ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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Resting Bingham graph supports nonzero stress

Physical scope

Worked model calculation/counterexample; leaf, no downstream suppliers.

Example

Use the Bingham graph on ${S y m}_{0} (3) = {D \in M_{3} (R) : D = D^{T}, tr D = 0}$ with Frobenius norm $∣ S ∣ = (tr (S^{T} S))^{1 / 2}$ , whose zero-rate value is $B (0) = {S \in {S y m}_{0} (3) : ∣ S ∣ \leq τ_{y}}$ . Let the strain rate be $D = 0$ , let the yield-ball radius satisfy $τ_{y} > 0$ in Pa, and choose

S = \frac{τ_{y}}{\sqrt{2}} (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \end{array}) .

This example tests membership in the constitutive graph only. It does not assert force balance, a realized material state, or an experimental observation.

Example references

J. N. Reddy, An Introduction to Continuum Mechanics, 2nd ed., Chapter 6, §6.6.2.3, PDF p. 27 (printed p. 247)

Chiang C. Mei, MIT 1.63J/2.21J Lecture Notes on Fluid Dynamics, §1.6.2, equations (1.6.14)–(1.6.16), complete 4-page PDF

Facts & Assumptions

Given: The Frobenius-normalized Bingham graph and a positive yield-ball radius $τ_{y}$ .

[F1]

Bingham yield graph and convex monotonicity: at zero strain rate, the Bingham graph value is the full set of trace-free symmetric stresses whose Frobenius norm is at most $τ_{y}$ .

[A1]

$τ_{y} > 0$ Pa and the strain rate is $D = 0$ .

Proof

1.1givenA1algebra

The displayed matrix $S$ is symmetric and has trace $\frac{τ_{y}}{\sqrt{2}} (1 - 1 + 0) = 0$ . Its squared Frobenius norm is the sum of its squared entries, $∣ S ∣^{2} = \frac{τ_{y}^{2}}{2} (1^{2} + (- 1)^{2}) = τ_{y}^{2}$ , so $∣ S ∣ = τ_{y}$ .

2.1F1step 1.1∎

By [F1] and $D = 0$ , the admissible rest stresses are exactly the trace-free symmetric matrices with norm at most $τ_{y}$ . Step 1.1 shows that $S$ meets these conditions; since $τ_{y} > 0$ , it is nonzero. Thus $S \in B (0)$ is a nonzero graph-admissible stress at rest.

Remarks

The figure shows the planar section of the five-dimensional space ${S y m}_{0} (3)$ spanned by

E_{1} = \frac{1}{\sqrt{2}} (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \end{array}), E_{2} = \frac{1}{\sqrt{2}} (\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}) .

These matrices are Frobenius-orthonormal, so stress coordinates $s_{1}, s_{2}$ in Pa give $∣ s_{1} E_{1} + s_{2} E_{2} ∣^{2} = s_{1}^{2} + s_{2}^{2}$ . The disk is the intersection of the full stress ball with this plane, and its boundary point $(τ_{y}, 0)$ represents the proved stress $S = τ_{y} E_{1}$ . The cross-section is explanatory; the proof above uses the matrix entries and the supplier's graph definition, not the drawing.

\begin{center} \begin{tikzpicture}[scale=1] \draw[->] (-1.45,0) -- (1.55,0) node[right] { $s_{1}$ }; \draw[->] (0,-1.35) -- (0,1.45) node[above] { $s_{2}$ }; \draw[thick] (0,0) circle [radius=1.1]; \fill (0,0) circle (1.2pt) node[below left] { $0$ }; \fill (1.1,0) circle (1.7pt) node[below right] { $(τ_{y}, 0)$ }; \draw[dashed] (0,0) -- (1.1,0) node[midway,above] { $τ_{y}$ }; \end{tikzpicture} \end{center}

Depends on

Bingham yield graph and convex monotonicity