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Angular-momentum balance without couples forces the stress matrix to be symmetric

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), as used by the divergence and surface-integral machinery. Let $O \subseteq R^{3}$ be open, let $σ : O \to M_{3} (R)$ be of class $C^{1}$ , and let $b : O \to R^{3}$ be continuous. Suppose that for every bounded $C^{1}$ domain $D$ with $\overline{D} \subseteq O$ (Bounded C1 domains and their outward normals), with outward unit normal $ν$ and the surface integral of Surface integration on compact C1 hypersurfaces:

$\begin{matrix} \int_{\partial D} σ ν d S + \int_{D} b d x = 0 & (LB) \end{matrix}$

and

$\begin{matrix} \int_{\partial D} y \times (σ ν) d S + \int_{D} y \times b d x = 0. & (AB) \end{matrix}$

Then

$σ (y)^{T} = σ (y) for every y \in O .$

Facts & Assumptions

Given: ${A C}_{ω}$ ; the open set $O$ ; the $C^{1}$ field $σ$ ; the continuous field $b$ ; the balances (LB) and (AB).

[F1]

Divergence theorem: for $F \in C^{1} (\overline{D}; R^{3})$ on a bounded $C^{1}$ domain $D$ , $\int_{D} div F d x = \int_{\partial D} F \cdot ν d S$ (Divergence on a bounded C1 Euclidean domain). Applying it to the rows $F_{i} = (σ_{i 1}, σ_{i 2}, σ_{i 3})$ gives $\int_{\partial D} σ ν d S = \int_{D} g d x$ with $g_{i} : = \sum_{j} \partial_{j} σ_{i j}$ .

[F2]

Cross product as a bilinear alternating map: $(a \times b)_{i} = \sum_{j, k} ε_{i j k} a_{j} b_{k}$ , where $ε_{i j k} = 0$ if indices repeat, $+ 1$ for an even permutation of $(1, 2, 3)$ , and $- 1$ for an odd permutation; and the product rule differentiates $y_{j} σ_{k l}$ as $\partial_{l} (y_{j} σ_{k l}) = δ_{j l} σ_{k l} + y_{j} \partial_{l} σ_{k l}$ (The cross product in $R^{3}$ , The cross product is bilinear, alternating, and orthogonal to both factors, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , A total derivative computes every directional derivative, and its matrix is the Jacobian, The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations).

[F3]

Localization lemma: a continuous function on an open set whose integral vanishes over every member of a neighbourhood-basis family is identically zero; the family of all small balls qualifies (Localization: a continuous function whose integrals vanish over a neighbourhood-basis family is identically zero, Open ball, closed ball and sphere in a metric space).

[F4]

The integrals in (LB) and (AB) are defined componentwise and are finite for $C^{1}$ data on a bounded $C^{1}$ domain (Integral over a measurable subset, The Lebesgue integral is linear on $L^{1} (μ)$ ).

Proof

technique · direct; differentiate the two integral balances by the divergence theorem and localize

1.1givenF1F3F4algebra

Fix a bounded $C^{1}$ domain $D$ with $\overline{D} \subseteq O$ . By [F1] applied to the three rows of $σ$ , (LB) becomes $\int_{D} (g + b) d x = 0, g_{i} : = \sum_{j} \partial_{j} σ_{i j} .$ Since this holds for every bounded $C^{1}$ domain with closure in $O$ , it holds in particular for every open ball $B (y_{0}, r)$ with $\overline{B} (y_{0}, r) \subseteq O$ ; the function $y \mapsto g (y) + b (y)$ is continuous, so [F3] gives $\begin{matrix} g + b = 0 on O . & (M) \end{matrix}$

2.1givenF1F2step 1.1algebra

Fix a bounded $C^{1}$ domain $D$ with $\overline{D} \subseteq O$ and a component index $i \in {1, 2, 3}$ . By [F2], and applying the divergence theorem [F1] to the components $F_{l}^{(i)} : = (ε_{i j k} y_{j} σ_{k l})_{l}$ of the vector field whose $l$ -th entry is $ε_{i j k} y_{j} σ_{k l}$ (summed over $j, k$ ), $(\int_{\partial D} y \times (σ ν) d S)_{i} = \int_{D} \sum_{j, k, l} ε_{i j k} (δ_{j l} σ_{k l} + y_{j} \partial_{l} σ_{k l}) d x = \int_{D} (\sum_{j, k} ε_{i j k} σ_{k j} + (y \times g)_{i}) d x .$ Hence (AB) reads $\int_{D} (\sum_{j, k} ε_{i j k} σ_{k j} + (y \times (g + b))_{i}) d x = 0.$ By (M), the second term vanishes identically, leaving $\int_{D} (\sum_{j, k} ε_{i j k} σ_{k j}) d x = 0$ for every bounded $C^{1}$ domain with closure in $O$ .

3.1step 2.1F3algebra∎

Apply step 2.1 to open balls with closure in $O$ and localize with [F3]: the continuous function $y \mapsto \sum_{j, k} ε_{i j k} σ_{k j} (y)$ vanishes on $O$ for each $i$ . For $i = 1$ this reads $σ_{32} - σ_{23} = 0$ , for $i = 2$ it reads $σ_{13} - σ_{31} = 0$ , and for $i = 3$ it reads $σ_{21} - σ_{12} = 0$ ; hence $σ^{T} = σ$ on $O$ .

Remarks

The content is conditional: (LB) and (AB) are adopted balance hypotheses for bodies without body couples or surface couples. If a couple density $c$ is present, (AB) must add $\int_{D} c d x$ and the conclusion weakens to the local relation $ε_{i j k} σ_{k j} + c_{i} = 0$ ; no symmetry follows in that case, and this item asserts symmetry only in the couple-free balance.
The orbital terms cancel exactly because both balances use the same $b$ : this is what the subtraction of (M) inside the angular integral expresses. Using a different body-force convention in the two balances would invalidate the cancellation and is not part of this statement.
The center of the angular balance is taken to be the origin of the fixed coordinates; translating the center adds to both integrals a term proportional to the total balance (LB) and therefore does not change the conclusion. No rotational frame or physical interpretation is used.

Angular-momentum balance without couples forces the stress matrix to be symmetric

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources