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Deflections in beams Dr Alessandro Palmeri Senior Lecturer in Structural Engineering <[email protected]>
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Learning Outcomes When we have completed this unit (3 lectures + 1
tutorial), you should be able to: ◦ Use the double integration technique to
determine transverse deflections in slender beams under distributed and/or concentrated loads
Schedule: ◦ Lecture #1: Double integration method ◦ Lecture #2: Macaulay’s notation ◦ Lecture #3: Numerical application ◦ Tutorial
DOUBLE INTEGRATION METHOD
Lecture #1
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Introduction
Structural members must have: ◦ Strength (ULS: Ultimate Limit State) ◦ Stiffness (SLS: Serviceability Limit State)
Need to limit deflection because: ◦ Cracking ◦ Appearance ◦ Comfort
Engineering Structures, Volume 56, 2013, 1346 - 1361
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Introduction Standards typically limit deflection of beams by fixing
the maximum allowable deflection in terms of span: ◦ e.g. span/360 for steel beams designed according to
Eurocode 3
Deflections in beams may occur under working loads, where the structure is usually in the linear elastic range
Theyare therefore checked using an elastic analysis ◦ no matter whether elastic or plastic theory has been used
in the design for strength We’ll introduce some basic concepts of plastic analysis for ductile
beams in bending later this semester
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Introduction Many methods are available for calculating
deflection in beams, but broadly speaking they are based on two different approaches
a) Differential equation of beams in bending This approach will be considered in this module
b) Energy methods e.g. Virtual Work Principle
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Curvature
From the simple theory of bending we have:
where ◦ E is the Young’s modulus of the material ◦ I is the second moment of area ◦ 1/R is referred as beam’s curvature
1R= ME I
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Curvature For a plane curve uz(x) in the xz plane, the curvature 1/Ry
(about the orthogonal axis y) is given by:
If duz/dx is small, then (duz/dx)2 can be considered negligible
Thus:
And so:
1Ry
=
d2uz
dx2
1+ duzdx
⎛⎝⎜
⎞⎠⎟
2⎡
⎣⎢⎢
⎤
⎦⎥⎥
M y
E I yy
=d2uz
dx2
1Ry
≅d2uz
dx2
x
z
Ry y
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Sign convention Mostly vertical loads act vertically ◦ Downward deflection uz is +ve
Already chosen bending moment convention ◦ Sagging moment My is +ve
We must reconcile these two choices:
x z
load
x z duz
dx> 0
slope
x z d2uz
dx2> 0
curvature
But this is the shape of hogging bending moment, i.e. My<0
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Differential equation of slender beams in bending Taking into account the correct sign convention
for deflection and bending moment, we have:
◦ This is the starting point of the double integration method, which enables one to evaluate slope duz/dx and deflection uz in a slender beam in bending
◦ Note that in the above equation: Iyy means second moment of area about the horizontal axis y My means bending moment about the same axis (depends on x) uz is the vertical deflection (also depends on x)
E I yy
d2uz (x)dx2 = −M y (x)
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Double integration method The differential equation of beams in bending
must be integrated twice with respect to the abscissa x ◦ The minus sign in the right-hand side is crucial
Since the bending moment My usually varies along the beam, therefore we need to write the mathematical expression of My=My(x)
As we are solving a 2nd-order differential equation, 2 integration constants, C1 and C2, will arise
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Boundary conditions The integration constants C1 and C2 are
determined from the known boundary conditions, i.e. conditions at the supports
Simple support No deflection
uz=0
Fixed support No deflection and no slope
uz=0 and duz/dx=0
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Worked example Determine deflection and slope at the free
end B of a cantilever beam of length L subjected to a uniformly distributed load qz ◦ subscript z means that the load acts vertically
x
RA
MA
qz
A B z
L
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Worked example 1st, determine the
support’s reactions:
ΣV = 0
⇒ RA − qz L = 0⇒ RA = qz L ()
Σ M (A) = 0
⇒ MA − qz LL2= 0
⇒ MA =qz L
2
2()
x
RA
MA
qz
A B z
L
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Worked example 2nd, write down the
expression of the bending moment My as a function of the abscissa x along the beam’s axis:
My = −MA + RA x −
qz x2
2
= −qz L
2
2+ qz L( )x − qz x
2
2
x
RA
MA
qz
A B z
L
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Worked example The differential equation for the beam’s
deflection reads:
3rd, we can integrate twice:
E I yy
d2uz
dx2 = −M y =qz L2
2− qz L x +
qz x2
2
E I yyduzdx
=qz L
2
2x −qz L x
2
2+qz x
3
6+C1
E I yy uz =qz L
2
4x2 −
qz L x3
6+qz x
4
24+C1 x +C2
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Worked example 4th, the known boundary conditions at the fixed support (i.e.
no deflection and no slope at left-hand side end A):
Substituting now the values of the integration constants C1 and C2, the expressions for slope and deflection throughout the beam become:
duzdx
= 0@ x = 0 ⇒ C1 = 0
uz = 0@ x = 0 ⇒ C2 = 0
duzdx
= 1E I yy
qz L2
2x −qz L x
2
2+qz x
3
6⎛
⎝⎜⎞
⎠⎟
uz =
1E I
qz L2
4x2 −
qz L x3
6+
qz x4
24⎛
⎝⎜⎞
⎠⎟
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Worked example 5th, intuitively we
know that slope and deflection in the cantilever beam take the maximum values at the free end B
By substituting x=L in the general expression of the slope along the beam, we get:
duzdx
⎛⎝⎜
⎞⎠⎟ B
= 1E I yy
qz L2
2L −
qz L L2
2+qz L
3
6
⎛
⎝⎜
⎞
⎠⎟ =
qz L3
6E I yy(> 0,)
x
RA
MA
qz
A B z
L
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Worked example
Similarly, by substituting x=L in the general expression of the deflection, we have:
yB =1E I yy
qz L2
4L2 −
qz L L3
6+qz L
4
24⎛
⎝⎜⎞
⎠⎟
= 6− 4+124
qz L4
E I yy= 18qz L
4
E I yy(> 0,)
x
RA
MA
qz
A B z
L
MACAULAY’S NOTATION
Lecture #2
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Beams under point loads
E.g. simply supported beam with a single concentrated load
x
z 6 m
4 m 2 m
Fz
RA RB
A B
C
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Beams under point loads
Support reactions
Σ M (A) = 0
⇒ − Fz 2+ RB 6 = 0
⇒ RB =2Fz6
=Fz3()
Σ M (B) = 0
⇒ − RA 6+ Fz 4 = 0
⇒ RA =4Fz6
= 23Fz ()
x
z 6 m
4 m 2 m
Fz
RA RB
A B
C
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Beams under point loads
In principle, we need two expression for the bending moment My:
◦ one for 0<x<2
◦ one for 2<x<6
My = RA x
My = RA x − Fz x − 2( )
0<x<2
RA
A
2<x<6 RA
Fz
2
A C
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Beams under point loads In principle, we need to integrate two
differential equations:
This is possible, but four integration constants arise, i.e. two for each differential equation ◦ For more than one points load, the procedure
becomes quite cumbersome
E I yyd2uzdx2
=RA x , 0 < x < 2
RA x − Fz x − 2( ) , 2 < x < 6⎧⎨⎪
⎩⎪
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Macaulay’s notation It would be much more effective to have a single
mathematical expression for the bending moment My along the beam
This is possible with the help of the so-called Macaulay’s notation, i.e. square brackets [ ] with a special meaning:
◦ If the term within square brackets is +ve, then it is evaluated
◦ If the term within square brackets is –ve, then it is ignored
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Macaulay’s notation
That is:
Let’s try the following examples:
[ ] , if 00 , if 0x x
xx>⎧
= ⎨ ≤⎩
2.3⎡⎣ ⎤⎦ = 2.3 0⎡⎣ ⎤⎦ = 0 −3 / 4⎡⎣ ⎤⎦ = 0
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Macaulay’s notation
It is possible now to write down a single expression for the bending moment along the beam:
My = RA x − Fz x − 2⎡⎣ ⎤⎦
x
z 6 m
4 m 2 m
Fz
RA RB
A B
C
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Macaulay’s method The differential equation of bending becomes:
This expression can be integrated twice, importantly, without expanding the term into square brackets:
E I yyd2uzdx2
= −My = −RA x + Fz x − 2⎡⎣ ⎤⎦
E I yy uz = −Fz x
3
9+ Fz
x − 2⎡⎣ ⎤⎦3
6+C1 x +C2
E Idydx
= −RAx2
2+W
x − 2⎡⎣ ⎤⎦2
2 E I yy
duz
dx= −
2 Fz
3x2
2+ Fz
x − 2⎡⎣ ⎤⎦2
2+C1
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Macaulay’s method Since we are integrating a single 2nd-order differential
equation, just 2 integration constants appear in the solution, C1 and C2: ◦ These quantities can be determined by using the boundary
conditions, i.e. conditions at the supports ◦ Importantly, the square bracket term is only included if the
quantity inside is +ve
0 = −0+ Fz−2⎡⎣ ⎤⎦
3
6+ 0+C2 ⇒ C2 = 0
0 = −Fz 6
3
9+ Fz
4⎡⎣ ⎤⎦3
6+C1 6 ⇒ 6C1 = 24− 32
3⎛⎝⎜
⎞⎠⎟Fz
⇒ C1 =1672− 323
⎛⎝⎜
⎞⎠⎟Fz =
20 Fz9
uz = 0@ x = 6 ⇒
uz = 0@ x = 0 ⇒
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Macaulay’s method Starting from a single expression of the bending
moment My, we obtained a single expression throughout the beam for the deflection uz, in which we have the Macaulay’s brackets:
We can now evaluate the deflection of the beam at the position of the point load uz(C), i.e. uz @ x= 2 m
uz =FzE I
− x3
9+x − 2⎡⎣ ⎤⎦
3
6+ 20 x9
⎛
⎝⎜⎜
⎞
⎠⎟⎟
uz (C) =FzE I yy
− 23
9+0⎡⎣ ⎤⎦
3
6+ 20× 2
9
⎛
⎝⎜⎜
⎞
⎠⎟⎟= 40−8
9FzE I yy
= 329FzE I yy
()
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Macaulay’s method We have also a single expression throughout the beam for
the slope duz/dx:
The slopes at the supports A and B, i.e. duz/dx @ x= 0 and x= 6 m take the values
duzdx
=FzE I yy
− x2
3+x − 2⎡⎣ ⎤⎦
2
2+ 209
⎛
⎝⎜⎜
⎞
⎠⎟⎟
duzdx
⎛⎝⎜
⎞⎠⎟ A
=FzE I yy
−0+−2⎡⎣ ⎤⎦
2
2+ 209
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟= 209FzE I yy
()
duzdx
⎛⎝⎜
⎞⎠⎟ B
=FzE I
− 62
3+4⎡⎣ ⎤⎦
2
2+ 209
⎛
⎝⎜⎜
⎞
⎠⎟⎟= −216+144+ 40
18FzE I yy
= −169FzE I yy
()
NUMERICAL APPLICATION
Lecture #3
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Numerical example Find position and value of the maximum deflection in
the simply supported beam shown below
The beam’s flexural rigidity is EIyy= 2.58×104 kN m2
x
z
2 m
60 k
N
RA RB
A B C
1 m 2 m
5 m
20 k
N
D
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Support reactions
The first step is to evaluate the support reactions at points A and B:
B
B
( ) 060 1 20 3 5 060 60 24 kN ( )5
M AR
R
Σ =⇒ − × − × + × =
+⇒ = =
Q
#
x
z
2 m
60 k
N
RA RB
A B C
1 m 2 m
5 m 20
kN
D
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Support reactions
The first step is to evaluate the support reactions at points A and B:
B
B
( ) 060 1 20 3 5 060 60 24 kN ( )5
M AR
R
Σ =⇒ − × − × + × =
+⇒ = =
Q
#
A
A
( ) 05 60 4 20 2 0240 40 56kN ( )5
M BR
R
Σ =⇒ − × + × + × =
+⇒ = =
Q
#
x
z
2 m
60 k
N
RA RB
A B C
1 m 2 m
5 m 20
kN
D
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Bending moment’s expression
Once we have all the external forces applied to the beam (external forces and support reaction), the second step is to write down the expression of the bending moment My along the beam
My = 56 x − 60 x −1⎡⎣ ⎤⎦ − 20 x − 3⎡⎣ ⎤⎦
x
z
2 m
60 k
N
RA RB
A B C
1 m 2 m
5 m 20
kN
D
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Double integration
EI yy
d2uz
dx2 = −M y = −56 x + 60 x −1⎡⎣ ⎤⎦ + 20 x − 3⎡⎣ ⎤⎦
EI yy
duz
dx= − 56
28 x2
2+ 60
30 x −1⎡⎣ ⎤⎦2
2+ 20
10 x − 3⎡⎣ ⎤⎦2
2+C1
EI yy uz = −28 x3
3+ 30
10 x −1⎡⎣ ⎤⎦3
3+10
x − 3⎡⎣ ⎤⎦3
3+C1 x +C2
uz = 0 @ x = 0 ⇒ C2 = 0uz = 0 @ x = 5 ⇒ C1 = 100
⎧⎨⎪
⎩⎪
Boundary conditions (simply supports at points A and B) gives:
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Abscissa of maximum deflection Within a span, the maximum deflection will occur where the
slope of the beam is zero. So to find the position of the maximum deflection, we can determine the value of the abscissa x that gives duz/dx=0.
We have the mathematical expression of the slope, but it contains two square brackets, and we must decide which of them should be retained.
As the position of maximum deflection is never very far away from the centre of the span, we can guess that it occurs between x=1 and x=3 m. In this region the expression for the slope becomes:
duz
dx= −28x2 + 30 x −1⎡⎣ ⎤⎦
2+ 10 x − 3⎡⎣ ⎤⎦
2+100
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Abscissa of maximum deflection We can now solve the quadratic equation:
duz
dx= 0 ⇒ − 28x2 + 30 x −1( )2
+100 = 0
−28x2 + 30x2 − 60x + 30+100 = 0
2x2 − 60x +130 = 0
x = 60 ± 602 − 4× 2×1304
= 15± 3,600−1,0404
xmax = 15±12.65 =27.65 → Root unacceptable
(outside the beam)
2.35 →Root consistent with the assumption 0 ≤ x ≤ 3
⎧
⎨
⎪⎪⎪
⎩
⎪⎪⎪
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Maximum deflection We can now evaluate the deflection at x=2.35 m:
uz ,max =1
EI yy
−28xmax
3
3+10 xmax −1⎡⎣ ⎤⎦
3+10
xmax − 3⎡⎣ ⎤⎦3
3+100 xmax
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
= 12.58×104 −28
2.353
3+10 2.35−1⎡⎣ ⎤⎦
3+ 10
2.35− 3⎡⎣ ⎤⎦3
3+ 235
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
= −121.13+ 24.60+ 2352.58×104 = 138.47
2.58×104 = 53.7 ×10−4 m = 5.4mm
So maximum deflection is 5.4mm at 2.35m from the left support
Now check that you can show that the deflections under the 60kN and 20kN loads are 3.5mm and 5.0mm, respectively.