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Continuum Mechanics

Part 3 · Theory of stress

3.6 The second Piola–Kirchhoff tensor

Source: Carboni’s notes for lesson 8 (20 Oct 2026), pp. 8–9

Objectives

  1. Vocabulary. Second Piola–Kirchhoff tensor .
  2. Theorem. is symmetric, and with it the material problem loses its awkward symmetry condition.
  3. Recipe. Move between the three stress tensors: , , .

The material system of Theorem 3.5.9 has a clumsy second line, , because is not symmetric. The fix is to factor out of .

Definition 3.6.1second Piola–Kirchhoff tensornotes L8, p. 8

The second Piola–Kirchhoff tensor is defined by

Theorem 3.6.2S is symmetricnotes L8, p. 8

The balance of angular momentum, , is equivalent to

Proof

. Substitute in the angular momentum condition: Multiply on the left by and on the right by : .

is symmetric. With it, the problem in the reference configuration reads:

Key observation 3.6.3the material problem with Snotes L8, p. 9

Three stress tensors

tensor force measured in per unit area of symmetric from
Cauchy current configuration current area yes
first Piola–Kirchhoff current configuration reference area no
second Piola–Kirchhoff pulled back by reference area yes
Key observation 3.6.4what S is forbeyond the lectures

is not a force per area of anything physical: the force has been pulled back to the reference configuration by . Its use is that it is symmetric and lives entirely on the reference configuration, like the Green–Lagrange strain (Definition 2.6.2). The constitutive law pairs the two (§3.7 Principal stresses and the linear elastic law).

Remarkbeyond the lectures

Under a rigid rotation, : rotates with the body, rotates on one side only, and does not change at all if the stress was carried along by the rotation. Try the “rotation 90°” preset in Figure 3.6.1: the Cauchy stress vector turns with the face, stays put. That is why , and not , is the natural partner of , which also ignores rotations.

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Figure 3.6.1 The three tensors for one state: simple shear with a vertical Cauchy tension. The readout gives (not symmetric) and (symmetric). Change and ; stays symmetric whatever you do.

Exercises

Exercise 3.6.1

For the bar of Exercise 3.5.1, and MPa, find . What does it pair with?

Solution

MPa. It divides the engineering stress once more by the stretch 2. It is not a force per area of anything; it is the tensor that pairs with .

Exercise 3.6.2

For simple shear with MPa (Exercise 3.5.2), find and check that it is symmetric.

Solution

and MPa, so symmetric.

Exercise 3.6.3

Given and , write the two steps that give the Cauchy stress, and apply them to , .

Solution
  1. (not symmetric).
  2. , with : , so , symmetric as it must be.

This is part of the practice sheet’s Problem 2 (Exercise 3.8.1).