Structural Engineering

Bending & Shear Stresses

Flexure formula, transverse shear distribution and torsion — converting beam moments and shears into the fibre stresses that govern design.

PART 1

Topic Breakdown & Traps

The Engineering Principle

When a beam bends, fibres above the neutral axis (which passes through the centroid) shorten and those below lengthen, producing a linear bending-stress distribution

σ = M y / I

— maximum at the extreme fibre. The transverse shear force is carried by a parabolic shear-stress distribution

τ = V Q / I b

that is maximum at the neutral axis and zero at the top/bottom. In circular shafts a torque produces shear

τ = T r / J

The Core Formula Matrix

Flexure formula:

\frac{σ}{y} = \frac{M}{I} = \frac{E}{R} \Rightarrow σ = \frac{M y}{I}

Rectangle:

I = \frac{b d ^{3}}{12}, Z = \frac{b d ^{2}}{6}, σ_{ma x} = \frac{M}{Z}

Transverse shear:

τ = \frac{V Q}{I b}

; max (rectangle)

τ_{ma x} = 1.5 \frac{V}{A}

, (circle)

\frac{4}{3} \frac{V}{A}

Torsion:

\frac{τ}{r} = \frac{T}{J} = \frac{Gθ}{L}, J = \frac{π d ^{4}}{32}

The ‘IIT Traps’

⚠
Bending max at extreme fibre, shear max at the neutral axis. They occur at different points of the section — never the same fibre.
⚠
**Section modulus uses $d^{2}$ , second moment uses $d^{3}$ .** Mixing $Z = b d^{2} /6$ with $I = b d^{3} /12$ is a frequent slip.
⚠
Shear factor. Average shear $V / A$ must be multiplied by $1.5$ (rectangle) or $4/3$ (circle) to get the peak.

📚 Standard references

Strength of Materials (Mechanics of Materials) — B.C. Punmia
Mechanics of Materials — Gere & Timoshenko

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · NAT

A rectangular beam

200 mm \times 400 mm

(b×d) carries a bending moment of

80 kN\cdotm

. The maximum bending stress is _____ MPa.

Q2MEDIUM2 Marks · NAT

A rectangular beam

200 mm \times 300 mm

carries a shear force of

90 kN

. The maximum transverse shear stress is _____ MPa.

Q3HARD2 Marks · MCQ

For a solid circular section under transverse shear, the ratio of maximum to average shear stress is: