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Mechanics of Materials · Chapter 4 of 10 · Intermediate

Pure Bending

Bend a beam and one face stretches while the other compresses, with a neutral axis in between. The flexure formula σ = My/I turns the bending moment into stress, and explains why beams are deep.

01

Readiness check

This chapter turns a bending moment into stress across a section. Tick only what you can do closed-notes.

  • Recall the bending moment as an internal couple.
  • Find the centroid of a section.
  • Compute a moment of inertia (bh³/12).
  • Use the parallel-axis theorem.
  • Work in N·mm and MPa.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsReview the moment of inertia of an area in Statics.
3 or more weak itemsRevisit centroids and second moments first.
02

The core idea

A bending moment makes the stress vary linearly across the section, zero at the neutral axis and greatest at the outer fibers: σ = My/I.

σ = My/I, σmax = Mc/Iσmax = M/S, S = I/cIrect = bh³/12

In pure bending, plane cross-sections rotate and stay plane, so the longitudinal strain is proportional to the distance y from the neutral axis, which passes through the centroid. With Hooke's law the stress is also linear in y. The moment of inertia I measures how the area is spread from the neutral axis, and the section modulus S = I/c packages it for the extreme fiber. Because I grows with depth cubed, putting material far from the neutral axis is the most effective way to resist bending.

The skill works when: you locate the neutral axis at the centroid and use σ = Mc/I (or M/S) at the extreme fiber.
The skill breaks down when: the neutral axis is misplaced, or I is taken about the wrong axis without the parallel-axis shift.
The concept. Bending stress varies linearly across the depth: maximum compression at the top fiber, maximum tension at the bottom, and zero at the neutral axis through the centroid.
03

The skills, taught in order

Bending is one formula plus the section properties it needs. Five skills cover the neutral axis, the flexure formula, the moment of inertia, the section modulus, and material efficiency.

4.1 Pure bending and the neutral axis

Under a bending moment, the beam curves; fibers on the convex side stretch, those on the concave side compress, and one layer, the neutral axis, neither. For a symmetric elastic section the neutral axis passes through the centroid. Strain varies linearly with distance from it.

4.2 The flexure formula

Combining linear strain with Hooke's law and moment equilibrium gives σ = My/I, the flexure formula. The largest stress is at the extreme fiber, y = c: σmax = Mc/I. Tension and compression are equal for a symmetric section.

4.3 The moment of inertia

I is the second moment of area about the neutral axis.

SectionI (about centroid)S = I/c
Rectangle (b×h)bh³/12bh²/6
Solid circle (d)πd⁴/64πd³/32

For built-up sections, sum the parts using the parallel-axis theorem I = I̅ + Ad².

4.4 The section modulus

Packaging I and c into S = I/c gives the compact form σmax = M/S. Design is then direct: the required section modulus is S = M/σallow, and you pick or size a section to meet it. Standard beams are tabulated by S.

4.5 Material efficiency in bending

Because Srect = bh²/6 grows with the square of the depth, a beam is far stronger oriented deep than flat. Better still, an I-beam puts most of its area in flanges far from the neutral axis, maximising I for the least material, which is why structural beams are I-shaped.

Engineering connection: floor joists, shafts in bending, aircraft spars, and machine frames are all sized with σ = Mc/I; the bending moment M comes from the diagrams of Chapter 5.

04

Worked example 1: bending stress in a beam

A rectangular beam 50 mm wide and 150 mm deep carries a bending moment of 12 kN·m. Find the maximum bending stress, and compare it with the same beam laid flat.

Figure 1. Oriented deep (h = 150), the beam has triple the section modulus of the flat orientation, so a third of the stress. Depth is the most valuable dimension in bending.
  1. ProblemFind the maximum bending stress for the beam in Figure 1, on edge and laid flat.
  2. Given / findb = 50 mm, h = 150 mm, M = 12 kN·m = 12×10⁶ N·mm. Find σmax both ways.
  3. AssumptionsPure bending, symmetric elastic section, neutral axis at the centroid.
  4. ModelCompute I and c (or S directly), then σmax = M/S for each orientation.
  5. EquationsI = bh³/12, c = h/2 S = bh²/6 σmax = M/S
  6. SolveOn edge: S = 50(150)²/6 = 187 500 mm³, so σ = 12×10⁶/187 500 = 64 MPa. Laid flat (h = 50, b = 150): S = 150(50)²/6 = 62 500 mm³, so σ = 192 MPa, three times higher.
  7. CheckThe stress ratio equals the depth ratio (150/50 = 3), as S ∝ h². This is exactly why joists and rafters stand on edge, not flat.
  8. ConclusionThe flexure formula gives the stress, and the section modulus shows orientation matters enormously. The same beam is three times stronger in bending when it is deep.
Result. σmax = 64 MPa on edge, 192 MPa laid flat (three times higher).
05

Worked example 2: designing a timber beam

A timber beam must carry a bending moment of 8 kN·m at an allowable stress of 10 MPa, with a depth twice the width (h = 2b). Find the required section modulus and the cross-section.

Figure 2. Design inverts the flexure formula: the required section modulus comes from the moment and allowable stress, then the depth-to-width rule fixes the dimensions.
  1. ProblemFind the required section modulus and dimensions for the beam in Figure 2.
  2. Given / findM = 8 kN·m = 8×10⁶ N·mm, σallow = 10 MPa, h = 2b. Find Sreq, b, h.
  3. AssumptionsPure bending, allowable stress at the extreme fiber, rectangular section.
  4. ModelRequired section modulus from M and σallow, then solve S = bh²/6 with h = 2b for the dimensions.
  5. EquationsSreq = M/σallow S = bh²/6 = 2b³/3 (h = 2b)
  6. SolveSreq = 8×10⁶/10 = 800 000 mm³. Then 2b³/3 = 800 000, so b³ = 1.2×10⁶, b = 106 mm and h = 213 mm; use 110 × 220 mm.
  7. CheckAt 110 × 220, S = 110(220)²/6 = 887 000 mm³ > 800 000, so the actual stress is below 10 MPa, the safe side. Rounding the dimensions up (not down) preserves the margin.
  8. ConclusionBending design is a section-modulus problem: compute Sreq, then choose a shape that provides it. For standard steel beams you would simply pick the lightest section with S ≥ Sreq from a table.
Result. Sreq = 800 000 mm³; cross-section about 106 × 213 mm, rounded to 110 × 220.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Neutral axis off the centroidWrong y and stress"Does the neutral axis pass through the centroid?"For symmetric elastic bending, it does; locate the centroid first.
Wrong axis for IStress far off"Is I about the neutral (bending) axis?"Use I about the bending axis, with the parallel-axis shift for parts.
Using diameter not radius for cStress halved or doubled"Is c the distance to the extreme fiber?"c = h/2 or the radius, the farthest fiber from the neutral axis.
Ignoring orientationBeam laid flat is overstressed"Is the depth the large dimension?"Orient deep; S ∝ h², so depth dominates strength.
07

Practice ladder

Level 1 · Direct skill

A solid circular shaft 60 mm in diameter carries a bending moment of 1.5 kN·m. Find the maximum bending stress.

Show answer

S = πd³/32 = π(60)³/32 = 21 200 mm³; σ = M/S = 1.5×10⁶/21 200 = 70.8 MPa. The circular section modulus is πd³/32.

Level 2 · Mixed concept

For the Worked Example 1 beam on edge, what moment would bring it to a 100 MPa allowable stress?

Show answer

M = σ·S = 100 × 187 500 = 1.875×10⁷ N·mm = 18.75 kN·m. Capacity scales directly with the section modulus.

Level 3 · Independent problem

Compare the moment of inertia of a 100 mm square with that of an I-shape of the same area (two 100×20 flanges and a 60×20 web). Which resists bending better?

Show answer

Square: I = 100⁴/12 = 8.3×10⁶ mm⁴. The I-shape concentrates area in flanges far from the axis, giving a much larger I for the same area (roughly 2 to 3 times), so it resists bending far better. This is why beams are I-shaped, not solid squares.

Level 4 · Transfer to real engineering

Find a real beam (a floor joist, a shelf bracket, a bike frame tube). Estimate the bending moment and section, compute the stress, and comment on the orientation or shape choice.

What good work looks like

The section modulus computed for the real shape, σ = M/S evaluated, and the depth or I-shape choice explained from S ∝ h² or flange placement.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my I is about the neutral axis and c is the extreme fiber."
"Give me five sections; I will rank them by section modulus before computing."
"Compute the bending stress." Finding I, c, and S yourself is the skill.
"What size beam?" Setting Sreq = M/σ and choosing a section is the point.

Portfolio task

Analyse one beam: compute its section modulus and bending stress, then size or select a section for a target moment and allowable stress.

Must include: I and c (or S), σ = Mc/I, and a section-modulus design check.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the flexure formula.

σ = My/I, with σmax = Mc/I at the extreme fiber.

2. Where is the neutral axis?

Through the centroid of the section (symmetric elastic bending).

3. Give I and S for a rectangle.

I = bh³/12; S = bh²/6.

4. What is the section modulus used for?

σmax = M/S, and design via Sreq = M/σallow.

5. Why are beams deep or I-shaped?

S ∝ h² and I grows with area placed far from the neutral axis, so depth and flanges maximise bending strength per unit material.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the bending stress and design from a blank page.
+3 daysOne analysis and one section-modulus design.
+7 daysFind M from shear-moment diagrams, Chapter 5.
+30 daysCombine bending with axial and torsion in Chapter 8.
10

Textbook mapping

ItemMapping
Primary sourceBeer, Johnston, DeWolf and Mazurek, Mechanics of Materials (6th ed), Chapter 4 (Pure Bending)
Cross-referenceHibbeler, Ch. 6 · Gere and Goodno, Ch. 5
Core topics4.1 Neutral axis · 4.2 Flexure formula · 4.3 Moment of inertia · 4.4 Section modulus · 4.5 Material efficiency
Engineering connectionJoists, spars, shafts in bending, and machine frames.
Read nextChapter 5: Beams, Shear and Moment Diagrams.