Elasticity Stress and Strain

01 / UNDERSTAND & APPLY

1. Stress describes loading; strain describes deformation

An ideal rigid body never deforms, but real solids change shape or size under forces. Stress measures force transmitted across an area, while strain measures fractional deformation. For uniform uniaxial loading of a slender rod, engineering normal stress is σ = F/A₀ and engineering longitudinal strain is ε = ΔL/L₀, using the original cross-sectional area and length.

Stress has units of pressure, Pa = N m⁻², while strain is dimensionless. A strain of 0.001 corresponds to a 0.1% change, not a 0.001% change. Tension and compression can be represented with signed stress and strain under a stated convention; many elementary questions use positive magnitudes and name the loading type separately.

The force F in F/A is the internal force across one cut section, not the sum of the equal forces applied at both ends. A rod pulled outward by 100 N at each end has 100 N tension, not 200 N. Net external force can be zero while the rod is stressed and stretched. Equilibrium of the whole body does not imply absence of internal forces.

For a small linear elastic uniaxial deformation, σ = Yε, where Y is Young’s modulus. Hence ΔL = FL₀/(A₀Y). A larger Y means less strain under the same stress. Y is a material parameter under specified conditions, while the stiffness of a particular specimen also depends on its length and area.

02 / UNDERSTAND & APPLY

2. A material modulus and a specimen stiffness are different

Comparing F = k_specimen ΔL with the uniaxial law gives k_specimen = YA/L. This effective spring constant has units N m⁻¹, unlike the modulus Y measured in Pa. Two wires made of the same material can have different stiffnesses because their areas and lengths differ, while their Young’s modulus is the same within the material model.

For a circular wire, A = πr² = πd²/4. Doubling its diameter at fixed length and material makes it four times stiffer and reduces extension under a fixed force to one quarter. Doubling length at fixed area halves stiffness and doubles extension. Check whether a comparison holds force, stress, length or volume fixed before applying a ratio.

If a wire is cut into n equal pieces, each piece has the original area and length L/n, so each has stiffness n times the original. Connecting all pieces end to end restores the original stiffness, whereas placing them in parallel between common supports produces n² times the original stiffness under the ideal load-sharing assumptions.

If a wire is redrawn to a different length while retaining its material and volume, area changes inversely with length. Then stiffness YA/L scales as 1/L², provided the drawing process is assumed not to change Y or the material state. Real processing can alter properties, so a textbook volume-preserving comparison must state its idealisation.

03 / UNDERSTAND & APPLY

3. Series and parallel rods have different compatibility conditions

For rods joined in series along the force direction, the same axial force passes through each massless junction in static equilibrium, and extensions add. Each segment contributes F Lᵢ/(AᵢYᵢ). Thus total compliance, extension per unit force, is ΣLᵢ/(AᵢYᵢ). Equal force does not imply equal stress if segment areas differ.

For parallel rods between rigid plates that remain parallel and translate without tilt, the rods undergo the same extension if they have equal initial lengths and no gaps. Their carried forces add, with Fᵢ = (YᵢAᵢ/Lᵢ)ΔL. Load sharing is proportional to stiffness, not automatically equal. Different initial lengths, slack or plate rotation require the actual geometry and compatibility equations.

For a nonuniform rod under axial force, divide it into small elements: dL_extension = F(x)dx/[A(x)Y(x)]. Integrate over the original length in the small-strain approximation. Both area and internal tension may vary. This general expression explains why one formula using an arbitrary average area can fail.

A hanging uniform wire stretched by its own weight has tension increasing upward. At a section a distance x above the bottom, tension is ρAgx. Integration gives total extension ρgL²/(2Y), or MgL/(2AY). Using the entire wire weight as the tension everywhere gives twice the correct result because the actual tension varies from zero to Mg.

04 / UNDERSTAND & APPLY

4. Compression of volume and change of shape need different moduli

Bulk modulus K relates a small uniform pressure increase to fractional volume decrease: K = −Δp/(ΔV/V₀). The minus sign makes K positive because increased pressure gives negative ΔV for an ordinary stable material. Compressibility is its reciprocal. Bulk modulus is relevant to compression from all directions, not simply the axial load divided by longitudinal strain.

Shear stress is tangential force divided by area. For a block of height h whose top face shifts sideways by Δx relative to the bottom, small shear strain is γ_s ≈ Δx/h, equal to the small shear angle in radians. Shear modulus G relates τ_s = Gγ_s. A larger G means less angular distortion for the same shear stress.

A fluid at rest cannot sustain a static shear stress indefinitely, though it has a finite bulk modulus. A solid can resist both volume change and shape change. Viscous resistance in a flowing fluid is a stress related to the rate of deformation, not a static solid shear modulus; the two ideas have different units and physical meaning.

For small homogeneous isotropic linear elasticity, the elastic constants are linked: Y = 2G(1 + ν) = 3K(1 − 2ν), where ν is Poisson’s ratio. These relations are not universal for anisotropic crystals or arbitrary nonlinear materials. A crystal may have direction-dependent stiffnesses that cannot be summarized by a single Y, G and K pair.

05 / UNDERSTAND & APPLY

5. Poisson’s ratio connects longitudinal and lateral changes

When an ordinary rod is stretched longitudinally, it usually contracts laterally. Poisson’s ratio is ν = −ε_lateral/ε_longitudinal under uniaxial loading in the small-strain regime. The minus sign gives a positive ν for the usual opposite-sign strains. A negative Poisson’s ratio is possible for auxetic materials, which widen when stretched; it is not an algebraic impossibility.

For an isotropic specimen with equal lateral strains, the first-order volume strain is ε_volume ≈ ε_longitudinal + 2ε_lateral = (1 − 2ν)ε_longitudinal. This sum neglects products of small strains. A value ν near 1/2 means nearly volume-preserving uniaxial deformation, not zero longitudinal strain.

For a stable homogeneous isotropic linear elastic model with positive bulk and shear moduli, −1 < ν < 1/2. The limiting incompressible idealization approaches ν = 1/2 with very large bulk modulus. Do not apply those bounds without qualification to every direction-dependent anisotropic definition of a Poisson ratio.

Engineering stress uses original area, whereas true stress uses the instantaneous area. They are close for small strains but diverge during large deformation and necking. The simple Y relation in this lesson consistently uses the small-strain approximation and cannot predict all stages of a tensile test by continuing one straight line indefinitely.

06 / UNDERSTAND & APPLY

6. The area under the loading curve is stored or dissipated work

For slow loading within the linear elastic range, force rises from zero to F as extension rises from zero to ΔL. Work stored is the triangular area U = FΔL/2 = k_specimen(ΔL)²/2. Dividing by original volume AL gives energy density u = σε/2 = σ²/(2Y) = Yε²/2. The factor one half comes from the increasing force, not from a universal rule for every loading history.

For a general reversible elastic law, stored energy per original volume is ∫σ dε along the loading path. A nonlinear elastic material can return to its original shape without obeying one constant Young’s modulus over the whole range. Elastic means recoverable deformation; Hookean means a linear stress–strain relation. These properties should not be treated as exact synonyms.

If a load is suddenly applied and left hanging on an ideal initially unstretched massless spring-like wire, the maximum extension can exceed its static value because the mass gains speed before stopping. For a mass released from rest just as an initially unstretched vertical element begins supporting it, Mgx = kx²/2 at the first maximum gives x = 2Mg/k, assuming no initial slack, negligible damping and sufficient elastic range.

If the mass first falls through a gap h before stretching the element, energy balance is Mg(h + x) = kx²/2 under the same ideal assumptions. Do not use the static force relation Mg = kx at the instant of greatest extension; acceleration generally remains nonzero even when velocity is momentarily zero.

07 / UNDERSTAND & APPLY

7. A stress–strain curve distinguishes stiffness, strength and ductility

In a tensile test, the initial straight portion of an engineering stress–strain curve has slope Y. The proportional limit marks the end of the approximately linear relation. The elastic limit marks the largest loading range from which the material returns without permanent strain. These limits can be close but are conceptually different.

Beyond yielding, plastic deformation produces a permanent set when the load is removed. A ductile material can undergo substantial plastic strain before fracture; a brittle material may fracture with little plastic deformation. High Young’s modulus means stiff, while high breaking stress means strong; neither automatically implies the other or guarantees large ductility.

The maximum engineering stress reached in a tensile test is the ultimate tensile strength, while fracture occurs at the end of the curve. Necking changes the local cross-sectional area, so engineering stress can fall even while the true local stress behaves differently. A schematic classroom curve describes a type of material behavior, not every material’s exact sequence of plateaus and peaks.

Loading and unloading along different paths create a hysteresis loop whose area represents energy dissipated per unit volume in a complete cycle, when the axes are stress and strain. This is distinct from the recoverable triangular energy under a purely linear loading curve. Repeated loading can also produce fatigue, so a static breaking-stress value alone is not an adequate real-world structural safety assessment.

08 / UNDERSTAND & APPLY

8. Original practice with geometry and loading assumptions

Use original dimensions and small linear strains unless a question explicitly specifies another model. Convert square millimetres carefully and distinguish total force from force per cross-sectional area.

09 / UNDERSTAND & APPLY

9. Choose the deformation and compatibility law first

Identify whether loading produces axial extension, bulk compression or shear. Choose the corresponding stress, strain and modulus. Then impose the actual compatibility: equal force in series, equal extension in the appropriate parallel arrangement, or a varying internal load under self-weight. Geometry determines specimen stiffness even when material properties stay fixed.

Check whether the requested event is gradual loading, a sudden dynamic maximum or unloading after plastic deformation. The same final force can lead to different histories and different energy storage or loss. Keep linear elastic formulas within their stated range and do not equate stiffness, strength and ductility.

• Stress has pressure units; strain is dimensionless.
• Young’s modulus is a material property, while YA/L is a specimen stiffness.
• Area scales with diameter squared.
• Bulk and shear moduli describe different deformations.
• Elastic recovery and linear proportionality are different claims.
• Energy under gradual linear loading includes the factor one half.
• Real safety margins require more than one ideal stress calculation.