Stress

Materials - OCR A-Level Physics

Key Definition
Stress
The force per unit cross-sectional area acting on a material. SI unit: pascal (Pa).
$$\sigma = \frac{F}{A}$$
Key Definition
Strain
The fractional change in length of a material. It is a ratio and has no units.
$$\varepsilon = \frac{\Delta L}{L}$$
Key Definition
Young modulus
The ratio of stress to strain for a material within the limit of proportionality. It measures the stiffness of the material itself.
$$\begin{aligned} E &= \frac{\sigma}{\varepsilon} \\ &= \frac{FL}{A\Delta L} \end{aligned}$$
  • The Young modulusThe ratio of stressThe force applied per unit cross-sectional area of a material. Measured in pascals (Pa). to strainThe fractional change in length of a material under stress. It is dimensionless (no units). for a material in the elastic region. A measure of stiffness. Measured in pascals (Pa). is a material propertyA property that depends only on the type of material (e.g. Young modulus, UTS), not on its dimensions.: it does not depend on the dimensions of the sample
  • StressThe force applied per unit cross-sectional area of a material. Measured in pascals (Pa). and strainThe fractional change in length of a material under stress. It is dimensionless (no units). allow comparison of materials regardless of their size or shape
  • The Young modulusThe ratio of stressThe force applied per unit cross-sectional area of a material. Measured in pascals (Pa). to strainThe fractional change in length of a material under stress. It is dimensionless (no units). for a material in the elastic region. A measure of stiffness. Measured in pascals (Pa). is the gradient of the linear region of a stress-strain graph
  • Typical values: steel ~200 GPa, copper ~120 GPa, glass ~70 GPa, rubber ~0.01 GPa
Common Mistake MEDIUM
Students often: Don't use the diameter instead of the radius when calculating cross-sectional area, or forget to halve the diameter.
Instead: $A = \pi r^2. If given a diameter d$, then $r = d/2$ and $A = \pi(d/2)^2 = \pi d^2/4. Always check whether$ the question gives radius or diameter.
Materials Overview