Slope & Deflection of Beams Notes

Last reviewed 16 Sept 2026 · 8 min read

Why deflection matters

A beam can be strong enough and still be unserviceable: excessive sag cracks plaster and partitions, causes ponding on roofs, misaligns machinery and alarms occupants. Design therefore checks deflection as well as stress. Deflection analysis is also the basis of solving statically indeterminate beams.

Deflection $y$ — vertical displacement of a point on the neutral axis.
Slope $θ = d y / d x$ — rotation of the tangent to the elastic curve (radians).
Elastic curve — the deflected shape of the neutral axis.

Differential equation of the elastic curve

From the bending equation, curvature $= 1/ R = M / E I$ . For small slopes $1/ R \approx d^{2} y / d x^{2}$ . Taking sagging moment positive and deflection $y$ positive upward:

FormulaGoverning equation

E I \frac{d ^{2} y}{d x ^{2}} = M, E I \frac{d ^{3} y}{d x ^{3}} = V, E I \frac{d ^{4} y}{d x ^{4}} = - w

Integrate once → slope $E I θ$ ; integrate twice → deflection $E I y$ . Constants come from boundary conditions.

Support	Boundary conditions
Simple support / hinge	$y = 0$
Fixed end	$y = 0$ and $θ = 0$
Free end	$M = 0$ and $V = 0$ (used when working from $w$ )
Point of symmetry (symmetric loading)	$θ = 0$

Double integration method

Write $M (x)$ for the whole span, integrate twice, apply boundary conditions. Straightforward when one expression for $M$ covers the beam (UDL over full span, end loads).

Cantilever with end load $W$ (fixed at $x = L$ , free at $x = 0$ ): $M = - W x$ .

$E I y^{'} = - \frac{W x ^{2}}{2} + C_{1}$ ; at $x = L$ , $y^{'} = 0$ → $C_{1} = \frac{W L ^{2}}{2}$

$E I y = - \frac{W x ^{3}}{6} + \frac{W L ^{2} x}{2} + C_{2}$ ; at $x = L$ , $y = 0$ → $C_{2} = - \frac{W L ^{3}}{3}$

At the free end ( $x = 0$ ): $y = - \frac{W L ^{3}}{3 E I}$ and $θ = \frac{W L ^{2}}{2 E I}$ .

Macaulay's method

When loads are discontinuous (point loads part-way along, partial UDLs), Macaulay writes one moment equation using brackets $[x - a]$ that are taken as zero when negative. Rules:

Measure $x$ from one end, taking a section in the last segment so all loads appear.
Write each term as $[x - a]^{n}$ and integrate the bracket as a whole: $\int [x - a] d x = [x - a]^{2} /2$ .
A UDL that stops before the end must be extended to the end and cancelled by an equal upward load.
A couple $M_{0}$ at $x = a$ is written $M_{0} [x - a]^{0}$ .

Moment-area method (Mohr's theorems)

Draw the $M / E I$ diagram.

FormulaMohr's theorems

Theorem I: The change in slope between A and B equals the area of the $M / E I$ diagram between A and B.

θ_{B} - θ_{A} = \int_{A}^{B} \frac{M}{E I} d x

Theorem II: The deviation of B from the tangent drawn at A equals the moment of the $M / E I$ diagram between A and B, taken about B.

t_{B / A} = \int_{A}^{B} \frac{M}{E I} x_{B} d x

Very efficient for cantilevers (tangent at the fixed end is horizontal) and symmetric simply supported beams (tangent at mid-span is horizontal).

Exam TipAreas and centroids you will need

Rectangle: area $bh$ , centroid at $b /2$ .
Triangle: area $bh /2$ , centroid at $b /3$ from the vertical side.
Parabolic spandrel (cantilever UDL diagram, vertex at the free end): area $bh /3$ , centroid at $3 b /4$ from the vertex ( $b /4$ from the largest ordinate).
Parabolic segment (simply supported UDL, half span): area $2 bh /3$ , centroid at $3 b /8$ from the maximum-ordinate end.

Conjugate beam method

Load an imaginary conjugate beam (same span) with the $M / E I$ diagram as its load. Then:

Slope at a point of the real beam = shear force at that point of the conjugate beam.
Deflection at a point of the real beam = bending moment at that point of the conjugate beam.

Real beam support	Conjugate beam support
Fixed end	Free end
Free end	Fixed end
Simple end support	Simple end support
Internal hinge	Internal support (roller)
Internal support	Internal hinge

Superposition

Within the elastic range, the slope or deflection due to several loads is the sum of those due to each load alone. Combined with the standard table below, this solves most exam problems in a few lines.

Standard results

Cantilever (span $L$ ; values at the free end)

Loading	Slope $θ$	Deflection $y$
Point load $W$ at free end	$\frac{W L ^{2}}{2 E I}$	$\frac{W L ^{3}}{3 E I}$
Point load $W$ at distance $a$ from fixed end	$\frac{W a ^{2}}{2 E I}$	$\frac{W a ^{3}}{3 E I} + \frac{W a ^{2}}{2 E I} (L - a)$
UDL $w$ over whole span	$\frac{w L ^{3}}{6 E I}$	$\frac{w L ^{4}}{8 E I}$
UVL, zero at free end, $w$ at fixed end	$\frac{w L ^{3}}{24 E I}$	$\frac{w L ^{4}}{30 E I}$
UVL, $w$ at free end, zero at fixed end	$\frac{w L ^{3}}{8 E I}$	$\frac{11 w L ^{4}}{120 E I}$
Couple $M$ at free end	$\frac{M L}{E I}$	$\frac{M L ^{2}}{2 E I}$

Simply supported beam (span $L$ )

Loading	End slope	Maximum deflection
Central point load $W$	$\frac{W L ^{2}}{16 E I}$	$\frac{W L ^{3}}{48 E I}$ at centre
UDL $w$ over whole span	$\frac{w L ^{3}}{24 E I}$	$\frac{5 w L ^{4}}{384 E I}$ at centre
Point load $W$ at $a$ from A, $b$ from B	$θ_{A} = \frac{W b ( L ^{2} - b ^{2} )}{6 E I L}$	under load: $\frac{W a ^{2} b ^{2}}{3 E I L}$
Couple $M$ at end A	$θ_{A} = \frac{M L}{3 E I}$ , $θ_{B} = \frac{M L}{6 E I}$	$\frac{M L ^{2}}{9 3 E I}$
Triangular load, zero at A to $w$ at B	—	$\approx 0.00652 \frac{w L ^{4}}{E I}$ (centre: $\frac{5 w L ^{4}}{768 E I}$ )

Fixed and propped beams

Beam and loading	Maximum deflection
Fixed both ends, central load $W$	$\frac{W L ^{3}}{192 E I}$
Fixed both ends, UDL $w$	$\frac{w L ^{4}}{384 E I}$
Propped cantilever, UDL $w$ (prop reaction $3 w L /8$ )	$\frac{w L ^{4}}{185 E I}$
Propped cantilever, central load $W$ (prop reaction $5 W /16$ )	$\frac{W L ^{3}}{107 E I}$ (approx.)

RememberRatios that are asked

Simply supported vs fixed, UDL: $5/384$ vs $1/384$ → fixed beam deflects 1/5.
Simply supported vs fixed, central load: $1/48$ vs $1/192$ → fixed beam deflects 1/4.
Cantilever UDL vs cantilever end load of same total $W = w L$ : $W L^{3} /8 E I$ vs $W L^{3} /3 E I$ .
Doubling the span of a simply supported beam with UDL multiplies deflection by 16.
Doubling the depth of a rectangular beam reduces deflection to 1/8.

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Slope & Deflection of Beams

Why deflection matters

Differential equation of the elastic curve

Double integration method

Macaulay's method

Moment-area method (Mohr's theorems)

Conjugate beam method

Superposition

Standard results

Cantilever (span $L$ ; values at the free end)

Simply supported beam (span $L$ )

Fixed and propped beams

This chapter is in the syllabus of

How ready are you for Strength of Materials?

Why deflection matters

Differential equation of the elastic curve

Double integration method

Macaulay's method

Moment-area method (Mohr's theorems)

Conjugate beam method

Superposition

Standard results

Cantilever (span L; values at the free end)

Simply supported beam (span L)

Fixed and propped beams

This chapter is in the syllabus of

How ready are you for Strength of Materials?

Cantilever (span $L$ ; values at the free end)

Simply supported beam (span $L$ )