Matakuliah
Tahun
: S0725 – Analisa Struktur
: 2009
Deflection of Indeterminate Structure
Session 03-04
Contents
•Elastic Beam Theory
•Moment area
•Conjugate Beam Method
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Introduction
What is deflection?
Deflection can occur from various causes,
such as loads, settlement,
temperature or fabrication error of
material.
Deflection must be limited in order to
prevent cracking and damaging of structure.
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Introduction
What is caused of structure
deflection?
In general It caused by its internal
loading such as Normal
force,
shear force or bending
moment
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Introduction
What is caused of Beam & Frames
deflection?
It caused by internal
bending
What is caused of truss
deflection?
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Introduction
Deflection Diagram represent the
Elastic Curve for the points at
the centroids of cross-sectional area
along the members
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Introduction
Deflection on supports :
(1) Roller 
D=0
A
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Introduction
Deflection on supports :
(2) Pin 
D=0
A
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Introduction
Deflection on supports :
(3) Fixed Support 
D=0 ;q=
0
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Introduction
Deflection on supports :
(4) Fixed Connected Joint  causes the
joint to rotate the members by the
same amount of q
q
q
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Introduction
Deflection on supports :
(5) Pin Connected Joint  the members will
have a different slope or rotation at pin,
since the pin can’t support moment
q1
q2
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Introduction
Sign Convention 
Bending Moment
M+
M+
Longitudinal axis
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Introduction
Sign Convention 
Bending Moment
M-
MLongitudinal axis
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Introduction
P2
Deflection curve :
A
B
P1
+
Moment
-
A
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B
Deflection curve
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P
x
P
y
Elastic curve
The deflection is measured from the original neutral axis to the neutral axis of the
deformed beam.
The displacement y is defined as the deflection of the beam.
It may be necessary to determine the deflection y for every value of x along the
beam. This relation may be written in the form of an equation which is frequently
called the equation of the deflection curve (or elastic curve) of the beam
Importance of Beam Deflections
A designer should be able to determine deflections, i.e.
In building codes ymax <=Lbeam/300
Analyzing statically indeterminate beams involve the use of various deformation relationships.
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Elastic Beam Theory
P
x
P
y
Elastic curve
The deflection is measured from the original neutral axis to the neutral axis of the
deformed beam.
The displacement y is defined as the deflection of the beam.
It may be necessary to determine the deflection y for every value of x along the
beam. This relation may be written in the form of an equation which is frequently
called the equation of the deflection curve (or elastic curve) of the beam
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Elastic Beam Theory
Importance of Beam Deflections
A designer should be able to determine deflections, i.e.
ymax <=Lbeam/300
Analyzing statically indeterminate beams involve
the use of various deformation relationships.
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Double-Integration Method
d2y
EI 2  M
The deflection curve of the bent beam isdx
In order to obtain y, above equation needs to be integrated twice.
y
r
y
M
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EI
r

Radius of
curvature
x
1 M

  (Curvature )
r EI
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Double-Integration Method
An expression for the curvature at any point along
the curve representing the deformed beam is
readily available from differential calculus. The
exact formula
2 for the curvature is
d y

dx 2
  dy 
1   
  dx 
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2



3
2
dy
is small
dx
d2y
  2
dx
d2y
EI 2  M
dx
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Double-Integration Method
The Integration Procedure
Integrating once yields to slope dy/dx at any point in the beam.
Integrating twice yields to deflection y for any value of x.
The bending moment M must be expressed as a function of the coordinate x
before the integration
Differential equation is 2nd order, the solution must contain two constants of
integration. They must be evaluated at known deflection and slope points
(i.e. at a simple support deflection is zero, at a built in support both slope
and deflection are zero)
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Double-Integration Method
Sign Convention
Positive Bending
Negative Bending
Assumptions and Limitations
Deflections caused by shearing action negligibly small compared to
bending
Deflections are small compared to the cross-sectional dimensions of
the beam
All portions of the beam are acting in the elastic range
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Beam is straight prior to the application of loads
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Moment Area
First Moment –Area Theorem
The first moment are theorem states
that: The angle between the tangents
at A and B is equal to the area of the
bending moment diagram between
these two points, divided by the
product EI.
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Moment Area
r
dq
A
B
ds
q
dx
dq
dq 
EI
r
M
ds
EI
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D
B
M
q
dx
EI
A
x
M
M
Mx
D
dx
A EI
B
ds  rdq  r 
B
M
dq 
dx integratin g will give
EI
ds
dq
Mx
xdq 
dx
EI
M
dx
EI
A
q   dq  
B

Mx
dx
EI
A
D
it is small lateral deflection s replace ds with dx
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Moment Area
Second Moment –Area Theorem
The second moment area theorem states that: The vertical distance of point B
on a deflection curve from the tangent drawn to the curve at A is equal to the
moment with respect to the vertical through B of the area of the bending
diagram between A and B, divided by the product EI.
Mx
dx
A EI
D
EI
M
r
ds
ds  rdq  r 
dq
M
dq 
dx integratin g will give
EI
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B
dq 
M
ds
EI
it is small lateral deflection s replace ds with dx
B
M
dx
EI
A
q   dq  
Mx
xdq 
dx
EI
B

Mx
dx
EI
A
D
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Procedure
Moment Area
1. The reactions of the beam are determined
2. An approximate deflection curve is drawn. This curve must
be consistent with the known conditions at the supports,
such as zero slope or zero deflection
3. The bending moment diagram is drawn for the beam.
Construct M/EI diagram
4. Convenient points A and B are selected and a tangent is
drawn to the assumed deflection curve at one of these
points, say A
5. The deflection of point B from the tangent at A is then
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calculated by the second moment area theorem
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Moment Area
Problem
PL
A
L
Tangent at A
q
P
P
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B
D?
Tangent at B
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Moment Area
PL
A
L
Tangent at A
q
D?
P
P
B
Tangent at B
M
PL
L
PL
 2L 
EI D   PL   
2
3
 3 
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EI q 
L
 PL 
2
3
PL 3
D
3EI
PL2
q
2 EI
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Moment Area
A
W N per unit length
Tangent A
D=?
WL2
2 WL
1 WL2
A
L
3 2
3
x L
4
L
B
x
WL2
2
L  W 2  3 
WL4
EI D    L   L   
3  2  4 
8
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WL 4
D
8EI
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Moment Area
Example
a
P
P
a
D=?
A
P
L
P
Tangent A
Pa
L
a
2
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a
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Moment Area
 L2 La La a 2  P 3
L
 L a
 Paa 2
EI D  Pa  a    a  
a  Pa 

   a
2 3
4
4
2 3
2
 4 2

8
PaL2 Pa3 PL3  3a 4a 3 



  3 
8
6
24  L
L 
PL3  3a 4a 3 
D
  3 
24 EI  L
L 
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Conjugate Beam
The method requires the same amount
of computation as the moment-area
theorems to determine a beam’s slope
or deflection. However, the method
relies only on the principles of
statics, its application will be more
familiar
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Dr Yan Zhuge lecturer notes
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Using the similarity of equations for
Beam Statics
Or integrating
V    wdx
M
q    ( ) dx
EI
Unit = kN·m2/EI
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Beam deflection


M     wdx dx
M


v     (
) dx  dx
EI


Unit = kN·m3/EI
Dr Yan Zhuge lecturer notes
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Conjugate Beam
Theorem 1:
The slope at a point in the real beam is
the shear at the corresponding point in the
Theorem 2:
The displacement of a point in the real
equal to the moment at the corresponding
conjugate beam.
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Dr Yan Zhuge lecturer notes
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Conjugate Beam
•Draw the conjugate beam for the real beam with a proper
boundary conditions
•Load the conjugate beam with the real beam’s M/EI
diagram. This loading is directed downward when M/EI is
positive and upward when M/EI is negative
•Determine the statics of the conjugate beam: reactions,
Shear force and moments
•Shear force V corresponds to the slope q of the real beam,
moment M corresponds to the displacement v of the real
beam.
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Dr Yan Zhuge lecturer notes
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Conjugate Beam
REAL BEAM
CONJUGATE BEAM
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Dr Yan Zhuge lecturer notes
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Conjugate Beam
REAL BEAM
CONJUGATE BEAM
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Conjugate Beam
REAL BEAM
CONJUGATE BEAM
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Conjugate Beam
REAL BEAM
+
+
CONJUGATE BEAM
+
+
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Conjugate Beam
Determine the maximum deflection of
the steel beam shown in the figure. E =
200 GPa, I = 60(106) mm4.
8 kN
B
A
x
9m
2 kN
3m
6 kN
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Dr Yan Zhuge lecturer notes
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Conjugate Beam
8 kN
B
A
x
Real Beam
18kNm
2 kN
81/EI
18/EI
27/EI 6 kN
B’
A’
x
45/EI
Maximum deflection
occurs at the point
where the slope is zero
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63/EI
9m
3m
Conjugate Beam
This corresponds to the
same point in the
conjugate beam where
the shear is zero
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