Mechanics of Materials I: Stress, Strain, Elastic Constants, Mohr’s Circle, Thin Cylinders, Thermal Stresses, Strain Rosettes and Material Testing
1. Stress, strain, elastic constants and Poisson’s ratio
Normal stress is force per unit area, σ = P/A, and normal strain is the fractional change in length, ε = δ/L. Within the elastic limit Hooke’s law gives σ = Eε, so an axially loaded bar extends by δ = PL/(AE). A 20 kN pull on a 1.5 m bar of 200 mm² and E = 200 GPa extends it 20 000 × 1.5/(200 × 10⁻⁶ × 200 × 10⁹) = 7.5 × 10⁻⁴ m = 0.75 mm. A stepped bar sums the extensions of its segments, and bars sharing one displacement share the load in proportion to AE. Shear stress τ = V/A causes shear strain γ, and τ = Gγ. The strain energy stored is U = σ²V/(2E) = P²L/(2AE) for an axial load.
An isotropic material has two independent elastic constants, and Young’s modulus E, shear modulus G, bulk modulus K and Poisson’s ratio ν (the ratio of lateral to axial strain, taken positive) are tied by E = 2G(1 + ν) = 3K(1 − 2ν). For E = 200 GPa and ν = 0.25, G = 200/(2 × 1.25) = 80 GPa. ν lies between −1 and 0.5 for a stable isotropic solid, 0.5 being incompressible and about 0.3 typical of metals. A bar stretched by ε has a volumetric strain ε_v = ε(1 − 2ν), which is why a material with ν = 0.5 does not change volume, and why K is finite only for ν < 0.5.
| Constant | Definition | In terms of E and ν |
|---|---|---|
| Young’s modulus E | σ/ε | E |
| Shear modulus G | τ/γ | E/[2(1 + ν)] |
| Bulk modulus K | p/ε_v (hydrostatic) | E/[3(1 − 2ν)] |
| Poisson’s ratio ν | −ε_lateral/ε_axial | ν |
2. Mohr’s circle for plane stress and plane strain
At a point in plane stress (σ_x, σ_y, τ_xy) the stresses on a plane whose normal is at θ to the x-axis are σ_n = (σ_x + σ_y)/2 + (σ_x − σ_y)/2 · cos 2θ + τ_xy sin 2θ and τ = −(σ_x − σ_y)/2 · sin 2θ + τ_xy cos 2θ. Plotted with σ across and τ up they trace a circle, Mohr’s circle, with centre (σ_x + σ_y)/2 and radius R = √[((σ_x − σ_y)/2)² + τ_xy²]. The principal stresses, where τ = 0, are σ₁,₂ = centre ± R, on planes 90° apart in the body (180° on the circle) at tan 2θ_p = 2τ_xy/(σ_x − σ_y). The maximum in-plane shear stress is R, on planes 45° from the principal planes.
Worked: σ_x = 80, σ_y = 20, τ_xy = 40 MPa. The centre is 50, R = √(30² + 40²) = 50, so σ₁ = 100 MPa, σ₂ = 0 and the in-plane maximum shear is 50 MPa. The absolute maximum shear stress must include the third principal stress, zero in plane stress: it is the largest of (σ₁ − σ₂)/2, σ₁/2 and σ₂/2 taken with the zero. For σ_x = 100, σ_y = 60, τ_xy = 30 MPa: centre 80, R = √(20² + 30²) = 36.06, σ₁ = 116.06, σ₂ = 43.94, both positive, so the absolute maximum is (116.06 − 0)/2 = 58.0 MPa, not the in-plane 36.06.
Plane strain has exactly the same geometry with strains: ε_x and ε_y on the σ axis and half the engineering shear strain, γ/2, on the vertical axis, centre (ε_x + ε_y)/2 and radius √[((ε_x − ε_y)/2)² + (γ_xy/2)²]. The two states differ in what is zero: plane stress has σ_z = 0 and ε_z = −ν(σ_x + σ_y)/E, and plane strain has ε_z = 0 and σ_z = ν(σ_x + σ_y). A thin plate loaded in its own plane is in plane stress; a long dam or a thick pipe section far from its ends is in plane strain.
3. Thin cylinders and spheres
A cylinder is thin when t is small against d (t < d/20), so the stress is uniform through the wall. Cutting the vessel along a diameter gives the hoop (circumferential) stress σ_h = pd/(2t), and cutting across a section gives the longitudinal stress σ_l = pd/(4t), exactly half. A thin sphere has σ = pd/(4t) in every direction. For d = 1 m, t = 10 mm and p = 2 MPa: σ_h = 2 × 10⁶ × 1/(0.02) = 100 MPa and σ_l = 50 MPa. That the hoop stress is twice the longitudinal one is why a boiler or a sausage splits along its length.
The strains follow from Hooke’s law in two dimensions, the radial stress being negligible: ε_h = (σ_h − νσ_l)/E = (pd/4tE)(2 − ν) and ε_l = (σ_l − νσ_h)/E = (pd/4tE)(1 − 2ν). The volumetric strain is ε_v = 2ε_h + ε_l = (pd/4tE)(5 − 4ν): with the numbers above, E = 200 GPa and ν = 0.3, (50/200 000)(5 − 1.2) = 2.5 × 10⁻⁴ × 3.8 = 9.5 × 10⁻⁴. The maximum shear stress in the wall’s plane is (σ_h − σ_l)/2 = pd/(8t); including the radial stress, which is −p at the inside and 0 outside, changes little for a thin wall.
4. Thermal stresses and composite bars
A free bar heated by ΔT grows by δ = αLΔT and carries no stress. If the ends are fully restrained, the thermal strain αΔT is cancelled by an equal and opposite elastic strain, so σ = −EαΔT, compressive on heating, and independent of the length and the area. Steel with E = 200 GPa and α = 12 × 10⁻⁶ per °C heated 50 °C between rigid walls develops 200 × 10³ × 12 × 10⁻⁶ × 50 = 120 MPa. With a gap g the bar expands freely by g first, and the stress arises from the remainder, E(αΔT − g/L).
In a composite bar of two materials joined and heated together, the bar as a whole is free, but each material tries to expand by its own α, so they push on each other: the forces are equal and opposite, P_1 = P_2, and the strains are compatible, so the total extension is common. The material with the larger α is put in compression and the other in tension when heated. A useful check for any thermal-stress problem is that a body free to expand, in a uniform temperature change, has none.
5. Strain gauges and rosettes, and testing of materials
A strain gauge measures the strain in one direction only, so the state of strain at a point on a free surface, which has three unknowns (ε_x, ε_y, γ_xy), needs three gauges: a rosette. For a rectangular (0°/45°/90°) rosette reading ε_a, ε_b and ε_c, the principal strains are ε₁,₂ = (ε_a + ε_c)/2 ± (1/√2)√[(ε_a − ε_b)² + (ε_b − ε_c)²], and the shear strain is γ_xy = 2ε_b − ε_a − ε_c. For 800, 400 and 200 microstrain, the centre is 500 and the radius √100 000 = 316.2, so ε₁ = 816 με and ε₂ = 184 με. Principal stresses follow from Hooke’s law in two dimensions: σ₁ = E(ε₁ + νε₂)/(1 − ν²). A delta (0°/60°/120°) rosette does the same job with different formulae.
The tension test on the universal testing machine draws the engineering stress–strain curve: a straight elastic line whose slope is E, the proportional and elastic limits, the yield point (for a material with no sharp yield, the stress at a 0.2% offset), the ultimate tensile strength at the peak load, necking, and fracture. Ductility is the percentage elongation and the percentage reduction of area; toughness is the area under the curve; resilience the area to the elastic limit. True stress, load over the current area, keeps rising through necking, unlike the engineering curve. Compression testing suits brittle materials such as cast iron and concrete.
| Test | Principle |
|---|---|
| Brinell | A hardened steel or carbide ball is pressed in; the hardness is load over the curved area from the diameter of the impression. |
| Vickers | A 136° square-based diamond pyramid; HV = 1.854 P/d² from the diagonal d of the impression. |
| Rockwell | The number is read from the depth of penetration under a major load after a minor load; different scales (B, C) for soft and hard metals. |
| Charpy and Izod impact | A notched bar is broken by a swinging hammer; the energy absorbed measures toughness. Charpy: simply supported; Izod: cantilever. |
For a Vickers test with P = 20 kgf and d = 0.25 mm, HV = 1.854 × 20/0.0625 = 593. For steels the ultimate strength in MPa is roughly 3.4–3.5 times the Brinell number, which is how hardness stands in for a tension test. Body-centred-cubic steels show a ductile-to-brittle transition as temperature falls: the impact energy drops over a narrow range, which the Charpy test exposes and the static tension test does not.
Key takeaways
- δ = PL/(AE); E = 2G(1 + ν) = 3K(1 − 2ν); volumetric strain under uniaxial strain ε is ε(1 − 2ν).
- Mohr’s circle: centre (σ_x + σ_y)/2, radius √[((σ_x − σ_y)/2)² + τ²]; include the zero third principal stress for the absolute maximum shear.
- Thin cylinder: σ_h = pd/2t, σ_l = pd/4t; sphere pd/4t; ε_v = (pd/4tE)(5 − 4ν).
- Fully restrained thermal stress σ = EαΔT, independent of length; a free body under uniform heating carries none.
- A rosette gives three strains, hence the full strain state; the 0.2% offset defines yield; Rockwell reads depth, Brinell and Vickers read an impression.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
For an isotropic elastic material, Young’s modulus E, shear modulus G and Poisson’s ratio ν are related by:
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Answer: A — E = 2G(1 + ν)
E = 2G(1 + ν) follows from a state of pure shear seen along its principal directions. E = 3K(1 − 2ν) is the relation with the bulk modulus, which is the source of option C.A steel bar of length 1.5 m and cross-sectional area 200 mm² carries an axial tensile load of 20 kN. Taking E = 200 GPa, its extension is ____ mm.
Numerical answer — type the value.
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Answer: 0.75
δ = PL/(AE) = 20 000 × 1.5/(200 × 10⁻⁶ × 200 × 10⁹) = 30 000/(4 × 10⁷) = 7.5 × 10⁻⁴ m = 0.75 mm. Forgetting to convert the area from mm² to m² is the usual slip.A material has E = 200 GPa and Poisson’s ratio 0.25. Its shear modulus is ____ GPa.
Numerical answer — type the value.
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Answer: 80
G = E/[2(1 + ν)] = 200/(2 × 1.25) = 80 GPa. Using E/(2ν) or E/(1 + ν) = 160 would ignore the factor 2.The theoretical range of Poisson’s ratio of a stable isotropic elastic solid is:
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Answer: A — −1 to 0.5
Positive G and K require −1 < ν < 0.5: 0.5 makes K infinite (incompressible) and below −1 would make G negative. Most engineering metals lie near 0.3, and the value 0 for cork and negative values for auxetic foams show that the range is real.At a point in a body under plane stress, σ_x = 80 MPa, σ_y = 20 MPa and τ_xy = 40 MPa. The maximum principal stress is ____ MPa.
Numerical answer — type the value.
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Answer: 100
The centre of Mohr’s circle is (80 + 20)/2 = 50 and the radius is √(30² + 40²) = 50, so σ₁ = 50 + 50 = 100 MPa and σ₂ = 0. Taking σ_x + τ = 120 adds quantities that do not combine linearly.A point in plane stress has σ_x = 100 MPa, σ_y = 60 MPa and τ_xy = 30 MPa, with σ_z = 0. The absolute maximum shear stress at the point, correct to one decimal place, is ____ MPa.
Numerical answer — type the value.
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Answer: 58.0
The centre is 80 and R = √(20² + 30²) = 36.06, so σ₁ = 116.06 and σ₂ = 43.94, both positive. With σ₃ = 0 the largest of the three circles is (σ₁ − σ₃)/2 = 58.03 MPa. The in-plane value R = 36.06 is only the middle circle.In Mohr’s circle for stress, the planes of maximum shear stress are inclined to the principal planes at:
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Answer: A — 45° in the body, 90° on the circle
Angles on the circle are twice the angles in the body. The top of the circle, where τ is greatest, is 90° from the σ-axis intercepts, the principal stresses, so the physical planes differ by 45°. On the principal planes the shear stress is zero.A thin cylindrical shell of internal diameter 1 m and wall thickness 10 mm carries an internal pressure of 2 MPa. The hoop stress is ____ MPa.
Numerical answer — type the value.
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Answer: 100
σ_h = pd/(2t) = 2 × 1000/(2 × 10) = 100 MPa (with p in MPa and d, t in mm). The longitudinal stress is half of it, 50 MPa; pd/4t = 50 MPa is the answer for a sphere.A thin cylindrical shell of internal diameter 1 m and wall thickness 10 mm carries an internal pressure of 2 MPa. It is made of a material with E = 200 GPa and ν = 0.3. Its volumetric strain is ____ × 10⁻⁴.
Numerical answer — type the value.
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Answer: 9.5
ε_v = (pd/4tE)(5 − 4ν) = (2 × 1000/(4 × 10 × 200 000))(5 − 1.2) = 2.5 × 10⁻⁴ × 3.8 = 9.5 × 10⁻⁴. Equivalently 2ε_h + ε_l with ε_h = 2.5 × 10⁻⁴ × 1.7 and ε_l = 2.5 × 10⁻⁴ × 0.4.In a thin-walled cylindrical pressure vessel, the ratio of the hoop stress to the longitudinal stress is:
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Answer: A — 2
σ_h = pd/2t and σ_l = pd/4t, so the hoop stress is twice the longitudinal stress; this is why a cylinder splits along a generator. A ratio of 1 belongs to a sphere.A steel bar with E = 200 GPa and α = 12 × 10⁻⁶ per °C is held between two rigid walls and heated by 50 °C. The compressive stress in the bar is ____ MPa.
Numerical answer — type the value.
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Answer: 120
σ = EαΔT = 200 000 MPa × 12 × 10⁻⁶ × 50 = 120 MPa. The length and area do not enter. A free bar would develop no stress at all.A rectangular strain rosette (0°/45°/90°) reads ε_a = 800, ε_b = 400 and ε_c = 200 microstrain. The maximum principal strain, to the nearest microstrain, is ____ με.
Numerical answer — type the value.
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Answer: 816
ε₁ = (ε_a + ε_c)/2 + (1/√2)√[(ε_a − ε_b)² + (ε_b − ε_c)²] = 500 + √(400² + 200²)/√2 = 500 + 316.2 = 816.2, so 816. Using the average alone, 500, ignores the shear that the 45° gauge reveals.A Vickers hardness test uses a load of 20 kgf and the mean diagonal of the impression is 0.25 mm. Taking HV = 1.854 P/d², the Vickers hardness number is ____.
Numerical answer — type the value.
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Answer: 593
HV = 1.854 × 20/0.25² = 37.08/0.0625 = 593.3, so 593. The diagonal is squared: using d alone gives 148, and the units are kgf and mm.For a metal with no clearly defined yield point, the yield strength is conventionally taken as the stress at which:
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Answer: A — a line parallel to the elastic slope, offset by a plastic strain of 0.2%, meets the curve
The 0.2% offset method draws a line with the slope E from ε = 0.002 and reads where it meets the curve. The peak load defines the ultimate strength and the end of the curve is fracture.Which of the following statements about material testing are true?
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Answer: A — The Charpy specimen is supported at both ends and struck at the middle, whereas an Izod specimen is held as a cantilever; B — Body-centred-cubic steels show a ductile-to-brittle transition on cooling; C — The Vickers test uses a square-based diamond pyramid indenter
(A) The two impact geometries. (B) The impact energy falls steeply over a temperature range. (C) The 136° pyramid gives the Vickers scale. (D) False: Rockwell reads the depth of penetration directly from the dial; the impression diameter is Brinell’s measurement.A bar of an isotropic material is stretched by a uniaxial strain ε = 0.001 with no lateral stress. If ν = 0.3, its fractional change in volume is:
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Answer: A — 0.0004
ε_v = ε(1 − 2ν) = 0.001 × 0.4 = 0.0004, because each of the two lateral strains is −νε = −0.0003. The volume still grows: it would stay constant only for ν = 0.5.