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Stress and strain

Forces and motion · level 8, lesson 5 of 6

A question to start

A crane on Mars lifts a 2,000 kg module on a steel cable. How thick must the cable be? And how much will it stretch?

The idea

A force on a solid squashes or stretches it. To compare materials and sizes fairly, engineers use:

  • Stress, force per unit area: \(\sigma = F/A\), in pascals (Pa), or MPa (a million pascals).
  • Strain, the fractional change in length: \(\varepsilon = \Delta L / L\), with no units.
  • Young's modulus, the stiffness of the material: \(E = \sigma/\varepsilon\), in Pa. It is Hooke's law for a material, not just one spring.

Below a material's yield strength, it springs back when unloaded. Above it, it deforms for good; at the ultimate strength, it breaks. Engineers keep working stresses well below the yield strength, dividing it by a safety factor.

Material (typical values) Young's modulus \(E\) Yield strength
Structural steel 200 GPa 250 to 350 MPa
Aluminium alloy (6061-T6) 69 GPa about 275 MPa
Concrete (in compression) about 30 GPa 20 to 40 MPa (crushing)
Nylon (solid) 2 to 4 GPa 50 to 80 MPa
strain εstress σ slope Eyieldultimatebreaks

Mars changes the loads, not the materials. A cable holding a mass \(m\) still needs to carry \(mg\), but \(g\) is 0.38 of Earth's, so the same cable can hold 2.6 times as much.

Worked examples

1. The crane cable. Load on Mars: \(2000 \times 3.71 = 7{,}420\) N. Steel with yield 300 MPa and a safety factor of 5: allowed stress 60 MPa. Area \(= 7420 / 60\times10^6 = 1.24\times10^{-4}\) m², a diameter of 12.6 mm. On Earth it would need \(19{,}600\) N, area \(3.27\times10^{-4}\) m², diameter 20.4 mm.

2. How much it stretches. 30 m of that cable on Mars: \(\varepsilon = \sigma / E = 60\times10^6 / 200\times10^9 = 3\times10^{-4}\). \(\Delta L = 30 \times 3\times10^{-4} = 9\) mm.

3. A column. A concrete column of 0.04 m² carries 400 kN. Stress \(= 10\) MPa, safe for concrete of 30 MPa with a safety factor of 3.

Try it

  1. A 5 mm-diameter steel rod carries 2,000 N. Find the stress.
  2. Find its strain, with \(E = 200\) GPa.
  3. If the rod is 2 m long, how much does it stretch?
  4. An aluminium strut has a working stress limit of 90 MPa. What area does it need to carry 18 kN?
Answers
  1. \(A = \pi \times 0.0025^2 = 1.96\times10^{-5}\) m². \(\sigma = 2000 / 1.96\times10^{-5} = 102\) MPa.
  2. \(102\times10^6 / 200\times10^9 = 5.1\times10^{-4}\).
  3. \(2 \times 5.1\times10^{-4} = 1.0\) mm.
  4. \(18{,}000 / 90\times10^6 = 2\times10^{-4}\) m², which is 2 cm².

Common mistake

Confusing stiffness and strength. Young's modulus says how much a material stretches under a stress; the yield strength says how much stress it can take. Glass is stiff but brittle; nylon is weak and stretchy.

Where it leads