Beam Handbook

Beam Handbook

Two-supported beam with symmetrical cantilevers

Diagrams. Shear forces & Bending moments


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

Shear forces

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}

Bending moments

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}


M_F = M_H = 0

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

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

Multi-supported beam with cantilevers

Characterization of a multi-supported beam under a distributed load with n spans, 2 cantilevers, n-2 inner spans and n-1 reactions.

Diagramas multi
Diagrams. Shear forces & Bending moments

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}

Shear forces

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}

Bending moments

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 


  x_k = d_k - \Large \frac{s_k}{2}

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