Why do we need theories of failure?

Strength data come from a uniaxial tensile test, but real parts carry combined stresses. A failure theory picks one quantity (maximum normal stress, maximum shear stress, strain, total or distortion energy) and assumes failure when it reaches the value it has at failure in the tensile test. That turns any multiaxial stress state into an equivalent stress that can be compared directly with σy or σut.

State the maximum shear stress (Tresca) theory and where it is used.

Yielding starts when the absolute maximum shear stress, half the largest difference between principal stresses, reaches σy/2, the value at yield in a tensile test. It suits ductile metals, is slightly conservative compared with von Mises, and gives simple design formulas such as the equivalent torque Te = √(M² + T²). It predicts a shear yield strength of 0.5σy.

What is the von Mises (distortion energy) criterion and why is it the most widely used for metals?

It assumes yielding is caused only by the strain energy of distortion (shape change), not by the volume-change part. The equivalent stress is σvm = √{½[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²]} and yield occurs when σvm = σy. It matches experiments on ductile metals best, correctly predicts no yield under hydrostatic stress, and gives τy = σy/√3 ≈ 0.577σy; FE codes report it by default.

Which theory of failure would you use for a cast iron component, and why?

The maximum principal stress (Rankine) theory, or a modified version such as Coulomb–Mohr. Brittle materials have little plastic capacity and fail by fracture on planes normal to the largest tensile stress, and their compressive strength is much higher than their tensile strength. Shear-based criteria like Tresca or von Mises describe yielding of ductile metals and do not capture that behaviour.

Compare the shear yield strength predicted by Tresca and von Mises.

In pure shear σ₁ = τ and σ₂ = −τ. Tresca gives σ₁ − σ₂ = 2τ = σy, so τy = 0.5σy. Von Mises gives √(τ² + τ² + τ²) = √3·τ = σy, so τy = 0.577σy. This is where the two theories differ most, by about 15 %; tests on ductile metals usually fall close to von Mises.

Why does a large hydrostatic pressure not cause yielding in a ductile metal?

Under equal triaxial stress all principal stresses are equal, so every shear stress is zero and there is no distortion, only a volume change. Tresca and von Mises therefore predict no yield, which agrees with experiments on metals under very high pressure. The total strain energy theory wrongly predicts yield in this case, which is one reason it was abandoned.

A shaft carries M = 3 kN·m and T = 4 kN·m. What is its equivalent torque by Tresca and its von Mises equivalent moment?

Tresca: Te = √(M² + T²) = √(9 + 16) = 5 kN·m, used with τ = 16Te/(πd³). Von Mises: Me = √(M² + 0.75T²) = √(9 + 12) = 4.58 kN·m, used with σ = 32Me/(πd³). For σy = 300 MPa and FoS 2 these give minimum diameters of about 69.8 mm and 67.8 mm.

Theories of failure

The five classical theories of failure, their equivalent stresses and yield loci, which suits ductile or brittle materials, and shaft design under combined bending and torsion.

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Why it matters

Material data come from a uniaxial tensile test, but real parts such as a rotor shaft under bending plus torque, a pressurised skin, or a lug with pin bearing see combined stresses. A theory of failure converts a multiaxial stress state into one equivalent number that can be compared with the tensile yield or ultimate strength. Choosing the right theory decides whether a design is safe, over-conservative or unsafe.

Key ideas

The idea. Each theory names one quantity (stress, strain or energy) that is assumed to cause failure, and says that failure under any stress state occurs when that quantity reaches the value it has at failure in the simple tensile test. All are written in terms of principal stresses σ₁ ≥ σ₂ ≥ σ₃. For plane stress one principal stress is zero; remember to include it.

1. Maximum principal stress theory (Rankine). Failure when the largest principal stress reaches the tensile strength (or the most compressive reaches the compressive strength). It ignores the other stresses. It suits brittle materials (cast iron, ceramics, some composites and glass), which fail by separation normal to the largest tension. It is unsafe for ductile metals in shear-dominated states.

2. Maximum shear stress theory (Tresca, Guest). Failure when the absolute maximum shear stress, (σmax − σmin)/2 over all three principal stresses, reaches the value at yield in tension, σy/2. It fits ductile metals well, is slightly conservative, and gives simple hand formulas. It predicts the shear yield strength τy = 0.5·σy.

3. Maximum principal strain theory (Saint-Venant). Failure when the largest principal strain reaches the yield strain σy/E. It gives σ₁ − ν(σ₂ + σ₃) = σy. Experiments do not support it well; it is mainly of historical and exam interest.

4. Total strain energy theory (Haigh, Beltrami). Failure when the strain energy per unit volume reaches its value at tensile yield, σy²/(2E). It predicts that hydrostatic pressure alone can cause yield, which contradicts experiment.

5. Distortion energy theory (von Mises, Hencky; also called maximum shear strain energy or octahedral shear stress theory). Only the energy of shape change causes yielding; the volume-change part does not. The equivalent (von Mises) stress is compared with σy. It agrees best with tests on ductile metals and is the default in design codes and finite-element post-processing. It predicts τy = σy/√3 ≈ 0.577·σy.

Comparing them. In the σ₁–σ₂ plane (plane stress) the yield loci are: Rankine a square; Tresca a hexagon; von Mises an ellipse that passes through the hexagon's corners; the strain theories are rhombus- and ellipse-shaped curves that depend on ν. All agree for uniaxial stress. The biggest difference is in pure shear (σ₁ = −σ₂), where Tresca gives τy = 0.5σy and von Mises 0.577σy. Tresca is never less conservative than von Mises.

Hydrostatic stress. Equal triaxial tension or compression has no shear and no distortion energy, so Tresca and von Mises predict no yielding however large it is, which matches experiments on ductile metals.

Factor of safety. FoS = strength/equivalent stress; or design with allowable stress σy/FoS.

Formulas

σ₁ = σut (Rankine) σ₁ = largest principal stress (Pa), σut = ultimate tensile strength (Pa); use σy for yield-based design.

max(|σ₁ − σ₂|, |σ₂ − σ₃|, |σ₃ − σ₁|) = σy (Tresca) Plane stress with σ₃ = 0: if σ₁, σ₂ have opposite signs, σ₁ − σ₂ = σy; if the same sign, max(|σ₁|, |σ₂|) = σy.

σ₁ − ν·(σ₂ + σ₃) = σy (Saint-Venant) ν = Poisson's ratio.

σ₁² + σ₂² + σ₃² − 2ν·(σ₁σ₂ + σ₂σ₃ + σ₃σ₁) = σy² (Haigh)

σvm = √{½·[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²]} (von Mises) Plane stress: σvm = √(σ₁² + σ₂² − σ₁·σ₂) or σvm = √(σx² + σy² − σx·σy + 3·τxy²).

σeq = √(σ² + 4·τ²) (Tresca), σeq = √(σ² + 3·τ²) (von Mises) For a normal stress σ combined with a shear stress τ (shafts, beams).

Te = √(M² + T²), Me = ½·[M + √(M² + T²)], Me,vm = √(M² + 0.75·T²) Equivalent torque (Tresca, use with τ = 16Te/(πd³)), equivalent moment (Rankine) and von Mises equivalent moment (use with σ = 32Me/(πd³)) for a solid shaft under M and T (N·m).

Worked examples

Example 1 (standard). At a critical point in a steel part, σ₁ = 120 MPa, σ₂ = −60 MPa, σ₃ = 0. σy = 250 MPa, ν = 0.3. Find the factor of safety against yield by each theory.

Rankine: FoS = 250/120 = 2.08.
Tresca: σ₁ and σ₂ have opposite signs, so the controlling difference is σ₁ − σ₂ = 180 MPa; FoS = 250/180 = 1.39.
Saint-Venant: σ₁ − νσ₂ = 120 + 18 = 138 MPa; FoS = 250/138 = 1.81.
Haigh: √(120² + 60² − 2 × 0.3 × 120 × (−60)) = √(14 400 + 3600 + 4320) = √22 320 = 149.4 MPa; FoS = 1.67.
von Mises: √(120² + 60² − 120 × (−60)) = √(14 400 + 3600 + 7200) = √25 200 = 158.7 MPa; FoS = 1.57.

Answer: FoS = 2.08 (Rankine), 1.39 (Tresca), 1.81 (Saint-Venant), 1.67 (Haigh), 1.57 (von Mises). For this ductile steel, design on von Mises (or Tresca for extra conservatism); Rankine would overestimate safety.

Example 2 (GATE level). A solid shaft carries a bending moment M = 3 kN·m and a torque T = 4 kN·m. The material has σy = 300 MPa and a factor of safety of 2 is required. Find the minimum diameter by (a) Tresca and (b) von Mises.

Allowable stresses: σ = 300/2 = 150 MPa; for Tresca the allowable shear is σy/(2·FoS) = 75 MPa.
(a) Te = √(M² + T²) = √(3² + 4²) = 5 kN·m = 5 × 10⁶ N·mm.
16·Te/(π·d³) = 75 gives d³ = 16 × 5 × 10⁶/(π × 75) = 3.395 × 10⁵ mm³, so d = 69.8 mm.
(b) Me,vm = √(M² + 0.75·T²) = √(9 + 12) = 4.583 kN·m.
32·Me,vm/(π·d³) = 150 gives d³ = 32 × 4.583 × 10⁶/(π × 150) = 3.112 × 10⁵ mm³, so d = 67.8 mm.

Answer: d ≈ 69.8 mm (Tresca), d ≈ 67.8 mm (von Mises). Tresca is about 3 % more conservative here; choose the next standard size, for example 70 mm.

Common mistakes

Forgetting σ₃ = 0 in plane stress, which makes Tresca unconservative when σ₁ and σ₂ are both tensile.
Using Rankine for ductile metals, or Tresca/von Mises for brittle ones.
Comparing the Tresca shear stress with σy instead of σy/2.
Mixing σx, σy, τxy into the principal-stress form of von Mises without converting.
Applying a factor of safety twice (once to the strength and again to the result).

For GATE AE

Expect factor-of-safety comparisons between theories for a given stress state, shear yield strength predicted by Tresca and von Mises, shaft diameter under combined bending and torsion, and conceptual questions on which theory suits ductile or brittle materials and why hydrostatic stress does not cause yield. Practise sketching the yield loci and spotting which theory governs at a given point.

Quick check

Which theory is best for cast iron?
What shear yield strength does von Mises predict for σy = 300 MPa?
For σ₁ = 100 MPa and σ₂ = 50 MPa (plane stress), what is the Tresca equivalent stress?
Does a large hydrostatic pressure cause yield according to von Mises?

Answers: 1. Maximum principal stress (Rankine). 2. 173 MPa. 3. 100 MPa (since σ₃ = 0). 4. No.

Interview questions

All Mechanics of Solids interview questions

Try answering each one aloud before you open it.

1.Why do we need theories of failure?Concept
Strength data come from a uniaxial tensile test, but real parts carry combined stresses. A failure theory picks one quantity (maximum normal stress, maximum shear stress, strain, total or distortion energy) and assumes failure when it reaches the value it has at failure in the tensile test. That turns any multiaxial stress state into an equivalent stress that can be compared directly with σy or σut.
2.State the maximum shear stress (Tresca) theory and where it is used.Concept
Yielding starts when the absolute maximum shear stress, half the largest difference between principal stresses, reaches σy/2, the value at yield in a tensile test. It suits ductile metals, is slightly conservative compared with von Mises, and gives simple design formulas such as the equivalent torque Te = √(M² + T²). It predicts a shear yield strength of 0.5σy.
3.What is the von Mises (distortion energy) criterion and why is it the most widely used for metals?Concept
It assumes yielding is caused only by the strain energy of distortion (shape change), not by the volume-change part. The equivalent stress is σvm = √{½[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²]} and yield occurs when σvm = σy. It matches experiments on ductile metals best, correctly predicts no yield under hydrostatic stress, and gives τy = σy/√3 ≈ 0.577σy; FE codes report it by default.
4.Which theory of failure would you use for a cast iron component, and why?Concept
The maximum principal stress (Rankine) theory, or a modified version such as Coulomb–Mohr. Brittle materials have little plastic capacity and fail by fracture on planes normal to the largest tensile stress, and their compressive strength is much higher than their tensile strength. Shear-based criteria like Tresca or von Mises describe yielding of ductile metals and do not capture that behaviour.
5.Compare the shear yield strength predicted by Tresca and von Mises.Concept
In pure shear σ₁ = τ and σ₂ = −τ. Tresca gives σ₁ − σ₂ = 2τ = σy, so τy = 0.5σy. Von Mises gives √(τ² + τ² + τ²) = √3·τ = σy, so τy = 0.577σy. This is where the two theories differ most, by about 15 %; tests on ductile metals usually fall close to von Mises.
6.Why does a large hydrostatic pressure not cause yielding in a ductile metal?Concept
Under equal triaxial stress all principal stresses are equal, so every shear stress is zero and there is no distortion, only a volume change. Tresca and von Mises therefore predict no yield, which agrees with experiments on metals under very high pressure. The total strain energy theory wrongly predicts yield in this case, which is one reason it was abandoned.
7.A shaft carries M = 3 kN·m and T = 4 kN·m. What is its equivalent torque by Tresca and its von Mises equivalent moment?Concept
Tresca: Te = √(M² + T²) = √(9 + 16) = 5 kN·m, used with τ = 16Te/(πd³). Von Mises: Me = √(M² + 0.75T²) = √(9 + 12) = 4.58 kN·m, used with σ = 32Me/(πd³). For σy = 300 MPa and FoS 2 these give minimum diameters of about 69.8 mm and 67.8 mm.

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