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Mechanics of materials / Sheet 01

Stress and strain

Formula reference for mechanical engineering interviews. Every relation assumes linear elastic, isotropic, homogeneous material unless the note says otherwise.

Scope: Uniaxial to 3D statesSections: 13Units: SI, MPa and mmRevision: A
47 formulas

Basic definitions

Normal stress
σ = P / A
A is the original area, so this is engineering stress
Normal strain
ε = δ / L0
Dimensionless, often reported in µε (10-6)
Direct shear stress
τ = V / A
Average over the sheared area, as in a pin or rivet
Shear strain
γ = τ / G
Angular change, in radians
Hooke's law
σ = Eε, τ = Gγ
Valid only below the proportional limit
Axial deflection
δ = PL / AE
Stiffness k = AE / L
Bulk relation
p = -K · εv
K is the bulk modulus, p is hydrostatic pressure

True vs engineering

True stress
σt = σ (1 + ε)
Equals P divided by instantaneous area; valid to fracture
True strain
εt = ln (1 + ε)
Valid only up to the onset of necking

The engineering curve falls after the ultimate point because area shrinks faster than the material hardens, while you keep dividing by the original area. True stress rises all the way to fracture.

Elastic constants

Poisson's ratio
ν = -εlateral / εaxial
Range 0 to 0.5; metals sit between 0.27 and 0.34
E and G
E = 2G (1 + ν)
So G = E / [2(1 + ν)], about 0.385E for steel
E and K
E = 3K (1 - 2ν)
So K = E / [3(1 - 2ν)]
Volumetric strain
εv = (1 - 2ν)(σx + σy + σz) / E
Zero at ν = 0.5, which is incompressible

ν approaches 0.5 for rubber and for metal undergoing plastic flow, because plastic deformation conserves volume.

Generalized Hooke's law, 3D

Plane stress
σz = 0, εz ≠ 0
Thin sheet, free surface
Plane strain
εz = 0, σz = ν(σx + σy)
Thick section, more constrained, more prone to brittle fracture

Principal stresses and Mohr's circle

Principal stresses, 2D
σ1,2 = (σx + σy)/2 ± √[((σx - σy)/2)² + τxy²]
Principal planes carry zero shear
Principal plane angle
tan 2θp = 2τxy / (σx - σy)
Two roots, 90° apart in real space
Max in-plane shear
τmax = (σ1 - σ2) / 2
Also equals the radius of Mohr's circle
Absolute max shear
τabs = (σ1 - σ3) / 2
Take σ3 = 0 in plane stress; easy to forget
Mohr's circle
centre (σx + σy)/2, radius τmax
Max shear planes sit 45° from the principal planes

Failure criteria

von Mises, 3D
σv = √{½[(σ12)² + (σ23)² + (σ31)²]}
Ductile metals, distortion energy
von Mises, 2D
σv = √(σx² - σxσy + σy² + 3τxy²)
The plane stress working form
von Mises, shaft
σv = √(σ² + 3τ²)
Bending plus torsion at a free surface
Tresca
σ1 - σ3 ≥ Sy
Ductile, conservative, up to 15% more so than Mises
Max normal stress
σ1 ≥ Sut
Brittle materials: cast iron, ceramics, glass
Shear yield strength
τy = 0.577 Sy | 0.5 Sy
von Mises | Tresca
Factor of safety
n = Sy / σv
Use Sut for brittle materials

Hydrostatic stress does not cause yielding, which is exactly why von Mises contains only stress differences.

Applied stress cases

Bending
σ = My / I, σmax = Mc / I = M / S
y is measured from the neutral axis, not along the beam
Transverse shear
τ = VQ / (It)
Rectangle 1.5V/A, circle 4V/3A, maximum at the neutral axis
Torsion, round shaft
τ = Tr / J, θ = TL / GJ
Linear from zero at the centre to max at the surface
Solid round properties
I = πd⁴/64, J = πd⁴/32 = 2I, c = d/2
Hollow: subtract the inner diameter term
Bending shortcut
σ = 32M / πd³
Solid round shaft, outer fibre
Torsion shortcut
τ = 16T / πd³
Solid round shaft, surface
Thin-wall cylinder
σhoop = pr / t, σlong = pr / 2t
Valid for t < r/10; hoop is why pipes split lengthwise
Thin-wall sphere
σ = pr / 2t
Equal in every direction

Because J = 2I, the same numerical moment and torque give a bending stress exactly twice the torsional stress.

Thermal

Free thermal strain
εth = α ΔT
No stress if the member is free to expand
Fully constrained stress
σ = -E α ΔT
Contains no length or area, so it is independent of geometry

For steel this works out to roughly 2.4 MPa per degree C, so a 100 degree excursion alone can yield mild steel. Watch this in press fits, bonded dissimilar materials, and motor housings.

Strain energy

Strain energy, axial
U = P²L / (2AE) = ½ P δ
Area under the load-deflection curve
Strain energy density
u = ½ σ ε = σ² / 2E
Energy per unit volume
Modulus of resilience
ur = Sy² / 2E
Area under the elastic region only
Toughness
area under the full curve
Strength times ductility; not the same as resilience

Stress concentration and measurement

Stress concentration
σmax = Kt · σnominal
Nominal is taken on the net section, not the gross
Strain gauge
GF = (ΔR / R) / ε
Typically about 2.0 for a foil gauge

Kt is purely geometric: independent of material and of load magnitude. It matters most in fatigue and in brittle materials, since ductile metals locally yield and blunt the peak under static load. Doubling a fillet radius is usually cheaper than upgrading the alloy.

Worked chain: shaft under bending and torsion

Step 1, bending stress
σ = 32M / πd³
Tensile, at the outer fibre
Step 2, shear stress
τ = 16T / πd³
At that same outer fibre
Step 3, combine
σv = √(σ² + 3τ²)
One 2D stress state, not two separate comparisons
Step 4, margin
n = Sy / σv
Then check deflection and twist separately

The critical element is on the surface, at the fibre farthest from the neutral axis in bending. Torsional shear is uniform around the circumference, so the point that matters is the one where the bending tensile stress is also at its maximum.

Numbers worth memorizing

MaterialE (GPa)ρ (kg/m³)Yield (MPa)να (µε/°C)
Steel, all alloys2007850250 to 900+0.2912
Aluminium 6061-T66927002760.3323
Aluminium 7075-T67128105030.3323
Titanium Ti-6Al-4V11444308800.348.6
ABS and PLA2 to 3.51050 to 125040 to 600.3570 to 90

Conversions: 1 ksi = 6.895 MPa, 1 GPa = 145 ksi, steel E = 29 Msi, aluminium E = 10 Msi.

Concepts that decide the interview

  1. Stiffness and strength are independent. Every steel is about 200 GPa, so alloy choice and heat treatment change yield but not deflection. Fix deflection with geometry (A, I); fix yielding with material.
  2. Specific stiffness E/ρ is roughly 25 MN·m/kg for steel, aluminium, magnesium and titanium alike. A material swap alone cannot lighten an axial member of equal stiffness. Aluminium wins in bending and buckling because you can afford more thickness, and I scales with t³ or d⁴.
  3. Necking starts at the ultimate point. That is also where εt = ln(1 + ε) stops being valid, because strain is no longer uniform along the gauge length.
  4. Mild steel shows a distinct upper and lower yield point. Aluminium and most non-ferrous alloys do not, so yield is defined by the 0.2 percent offset.
  5. Cast iron in torsion fractures on a 45 degree helix, because brittle materials fail on the plane of maximum tensile principal stress.
  6. Thick sections approach plane strain, are more constrained, and are more prone to brittle fracture.
  7. Support configuration only exists to produce the bending moment. Once M is given at a section, the layout no longer affects the stress calculation.
Sheet 01 of 07 / Rev AAssumes linear elastic isotropic material

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