Solid materials: Hooke's law, stress and strain

WPH11Topic 1.64 questions

Hooke's law, stress, strain and the Young modulus, force-extension and stress-strain graphs, and elastic strain energy.

Specification points

Be able to use the Hooke's law equation, $\Delta F = k\Delta x$ , where $k$ is the stiffness of the object
Understand how to use the relationships (tensile or compressive) stress = force / cross-sectional area, (tensile or compressive) strain = change in length / original length, and Young modulus = stress / strain

E

Be able to draw and interpret force-extension and force-compression graphs

Understand the terms limit of proportionality, elastic limit, yield point, elastic deformation and plastic deformation and be able to apply them to force-extension and force-compression graphs

Be able to draw and interpret tensile or compressive stress-strain graphs, and understand the term breaking stress

CORE PRACTICAL 3: Determine the Young modulus of a material

Be able to calculate the elastic strain energy $E_{el}$ in a deformed material sample, using the equation $\Delta E_{el} = \frac{1}{2}F\Delta x$ , and from the area under the force-extension graph (including estimating the area for linear and non-linear graphs)

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Sample questions

Original practice questions written in the style of the exam.

Question 13 marks
Explain what is meant by each of the following terms for a material under tension.

(a) elastic deformation

(b) plastic deformation

(c) brittle
Mark scheme
- Elastic deformation: the material returns to its original shape/length when the force is removed (1)
- Plastic deformation: the material does not return to its original shape; it is permanently deformed when the force is removed (1)
- Brittle: the material breaks with little or no plastic deformation (1)
Question 24 marks
A steel wire of original length $2.50\,\text{m}$ and diameter $0.50\,\text{mm}$ hangs vertically. When a load of $40\,\text{N}$ is attached, the wire extends by $2.4\,\text{mm}$ .

Calculate the Young modulus of the steel.
Mark scheme
- Cross-sectional area $A = \pi (0.25 \times 10^{-3})^2 = 1.96 \times 10^{-7}\,\text{m}^2$ (1)

1.96

×

10−7m2

Stress

= 40/1.96 \times 10^{-7} = 2.04 \times 10^{8}\,\text{Pa}

(1)

Strain

= 2.4 \times 10^{-3}/2.50 = 9.6 \times 10^{-4}

(1)

E = \text{stress}/\text{strain} = 2.1 \times 10^{11}\,\text{Pa}

(1)