Principal stresses and theories of failure

Principal stresses and the classical failure theories (Rankine, St Venant, Tresca, Haigh, von Mises), when each applies, and equivalent torque and moment for shafts, with a four-theory check and a shaft-sizing example.

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Why it matters

A tensile test gives one number, the yield or ultimate strength under uniaxial stress. A motor shaft, a gearbox output flange or a robot wrist sees bending and torsion together, so its stress state is biaxial. Theories of failure turn that combined state into one equivalent stress you can compare with the tensile-test strength, which is how every combined-loading part is sized and how FEA results ("von Mises stress") are judged.

Key ideas

Principal stresses. At any point there are three mutually perpendicular planes with zero shear stress; the normal stresses on them are the principal stresses σ₁ ≥ σ₂ ≥ σ₃. In plane stress one of them is zero. Every failure theory is written in terms of principal stresses, so the first step is always to find them (formula or Mohr's circle).

Failure means different things. For a ductile material (steel, aluminium) failure means the onset of yielding, so the limit is the yield strength S_y. For a brittle material (cast iron, ceramics) failure means fracture, so the limit is the ultimate strength S_ut in tension (and S_uc in compression, often much larger).

The theories (static loading, isotropic material).

  • Maximum principal stress theory (Rankine): failure when the largest principal stress reaches the uniaxial strength. Good for brittle materials, especially in tension. It ignores the other principal stresses, so it is unsafe for ductile materials in shear.
  • Maximum principal strain theory (St Venant): failure when the largest principal strain reaches the uniaxial strain at failure, giving σ₁ − ν·(σ₂ + σ₃) = S. Rarely used now; it does not agree well with tests.
  • Maximum shear stress theory (Tresca, Guest): yielding when the maximum shear stress reaches the value at yield in a tensile test, S_y/2. In terms of principal stresses, σ_max − σ_min = S_y, using σ₃ = 0 in plane stress. Simple and conservative for ductile materials.
  • Maximum strain energy theory (Haigh): failure when the total strain energy per unit volume equals that at yield in tension. It wrongly predicts yielding under high hydrostatic pressure, so it is not used for design.
  • Maximum distortion energy theory (von Mises, Hencky): yielding when the distortion (shape-change) energy per unit volume equals that at yield in tension. It matches tests on ductile metals best and is the basis of the "von Mises stress" in FEA. It is equivalent to the octahedral shear stress theory.

Comparing them. In the σ₁-σ₂ plane (plane stress), Tresca's boundary is a hexagon inscribed in the von Mises ellipse; they agree for uniaxial and equal biaxial stress and differ most in pure shear. For pure shear τ: Rankine predicts yield at τ = S_y, St Venant at τ = S_y/(1 + ν), Tresca at τ = 0.5·S_y, von Mises at τ = 0.577·S_y. Tests on ductile steels give about 0.55 to 0.6, closest to von Mises.

Hydrostatic stress. Equal stress in all three directions changes volume but not shape, so Tresca and von Mises predict no yielding however large it is. That is correct for ductile metals.

Shafts under bending and torsion. For a solid circular shaft with bending moment M and torque T at a section, the surface stresses are σ = 32M/(πd³) and τ = 16T/(πd³). Tresca leads to the equivalent torque T_e = √(M² + T²); von Mises leads to the equivalent moment M_e = √(M² + 0.75·T²). (The older Rankine equivalent moment ½[M + √(M² + T²)] is also met in textbooks.) Fatigue, keyways and stress concentrations need further factors from your design data book.

Formulas

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

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

σ₁ = S_ut / N (Rankine)

  • S_ut: ultimate tensile strength (Pa); N: factor of safety. Use S_y for a ductile part if Rankine is required.

σ₁ − ν·(σ₂ + σ₃) = S_y / N (St Venant)

  • ν: Poisson's ratio.

σ_max − σ_min = S_y / N (Tresca)

  • Use the largest difference among σ₁, σ₂ and σ₃ = 0. Equivalent to τ_max = S_y/(2N).

σ_e = √(σ₁² + σ₂² − σ₁·σ₂) = S_y / N (von Mises, plane stress) σ_e = √{½·[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²]} (general) σ_e = √(σ² + 3τ²) (one normal stress plus shear, as in a shaft)

  • σ_e: von Mises equivalent stress (Pa).

σ_Tresca = √(σ² + 4τ²)

  • Tresca equivalent stress for one normal stress σ plus shear τ.

T_e = √(M² + T²), M_e = √(M² + 0.75·T²)

  • Equivalent torque (Tresca, used with τ = 16T_e/(πd³)) and equivalent moment (von Mises, used with σ = 32M_e/(πd³)); M, T in N·m; d: shaft diameter (m).

Worked examples

Example 1 (standard): checking a point by four theories. Given: at the surface of a motor shaft, bending stress σ = 80 MPa and torsional shear τ = 45 MPa. Steel with S_y = 250 MPa, ν = 0.3. Find the factor of safety by each theory.

  1. Principal stresses: centre 40 MPa, radius √(40² + 45²) = 60.21 MPa. σ₁ = 100.21 MPa, σ₂ = −20.21 MPa, σ₃ = 0.
  2. Rankine: N = 250/100.21 = 2.49.
  3. St Venant: σ₁ − ν·σ₂ = 100.21 + 0.3 × 20.21 = 106.27 MPa, so N = 2.35.
  4. Tresca: σ₁ − σ₂ = 120.42 MPa (the largest difference, since σ₂ < 0 < σ₁), so N = 250/120.42 = 2.08. Same as √(σ² + 4τ²) = √(6400 + 8100) = 120.4.
  5. von Mises: σ_e = √(80² + 3 × 45²) = √12 475 = 111.69 MPa, so N = 250/111.69 = 2.24. (Check: √(100.21² + 20.21² + 100.21 × 20.21) gives the same.) Answer: N = 2.49 (Rankine), 2.35 (St Venant), 2.08 (Tresca), 2.24 (von Mises). Tresca is the most conservative; Rankine overestimates safety for this ductile shaft.

Example 2 (GATE level): sizing a gearbox shaft. Given: a solid steel shaft carries M = 1.2 kN·m and T = 1.8 kN·m at the critical section; S_y = 300 MPa, factor of safety 2. Find the minimum diameter by Tresca and by von Mises.

  1. Tresca allowable shear: S_y/(2N) = 300/4 = 75 MPa.
  2. T_e = √(M² + T²) = √(1.2² + 1.8²) = 2.163 kN·m.
  3. 16·T_e/(π·d³) = 75 gives d³ = 16 × 2.163 × 10⁶/(π × 75) = 146 900 mm³, so d = 52.8 mm.
  4. von Mises allowable: S_y/N = 150 MPa.
  5. M_e = √(M² + 0.75·T²) = √(1.44 + 2.43) = 1.967 kN·m.
  6. 32·M_e/(π·d³) = 150 gives d³ = 32 × 1.967 × 10⁶/(π × 150) = 133 600 mm³, so d = 51.1 mm. Answer: d ≈ 52.8 mm (Tresca), d ≈ 51.1 mm (von Mises). You would choose the next standard size, for example 55 mm, after adding allowances for keyways and fatigue from your design data book.

Common mistakes

  • Forgetting σ₃ = 0 in plane stress. With σ₁ and σ₂ both positive, Tresca uses σ₁ − 0, not σ₁ − σ₂.
  • Using Rankine for ductile materials; it is unsafe in shear-dominated loading.
  • Using S_ut for a ductile part's yield check, or S_y for a brittle part.
  • Plugging x-y stresses into σ_e = √(σ₁² + σ₂² − σ₁σ₂) instead of principal stresses (the x-y form needs + 3τ_xy²).
  • Mixing up 16 (torsion) and 32 (bending) in shaft formulas.
  • Thinking the theories cover fatigue; they are for static loading.

For GATE ME

Expect factor of safety or diameter by a named theory, comparison of theories in pure shear (ratios 0.5 and 0.577), shapes of failure envelopes in the σ₁-σ₂ plane, and equivalent torque or moment for shafts. Practise finding principal stresses quickly and remembering when σ₃ = 0 controls Tresca.

Quick check

  1. According to Tresca, a material with S_y = 240 MPa yields in pure shear at what τ?
  2. And according to von Mises?
  3. σ₁ = 120 MPa, σ₂ = 40 MPa, σ₃ = 0. What is the von Mises stress?
  4. Which theory is preferred for cast iron in tension?
  5. Why does hydrostatic pressure not cause yielding by von Mises?

Answers: 1. 120 MPa. 2. 240/√3 = 138.6 MPa. 3. √(14 400 + 1600 − 4800) = 105.8 MPa. 4. Maximum principal stress (Rankine). 5. It causes no distortion (shape change), only volume change.

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 occur at specific orientations and are the maximum and minimum normal stresses that a material element can experience.

  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 typically used for brittle materials where failure is more likely to occur due to normal stresses rather than shear stresses.

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

    Tresca's theory says a ductile material yields when the maximum shear stress at a point reaches the maximum shear stress at yield in a simple tension test, which is S_y/2. In principal stresses that is σ_max − σ_min = S_y, remembering σ₃ = 0 in plane stress. It is simple and slightly conservative compared with von Mises (by up to about 15 %, in pure shear), so it is widely used for ductile shaft and machine-part design, giving the equivalent torque √(M² + T²).

  4. 4.Why is the von Mises Stress Theory preferred for ductile materials?Application

    Yielding in ductile metals is caused by slip, which is driven by shape change, not by volume change. The von Mises theory uses the distortion energy, which depends on all three principal stress differences and ignores the hydrostatic part, so it predicts no yielding under pure hydrostatic pressure and gives τ_yield = 0.577·S_y in pure shear. Both agree well with tests on ductile metals, better than Tresca, whose hexagon lies inside the von Mises ellipse.

  5. 5.What happens if the stresses at a point exceed what the material can take?Application

    For a ductile material, yielding starts when an equivalent stress, the von Mises or Tresca stress built from all the principal stresses, reaches the yield strength; an individual principal stress exceeding S_y is not by itself the criterion. After yielding the part takes a permanent set, which may misalign a mechanism even if it does not break. For a brittle material, the limit is fracture, usually governed by the largest principal tensile stress reaching the ultimate strength.

  6. 6.A material has principal stresses of 100 MPa and 50 MPa. Calculate the maximum shear stress.Numerical

    The maximum in-plane shear stress is (σ₁ − σ₂)/2 = (100 − 50)/2 = 25 MPa. If this is a plane-stress state, the third principal stress is σ₃ = 0, so the absolute maximum shear stress is (σ₁ − σ₃)/2 = 100/2 = 50 MPa, acting on planes inclined out of the plane of loading. The Tresca criterion must use the 50 MPa value.

  7. 7.Given a state of stress with σx = 80 MPa, σy = 40 MPa, and τxy = 30 MPa, find the principal stresses.Numerical

    The principal stresses can be found using the formulas: σ1,2 = [(σx + σy) / 2] ± √[((σx - σy) / 2)² + τxy²]. Substituting the given values: σ1,2 = [(80 + 40) / 2] ± √[((80 - 40) / 2)² + 30²] = 60 ± √[400 + 900] = 60 ± √1300. Therefore, σ1 = 96.06 MPa and σ2 = 23.94 MPa.

  8. 8.Explain why brittle materials are more likely to fail under tensile stresses rather than shear stresses.Application

    Brittle materials such as cast iron contain many small flaws, and tensile stress opens them into cracks that grow without much plastic deformation, so fracture is governed by the maximum principal tensile stress. They are typically much stronger in compression and in shear than in tension. That is why a cast-iron bar in torsion breaks on a 45° helix, the plane of maximum tensile stress, rather than on the plane of maximum shear as a ductile steel bar does.

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