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Stress, strain and the Young modulus questions
The spring constant describes one object; reshape the object and k changes even though the material has not. Divide out the geometry and what remains is the Young modulus, a stiffness that belongs to the material itself.
18 original questions · 54 marks · the stress, strain and the young modulus notes · Materials
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Define tensile stress and tensile strain, stating the unit of each.
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Tensile stress = force / cross-sectional area (σ = F/A), unit pascal (Pa) (1). Tensile strain = extension / original length (ε = ΔL/L), which has no unit (1).A wire of cross-sectional area 2.0 × 10−6 m2 carries a tension of 40 N. Calculate the tensile stress.
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σ = F/A = 40/(2.0 × 10−6) (1)
σ = 2.0 × 107 Pa (1)A wire of original length 2.5 m extends by 1.0 mm. Calculate the tensile strain.
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ε = ΔL/L = 1.0 × 10−3/2.5 (1)
ε = 4.0 × 10−4 (1)State what is meant by a brittle material, and describe how its stress–strain graph shows this behaviour.
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A brittle material fractures with little or no plastic deformation (1). Its stress–strain graph is a straight line that ends suddenly at the fracture point, with no extended plastic region (1).The straight initial region of a material's stress–strain graph passes through a stress of 8.0 × 107 Pa at a strain of 4.0 × 10−4. Determine the Young modulus of the material.
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E is the gradient of the straight region = σ/ε = 8.0 × 107/(4.0 × 10−4) (1) = 2.0 × 1011 Pa (1).A wire of length 2.0 m and cross-sectional area 1.5 × 10−6 m2 extends by 0.80 mm when a force of 60 N is applied. Calculate (a) the stress, (b) the strain and (c) the Young modulus of the wire.
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(a) σ = 60/(1.5 × 10−6) = 4.0 × 107 Pa (1)
(b) ε = 0.80 × 10−3/2.0 = 4.0 × 10−4 (1)
(c) E = σ/ε (1)
E = 1.0 × 1011 Pa (1)A steel wire (E = 2.0 × 1011 Pa) has length 3.0 m and cross-sectional area 1.0 × 10−6 m2. Calculate its extension under a tension of 50 N.
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ΔL = FL/(AE) (1)
= (50 × 3.0)/(1.0 × 10−6 × 2.0 × 1011) (1)
ΔL = 7.50 × 10−4 m (0.75 mm) (1)Sketch a typical stress–strain graph for a ductile metal wire and mark the limit of proportionality and the elastic limit. State how the Young modulus is found from the graph.
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The graph is a straight line from the origin up to the limit of proportionality (1), then curves; the elastic limit lies just beyond it (1). The Young modulus is the gradient of the straight, initial part (stress ÷ strain) (1).A wire of length 2.0 m and cross-sectional area 1.5 × 10−6 m2 extends by 0.80 mm under a tension of 60 N. Calculate (a) the elastic strain energy stored in the wire and (b) the energy stored per unit volume.
Wires P and Q are made of the same material and carry the same load. P has twice the original length and twice the diameter of Q. Determine the ratio of the extension of P to the extension of Q.
Explain why the wire used in an experiment to measure the Young modulus should be long and thin.
A steel wire of density ρ and length L hangs vertically from a fixed clamp. Show that the stress at the top of the wire due to its own weight is ρgL, and go on to calculate this stress for a 120 m wire. (Density of steel = 7800 kg m−3; g = 9.81 m s−2.)
A wire supporting a hanging sign has a breaking stress of 4.0 × 108 Pa and a diameter of 0.50 mm. Calculate the maximum tension the wire can support before breaking.
A steel wire (E = 2.0 × 1011 Pa) of length 1.5 m must support a 100 N load while extending by no more than 0.50 mm. Calculate the minimum cross-sectional area required and the corresponding minimum diameter.
Describe a simple laboratory method for measuring the Young modulus of a metal in the form of a wire, naming the measurements you would take.
A stage rigging cable of diameter 6.0 mm must carry a load of 8.0 kN. The breaking stress of the cable material is 5.0 × 108 Pa, and safety rules require the working stress to be no more than half the breaking stress. Deduce whether this cable may be used.
A composite wire is made by joining a 1.0 m steel wire (E = 2.0 × 1011 Pa) end to end with a 1.0 m brass wire (E = 1.0 × 1011 Pa). Both have cross-sectional area 2.0 × 10−7 m2. The composite wire hangs vertically and supports a load of 40 N. Ignoring the weight of the wire itself, calculate the total extension.
Compare the behaviour of a ductile metal and a brittle material as each is stretched to fracture, with reference to the features of their stress–strain graphs.
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