Stress and Strain, Mohr’s Circle, Elastic Constants, Thermal Stress and Theories of Failure

This chapter belongs to the Engineering Sciences (XE) paper and answers the first half of Section XE3.2, Mechanics of Deformable Bodies, of XE3 Solid Mechanics, one of the nine optional sections, any two of which a candidate chooses in the exam. The lines it covers read: “Definition of stress and strain; Transformation of stresses and strains; Principal Stresses; Mohr’s circle for plane stress and plane strain; Elastic Constants; Generalized Hooke’s Law; Thermal Stresses; Theories of Failure - von Mises, Tresca and maximum principal stress theories.” It teaches stress and strain with their sign conventions, the transformation equations, principal stresses and maximum shear, Mohr’s circle for stress and for strain (with strain rosettes), the elastic constants and Hooke’s law in 3-D, plane stress and plane strain, thermal stress in restrained and composite bars, and the three failure theories with their yield loci. The numericals are principal stresses and stresses on inclined planes, principal strains from gauge readings, strains from Hooke’s law, thermal stress with and without a gap, composite bars and factors of safety.

1. Stress and strain: definitions, sign convention and axial deformation

Stress is internal force per unit area on an imaginary cut. Normal stress σ = F/A acts perpendicular to the cut (tension positive); shear stress τ = V/A acts along it. At a point the 3-D state is six numbers, σx, σy, σz, τxy, τyz, τzx, because moment equilibrium of a small element forces τxy = τyx. Sign convention in this chapter: tension positive, and a shear stress is positive when, on a face whose outward normal points along +x, it points along +y (and likewise on the other faces). Strain is deformation per unit length: normal strain ε = ΔL/L (extension positive, dimensionless); shear strain γ is the decrease of an originally right angle, in radians (engineering shear strain, twice the tensor shear strain).

In the elastic range σ = Eε, and the lateral strain is ε_lat = −ν ε_axial, where ν is Poisson’s ratio. A bar of area A and length L under axial force P extends by δ = PL/(AE); for steps or a varying force, δ = Σ P_iL_i/(A_iE_i) or ∫P dx/(AE). Worked example. A steel bar (E = 200 GPa) is 800 mm long: the first 500 mm has area 100 mm², the remaining 300 mm has 50 mm², and a 10 kN tension acts throughout. δ = 10 000 × 500/(100 × 200 000) + 10 000 × 300/(50 × 200 000) = 0.25 + 0.30 = 0.55 mm. The stress is 100 MPa in the thick part and 200 MPa in the thin part: the same force, different stress, so always find σ section by section.

⚠️ Keep the units consistent
With force in N, length in mm and area in mm², stress comes out in N/mm² = MPa, so E must also be in MPa (1 GPa = 1000 MPa). A slip of 1000 is the commonest error in these numericals. Shear strain is in radians, so τ = Gγ needs γ as a pure number.

2. Transformation of stress, principal stresses and maximum shear

For plane stress (σz = 0) let the x and y faces carry σx, σy and τxy. On a plane whose normal is rotated by θ counter-clockwise from the x axis:

  • Normal stress: σθ = (σx + σy)/2 + [(σx − σy)/2] cos 2θ + τxy sin 2θ
  • Shear stress: τθ = −[(σx − σy)/2] sin 2θ + τxy cos 2θ
  • Invariant: σθ + σθ+90° = σx + σy, the sum of the normal stresses on any two perpendicular planes
  • Principal planes (τθ = 0): tan 2θp = 2τxy/(σx − σy), two planes 90° apart; principal stresses σ1,2 = (σx + σy)/2 ± R with R = √[((σx − σy)/2)² + τxy²]
  • Maximum in-plane shear: τmax = R = (σ1 − σ2)/2, on planes 45° from the principal planes, where the normal stress is (σx + σy)/2

Worked example. σx = 100 MPa, σy = −50 MPa, τxy = 40 MPa. Centre = (100 − 50)/2 = 25; half-difference = 75; R = √(75² + 40²) = 85. So σ1 = 25 + 85 = 110 MPa, σ2 = 25 − 85 = −60 MPa and the in-plane τmax = 85 MPa. tan 2θp = 80/150 = 0.533, so 2θp = 28.07° and θp = 14.04°. Check: σθ at 14.04° = 25 + 75 cos 28.07° + 40 sin 28.07° = 25 + 66.2 + 18.8 = 110 MPa, so σ1 acts on the plane at 14.04° and σ2 on the plane at 104.04°.

⚠️ The absolute maximum shear includes the zero
In plane stress the third principal stress is σ3 = 0, so the absolute maximum shear stress is half the largest difference among σ1, σ2 and 0, which is not always R. For σx = 100 MPa, σy = 20 MPa, τxy = 0: R = 40 MPa, but the largest shear is (100 − 0)/2 = 50 MPa, on planes inclined 45° to the free z direction. Whenever σ1 and σ2 have the same sign, quoting R underestimates the shear.

3. Mohr’s circle for plane stress and plane strain

Mohr’s circle draws the transformation equations as a circle on axes σ (horizontal) and τ (vertical). With τxy plotted upward under the convention above, X = (σx, τxy) and Y = (σy, −τxy) are the ends of a diameter; the centre is C = (σx + σy)/2 and the radius R = √[((σx − σy)/2)² + τxy²]. The points where the circle cuts the σ axis are σ1 = C + R and σ2 = C − R; the top and bottom points give τmax = ±R; an angle on the circle is twice the angle on the element, and perpendicular planes lie at opposite ends of a diameter. The angle from X to the σ1 point is 2θp. With τxy plotted upward a counter-clockwise turn θ of the plane is a clockwise turn 2θ on the circle; books that plot τxy downward remove this reversal, so fix the convention before reading an angle.

Mohr’s circle for strain has the same form with σ replaced by ε and τ by γ/2 (never γ). Centre (εx + εy)/2, radius R = √[((εx − εy)/2)² + (γxy/2)²]; principal strains ε1,2 = C ± R; maximum in-plane shear strain γmax = 2R = ε1 − ε2. The transformation is ε_θ = (εx + εy)/2 + [(εx − εy)/2] cos 2θ + (γxy/2) sin 2θ. For an isotropic material the principal axes of strain and of stress coincide.

Strain rosette. A gauge reads only the normal strain along its own axis, so three gauges are needed to fix a plane strain state. For gauges at 0°, 45° and 90° to the x axis, εx = ε0, εy = ε90 and, since ε45 = (εx + εy)/2 + γxy/2, γxy = 2ε45 − ε0 − ε90. Worked example: ε0 = 800, ε45 = 600, ε90 = −200 (all × 10⁻⁶). γxy = 1200 − 800 + 200 = 600 × 10⁻⁶. C = 300; R = √(500² + 300²) = 583.1; ε1 = 883.1 × 10⁻⁶, ε2 = −283.1 × 10⁻⁶, γmax = 1166 × 10⁻⁶.

ℹ️ Plane stress is not plane strain
Plane stress: σz = 0 but εz = −ν(σx + σy)/E ≠ 0 (a thin plate loaded in its plane). Plane strain: εz = 0 but σz = ν(σx + σy) ≠ 0 (a long dam, tunnel or pipe restrained along its axis). The in-plane circle is drawn the same way for both, but in plane strain σz is a third principal stress that must enter the absolute maximum shear and the von Mises stress.

4. Elastic constants and the generalised Hooke’s law

A linear elastic isotropic material has only two independent constants. Young’s modulus E = σ/ε (uniaxial), shear modulus G = τ/γ, bulk modulus K = p/ε_v (p the hydrostatic pressure, ε_v the volumetric strain) and Poisson’s ratio ν are linked by:

  • G = E/[2(1 + ν)]; K = E/[3(1 − 2ν)]; E = 9KG/(3K + G); ν = (3K − 2G)/[2(3K + G)]
  • Limits: G > 0 and K > 0 require −1 < ν < 0.5; ν = 0.5 is incompressible (K → ∞); most metals have ν ≈ 0.25 to 0.35; E/G = 2(1 + ν) lies between 2 and 3 for 0 < ν < 0.5

Generalised Hooke’s law (3-D, isotropic): εx = [σx − ν(σy + σz)]/E, with εy and εz by cyclic change, and γxy = τxy/G, γyz = τyz/G, γzx = τzx/G. Adding the three normal strains gives the volumetric strain ε_v = (1 − 2ν)(σx + σy + σz)/E = (σx + σy + σz)/(3K): only the mean stress changes the volume, while shear stress changes shape alone.

  • Plane stress (σz = 0): εx = (σx − νσy)/E, εy = (σy − νσx)/E, εz = −ν(σx + σy)/E
  • Plane strain (εz = 0): σz = ν(σx + σy) and εx = [(1 − ν²)σx − ν(1 + ν)σy]/E

Worked example. Aluminium plate in plane stress, E = 70 GPa, ν = 0.33, σx = 140 MPa, σy = 70 MPa. εx = (140 − 0.33 × 70)/70 000 = 1.67 × 10⁻³; εy = (70 − 0.33 × 140)/70 000 = 3.4 × 10⁻⁴; εz = −0.33 × 210/70 000 = −9.9 × 10⁻⁴, so the plate thins although no stress acts through the thickness.

⚠️ Uniaxial stress is not uniaxial strain
A block compressed along x while its y and z faces are held rigid (εy = εz = 0) is stiffer: σx = E(1 − ν)εx/[(1 + ν)(1 − 2ν)], which is 1.346 E for ν = 0.3. Using σ = Eε for such a confined block overestimates the strain by that factor.

5. Thermal stresses in restrained and composite bars

A free temperature change ΔT produces strain αΔT (α is the coefficient of linear thermal expansion) and no stress; stress appears only when expansion is restrained. Total strain = mechanical + thermal: ε = σ/E + αΔT. A statically determinate body heated uniformly is stress-free.

  • Fixed between rigid walls: ε = 0, so σ = −EαΔT (compressive on heating) and force −EAαΔT. Steel, E = 200 GPa, α = 12 × 10⁻⁶ per °C, ΔT = 50 °C: σ = −200 000 × 12 × 10⁻⁶ × 50 = −120 MPa. Neither L nor A enters the stress.
  • With a gap g: free expansion αLΔT; stress arises only if it exceeds g, and σ = E(αLΔT − g)/L. For L = 1000 mm, g = 0.3 mm, ΔT = 50 °C: free expansion 0.6 mm, σ = 200 000 × 0.3/1000 = 60 MPa. Here L does matter.
  • Biaxial and triaxial restraint: a plate held in both in-plane directions has σ = −EαΔT/(1 − ν); a block held in all three directions has σ = −EαΔT/(1 − 2ν). Both exceed the uniaxial value because the Poisson contraction is blocked too.

Composite bar. Two bars of equal length L, joined rigidly at both ends (or one inside a tube), heated by ΔT with no external load. Equilibrium: F1 + F2 = 0. Compatibility, equal total strain: σ1/E1 + α1ΔT = σ2/E2 + α2ΔT. Taking F as the tension in material 1, F = (α2 − α1)ΔT/[1/(A1E1) + 1/(A2E2)], and the material of larger α, held back, ends in compression. Worked example: steel (A = 300 mm², E = 200 GPa, α = 12 × 10⁻⁶) joined to copper (A = 600 mm², E = 100 GPa, α = 17 × 10⁻⁶), ΔT = 60 °C. 1/(300 × 200 000) + 1/(600 × 100 000) = 3.333 × 10⁻⁸ mm/N, so F = 5 × 10⁻⁶ × 60/3.333 × 10⁻⁸ = 9000 N. Steel carries +30 MPa (tension) and copper −15 MPa (compression).

⚠️ Check what is really restrained
σ = −EαΔT holds only for a bar whose both ends are rigidly fixed. A bar with one free end, or a determinate frame, moves but stays unstressed. Cooling a restrained bar gives tension, since the sign of ΔT flips. For a bar inside a sleeve or against a flexible support, write compatibility for the whole assembly.

6. Theories of failure: maximum principal stress, Tresca and von Mises

A failure theory compares a multiaxial state with the single number from a tension test: the yield strength σY for ductile metals, the tensile strength for brittle ones. Order the principal stresses σ1 ≥ σ2 ≥ σ3, with σ3 = 0 in plane stress. Factor of safety n = σY/σeq, where σeq is the equivalent stress of the chosen theory and all loads scale together.

  • Maximum principal stress (Rankine): failure when σ1 reaches the tensile strength (or σ3 the compressive strength). Suited to brittle materials such as cast iron, ceramics and concrete in tension; poor for ductile metals.
  • Tresca (maximum shear stress): yield when τmax = σY/2, i.e. max(|σ1 − σ2|, |σ2 − σ3|, |σ3 − σ1|) = σY.
  • von Mises (distortion energy): yield when σvm = √{½[(σ1 − σ2)² + (σ2 − σ3)² + (σ3 − σ1)²]} = σY. In plane stress σvm = √(σ1² − σ1σ2 + σ2²) = √(σx² − σxσy + σy² + 3τxy²).
Yield loci in the σ1–σ2 plane (plane stress)
TheoryBoundaryShapePure-shear yield τ
Maximum principal stress|σ1| = σY, |σ2| = σYSquare with corners (±σY, ±σY)σY
Tresca|σ1| = σY, |σ2| = σY, and |σ1 − σ2| = σY when signs differHexagon inscribed in the ellipse0.5σY
von Misesσ1² − σ1σ2 + σ2² = σY²Ellipse through the six hexagon vertices0.577σY

In pure shear (σ1 = τ, σ2 = −τ) Tresca gives τy = 0.5σY and von Mises τy = σY/√3 = 0.577σY. Under hydrostatic stress (σ1 = σ2 = σ3) both give zero equivalent stress, so neither predicts yield at any pressure, whereas the maximum-principal-stress theory would: one reason it is unfit for ductile metals. Tresca is never less conservative than von Mises; they agree in uniaxial stress and in equal biaxial stress (σ1 = σ2) and differ most, by 2/√3 = 1.155 (15.5%), in pure shear. Worked example: the state of section 2 with σY = 300 MPa has principal stresses 110, −60 and 0 MPa. Tresca: max(170, 60, 110) = 170 MPa, n = 300/170 = 1.76. von Mises: √(110² + 110 × 60 + 60²) = √22 300 = 149.3 MPa, n = 2.01. Maximum principal stress: 110 MPa, n = 2.73, the least conservative estimate for a ductile metal.

⚠️ Tresca needs all three principal stresses
In plane stress with σ1 and σ2 both tensile, the largest difference is σ1 − σ3 = σ1 (as σ3 = 0), not σ1 − σ2, so Tresca then reduces to σ1 = σY. Forgetting σ3 = 0 makes Tresca look safer than it is.

Key takeaways

  • Stress is tension-positive and σx + σy is invariant; σ1,2 = (σx + σy)/2 ± R, and in plane stress σ3 = 0 must be included when finding the absolute maximum shear.
  • Mohr’s circle has centre (σx + σy)/2, radius R and doubled angles; for strain plot γ/2, not γ, and a three-gauge rosette gives γxy = 2ε45 − ε0 − ε90.
  • Only two elastic constants are independent: G = E/[2(1 + ν)] and K = E/[3(1 − 2ν)]. Plane stress has εz ≠ 0; plane strain has σz = ν(σx + σy).
  • Thermal stress needs restraint: a fixed bar has σ = −EαΔT, independent of L; a gap gives σ = E(αLΔT − g)/L; a composite bar has F1 + F2 = 0 and equal total strain.
  • In pure shear Tresca yields at 0.5σY and von Mises at 0.577σY; the Tresca hexagon is inscribed in the von Mises ellipse, and the maximum principal stress theory suits brittle fracture only.
  • In every numerical fix the convention and units first, then the principal stresses (with σ3), and only then the equivalent stress and the factor of safety.

Practice questions (18)

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.

  1. For a plane stress state at a point, the sum of the normal stresses on any two mutually perpendicular planes through that point:

    1. equals the difference σ1 − σ2 of the two principal stresses, whatever the planes chosen
    2. is the same for every such pair of planes and equals σx + σy, which is also σ1 + σ2
    3. is largest on the principal planes and smallest on planes inclined at 45° to them
    4. equals twice the maximum in-plane shear stress, 2τmax
    Show answer

    Answer: B — is the same for every such pair of planes and equals σx + σy, which is also σ1 + σ2

    Adding σθ and σθ+90° from the transformation equation, the cos 2θ and sin 2θ terms cancel and the sum is σx + σy = 2C, twice the abscissa of the centre of Mohr’s circle, which equals σ1 + σ2. The difference σ1 − σ2 = 2R is also fixed, but it is the difference, not the sum, and it is the tempting wrong choice because it equals 2τmax; the sum is not changed by rotation at all.

  2. At a point in plane stress, σx = 120 MPa, σy = 40 MPa and τxy = −30 MPa (tension positive; τxy is positive when, on the +x face, it acts along +y). Find the normal stress, in MPa, on the plane whose normal is rotated 30° counter-clockwise from the x axis. Give the answer to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 74.0

    σθ = (σx + σy)/2 + [(σx − σy)/2] cos 2θ + τxy sin 2θ = 80 + 40 cos 60° + (−30) sin 60° = 80 + 20 − 25.98 = 74.02, i.e. 74.0 MPa. The usual slip is to drop the negative sign of τxy, which gives 125.98 MPa.

  3. The principal stresses at a point on the free surface of a body are σ1 = 100 MPa and σ2 = 20 MPa, both in the plane of the surface, with no stress normal to the surface. The absolute maximum shear stress at the point is:

    1. 40 MPa, half of σ1 − σ2, as found from the radius of the in-plane Mohr’s circle
    2. 50 MPa, half of σ1 − σ3 with the zero normal stress σ3 = 0 included
    3. 60 MPa, the mean of the two principal stresses
    4. 80 MPa, the difference σ1 − σ2
    Show answer

    Answer: B — 50 MPa, half of σ1 − σ3 with the zero normal stress σ3 = 0 included

    With σ3 = 0 the three Mohr’s circles have diameters 100 − 20, 20 − 0 and 100 − 0, so the largest shear is (100 − 0)/2 = 50 MPa. The in-plane radius of 40 MPa is the tempting answer but it ignores the third principal stress, which matters because σ1 and σ2 have the same sign. The mean 60 MPa is a normal stress, not a shear stress.

  4. In Mohr’s circle for plane strain, drawn with the normal strain ε on the horizontal axis, the vertical axis represents:

    1. the engineering shear strain γ, exactly as τ is used for stress
    2. the shear stress τ divided by Young’s modulus E
    3. twice the engineering shear strain, 2γ
    4. half the engineering shear strain, γ/2
    Show answer

    Answer: D — half the engineering shear strain, γ/2

    The strain transformation equations have the same form as the stress ones only when τ is replaced by γ/2 (the tensor shear strain). Plotting γ itself, the tempting choice, would distort the circle into an ellipse and give wrong radii. The quantity τ/E is not a strain at all.

  5. A 0°–45°–90° rectangular strain rosette on a thin plate reads ε0 = 500, ε45 = 700 and ε90 = 100 microstrain (1 microstrain = 10⁻⁶). The 0° gauge lies along x, the 90° gauge along y, and the 45° gauge lies between +x and +y. The largest principal strain in the plane of the plate, in microstrain, to the nearest integer, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 747

    εx = 500, εy = 100 and γxy = 2ε45 − ε0 − ε90 = 1400 − 500 − 100 = 800 (× 10⁻⁶). Centre C = (500 + 100)/2 = 300; R = √[((500 − 100)/2)² + (800/2)²] = √(200² + 400²) = 447.2. ε1 = C + R = 747.2, i.e. 747 microstrain. Using γ = 800 instead of γ/2 = 400 in the radius is the usual error.

  6. Which of the following statements about Mohr’s circle for plane stress are correct?

    1. Two points at opposite ends of a diameter represent two mutually perpendicular planes of the element.
    2. An angle θ between two planes of the element appears as an angle θ/2 between the corresponding points on the circle.
    3. The shear stress is zero at the two points where the circle cuts the σ axis, and these points give the principal stresses.
    4. The centre of the circle lies at (σx − σy)/2 on the σ axis.
    Show answer

    Answer: A — Two points at opposite ends of a diameter represent two mutually perpendicular planes of the element.; C — The shear stress is zero at the two points where the circle cuts the σ axis, and these points give the principal stresses.

    Opposite ends of a diameter are 180° apart on the circle, i.e. 90° apart on the element, and the σ-axis crossings are the zero-shear points, which are by definition the principal stresses. Angles on the circle are doubled, not halved, and the centre is the average (σx + σy)/2, not the half-difference, which is the horizontal half-extent of the diameter XY.

  7. For an isotropic, linear elastic material with Poisson’s ratio ν = 0.25, the ratio E/G of Young’s modulus to the shear modulus is:

    1. 1.25, since G exceeds E by the factor (1 + ν)
    2. 2.00, the value obtained when ν = 0
    3. 2.50, from G = E/[2(1 + ν)]
    4. 3.00, the value obtained for an incompressible material
    Show answer

    Answer: C — 2.50, from G = E/[2(1 + ν)]

    G = E/[2(1 + ν)] gives E/G = 2(1 + 0.25) = 2.5. The value 3.00 belongs to ν = 0.5 and 2.00 to ν = 0, which are the two ends of the range, so a mid-range ν = 0.25 must lie between them. G is always smaller than E, so a ratio below 2 cannot occur for ν ≥ 0.

  8. A thin steel plate (E = 200 GPa, ν = 0.3) is in plane stress with σx = 150 MPa and σy = −60 MPa (tension positive), and no shear stress. The normal strain εx, in microstrain (10⁻⁶), to the nearest integer, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 840

    εx = (σx − νσy)/E = (150 − 0.3 × (−60))/200 000 = (150 + 18)/200 000 = 8.4 × 10⁻⁴ = 840 microstrain. The compressive σy adds to εx through the Poisson effect; omitting its sign gives 660.

  9. A long tunnel lining is in plane strain (εz = 0 along the axis). At a point σx = 100 MPa and σy = 40 MPa, and ν = 0.3. The axial stress σz at that point is:

    1. 0 MPa, because no load is applied along the axis
    2. 18 MPa, the value νσ1 − νσ2 from the in-plane difference
    3. 42 MPa, from σz = ν(σx + σy)
    4. 140 MPa, the sum σx + σy carried along the axis
    Show answer

    Answer: C — 42 MPa, from σz = ν(σx + σy)

    For εz = 0: [σz − ν(σx + σy)]/E = 0, so σz = 0.3 × 140 = 42 MPa. Zero is the plane stress answer and is tempting because no axial load is applied, but the restraint of the axial strain itself generates σz. 18 MPa has no basis in the equations.

  10. A steel bar (A = 200 mm², E = 200 GPa, α = 12 × 10⁻⁶ per °C) and an aluminium bar (A = 400 mm², E = 70 GPa, α = 23 × 10⁻⁶ per °C), both of the same length, are joined rigidly at both ends and heated uniformly by 50 °C with no external load. The magnitude of the stress in the aluminium bar, in MPa, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 22.6

    Equilibrium: F_st + F_al = 0; compatibility: σst/Est + αst ΔT = σal/Eal + αal ΔT. With F the steel tension, F = (αal − αst)ΔT/[1/(AstEst) + 1/(AalEal)] = (11 × 10⁻⁶ × 50)/[1/(200 × 200 000) + 1/(400 × 70 000)] = 5.5 × 10⁻⁴/(2.5 × 10⁻⁸ + 3.571 × 10⁻⁸) = 9058.8 N. Aluminium (larger α) is compressed: σal = 9058.8/400 = 22.6 MPa (the steel carries 45.3 MPa tension).

  11. A steel rod 1.2 m long (E = 200 GPa, α = 12 × 10⁻⁶ per °C) is placed with a gap of 0.5 mm between its free end and a rigid wall, the other end being rigidly fixed. The rod is heated uniformly by 80 °C. The compressive stress in the rod after it touches the wall, in MPa, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 108.7

    Free expansion = αLΔT = 12 × 10⁻⁶ × 1200 × 80 = 1.152 mm. The wall stops the rod after 0.5 mm, so 0.652 mm is suppressed over the 1200 mm length: σ = E × 0.652/1200 = 200 000 × 5.433 × 10⁻⁴ = 108.7 MPa. Using the full 1.152 mm (ignoring the gap) gives 192 MPa.

  12. A steel rod is clamped between two rigid walls and heated uniformly by ΔT. If the length of the rod is doubled, keeping the same cross-section, material and ΔT, the thermal stress in it:

    1. remains the same, because σ = −EαΔT does not contain the length
    2. doubles, because the longer rod tries to expand twice as much
    3. halves, because the same wall spacing is now shared among more material
    4. doubles, because the compressive force is proportional to the length
    Show answer

    Answer: A — remains the same, because σ = −EαΔT does not contain the length

    Free expansion αLΔT doubles, but the length over which it is suppressed also doubles, so the strain αΔT, and hence the stress EαΔT, is unchanged; the force EAαΔT is also unchanged. The expansion doubling is the tempting argument, but the strain, not the elongation, produces stress. Length enters only when there is a gap.

  13. A steel block (E = 200 GPa, ν = 0.3) is compressed along x by σx = −100 MPa while its faces normal to y and to z are held between rigid platens so that εy = εz = 0. Its axial strain εx is:

    1. −5.00 × 10⁻⁴, from σx/E as for a free bar
    2. −4.55 × 10⁻⁴, from (1 − ν²)σx/E with only one lateral direction held
    3. −3.71 × 10⁻⁴, with σy = σz = νσx/(1 − ν) = −42.9 MPa developing
    4. −2.00 × 10⁻⁴, from (1 − 2ν)σx/E
    Show answer

    Answer: C — −3.71 × 10⁻⁴, with σy = σz = νσx/(1 − ν) = −42.9 MPa developing

    Setting εy = 0 and εz = 0 gives σy = σz = νσx/(1 − ν) = −42.86 MPa. Then εx = [σx − ν(σy + σz)]/E = [−100 + 0.3 × 85.71]/200 000 = −3.71 × 10⁻⁴. The first value ignores the lateral reactions altogether, and −4.55 × 10⁻⁴ is the plane-strain value with σy = 0, which holds the z faces only.

  14. A ductile metal of yield strength σY = 240 MPa is loaded in pure shear. The shear stress at which yielding starts is predicted by the Tresca and the von Mises theories, respectively, as:

    1. 138.6 MPa and 120 MPa
    2. 120 MPa and 138.6 MPa
    3. 120 MPa and 120 MPa
    4. 240 MPa and 138.6 MPa
    Show answer

    Answer: B — 120 MPa and 138.6 MPa

    In pure shear σ1 = τ, σ2 = −τ, σ3 = 0. Tresca: σ1 − σ2 = 2τ = σY gives τ = σY/2 = 120 MPa. von Mises: √3 τ = σY gives τ = σY/√3 = 138.6 MPa. Tresca yields earlier, hence is the more conservative; the reversed pair is the tempting error.

  15. A ductile steel sphere lies deep in the ocean so that the stress at every point is hydrostatic compression, σ1 = σ2 = σ3 = −p, with p large. Which statement is correct?

    1. Tresca predicts yield at p = σY/2, since every plane carries a shear stress of p/2
    2. von Mises predicts yield at p = σY, since the largest normal stress then equals σY
    3. All three theories predict yield at p = σY, since every principal stress has magnitude p
    4. Tresca and von Mises predict no yield at any p, whereas the maximum principal stress theory would predict failure at p = σY
    Show answer

    Answer: D — Tresca and von Mises predict no yield at any p, whereas the maximum principal stress theory would predict failure at p = σY

    In a hydrostatic state all principal-stress differences vanish, so the Tresca and von Mises equivalent stresses are zero; no plane carries any shear stress. The maximum principal stress theory looks only at the magnitude of the largest stress, so it would wrongly predict failure at p = σY, one of the reasons it is not used for ductile metals. The claim of shear p/2 on every plane is false: shear is zero on every plane.

  16. A point on a machine part is in plane stress with σx = 150 MPa, σy = 30 MPa and τxy = 50 MPa (tension positive). The material is ductile with yield strength 400 MPa. The factor of safety against yielding by the von Mises theory, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 2.46

    σvm = √(σx² − σxσy + σy² + 3τxy²) = √(22 500 − 4 500 + 900 + 7 500) = √26 400 = 162.5 MPa. Factor of safety = 400/162.5 = 2.46. Forgetting the factor 3 on τxy² (using τxy² alone) gives 2.73, the usual slip.

  17. Which of the following statements about the plane-stress yield loci of a ductile metal with uniaxial yield strength σY are correct?

    1. In pure shear, the Tresca theory predicts a larger yield shear stress than the von Mises theory.
    2. The Tresca hexagon meets the von Mises ellipse at the points (σY, 0) and (σY, σY) of the σ1–σ2 plane.
    3. The square of the maximum principal stress theory lies wholly inside the Tresca hexagon.
    4. For every plane stress state the von Mises equivalent stress is less than or equal to the Tresca equivalent stress.
    Show answer

    Answer: B — The Tresca hexagon meets the von Mises ellipse at the points (σY, 0) and (σY, σY) of the σ1–σ2 plane.; D — For every plane stress state the von Mises equivalent stress is less than or equal to the Tresca equivalent stress.

    At (σY, 0) both theories reduce to σ1 = σY, and at (σY, σY) both give σY, so the hexagon vertices lie on the ellipse; elsewhere the hexagon is inside, which is the same as σvm ≤ σTresca. In pure shear Tresca gives 0.5σY, less than von Mises’ 0.577σY, so the statement about pure shear is false. The square circumscribes the hexagon (it is outside it in the second and fourth quadrants), so it is not inside.

  18. Which of the following statements about the elastic constants of an isotropic, linear elastic solid are correct?

    1. Positive shear modulus and positive bulk modulus together require −1 < ν < 0.5.
    2. A material with ν = 0.5 has a bulk modulus K = 0 and changes volume freely under pressure.
    3. Only the mean stress (σx + σy + σz)/3 changes the volume of an isotropic element; shear stresses change its shape only.
    4. E, G and ν are three independent constants of an isotropic material.
    Show answer

    Answer: A — Positive shear modulus and positive bulk modulus together require −1 < ν < 0.5.; C — Only the mean stress (σx + σy + σz)/3 changes the volume of an isotropic element; shear stresses change its shape only.

    From G = E/[2(1 + ν)] > 0 and K = E/[3(1 − 2ν)] > 0 the limits −1 < ν < 0.5 follow, and ε_v = (σx + σy + σz)(1 − 2ν)/E contains only the sum of the normal stresses. At ν = 0.5 the bulk modulus becomes infinite (incompressible), not zero. Only two of E, G, ν (or K) are independent because G = E/[2(1 + ν)].