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Lesson 3 of 7

Mechanical Properties of Solids · Lesson 3 of 7

Young’s Modulus and Deformation of Wires

“Derive and apply Young’s modulus to rods, connected wires, compression, shared supports and a moving load.”

Learning Objectives

• Derive ΔL = FL/(AY) and separate material stiffness from specimen stiffness. • Solve the steel-rod, connected-wire and femur examples with explicit load diagrams. • Calculate forces in hanging segments and parallel supports before finding strains. • Use annular areas, rectangular areas and circular-motion forces correctly.

From stress and strain to Young’s modulus

Young’s modulus Y describes the linear longitudinal response of a solid. Apply an axial tensile stress σ to a uniform specimen and measure longitudinal strain ε. In the proportional range, their ratio is constant. Small compression can be treated in the same way using signed stress/strain consistently or using both magnitudes. A slender compressed member can buckle before its material fails; the axial formula alone does not rule out that instability.

Young’s modulus and the axial-extension equationLaTeX
Assume a uniform member, axial loading, small linear elastic deformation and approximately uniform stress. Use original dimensions.

The algebra has a physical message. A longer wire extends more because the same small fractional stretch acts over a longer original length. A larger area shares the force among more material and reduces the stress. A larger Y gives a smaller strain at the same stress. For a circular wire A = πr² = πd²/4, so doubling the radius quarters the extension at fixed force and length. Never substitute a diameter as though it were a radius.

Effective force–extension stiffness of a uniform wireLaTeX
Y has units Pa and characterizes material response; kwire has units N/m and also depends on specimen geometry.

Cutting a wire into two equal lengths does not change its material’s Young’s modulus, but each half is twice as stiff in force–extension response. The distinction is useful whenever two specimens of the same material seem to behave differently. Material properties must also be compared under stated conditions; table values are representative rather than exact constants for every grade or specimen.

MaterialDensity / kg m⁻³Y / GPaUltimate strength / MPaYield strength / MPa
Aluminium27107011095
Copper8890110400200
Wrought iron7800–7900190330170
Steel7860200400250
Glass (compression test)21906550Not listed
Concrete (compression test)23203040Not listed
Wood (compression test)5251350Not listed
Bone (compression test)19009.4170Not listed
Polystyrene1050348Not listed

Read the columns separately. Young’s modulus describes the initial elastic response, whereas yield and ultimate strengths concern departure from elasticity and failure. The compression-test marks matter: a listed strength for a compression specimen is not automatically its tensile strength. One gigapascal is 10⁹ Pa; one megapascal is 10⁶ Pa. Use the table value Y = 200 GPa for the steel models below and Y = 110 GPa for copper.

How stiff metals compare

Problem
Estimate the force needed to produce strain 0.1% in a wire of area 0.1 cm².

  1. 1.Convert strain to 0.001 and area to 1.0 × 10⁻⁵ m².
  2. 2.F = AYε. Steel with Y = 200 GPa needs 2000 N; copper with Y = 110 GPa needs 1100 N.
  3. 3.Brass with Y = 90 GPa needs 900 N. Aluminium at 70 GPa needs about 700 N; using the nearby rounded value 69 GPa gives 690 N.
  4. 4.These comparisons assume elastic operation. High modulus means more force for the same strain, not unlimited load capacity.

A loaded rod: stress, extension and strain

A structural-steel rod

Problem
A 1.0 m rod has radius 10 mm and Y = 2.0 × 10¹¹ Pa. Calculate the linear-model stress, extension and strain under 100 kN.

  1. 1.Convert F = 1.00 × 10⁵ N and r = 0.010 m. A = πr² = 3.142 × 10⁻⁴ m².
  2. 2.σ = F/A = 3.18 × 10⁸ Pa.
  3. 3.ΔL = σL/Y = 1.59 × 10⁻³ m = 1.59 mm.
  4. 4.ε = ΔL/L = 1.59 × 10⁻³ = 0.159%, approximately 0.16%.
  5. 5.These are the results of the assumed linear elastic model. Their validity must be checked against the actual steel grade’s proportional/yield limits.
Check the material limit

The calculated 318 MPa exceeds the representative 250 MPa yield stress in the steel row above. Those generic table data therefore do not establish that this particular rod remains elastic. The computed extension is conditional on a steel grade remaining in its proportional range at this stress; otherwise a full stress–strain curve is needed.

Connected wires: the same tension is not the same extension

For two light wires connected end to end, an interior joint with no additional load must be in equilibrium. The axial force is the same in both segments, though the extensions can differ. The total extension is their sum. Under a common force F, each segment contributes FL/(AY), so the more compliant segment contributes the larger extension. This is a series arrangement.

Two axial segments in seriesLaTeX
The same force acts in the segments only when no extra axial load acts at their junction and their own weights are negligible.
Copper and steel connected end to end

Problem
Copper of length 2.2 m and steel of length 1.6 m both have diameter 3.0 mm. Their total extension is 0.70 mm. Find the load using Yc = 1.1 × 10¹¹ Pa and Ys = 2.0 × 10¹¹ Pa.

  1. 1.Both areas are π(1.5 × 10⁻³)² = 7.069 × 10⁻⁶ m². The common tension is F.
  2. 2.ΔLc/ΔLs = (Lc/Yc)/(Ls/Ys) = 2.5. Together with ΔLc + ΔLs = 0.70 mm, this gives ΔLc = 0.50 mm and ΔLs = 0.20 mm.
  3. 3.F = AYcΔLc/Lc = 176.7 N, or about 1.8 × 10² N.
  4. 4.The common force and area mean equal stresses, but the lower copper modulus produces a greater strain. Its greater length increases its extension further.

Compression and sharing the load

220 kg × g = 2156 N1078 N1078 NEach femur: L = 0.50 m, effective radius = 0.020 mSymmetric axial-load model, not to scale
A symmetric model of the circus load on two femurs— The supported mass is 280 − 60 = 220 kg. Each leg carries half its weight in the simplified model; no anatomical safety claim is implied.
Compression of each thighbone

Problem
A circus group and its equipment total 280 kg, including the 60 kg performer at the bottom. Each of that performer’s femurs is 0.50 m long with effective radius 0.020 m. Use Y = 9.4 × 10⁹ Pa.

  1. 1.The extra load transmitted through the legs is (280 − 60)g = 2156 N for g = 9.8 m/s².
  2. 2.With symmetric loading, each femur carries 1078 N. Its effective area is π(0.020)² = 1.257 × 10⁻³ m².
  3. 3.Compression magnitude = FL/(AY) = 4.56 × 10⁻⁵ m = 0.0456 mm.
  4. 4.The strain magnitude is 9.13 × 10⁻⁵, or 0.00913%. Signed ΔL and longitudinal strain are negative because the bone shortens.
  5. 5.The calculation treats each femur as a uniform axially loaded member and uses the stated effective radius; real bone shape and loading are more complex.
Four hollow supporting columns

Problem
Four identical mild-steel columns share a 50,000 kg load equally. Each has outer radius 0.60 m and inner radius 0.30 m. Find compression strain using Y = 2.0 × 10¹¹ Pa.

  1. 1.Each column supports F = Mg/4 = 122,500 N for g = 9.8 m/s².
  2. 2.The load-bearing material area is the annulus A = π(R² − r²) = π(0.60² − 0.30²) = 0.8482 m². The hollow region carries no axial stress.
  3. 3.Strain magnitude = F/(AY) = 7.22 × 10⁻⁷. No column length is needed because strain, rather than shortening, was requested.
  4. 4.This checks axial material strain only; a real column also needs a buckling check. The given dimensions are radii, not diameters.

Hanging masses and compatibility in parallel supports

Steel: 1.5 m4 kgBrass: 1.0 m6 kg6g10gUpper segment supports both masses; lower supports only 6 kg.
Find each segment’s supported mass before computing extension— Both wires have diameter 0.25 cm. The upper steel wire carries 4 kg plus 6 kg, while the lower brass wire carries only the 6 kg load.
Different tensions in vertically connected wires

Problem
A 1.5 m steel wire and 1.0 m brass wire have diameter 0.25 cm. A 4 kg mass hangs at their junction and a 6 kg mass at the bottom. Use Ys = 200 GPa and Yb = 90 GPa.

  1. 1.For the bottom mass, brass tension is 6g = 58.8 N. At the junction the steel must support that tension plus 4g, giving 98 N.
  2. 2.A = π(0.00125)² = 4.909 × 10⁻⁶ m² for either wire.
  3. 3.Steel extension = 98(1.5)/(A × 200 × 10⁹) = 1.50 × 10⁻⁴ m.
  4. 4.Brass extension = 58.8(1.0)/(A × 90 × 10⁹) = 1.33 × 10⁻⁴ m.
  5. 5.Using one common tension would be wrong because a real load acts at the junction.

Parallel supports require both force balance and compatibility of displacement. A rigid horizontal bar supported symmetrically by three equal-length wires remains horizontal when the two end wires match. All three lower attachment points move down by the same amount, so their extensions are equal. If the design also requires equal tension, FL/(AY) demands equal AY. For the copper end wires and iron centre wire, AcYc = AiYi. Since A is proportional to diameter squared, dc/di = √(Yi/Yc). With the table’s 190 GPa for wrought iron and 110 GPa for copper, the ratio is about 1.31. Equal tension is then Mg/3 = 49 N for the 15 kg bar. The ratio depends on the chosen material moduli, not on the bar’s mass.

Geometry, comparisons and a moving load

Equal extensions under the same force

Problem
A steel wire has Ls = 4.7 m and As = 3.0 × 10⁻⁵ m². Copper has Lc = 3.5 m and Ac = 4.0 × 10⁻⁵ m². Their extensions under the same load are equal. Find Ys/Yc.

  1. 1.Set FLs/(AsYs) = FLc/(AcYc) and cancel the common force.
  2. 2.Rearrange before substituting: Ys/Yc = LsAc/(LcAs).
  3. 3.The ratio is 4.7 × 4.0/(3.5 × 3.0) = 1.79, approximately 1.8. Equal extension is not equal strain when lengths differ.
A rectangular copper specimen

Problem
A copper strip has cross-section 15.2 mm × 19.1 mm and is pulled by 44,500 N. Find its elastic strain using Y = 110 GPa.

  1. 1.A = 0.0152 × 0.0191 = 2.9032 × 10⁻⁴ m².
  2. 2.σ = F/A = 1.533 × 10⁸ Pa. This is below the table’s representative 200 MPa copper yield strength, consistent with the stated elastic assumption.
  3. 3.ε = σ/Y = 1.393 × 10⁻³, or 0.1393%. Do not confuse a fractional strain with its percentage.
A mass at the bottom of a vertical circle

Problem
A 14.5 kg mass on a 1.0 m steel wire moves at 2 revolutions per second at the lowest point. The wire area is 0.065 cm² and Y = 200 GPa. Estimate its extension.

  1. 1.At the bottom, acceleration points upward toward the circle’s centre. The radial equation is T − mg = mω²r, so T = m(ω²r + g). Weight alone is not the tension.
  2. 2.Convert frequency: ω = 2πf = 4π rad/s. Convert area: 0.065 cm² = 6.5 × 10⁻⁶ m².
  3. 3.Use r ≈ L = 1.0 m because the extension is small. Then T = 2432 N and ΔL = TL/(AY) = 1.871 × 10⁻³ m = 1.87 mm.
  4. 4.The computed strain is about 0.00187, confirming that the geometric correction is small. Using r = L + ΔL consistently gives about 1.874 mm.
  5. 5.This is an ideal elastic-wire estimate. The required stress is about 374 MPa, so a real wire grade must be checked for elastic operation; the generic 250 MPa yield row does not guarantee it.

Quick Quiz

Quiz

Quick check

A wire’s radius doubles while its material, length and applied force stay fixed. What happens to its small elastic extension?

Explanation

Area grows by four and ΔL = FL/(AY), so extension falls by four.

Quick check

Two equal-area wires in series have no intermediate load. Which statement is necessarily true?

Explanation

Force balance at a light junction fixes equal tensions; each extension still depends on L/(AY).

Quick check

In the hanging steel–brass arrangement, why is the upper tension larger?

Explanation

The load at the junction breaks the equal-tension condition of an unloaded series joint.

Quick check

At the bottom of a vertical circle, the force equation for a mass attached to a wire is

Explanation

The inward direction is upward there: tension acts inward and gravity outward relative to the centre.

Quick check

Which changes the Young’s modulus of an otherwise identical material specimen at the same physical conditions?

Explanation

Geometry changes force–extension stiffness AY/L, not the material parameter Y.

Practice Problems

Practice Problems
  1. A uniform wire has length 1.5 m, radius 0.50 mm and Young’s modulus 2.0 × 10¹¹ Pa. It is stretched by 40 N. Find its stress and extension, neglecting its weight. The material is proportional up to 100 MPa: check whether the calculated stress is within that stated range.
  2. A 1.0 m copper wire and a 2.0 m steel wire, each of area 1.0 mm², are connected end to end with no junction load. Find each extension and the total extension under 50 N. Use Ycopper = 110 GPa and Ysteel = 200 GPa. Explain why their tensions are equal while their extensions need not be.
  3. Two steel wires of diameter 1.0 mm hang vertically in series. The upper wire is 1.0 m long and the lower wire 0.50 m long. A 2.0 kg mass hangs at the junction and a 3.0 kg mass at the bottom. Neglect the wire weights and use Y = 200 GPa and g = 9.8 m/s². Find both tensions and both extensions.
  4. Two identical hollow steel columns share a 12,000 kg static load equally. Each has outer radius 0.20 m and inner radius 0.10 m. Calculate the axial compression strain using Y = 200 GPa and g = 9.8 m/s². State why this result alone does not establish resistance to buckling.
  5. A 2.0 kg mass attached to a steel wire of unstretched length 0.80 m moves at 1.0 revolution per second at the bottom of a vertical circle. The wire area is 1.0 mm² and Y = 200 GPa. Use r ≈ 0.80 m and g = 9.8 m/s² to find tension and extension. Check against a stated proportional limit of 150 MPa and explain why tension cannot be replaced by mg.

Key Takeaways

Key Takeaways

• Derive ΔL = FL/(AY) only after identifying the actual axial force and material’s linear range. • Use equal tension for an unloaded series joint, and equal displacement when rigid parallel supports impose it. • Use the material area: πr² for a solid wire, π(R² − r²) for a hollow column, and breadth × thickness for a strip. • Moving loads require a force equation first; small calculated strain does not by itself prove stress is below yield.

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Hooke’s Law and Stress–Strain Curves

Next · Lesson 4

Shear Modulus and Bulk Compression