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Machine Elements · Chapter 2 of 10 · Intermediate

Load and Stress Analysis

Real parts rarely carry one clean load. Bending, torsion, and axial force act together, and the design question is always the same: what is the worst stress, and where?

01

Readiness check

This chapter combines stress formulas. Tick only what you can do closed-notes.

  • Write bending stress σ = Mc/I and torsion τ = Tr/J.
  • Find I and J for a round section.
  • Recall Mohr's circle for plane stress.
  • Add stresses from different loads by superposition.
  • Identify where on a section a stress peaks.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsReview transformation in Mechanics of Materials, Chapter 7.
3 or more weak itemsRevisit bending and torsion in Mechanics of Materials first.
02

The core idea

Each load produces its own stress; superposition adds them on a stress element, and Mohr's circle turns that element into the principal and maximum-shear stresses a failure theory needs.

σ = 32M/(πd³), τ = 16T/(πd³)σ1,2 = σ/2 ± √((σ/2)² + τ²)τmax = √((σ/2)² + τ²)

For a round shaft, bending gives a normal stress σ = 32M/πd³ at the surface, and torsion gives a shear stress τ = 16T/πd³. Acting together they define a plane-stress element. The principal stresses and the maximum shear stress come from Mohr's circle: the element's normal stresses set the centre and the shear sets the radius. These transformed stresses, not the raw σ and τ, are what the failure theories of the next two chapters use.

The skill works when: you build the stress element from every load, then transform it to principal and maximum-shear stresses.
The skill breaks down when: stresses from different loads are compared separately instead of combined on one element.
The concept. Bending and torsion build one stress element (left). Mohr's circle (right) converts it into principal stresses σ₁, σ₂ and the maximum shear, the quantities a failure theory needs.
03

The skills, taught in order

Five skills cover the basic stress formulas, how to combine them, how to transform to principal stresses, and how stress concentration raises the local peak.

2.1 Stresses from each load

Axial force gives σ = F/A; bending gives σ = Mc/I, peaking at the outer fibre; torsion of a round shaft gives τ = Tr/J at the surface; transverse shear adds a smaller τ. For a solid round shaft these reduce to the compact forms σ = 32M/πd³ and τ = 16T/πd³.

LoadStressRound shaft form
Axialσ = F/A4F/πd²
Bendingσ = Mc/I32M/πd³
Torsionτ = Tr/J16T/πd³
Transverse shear (max)τ = VQ/Ib4V/3A

2.2 Combined loading and superposition

When several loads act, their stresses add at each point as long as the material stays linear. Find the critical point (often the surface where bending and torsion both peak), and assemble the normal and shear stresses there onto a single element. Combining first and transforming second is the disciplined order.

2.3 Principal stresses and maximum shear

From a plane-stress element with normal σ and shear τ, Mohr's circle gives the principal stresses σ1,2 = σ/2 ± √((σ/2)² + τ²) and the maximum shear τmax = √((σ/2)² + τ²). These are the extreme stresses that decide failure, independent of the axes you started with.

2.4 Stress concentration

Holes, fillets, and grooves raise the local stress above the nominal value by a factor Kt, read from charts: σmax = Kt σnom. For static loading of ductile materials, local yielding relieves the peak so Kt is often set to one; for brittle materials and for fatigue, it must always be applied.

2.5 The design stress element

The output of stress analysis is a stress element at the critical point, reduced to its principal and maximum-shear stresses. This element is the input to every static and fatigue failure check that follows, so getting it right is the foundation of the whole course.

Engineering connection: the von Mises and maximum-shear theories of Chapter 4 act directly on the principal stresses computed here.

04

Worked example 1: combined bending and torsion

A solid round shaft of diameter 30 mm carries a bending moment of 200 N·m and a torque of 150 N·m at the same section. Find the bending and torsional stresses, then the principal stresses and the maximum shear stress at the surface.

Figure 1. Bending and torsion act together at the surface of the shaft. Their stresses build one element; Mohr's circle gives the principal and maximum-shear values.
  1. ProblemFind σ, τ, the principal stresses, and τmax for the shaft in Figure 1.
  2. Given / findd = 30 mm, M = 200 N·m = 200 000 N·mm, T = 150 N·m = 150 000 N·mm. Find σ, τ, σ1,2, τmax.
  3. AssumptionsLinear elastic; the critical point is the surface where both stresses peak; transverse shear negligible there.
  4. ModelUse the round-shaft forms for σ and τ, then Mohr's circle for the transformed stresses.
  5. Equationsσ = 32M/(πd³), τ = 16T/(πd³)σ1,2 = σ/2 ± √((σ/2)² + τ²)
  6. Solveσ = 32(200 000)/(π·30³) = 75.5 MPa; τ = 16(150 000)/(π·30³) = 28.3 MPa. Centre σ/2 = 37.7, radius √(37.7² + 28.3²) = 47.2. So σ1 = 84.9 MPa, σ2 = −9.4 MPa, and τmax = 47.2 MPa.
  7. CheckThe principal stresses straddle zero, as a bending-plus-torsion element should, and τmax equals the circle radius. All consistent.
  8. ConclusionThe transformed stresses, not the raw σ and τ, feed the failure theories. This element is exactly what Chapter 4 needs.
Result. σ = 75.5 MPa, τ = 28.3 MPa, σ₁ = 84.9 MPa, σ₂ = −9.4 MPa, τmax = 47.2 MPa.
05

Worked example 2: stress concentration at a hole

A flat bar 30 mm wide and 6 mm thick has a 10 mm transverse hole and carries an axial load of 12 kN. With a geometric stress-concentration factor Kt = 2.35 (from the chart for d/w = 0.33), find the nominal and maximum stresses at the hole.

Figure 2. The hole removes material and crowds the stress lines, so the local peak is Kt times the nominal stress computed on the reduced net section.
  1. ProblemFind the nominal and peak stress at the hole in Figure 2.
  2. Given / findw = 30 mm, t = 6 mm, hole d = 10 mm, F = 12 kN, Kt = 2.35. Find σnom and σmax.
  3. AssumptionsNominal stress on the net (reduced) section; the chart Kt applies; the material is loaded statically.
  4. ModelCompute σnom on the net area, then multiply by Kt for the local peak.
  5. EquationsAnet = (w − d)tσnom = F/Anetσmax = Kt σnom
  6. SolveAnet = (30 − 10)(6) = 120 mm². σnom = 12 000/120 = 100 MPa. σmax = 2.35 × 100 = 235 MPa.
  7. CheckThe peak is well above the nominal, which is why holes and fillets drive fatigue cracks. For a static ductile part, local yielding would relieve this peak; for fatigue or a brittle material, the full 235 MPa must be used.
  8. ConclusionStress concentration is a local effect tied to geometry. Whether it matters depends on the failure mode, which is the recurring lesson of design.
Result. σnom = 100 MPa and σmax = 235 MPa at the hole.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Comparing loads separatelyChecking bending and torsion one at a time"Are these stresses on the same element?"Combine them, then transform to principal stresses.
Nominal stress on gross areaKt applied to the wrong base stress"Is the nominal stress on the net section?"Compute σnom on the reduced area, then multiply by Kt.
Kt always or neverConcentration applied or dropped blindly"Is this static-ductile, or fatigue or brittle?"Drop Kt for static ductile; keep it for fatigue and brittle.
Wrong critical pointStress evaluated away from the peak"Where do the stresses actually peak?"Find the point where the combined stress is largest, usually the surface.
07

Practice ladder

Level 1 · Direct skill

A 25 mm shaft carries a torque of 120 N·m. Find the surface shear stress.

Show answer

τ = 16T/πd³ = 16(120 000)/(π·25³) = 1 920 000/49 087 = 39.1 MPa.

Level 2 · Mixed concept

The shaft of Worked Example 1 keeps M = 200 N·m but the torque rises to 250 N·m. What is the new maximum shear stress?

Show answer

σ = 75.5 MPa (unchanged), τ = 16(250 000)/(π·30³) = 47.2 MPa. τmax = √(37.7² + 47.2²) = √(1421 + 2228) = 60.4 MPa.

Level 3 · Independent problem

A flat bar 40 mm wide, 8 mm thick, with a 12 mm hole carries 20 kN. With Kt = 2.5, find the peak stress, and state whether you would apply Kt for a static ductile part.

Show answer

Anet = (40 − 12)(8) = 224 mm². σnom = 20 000/224 = 89.3 MPa. σmax = 2.5 × 89.3 = 223 MPa. For a static ductile part, local yielding relieves the peak, so Kt is usually set to one and 89.3 MPa governs.

Level 4 · Transfer to real engineering

Find a part with a fillet or hole near a high-stress region. Explain where it would crack first and why, using stress concentration and the failure mode.

What good work looks like

The crack located at the concentration, an estimate of Kt from the geometry, and a statement that fatigue (cyclic) loading is what makes the concentration decisive.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I combined the stresses on one element before transforming."
"Give me four geometries; I will say whether Kt matters for each load type."
"What are the principal stresses?" Building and transforming the element yourself is the skill.
"Where will it fail?" Finding the critical point and stress is the point.

Portfolio task

Take a part under combined loading, build the stress element at its critical point, transform it to principal and maximum-shear stresses, and note any stress concentration and whether it applies.

Must include: a critical point, a combined stress element, principal and maximum-shear stresses, and a Kt decision.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write bending and torsion stress for a round shaft.

σ = 32M/πd³ and τ = 16T/πd³, both at the surface.

2. Give the principal stresses for a plane-stress element.

σ1,2 = σ/2 ± √((σ/2)² + τ²).

3. What is the maximum shear stress?

τmax = √((σ/2)² + τ²), the radius of Mohr's circle.

4. When is Kt set to one?

For static loading of ductile materials, where local yielding relieves the peak.

5. What is the output of stress analysis?

A stress element at the critical point, reduced to principal and maximum-shear stresses.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the combined-stress element from a blank page.
+3 daysTransform two new stress elements.
+7 daysCarry the stress element into deflection, Chapter 3.
+30 daysFeed the principal stresses into the failure theories of Chapter 4.
10

Textbook mapping

ItemMapping
Primary sourceBudynas and Nisbett, Shigley's Mechanical Engineering Design, Chapter 3 (Load and Stress Analysis)
Cross-referenceNorton, Ch. 4 · Mechanics of Materials, Ch. 7
Core topics2.1 Stresses from each load · 2.2 Combined loading · 2.3 Principal stresses · 2.4 Stress concentration · 2.5 The design stress element
Engineering connectionThe principal stresses feed the static and fatigue failure theories.
Read nextChapter 3: Deflection and Stiffness.