Interactive Lecture
Principal Stresses, Invariants, and Failure Criteria in 3D
by Dr Arnaldo Delli Carri
The Cauchy stress tensor is a second-order tensor used to represent the state of stress at a point within a material. In three dimensions, it is represented as a 3×3 matrix:
Where:
Due to moment equilibrium, the stress tensor is symmetric, meaning τij = τji.
The stress tensor completely defines the state of stress at a point. Each component represents the stress (force per unit area) acting on a plane perpendicular to one axis in the direction of another axis.
The Green-Lagrange strain tensor describes the deformation of a material from an initial to a deformed state. For small strains, it simplifies to the infinitesimal strain tensor:
Where ε represents normal strains and γ represents engineering shear strains.
Principal stresses are the normal stresses acting on planes where the shear stress is zero. These are found by solving the eigenvalue problem for the stress tensor:
This characteristic equation yields three principal stresses: σ₁ (maximum), σ₂ (intermediate), and σ₃ (minimum).
The principal directions (eigenvectors) indicate the orientations of the planes on which the principal stresses act. For each principal stress σᵢ, the corresponding principal direction nᵢ is found by solving:
The three principal directions are mutually perpendicular (orthogonal) and form a natural coordinate system for the stress state.
At the principal planes, only normal stresses act and shear stresses vanish. That simplification is what makes failure analysis tractable — the principal stress coordinate system is the most natural way to describe the stress state at a point.
The principal stresses can be used to calculate stress invariants, which remain constant regardless of the coordinate system:
Because the invariants don't change when you rotate the coordinate axes, a failure criterion written in terms of them gives the same verdict in every frame — you can test for yielding without first finding the principal directions.
Several criteria predict material failure under complex stress states:
Failure occurs when σVM ≥ σy (yield strength)
Failure occurs when τmax ≥ σy/2
The von Mises criterion is based on distortion energy theory and is widely used for ductile materials. The Tresca criterion is more conservative than von Mises but simpler to calculate.
Full 3D stress analysis is needed wherever a component carries load along more than one axis at once, so the state at a point can't be reduced to a single direction:
| Concept | Equation | Description |
|---|---|---|
| Stress Tensor | \(\boldsymbol{\sigma} = \begin{bmatrix} \sigma_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_{yy} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_{zz} \end{bmatrix}\) | 3D stress state |
| Principal Stresses | \(\det(\boldsymbol{\sigma} - \sigma_p\mathbf{I}) = 0\) | Eigenvalue problem |
| Hydrostatic Stress | \(\sigma_m = \frac{\sigma_1 + \sigma_2 + \sigma_3}{3}\) | Mean normal stress |
| Von Mises Stress | \(\sigma_{VM} = \sqrt{\frac{1}{2}[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2]}\) | Equivalent stress |
| Octahedral Shear | \(\tau_{oct} = \frac{1}{3}\sqrt{(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2}\) | Shear stress on octahedral plane |
A stress state at a point in a material is given by the following stress tensor:
\[[\sigma] = \begin{bmatrix} 60 & 20 & 0 \\ 20 & 40 & 0 \\ 0 & 0 & 30 \end{bmatrix} \text{ MPa}\]
Determine:
Show Solution
(a) Principal stresses:
For the given stress tensor, since σ₃₃ = 30 MPa has no coupling with the other components (σ₁₃ = σ₂₃ = 0), one principal stress is σ₃ = 30 MPa.
For the remaining 2×2 submatrix in the xy-plane:
\[\begin{bmatrix} 60 & 20 \\ 20 & 40 \end{bmatrix}\]
Average stress: \(\sigma_{avg} = \frac{60 + 40}{2} = 50\) MPa
Radius: \(R = \sqrt{(60-40)^2/4 + 20^2} = \sqrt{100 + 400} = \sqrt{500} = 22.36\) MPa
\[\sigma_1 = 50 + 22.36 = 72.36 \text{ MPa}\]
\[\sigma_2 = 50 - 22.36 = 27.64 \text{ MPa}\]
\[\sigma_3 = 30 \text{ MPa}\]
Ordered: σ₁ = 72.36 MPa, σ₂ = 30 MPa, σ₃ = 27.64 MPa
(b) Maximum shear stress:
\[\tau_{max} = \frac{\sigma_1 - \sigma_3}{2} = \frac{72.36 - 27.64}{2} = 22.36 \text{ MPa}\]
(c) Hydrostatic stress:
\[\sigma_m = \frac{\sigma_1 + \sigma_2 + \sigma_3}{3} = \frac{72.36 + 30 + 27.64}{3} = \frac{130}{3} = 43.33 \text{ MPa}\]
Given the stress tensor from Problem 1, calculate:
Show Solution
(a) Stress invariants:
First invariant (trace):
\[I_1 = \sigma_{11} + \sigma_{22} + \sigma_{33} = 60 + 40 + 30 = 130 \text{ MPa}\]
Using principal stresses (easier): \(I_1 = 72.36 + 30 + 27.64 = 130\) MPa ✓
Second invariant:
\[I_2 = \sigma_1\sigma_2 + \sigma_2\sigma_3 + \sigma_3\sigma_1 = 72.36 \times 30 + 30 \times 27.64 + 27.64 \times 72.36\]
\[I_2 = 2170.8 + 829.2 + 2000 = 5000 \text{ MPa}^2\]
Third invariant (determinant):
\[I_3 = \sigma_1\sigma_2\sigma_3 = 72.36 \times 30 \times 27.64 = 60,000 \text{ MPa}^3\]
(b) Deviatoric stress tensor:
Hydrostatic component: \(\sigma_m = I_1/3 = 43.33\) MPa
\[[s] = [\sigma] - \sigma_m[I] = \begin{bmatrix} 60-43.33 & 20 & 0 \\ 20 & 40-43.33 & 0 \\ 0 & 0 & 30-43.33 \end{bmatrix}\]
\[[s] = \begin{bmatrix} 16.67 & 20 & 0 \\ 20 & -3.33 & 0 \\ 0 & 0 & -13.33 \end{bmatrix} \text{ MPa}\]
(c) Von Mises equivalent stress:
\[\sigma_{VM} = \sqrt{\frac{1}{2}[(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2]}\]
\[\sigma_{VM} = \sqrt{\frac{1}{2}[(72.36-30)^2 + (30-27.64)^2 + (27.64-72.36)^2]}\]
\[\sigma_{VM} = \sqrt{\frac{1}{2}[1794.5 + 5.57 + 1999.7]} = \sqrt{1899.9} = 43.59 \text{ MPa}\]
A material is subjected to the following principal stresses:
Using the concept of 3D Mohr's circles, determine:
Show Solution
(a) Radii of three Mohr's circles:
Circle 1 (largest, between σ₁ and σ₃):
\[R_1 = \frac{\sigma_1 - \sigma_3}{2} = \frac{100 - (-20)}{2} = 60 \text{ MPa}\]
Circle 2 (between σ₁ and σ₂):
\[R_2 = \frac{\sigma_1 - \sigma_2}{2} = \frac{100 - 40}{2} = 30 \text{ MPa}\]
Circle 3 (between σ₂ and σ₃):
\[R_3 = \frac{\sigma_2 - \sigma_3}{2} = \frac{40 - (-20)}{2} = 30 \text{ MPa}\]
(b) Absolute maximum shear stress:
\[\tau_{max} = R_1 = 60 \text{ MPa}\]
This occurs on planes at 45° to the σ₁ and σ₃ principal directions.
(c) Minimum normal stress:
The minimum normal stress on any plane is the smallest principal stress:
\[\sigma_{min} = \sigma_3 = -20 \text{ MPa (compression)}\]
(d) Factor of safety:
Tresca (Maximum Shear Stress) Criterion:
\[\sigma_{eq,Tresca} = \sigma_1 - \sigma_3 = 100 - (-20) = 120 \text{ MPa}\]
\[FS_{Tresca} = \frac{\sigma_y}{\sigma_{eq}} = \frac{200}{120} = 1.67\]
Von Mises Criterion:
\[\sigma_{VM} = \sqrt{\frac{1}{2}[(100-40)^2 + (40-(-20))^2 + ((-20)-100)^2]}\]
\[\sigma_{VM} = \sqrt{\frac{1}{2}[3600 + 3600 + 14400]} = \sqrt{10800} = 103.92 \text{ MPa}\]
\[FS_{vonMises} = \frac{200}{103.92} = 1.92\]
Interpretation: The von Mises criterion predicts a higher factor of safety (1.92 vs 1.67) because it combines all three principal-stress differences quadratically, whereas Tresca keys only on the largest difference σ₁ − σ₃. Tresca therefore ignores the moderating effect of the intermediate stress σ₂ and gives the more conservative (lower) factor of safety.