Is stress a vector or scalar?

Question

I have seen some theories saying stress is a scalar. I don't understand why. Anyone can explain.

Answer 1

As a mechanical engineer, I've come across the stress being represented mainly as a second order tensor, the Cauchy stress tensor, however vector representations are also common. As for scalar representations? This might be common in scenarios of uniaxial stress, such as a bar under uniform tensile stress, where there is only one component of stress.

Cauchy stress tensor

In general, for a given coordinate system $O(x,y,z)$, there are 6 components of stress:

$$\sigma_{xx}\quad \sigma_{yy}\quad \sigma_{zz}\quad \sigma_{xy}\quad \sigma_{xz}\quad \sigma_{yz}$$

Let's consider a solid object of arbitrary shape. It can be a beam, a brick, a gear, etc. In general, the stress will vary throughout the solid.

To consider the stresses at a point in the solid body, let's surgicaly remove an infinitesimally small cube from within the solid at the point of interest, to observe the stress acting on the faces of the cube. The faces of the cube is aligned with the $x$-, $y$- and $z$-axes of our coordinate system. On each face of the six faces of the cube, three components of stress act: two components lie tangential to the face (these components are called shear stresses), and one component is perpendicular to the face (this component is called the normal stress). The diagram below shows the stresses acting on three of the faces.

Six faces, each with three stress components, might suggest a total of 18 components. However, balance of forces (stress $\times$ infinitesimal face area) onto the cube means pairs of stress components on opposite faces are equal and opposite ($\rightarrow$ 9 unique components).

The convention for defining stress components is to use the form $\sigma_{ij}$, where subscript $i$ indicates the face on the cube which the stress component acts, and $j$ indicates the direction of the stress. ($\sigma_{xy}$ acts in the $y$ direction on the $x$-face of the cube).

By balance of moments, it can be shown that $$\sigma_{xy} = \sigma_{yx}, \qquad \sigma_{xz} = \sigma_{zx}, \qquad \sigma_{yz} = \sigma_{zy}$$ reducing the number of unique stress components to six.

Therefore, at any point in a 3D solid, the stress state is fully described by six numbers:

$\sigma_{xx}\quad \sigma_{yy}\quad \sigma_{zz}$ are the normal stress components

$\sigma_{xy}\quad \sigma_{xz}\quad \sigma_{yz}$ are the shear stress components.

(Frequently, shear stresses (but not normal stresses) are alternatively expressed as $\tau_{ij}$ instead of $\sigma_{ij}$)

A natural means of expressing this stress state is as a Cauchy stress tensor:

$$[\sigma] = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \\ \end{bmatrix}$$

or, equivalently, to highlight the symmetric nature of the stress tensor as well as using only the 6 unique components:

$$[\sigma] = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \\ \end{bmatrix}$$

In 2D stress problems where, say, stresses in the $z$-direction are not of concern, this tensor is often reduced to

$$[\sigma] = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} \\ \sigma_{xy} & \sigma_{yy} \\ \end{bmatrix}$$

The advantage of the Cauchy tensor is that it is possible to re-express the stress components in a different frame of reference $O(x',y',z')$ by performing a change of basis on the tensor. This is done by performing a matrix transformation on the array of elements $[\sigma]$ to obtain the array of stress components in the new coordinate system $[\sigma']$. That is, if a vector with components $[a]$ expressed in $O(x,y,z)$ is related to the components $[a']$ re-expressed in $O(x',y',z')$ via the equation

$$[a'] = [Q][a]$$

where $[Q]$ is an orthogonal matrix, then the Cauncy stress tensor elements in both systems are interrelated by

$$[\sigma'] = [Q][\sigma][Q]^T$$

There is nothing stopping you from re-arranging the 6 stress components into a 6-dimension column vector, such as

$[\sigma] = \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{zz} \\ \sigma_{xy} \\ \sigma_{xz} \\ \sigma_{yz} \\ \end{bmatrix}$

This is convenient for expressing material laws in linear elasticity problems, as in this form, the stress vector is related to a similarly defined strain vector via multiplication with a $6\times 6$ constitution matrix (as opposed to the stress tensor being related with the strain tensor via appropriate multiplication of a fourth order tensor!). However, re-expressing this vector in different coordinate systems is not a readily available action. A transformation matrix $[Q]$ cannot be used directly here, least to say that $[Q]$ is a $3\times 3$ matrix, and the stress vector here is $6 \times 1$!

The Cauchy stress vector

In the first case, we used the unit cube to describe the stress state at a point by observing the faces of an infinitesimal cube. An alternative way to examine stress is to analyse how the stress act on an imaginary surface at that point. Imagine now that we cut the solid body about a plane that passes through the point, and the plane has a normal vector

$$[n] = \begin{bmatrix}n_x \\ n_y \\ n_z\end{bmatrix}$$

Now let's analyse the stress that act on this plane at the point of interest.

Like for the case of the cube, three perpendicular components of stress act on the cutting plane, and these are

$$t_x \quad t_y \quad t_z$$

In fact, the total stress acting on this surface can be represented as a vector, in the form

$$[t] = \begin{bmatrix}t_x \\ t_y \\ t_z\end{bmatrix}$$

This vector is a true vector is the sense that it can be transformed to re-expressed its elements in a new coordinate system. Considering the transformation matrix $[Q]$ from earlier, the stress vector representations in the two coordinate systems are interrelated by

$$[t'] = [Q][t]$$

This stress vector is referred to as the Cauchy stress vector. Note that expression of stress as a vector requires a reference surface of a given normal $[n]$ (The Cauchy stress tensor can be defined without providing a reference surface).

The Cauchy stress vector is related to the Cauchy stress tensor by the Cauchy stress equation:

$$[\sigma][n] = [t]$$

In full:

$$\begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \\ \end{bmatrix} \begin{bmatrix}n_x \\ n_y \\ n_z\end{bmatrix} = \begin{bmatrix}t_x \\ t_y \\ t_z\end{bmatrix} $$

Stress vs traction

To define our stress vector, we had to cut at a plane to obtain a surface on which to describe stress. However, what happens if we consider a point on the outer surface of our solid? A similar vector can be computed by considering the force per unit acting on the outer surface of the solid, but, as a mechanical engineer, I would argue that this vector is called a traction vector instead of a stress vector at this point, since now this vector arises due to external forces rather that internal. Stresses are internal forces per unit area, whereas tractions are external forces per unit area. However, this is just a matter of terminology, and other disciplines may use different terminologies.

To summarise

Stress can be expressed as a vector or a tensor. Stress can only really be described by a scalar for simple uniaxial stress problems.

Answer 2

It is neither a scalar, vector, nor tensor, but a tensor density; which is seen directly by virtue of its involvement in the transport law for momentum.

In vector-dyad notation, the relevant differential equations for mass and momentum transport for fluid dynamics are, respectively: $$\frac{∂ρ}{∂t} + ∇·(𝐮ρ) = 0, \hspace 1em \frac{∂𝝿}{∂t} + ∇·(𝐮𝝿 + 𝗣) = 𝕱,$$ where $ρ$ represents the volume density of mass, $𝝿 = \left(π_1, π_2, π_3\right)$ the volume density of mass, $𝐮 = \left(u^1, u^2, u^3\right)$ the fluid flow velocity, $𝕱 = \left(𝔉_1, 𝔉_2, 𝔉_3\right)$ the volume density of force, and $𝗣 = \left(𝔓^i_j: i,j = 1,2,3\right)$ the stress. This is expressed in Cartesian coordinates $\left(x^1,x^2,x^3\right) = (x,y,z)$ with $∇ = \left(∂/∂x, ∂/∂y, ∂/∂z\right)$ being the gradient operator. The fluid flow velocity is defined in such a way that mass and momentum density are related by $𝝿 = ρ𝐮$. The component $𝔓^i_j$ indicates how much force in the $x^j$ direction there is per unit area perpendicular to the $x^i$ direction. All of the quantities, other than $𝐮$ are densities.

In component form, the relevant equations are $$\frac{∂ρ}{∂t} + \sum_{i=1,2,3} \frac{∂\left(u^iρ\right)}{∂x^i} = 0, \hspace 1em \frac{∂π_j}{∂t} + \sum_{i=1,2,3} \frac{∂\left(u^iπ_j + 𝔓^i_j\right)}{∂x^i} = 𝔉_j \hspace 1em (j = 1, 2, 3).$$ It is also usual to express the force density on a per-unit-mass basis as $𝕱 = ρ𝐟$, in which case the vector density $𝕱$ divides out by $ρ$ to an ordinary vector $𝐟 = \left(f_1, f_2, f_3\right)$.

In integral form, if you treat the body as a blob occupying a time-variable volume $V$ that moves with the fluid (in which case, it's called a "moving control volume"), then the time rate of change for the various densities reduces to the form: $$\frac{d}{dt} \int_V (⋯) dV = \int_V \frac{∂}{∂t} (⋯) dV + \int_{∂V} d𝐒·𝐮(⋯) $$ where $∂V$ denotes the boundary of the region $V$, $dV = dx dy dz$ a unit volume element in $V$, $d𝐒 = \left(dS_1, dS_2, dS_3\right) = (dy dz, dz dx, dz dy)$ a unit area element on the boundary $∂V$. The boundary integral accounts for the movement of the region $V$. Alternatively, if you think of $V$ as being fixed, then the boundary integral gives you the flux of the relevant quantity across the boundary $∂V$. By Gauss' Theorem $$\int_{∂V} d𝐒·(⋯) = \int_V ∇·(⋯),$$ consequently, you can rewrite the time rate of change of the volume total as a volume integral $$\frac{d}{dt} \int_V (⋯) dV = \int_V \left(\frac{∂}{∂t} (⋯) + ∇·(𝐮(⋯))\right) dV.$$

Thus, the integral forms of the relevant equations are: $$\frac d {dt} \int_V ρ dV = 0, \hspace 1em \frac d {dt} \int_V 𝝿 dV + \int_{∂V} d𝐒·𝗣 = \int_V 𝕱 dV.$$

The stress tensor density is normally assumed to be symmetric: $𝔓^i_j = 𝔓^j_i$. A scalar (density) part $p = ⅓ \left(𝔓^1_1 + 𝔓^2_2 + 𝔓^3_3\right)$ can be subtracted out from it. That plays the role of the pressure. The remainder ${\left(𝔓_0\right)}^i_j = 𝔓^i_j - δ^i_j p$ (where $δ^i_j = 1$ if $i = j$, $δ^i_j = 0$ if $i ≠ j$, denotes - as usual - the components of the Kronecker delta) is then the "trace-free" part of $𝔓$.

For example, for the Navier-Stokes equation, one has $$𝔓^i_j = p δ^i_j - μ \left(σ^i_j - ⅔ ∇·𝐮 δ^i_j\right) - ζ ∇·𝐮 δ^i_j,$$ where $σ^i_j = ∂u^i/∂x^j + ∂u^j/∂x^i$, where $ζ$ is the second viscosity and $μ$ is the dynamic viscosity; and $𝐟$ is set to $𝐠$, the gravity vector. And actually, the way they set up the equation, their "pressure" $p$ (called the "mechanical pressure") differs from mine, because they separated part of the trace of $𝔓$ into the term with the dynamic viscosity.