Stress and Strain Fundamentals

Key Takeaways

  • Normal stress σ = P/A (force perpendicular to cross-section); shear stress τ = V/A (force parallel to cross-section).
  • Normal strain ε = ΔL/L (change in length divided by original length); shear strain γ = angular deformation in radians.
  • Hooke's Law: σ = Eε for normal stress (E = Young's modulus) and τ = Gγ for shear stress (G = shear modulus).
  • Poisson's ratio ν = -εlateral/εaxial; for most metals ν ≈ 0.25-0.35.
  • Relationship: G = E / [2(1+ν)] connects the three elastic constants.
  • The stress-strain diagram shows elastic region, yield point, ultimate strength, and fracture point.
Last updated: March 2026

Stress and Strain Fundamentals

FE Exam Weight: Strength of Materials accounts for 9-14 questions (~10% of the exam). This is a high-weight topic closely connected to Statics.

Types of Stress

Normal Stress (σ)

Force perpendicular to the cross-sectional area: σ=PA\sigma = \frac{P}{A}

  • Tensile stress (+): member is being pulled apart
  • Compressive stress (-): member is being pushed together

Shear Stress (τ)

Force parallel to the cross-sectional area: τ=VA\tau = \frac{V}{A}

Bearing Stress

Contact pressure between two surfaces: σb=PAb\sigma_b = \frac{P}{A_b}

where Ab is the projected area of contact (diameter × thickness for a pin in a plate).

Types of Strain

Normal Strain (ε)

ϵ=ΔLL=δL\epsilon = \frac{\Delta L}{L} = \frac{\delta}{L}

Dimensionless (often expressed as in/in, mm/mm, or percentage).

Shear Strain (γ)

The angular deformation in radians: γ=ΔxL\gamma = \frac{\Delta x}{L}

Hooke's Law

In the elastic (linear) region:

σ=Eϵτ=Gγ\sigma = E\epsilon \qquad \tau = G\gamma

PropertySymbolTypical Values (Steel)
Young's Modulus (Elastic Modulus)E200 GPa (29,000 ksi)
Shear ModulusG77 GPa (11,500 ksi)
Poisson's Ratioν0.30

Relationships between elastic constants: G=E2(1+ν)G = \frac{E}{2(1 + \nu)} K=E3(12ν)K = \frac{E}{3(1 - 2\nu)}

where K = bulk modulus.

Axial Deformation

For a prismatic bar under axial load: δ=PLAE\delta = \frac{PL}{AE}

For multiple segments or varying loads: δ=iPiLiAiEi\delta = \sum_i \frac{P_i L_i}{A_i E_i}

Thermal deformation: δT=αLΔT\delta_T = \alpha L \Delta T

where α is the coefficient of thermal expansion.

The Stress-Strain Diagram

Point/RegionDescription
Proportional limitEnd of linear (Hooke's Law) region
Elastic limitMaximum stress for full strain recovery
Yield point (σy)Onset of permanent deformation
Strain hardeningStress increases after yielding due to dislocation pileup
Ultimate tensile strength (σu)Maximum stress the material can withstand
NeckingCross-section reduces; stress appears to decrease
FractureMaterial breaks

Ductile vs. Brittle Materials

PropertyDuctile (Steel)Brittle (Cast Iron, Concrete)
Yield pointWell-definedNot clearly defined
Fracture strainLarge (>5%)Small (<5%)
NeckingSignificantLittle to none
Warning before failureYesNo
Failure modeShear (45° surface)Tensile (flat surface)

Factor of Safety

FS=Failure StressAllowable Stress=σfailureσallowFS = \frac{\text{Failure Stress}}{\text{Allowable Stress}} = \frac{\sigma_{failure}}{\sigma_{allow}}

  • For ductile materials: FS = σy / σallow (based on yield strength)
  • For brittle materials: FS = σu / σallow (based on ultimate strength)
  • Typical values: 1.5-3.0 for structural applications
Test Your Knowledge

A steel rod with a cross-sectional area of 500 mm² carries a tensile load of 100 kN. What is the normal stress?

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Test Your Knowledge

A 2 m steel bar (E = 200 GPa, A = 400 mm²) is subjected to a 80 kN tensile force. What is the elongation?

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Test Your Knowledge

If E = 200 GPa and ν = 0.3 for steel, what is the shear modulus G?

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