Stress
Materials - OCR A-Level Physics
Key Definition
Stress
The force per unit cross-sectional area acting on a material. SI unit: pascal (Pa).
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.
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.
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
Wrong: Using the diameter as the radius, or forgetting to halve it. This makes $A$ four times too big and $E$ four times too small.
Right: $A = \pi r^2$. If the question gives diameter $d$, take $r = d/2$ so $A = \pi (d/2)^2 = \pi d^2 / 4$. Always check whether the value given is radius or diameter before substituting.
Right: $A = \pi r^2$. If the question gives diameter $d$, take $r = d/2$ so $A = \pi (d/2)^2 = \pi d^2 / 4$. Always check whether the value given is radius or diameter before substituting.