Part 3 · Theory of stress

3.5 The problem in two configurations, and the first Piola–Kirchhoff tensor

Source: Carboni’s notes for lesson 7, p. 15, and lesson 8 (20 Oct 2026), pp. 1–7

Objectives

Recipe. The boundary-value problem in the current (Eulerian) configuration, and why it is awkward.
Vocabulary. Nominal stress vector, first Piola–Kirchhoff tensor $T$ .
Theorem. $T = det F \overset{˘}{T} F^{- T}$ , $\overset{˘}{T} = \frac{1}{d e t F} T F^{T}$ , and $\overset{˘}{\nabla} \cdot \overset{˘}{T} = \frac{1}{d e t F} \nabla \cdot T^{T}$ .
Recipe. The same problem on the reference configuration, where $T$ is not symmetric.

The problem in the current configuration

Lesson 7 ends by putting together everything the balance laws say.

Definition 3.5.1Eulerian formulationnotes L7, p. 15; L8, p. 1

The Eulerian (spatial) formulation of the problem, with fields over the current configuration $\overset{˘}{B}$ : $⎩ ⎨ ⎧ \overset{˘}{\nabla} \cdot \overset{˘}{T} + \overset{˘}{b} = \overset{ρ}{˘} \overset{˘}{a} \overset{˘}{T} = \overset{˘}{T}^{T} \overset{˘}{T} \overset{˘}{n} = \overset{˘}{f} p = \overset{ˉ}{p} \forall y \in \overset{˘}{B} \forall y \in \overset{˘}{B} \forall y \in \partial \overset{˘}{B}_{f} \forall y \in \partial \overset{˘}{B}_{\overset{p}{ˉ}}$

the first line is Cauchy’s equation of motion, $\partial \overset{˘}{T}_{ij} / \partial y_{j} + \overset{˘}{b}_{i} = \overset{ρ}{˘} \overset{a}{˘}_{i}$ ;
the second is the balance of angular momentum;
the third is the traction boundary condition: on the part of the boundary where surface forces are prescribed, the stress vector on the boundary’s normal equals the applied force per unit current area;
the fourth prescribes the position on the rest of the boundary.

Remark

Divergence convention. Carboni writes the divergence of a tensor as $(\nabla \cdot A)_{i} = \partial A_{ji} / \partial X_{j}$ : it differentiates along the columns. So $(\nabla \cdot A^{T})_{i} = \partial A_{ij} / \partial X_{j}$ runs along the rows. For the symmetric $\overset{˘}{T}$ the two are the same, and $(\overset{˘}{\nabla} \cdot \overset{˘}{T})_{i} = \partial \overset{˘}{T}_{ij} / \partial y_{j}$ , the row-wise form of Key observation 3.4.2. It matters for $T$ below, which is not symmetric.

Lesson 8 restates the system with $\overset{˘}{m}$ for the boundary normal, $\overset{˘}{g}$ for the surface force, and $p = p_{0}$ on $\partial \overset{˘}{B}_{u}$ . It also writes the acceleration with the displacement: $p = X + U$ and $X$ does not depend on time, so $\overset{˘}{a} = \frac{\partial ^{2} p}{\partial t ^{2}} = \frac{\partial ^{2} X}{\partial t ^{2}} + \frac{\partial ^{2} U}{\partial t ^{2}} = \ddot{U} .$

Key observation 3.5.2the drawback

$\overset{˘}{B}$ , $\partial \overset{˘}{B}_{f}$ and the current normals are not known until the problem is solved: the equations live on a region that is part of the answer. The way out is to write everything on the reference configuration $B$ , which is known.

Nominal stress and the first Piola–Kirchhoff tensor

The Lagrangian (material) formulation uses the nominal quantities of Proposition 3.1.2, written as in lesson 8: $\overset{˘}{b} = \frac{1}{det F} b, \overset{ρ}{˘} = \frac{1}{det F} ρ, \overset{˘}{g} = \frac{1}{Λ _{m}} g, Λ_{m} = det F m^{T} F^{- 1} F^{- T} m .$ Do the same with the contact force. On an element with reference normal $m$ and area $d A$ (current normal $\overset{˘}{m}$ , area $d \overset{˘}{A}$ ), the contact force is the same in both descriptions: $t_{m} d A = \overset{˘}{t}_{m} d \overset{˘}{A} ⟹ t_{m} = \frac{d A ˘}{d A} \overset{˘}{t}_{m} = Λ_{m} \overset{˘}{t}_{m} .$

Definition 3.5.3nominal stress vectornotes L8, p. 2

$t_{m} = Λ_{m} \overset{˘}{t}_{m}$ is the nominal stress vector: the actual force on the element, per unit of its reference area.

With $\overset{˘}{t}_{m} = \overset{˘}{T} \overset{˘}{m}$ and the areal stretch vector $λ_{m} = Λ_{m} \overset{˘}{m} = det F F^{- T} m$ (Theorem 2.5.5): $t_{m} = Λ_{m} \overset{˘}{T} \overset{˘}{m} = \overset{˘}{T} λ_{m} = det F \overset{˘}{T} F^{- T} m .$ The nominal stress vector is linear in the reference normal $m$ , so it too comes from a tensor.

Definition 3.5.4first Piola–Kirchhoff tensorEssentialnotes L8, pp. 2–3

The first Piola–Kirchhoff tensor (nominal stress tensor) $T$ maps the reference normal of an element to its nominal stress vector: $t_{m} = T m, T = det F \overset{˘}{T} F^{- T} .$

Proposition 3.5.5Cauchy stress from the first Piola–Kirchhoff tensornotes L8, p. 3

$\overset{˘}{T} = \frac{1}{det F} T F^{T} = \frac{1}{det F} F T^{T} .$

Proof

Multiply $T = det F \overset{˘}{T} F^{- T}$ on the right by $F^{T}$ : $T F^{T} = det F \overset{˘}{T} F^{- T} F^{T} = det F \overset{˘}{T}$ . Since $\overset{˘}{T}$ is symmetric, $\overset{˘}{T} = \overset{˘}{T}^{T} = \frac{1}{d e t F} (T F^{T})^{T} = \frac{1}{d e t F} F T^{T}$ .

Key observation 3.5.6what T meansbeyond the lectures

Column $J$ of $T$ is the force, per unit reference area, on the element whose reference normal was $e_{J}$ . The force is a vector of the current configuration; the area and the normal belong to the reference one. $T$ is a “two-point” tensor, and in general not symmetric. In a tension test it is the engineering stress: force over original cross-section.

Loading the demo…

Figure 3.5.1 One force, two stress vectors. Left, a reference block and a face with normal

m

(turn it). Right, the deformed block, the same face with its new normal

\overset{˘}{m}

, and the Cauchy stress vector

\overset{˘}{t}_{m} = \overset{˘}{T} \overset{˘}{m}

per unit current area. On the left,

t_{m} = Tm

per unit reference area. Change

F

(drag

ν_{1}

ν_{2}

, or a preset) and

\overset{˘}{T}

(sliders); in simple shear

T

is not symmetric.

The boundary condition

Proposition 3.5.7traction condition on the reference boundarynotes L8, p. 4

$\overset{˘}{T} \overset{˘}{m} = \overset{˘}{g} ⟺ T m = g .$ The nominal stress on the reference normal equals the nominal surface force.

Proof

$\overset{˘}{T} \overset{˘}{m} = \overset{˘}{t}_{m} = \frac{1}{Λ _{m}} t_{m} = \frac{1}{Λ _{m}} Tm$ and $\overset{˘}{g} = \frac{1}{Λ _{m}} g$ . Multiply both sides by $Λ_{m}$ .

The divergence

Theorem 3.5.8the divergence of the stressnotes L8, pp. 4–5

$\overset{˘}{\nabla} \cdot \overset{˘}{T} = \frac{1}{det F} \nabla \cdot T^{T}, that is \frac{\partial T ˘ _{ij}}{\partial y _{j}} = \frac{1}{det F} \frac{\partial T _{i J}}{\partial X _{J}},$ where $\overset{˘}{\nabla}$ differentiates in $y$ (current) and $\nabla$ in $X$ (reference).

The notes’ route: by the chain rule the two gradients are related through $F$ , written $\overset{˘}{\nabla} = F^{- 1} \nabla$ ; then substitute $\overset{˘}{T} = \frac{1}{d e t F} F T^{T}$ : $\overset{˘}{\nabla} \cdot \overset{˘}{T} = F^{- 1} (\nabla \cdot (\frac{1}{det F} F T^{T})),$ and the result follows. That last step hides a real argument (warning below), so here is a complete proof.

Proofbeyond the lectures

Take any part $B_{0}$ of the reference configuration and its image $\overset{˘}{B}_{0}$ . The total contact force on its boundary is the same force, measured either way: $\int_{\partial \overset{˘}{B}_{0}} \overset{˘}{T} \overset{˘}{m} d \overset{˘}{A} = \int_{\partial B_{0}} T m d A,$ because $\overset{˘}{T} \overset{˘}{m} d \overset{˘}{A} = \overset{˘}{t}_{m} d \overset{˘}{A} = t_{m} d A = Tm d A$ element by element. Apply the divergence theorem on each side, in its own variables: $\int_{\partial B_{0}} T_{i J} m_{J} d A = \int_{B_{0}} \partial T_{i J} / \partial X_{J} d V$ , and likewise in $y$ . With $d \overset{˘}{V} = det F d V$ (Theorem 2.5.2): $\int_{B_{0}} \frac{\partial T ˘ _{ij}}{\partial y _{j}} det F d V = \int_{B_{0}} \frac{\partial T _{i J}}{\partial X _{J}} d V .$ This holds for every $B_{0}$ , so $det F \partial \overset{˘}{T}_{ij} / \partial y_{j} = \partial T_{i J} / \partial X_{J}$ at every point.

Warningslip in the notes

The step from the notes’ second line to the result is not shown, and it is not a simple cancellation: $F$ and $det F$ depend on the point, so the divergence also acts on them. The result is right, because the extra terms add up to zero (the Piola identity, $\partial (det F F_{J j}^{- 1}) / \partial X_{J} = 0$ ), but the notes do not show it. The chain rule is also written loosely: in components $\partial / \partial X_{J} = F_{i J} \partial / \partial y_{i}$ , i.e. $\nabla = F^{T} \overset{˘}{\nabla}$ and $\overset{˘}{\nabla} = F^{- T} \nabla$ ; the notes drop the transposes.

The material system

Rewrite the Eulerian system with the nominal quantities: $\overset{˘}{\nabla} \cdot \overset{˘}{T} = \frac{1}{d e t F} \nabla \cdot T^{T}$ , $\overset{˘}{b} = \frac{1}{d e t F} b$ , $\overset{ρ}{˘} = \frac{1}{d e t F} ρ$ , and $\overset{˘}{T} = \frac{1}{d e t F} T F^{T} = \frac{1}{d e t F} F T^{T}$ : $⎩ ⎨ ⎧ \frac{1}{det F} \nabla \cdot T^{T} + \frac{1}{det F} b = \frac{1}{det F} ρ a \frac{1}{det F} F T^{T} = T F^{T} \frac{1}{det F} T m = g U = U_{0}$

Theorem 3.5.9the problem in the reference configurationnotes L8, pp. 6–7

Every $\frac{1}{d e t F}$ cancels, leaving the Lagrangian (material) formulation: $⎩ ⎨ ⎧ \nabla \cdot T^{T} + b = ρ a F T^{T} = T F^{T} T m = g U = U_{0} \forall X \in B \forall X \in B \forall X \in \partial B_{f} \forall X \in \partial B_{u}$

The first line is $\partial T_{i J} / \partial X_{J} + b_{i} = ρ a_{i}$ , with the nominal body force, the nominal density and $a = \ddot{U}$ .
Everything is on $B$ , which is known.
The second line replaces the symmetry of $\overset{˘}{T}$ . Since $F \neq = F^{T}$ in general, $T^{T} \neq = T$ : the first Piola–Kirchhoff tensor is not symmetric.

Exercises

Exercise 3.5.1

A bar is stretched to $F = diag (2, \frac{1}{2}, \frac{1}{2})$ and carries the uniaxial Cauchy stress $\overset{˘}{T} = 100 e_{1} \otimes e_{1}$ MPa (only $\overset{˘}{T}_{11} = 100$ MPa). Find $T$ and explain its value.

Solution

$det F = 1$ : no volume change. $F^{- T} = diag (\frac{1}{2}, 2, 2)$ .
$T = det F \overset{˘}{T} F^{- T} = 50 e_{1} \otimes e_{1}$ MPa.
The cross-section’s area is $Λ_{e_{1}} = det F ∣ F^{- T} e_{1} ∣ = \frac{1}{2}$ of the original. The force on it is $100 \cdot \frac{1}{2} = 50$ per unit original area: the engineering stress, $T_{11}$ . The Cauchy stress, per unit current area, is the true stress.

Exercise 3.5.2

Simple shear $F = 100 \frac{1}{2} 10 001$ with $\overset{˘}{T} = 10 e_{2} \otimes e_{2}$ MPa (vertical tension). Find $T$ ; check $F T^{T} = T F^{T}$ ; find the nominal force on the face whose reference normal is $e_{1}$ , and check it from the current configuration.

Solution

$det F = 1$ , $F^{- T} = 1 - \frac{1}{2} 0 010001$ .
$T = \overset{˘}{T} F^{- T} = 0 - 5 0 0100000$ MPa: not symmetric.
$T F^{T} = 10 e_{2} \otimes e_{2} = F T^{T}$ : the angular balance in the material form holds.
Face with reference normal $e_{1}$ : $T e_{1} = (0, - 5, 0)$ MPa per unit reference area.
Check: the face is tilted, $\overset{˘}{m} \propto F^{- T} e_{1} = (1, - \frac{1}{2}, 0)$ , $Λ = 1 + \frac{1}{4}$ . Then $Λ \overset{˘}{T} \overset{˘}{m} = \overset{˘}{T} (1, - \frac{1}{2}, 0) = (0, - 5, 0)$ . ✓ The vertical side of the block, once sheared, carries a vertical force although $\overset{˘}{T}$ has no shear in the $e_{1}, e_{2}$ axes.

Exercise 3.5.3

Show that for a rigid rotation $F = R$ , $T = \overset{˘}{T} R$ . If $\overset{˘}{T}$ is a uniaxial tension along the rotated axis $R e_{1}$ , what is $T$ ?

Solution

$det R = 1$ and $R^{- T} = R$ , so $T = \overset{˘}{T} R$ . With $\overset{˘}{T} = σ (R e_{1}) (R e_{1})^{T} = σ R e_{1} e_{1}^{T} R^{T}$ : $T = σ R e_{1} e_{1}^{T} R^{T} R = σ (R e_{1}) e_{1}^{T}$ . Its only non-zero column is the first, $σ R e_{1}$ : the face whose reference normal was $e_{1}$ carries the force $σ$ along the rotated axis. Not symmetric unless $R e_{1} = \pm e_{1}$ .