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Questions on stress, strain and the Young modulus
The Young modulus, tensile stress over tensile strain, is stiffness with the specimen's dimensions divided out, so it belongs to the material and does not change when the material is reshaped. Strain has no unit, so the modulus is measured in pascals like the stress. The gradient of the straight part of a stress-strain graph gives it, and the curves for ductile, brittle and polymeric materials differ beyond the straight part.
19 original questions · 58 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 = 7.5 × 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.
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(a) E = ½FΔL = ½ × 60 × 0.80 × 10−3 (1)
E = 0.024 J (1)
(b) energy density = ½σε = ½ × 4.0 × 107 × 4.0 × 10−4 = 8000 J m−3 (1)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.
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ΔL = FL/(AE), and A ∝ d2, so doubling the diameter makes AP = 4AQ (1). ΔLP/ΔLQ = (LP/AP)/(LQ/AQ) = 2/4 (1) = 0.5: P extends half as much as Q (1).Explain why the wire used in an experiment to measure the Young modulus should be long and thin.
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A long wire, and the larger stress a small cross-section gives, both make the extension larger for a given load (1). A larger extension carries a smaller percentage uncertainty, so the value of E is more precise (1).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.)
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Weight of wire = mass × g = ρALg, and the top of the wire supports all of it (1). Stress = weight/A = ρALg/A = ρgL, independent of the area (1). σ = 7800 × 9.81 × 120 = 9.2 × 106 Pa (1).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.
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A = π(d/2)2 = π × (0.25 × 10−3)2 = 1.963 × 10−7 m2 (1)
Maximum force = σbreakA = 4.0 × 108 × 1.963 × 10−7 (1)
F = 79 N, to the two significant figures the stress and the diameter carry (1)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.
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A = FL/(EΔL) (1)
= (100 × 1.5)/(2.0 × 1011 × 0.50 × 10−3) (1)
A = 1.50 × 10−6 m2 (1)
d = √(4A/π) = 1.38 mm (1)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.
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Clamp a long thin wire, hang increasing known loads and measure the extension with a marker and ruler (or vernier) (1), plus the original length with a rule and the diameter with a micrometer (to get area) (1). Plot stress against strain; the gradient of the straight region gives E (1).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.
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A = π(d/2)2 = π × (3.0 × 10−3)2 = 2.83 × 10−5 m2 (1). Working stress = F/A = 8000/(2.83 × 10−5) = 2.8 × 108 Pa (1). Maximum permitted stress = ½ × 5.0 × 108 = 2.5 × 108 Pa (1). The working stress exceeds this, so the cable may not be used; a larger diameter is needed (1).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.
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The same tension of 40 N acts throughout both sections (1). ΔLsteel = FL/(AE) = (40 × 1.0)/(2.0 × 10−7 × 2.0 × 1011) = 1.0 mm (1). ΔLbrass = (40 × 1.0)/(2.0 × 10−7 × 1.0 × 1011) = 2.0 mm (1). Total extension = 3.0 mm (1).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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Both graphs begin with a straight region through the origin whose gradient is the Young modulus (1). The ductile metal passes a yield point, beyond which it undergoes a large plastic strain at little extra stress (1); the maximum stress it withstands is its breaking stress, and fracture occurs at large strain (1). The brittle material's graph stays straight until it fractures, with little or no plastic region (1). The area under the brittle graph is far smaller, so the brittle material absorbs little energy and fails suddenly, without warning (1).The graph is the tensile stress–strain curve for a metal, obtained from a test specimen. A wire made of the same metal has a diameter of 0.80 mm and an original length of 1.60 m. Determine the Young modulus of the metal, and hence the extension of the wire when it carries a load of 60 N.
The stress–strain curve for the metal, straight at first and then bending over. The Young modulus is not stated anywhere; it has to come from the straight part. Mark scheme
The Young modulus is the gradient of the straight section, so read one point on it and divide the stress by the strain, for example 3.0 × 108 Pa at a strain of 2.0 × 10−3 (1)
E = 3.0 × 108/(2.0 × 10−3) = 1.5 × 1011 Pa (1)
A = πd2/4 = π × (0.80 × 10−3)2/4 = 5.03 × 10−7 m2, so the stress in the wire is 60/(5.03 × 10−7) = 1.19 × 108 Pa, which lies on the straight section, so the Young modulus may be used (1)
ΔL = σL/E = 1.19 × 108 × 1.60/(1.5 × 1011) = 1.3 × 10−3 m, that is 1.3 mm (1)
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