Stress, Strain and Hooke's Law in Strength of Materials

Statics tells you how large the forces in a structure are, but it cannot tell you whether a part will stretch too much or break. That is the job of strength of materials, and its starting point is three simple ideas: stress, strain and Hooke's law. Once you can connect load, area, stiffness and deformation, you can size real components such as tie rods, bolts and columns.

Stress: Force Spread Over an Area

When a rod is pulled, every cross-section inside it carries the load. Normal stress is the internal force per unit area acting perpendicular to that section:

$$ \sigma = \frac{F}{A} $$

The SI unit is the pascal (\( 1 \ \text{Pa} = 1 \ \text{N/m}^2 \)). Because engineering stresses are large, we usually work in megapascals, and a very handy fact is that

$$ 1 \ \text{MPa} = 1 \ \text{N/mm}^2 $$

Tensile stress (pulling) is taken as positive and compressive stress (pushing) as negative. There is also shear stress, \( \tau = V/A \), when the force acts parallel to the surface, as in a bolt being sliced sideways by two plates.

Strain: Deformation Without Units

Stress alone does not tell us how much a part moves. Normal strain measures the change in length relative to the original length:

$$ \varepsilon = \frac{\delta}{L} $$

where \( \delta \) is the elongation and \( L \) is the original length. Strain is dimensionless, but it is often written as mm/mm or in "microstrain" (\( 10^{-6} \)). Typical strains in steel structures are tiny, often well below 0.001.

Hooke's Law and Young's Modulus

In 1660 Robert Hooke discovered that, for small loads, the extension of a spring is proportional to the force. For materials, the same idea becomes

$$ \sigma = E \varepsilon $$

The constant \( E \) is Young's modulus (modulus of elasticity), a measure of stiffness. Typical values:

• Structural steel: about 200 GPa
• Aluminium alloys: about 70 GPa
• Concrete: roughly 25–35 GPa
• Timber (along the grain): roughly 10–12 GPa

Combining the three equations gives the most useful formula for axially loaded members:

$$ \delta = \frac{F L}{A E} $$

Hooke's law is only valid in the elastic region, below the proportional limit. Beyond the yield stress (about 250 MPa for mild steel), the material deforms permanently, and a designer usually keeps working stresses well below yield using a factor of safety.

Poisson's Ratio

A rod that stretches also gets thinner. The ratio of lateral strain to axial strain is Poisson's ratio:

$$ \nu = -\frac{\varepsilon_{\text{lateral}}}{\varepsilon_{\text{axial}}} $$

For most metals \( \nu \) is about 0.3; for rubber it approaches 0.5, meaning its volume hardly changes.

Worked Example 1: Steel Tie Rod in Tension

A steel tie rod in a roof structure is 2 m long with a diameter of 20 mm. It carries a tensile load of 50 kN. Take \( E = 200 \ \text{GPa} \) and yield stress 250 MPa. Find the stress, strain and elongation, and the factor of safety against yield.

Step 1: Cross-sectional area.

$$ A = \frac{\pi d^2}{4} = \frac{\pi (20 \ \text{mm})^2}{4} = 314.16 \ \text{mm}^2 $$

Step 2: Stress.

$$ \sigma = \frac{F}{A} = \frac{50\,000 \ \text{N}}{314.16 \ \text{mm}^2} = 159.15 \ \text{N/mm}^2 = 159.15 \ \text{MPa} $$

Step 3: Strain. Convert \( E \) to MPa: \( 200 \ \text{GPa} = 200\,000 \ \text{MPa} \).

$$ \varepsilon = \frac{\sigma}{E} = \frac{159.15 \ \text{MPa}}{200\,000 \ \text{MPa}} = 7.96 \times 10^{-4} $$

Step 4: Elongation.

$$ \delta = \varepsilon L = 7.96 \times 10^{-4} \times 2000 \ \text{mm} = 1.59 \ \text{mm} $$

Step 5: Factor of safety.

$$ \text{FoS} = \frac{250 \ \text{MPa}}{159.15 \ \text{MPa}} = 1.57 $$

Sanity check. Using \( \delta = FL/(AE) = (50\,000 \times 2000)/(314.16 \times 200\,000) = 1.59 \) mm gives the same result. A 2 m steel rod stretching by about 1.6 mm (less than 0.1%) is typical of elastic behaviour, and the stress is below yield, so Hooke's law applies.

Worked Example 2: Aluminium Tube in Compression

A short hollow aluminium tube, 400 mm long, has an outer diameter of 60 mm and an inner diameter of 50 mm. It supports a compressive load of 80 kN. Take \( E = 70 \ \text{GPa} \) and \( \nu = 0.33 \). Find the shortening of the tube and the increase in its outer diameter.

Step 1: Area of the annulus.

$$ A = \frac{\pi}{4}\left(60^2 - 50^2\right) \ \text{mm}^2 = \frac{\pi}{4}(1100) \ \text{mm}^2 = 863.94 \ \text{mm}^2 $$

Step 2: Compressive stress.

$$ \sigma = \frac{80\,000 \ \text{N}}{863.94 \ \text{mm}^2} = 92.60 \ \text{MPa} $$

Step 3: Axial strain.

$$ \varepsilon = \frac{92.60 \ \text{MPa}}{70\,000 \ \text{MPa}} = 1.323 \times 10^{-3} $$

Step 4: Shortening.

$$ \delta = 1.323 \times 10^{-3} \times 400 \ \text{mm} = 0.529 \ \text{mm} $$

Step 5: Lateral strain and diameter change. In compression the tube bulges outward:

$$ \varepsilon_{\text{lat}} = \nu \varepsilon = 0.33 \times 1.323 \times 10^{-3} = 4.37 \times 10^{-4} $$

$$ \Delta d = 4.37 \times 10^{-4} \times 60 \ \text{mm} = 0.026 \ \text{mm} $$

Sanity check. The diameter change (0.026 mm) is far smaller than the length change (0.529 mm), as expected because the diameter is much smaller than the length and \( \nu < 0.5 \). The stress of about 93 MPa is safely below typical aluminium alloy yield strengths (roughly 200–275 MPa for common structural grades). Because the tube is short, buckling is not a concern here; a long slender tube would need a separate buckling check.

Common Mistakes to Avoid

• Using diameter instead of radius (or forgetting the 4 in \( \pi d^2/4 \)).
• Mixing GPa with MPa: always convert to one system before dividing.
• Applying Hooke's law beyond yield, where stress and strain are no longer proportional.
• Forgetting that hollow sections use the area of material, not the outer area.

Did You Know?

• Hooke first published his law in 1676 as the anagram "ceiiinosssttuv", revealing two years later that it stood for the Latin ut tensio, sic vis: "as the extension, so the force".
• Young's modulus is named after Thomas Young, who described it in lectures published in 1807, although Leonhard Euler had used a similar idea decades earlier.
• Some engineered "auxetic" materials have a negative Poisson's ratio: they get fatter when you stretch them.

Key Takeaways

• Stress is force per area (\( \sigma = F/A \)); strain is change in length per original length (\( \varepsilon = \delta/L \)).
• In the elastic range, \( \sigma = E\varepsilon \), giving the axial deformation formula \( \delta = FL/(AE) \).
• Poisson's ratio links axial stretching to lateral shrinking, typically about 0.3 for metals.
• Keep units consistent; 1 MPa equals 1 N/mm², which makes hand calculations easy.
• Always compare working stress with yield stress to check the design is safe.

If you need help with strength of materials problems or a full assignment, Amathify's experts and PhD scholars provide step-by-step solutions; just reach us on WhatsApp or through the Ask a Question form.