Derivation of shear stress formula in beams

Assumptions

1. Elastic, homogeneous material; the flexure formula holds.
2. Shear stress is uniform across the width $b$ at the level considered, and acts parallel to the shear force.
3. The beam is prismatic and the load acts through the shear centre in a plane of symmetry.

Derivation

  M          M+dM
  |<--- dx --->|        cut at y1; shaded area A* above it
  +-------------+  --- top fibre
  |    A*       |
  |-------------|  y1   width b
  |      NA     |
  +-------------+

Take a small element of length $dx$ with bending moments $M$ and $M + dM$ at its faces. The bending stress at a distance $y$ from the NA on the left face is $\sigma = My/I$. Consider the part of the element above the level $y_1$, with area $A^*$.

Force on the left face: $F_1 = \int_{A^*}\dfrac{My}{I}\,dA$. Force on the right face: $F_2 = \int_{A^*}\dfrac{(M+dM)y}{I}\,dA$.

Net unbalanced horizontal force:

$$F_2 - F_1 = \frac{dM}{I}\int_{A^*}y\,dA = \frac{dM}{I}\,Q$$

where $Q = \int_{A^*}y\,dA = A^*\bar y^*$ is the first moment of the area above the level about the NA.

This force is balanced by the horizontal shear force on the cut face of area $b\,dx$:

$$\tau\,b\,dx = \frac{dM}{I}Q \;\Rightarrow\; \tau = \frac{Q}{I\,b}\,\frac{dM}{dx}$$

Since $dM/dx = V$:

$$\boxed{\tau = \frac{V\,Q}{I\,b}}$$

By complementary shear, the same stress acts on the vertical plane. The stress is zero at the top and bottom fibres (where $Q = 0$) and is usually greatest at the neutral axis.