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Machine Elements · Chapter 4 of 10 · Intermediate

Failure from Static Loading

A combined stress state is not a single number, so we need a rule that turns it into one and compares it to a strength. Ductile and brittle materials need different rules.

01

Readiness check

This chapter turns a stress state into a failure check. Tick only what you can do closed-notes.

  • Find principal stresses from σ and τ.
  • Distinguish ductile from brittle behaviour.
  • Recall yield and ultimate strength.
  • Form a factor of safety as strength over stress.
  • Handle a compressive (negative) principal stress.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsRevisit principal stresses in Chapter 2.
3 or more weak itemsReview ductile and brittle behaviour in Materials, Chapter 5.
02

The core idea

A failure theory collapses a multiaxial stress state into one equivalent stress to compare with a strength. Ductile parts fail by yielding (shear); brittle parts fail by fracture (normal stress).

σ' = √(σA² − σAσB + σB²)n = Sy/σ' (ductile, DE)σ1/Sut − σ3/Suc = 1/n (brittle)

Ductile materials yield when distortion energy reaches a limit, summarised by the von Mises stress σ', and the distortion-energy theory compares σ' to the yield strength. A simpler, slightly more conservative rule, the maximum-shear-stress theory, limits the largest shear. Brittle materials instead fracture from the largest tensile stress, so the maximum-normal-stress and Coulomb-Mohr theories use the ultimate strengths in tension and compression. Choosing the right theory starts with asking whether the material is ductile or brittle.

The skill works when: you match the theory to the material, reduce the state to an equivalent stress, and divide into the strength.
The skill breaks down when: a ductile theory is used on a brittle casting, or a negative principal stress is dropped.
The concept. Ductile failure envelopes in the principal-stress plane: the distortion-energy ellipse and the inscribed maximum-shear-stress hexagon. Any state inside is safe; the hexagon is the more conservative boundary.
03

The skills, taught in order

Five skills separate ductile from brittle, state the two ductile theories and the two brittle ones, and tie each to a factor of safety.

4.1 Ductile versus brittle

Ductile materials (most steels, aluminium) stretch noticeably before breaking and fail by yielding, governed by shear. Brittle materials (gray cast iron, ceramics) break with little warning and fail in tension, governed by normal stress. The split decides which family of theory applies.

4.2 Maximum-shear-stress theory (ductile)

The MSS or Tresca theory says yielding begins when the largest shear stress reaches the shear at yield in tension, Sy/2. With ordered principal stresses, n = Sy/(σ1 − σ3). It is simple and conservative, a safe first check.

4.3 Distortion-energy theory (ductile)

The DE or von Mises theory compares the von Mises stress σ' = √(σA² − σAσB + σB²) with the yield strength, giving n = Sy/σ'. It matches test data best and is the default for ductile parts; it predicts a slightly higher capacity than MSS.

TheoryMaterialFactor of safety
Distortion energy (von Mises)ductilen = Sy/σ'
Maximum shear (Tresca)ductilen = Sy/(σ1 − σ3)
Maximum normal stressbrittlen = Sut/σ1
Coulomb-Mohrbrittleσ1/Sut − σ3/Suc = 1/n

4.4 Comparing the ductile theories

For the same state, MSS gives a smaller factor of safety than DE, by up to about 15 percent. Using MSS is never unsafe; using DE uses the material more fully. Many designers check with DE and keep MSS in mind as the conservative bound.

4.5 Brittle theories

Brittle materials are far stronger in compression than tension, so their criteria use both ultimate strengths. The maximum-normal-stress theory simply limits σ1 to Sut; the Coulomb-Mohr theory, σ1/Sut − σ3/Suc = 1/n, handles a tensile and a compressive principal stress together and is preferred.

Engineering connection: the von Mises stress reappears in the fatigue and shaft chapters, where it is built from alternating and mean components.

04

Worked example 1: a ductile part, two theories

A ductile steel part (Sy = 370 MPa) carries a plane-stress state σx = 120 MPa, σy = 0, τxy = 50 MPa. Find the factor of safety by the distortion-energy and maximum-shear-stress theories.

Figure 1. The plane-stress element reduces to a von Mises stress and an ordered set of principal stresses; each theory turns it into a factor of safety.
  1. ProblemFind n by DE and by MSS for the element in Figure 1.
  2. Given / findSy = 370 MPa, σx = 120, σy = 0, τxy = 50 MPa. Find nDE and nMSS.
  3. AssumptionsDuctile material; static load; plane stress.
  4. ModelFor DE use σ' = √(σx² + 3τ²); for MSS find principal stresses, then n = Sy/(σ1 − σ3).
  5. Equationsσ' = √(σx² + 3τ²)σ1,2 = σx/2 ± √((σx/2)² + τ²)
  6. Solveσ' = √(120² + 3·50²) = √21 900 = 148 MPa, so nDE = 370/148 = 2.50. Principal stresses: 60 ± √(60² + 50²) = 60 ± 78.1, so σ1 = 138.1, σ3 = −18.1 MPa. Then nMSS = 370/(138.1 − (−18.1)) = 370/156.2 = 2.37.
  7. CheckMSS gives the smaller factor of safety, as it must, by about 5 percent here. Because the principal stresses straddle zero, the out-of-plane stress does not change the shear.
  8. ConclusionBoth theories pass with a comfortable margin. DE is the default; MSS is the conservative cross-check.
Result. σ' = 148 MPa, nDE = 2.50, nMSS = 2.37.
05

Worked example 2: a brittle casting

A gray cast-iron part (Sut = 214 MPa, Suc = 752 MPa) carries σx = 80 MPa, σy = 0, τxy = 40 MPa. Find the factor of safety by the Coulomb-Mohr theory.

Figure 2. A brittle material is far stronger in compression, so the Coulomb-Mohr theory weights the tensile and compressive principal stresses against their separate ultimate strengths.
  1. ProblemFind n by Coulomb-Mohr for the brittle element in Figure 2.
  2. Given / findSut = 214 MPa, Suc = 752 MPa, σx = 80, τxy = 40 MPa. Find σ1, σ2, and n.
  3. AssumptionsBrittle material; static load; plane stress with a tensile and a compressive principal stress.
  4. ModelFind the principal stresses, then apply σ1/Sut − σ2/Suc = 1/n.
  5. Equationsσ1,2 = σx/2 ± √((σx/2)² + τ²)1/n = σ1/Sut − σ2/Suc
  6. Solveσ1,2 = 40 ± √(40² + 40²) = 40 ± 56.6, so σ1 = 96.6 MPa, σ2 = −16.6 MPa. Then 1/n = 96.6/214 − (−16.6)/752 = 0.451 + 0.022 = 0.473, so n = 2.1.
  7. CheckThe compressive term adds only a little, because cast iron's large Suc makes compression nearly harmless. A simpler maximum-normal-stress check, n = 214/96.6 = 2.2, is close, as expected when compression is small.
  8. ConclusionFor brittle parts, tension governs and the two ultimate strengths enter separately. Coulomb-Mohr is the safe, standard choice.
Result. σ1 = 96.6 MPa, σ2 = −16.6 MPa, and n = 2.1.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Ductile theory on a castingvon Mises used on gray cast iron"Is this material ductile or brittle?"Use a brittle theory (Coulomb-Mohr) for cast iron and ceramics.
Dropping a negative principal stressMSS or Coulomb-Mohr too optimistic"Did I include the compressive principal stress?"Keep σ3 (or σ2); the shear or the compressive term needs it.
Forgetting the out-of-plane stressMSS shear span too small"Is the third principal stress zero or not?"Include σ = 0 out of plane when ordering the principals.
Mixing yield and ultimateDuctile check uses Sut"Which strength matches the failure mode?"Yield for ductile theories, ultimate for brittle ones.
07

Practice ladder

Level 1 · Direct skill

A ductile bar has σx = 150 MPa, τxy = 0. With Sy = 370 MPa, find the von Mises factor of safety.

Show answer

With no shear, σ' = σx = 150 MPa, so n = 370/150 = 2.47. Uniaxial states make σ' equal the applied stress.

Level 2 · Mixed concept

The ductile element of Worked Example 1 keeps σx = 120 MPa but the shear rises to 70 MPa. What is the new von Mises factor of safety?

Show answer

σ' = √(120² + 3·70²) = √(14 400 + 14 700) = √29 100 = 170.6 MPa. n = 370/170.6 = 2.17. More shear lowers the margin.

Level 3 · Independent problem

A gray cast-iron part (Sut = 214, Suc = 752 MPa) has principal stresses σ1 = 70 MPa, σ2 = −120 MPa. Find n by Coulomb-Mohr.

Show answer

1/n = 70/214 − (−120)/752 = 0.327 + 0.160 = 0.487, so n = 2.05. Here the compressive stress contributes meaningfully because it is large.

Level 4 · Transfer to real engineering

Pick one ductile and one brittle component. State which failure theory applies to each and why, naming the strength each uses.

What good work looks like

The ductile part assigned a shear-based theory with Sy, the brittle part a normal-stress theory with Sut and Suc, each justified by how the material fails.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I picked a ductile or brittle theory to match this material."
"Give me four materials; I will choose the right failure theory for each."
"What is the factor of safety?" Reducing the state and dividing into strength is the skill.
"Is this part safe?" Selecting and applying the theory is the point.

Portfolio task

Take a part under a known stress state, classify the material, choose and apply the matching failure theory, and report a factor of safety with the theory named.

Must include: a material classification, the matching theory, an equivalent stress, and a factor of safety.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. How do ductile and brittle materials fail?

Ductile by yielding (shear); brittle by fracture (normal stress).

2. Write the von Mises stress for plane stress.

σ' = √(σA² − σAσB + σB²), or √(σx² + 3τ²) for σy = 0.

3. Give the MSS factor of safety.

n = Sy/(σ1 − σ3), using the ordered principal stresses.

4. State the Coulomb-Mohr criterion.

σ1/Sut − σ3/Suc = 1/n for a brittle material.

5. Which is more conservative, DE or MSS?

MSS; it gives a smaller factor of safety than DE for the same state.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive both worked examples from a blank page.
+3 daysApply a theory to two new stress states.
+7 daysCarry the von Mises stress into fatigue, Chapter 5.
+30 daysReuse the failure theories in shaft and fastener design.
10

Textbook mapping

ItemMapping
Primary sourceBudynas and Nisbett, Shigley's Mechanical Engineering Design, Chapter 5 (Failures Resulting from Static Loading)
Cross-referenceNorton, Ch. 5 · Mechanics of Materials, Ch. 7
Core topics4.1 Ductile vs brittle · 4.2 Maximum shear · 4.3 Distortion energy · 4.4 Comparing the theories · 4.5 Brittle theories
Engineering connectionThe von Mises stress carries into fatigue and shaft design.
Read nextChapter 5: Fatigue Failure from Variable Loading.