# Stress tensor of an elastic medium

I don't understand a passage from the book I'm reading about tensor analysis. The state of stress of an elastic medium can be expressed by the stress function $$\mathbf{p}(r,n)$$ so that the force acting on an arbitrary element of area $$d\sigma$$ is $$\mathbf{p}(r,n)$$ $$\cdot d\sigma$$, with $$\mathbf{n}$$ the normal of $$d\sigma$$ and $$\mathbf{r}$$ the radius vector of the point M, of application of the force.

In the limit of the body shrinking to the point M, we find that:

$$\mathbf{p_n} d\sigma = \mathbf{p_1} d\sigma_1 + \mathbf{p_2} d\sigma_2 + \mathbf{p_3} d\sigma_3$$

I don't understand why there are there $$\mathbf{p}$$ vectors and why their sum is equal to $$\mathbf{p_n}$$ acting on the surface $$d\sigma_n$$ • This is just an expression of force equilibrium for the pyramidal body under consideration. – Chet Miller Feb 2 '19 at 11:43

I guess that $$p$$ has units of force per unit area, like pressure. Consequently, $$p_1\cdot d\sigma_1$$ is the force on the plane $$1$$, and your equation is then

$$\vec{F}_{Total}=\vec{F}_1+\vec{F}_2+\vec{F}_3$$

Which makes sense: the total force is the sum of three cartesian components. Such net force will depend (in magnitude and direction) on the proportions in each component.

$$\let\s=\sigma \def\bF{\mathbf F} \def\ba{\mathbf a} \def\be{\mathbf e} \def\bn{\mathbf n} \def\bp{\mathbf p} \def\br{\mathbf r}$$ The figure has a defect: whereas $$\bn$$ is shown $$\bn_1$$, $$\bn_2$$, $$\bn_3$$ don't. We should assume they coincide with $$\be_1$$, $$\be_2$$, $$\be_3$$ (unit vectors along $$x_i$$ axes). If it's so, note that whereas $$\be_1$$, $$\be_2$$, $$\be_3$$ point inwards $$\bn$$ points outwards. This means that $$\bp(\br,\be_1)\,d\s_1$$ is the force acted on tetrahedron from outside, and the same for the other orthogonal faces, $$\bp(\br,\bn)\,d\s$$ is the force the tetrahedron is acting on the outside.

Then if we wanted to compute the net force acting on tetrahedron we should take the last with the $$-$$ sign: $$\bF = \bp(\br,\be_1)\,d\s_1 + \bp(\br,\be_2)\,d\s_2 + \bp(\br,\be_3)\,d\s_3 - \bp(\br,\bn)\,d\s.\tag1$$

I don't know the exact context wherein you're studying the matter. In particular, if it's statics or dynamics. But we may work on both for the same price. In statics we'll require $$\bF=0$$. In dynamics we would write $$\bF = m \ba$$ with $$m$$ the tetrahedron's mass, $$\ba$$ the acceleration of its com.

Now take the limit, i.e. shrink all tetrahedron's sides by the same factor $$k$$ and let $$k\to0$$. You may see from (1) that $$\bF$$ goes to 0 as $$k^2$$, whereas $$m$$ goes as $$k^3$$ and acceleration must stay finite. This is only possible if limit of $$\bF$$ is 0, and we get $$\bp(\br,\bn)\,d\s = \bp(\br,\be_1)\,d\s_1 + \bp(\br,\be_2)\,d\s_2 + \bp(\br,\be_3)\,d\s_3$$ which is the formula you're asking about.