Stress, Strain and Young's Modulus: Extension of Wires and Safe Loads

Understand Stress, Strain and Young's Modulus

Stress, Strain and Young's Modulus: formulas and key facts

Stress and strain

$Stress=F/A|Strain=\Delta L/L$

Stress in Pa; strain has no unit.

Young's modulus

$Y=(F/A)/(\Delta L/L)=FL/(A\Delta L)$

For a wire A = πr² = πd²/4.

Extension

$\Delta L=FL/(AY)$

ΔL ∝ L/r² for a given load and material.

Safe load

$F_{s}afe=(breakingstress\times A)/factorofsafety$

The factor of safety is the ratio of breaking stress to working stress.

Stress, Strain and Young's Modulus in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Tricks to remember Stress, Strain and Young's Modulus

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Strain is a pure number

  Strain = change/original, so it has no unit; Y has the unit of stress.

  Use it when: Checking units and dimensions.

• Thin and long stretches more

  ΔL ∝ L/r²: halving the radius gives four times the extension.

  Use it when: Comparing two wires under the same load.

Stress, Strain and Young's Modulus: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Where students lose marks in Stress, Strain and Young's Modulus

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using diameter as radius

  Taking A = π d² in a calculation.

  ✓Do this: Use A = π(d/2)² = πd²/4.

  Usually costs: 2 marks.

• ✗Mistake: Not converting units

  Using mm or kg without conversion.

  ✓Do this: Convert to m and N; F = mg.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Confusing stress and force

  Writing stress = F.

  ✓Do this: Stress = F/A; give the unit as Pa (N m⁻²).

  Usually costs: 1 mark.

Practice questions on Stress, Strain and Young's Modulus

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   A wire extends by x under a load. A wire of the same material and radius but twice as long, under the same load, extends by:
   (a) x/2
   (b) x
   (c) 2x
   (d) 4x

   Hint

   ΔL ∝ L.

   Answer

   (c) 2x

2. Q2 · 2 marks · Short answerBoard level

   A rod is under a tensile stress of 4 × 10⁶ Pa and Y = 2 × 10¹¹ Pa. Find the strain and the extension in a 1.5 m length.

   Hint

   Strain = stress/Y.

   Answer

   Strain = 2 × 10⁻⁵. ΔL = 2 × 10⁻⁵ × 1.5 = 3 × 10⁻⁵ m.

3. Q3 · 3 marks · Short answerHarder board level

   A wire stretches by 1 mm under a load. What is the extension if the load is doubled and the radius halved (same length and material)?

   Hint

   ΔL ∝ F/r².

   Answer

   ΔL ∝ F/r²: 2 × 4 = 8 times, so 8 mm (if still within the elastic limit).

4. Q4 · 3 marks · Short answerHarder board level

   A wire of length 1.5 m and cross-section 1 × 10⁻⁶ m² extends by 0.25 mm under 40 N and by 0.50 mm under 80 N. Find Y and the force constant F/ΔL.

   Hint

   F/ΔL is the slope of the load-extension graph.

   Answer

   F/ΔL = 40/2.5 × 10⁻⁴ = 1.6 × 10⁵ N m⁻¹. Y = (F/ΔL)(L/A) = 1.6 × 10⁵ × 1.5/10⁻⁶ = 2.4 × 10¹¹ Pa.

5. Q5 · 3 marks · Long answerHarder board level

   Identify the proportional limit, elastic limit, yield point and breaking point on the stress-strain curve of a ductile metal, and explain what happens between the yield point and the breaking point.

   Hint

   Describe the shape of the curve in order.

   Answer

   Straight line to the proportional limit (Hooke's law holds); the elastic limit lies just beyond (the body still recovers); at the yield point the strain grows rapidly with little rise in stress; between yield and breaking point the wire flows plastically and shows a permanent set, thinning until it breaks at the breaking point.

6. Q6 · 4 marks · Case-basedHOTS / competency

   Case study: Suppose a student measures Y by stretching a wire of length 2.00 m and diameter 0.50 mm. For each extra load of 1.00 kg the extension is 0.50 mm (g = 9.8 m s⁻²).
   (i) Find the cross-sectional area. (1 mark)
   (ii) Find the stress for a 1 kg load. (1 mark)
   (iii) Find the strain for a 1 kg load. (1 mark)
   (iv) Find Y, and the percentage error caused by a diameter error of ±0.01 mm. (1 mark)

   Hint

   A = πd²/4; the error in A is twice the relative error in d.

   Answer

   (i) A = π(0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m². (ii) 9.8/1.96 × 10⁻⁷ = 5.0 × 10⁷ Pa. (iii) 0.50 × 10⁻³/2.00 = 2.5 × 10⁻⁴. (iv) Y = 5.0 × 10⁷/2.5 × 10⁻⁴ = 2.0 × 10¹¹ Pa; diameter error gives 2 × 0.01/0.50 = 4%.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define Young's modulus.Show answer

   Ratio of longitudinal stress to longitudinal strain within the elastic limit.

2. 2.SI unit of Y?Show answer

   Pascal (N m⁻²).

3. 3.Extension depends on L and r how?Show answer

   ΔL ∝ L/r².

4. 4.What is the breaking stress?Show answer

   The stress at which the wire breaks.

Stress, Strain and Young's Modulus: frequently asked questions