Young modulus and the wire practical

Material modulus and specimen stiffness

Young modulus E is stress divided by strain in the linear elastic region. It is measured in pascals because strain has no unit. A large modulus means a large stress is needed for a given small strain. It does not directly tell you the load at which an object breaks. In this lesson E denotes modulus; energy is written out or given a subscript to avoid confusing the two meanings.

Substitute stress F/A and strain x/L₀ to obtain E = FL₀/(Ax). Rearranging gives x = FL₀/(AE). For a uniform wire at fixed temperature, doubling its length doubles its extension under the same force; doubling its cross-sectional area halves extension. A doubled diameter gives four times the area, so extension becomes a quarter as large. These predictions assume small axial deformation in the same linear material state, without slipping, significant heating or bending.

The sample stiffness is k = F/x = EA/L₀. Therefore shortening a wire or increasing its area changes stiffness without changing its material’s Young modulus. A stress–strain gradient gives E directly. A force–extension gradient gives k and must be multiplied by L₀/A to obtain E. If extension is the vertical axis and force the horizontal axis, the gradient is 1/k; identify the plotted quantities before using the slope.

E = σ/ε = FL₀/(Ax) x = FL₀/(AE) k = EA/L₀

Symbols and units

• E is Young modulus in Pa; F is tensile force in N; L₀ is original length in m.
• A is cross-sectional area in m²; x is extension in m; k is stiffness in N m⁻¹.

How the equation explains the idea

1. Substitute stress F/A and strain x/L₀ into E = σ/ε: E = FL₀/(Ax).
2. Rearrange to x = FL₀/(AE). Comparing F = kx gives k = EA/L₀.
3. Doubling original length halves specimen stiffness at the same material and area, while doubling area doubles stiffness. Young modulus itself is unchanged.

Worked example

Predict the extension of a longer wire

1. A wire has E = 2.0 × 10¹¹ Pa, area 2.0 × 10⁻⁷ m² and original length 2.0 m.
2. With force 20 N, x = 20 × 2.0 / [(2.0 × 10⁻⁷)(2.0 × 10¹¹)] = 0.0010 m.
3. A 4.0 m length of the same wire under the same force extends 0.0020 m; its modulus is unchanged.

The first wire extends 1.0 mm and the doubled-length wire extends 2.0 mm.

Keep this idea. Young modulus describes the material; force–extension stiffness also depends on area and length.

Core practical 3: measure a wire’s modulus

Use a long, straight wire securely supported so a controlled axial load stretches a defined gauge length. Measure that length between the two reference points over which extension is observed; do not use a length that includes an unmeasured or clamped segment. Check micrometer zero, then measure diameter at several positions and in two perpendicular directions. Record the readings, inspect variation and calculate a representative area A = πd²/4 for a sufficiently uniform round wire. Nonuniform wire requires caution because its thinnest parts extend disproportionately.

Use a small initial load to remove slack, then record the corresponding reference position. Apply known load increments gently and allow the wire to settle before recording extension with a suitable micrometer, travelling microscope or displacement sensor. Calculate force increments from added masses using ΔF = Δmg, with the hanger and preload treated consistently. Use several well-spaced readings in the linear region rather than relying on a single pair. Read a pointer and scale at eye level if used, and ensure measured displacement comes from the gauge length rather than support movement.

Plot force vertically against extension horizontally, fit the linear region and calculate a large-triangle gradient k = ΔF/Δx. The preload offset does not affect this gradient when force and extension differences are consistent. Calculate E = kL₀/A, or use a stress–strain gradient after converting every point. Reduce the load in stages and check whether the reference length is recovered within uncertainty. Loading and unloading agreement supports elastic operation; permanent extension, clamp slip or curvature requires investigation before treating the data as a linear modulus measurement.

A reference wire beside the test wire can help compensate for shared support movement or thermal expansion when an appropriate differential apparatus compares their extensions; it does not remove every error. Keep temperature steady, choose a measuring instrument that resolves the small extension, and keep within a teacher-set safe load. Protect eyes from a snapping wire, secure clamps and masses, keep faces out of the wire’s recoil line, and place a catch tray close below hanging masses. A supervisor must assess the apparatus and stop the run before unsafe loading; these notes are a planning resource rather than proof that a practical has been performed.

A = πd²/4 k = ΔF/Δx from a linear fit E = kL₀/A

Symbols and units

• d is circular-wire diameter in m; A = πd²/4 is area in m².
• k is best-fit force–extension gradient in N m⁻¹; L₀ is original length in m; E is modulus in Pa.

How the equation explains the idea

1. Obtain k = ΔF/Δx from the linear force–extension region, using extension in metres rather than unconverted millimetres.
2. Calculate area from diameter, then E = kL₀/A. The unit is (N m⁻¹)m/m² = Pa.
3. An intercept from an offset does not change the gradient of a valid straight-line fit; it must still be investigated rather than blindly forcing the fit through zero.

Worked example

Convert a measured gradient to modulus

1. Original gauge length is 1.80 m and mean diameter is 0.400 mm = 4.00 × 10⁻⁴ m.
2. Area = π(4.00 × 10⁻⁴)²/4 = 1.2566 × 10⁻⁷ m².
3. The fitted force–extension gradient is 14.0 N mm⁻¹ = 1.40 × 10⁴ N m⁻¹.
4. E = (1.40 × 10⁴ × 1.80)/(1.2566 × 10⁻⁷) = 2.005 × 10¹¹ Pa.

The measured modulus is about 2.01 × 10¹¹ Pa, subject to the stated experimental checks and uncertainty.

Keep this idea. Measure area and gauge length, determine a linear gradient, check unloading, and control support, temperature and loading.

Find the measurement that limits the result

The area of a circular wire depends on diameter squared. For a small diameter uncertainty, the fractional area uncertainty is approximately twice the fractional diameter uncertainty. In E = 4kL₀/(πd²), an approximate worst-case fractional uncertainty is δk/k + δL₀/L₀ + 2δd/d. This additive estimate follows the course’s simple uncertainty convention; it is not a statistical confidence interval. The numerical value of π is not a measured contribution.

Extension is a difference between two position readings. Under a worst-case estimate, add their absolute uncertainties before dividing by extension. If each reading has uncertainty ±0.01 mm and extension is only 0.10 mm, the extension uncertainty is ±0.02 mm, or 20%. A longer gauge length gives a larger extension at the same stress and may reduce this fractional contribution. Increasing load also increases extension within the linear region, but safety and the material limit constrain that strategy.

For a graph, use measurement uncertainties or scatter to judge plausible steepest and shallowest fitted gradients over the linear region. A large gradient triangle on the best-fit line avoids the extra error of using adjacent noisy points. Repeated diameter readings reveal irregularity and scatter, but all readings can share a micrometer zero error. A consistent diameter overestimate makes the calculated area too large and modulus too small. Record corrections explicitly, retain raw readings, and improve the dominant uncertainty rather than reporting more calculator digits.

Approximate worst-case: δE/E ≈ δk/k + δL₀/L₀ + 2δd/d δx ≈ δLloaded + δLreference

Symbols and units

• δE/E, δk/k, δL₀/L₀ and δd/d are positive fractional uncertainties, each dimensionless.
• δx, δLloaded and δLreference are absolute uncertainties in the same length unit.

How the equation explains the idea

1. Since E = 4kL₀/(πd²), small worst-case fractional uncertainties add with the magnitude of each power: δE/E ≈ δk/k + δL₀/L₀ + 2δd/d.
2. The diameter term is doubled because area depends on d². A 2% diameter uncertainty contributes approximately 4% to the modulus uncertainty by this rule.
3. For x = Lloaded − Lreference, add absolute uncertainties for a worst-case bound. Do not apply quadrature unless a justified statistical uncertainty model is being used.

Worked example

Combine an uncertainty budget

1. Gradient uncertainty is 3.0%. Gauge length is 1.80 ± 0.01 m, giving about 0.56%.
2. Diameter is 0.400 ± 0.005 mm, giving 1.25%; its area contribution is 2.50%.
3. Total approximate percentage uncertainty is 3.0 + 0.56 + 2.50 = 6.06%, about 6%.
4. For E = 2.0 × 10¹¹ Pa, a 6% uncertainty is about 0.12 × 10¹¹ Pa.

A suitable report is E ≈ (2.00 ± 0.12) × 10¹¹ Pa under this uncertainty model.

Keep this idea. Square the diameter contribution, account for two-reading extension uncertainty, and separate scatter from systematic bias.