Principal stresses and theories of failure

Principal stresses and the classical failure theories (Rankine, St Venant, Tresca, Haigh, von Mises, Coulomb–Mohr), when each applies, and their use for factors of safety and shaft sizing under bending and torsion.

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Why it matters

A tensile test tells you when a material yields or fractures under one stress. Real parts such as axle shafts, crankshafts and steering knuckles carry bending, torsion and pressure together. Theories of failure translate a multi-axial stress state into one equivalent stress you can compare with the tensile-test strength. Picking the right theory, Tresca or von Mises for ductile steels and maximum normal stress or Mohr–Coulomb for cast iron, is a routine design decision and a favourite GATE question.

Key ideas

Principal stresses. At every point there are three mutually perpendicular planes with no shear stress; the normal stresses on them are the principal stresses σ₁ ≥ σ₂ ≥ σ₃. They are the largest and smallest normal stresses at that point. For plane stress, two come from the Mohr's-circle formula and the third is zero. Failure theories are written in terms of principal stresses, so finding them is always step one.

Why theories are needed. Yielding of ductile metals is driven by shear (slip on crystal planes), while fracture of brittle materials is driven by the largest tensile stress at flaws. No single quantity matches every material, so several theories exist, each assuming that failure occurs when one chosen quantity reaches the value it has at failure in a simple tension test.

The five classical theories (static loading, isotropic material).

  1. Maximum principal (normal) stress theory, Rankine: failure when σ₁ reaches the tensile strength (or |σ₃| reaches the compressive strength). Suits brittle materials; badly unsafe for ductile materials in shear.
  2. Maximum principal strain theory, St Venant: failure when the largest principal strain reaches the strain at yield in tension. Rarely used now.
  3. Maximum shear stress theory, Tresca (Guest): yielding when τ_max = (σ_max − σ_min)/2 reaches S_y/2. Simple and slightly conservative for ductile metals; predicts shear yield strength S_sy = 0.5·S_y.
  4. Total strain energy theory, Haigh: failure when total strain energy per unit volume reaches its value at yield in tension. Wrongly predicts yielding under hydrostatic pressure.
  5. Maximum distortion energy theory, von Mises (Hencky): yielding when the distortion (shape-change) energy reaches its value at yield in tension. Best agreement with tests on ductile metals; predicts S_sy = 0.577·S_y.

Hydrostatic stress. Equal all-round stress changes volume, not shape. Tresca and von Mises ignore it, which matches experiments for ductile metals. Brittle materials under hydrostatic tension still fracture when the stress reaches the tensile strength.

Graphical view (plane stress, σ₃ = 0). In the σ₁–σ₂ plane, Rankine gives a square, Tresca a hexagon inscribed in the von Mises ellipse; they agree in uniaxial tension and equal biaxial tension and differ most in pure shear (by about 15 %).

Brittle materials. Use maximum normal stress or the Coulomb–Mohr theory, which uses different tensile and compressive strengths: σ₁/S_ut − σ₃/S_uc = 1/N.

Shaft design under bending and torsion. On the surface of a solid circular shaft, σ = 32·M/(π·d³) and τ = 16·T/(π·d³). Substituting into Tresca gives an equivalent torque, into von Mises an equivalent bending moment form. Fatigue, stress concentration and shock factors are added in machine design according to the code or data book used.

Formulas

σ₁,₂ = (σ_x + σ_y)/2 ± √[((σ_x − σ_y)/2)² + τ_xy²]

  • Principal stresses in plane stress (Pa); σ₃ = 0 out of plane.

Rankine: σ₁ = S_ut / N

  • S_ut: ultimate tensile strength (Pa); N: factor of safety.

Tresca: σ_max − σ_min = S_y / N

  • S_y: tensile yield strength (Pa). Include σ₃ = 0 when choosing σ_max and σ_min.

von Mises: σ_e = √{[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²] / 2} = S_y / N

  • Plane stress: σ_e = √(σ₁² − σ₁·σ₂ + σ₂²); in terms of components σ_e = √(σ_x² − σ_x·σ_y + σ_y² + 3·τ_xy²).

Axial or bending stress σ with shear τ (σ_y = 0): τ_max = √[(σ/2)² + τ²] (Tresca), σ_e = √(σ² + 3·τ²) (von Mises)

Solid shaft with bending moment M and torque T: T_e = √(M² + T²), d³ = 16·T_e / (π·τ_allow) with τ_allow = S_y / (2N) (Tresca) σ_e = (32 / (π·d³))·√(M² + 0.75·T²) (von Mises)

  • M, T in N·m; d: shaft diameter (m).

Max principal strain: σ₁ − ν·(σ₂ + σ₃) = S_y / N

Worked examples

Example 1 (standard): factor of safety by three theories. Given: a point on a shaft surface has σ = 80 MPa (bending) and τ = 50 MPa (torsion). The steel has S_y = 250 MPa. Find the factor of safety by Rankine, Tresca and von Mises.

  1. Principal stresses: σ₁,₂ = 40 ± √(40² + 50²) = 40 ± 64.03, so σ₁ = 104.03 MPa, σ₂ = −24.03 MPa, σ₃ = 0.
  2. Rankine (using S_y as the limit for comparison): N = 250 / 104.03 = 2.40.
  3. Tresca: σ_max − σ_min = 104.03 − (−24.03) = 128.06 MPa; N = 250 / 128.06 = 1.95.
  4. von Mises: σ_e = √(80² + 3 × 50²) = √13 900 = 117.9 MPa; N = 250 / 117.9 = 2.12. Answer: N = 2.40 (Rankine), 1.95 (Tresca), 2.12 (von Mises). For a ductile steel the Rankine value is not safe to rely on; Tresca is the most conservative.

Example 2 (GATE level): shaft diameter under combined loading. Given: a solid steel shaft carries M = 1.5 kN·m and T = 2.0 kN·m. S_y = 300 MPa, factor of safety 2. Find the minimum diameter by Tresca and by von Mises.

  1. Tresca: T_e = √(M² + T²) = √(1.5² + 2²) = 2.5 kN·m. τ_allow = 300 / (2 × 2) = 75 MPa.
  2. d³ = 16 × 2.5 × 10⁶ / (π × 75) = 1.698 × 10⁵ mm³, so d = 55.4 mm.
  3. von Mises: √(M² + 0.75·T²) = √(2.25 + 3.0) = 2.291 kN·m; allowable σ_e = 300 / 2 = 150 MPa.
  4. d³ = 32 × 2.291 × 10⁶ / (π × 150) = 1.556 × 10⁵ mm³, so d = 53.8 mm. Answer: d ≈ 55.4 mm (Tresca), 53.8 mm (von Mises). You would then round up to a standard size.

Common mistakes

  • Ignoring σ₃ = 0 in plane stress when both in-plane principal stresses are tensile; Tresca then uses σ₁ − 0.
  • Using the maximum normal stress theory for ductile parts in torsion; it overestimates strength.
  • Writing von Mises as √(σ² + 4τ²) (that is Tresca's form, 2·τ_max) or Tresca as √(σ² + 3τ²).
  • Comparing σ_e with ultimate strength for ductile yielding, or with yield strength for brittle fracture.
  • Forgetting that the shear yield strength is 0.5·S_y (Tresca) or 0.577·S_y (von Mises), not S_y.
  • Applying static theories to fluctuating loads, which need fatigue criteria.

For GATE ME

Expect: equivalent stress or factor of safety for given principal stresses, ratio of shear yield to tensile yield under each theory, shaft diameter under combined bending and torsion, identification of which theory suits which material, and the shapes of failure envelopes. Practise computing all three theories for the same stress state and know which is most conservative where.

Quick check

  1. By von Mises, what is the shear yield strength of a steel with S_y = 300 MPa?
  2. σ₁ = 150 MPa, σ₂ = 50 MPa, σ₃ = 0. What equivalent stress does von Mises give?
  3. Which theory suits grey cast iron?
  4. In pure shear τ, what are the principal stresses?

Answers: 1. 0.577 × 300 = 173 MPa. 2. √(22 500 − 7500 + 2500) = 132.3 MPa. 3. Maximum normal stress (or Coulomb–Mohr). 4. +τ, −τ and 0.

Interview questions

All Engineering Mechanics and Strength of Materials interview questions

Try answering each one aloud before you open it.

  1. 1.What are principal stresses?Concept

    Principal stresses are the normal stresses acting on a particular plane where the shear stress is zero. These stresses are the maximum and minimum normal stresses that occur at a point in a material. They are important in determining the failure criteria of materials.

  2. 2.What is the Maximum Normal Stress Theory of failure, and when is it used?Concept

    The Maximum Normal Stress Theory, also known as Rankine's Theory, states that failure occurs when the maximum principal stress in a material reaches the ultimate tensile stress of the material. It is used for brittle materials where failure is primarily due to normal stresses rather than shear stresses.

  3. 3.Describe the Maximum Shear Stress Theory and its application.Concept

    The Maximum Shear Stress Theory, also known as Tresca's Theory, suggests that failure occurs when the maximum shear stress in a material reaches the shear stress at yield in a simple tension test. It is commonly used for ductile materials where shear stresses are more critical in causing failure.

  4. 4.Why is the von Mises Stress used in the Distortion Energy Theory of failure?Application

    The distortion energy theory says a ductile material yields when the strain energy associated with change of shape (not change of volume) reaches the value it has at yield in a simple tension test. Working that energy out in terms of principal stresses gives the von Mises equivalent stress σ_e = √{[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²]/2}, which can be compared directly with S_y. It ignores hydrostatic stress, as experiments on metals show it should, and predicts a shear yield strength of 0.577·S_y, close to test data for steels.

  5. 5.What happens if a brittle material is subjected to a stress state where the principal stresses are equal?Application

    Equal principal stresses form a hydrostatic state with no shear stress on any plane. Brittle materials fail by tensile fracture, not by shear, so under equal hydrostatic tension a brittle material will still fracture when that stress reaches its tensile strength, as the maximum normal stress theory predicts. Under hydrostatic compression it can carry very high stress, because there is no tension to open flaws. Ductile metals do not yield under pure hydrostatic stress at all, which is why Tresca and von Mises ignore it.

  6. 6.How does the presence of a notch or crack affect the principal stresses in a material?Application

    The presence of a notch or crack in a material creates stress concentrations, which significantly increase the local principal stresses around the defect. This can lead to premature failure, especially in brittle materials, as the stress concentration factor amplifies the applied stress, potentially exceeding the material's strength.

  7. 7.A cylindrical rod is subjected to an axial tensile load of 10 kN and a torque of 200 Nm. Determine the maximum shear stress in the rod if its diameter is 50 mm.Numerical
    1. Torsional shear stress: τ = 16·T/(π·d³) = 16 × 200 / (π × 0.05³) = 8.15 MPa (J = π·d⁴/32 = 6.14 × 10⁻⁷ m⁴).
    2. Axial stress: σ = P/A = 10 000 / (π × 0.05²/4) = 5.09 MPa.
    3. Maximum shear stress: τ_max = √[(σ/2)² + τ²] = √(2.55² + 8.15²) = 8.54 MPa. So τ_max ≈ 8.5 MPa on the surface of the rod.
  8. 8.Explain why the Maximum Normal Stress Theory is not suitable for ductile materials.Application

    The Maximum Normal Stress Theory is not suitable for ductile materials because it only considers the maximum principal stress and ignores shear stresses, which are critical in ductile material failure. Ductile materials typically fail due to yielding, which is more accurately predicted by considering shear stresses, as done in the Maximum Shear Stress Theory or the Distortion Energy Theory.

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