AcademyEquilibrium and Materials
Academy
Stress and Strain
Level 1 - Physics topic page in Equilibrium and Materials.
Principle
Stress measures internal load; strain measures fractional deformation.
Notation
\(\sigma,\tau\)
normal and shear stress
\(\mathrm{Pa}\)
\(\epsilon,\gamma\)
normal and shear strain
none
\(F_{\perp},F_{\parallel}\)
normal and parallel force components
\(\mathrm{N}\)
\(A\)
loaded cross-sectional area
\(\mathrm{m^{2}}\)
\(\Delta L,L_0\)
length change and original length
\(\mathrm{m}\)
\(\Delta x,h\)
sideways shift and specimen height
\(\mathrm{m}\)
Method
Derivation 1: Normalize load by area
Stress uses the force component that acts on a chosen area, so it compares loading without depending on specimen size.
Normal stress
\[\sigma=\frac{F_{\perp}}{A}\]
Shear stress
\[\tau=\frac{F_{\parallel}}{A}\]
Derivation 2: Normalize deformation by original size
Strain compares the change in shape with the original dimension that is being distorted.
Normal strain
\[\epsilon=\frac{\Delta L}{L_0}\]
Shear strain
\[\gamma=\frac{\Delta x}{h}\]
Derivation 3: Separate geometry from material response
The stress-strain diagram shows material behavior after load and deformation have been normalized.
Rules
These are the compact results from the derivations above.
Normal stress
\[\sigma=\frac{F_\perp}{A}\]
Normal strain
\[\epsilon=\frac{\Delta L}{L_0}\]
Shear stress
\[\tau=\frac{F_\parallel}{A}\]
Shear strain
\[\gamma=\frac{\Delta x}{h}\]
Examples
Question
A
\[2000\,\mathrm{N}\]
tensile force acts on area \[4.0\times10^{-4}\,\mathrm{m^2}\]
Find stress.Answer
\[\sigma=F/A=2000/(4.0\times10^{-4})=5.0\times10^6\,\mathrm{Pa}\]
Checks
- Stress has pressure units.
- Strain is dimensionless.
- Use loaded cross-sectional area.
- Large deformation breaks small-strain models.