Page 223 - Intro to Tensor Calculus

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Let
∆S 1 = the surface area 0BC
∆S 2 = the surface area 0AC
∆S 3 = the surface area 0AB

∆S = the surface area ABC .
These surface areas are related by the relations

∆S 1 = n 1 ∆S, ∆S 2 = n 2 ∆S, ∆S 3 = n 3 ∆S (2.3.9)

which can be thought of as projections of ∆S upon the planes x i =constant for i =1, 2, 3.

Cauchy Stress Law

Let t j (n) denote the components of the surface traction on the surface ABC. That is, we let

~ t (n) = t 1(n) ˆ e 1 + t 2(n) ˆ e 2 + t 3(n) ˆ e 3 = t j (n) ˆ e j . (2.3.10)

It will be demonstrated that the components t j (n) of the surface traction forces ~ t (n) associated with a plane
through P and having the unit normal with direction cosines n 1 ,n 2 and n 3 , must satisfy the relations

ij
t j (n) = n i σ , i, j =1, 2, 3. (2.3.11)

This relation is known as the Cauchy stress law.
Proof: Sum the forces acting on the elemental tetrahedron in the figure 2.3-5. If the body is in equilibrium,
then the sum of these forces must equal zero or

1 2 3 (n)
(− ~ t ∆S 1 )+ (− ~ t ∆S 2 )+(− ~ t ∆S 3 )+ ~ t ∆S =0. (2.3.12)

The relations in the equations (2.3.9) are used to simplify the sum of forces in the equation (2.3.12). It is
readily verified that the sum of forces simplifies to

1
2
i
3
~
~
~
~ t (n) = n 1 t + n 2 t + n 3 t ~ = n i t . (2.3.13)
Substituting in the relations from equation (2.3.8) we find
ij
~ t (n) = t j (n) ˆ e j = n i σ ˆ e j , i, j =1, 2, 3 (2.3.14)
or in component form
t j (n) = n i σ ij (2.3.15)

which is the Cauchy stress law.