Structural Engineering

Deflection of Beams

Double-integration, moment-area and unit-load methods plus the standard deflection results examiners expect you to recall instantly.

PART 1

Topic Breakdown & Traps

The Engineering Principle

Beam deflection follows from

E I \frac{d ^{2} y}{d x ^{2}} = M (x)

: integrate the bending-moment expression twice and apply boundary conditions for slope and deflection. For exams the standard cases (cantilever and simply-supported under point load or UDL) should be memorised, while the unit-load (virtual-work) method handles deflection at any point of any determinate structure.

The Core Formula Matrix

Governing equation:

E I y^{''} = M (x)

**Cantilever, end load

P

**:

δ = \frac{P L ^{3}}{3 E I}, θ = \frac{P L ^{2}}{2 E I}

**Cantilever, UDL

w

**:

δ = \frac{w L ^{4}}{8 E I}

**SS beam, central load

P

**:

δ = \frac{P L ^{3}}{48 E I}

**SS beam, UDL

w

**:

δ = \frac{5 w L ^{4}}{384 E I}

Unit-load method:

δ = \int \frac{M m}{E I} d x

The ‘IIT Traps’

⚠
Unit consistency. Working in $kN$ and $m$ gives $δ$ in metres — convert to mm at the end, not midway.
⚠
Right formula for the load. $P L^{3} /48 E I$ is a central point load; $5 w L^{4} /384 E I$ is a UDL — don't interchange.
⚠
Cantilever vs simply supported. A cantilever is far more flexible ( $L^{3} /3 E I$ vs $L^{3} /48 E I$ ) for the same span and load.

📚 Standard references

Elementary Structural Analysis — Wilbur & Norris
Basic Structural Analysis — C.S. Reddy

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · NAT

A cantilever of span

3 m

(

E I = 8000 kN\cdotm^{2}

) carries an end point load of

10 kN

. The free-end deflection is _____ mm.

Q2MEDIUM2 Marks · NAT

A simply supported beam of span

4 m

(

E I = 10000 kN\cdotm^{2}

) carries a central point load of

20 kN

. The mid-span deflection is _____ mm.

Q3HARD2 Marks · MCQ

Which method is most directly suited to finding the deflection at a single specified point of a determinate truss?