**1**Beam of L= 20 ft span, with uniformly distributed load w = 100 plf

**2**Free-body diagram of partial beam x units long

**3**Shear diagram

**4**Bending moment diagram

To find the distribution of shear and bending along the beam, we investigate the beam at intervals of 5’, from left to right. This is not normally required.

**Reactions R**are half the load on each support due to symmetry

Shear force Vx at any distance x from left is found using ∑ V = 0

**V = R - w x**

**Bending moment**Mx at any distance x from left is found by ∑ M = 0.

∑ M = 0; R x – w x (x/2) - Mx = 0; solving for Mx

Bending is zero at both supports since pins and rollers have no moment resistance. Since the bending formula Mx= Rx-wx2/2 is quadratic, bending increase is quadratic (parabolic curve) toward maximum at center, and decreases to zero at the right support. For simple beams with uniform load the maximum shear force is at the supports and the maximum bending moment at mid-span (x= L/2) are:

**Vmax = R = w L / 2**

**Mmax = wL^2 / 8**

This formula is only for simple beams with uniform load. Verifying example: