---
title: "Derivation of shear stress formula in beams"
subject: "Strength of Materials"
chapter: 5
key: "semester_4/strength_of_materials_mechanical:beam-shear-stress-formula-derivation"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/beam-shear-stress-formula-derivation
site: BCT Engineering
---

# Derivation of shear stress formula in beams

Strength of Materials · Chapter 5: Transverse loading

- Asked in: year not given, 6 marks
- Video: [AE 204 Derivation of the Shear Formula](https://www.youtube.com/watch?v=rHA1nRSSvZc) — UW Mechanics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/beam-shear-stress-formula-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-5

## Question

Derive the expression tau = V*Q/(I*b) for the shear stress at a level y1 above the neutral axis in a beam of any cross-section carrying a shear force V. State the assumptions used.

## Short answer

Horizontal force equilibrium on an element above a level y1 gives τ b dx = (dM/I)Q. With dM/dx=V, τ=VQ/(Ib). Assumptions: elastic homogeneous material, shear stress uniform across width, prismatic beam, load applied through shear centre in a symmetry plane.

## Answer

#### 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
```text
  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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What causes the net horizontal force on the part of the element above y1?
   A. The difference in bending moment dM between the two faces
   B. The shear force V itself
   C. The self weight
   D. The axial load
   Answer: A. Bending stresses differ on the two faces by dM y / I, so F2 - F1 = (dM/I) Q. (Key point: Unbalanced horizontal force)

2. What is Q in tau = VQ/(Ib)?
   A. Second moment of the whole section
   B. First moment of the area above the level about the NA
   C. Section modulus
   D. Width of the section
   Answer: B. Q = integral of y dA over A* = A* y_bar*. (Key point: Meaning of Q)

3. Which relation gives tau = VQ/(Ib) from tau b dx = (dM/I) Q?
   A. dM/dx = V
   B. dM/dx = w
   C. dV/dx = M
   D. dM/dx = 0
   Answer: A. The shear force equals the rate of change of bending moment, so tau = (Q/Ib) V. (Key point: Final step)

## Related in Strength of Materials

- [Shear stress distribution in an I-section](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Ashear-stress-i-section-distribution) — Strength of Materials, chapter 5
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## Related in other subjects

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