What is Young's modulus?

How it works

Young's modulus E (OpenStax writes Y) is stress divided by strain:

• Stress: σ = F ÷ A, the force F along the bar over the cross-section area A.
• Strain: ε = ΔL ÷ L₀, the change in length ΔL over the original length L₀.
• Young's modulus: E = σ ÷ ε = F × L₀ ÷ (A × ΔL).

Choose what to find:

• Young's modulus from a load test: E = F × L₀ ÷ (A × ΔL).
• Change in length: ΔL = F × L₀ ÷ (A × E).
• Force: F = E × A × ΔL ÷ L₀.
• E from stress and strain: E = σ ÷ ε, from a stress and a strain you type.

For a round bar or wire, type the diameter d instead of the area: A = π × d² ÷ 4.

The page shows E in Pa, GPa and psi; the stress σ = F ÷ A in MPa and psi and the strain ε = ΔL ÷ L₀ (with E from stress and strain, the ones you typed), with the strain as a percentage (ε × 100); and, except with E from stress and strain, ΔL in mm and in, F in N and lbf, and A in mm².

Materials (Young's modulus, OpenStax University Physics Table 12.1): steel 200 GPa; aluminum 70; brass 90; copper 110; iron 210; nickel 210; lead 16; crown glass 60; granite 45; marble 60; concrete 20; brick 15; hardwood 15; bone in tension 16; bone in compression 9; human hair 10 GPa. Table 12.1 lists these in units of 10¹⁰ Pa: 20 × 10¹⁰ Pa = 200 GPa. Polystyrene 3, silk 6 and spider thread 3 GPa come from OpenStax College Physics 2e Table 5.3 (in 10⁹ N/m²); Table 12.1 prints them as 3.0, 6.0 and 3.0 × 10¹⁰ Pa, ten times larger. Or type your own E.

Units (exact): 1 kN = 1,000 N; 1 lbf = 4.4482216152605 N; 1 µm = 10⁻⁶ m; 1 mm = 0.001 m; 1 cm = 0.01 m; 1 in = 0.0254 m; 1 ft = 0.3048 m; areas are those lengths squared (1 in² = 0.00064516 m², 1 ft² = 0.09290304 m²); 1 kPa = 1,000 Pa; 1 MPa = 10⁶ Pa; 1 GPa = 10⁹ Pa; 1 psi = 4.4482216152605 ÷ 0.00064516 Pa (6,894.757 Pa); 1 ksi = 1,000 psi; 1 Msi = 10⁶ psi.

Rules

• Force from 10⁻⁶ N to 10¹² N; area from 10⁻¹² m² to 10⁶ m²; diameter from 10⁻⁶ m to 1,000 m; original length from 10⁻⁶ m to 10⁶ m; change in length from 10⁻¹² m to 10⁶ m. Limits are in SI units and include their ends; a value typed in another unit is converted first.
• A typed stress or modulus is a number from 10⁻⁶ to 10¹² in the unit chosen beside it. Strain is a number from 10⁻¹² to 10.
• Every value is a size, so it is positive: a shortening under a push is typed as a positive ΔL.
• A result too large or too small to show as a number gives no answer. Within these limits it does not happen.

Output format

The answer line reads E = a GPa (b psi), ΔL = a mm (b in) or F = a N (b lbf), for the value found. It and every number show to 6 significant figures, rounded half up, with trailing zeros dropped. Numbers in text are written with thousands commas, and below 10⁻⁶ or from 10¹⁵ up as a × 10ⁿ. Typed values are read back as the exact decimals you typed, and every unit size is exact, so each result is an exact fraction rounded once; a round bar's area brings in π, so its results are 64-bit floats.

Worked examples by hand

How far a steel rod stretches. OpenStax Example 12.8: a steel rod 2.0 m long with A = 0.30 cm² = 0.00003 m² holds a 550 kg platform, F = 550 × 9.8 = 5,390 N. With E = 2 × 10¹¹ Pa: ΔL = 5,390 × 2 ÷ (0.00003 × 2 × 10¹¹) = 10,780 ÷ 6,000,000 = 0.00179667 m = 1.79667 mm (0.0707349 in); OpenStax rounds this to 1.8 mm. The stress is 5,390 ÷ 0.00003 = 1.79667 × 10⁸ Pa = 179.667 MPa.

Young's modulus from a test. The same rod measured at ΔL = 1.8 mm: E = 5,390 × 2 ÷ (0.00003 × 0.0018) = 10,780 ÷ 0.000000054 = 1.99630 × 10¹¹ Pa = 199.63 GPa (28,953,800 psi). The strain is 0.0018 ÷ 2 = 0.0009.

From stress and strain. σ = 180 MPa and ε = 0.0009: E = 180,000,000 ÷ 0.0009 = 2 × 10¹¹ Pa = 200 GPa.

The force for a stretch. Steel (200 GPa), A = 0.00003 m², L₀ = 2 m, ΔL = 1.8 mm: F = 2 × 10¹¹ × 0.00003 × 0.0018 ÷ 2 = 5,400 N (1,213.97 lbf).

A round bar. Aluminum (70 GPa), d = 10 mm, L₀ = 1 m, F = 10 kN: A = π × 0.01² ÷ 4 = 7.85398 × 10⁻⁵ m² (78.5398 mm²), and ΔL = 10,000 × 1 ÷ (7.85398 × 10⁻⁵ × 7 × 10¹⁰) = 0.00181891 m = 1.81891 mm.