large \frac{1}{2} \normalsize q(x^2-Lx+L\Large\frac{L-s}{2}\normalsize) & \text{for}   \large \frac{L-s}{2} \normalsize < x \le \small L-\large \frac{L-s}{2} \\ -\large \frac{1}{2} \normalsize q(x^2-2Lx+L^2) & \text{for}   x > \small L-\large \frac{L-s}{2} \end{cases}

M_E = M_I = -\Large \frac{1}{8}\normalsize qL(L-s);  \text{for}  x = \large \frac{L-s}{2} \normalsize = \small L- \large \frac{L-s}{2}

M_G = \Large \frac{1}{2} \normalsize qL \Large (\frac{Ls}{2} \normalsize - \Large \frac{L^2}{4}) \normalsize;  \text{for}  x = \large \frac{L}{2}

It can be proved that the relation between L and s that minimize the 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 null moments at points F and H on 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 multi-supported beam under a distributed load with n spans, 2 cantilevers, n-2 inner 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 hyperstatic problem, it will be assumed, approximately, that each reaction assumes half of the load on each adjacent span, except for the reactions next to the cantilevers, which will take the total load of the cantilever plus half of that on the adjacent span.

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 subindex k expresses the calculation span.

k = \begin{cases} 0 & \text{for left cantilever}   \\ \in [1, n-1] \in \mathbb{N} & \text{for inner span}   \\ n & \text{for right cantilever}   \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 produced at the middle of each inner span can be calculated as follows:

  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}