---
title: "Cauchy's formula and equilibrium equations"
subject: "Mechanics of Solids"
chapter: 2
key: "semester_5/mechanics_of_solids:cauchy-formula-equilibrium"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/cauchy-formula-equilibrium
site: BCT Engineering
---

# Cauchy's formula and equilibrium equations

Mechanics of Solids · Chapter 2: Stress Tensor

- Asked in: year not given, 6 marks
- Video: [Stress – 6: Cauchy's Formula](https://www.youtube.com/watch?v=iJox70itcr4) — jc
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/cauchy-formula-equilibrium
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-2

## Question

Derive Cauchy's formula relating the traction on an arbitrary plane to the stress tensor, using an elemental tetrahedron. State the equilibrium equations for a deformable body in terms of stress components.

## Short answer

Traction on a plane with normal (l,m,n) is t_i = σ_ij n_j, derived from force equilibrium of a tetrahedron. Equilibrium equations: ∂σ_x/∂x + ∂τ_xy/∂y + ∂τ_xz/∂z + X = 0, and similarly for y and z; τ_ij = τ_ji.

## Answer

#### Cauchy's formula

Take a tetrahedron at a point with three faces on the coordinate planes and a fourth inclined face of area $dA$ and unit normal $\mathbf{n} = (l, m, n)$ (direction cosines). The coordinate-plane faces then have areas

$$
dA_x = l\,dA,\quad dA_y = m\,dA,\quad dA_z = n\,dA
$$

Let $\mathbf{t} = (t_x, t_y, t_z)$ be the traction (stress vector) on the inclined face. The faces on the coordinate planes have outward normals along $-x, -y, -z$, so their stresses act opposite to the positive components.

```text
        z
        |  C
        | /|\
        |/ | \
        A--+--B  (inclined face ABC)
       /  O
      y       x
```

Force equilibrium in the x-direction (body force and inertia are of higher order, since volume $\propto h^3$ while area $\propto h^2$ as the tetrahedron shrinks):

$$
\begin{aligned}
t_x\, dA - \sigma_x (l\,dA) - \tau_{yx}(m\,dA) - \tau_{zx}(n\,dA) &= 0 \\
t_x &= \sigma_x l + \tau_{yx} m + \tau_{zx} n
\end{aligned}
$$

Similarly for $y$ and $z$. In matrix form (symmetric tensor):

$$
\begin{Bmatrix} t_x \\ t_y \\ t_z \end{Bmatrix} =
\begin{bmatrix} \sigma_x & \tau_{xy} & \tau_{xz} \\ \tau_{xy} & \sigma_y & \tau_{yz} \\ \tau_{xz} & \tau_{yz} & \sigma_z \end{bmatrix}
\begin{Bmatrix} l \\ m \\ n \end{Bmatrix}
$$

or $t_i = \sigma_{ij} n_j$. The normal and shear stress on the plane are then

$$
\sigma_n = \mathbf{t}\cdot\mathbf{n},\qquad \tau = \sqrt{|\mathbf{t}|^2 - \sigma_n^2}
$$

So the six components of $\sigma_{ij}$ determine the traction on every plane through the point.

#### Equilibrium equations

For a body force per unit volume $(X, Y, Z)$:

$$
\begin{aligned}
\frac{\partial \sigma_x}{\partial x} + \frac{\partial \tau_{xy}}{\partial y} + \frac{\partial \tau_{xz}}{\partial z} + X &= 0 \\
\frac{\partial \tau_{xy}}{\partial x} + \frac{\partial \sigma_y}{\partial y} + \frac{\partial \tau_{yz}}{\partial z} + Y &= 0 \\
\frac{\partial \tau_{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} + \frac{\partial \sigma_z}{\partial z} + Z &= 0
\end{aligned}
$$

Moment equilibrium gives the symmetry $\tau_{ij}=\tau_{ji}$. At the boundary, the traction must satisfy $t_i = \sigma_{ij}n_j$ equal to the applied surface load.

## Check yourself

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

1. What does Cauchy's formula state?
   A. t_i = sigma_ij n_j: the traction on any plane follows from the stress tensor and the plane's normal
   B. t = sigma/A
   C. sigma_n = E epsilon
   D. t_i = n_i / sigma_ij
   Answer: A. The traction on an arbitrary plane is the stress tensor times the unit normal, so six components determine traction on every plane. (Key point: Cauchy's formula)

2. How are the areas of the coordinate-plane faces related to the inclined face dA?
   A. dA_x = dA, dA_y = dA, dA_z = dA
   B. dA_x = l dA, dA_y = m dA, dA_z = n dA using the direction cosines
   C. dA_x = dA/l
   D. They are equal to each other
   Answer: B. Projected areas equal dA times the direction cosines of the normal. (Key point: Tetrahedron face areas)

3. Why can body force and inertia be neglected as the tetrahedron shrinks?
   A. They are zero in all bodies
   B. They act only on the inclined face
   C. They balance the shear stresses
   D. Volume varies as h^3 while area varies as h^2, so they are of higher order
   Answer: D. As h tends to 0, volume forces vanish faster than surface forces. (Key point: Body force neglected)

4. What is the x-equilibrium equation of a deformable body?
   A. d(sigma_x)/dx + d(tau_xy)/dy + d(tau_xz)/dz + X = 0
   B. sigma_x + tau_xy + tau_xz = 0
   C. d(sigma_x)/dx = X only
   D. d(sigma_x)/dy + d(tau_xy)/dx = 0
   Answer: A. Stress gradients plus the body force X per unit volume sum to zero. (Key point: Equilibrium equation)

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