Stress, Strain and Hooke's Law Explained with Worked Examples
Every machine part, from a bicycle spoke to a crane hook, is sized using two ideas: how hard the material is being pulled or pushed inside (stress) and how much it stretches or shrinks as a result (strain). Hooke's law links the two for the elastic range. Once you are comfortable with these three ideas you can check whether a rod will yield, how much a bolt will stretch, or what force a pin can carry. This article builds the definitions and then works through four numerical problems of the kind asked in university exams and used in real design.
Stress, Strain and Their Units
Normal stress is the internal force per unit area acting perpendicular to a cross-section. It is tensile when the member is pulled and compressive when it is pushed. Shear stress acts parallel to the section, as in a pin or rivet that is being sliced. Strain is the change in length divided by the original length, so it has no unit. Engineers often quote it in percent or in microstrain (one millionth).
Normal stress sigma = P / A (N/mm^2 = MPa) Shear stress tau = F / A_shear (MPa) Normal strain eps = delta_L / L (no unit) Hooke's law sigma = E * eps Elongation delta_L = P * L / (A * E) Poisson's ratio nu = - lateral strain / longitudinal strain Shear modulus G = E / (2 * (1 + nu))
A useful habit is to work in newtons and millimetres. Then stress comes out directly in N/mm^2, which is the same as megapascal (MPa), and Young's modulus for steel is written as 200,000 N/mm^2 instead of 200 GPa.
Typical Elastic Properties of Common Materials
The table lists approximate room-temperature values used in textbook problems. Real values depend on the exact grade and heat treatment, so always take design values from the material standard or supplier data sheet.
| Material | Young's modulus E (GPa) | Poisson's ratio | Typical use |
|---|---|---|---|
| Structural / mild steel | 200 to 210 | 0.27 to 0.30 | Shafts, frames, bolts |
| Aluminium alloys | about 70 | about 0.33 | Aircraft parts, light frames |
| Copper | about 110 to 120 | about 0.34 | Busbars, heat exchangers |
| Titanium alloys | about 110 to 115 | about 0.32 | Aerospace, implants |
| Grey cast iron | varies widely, roughly 80 to 140 | about 0.26 | Machine beds, housings |
Notice that steel is roughly three times stiffer than aluminium. For the same load and size, an aluminium rod stretches about three times as much, which is why aluminium parts are often made thicker.
Worked Examples
Example 1: Steel rod in tension
A mild steel rod of diameter 20 mm and length 2 m carries an axial tensile load of 50 kN. Take E = 200 GPa and yield strength 250 MPa. Find the stress, strain, elongation and factor of safety against yielding.
A = pi/4 * 20^2 = 314.16 mm^2 sigma = 50,000 / 314.16 = 159.15 MPa eps = 159.15 / 200,000 = 7.96 x 10^-4 (796 microstrain) delta_L = eps * L = 7.96e-4 * 2000 = 1.59 mm FoS = 250 / 159.15 = 1.57
The rod stretches about 1.6 mm. A factor of safety of 1.57 is acceptable for a static, well-defined load but too low if the load is shock or fatigue loading.
Example 2: Lateral contraction with Poisson's ratio
For the same rod take Poisson's ratio 0.3. The lateral strain is 0.3 x 7.96 x 10^-4 = 2.39 x 10^-4, so the diameter reduces by 20 x 2.39 x 10^-4 = 0.0048 mm. This is tiny, but it is the reason a tight-fitting bolt becomes slightly easier to pull when it is loaded in tension.
Example 3: Pin in single and double shear
A 10 mm diameter pin transmits a force of 10 kN. In single shear only one cross-section resists the force; in a clevis joint (double shear) two sections share it.
A_pin = pi/4 * 10^2 = 78.54 mm^2 Single shear tau = 10,000 / 78.54 = 127.3 MPa Double shear tau = 10,000 / (2 * 78.54) = 63.7 MPa
Using a clevis instead of a single lap halves the shear stress, which is why fork-and-eye joints are standard on hydraulic cylinders.
Example 4: Thermal stress in a fixed bar
A steel rail segment is held between rigid supports and heated by 50 degrees C. With coefficient of expansion 12 x 10^-6 per degree C and E = 200 GPa, the free expansion is prevented, so a compressive stress develops.
sigma_thermal = E * alpha * delta_T
= 200,000 * 12e-6 * 50
= 120 MPa (compressive)The stress does not depend on the length of the bar. This is why expansion gaps are left in rails, bridges and long pipelines.
Using These Results in Design
In practice you usually run the calculation backwards: you know the load and the allowable stress, and you need the size. Allowable stress is the yield strength divided by a factor of safety. For the 50 kN load with a factor of safety of 2, the allowable stress is 125 MPa, so the required area is 50,000 / 125 = 400 mm^2 and the required diameter is the square root of (4 x 400 / pi), which is 22.6 mm. You would then choose the next standard size, 24 or 25 mm.
- Find the load and decide whether it is tension, compression or shear.
- Pick the material and note its yield strength and modulus.
- Choose a factor of safety suited to the loading and consequences of failure.
- Calculate the required area and round up to a standard size.
- Check the deflection with delta_L = PL/AE if stiffness matters.
Common Mistakes
- Mixing units: GPa with N and mm without converting. 200 GPa is 200,000 N/mm^2.
- Using the diameter instead of the radius in the area formula, which gives four times the area.
- Forgetting double shear in clevis joints, which doubles the calculated stress.
- Applying Hooke's law after yielding. Plastic deformation needs the full stress-strain curve.
- Ignoring stress concentration at holes, threads and shoulders, where local stress can be two to three times the nominal value.
Summary
Stress is force per area, strain is change in length per length, and within the elastic range they are tied together by Young's modulus. With the five formulas in this article you can size a rod, check a pin, estimate deflection and predict thermal stress. The next step is to see what happens beyond the elastic limit, which is the subject of the stress-strain curve of mild steel.