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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 $d\sigma_n$

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  • $\begingroup$ This is just an expression of force equilibrium for the pyramidal body under consideration. $\endgroup$ – Chet Miller Feb 2 '19 at 11:43
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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.

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$\let\s=\sigma \def\bF{\mathbf F} \def\ba{\mathbf a} \def\be{\mathbf e} \def\bn{\mathbf n} \def\bp{\mathbf p} \def\br{\mathbf r}$ 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.

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