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| Topic | Main idea | Key equation |
|---|---|---|
| Stress tensor | Traction depends on plane orientation | $\mathbf t^{(\mathbf n)}=\boldsymbol\sigma\mathbf n$ |
| Strain tensor | Small deformation from displacement gradients | $\boldsymbol\varepsilon=\tfrac12(\nabla\mathbf u+\nabla\mathbf u^T)$ |
| Hooke's law | Linear elastic stress--strain relation | $\boldsymbol\sigma=2G\boldsymbol\varepsilon+\lambda\operatorname{tr}(\boldsymbol\varepsilon)\mathbf I$ |
| Principal stress | Eigenvalues of the stress tensor | $\boldsymbol\sigma\mathbf n=\sigma\mathbf n$ |
| Plane stress | Out-of-plane stresses vanish | $\sigma_z=\tau_{xz}=\tau_{yz}=0$ |
| Plane strain | Out-of-plane strains vanish | $\varepsilon_z=\gamma_{xz}=\gamma_{yz}=0$ |
| Energy methods | Work and strain energy give displacement | $\delta=\partial U/\partial P$ |
| Tresca | Yield controlled by maximum shear stress | $\sigma_1-\sigma_3=\sigma_Y$ |
| von Mises | Yield controlled by distortion energy | $\sigma_{\mathrm{vm}}=\sigma_Y$ |
| Plasticity | Permanent strain remains after unloading | $\boldsymbol\varepsilon=\boldsymbol\varepsilon^e+\boldsymbol\varepsilon^p$ |
Difficulty
For the plane-stress tensor
find the principal stresses.
Solution
The principal stresses are the eigenvalues of the matrix:
Therefore,
An isotropic material has $E=210\ \text{GPa}$ and $\nu=0.30$. Find its shear modulus.
Solution
Use
For the plane-stress tensor
and unit normal $\mathbf n=(1,0)^T$, find the traction vector.
Solution
Use $\mathbf t^{(\mathbf n)}=\boldsymbol\sigma\mathbf n$:
The normal component is $60\ \text{MPa}$ and the in-plane shear component is $20\ \text{MPa}$.
A strain tensor has $\varepsilon_{xy}=0.002$. What is the engineering shear strain $\gamma_{xy}$?
Solution
Tensor shear strain is half the engineering shear strain:
A ductile material is in plane stress with $\sigma_x=80\ \text{MPa}$, $\sigma_y=20\ \text{MPa}$, and $\tau_{xy}=30\ \text{MPa}$. Estimate the von Mises stress.
Solution
For plane stress,