Stress is internal force per unit area; strain is deformation per unit length. Stress tells you whether a part will yield or break, strain tells you how much it stretches, and for most metals in normal service the two are tied together by Hooke’s law, σ = Eε.
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Normal and shear stress
Cut a loaded member and look at the internal force on the cut. The part acting perpendicular to the cut, divided by the area, is the normal stress σ = P/A. Tension is positive, compression negative. The part acting along the cut, divided by the area, is the average shear stress τ = V/A.
Both are in pascals (N/m²). Engineering values are usually in MPa, and 1 MPa = 1 N/mm², which makes hand calculations in millimetres easy. To find P or V you first need the internal force from equilibrium.
Strain, Hooke’s law and Poisson’s ratio
Normal strain is the change in length divided by the original length, ε = δ/L. It has no units, and is often written in microstrain (µε, 10⁻⁶). Shear strain γ is the change in a right angle, in radians.
In the linear-elastic range, stress is proportional to strain: σ = Eε, where E is the modulus of elasticity. Combining it with σ = P/A and ε = δ/L gives the axial deformation of a bar, δ = PL/(AE).
A bar that stretches also gets thinner. Poisson’s ratio ν is the ratio of lateral strain to axial strain, with a minus sign so that ν is positive: ε_lat = −ν·ε.
Reading a stress–strain diagram
A tensile test pulls a standard specimen and plots stress against strain. The diagram shows the regions every design check refers to.
Linear-elastic region: a straight line whose slope is E. Unloading returns the part to its original size.
Yield: the point where permanent deformation begins. Many metals have no sharp yield point, so the yield strength is defined by the 0.2% offset method.
Strain hardening: stress keeps rising with strain up to the ultimate tensile strength.
Necking and fracture: a ductile specimen thins locally and breaks. Brittle materials fracture with little plastic strain.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Average normal stress
σ = P / A
P = internal axial force (N), A = cross-sectional area (m²)
Gere & Goodno, Mechanics of Materials
Average shear stress
τ_avg = V / A
V = internal shear force (N), A = sheared area (m²)
Hibbeler, Mechanics of Materials
Normal strain
ε = δ / L
δ = change in length, L = original length
Gere & Goodno, Mechanics of Materials
Hooke’s law (uniaxial)
σ = E·ε
E = modulus of elasticity (Pa), linear-elastic range only
Beer & Johnston, Mechanics of Materials
Axial deformation
δ = P·L / (A·E)
constant P, A and E along the length
Hibbeler, Mechanics of Materials
Poisson’s ratio
ν = −ε_lat / ε_axial
both strains from the same uniaxial load
Gere & Goodno, Mechanics of Materials
Worked examples
Example: Stress, strain and stretch in a steel tie rod
Given
A solid round steel tie rod has a diameter of 20 mm and a length of 1.8 m.
It carries an axial tensile load of 45 kN.
Assume E = 200 GPa and ν = 0.30, typical textbook values for steel. Use the values for your actual material in practice.
Find
The normal stress σ, the axial strain ε, the elongation δ, and the change in diameter.
Normal stress (N/mm² = MPa):σ = P / A = 45 000 N / 314.16 mm² = 143.2 MPa
Strain from Hooke’s law, with E = 200 GPa = 200 000 MPa:ε = σ / E = 143.2 / 200 000 = 7.16 × 10⁻⁴ (716 µε)
Elongation:δ = ε·L = 7.16 × 10⁻⁴ × 1800 mm = 1.29 mm
Check with the combined formula:δ = P·L / (A·E) = (45 000 × 1800) / (314.16 × 200 000) = 1.29 mm ✓
Lateral strain and change in diameter:ε_lat = −0.30 × 7.16 × 10⁻⁴ = −2.15 × 10⁻⁴; Δd = −2.15 × 10⁻⁴ × 20 mm = −0.0043 mm
Answer: σ ≈ 143 MPa, ε ≈ 716 µε, the rod stretches about 1.29 mm, and its diameter shrinks about 0.004 mm. The stress must stay below the steel’s yield strength for Hooke’s law to apply.
Example: Shear and bearing stress in a clevis pin
Given
A clevis joint carries an 18 kN tensile load through a 12 mm diameter pin.
The pin passes through a 10 mm thick center plate held between two outer plates, so it is in double shear.
Find
The average shear stress in the pin and the bearing stress between the pin and the center plate.
Solution
FBD of the pin: the center plate pushes with 18 kN, and each outer plate resists with half of it. The pin is sheared on two planes.
Shear force on each plane:V = P / 2 = 18 kN / 2 = 9 kN
Area of one shear plane:A = π (12 mm)² / 4 = 113.1 mm²
Average shear stress:τ_avg = V / A = 9000 N / 113.1 mm² = 79.6 MPa
Bearing stress uses the projected area, pin diameter times plate thickness:σ_b = P / (d·t) = 18 000 N / (12 mm × 10 mm) = 150 MPa
Answer: τ_avg ≈ 79.6 MPa in the pin and σ_b = 150 MPa on the center plate. Compare each with the allowable value for the pin and plate materials.
Common mistakes
Using the diameter where the formula needs the area. Compute A = πd²/4 first and write it down with units.
Mixing MPa, GPa and N/m² in one line. Work in N and mm (so stress is in MPa) or in N and m (so stress is in Pa), and convert E to match: 200 GPa = 200 000 MPa.
Treating a pin in double shear as single shear. Count the planes the pin would break on. Two planes halve the shear force on each.
Applying σ = Eε above the yield strength. Hooke’s law only holds in the linear-elastic region. Check σ against yield before trusting a deformation result.
Calling strain a percentage without saying so. Strain is dimensionless. Write 7.16 × 10⁻⁴, 716 µε or 0.0716%, and label which one you mean.
In industry
Frequently asked questions
What is the difference between stress and pressure?
Both are force per area with the same units. Pressure is a load applied to a surface from outside; stress is the internal force per area inside the material, and it can be normal or shear.
Why is strain unitless?
It is a length divided by a length. Writing it as mm/mm or µε is common, but the units cancel.
Is E the same in tension and compression?
For most metals in the elastic range, yes, to a good approximation. Many non-metals, such as concrete and some polymers, behave differently in tension and compression.
Is average shear stress the maximum shear stress?
No. τ = V/A assumes the shear is spread evenly, which is a simplification. In a beam, for example, shear stress varies across the section and peaks at the neutral axis.