Tensile and compressive stress, σ= F/A, including the area of a wire from its diameter

Tensile and compressive strain, ε= Δx/x

The Young modulus, E = σ/ε, and using it to predict extensions

Stress–strain graphs for a ductile metal; breaking stress (ultimate tensile stress)

Describing materials as stiff, strong, brittle, ductile, malleable, hard or tough, using graphs as evidence

Core Practical 3: determining the Young modulus of a material in the form of a wire

1. Dividing force by area and extension by original length turns a force–extension graph for one sample into a stress–strain graph for the material, whose gradient is the Young modulus.

2. Stress σ= F/A (unit Pa = N m⁻²); strain ε= Δx/x (no unit). Both can be tensile (stretching) or compressive (squashing).

3. Young modulus E = σ/ε= Fx/(AΔx), unit Pa. A thicker or longer wire of the same metal has the same E.

4. On a stress–strain graph the gradient of the straight section is E; the highest point is the breaking stress (ultimate tensile stress).

5. Stiff = large E; strong = large breaking stress; brittle = breaks with little or no plastic deformation; ductile = large plastic deformation in tension; tough = absorbs a lot of energy before breaking.

6. Core Practical 3 uses a long thin wire, a micrometer for the diameter at several places, a fixed scale for the extension, and the gradient of a stress–strain graph.

A force–extension graph describes one particular wire. Hang the same load on a wire twice as thick and it stretches a quarter as much; use a wire twice as long and it stretches twice as much. To compare materials rather than objects, we need quantities that take out the size of the sample. Stress does that for the force.

Tensile stress: the forces pull outwards and tend to stretch the sample (a cable, a rope, a guitar string).

Compressive stress: the forces push inwards and tend to squash it (a column, a table leg, the sole of a shoe).

The area A is the cross-section at right angles to the force.

Stresses in materials are large numbers: 1 MPa = 10⁶ Pa, 1 GPa = 10⁹ Pa. Areas of wires are tiny: 1 mm² = 10⁻⁶ m².

WORKED EXAMPLE · QUESTION

A lamp of mass 42 kg hangs at rest from a single steel wire of diameter 2.4 mm. Calculate the tensile stress in the wire.

SOLUTION

The tension equals the weight: F = mg = 42×9.81 = 412 N.

Area from the diameter, in metres: A = π×(2.4×10⁻³)²÷4 = 4.52×10⁻⁶ m².

σ= 412÷4.52×10⁻⁶ → σ = 9.1 × 10⁷ Pa (91 MPa).

Strain is a ratio of two lengths, so it has no unit. It may be given as a decimal or as a percentage: a strain of 0.002 is 0.2%. Tensile strain is an extension; compressive strain is a reduction in length.

Strain compares the change with the size of the sample, so a 1 mm stretch is a big strain for a 10 cm wire but a tiny one for a 10 m cable.

Metals in their elastic region usually show strains well below 0.01 (1%). This is why the Young modulus practical needs a long wire.

Use the original length, not the stretched length.

E measures the stiffness of a material: a large E means a large stress is needed for a small strain.

It does not depend on the shape or size of the sample. Compare the stiffness of an object (1.8): k = F/Δx = EA/x, which does depend on A and x.

Typical values: steel about 200 GPa, copper about 120 GPa, aluminium about 70 GPa, rubber only a few MPa.

WORKED EXAMPLE · QUESTION

A granite column of square cross-section, 0.30 m × 0.30 m, supports a compressive load of 1.8 × 10⁵ N. The column is 4.5 m tall and the load shortens it by 0.21 mm. Calculate (a) the compressive stress, (b) the compressive strain and (c) the Young modulus of the granite.

SOLUTION

(a) σ= 1.8×10⁵÷(0.30×0.30) = 1.8×10⁵÷0.090 → σ = 2.0 × 10⁶ Pa.

(b) Both lengths in metres: ε= 2.1×10⁻⁴÷4.5 → ε = 4.67 × 10⁻⁵ (no unit).

(c) E = 2.0×10⁶÷4.67×10⁻⁵ → E = 4.3 × 10¹⁰ Pa (43 GPa).

WORKED EXAMPLE · QUESTION

A copper wire is 1.6 m long and 0.50 mm in diameter. A load of 25 N is hung from it. The Young modulus of copper is 1.2 × 10¹¹ Pa; assume the wire obeys Hooke's law. (a) Calculate the extension. (b) A second wire of the same copper has the same length but twice the diameter. State its extension under the same load.

SOLUTION

(a) A = π×(0.50×10⁻³)²÷4 = 1.96×10⁻⁷ m², so σ= 25÷1.96×10⁻⁷ = 1.27×10⁸ Pa.

ε= σ/E = 1.27×10⁸÷1.2×10¹¹ = 1.06×10⁻³, so Δx = εx = 1.06×10⁻³×1.6 → Δx = 1.7 mm.

(b) Doubling d multiplies A by 4, so the stress and the strain are quartered: 1.70÷4 → 0.42 mm. The Young modulus is unchanged because it is the same material; only the object's stiffness changes.

A stress–strain graph has the same shape as the force–extension graph of the same sample, but it describes the material: any sample of it gives the same curve. The points on it are the ones you met in 1.8.

Gradient of the straight section = Young modulus E.

P, limit of proportionality: the end of the straight line.

E, elastic limit: beyond this the sample does not return to its original length when the load is removed.

Y, yield point: the material starts to deform plastically with little or no increase in stress.

Breaking stress (ultimate tensile stress, UTS): the greatest stress the material can withstand, the highest point of the curve.

For a ductile metal, beyond the UTS the wire forms a narrow "neck"; the force it can support falls and it breaks soon after.

WORKED EXAMPLE · QUESTION

Use graphs A and B for this metal. (a) Determine the Young modulus. (b) A wire of this metal has diameter 0.80 mm. Calculate the greatest load it can support. (c) A 1.2 m length of the wire is stretched to a strain of 2.0 × 10⁻³ and the load is removed. Estimate its permanent extension.

SOLUTION

(a) Gradient from the origin to P: E = 90×10⁶÷0.75×10⁻³ → E = 1.2 × 10¹¹ Pa.

(b) Greatest load = breaking stress × area: A = π×(0.80×10⁻³)²÷4 = 5.03×10⁻⁷ m², F = 220×10⁶×5.03×10⁻⁷ = 111 N → 1.1 × 10² N.

(c) At ε= 2.0×10⁻³ the stress is about 116 MPa. On unloading the elastic part is recovered along a line of gradient E: elastic strain = 116×10⁶÷1.2×10¹¹ = 9.7×10⁻⁴. Permanent strain ≈2.0×10⁻³ - 0.97×10⁻³ = 1.0×10⁻³, so permanent extension ≈1.0×10⁻³×1.2 → ≈ 1.2 mm.

WORKED EXAMPLE · QUESTION

A hoist must hold a load of mass 2500 kg on a single steel cable. The breaking stress of the steel is 1.6 × 10⁹ Pa and, for safety, the stress in the cable must not exceed one fifth of this. The Young modulus of the steel is 2.0 × 10¹¹ Pa. (a) Calculate the minimum diameter of the cable. (b) Calculate the extension of a 32 m length of this cable when it carries the load.

SOLUTION

Greatest allowed stress σ= 1.6×10⁹÷5 = 3.2×10⁸ Pa; load F = 2500×9.81 = 2.45×10⁴ N.

(a) A = F/σ= 2.45×10⁴÷3.2×10⁸ = 7.66×10⁻⁵ m², so d = √(4A/π) → d = 9.9 mm.

(b) ε= σ/E = 3.2×10⁸÷2.0×10¹¹ = 1.6×10⁻³, so Δx = 1.6×10⁻³×32 → Δx = 51 mm.

The cable diameter must be rounded up to a size that is available, so the real stress is a little lower.

Everyday words like "strong" and "tough" have precise meanings in physics, and each one is linked to a different feature of the stress–strain graph.

The properties are independent: glass is stiff and strong but brittle; copper is ductile but not very strong; a material can be strong without being tough. The opposite of brittle is tough or ductile, not "strong"; the opposite of stiff is flexible, not "weak".

WORKED EXAMPLE · QUESTION

Use the graph of materials P, Q and R. Identify, with a reason from the graph, the material that is (a) the stiffest, (b) the strongest, (c) brittle, (d) the most ductile and (e) the toughest.

SOLUTION

(a) Gradients of the straight sections: P: 480 MPa ÷ 0.008 = 6.0 × 10¹⁰ Pa; Q: 280 MPa ÷ 0.007 = 4.0 × 10¹⁰ Pa; R: 80 MPa ÷ 0.010 = 8.0 × 10⁹ Pa → P is the stiffest.

(b), (c) P reaches 480 MPa, above Q (400 MPa) and R (120 MPa) → P is the strongest; its line ends where it stops being straight → P is brittle.

(d) R breaks at a strain of 0.090, Q at 0.080, P at 0.008 → R is the most ductile.

(e) Area under P = ½×480×10⁶×0.008 ≈1.9×10⁶ J m⁻³; Q's area is about 15 times this and R's about 5 times → Q is the toughest. P is the strongest material but the least tough: strength and toughness are different properties.

Idea: load a wire, measure its extension, and plot stress against strain; the gradient of the straight section is E.

Why long and thin: metal strains are tiny. A long wire (2–3 m) gives a larger extension for the same strain; a thin wire gives a large stress for a modest load. Both make the extension big enough to measure with a small percentage uncertainty.

Set-up: clamp one end firmly; run the wire along the bench over a pulley; attach a small pointer (a tape flag) near the pulley, against a fixed scale; start with a small load to straighten the wire.

Original length x: from the clamp to the pointer, with a metre rule or tape measure.

Diameter d: a micrometer at several points along the wire and in two directions at right angles at each point; take the mean. This finds the average size and allows for a wire that is not perfectly round.

Extension: add masses in equal steps; record the pointer position each time and subtract the starting reading. A vernier scale (or a travelling microscope) gives a finer reading than a millimetre scale.

Check: remove the masses one at a time and read again; returning readings show the wire stayed elastic.

Safety: eye protection in case the wire snaps; a box of padding under the masses; keep feet clear.

WORKED EXAMPLE · QUESTION

The micrometer readings of the wire's diameter were 0.30, 0.29, 0.31, 0.30 and 0.30 mm. The original length was 2.40 m. (a) Calculate the mean diameter and its percentage uncertainty. (b) Show how the point for a 2.0 kg load was calculated: its extension was 3.4 mm. (c) Use the graph to determine the Young modulus of the steel.

SOLUTION

(a) Mean, with half the range as the uncertainty: d = 0.300 mm ± 0.01 mm, percentage uncertainty = 0.01÷0.30×100 → 3.3%.

(b) A = π×(0.300×10⁻³)²÷4 = 7.07×10⁻⁸ m²; σ= 19.62÷7.07×10⁻⁸ = 2.78×10⁸ Pa; ε= 3.4×10⁻³ m÷2.40 = 1.42×10⁻³.

(c) Large triangle on the best-fit line, from (0.50×10⁻³, 0.99×10⁸ Pa) to (2.50×10⁻³, 5.01×10⁸ Pa): E = 4.02×10⁸÷2.0×10⁻³ = 2.01×10¹¹ Pa → E = 2.0 × 10¹¹ Pa.

For comparison: the largest extension, 6.6 mm, read to ± 0.05 mm, has a percentage uncertainty of 0.76%, and the length (2.40 m ± 0.5 mm) only 0.02%. The diameter dominates, and it is squared in the area, so measure it with the most care.

8 Hooke's law and elastic strain energy

The force–extension graph of a sample becomes a stress–strain graph when you divide by A and x; the object's stiffness is k = EA/x, and the area under the force–extension graph is the elastic strain energy.

2 Forces and Newton's laws

The force in stress = F/A is usually a tension or a weight found from equilibrium: the same free-body diagrams as for ropes and hanging loads.

8 Resistivity and 3.3 The AS core practicals

Finding the resistivity of a wire uses the same micrometer readings and the same area πd²/4; both experiments are most sensitive to the diameter. Mean and half-range uncertainties and gradients from a large triangle are practised here too.

1. Express the pascal in SI base units.

2. A nylon cord 2.50 m long stretches by 45 mm. Calculate the strain.

3. Wire B is made of the same metal as wire A but is twice as long and has twice the diameter. Both carry the same load. Compare their extensions.

4. Explain the difference between a ductile and a malleable material.

5. Explain why a long wire is used in the Young modulus core practical.

6. A steel wire (E = 2.0 × 10¹¹ Pa) of length 3.0 m and cross-sectional area 1.5 × 10⁻⁷ m² carries 30 N. Calculate its extension.

1. 1 Pa = 1 N m⁻² = 1 kg m s⁻² ÷ m² = 1 kg m⁻¹ s⁻².

2. ε= 0.045÷2.50 = 0.018 (1.8%), with no unit.

3. Δx = Fx/(EA). Doubling x doubles Δx; doubling d quadruples A. B's extension is 2÷4 = 0.5 times A's.

4. Both undergo large plastic deformation without breaking. Ductile describes this under tension (the material can be drawn into a wire); malleable describes it under compression (it can be hammered or rolled into sheets).

5. For a given stress the strain is fixed, so the extension is proportional to the original length. A long wire gives a larger extension, which reduces its percentage uncertainty.

6. Δx = Fx/(EA) = 30×3.0÷(2.0×10¹¹×1.5×10⁻⁷) = 3.0×10⁻³ m = 3.0 mm.