State the flexure (bending) formula and identify each term.

. bending moment, second moment of area about neutral axis, bending stress at distance, radius of curvature, Young’s modulus.

What is the neutral axis? Where does it pass for a homogeneous symmetric section?

The locus of zero longitudinal strain/stress in pure bending. For homogeneous sections it passes through the centroid. Above NA: compression or tension depending on moment sense; below: opposite.

List the assumptions of the simple theory of bending.

Material homogeneous isotropic and linearly elastic; plane sections remain plane; beam initially straight with constant cross-section; pure bending (or approx. for slender beams); same in tension/compression; deflections small.

Define section modulus. How is it used in design?

. Then. Larger means lower bending stress for given — used to select beam sections.

Why are I-beams efficient in bending?

Most material is placed in flanges far from the NA, maximizing and for given area. The web mainly carries shear.

How do you find bending stress in a beam with unsymmetric section about the loading plane?

Use principal centroidal axes: resolve into and (signs by inspection). Neutral axis is generally inclined.

What is the difference between pure bending and ordinary bending?

Pure bending: constant, zero shear — exact flexure theory. Ordinary bending: varies with shear present; flexure formula is still used approximately for slender beams (Saint-Venant).

True/False: Bending stress is maximum at the neutral axis.

False — bending stress is zero at the NA and maximum at extreme fibres.

For the same cross-sectional area, which is stronger in bending: square or circular section?

Square section has larger than the circle of equal area — higher elastic moment capacity.

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Study notes

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Qwestrum Engineering360 · Mechanical Engineering · Strength of Materials (SOM)

Bending Stress in Beams

Q: Bending stress in beams

Pure bending of elastic beams gives the flexure formula. Bending stress is zero at the neutral axis and maximum at the extreme fibres. Section modulus governs strength design (RK Bansal).

Pure bending of elastic beams gives the flexure formula

\dfrac{M}{I}=\dfrac{\sigma}{y}=\dfrac{E}{R}

. Bending stress

\sigma=My/I

is zero at the neutral axis and maximum at the extreme fibres. Section modulus

Z=I/y_{\max}

governs strength design (RK Bansal).

Exam tip: keep units consistent — N with mm² gives MPa directly (1 MPa = 1 N/mm²).

Key formulas & points

Skim these first — then read the full notes below.

Flexure formula: $\dfrac{M}{I}=\dfrac{\sigma}{y}=\dfrac{E}{R}$ .
Bending stress: $\sigma=\dfrac{My}{I}$ ; $\sigma_{\max}=\dfrac{M}{Z}$ with $Z=I/y_{\max}$ .
Neutral axis passes through centroid for pure bending (no axial force).
Moment of inertia: rectangle $I = \frac{bd^3}{12}$ ; circular $I=\pi d^4/64$ .
Curvature: $\dfrac{1}{R}=\dfrac{M}{EI}$ .
Shear force $V$ and bending moment $M$ from beam equilibrium / SFD–BMD.
Combined axial + bending: $\sigma=\dfrac{P}{A}\pm\dfrac{My}{I}$ .

Topic details

Definition and physical meaning

Bending produces curvature of the beam axis. Longitudinal fibres above/below the neutral axis (NA) shorten or elongate, causing normal bending stresses.

Flexure (bending) formula for elastic pure bending:

\frac{M}{I}=\frac{\sigma}{y}=\frac{E}{R}

Hence

\sigma=\frac{My}{I},\qquad\frac{1}{R}=\frac{M}{EI}

Physical parameters

Symbol	Meaning	SI unit
$M$	Bending moment	$\mathrm{N\cdot m}$
$I$	Second moment of area about NA	$\mathrm{m^4}$
$y$	Distance from NA	$\mathrm{m}$
$\sigma$	Bending stress	$\mathrm{Pa}$
$Z$	Section modulus $I/y_{\max}$	$\mathrm{m^3}$
$E$	Young’s modulus	$\mathrm{Pa}$
$R$	Radius of curvature	$\mathrm{m}$
$EI$	Flexural rigidity	$\mathrm{N\cdot m^2}$

Sign convention (common textbook): sagging

M

→ compression on top fibres, tension on bottom (simply supported beam with downward load).

Fig 3.1 — Sagging moment: top fibre in compression (−σ), bottom in tension (+σ). σ = My/I, σ_max = M/Z at y = ±c.

Schematic diagram for study — aligned with standard B.Tech / GATE syllabus.

Core assumptions (theory of pure bending)

1. Beam is initially straight with a symmetric cross-section about the plane of bending (or load in a principal plane).
2. Material is homogeneous, isotropic, linear elastic.
3. Plane sections remain plane and normal to the deflected axis (Euler–Bernoulli).
4. Each longitudinal fibre is in uniaxial stress (lateral stresses neglected).
5. Young’s modulus same in tension and compression.
6. Beam is subjected to pure bending (or

M

varies slowly — local application of flexure formula still used in strength of materials).
7. Deflections are small.

If axial force is also present, superpose

\sigma_{\mathrm{axial}}=P/A

Derivation summary

Kinematics. For curvature

\frac{1}{R}

, longitudinal strain at distance

y

from NA:

\varepsilon=\frac{y}{R}

(NA fibre undeformed:

\varepsilon=0

Constitutive.

\sigma=E\varepsilon=Ey/R

— linear stress distribution.

Equilibrium. Axial force resultant zero for pure bending:

\int_A\sigma\, dA=0\Rightarrow\text{NA through centroid}

Moment resultant:

M=\int_A\sigma\, y\, dA=\frac{E}{R}\int_A y^2\, dA=\frac{E I}{R}

Therefore

\frac{M}{I}=\frac{E}{R}=\frac{\sigma}{y}

Section modulus and common I values

Section modulus

Z=\frac{I}{y_{\max}}\qquad\sigma_{\max}=\frac{M}{Z}

Larger

Z

→ stronger section for same material and

M

**Useful

I_{NA}

Section	$I$ about centroidal axis
Rectangle $b\times d$ (bend about axis ∥ $b$ )	$\frac{bd^3}{12}$
Circular diameter $d$	$\pi d^4/64$
Hollow circular	$\pi(D^4-d^4)/64$
Triangular (base $b$ , height $h$ )	$\frac{bh^3}{36}$

Parallel-axis theorem for built-up sections:

I_{\mathrm{NA}}=\sum\big(I_{\mathrm{own}}+A\bar{y}^2\big)

after locating the composite centroid.

Shear force, BM, and combined loading

Relate load

w

, shear

V

, moment

M

\frac{dV}{dx}=-w,\qquad\frac{dM}{dx}=V

Draw SFD and BMD to find

M_{\max}

at the critical section, then apply

\sigma=My/I

Combined axial and bending (eccentric load

P

at eccentricity

e

\sigma=\frac{P}{A}\pm\frac{(Pe)y}{I}

Kern of section: region of

e

for which entire cross-section stays compressive (important for columns/foundations).

Flitched beams (steel plates + timber): same curvature/strain distribution; stress in each material

\sigma=E y/R

; moment shared by

\sum E_i I_i

Step-by-step problem approach

1. Find support reactions; draw SFD/BMD; identify

M

at the section of interest.
2. Locate centroid (NA) of the cross-section.
3. Compute

I

about NA (use parallel-axis theorem if needed).
4.

\sigma=My/I

at required

y

; report tension/compression side.
5. For design:

Z_{\mathrm{req}}=M/\sigma_{\mathrm{allow}}

; choose section.
6. Keep units consistent (N·mm and mm⁴ → N/mm² = MPa).
7. State pure-bending assumptions in exam answers.

Common mistakes in exams

• Taking

I

about the wrong axis (not the NA / not the bending axis).
• Using

y

from the bottom fibre when NA is not at mid-depth (unsymmetric sections).
• Confusing

Z=I/y_{\max}

with

I

itself in

\sigma=M/Z

.
• Forgetting parallel-axis

A\bar{y}^2

for I-beams / built-up sections.
• Mixing

J

(polar) with

I

(bending).
• Wrong sign of tension/compression relative to BMD convention.

Calculator

Bending stress (circular section)

\sigma_b = \frac{32M}{\pi d^{3}}

Moment MN·mmDiameter dmm

Result

162.9747N/mm² (MPa)

σ_b = 32M/(πd³) = 32×2.000e+6 / (π×50³) = 163 N/mm²

Worked examples

Try the problem first — open the solution when you are ready to check.

Rectangular beam — max bending stress

Problem

A simply supported beam

L=4\,\mathrm{m}

carries UDL

w=5\,\mathrm{kN/m}

. Section

b=100\,\mathrm{mm}

d=200\,\mathrm{mm}

. Find

\sigma_{\max}

Solution

Formulas (Indian textbook notation)

$M_{\max}=\frac{wL^2}{8}=\frac{5\times 4^2}{8}=10\,\mathrm{kN\cdot m}=10\times 10^6\,\mathrm{N\cdot mm}$

Formulas (Indian textbook notation)

$I=\frac{bd^3}{12}=\frac{100\times 200^3}{12}=6.667\times 10^7\,\mathrm{mm^4}$

Formulas (Indian textbook notation)

$y_{\max}=100\,\mathrm{mm}$

Formulas (Indian textbook notation)

$\sigma_{\max}=\frac{My}{I}=\frac{(10\times 10^6)(100)}{6.667\times 10^7}=15.0\,\mathrm{N/mm^2}=15\,\mathrm{MPa}$

Section modulus design

Problem

A beam must resist

M=40\,\mathrm{kN\cdot m}

with

\sigma_{\mathrm{allow}}=150\,\mathrm{MPa}

. Find required

Z

. If a circular section is used, find diameter.

Eccentric axial load

Problem

A short column

100\,\mathrm{mm}\times 150\,\mathrm{mm}

carries

P=180\,\mathrm{kN}

with eccentricity

e=20\,\mathrm{mm}

about the major axis (bending about axis parallel to 100 mm side). Find max and min normal stresses.

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

1
State the flexure (bending) formula and identify each term.
Model answer
$\dfrac{M}{I}=\dfrac{\sigma}{y}=\dfrac{E}{R}$ . $M$ bending moment, $I$ second moment of area about neutral axis, $\sigma$ bending stress at distance $y$ , $R$ radius of curvature, $E$ Young’s modulus.
2
What is the neutral axis? Where does it pass for a homogeneous symmetric section?
Model answer
The locus of zero longitudinal strain/stress in pure bending. For homogeneous sections it passes through the centroid. Above NA: compression or tension depending on moment sense; below: opposite.
3
List the assumptions of the simple theory of bending.
Model answer
Material homogeneous isotropic and linearly elastic; plane sections remain plane; beam initially straight with constant cross-section; pure bending (or approx. for slender beams); $E$ same in tension/compression; deflections small.
4
Define section modulus. How is it used in design?
Model answer
$Z=I/y_{\max}$ . Then $\sigma_{\max}=M/Z$ . Larger $Z$ means lower bending stress for given $M$ — used to select beam sections.
5
Why are I-beams efficient in bending?
Model answer
Most material is placed in flanges far from the NA, maximizing $I$ and $Z$ for given area. The web mainly carries shear.
6
How do you find bending stress in a beam with unsymmetric section about the loading plane?
Model answer
Use principal centroidal axes: resolve $M$ into $M_{u},M_{v}$ and $\sigma=M_u v/I_u+M_v u/I_v$ (signs by inspection). Neutral axis is generally inclined.
7
What is the difference between pure bending and ordinary bending?
Model answer
Pure bending: constant $M$ , zero shear — exact flexure theory. Ordinary bending: $M$ varies with shear present; flexure formula is still used approximately for slender beams (Saint-Venant).
8
Write $\sigma_{\max}$ for a rectangular section $b\times d$ under moment $M$ .
Model answer
$I = \frac{bd^3}{12}$ , $y_{\max}=d/2$ , $Z = \frac{bd^2}{6}$ , so $\sigma_{\max}=6M/(bd^2)$ .
9
How is radius of curvature related to bending moment?
Model answer
$\dfrac{1}{R}=\dfrac{M}{EI}$ . Larger $EI$ means smaller curvature for the same $M$ .
10
What is flitched (composite) beam? How is it analysed?
Model answer
Beam of two materials rigidly connected (e.g. timber with steel plates). Transform one material using modular ratio $m = \frac{E_{1}}{E_{2}}$ into an equivalent section, then apply $\sigma=My/I_{\mathrm{eq}}$ and convert stresses back.
11
Explain bending stress distribution over the depth of a beam.
Model answer
Linear with $y$ : zero at NA, maximum at extreme fibres. Tension on one side, compression on the other for sagging/hogging as appropriate.
12
What is the moment of resistance of a section?
Model answer
Maximum moment the section can resist at allowable stress: $M_r=\sigma_{\mathrm{all}} Z$ . For elastic design of symmetric sections.
13
How does beam strength change if depth is doubled keeping width same?
Model answer
$Z\propto d^2$ for rectangle, so strength (moment capacity) becomes four times for same $\sigma_{\mathrm{all}}$ .
14
Can the flexure formula be used at a sudden change of section or near supports?
Model answer
Locally, stress concentrations and constraint effects appear; the formula gives nominal stress. Use stress concentration factors or detailed analysis near discontinuities.
15
Relate shear force and bending moment to loading (brief).
Model answer
$\frac{dV}{dx} = -w$ , $\frac{dM}{dx} = V$ (sign convention dependent). Bending stress depends on $M(x)$ ; transverse shear stress depends on $V(x)$ .

Exams & GATE

1
Textbook: RK Bansal — bending stresses / shear force & BM.
2
Always locate the NA (centroid), use correct $I$ about NA, and take $y$ to the fibre of interest.
3
GATE favourites: flitched beams, unsymmetric sections about NA, and which fibre has max tensile/compressive stress.

📖 Standard books (India)

Strength of Materials — RK Bansal
Read: Ch. 9–10
SOM — beams, torsion, columns, and deflection

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📚Key formulas & points

Topic details

Definition and physical meaning

Core assumptions (theory of pure bending)

Derivation summary

Section modulus and common I values

Shear force, BM, and combined loading

Step-by-step problem approach

Common mistakes in exams

Bending stress (circular section)

Worked examples

Rectangular beam — max bending stress

Section modulus design

Eccentric axial load

Practice questions

📝Exams & GATE

📖 Standard books (India)

Explore related topics

See real mechanical engineering careers

Key formulas & points

Exams & GATE