Beam Deflection:

Double Integration Method

Dosen: Dr. Djwantoro Hardjito
Jurusan Teknik Sipil
Fakultas Teknik Sipil & Perencanaan
Universitas Kristen Petra Surabaya

Materials and figures are mostly taken from “Mechanics of Materials”, by R.C. Hibbeler, 7 th SI Edition, Prentice Hall.

WHAT are Elastic Deformations

 Elastic deformations (only) of structures will
be considered
 Plastic deformation or inelastic deformations
are permanent deformations after loads are
removed
 Linear elastic refers to:
 Stresses proportional to strain
 Linear variation in stress from the neutral axis of
the beam to its extreme fibres
 Members will return to their original geometry
after loads are removed

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Sources of Deflection

 Deflections may be caused by :

 Loads
 Temperature
 Fabrication errors
 Settlement

Sources of Deflection
 Bending moment induced deflection
 Largest for beams and frames
 Shear force induced deflection
 For most beams shear force deflections are small and
ignored
 Span to depth ratio of 1/12 to 1/6 typically have deflections due to
shear around 1% to 8%
 Deeper beams, say span to depth ratio of ¼, have more significant
shear deflection, maybe up to 18%1
 Axial force induced deflection
 Largest for trusses

1. Reference Wang, in Nelson and McCormac

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WHY calculate deflections?

 Excessive deflection is undesirable:

 Unsightly and impede use
 Vibrations or sway may be excessive causing user fear or
physical effects such as nausea
 Attached materials, especially if brittle, may crack or be
damaged (ceilings, floors, tiles, plaster, windows, cladding)
 Large deflections may increase the load on the
structure
 Water ponding
 Increased eccentricity of loads
 Necessary for analysis of indeterminate structures

HOW Methods of Computation

 Geometric
 Based on geometry of the deflected shapes of structures
 Direct/Double integration
 Moment area method
 Conjugate beam method
 Work-energy
 Principles of work and energy (and conservation of energy)
 Virtual work – Unit load method
 Castigliano’s 1st and 2nd theorems

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 Considering the flexural deformation of a
beam element as before:

 The curvature k of the beam can be expressed in relationship

to the radius of curvature  by a calculus function:

1 M
Substituting  gives the expression:
 EI

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 This equation is non linear second order differential, the
solution of which gives v =f(x) where v is the vertical
displacement thereby giving the exact shape of the elastic
deflection curve (for all deflections caused by bending alone)
note y = v in some texts
 Since slope  for most structures is very small compared to
unity, the denominator is simplified to 1.0, and the equation is :

d dv d
 [ ]
dx dx dx
M
dO  dx
EI

Integrating:

Gives:
Where  is the slope of
the deflected structure
Since slope is related to
deflection by:

Integrating again gives:

Where y is the deflection
of the structure

5
 West and Geschwindner

West and Geschwindner

-M
If drawn on tension
side of beam –
Negative BM on top
surface

6
Positive sign conventions

Positive sign conventions- the negative

sign (in front of M) is introduced to
account for the sign convention of
downwards deflection positive and
sagging moment positive

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Direct Integration Method
 Write the expression for M/EI (bending moment
divided by the flexural rigidity of the beam) in terms
of distance x along the axis of the beam
 Integrate to obtain equations for the slope of the
beam
 Integrate to obtain the deflection of the beam
 Use boundary conditions to solve for the constants
of integration

Integration Method

Convenient if M/EI can be expressed as single

continuous function over the length of the beam
Can be complicated for structures for which M/EI is
not continuous
At each discontinuity (due to change in loading or EI) two
additional constants of integration in the analysis are
introduced which must be determined from boundary
conditions
Can be overcome by singularity functions – Macaulay’s Method

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 Hibbeler
Boundary conditions

 = 0 deflection
  0 slope

 = 0 deflection
  0 slope

 = 0 deflection
 = 0 slope

 = 0 deflection
  0 slope

 = 0 deflection
  0 slope

Continuity of slope at roller or pin can be used

to find constants of integration

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Example: Determine displacement in the centre of the
beam at B, assume A is a pin and C is a roller and EI
is constant.

Hibbeler

Example: find the slope and deflection at the

free end of the cantilever

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Determine the deflection and slope at B

90 kN

Kassimali

Determine the slope and deflection at B by

the direct integration method

Kassimali

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To do: use the integration method to
determine the expressions for slope and
deflection throughout each of following
structures for the indicated loading, EI =
constant

West and Geschwindner

Kassimali

12
 determine the slope and displacement at A, assume B
is a fixed support and EI is constant

Hibbeler

13
 Determine the equation of the elastic curve for the
simply supported beam which is subject to the couple
moments as shown. Find the maximum slope and
deflection of the beam. EI is constant.

Hibbeler

Macaulay’s Method
 For discontinuous loading need a separate bending
moment expression for each part of the beam
 Therefore to obtain the deflection function double
integration has to be performed on the BM
expression for each section of the beam
 Constants of integration need to be found for each
section, for each integration!
 For example:

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15
 Process is tedious therefore use mathematical technique
termed step function or singularity function
 Known as Macaulay’s method when applied to beam
solutions
One bending moment expression is written for the right
hand end of the beam to cover the bending moment
conditions for the whole of the beam
Only two unknown constants of integration need to be
found

Step Function
 The step function is written in the form:
 fn(x) = [x – a ] n
 Such that for
 x < a, fn(x) = 0 and
 x > a, fn(x) = (x - a) n

 Square bracket used to indicate step function

 Curved bracket indicates normal mathematical function

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Important features of the step
function:
 If on substituting a value for x, the quantity inside the
square brackets becomes negative, it is omitted from further
analysis
 Square bracket terms must be integrated to preserve the
identity of the bracket, e.g.

1
 [ x  a]2dx  3 [ x  a]3
[ ]n + 1
 [ ]ndx 
(n  1)

 For mathematical continuity, distributed loading that

does not extend to the right hand end of the beam
must be arranged to continue to x = L (where L is
the total length of the beam), whether the distributed
loading starts from x = 0 (left hand end) or x = a
(arbitrary point along the beam)
 The continuity of the UDL to the end can be effected
by superposition of loading which cancel each other
in the required portions of the beam.
 For example:

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 Zero UDL in CD

 If an externally applied moment Mo exisits, it must be

expressed as a step function:
 Mo[x-a]0

The three common step functions for bending

moment are:

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Example beam with central point
load
 Reactions from symmetry RA = RC
= P/2
 Section at “x” at far right hand end
of beam
 Take moments about x clockwise
x positive:

P L
( x)  P( x  )  Mx  0
2 L
Standard step function
Mx  RA.x  P[ x  a]1
P L
Mx  [ x]1  P[ x  ]1
Sub for RA 2 2

d 2v
EI 2   M
dx
d 2v  P 1 L
EI 2  [ x]  P[ x  ]1 integrate
dx 2 2
dv  P 2 P L 2
EI  [ x]  [ x  ]  c1 integrate
dx 4 2 2
P 3 P L
EIv  [ x]  [ x  ]3  c1 x  c2
12 6 2

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 Deflection equation:
P 3 P L
EIv  [ x]  [ x  ]3  c1 x  c2
12 6 2
Boundary condition at pin A:
x  0, v  0
0
P 3 P L 3
0 [ x]  [ x  ]  c1 x  c2 First two terms
12 6 2
become negative
c2  0
therefore zero

P 3 P L
EIv  [ x]  [ x  ]3  c1 x  0
12 6 2
Boundary condition at pin B
x  L, v  0
P 3 P L
0 [ L]  [ L  ]3  c1 L  0
12 6 2
3
P PL
c1 L  [ L]3  Take care with signs
12 6 8 when re-arranging
3PL 3

c1 L 
48
3PL2
c1 
48

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FOR DEFLECTION AT ANY POINT ALONG BEAM, sub constants of integration

P 3 P L 3PL2
EIv  [ x]  [ x  ]3  x
12 6 2 48

L
sub x  Location of max. deflection by inspection (mid span) or where
2 slope =0
P L 3 3PL2 L
EIv  [ ] 0 .
12 2 48 2
PL3
EIv  Standard formula - check
48

Example – simply supported beam with

concentrated load

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Example- simply supported beam with
distributed load on part of the span

Example: calculate the magnitude of

deflection at the free end of the cantilever
take EI = 1000 kNm2

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Find the deflection function for the following beam
and determine the deflection value at a location 1.8 m
from the left hand end; take E = 70 GPa and I = 8.32
x 106 mm4

solution: deflection is 25 mm  and deflection function is:

y
1
{ 11 3x  1.5[ x  1.5]2  8 [ x  1.5] 4 10 [ x  2.4] 3 14 x}
EI 6 12 3

Determine the deflection at D, E = 200 kN/mm2 and I

= 1000 x 10 6 mm4
solution (-0.45mm ↑)

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Calculate the slope and the deflection at C, E = 200
kN/mm2 and I = 500 x 10 6 mm4
solution (0.0045 radians, 22 mm ↓)

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