A metal wire of diameter \(0.50\,\mathrm{mm}\) and original length \(2.00\,\mathrm{m}\) is stretched by \(1.2\,\mathrm{mm}\). Calculate the strain in the wire.
Strain \(= \Delta L / L = 1.2\times10^{-3}/2.00 = 6.0\times10^{-4}\) (dimensionless).
Mechanics & Materials
10 exam-style questions on density, stress, strain, Young modulus and elastic energy. Tap Show answer for full working.
3 Easy 4 Medium 3 Hard
A steel bar has cross-sectional area \(2.5\times10^{-4}\,\mathrm{m}^{2}\) and is subjected to a tensile force of \(15\,\mathrm{kN}\). Calculate the stress in the bar.
Stress \(\sigma = F/A = 15000 / 2.5\times10^{-4} = 6.0\times10^{7}\,\mathrm{Pa}\) (\(60\,\mathrm{MPa}\)).
Define density and state its SI unit. A cube of side \(4.0\,\mathrm{cm}\) has mass \(0.48\,\mathrm{kg}\). Calculate its density.
Density \(\rho = m/V\), unit \(\mathrm{kg\,m}^{-3}\).
\(V = (0.04)^{3} = 6.4\times10^{-5}\,\mathrm{m}^{3}\); \(\rho = 0.48/6.4\times10^{-5} = 7500\,\mathrm{kg\,m}^{-3}\).
A wire of length \(1.50\,\mathrm{m}\) and cross-sectional area \(0.80\,\mathrm{mm}^{2}\) is stretched by \(0.90\,\mathrm{mm}\) when a load of \(25\,\mathrm{N}\) is applied. Calculate the Young modulus of the material.
\(E = \sigma/\varepsilon = (F/A)/(\Delta L/L) = FL/(A\Delta L)\).
\(A = 0.80\times10^{-6}\,\mathrm{m}^{2}\); \(E = 25\times1.50/(0.80\times10^{-6}\times0.90\times10^{-3}) = 5.21\times10^{10}\,\mathrm{Pa}\).
A spring of natural length \(0.40\,\mathrm{m}\) and spring constant \(120\,\mathrm{N\,m}^{-1}\) is stretched to \(0.55\,\mathrm{m}\). Calculate the elastic potential energy stored and the force required to hold it at this extension.
Extension \(x = 0.15\,\mathrm{m}\).
\(E = \frac12 kx^{2} = 0.5\times120\times0.0225 = 1.35\,\mathrm{J}\).
\(F = kx = 120\times0.15 = 18\,\mathrm{N}\).
A copper wire of diameter \(1.2\,\mathrm{mm}\) and length \(2.5\,\mathrm{m}\) is subjected to a tensile stress of \(1.5\times10^{8}\,\mathrm{Pa}\). Young modulus for copper is \(1.2\times10^{11}\,\mathrm{Pa}\). Calculate the extension of the wire and the strain energy stored if the deformation is elastic.
Strain \(\varepsilon = \sigma/E = 1.5\times10^{8}/1.2\times10^{11} = 1.25\times10^{-3}\).
\(\Delta L = \varepsilon L = 1.25\times10^{-3}\times2.5 = 3.125\times10^{-3}\,\mathrm{m}\).
\(A = \pi(0.6\times10^{-3})^{2} = 1.13\times10^{-6}\,\mathrm{m}^{2}\).
Energy \(= \frac12 \sigma\varepsilon\times\text{volume} = 0.5\times1.5\times10^{8}\times1.25\times10^{-3}\times(1.13\times10^{-6}\times2.5) = 0.265\,\mathrm{J}\).
Sketch a typical stress–strain graph for a ductile metal up to fracture. Label the limit of proportionality, yield point and ultimate tensile strength. Explain what is meant by plastic deformation.
Graph rises linearly to the limit of proportionality, continues almost linearly to the yield point, then curves, reaches a maximum (UTS) and falls to fracture.
Plastic deformation: permanent deformation that remains after the stress is removed; the material does not return to its original length.
Two wires of the same material and length are connected in parallel and support a total load of \(200\,\mathrm{N}\). Wire A has cross-sectional area \(1.0\,\mathrm{mm}^{2}\) and wire B has area \(3.0\,\mathrm{mm}^{2}\). Calculate the stress in each wire, assuming they stretch by the same amount.
Same material and length \(\Rightarrow\) same strain \(\Rightarrow\) stress proportional to area? No: same strain \(\Rightarrow\) same stress (Hooke’s law). Force shared in proportion to area.
\(F_A/F_B = A_A/A_B = 1/3\). \(F_A + F_B = 200 \Rightarrow F_A = 50\,\mathrm{N}\), \(F_B = 150\,\mathrm{N}\).
Stress in both: \(\sigma = 50/(1.0\times10^{-6}) = 5.0\times10^{7}\,\mathrm{Pa}\) (same for B).
A steel cable of unstretched length \(25\,\mathrm{m}\) and cross-sectional area \(4.0\,\mathrm{cm}^{2}\) supports a lift of mass \(1200\,\mathrm{kg}\). Young modulus for steel \(= 2.0\times10^{11}\,\mathrm{Pa}\). Calculate the extension of the cable when the lift is at rest, and the additional extension when the lift accelerates upwards at \(1.5\,\mathrm{m\,s}^{-2}\). (\(g=9.8\,\mathrm{m\,s}^{-2}\).)
At rest: \(F = mg = 11760\,\mathrm{N}\); \(A = 4.0\times10^{-4}\,\mathrm{m}^{2}\).
\(\Delta L = FL/(AE) = 11760\times25/(4.0\times10^{-4}\times2.0\times10^{11}) = 3.68\times10^{-3}\,\mathrm{m}\).
Accelerating: \(T - mg = ma \Rightarrow T = m(g+a) = 1200\times11.3 = 13560\,\mathrm{N}\).
New extension \(= 13560\times25/(4.0\times10^{-4}\times2.0\times10^{11}) = 4.24\times10^{-3}\,\mathrm{m}\).
Additional extension \(= 0.56\,\mathrm{mm}\).
A rubber cord of natural length \(0.80\,\mathrm{m}\) and cross-sectional area \(4.0\,\mathrm{mm}^{2}\) is stretched to \(1.20\,\mathrm{m}\). The force–extension graph is linear up to this point with gradient \(150\,\mathrm{N\,m}^{-1}\). Calculate the Young modulus of the rubber and the energy stored in the cord.
Extension \(x = 0.40\,\mathrm{m}\); \(k = 150\,\mathrm{N\,m}^{-1}\).
\(E = (F/A)/(\Delta L/L) = (kx\,L)/(A\,x) = kL/A = 150\times0.80/(4.0\times10^{-6}) = 3.0\times10^{7}\,\mathrm{Pa}\).
Energy \(= \frac12 kx^{2} = 0.5\times150\times0.16 = 12\,\mathrm{J}\).