A note accompanying Chapter 14

A note on the physical interpretation of the second Piola–Kirchhoff stress

By Chiara

01

Interpreting the stress measures

The second Piola–Kirchhoff stress tensor S\boldsymbol{S} is extensively used in finite-deformation mechanics, especially in Total Lagrangian formulations, but its physical interpretation requires care.

The Cauchy stress tensor σ\boldsymbol{\sigma} has the most direct physical interpretation: it represents the actual traction on a surface of the current configuration. The first Piola–Kirchhoff stress tensor P\boldsymbol{P} relates forces in the current configuration to areas defined in the initial configuration.

Let WW denote the strain-energy density, i.e. the strain energy per unit volume of the initial configuration.

The Green–Lagrange strain E\boldsymbol{E} and the second Piola–Kirchhoff stress S\boldsymbol{S} form a conjugate strain–stress pair. In engineering notation, the variation of the strain-energy density may be written as:

δW=STδE\boxed{\delta W=\boldsymbol{S}^{T}\,\delta\boldsymbol{E}}

Here, the engineering-component vectors are:

E={ExEyγxy}S={SxSySxy}\boldsymbol{E}=\begin{Bmatrix}E_x\\E_y\\\gamma_{xy}\end{Bmatrix}\qquad\boldsymbol{S}=\begin{Bmatrix}S_x\\S_y\\S_{xy}\end{Bmatrix}

This relation expresses the conjugate character of the Green–Lagrange strain E\boldsymbol{E} and the second Piola–Kirchhoff stress S\boldsymbol{S}.

For the linear constitutive law used on this site:

S=DE\boxed{\boldsymbol{S}=\boldsymbol{D}\boldsymbol{E}}

where D\boldsymbol{D} is the constitutive matrix of Hooke’s law.

In one dimension:

Sx=EEx\boxed{S_x=E\,E_x}

This association makes PK2 very convenient in the Total Lagrangian formulation. However, S\boldsymbol{S} should not be interpreted as the physical traction acting directly on a current surface.

02

One-dimensional example

Consider the data of Example 1 in Chapter 14, using N, mm, and MPa:

F=24000E=1000A0=5×60=300L0=400\begin{aligned}F&=24000\qquad E=1000\\ A_0&=5\times60=300\qquad L_0=400\end{aligned}

Let Δ\Delta denote the elongation. The stretch ratio is:

λ=LL0=1+ΔL0\lambda=\frac{L}{L_0}=1+\frac{\Delta}{L_0}

The Green–Lagrange strain is:

EGL=12(λ21)E_{GL}=\frac12\left(\lambda^2-1\right)

or:

EGL=ΔL0+12(ΔL0)2E_{GL}=\frac{\Delta}{L_0}+\frac12\left(\frac{\Delta}{L_0}\right)^2

The second Piola–Kirchhoff stress, denoted here by the scalar SS, is:

S=EEGLS=E\,E_{GL}

03

From PK2 to the applied force

At finite deformation, it is not correct to identify the applied force directly with the initial area multiplied by PK2 stress:

F=A0SF=A_0S

The force referred to the initial area is instead related to the first Piola–Kirchhoff stress:

F=A0PF=A_0P

The two Piola–Kirchhoff stress measures satisfy:

P=FS\boxed{\boldsymbol{P}=\boldsymbol{F}\boldsymbol{S}}

Here, bold F\boldsymbol{F} denotes the deformation gradient, while the non-bold FF used in the numerical example denotes the applied force. Recall:

F=I+u\boldsymbol{F}=\boldsymbol{I}+\nabla\boldsymbol{u}

In one dimension, the deformation gradient reduces to the stretch ratio:

Fλ\boldsymbol{F}\longrightarrow\lambda

Therefore:

P=λS\boxed{P=\lambda S}

is the one-dimensional counterpart of P=FS\boldsymbol{P}=\boldsymbol{F}\boldsymbol{S}. Consequently:

F=A0λS\boxed{F=A_0\lambda S}

and:

F=EA0λEGL\boxed{F=EA_0\lambda E_{GL}}

04

Numerical solution

For the specified load:

FEA0=240001000×300=0.08\frac{F}{EA_0}=\frac{24000}{1000\times300}=0.08

Let:

x=ΔL0x=\frac{\Delta}{L_0}

Then:

λ=1+xEGL=x+12x2\lambda=1+x\qquad E_{GL}=x+\frac12x^2

The correct equilibrium equation is:

0.08=(1+x)(x+12x2)0.08=(1+x)\left(x+\frac12x^2\right)

Therefore:

x+32x2+12x3=0.08x+\frac32x^2+\frac12x^3=0.08

or:

x3+3x2+2x0.16=0\boxed{x^3+3x^2+2x-0.16=0}

The positive physical root is approximately:

x0.072x\approx0.072

Hence:

Δ28.8 mm\boxed{\Delta\approx28.8\ \mathrm{mm}}

This agrees with the finite-element result obtained with the program used in Chapter 12, as reported in Example 1 of Chapter 14.

05

The incorrect shortcut

If PK2 were incorrectly used directly in F=A0SF=A_0S, the equation would be:

0.08=x+12x20.08=x+\frac12x^2

This would give approximately:

x0.077Δ30.8 mmx\approx0.077\qquad\Delta\approx30.8\ \mathrm{mm}

The difference between 30.8 mm and 28.8 mm is explained by the missing stretch factor:

λ=1+ΔL0\boxed{\lambda=1+\frac{\Delta}{L_0}}

which converts PK2 into PK1 in the one-dimensional relation:

P=λSP=\lambda S

06

Final remark

The second Piola–Kirchhoff stress has a precise and important mechanical role, but it is not the physical traction measure acting directly on a surface of the current configuration. Its usefulness in the Total Lagrangian formulation comes from its natural association with the Green–Lagrange strain tensor.

In one dimension, the conversion is:

S  ×λ  P  ×A0  F\boxed{S\;\xrightarrow{\times\lambda}\;P\;\xrightarrow{\times A_0}\;F}