Strength of Materials · Stress and Strain

Normal and shear stress, strain

Define normal and shear stress and strain, use Hooke's law, and apply a factor of safety to an axially loaded member.

40 minYear 1-2Prerequisites: Equilibrium and support reactions

By the end of this topic you will be able to

  • Calculate normal stress and shear stress from a force and an area.
  • Calculate normal strain and relate stress and strain with Hooke's law.
  • Apply a factor of safety to decide whether a member is acceptable.

Stress: force spread over an area

A force tells you how hard something is pushed or pulled; it does not tell you whether the material will survive. A 10 kN load that is fine for a thick steel bar can snap a thin wire. What matters is how concentrated the force is, which is stress:

σ=PA\sigma = \frac{P}{A}

with PP the internal axial force and AA the cross-sectional area. The SI unit is the pascal (Pa); in practice engineers use megapascals, and 1 MPa=1 N/mm21 \text{ MPa} = 1 \text{ N/mm}^2, which makes calculations in newtons and millimetres convenient.

Normal stress acts perpendicular to the area (tension is positive, compression negative). Shear stress acts along the area, as in a bolt or a rivet being sliced across:

τ=VA\tau = \frac{V}{A}

Strain: how much it stretches

Strain is the change in length per unit of original length. It has no units.

ε=δL\varepsilon = \frac{\delta}{L}

For most metals the strain at working loads is tiny (a few hundred micro-strain), so it is often quoted in με\mu\varepsilon (10−610^{-6}).

Hooke's law

Within the elastic range, stress and strain are proportional:

σ=Eε\sigma = E\varepsilon

The constant EE is the elastic modulus (Young's modulus). For steel it is about 200 GPa, for aluminium about 70 GPa and for timber roughly 10 GPa along the grain.

Factor of safety

Real loads are uncertain and real materials vary, so members are designed to be stronger than strictly needed. The factor of safety compares the failure stress to the actual stress:

FoS=σyσactual\text{FoS} = \frac{\sigma_{\text{y}}}{\sigma_{\text{actual}}}

A FoS of 1 means the member is just yielding. Typical design values range from about 1.5 to 4 depending on the consequences of failure.

Interactive modelChange the load and diameter. The colour of the bar shows the factor of safety.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

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Practice problems

  1. A 12 mm diameter steel bar carries 25 kN in tension. Find the stress.
    Show answer
    A=π×122/4=113.1A = \pi\times12^2/4 = 113.1 mm². σ=25000/113.1=221\sigma = 25\,000/113.1 = 221 MPa.
  2. A rod of length 2.5 m stretches 1.2 mm. What is the strain?
    Show answer
    ε=1.2/2500=4.8×10−4\varepsilon = 1.2 / 2500 = 4.8\times10^{-4} (480 με\mu\varepsilon).
  3. A steel bar has σy=250\sigma_y = 250 MPa and must have FoS = 2. What is the largest permitted stress?
    Show answer
    σallow=250/2=125\sigma_{\text{allow}} = 250/2 = 125 MPa.
  4. Using the model with mild steel, what load can a 25 mm bar carry with FoS = 2?
    Show answer
    A=490.9A = 490.9 mm², so P=σallowA=125×490.9=61.4P = \sigma_{\text{allow}} A = 125\times490.9 = 61.4 kN. The model's last readout agrees.

Key formulae

Normal stress
σ=PA\sigma = \frac{P}{A}
Shear stress (average)
τ=VA\tau = \frac{V}{A}
Normal strain
ε=δL\varepsilon = \frac{\delta}{L}
Hooke's law
σ=Eε\sigma = E\varepsilon
Factor of safety
FoS=σyσactual\text{FoS} = \frac{\sigma_{\text{y}}}{\sigma_{\text{actual}}}
Lecturer notes: suggested plan, activity and assessment

Suggested plan (40 min). 5 min: hold up a thin and a thick rubber band loaded equally; 10 min: definitions; 10 min: Hooke's law with typical E values; 15 min: worked examples and the model.

Activity. Give each group a bar diameter and ask them to find the load at which the model's bar turns red (FoS < 1).

Discussion. Why is the factor of safety larger for passenger lifts than for garden fences?

Assessment idea. Ask for the minimum diameter of a hanger rod for a given load and FoS.