Principal stresses and theories of failure
Principal stresses and the main failure theories (Rankine, Tresca, von Mises), when each applies, factors of safety, and shaft design under combined bending and torsion.
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Why it matters
Material strengths come from a simple tension test, but real parts (shafts, crank pins, pressure vessels, brackets) carry combined stresses. A theory of failure is the rule that compares a combined stress state with the tension-test value. Choosing the right theory decides the shaft diameter or plate thickness, and the wrong one can be either unsafe or wasteful.
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. They are the largest and smallest normal stresses at the point. In plane stress one of them is zero (the stress normal to the free surface), and the other two follow from the transformation formulas or Mohr's circle (previous topic). Failure theories are written in terms of principal stresses because they do not depend on the chosen axes.
Ductile vs brittle failure. Ductile metals (mild steel, aluminium alloys) fail by yielding, which is driven by shear and distortion; the limiting value is the yield strength Syt. Brittle materials (cast iron, ceramics) fracture on the plane of largest tension; the limiting value is the ultimate tensile strength Sut, and the compressive strength is usually much higher.
1. Maximum principal stress theory (Rankine). Failure when the largest principal stress reaches the tensile strength (or the most negative one reaches the compressive strength). Good for brittle materials; unsafe for ductile materials in shear (it predicts pure-shear failure at τ = Syt, but ductile metals yield near 0.5–0.58 Syt).
2. Maximum shear stress theory (Tresca, Guest). Yielding starts when the maximum shear stress reaches the value at yield in a tension test, Syt/2. Remember to use the absolute maximum shear including σ3 = 0 in plane stress. It is slightly conservative and simple; its locus in the σ1–σ2 plane is a hexagon. Shear yield strength predicted: Ssy = 0.5 Syt.
3. Distortion energy theory (von Mises, Hencky). Yielding starts when the strain energy of distortion (shape change) per unit volume reaches its value at yield in a tension test. Equivalent to octahedral shear stress theory. Best agreement with tests on ductile metals; locus is an ellipse that encloses the Tresca hexagon. Predicts Ssy = Syt/√3 = 0.577 Syt.
4. Maximum principal strain theory (Saint-Venant) and 5. total strain energy theory (Haigh) are of historical interest; they depend on ν and do not match tests as well.
Comparison. For the same stress state, Tresca gives a larger equivalent stress than von Mises (by at most about 15.5%, in pure shear), so Tresca is the more conservative of the two. For equal biaxial tension the theories agree. Rankine and Tresca agree when σ1 and σ2 have the same sign.
Factor of safety. N = strength / equivalent stress. Take strengths and factors of safety from your data book or code.
Shafts under bending and torsion. On the surface, σ = 32M/(πd³) and τ = 16T/(πd³). These give the equivalent torque and equivalent bending moment formulas below.
Formulas
σ1,2 = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²] (plane stress, σ3 = 0)
Rankine: σ1 = Sut / N (and |σ3| = Suc / N in compression)
Tresca: max(|σ1 − σ2|, |σ2 − σ3|, |σ3 − σ1|) = Syt / N
- Plane stress with σ1, σ2 of opposite sign:
σ1 − σ2 = Syt/N; same sign:max(|σ1|, |σ2|) = Syt/N.
von Mises: σv = √{[(σ1 − σ2)² + (σ2 − σ3)² + (σ3 − σ1)²] / 2} = Syt / N
- Plane stress:
σv = √(σ1² − σ1σ2 + σ2²); for σx, σy, τxy:σv = √(σx² − σxσy + σy² + 3τxy²). - Bending plus shear (σ, τ): Tresca
√(σ² + 4τ²) = Syt/N, von Mises√(σ² + 3τ²) = Syt/N.
Shaft: Te = √(M² + T²), τmax = 16 Te/(π d³) (Tresca); Me = √(M² + 0.75 T²), σv = 32 Me/(π d³) (von Mises)
- M = bending moment, T = torque (N·mm), d = diameter (mm), stresses in MPa.
Worked examples
Example 1 (standard). At the surface of a shaft σ = 80 MPa (bending) and τ = 40 MPa (torsion). The material is ductile with Syt = 250 MPa. Find the principal stresses and the factor of safety by Rankine, Tresca and von Mises.
σ1,2 = 40 ± √(40² + 40²) = 40 ± 56.57:σ1 = 96.57 MPa,σ2 = −16.57 MPa, σ3 = 0.- Rankine:
N = 250 / 96.57 = 2.59. - Tresca: σ1 and σ2 have opposite signs, so
τmax = (96.57 + 16.57)/2 = 56.57 MPa;N = (250/2) / 56.57 = 2.21. - von Mises:
σv = √(80² + 3 × 40²) = √11 200 = 105.83 MPa(check:√(96.57² + 96.57 × 16.57 + 16.57²) = 105.83✓);N = 250 / 105.83 = 2.36.
Answer: σ1 = 96.6 MPa, σ2 = −16.6 MPa; N = 2.59 (Rankine, not appropriate for a ductile material), 2.21 (Tresca), 2.36 (von Mises).
Example 2 (GATE level). A solid shaft carries M = 2 kN·m and T = 3 kN·m. Syt = 300 MPa, factor of safety 2. Find the diameter by Tresca and by von Mises.
- Tresca: allowable shear
= Syt/(2N) = 300/4 = 75 MPa. Te = √(2² + 3²) = 3.606 kN·m = 3.606 × 10⁶ N·mm.d³ = 16 Te / (π × 75) = 16 × 3.606 × 10⁶ / 235.6 = 2.448 × 10⁵ mm³, sod = 62.6 mm.- von Mises: allowable
= 300/2 = 150 MPa;Me = √(2² + 0.75 × 3²) = √10.75 = 3.279 kN·m. d³ = 32 Me / (π × 150) = 32 × 3.279 × 10⁶ / 471.2 = 2.226 × 10⁵ mm³, sod = 60.6 mm.
Answer: d ≈ 62.6 mm (Tresca), d ≈ 60.6 mm (von Mises); Tresca is more conservative. Choose the next standard size from the data book.
Common mistakes
- Using the in-plane shear R in Tresca when σ1 and σ2 have the same sign; the absolute maximum includes σ3 = 0.
- Applying Rankine to a ductile shaft in torsion (unsafe).
- Comparing Tresca shear stress with Syt instead of Syt/2.
- Forgetting the 0.75 in the von Mises equivalent moment, or the 3 in
√(σ² + 3τ²). - Writing that von Mises is more conservative than Tresca; it is the other way round.
- Mixing N·m and N·mm in shaft formulas.
For GATE PI
Expect factor-of-safety calculations by two or three theories for a given stress state, shaft diameter under combined bending and torsion, ratio of shear yield to tensile yield predicted by each theory, and identification of loci (hexagon, ellipse, square). Practise doing the plane-stress formulas quickly and checking the signs of σ1 and σ2 first.
Quick check
- According to von Mises, what is the shear yield strength of a material with Syt = 300 MPa?
- Which theory suits cast iron?
- In pure shear τ, what are σv and the Tresca equivalent stress?
- σ1 = 120 MPa, σ2 = 40 MPa (plane stress). What is σv?
Answers: 1. 173.2 MPa (300/√3). 2. Maximum principal stress (Rankine). 3. σv = √3 τ = 1.732 τ; Tresca 2τ. 4. √(14 400 − 4800 + 1600) = 105.8 MPa.
Interview questions
All Engineering Mechanics and Strength of Materials interview questionsTry answering each one aloud before you open it.
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 can experience at a point.
2.What is the significance of the maximum shear stress theory in failure analysis?Concept
The maximum shear stress theory, also known as Tresca's criterion, is used to predict the failure of ductile materials. It states that yielding begins when the maximum shear stress in a material reaches the shear stress at yielding in a simple tension test. This theory is significant because it helps engineers design components that can withstand applied loads without failing.
3.Why is the von Mises stress used in failure theories for ductile materials?Application
Ductile metals yield by shape change (slip), not by volume change, so the von Mises criterion compares the distortion energy per unit volume with its value at yield in a tension test. This reduces a three-dimensional stress state to one equivalent stress, σv = √{[(σ1 − σ2)² + (σ2 − σ3)² + (σ3 − σ1)²]/2}, that can be compared directly with the yield strength. Hydrostatic stress does not change σv, which agrees with experiments on metals.
4.What does the maximum principal stress theory predict, and when should it be used?Application
The Rankine theory predicts failure when the largest principal stress reaches the strength found in a simple tension test (or the most compressive principal stress reaches the compressive strength). It fits brittle materials such as cast iron, which fracture on the plane of maximum tension. For ductile materials it is unsafe in shear-dominated loading, because it predicts failure in pure shear at τ = Syt, whereas ductile metals actually yield at about 0.5–0.58 Syt.
5.How do you determine the principal stresses from a given stress state using analytical methods?Numerical
To determine the principal stresses from a given stress state, you solve the characteristic equation derived from the stress tensor. The equation is: σ^2 - (σ_x + σ_y)σ + (σ_xσ_y - τ_xy^2) = 0, where σ_x and σ_y are normal stresses and τ_xy is the shear stress. Solving this quadratic equation gives the principal stresses.
6.Explain the difference between the maximum principal stress theory and the maximum shear stress theory.Concept
The maximum principal stress theory, also known as Rankine's criterion, predicts failure when the maximum principal stress reaches the material's tensile strength. In contrast, the maximum shear stress theory, or Tresca's criterion, predicts failure when the maximum shear stress reaches the shear yield strength. The former is more applicable to brittle materials, while the latter is used for ductile materials.
7.Why might engineers choose the von Mises criterion over the Tresca criterion for certain applications?Application
For ductile metals, von Mises (distortion energy) matches test data more closely, predicting shear yield at about 0.577 Syt, while Tresca predicts 0.5 Syt. Tresca is therefore the more conservative of the two, by up to about 15% in pure shear, and von Mises allows a slightly lighter design with the same reliability. von Mises is also a smooth function of all stress components, which suits finite-element post-processing; Tresca is kept where simplicity or extra conservatism is wanted.
8.What is the role of safety factors in the context of principal stresses and failure theories?Application
A failure theory turns the principal stresses into one equivalent stress; the factor of safety is the material strength divided by that equivalent stress (or the allowable stress is the strength divided by N). It covers uncertainty in loads, material properties, stress concentrations, manufacturing flaws and the approximations of the theory itself. Values come from design codes or data books and are higher for brittle materials, shock loads and where failure is dangerous.
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