Mechanical Properties of Solids · Lesson 5 of 7
Lateral Strain and Stored Elastic Energy
“Explain the lateral response of a stretched wire and derive the elastic energy stored during gradual linear loading.”
• Calculate Poisson’s ratio using either signed lateral strain or contraction magnitude. • Explain why uniaxial stretching need not preserve volume. • Derive stored energy from variable-force work and a force–extension graph. • Use energy density and recognize when the triangular-area shortcut is valid.
A longitudinal force also changes transverse dimensions
Stretch a soft rubber band and notice that it becomes thinner. In a metal wire the same tendency is usually far less visible, but the diameter still changes. Longitudinal strain alone therefore does not describe the complete change of shape. A second material response connects the transverse contraction with the longitudinal extension in the small linear elastic range.
Let d be the original diameter and Δd the signed change in diameter. Lateral strain is Δd/d. In ordinary tensile extension, ΔL/L is positive while Δd/d is negative. Poisson’s ratio ν is conventionally the negative of the ratio of signed lateral strain to signed longitudinal strain. The minus sign makes ν positive for the usual sideways contraction. The source instead describes Δd as the positive amount of diameter contraction; then the same physical ratio can be written without the explicit minus sign.
Because it is a ratio of strains, ν has no unit or dimensions. Within the stated material model and conditions it is a material property rather than a geometrical property of the specimen. Representative values quoted for steel are 0.28–0.30 and for aluminium alloys about 0.33. These numbers connect the axial and lateral responses; they do not replace Young’s modulus, which is still needed to connect the axial stress to its strain.
Problem
A wire of original diameter 1.0 mm has axial strain 0.0020 and Poisson’s ratio 0.30. Find the diameter change.
- 1.Lateral strain = −νε = −0.30 × 0.0020 = −0.00060.
- 2.Δd = d × lateral strain = (1.0 mm)(−0.00060) = −0.00060 mm = −0.60 μm.
- 3.The negative sign means the diameter decreases. The positive amount of contraction is 0.60 μm.
- 4.Its length has increased by 0.20%, but its diameter has decreased by 0.060%; do not assume these percentages are equal.
Does stretching preserve the volume?
A long wire has volume V = AL. If both transverse linear dimensions acquire a small strain εlat, the area changes by approximately 2εlat times its original value. Multiplying the small changes in area and length and neglecting products of small strains gives volume strain approximately ε + 2εlat. With εlat = −νε, this becomes (1 − 2ν)ε. This short geometrical check explains why a length change cannot normally be treated as having exactly zero volume change.
For ν = 0.30 and ε = 0.0020, the estimated volume increase is 0.00080, or 0.080%. The wire lengthens and gets thinner, but the two effects need not cancel exactly. The comparison table for deformation modes is useful for identifying the principal response; a blanket statement that longitudinal loading never changes volume would conflict with the lateral-strain analysis.
Work must be done to stretch an elastic wire
As the wire is gradually stretched, the external agent does work against its restoring response. In a reversible elastic model this work is stored as elastic potential energy and can be released during unloading. The restoring force is not constant during a Hookean extension: it starts at zero for the unloaded wire and grows with extension. Using final force times total extension would therefore count too much work.
Take a very small extra displacement dx. Across such a small step the force is almost constant, so the work is dW = F(x) dx. To add all the steps from zero extension to the final extension ΔL, integrate. The integral of x is x²/2, producing the factor one half. The same result is the area of a triangle under the straight force–extension graph.
Energy per unit volume
Write ΔL = εL and V = AL in the work expression. The energy becomes U = ½Yε²V. Since σ = Yε, it can also be written ½σεV. Dividing by original volume gives energy density u. This is useful when comparing the amount of elastic energy stored in different-sized specimens under equivalent stress or strain conditions.
Problem
A wire has L = 2.0 m, A = 1.0 × 10⁻⁶ m² and Y = 2.0 × 10¹¹ Pa. It is gradually loaded to 100 N within its linear elastic range. Find extension, stored energy and energy density.
- 1.ΔL = FL/(AY) = 100 × 2/(1.0 × 10⁻⁶ × 2.0 × 10¹¹) = 1.0 × 10⁻³ m.
- 2.U = ½FΔL = ½ × 100 × 0.0010 = 0.050 J.
- 3.Original volume V = AL = 2.0 × 10⁻⁶ m³, so u = 0.050/V = 2.5 × 10⁴ J/m³.
- 4.Independently, σ = 1.0 × 10⁸ Pa and ε = 5.0 × 10⁻⁴. Then ½σε gives the same 2.5 × 10⁴ J/m³.
Conditions behind the energy formulas
The triangular-area shortcut requires a straight loading graph from the chosen unloaded reference. For nonlinear elastic response the work is still the area under the actual force–extension curve, but it is not generally ½FfinalΔL. Plastic deformation or dissipative unloading can prevent all input work being recovered as elastic energy. For small reversible compression, signed normal stress and signed longitudinal strain are both negative, so their product and the stored energy are positive. Check the loading model rather than changing an energy sign by intuition.
Quick Quiz
Quiz
A wire lengthens while its diameter decreases. With signed final-minus-initial changes, why is Poisson’s ratio usually defined with a minus sign?
For ordinary tensile response, the lateral strain is negative and longitudinal strain positive; the extra minus gives positive ν.
A small axial strain ε is accompanied by lateral strain −0.30ε in both transverse directions. What is the first-order volume strain?
Add one axial strain and two lateral strains: ε − 0.30ε − 0.30ε = 0.40ε.
Why is the stored work ½FfinalΔL rather than FfinalΔL for a Hookean wire loaded from zero?
The relevant work is the area under F(x); a straight rise from zero gives a triangular area.
For the same wire within its Hookean range, doubling the final extension multiplies stored elastic energy by
U = AY(ΔL)²/(2L), so stored energy grows with the square of extension.
An elastomer has a curved force–extension graph. Which work method remains valid for gradual loading?
Work is the integral of force over displacement; the triangular expression is a special linear case.
Practice Problems
- A wire of original diameter 2.0 mm develops longitudinal strain 0.0015. Its Poisson’s ratio is 0.28. Find the signed lateral strain, the positive amount of diameter contraction in μm and the final diameter in mm.
- A specimen has longitudinal strain 0.0025 and Poisson’s ratio 0.25. Use the first-order geometrical relation to estimate its volume strain and percentage volume change. Explain why lateral contraction does not necessarily make its total volume change zero.
- A wire has length 1.0 m, area 0.50 mm² and Young’s modulus 200 GPa. It is loaded gradually from zero to 30 N in its linear elastic range. Calculate its extension, stored elastic energy and energy density. Check the energy density using stress and strain independently.
- The same Hookean wire is first loaded gradually from zero to 20 N and, in a separate trial from the unloaded state, gradually from zero to 60 N. Compare the final extensions and stored elastic energies. Explain the role of the increasing force during loading.
- An idealized recoverable loading graph consists of a straight segment from (extension 0, force 0) to (1.0 mm, 10 N), followed by a straight segment to (2.0 mm, 30 N). Calculate the work from the two areas under the graph. Compare it with ½ × final force × final extension and explain why that shortcut is not valid over the complete graph.
Key Takeaways
• Poisson’s ratio connects lateral and longitudinal strain; state whether transverse contraction is signed or a positive magnitude. • Axial stretching generally changes both shape and volume; lateral contraction need not exactly cancel elongation. • Integrating the gradually increasing restoring force gives U = ½FΔL for linear elastic loading. • Energy density is ½σε = ½Yε²; the actual loading curve determines work outside the linear range.