Stress and strain

AS Level Physics · Deformation of solids. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

A pair of tensile forces stretches an object and a pair of compressive forces squashes it. The load is the force applied and the extension is the increase in length. Hooke's law says the extension is proportional to the load, up to the limit of proportionality. The spring constant k tells you how stiff one particular object is.

To compare materials, not objects, we use stress (force per unit area) and strain (extension per unit length). Their ratio is the Young modulus, which is the same for every wire of that material.

In the experiment, a long thin wire is loaded in steps. Measure its diameter with a micrometer, its length with a metre rule and its small extension with a vernier scale.

Rules to remember

Common mistake

Students leave the extension in mm or use the diameter as the radius when finding the area. Change every length to metres and use A = πd2/4 before you calculate stress or strain.

Worked example

A wire of length 2.5 m and diameter 0.80 mm extends by 1.2 mm under a tension of 50 N. What is the Young modulus of the metal?

  1. Area = πd2/4 = π × (0.80 × 10−3)2 ÷ 4 = 5.03 × 10−7 m2
  2. Stress = F ÷ A = 50 ÷ (5.03 × 10−7) = 9.95 × 107 Pa
  3. Strain = x ÷ L = (1.2 × 10−3) ÷ 2.5 = 4.8 × 10−4
  4. Young modulus = stress ÷ strain = (9.95 × 107) ÷ (4.8 × 10−4) = 2.1 × 1011 Pa

Answer: 2.1 × 1011 Pa

Practice questions

Try each one, then open the answer.

1. A wire of cross-sectional area 2.0 × 10−7 m2 is under a tension of 40 N. What is the stress in the wire?

  1. A
    8.0 × 10−6 Pa
  2. B
    2.0 × 106 Pa
  3. C
    2.0 × 108 Pa
  4. D
    5.0 × 10−9 Pa
Show answer

Answer: C. Stress = F/A = 40 ÷ 2.0 × 10−7 = 2.0 × 108 Pa.

2. Two identical springs, each of spring constant 400 N m−1, are joined end to end and hung vertically. A load of 20 N hangs from the bottom. What is the total extension?

  1. A
    0.025 m
  2. B
    0.050 m
  3. C
    0.10 m
  4. D
    0.20 m
Show answer

Answer: C. Each spring carries the full 20 N, so each extends 20 ÷ 400 = 0.050 m. Total extension = 0.10 m (the combination has k = 200 N m−1).

3. In which example is the object deformed by compressive forces?

  1. A
    the legs of a table supporting a heavy load
  2. B
    the cable holding up a lift
  3. C
    a rope in a tug of war
  4. D
    a stretched rubber band
Show answer

Answer: A. The table legs are squeezed between the load and the floor, so they are compressed. The rope, the cable and the rubber band are all in tension.

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