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Define stress and strain, and state the SI unit of each.
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Stress is the force applied per unit cross-sectional area; strain is the extension per unit original length; the SI unit of stress is the pascal (Pa) / N m⁻², and strain has no unit (dimensionless).
A steel wire of original length 1.50 m and diameter 0.30 mm extends by 0.80 mm under a tensile force of 8.0 N. Show that the Young modulus of the steel is approximately 2.1 × 10¹¹ Pa.
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Finding the cross-sectional area $$A=\pi {\left(\frac{d}{2}\right)}^{2}=\pi {\left(\frac{0.30\times {10}^{-3}}{2}\right)}^{2}$$ $$A=7.07\times {10}^{-8}{\text{ m}}^{2}$$ Finding the stress $$\sigma =\frac{F}{A}=\frac{8.0}{7.07\times {10}^{-8}}$$ $$\sigma =1.13\times {10}^{8}\text{ Pa}$$ Finding the strain $$\epsilon =\frac{\Delta x}{L}=\frac{0.80\times {10}^{-3}}{1.50}$$ $$\epsilon =5.33\times {10}^{-4}$$ Finding the Young modulus $$E=\frac{\sigma }{\epsilon }=\frac{1.13\times {10}^{8}}{5.33\times {10}^{-4}}$$ $$E=2.12\times {10}^{11}\text{ Pa}$$ This is approximately 2.1 × 10¹¹ Pa as required.
Explain why diameter is measured at several points along the wire and the readings averaged in Core Practical 3.
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The wire may not be perfectly uniform along its length / the cross-section may vary; averaging reduces random error in the calculated cross-sectional area.
A copper wire breaks when the stress exceeds 2.2 × 10⁸ Pa. The wire has a cross-sectional area of 4.0 × 10⁻⁷ m². Deduce whether a tensile load of 75 N will break the wire.
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Finding the stress in the wire $$\sigma =\frac{F}{A}=\frac{75}{4.0\times {10}^{-7}}$$ $$\sigma =1.875\times {10}^{8}\text{ Pa}$$ This is less than the breaking stress of 2.2 × 10⁸ Pa; therefore the wire will not break under this load.
A student carries out Core Practical 3 to determine the Young modulus of a metal wire. Discuss how the student should plan and process the experiment to obtain an accurate value of the Young modulus. In your answer you should: describe the key measurements and the instruments used; explain how the data should be processed to obtain the Young modulus; identify two sources of error and how each is reduced.
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Measure the original length L using a metre ruler and the diameter d at several points using a micrometer screw gauge; calculate the cross-sectional area from $A=\pi (d/2{)}^{2}$; apply increasing known loads and record the extension Δx for each load using a fixed marker and ruler. Calculate the stress $\sigma =F/A$ and strain $\epsilon =\Delta x/L$ for each reading; plot stress against strain; the gradient of the linear region gives the Young modulus. Random error in the diameter is reduced by averaging readings at several positions; systematic error from a kinked wire is reduced by applying a small preload before taking the zero reading; use of a long thin wire produces a larger and more easily measured extension, improving accuracy.
State the equation for elastic strain energy stored in a material obeying Hooke’s law, and define each symbol.
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${E}_{el}=1/2F\Delta x$; where Eel = elastic strain energy (J), F = applied force (N), and Δx = extension (m).
A spring of stiffness 250 N m⁻¹ is stretched by 0.080 m within its elastic limit. Calculate the elastic strain energy stored.
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$${E}_{el}=1/2k{x}^{2}=1/2\times 250\times (0.080{)}^{2}$$ $${E}_{el}=0.80\text{ J}$$ The spring stores 0.80 J of elastic strain energy.
A wire obeying Hooke’s law stores 0.20 J of elastic strain energy when stretched by 4.0 mm. The extension is now doubled while the wire remains within its elastic limit. Calculate the new elastic strain energy stored.
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Since ${E}_{el}∝{x}^{2}$ when Hooke’s law holds, doubling the extension multiplies the energy by 4: $${E}_{el}=4\times 0.20=0.80\text{ J}$$
A rubber band produces a non-linear force–extension graph. Explain why ${E}_{el}=1/2F\Delta x$ cannot be used to find the energy stored, and describe the correct method.
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The equation ${E}_{el}=1/2F\Delta x$ only applies when the graph is a straight line through the origin / when Hooke’s law is obeyed; for a non-linear graph the area under the curve is no longer a triangle; the elastic strain energy is found by estimating the area under the curve, for example by counting squares on the gridded graph or by dividing the area into trapeziums and summing them.