PowerPoint slides of 3 lectures given to 1st-year Civil Engineering students

1. Deflections in beams
Dr Alessandro Palmeri
Senior Lecturer in Structural Engineering
<A.Palmeri@lboro.ac.uk>
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39

2. 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
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39

3. Lecture #1
DOUBLE INTEGRATION
METHOD

4. Introduction
 Structural
members must have:
◦ Strength (ULS: Ultimate Limit State)
◦ Stiffness (SLS: Serviceability Limit State)
 Need
to limit deflection because:
◦ Cracking
◦ Appearance
◦ Comfort
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Engineering Structures,Volume 56, 2013, 1346 - 1361

5. 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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39

6. 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

6/
39
e.g.Virtual Work Principle

7. Curvature
 From
the simple theory of bending we
have:
1
M
=
R EI
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
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39

8. Curvature

For a plane curve uz(x) in the xz plane, the curvature 1/Ry
(about the orthogonal axis y) is given by:
x
d 2uz
dx 2
1
=
Ry ⎡ ⎛ du ⎞ 2 ⎤
⎢1+ ⎜ z ⎟ ⎥
dx ⎠ ⎥
⎢ ⎝
⎣
⎦


8/
39

y
Ry
z
If duz/dx is small, then (duz/dx)2 can be considered negligible
d 2uz
1
Thus:
≅
Ry
dx 2
And so:
My
E I yy
d 2uz
=
dx 2

9. 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:
load
slope
x
z
9/
39
z
duz
>0
dx
curvature
x
x
z
d 2uz
2
>0
dx
But this is the
shape of hogging
bending moment,
i.e. My<0

10. Differential equation of slender
beams in bending

Taking into account the correct sign convention
for deflection and bending moment, we have:
d 2uz (x)
E I yy
= − M y (x)
2
dx
◦ 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:
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39
 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)

11. 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


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39
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

12. 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
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39
uz=0 and duz/dx=0

13. 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
qz
MA
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39
A
z
RA
B
L
x

14. Worked example

qz
MA
A
z
RA
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39
B
L
x
1st, determine the
support’s reactions:
ΣV = 0 
⇒ RA − qz L = 0
⇒ RA = qz L ()
Σ M (A) = 0 
L
⇒ M A − qz L = 0
2
qz L2
⇒ MA =
()
2

15. Worked example

qz
MA
A
z
RA
B
L
x
2nd, write down the
expression of the
bending moment My as
a function of the
abscissa x along the
beam’s axis:
qz x 2
M y = −M A + RA x −
2
qz L2
qz x 2
=−
+ qz L x −
2
2
(
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39
)

16. Worked example
 The
differential equation for the beam’s
deflection reads:
d 2uz
qz L2
qz x 2
E I yy 2 = − M y =
− qz L x +
2
2
dx
 3rd, we
can integrate twice:
2
2
3
duz qz L
qz L x
qz x
E I yy
=
x−
+
+ C1
dx
2
2
6
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39
qz L2 2 qz L x 3 qz x 4
E I yy uz =
x −
+
+ C1 x + C2
4
6
24

17. Worked example

4th, the known boundary conditions at the fixed support (i.e.
no deflection and no slope at left-hand side end A):
duz
= 0 @ x = 0 ⇒ C1 = 0
dx
u z = 0 @ x = 0 ⇒ C2 = 0

Substituting now the values of the integration constants C1
and C2, the expressions for slope and deflection throughout
the beam become:
2
duz
qz L x 2 qz x 3 ⎞
1 ⎛ qz L
=
⎜ 2 x− 2 + 6 ⎟
dx E I yy ⎝
⎠
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39
2
3
4
1 ⎛ qz L 2 qz L x qz x ⎞
uz =
⎜ 4 x − 6 + 24 ⎟
EI⎝
⎠

18. 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:
qz
MA
A
z
RA
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39
B
L
x
2
⎛ duz ⎞
qz L L2 qz L3 ⎞
qz L3
1 ⎛ qz L
(> 0, )
⎜ dx ⎟ = E I ⎜ 2 L − 2 + 6 ⎟ = 6 E I
⎝
⎠B
yy ⎝
⎠
yy

19. Worked example
qz
MA
A
z
RA
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39
B
L

x
Similarly, by
substituting x=L in the
general expression of
the deflection, we have:
2
3
4
1 ⎛ qz L 2 qz L L qz L ⎞
yB =
⎜ 4 L − 6 + 24 ⎟
E I yy ⎝
⎠
qz L4 1 qz L4
6 − 4 +1
=
=
(> 0, )
24 E I yy 8 E I yy

20. Lecture #2
MACAULAY’S
NOTATION

21. Beams under point loads
 E.g. simply
supported beam with a single
concentrated load
2m
4m
Fz
A
B
C
z
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39
RA
6m
RB
x

22. Beams under point loads

2m
Σ M (A) = 0 
⇒ − Fz 2 + RB 6 = 0
4m
Fz
A
B
C
z
RA
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39
Support reactions
6m
RB
x
2 Fz Fz
⇒ RB =
=
()
6
3
Σ M (B) = 0 
⇒ − RA 6 + Fz 4 = 0
4 Fz 2
⇒ RA =
= Fz ()
6
3

23. Beams under point loads

0<x<2
A
RA
2
Fz
A
RA
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39
In principle, we need
two expression for
the bending moment
My:
◦ one for 0<x<2
M y = RA x
C
2<x<6
◦ one for 2<x<6
(
M y = RA x − Fz x − 2
)

24. Beams under point loads
 In
principle, we need to integrate two
differential equations:
⎧ RA x , 0 < x < 2
⎪
=⎨
2
dx
⎪ RA x − Fz x − 2 , 2 < x < 6
⎩
2
E I yy
d uz
(
)
 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
24/
39

25. 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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39

26. Macaulay’s notation
 That
is:
⎧ x , if x > 0
[ x] = ⎨
⎩0 , if x ≤ 0
 Let’s
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39
try the following examples:
⎡ 2.3⎤ = 2.3
⎣ ⎦
⎡0 ⎤ = 0
⎣ ⎦
⎡ −3 / 4 ⎤ = 0
⎣
⎦

27. Macaulay’s notation

2m
4m
Fz
A
B
C
z
RA
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39
x
It is possible now to
write down a single
expression for the
bending moment
along the beam:
M y = RA x − Fz ⎡ x − 2 ⎤
⎣
⎦
6m
RB

28. Macaulay’s method

The differential equation of bending becomes:
E I yy

d 2uz
dx
2
= −M y = −RA x + Fz ⎡ x − 2 ⎤
⎣
⎦
This expression can be integrated twice,
importantly, without expanding the term into
square brackets:
2
2
du
2 Fz x ⎡ x − 2 ⎤ − 2 ⎤
dy z
x
⎣x ⎦ + C
⎣ F ⎡⎦
=−
E I yy = −RA + W + z
1
2
dxdx
23 2
2
2
2
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39
3
⎡ x − 2⎤
Fz x
⎦ +C x +C
E I yy uz = −
+ Fz ⎣
1
2
9
6
3

29. 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
uz = 0 @ x = 0 ⇒
3
⎡ −2 ⎤
0 = −0 + Fz ⎣ ⎦ + 0 + C2 ⇒ C2 = 0
6
uz = 0 @ x = 6 ⇒
3
3
⎡4⎤
Fz 6
⎛
32 ⎞
0=−
+ Fz ⎣ ⎦ + C1 6 ⇒ 6 C1 = ⎜ 24 − ⎟ Fz
9
6
3⎠
⎝
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39
20 Fz
1 ⎛ 72 − 32 ⎞
⇒ C1 = ⎜
⎟ Fz = 9
6⎝ 3 ⎠

30. Macaulay’s method


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39
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:
⎛ 3 ⎡ x − 2⎤3
⎞
Fz
x ⎣
20 x
⎦ +
⎜− +
⎟
uz =
EI⎜ 9
6
9 ⎟
⎝
⎠
We can now evaluate the deflection of the beam at
the position of the point load uz(C), i.e. uz @ x= 2 m
⎛ 3 ⎡0 ⎤ 3
⎞
Fz
2 ⎣ ⎦ 20 × 2
40 − 8 Fz
⎜− +
⎟=
uz (C) =
+
E I yy ⎜ 9
6
9 ⎟
9 E I yy
⎝
⎠
32 Fz
=
()
9 E I yy

31. Macaulay’s method


We have also a single expression throughout the beam for
the slope duz/dx:
⎛ 2 ⎡ x − 2⎤2
⎞
duz
Fz
x
⎦ + 20 ⎟
⎜− + ⎣
=
dx E I yy ⎜ 3
2
9⎟
⎝
⎠
The slopes at the supports A and B, i.e. duz/dx @ x= 0 and
x= 6 m take the values
2
⎛
⎞
⎡ −2 ⎤
⎛ duz ⎞
Fz
20 ⎟ 20 Fz
⎜ −0 + ⎣ ⎦ +
=
=
()
⎜ dx ⎟
2
9 ⎟ 9 E I yy
⎝
⎠ A E I yy ⎜
⎝
⎠
⎛ 2 ⎡4⎤2
⎞
⎛ duz ⎞
Fz
6 ⎣ ⎦ 20
−216 + 144 + 40 Fz
⎜− +
=
+ ⎟=
⎜ dx ⎟
2
9⎟
18
E I yy
⎝
⎠B E I ⎜ 3
⎝
⎠
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39
=−
16 Fz
()
9 E I yy

32. Lecture #3
NUMERICAL
APPLICATION

33. Numerical example

Find position and value of the maximum deflection in
the simply supported beam shown below
z
RA
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39

C
20 kN
A
2m
60 kN
1m 2m
B
D
x
5m
RB
The beam’s flexural rigidity is EIyy= 2.58×104 kN m2

34. Support reactions
z
RA
C
20 kN
A
2m
60 kN
1m 2m

B
D
5m
x
The first step is to
evaluate the support
reactions at points A
and B:
RB
Σ M ( A) = 0 Q
⇒ − 60 × 1 − 20 × 3 + RB × 5 = 0
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39
60 + 60
⇒ RB =
= 24 kN (#)
5

35. Support reactions
z
RA
C
20 kN
A
2m
60 kN
1m 2m

B
D
5m
x
The first step is to
evaluate the support
reactions at points A
and B:
RB
A
Σ M ( B) = 0 Q
⇒ − 60 × 1 − 20 × 34 +RB × 52= 00
RA × 5 + 60 × + 20 × =
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39
60 ++ 40
240 60
= = 56kN ##)
24 kN ( ()
⇒ RB =
A
55

36. Bending moment’s expression
z
RA
35/
39
C
20 kN
A
2m
60 kN
1m 2m

B
D
5m
RB
x
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
M y = 56 x − 60 ⎡ x −1⎤ − 20 ⎡ x − 3⎤
⎣
⎦
⎣
⎦

37. Double integration
d 2uz
EI yy 2 = − M y = −56 x + 60 ⎡ x − 1⎤ + 20 ⎡ x − 3⎤
⎣
⎦
⎣
⎦
dx
2
2
⎡ x − 1⎤
duz
x
⎣
⎦ + 20 ⎡ x − 3⎤ + C
⎣
⎦
EI yy
= − 56
+ 60
1
dx
2
2
2
28
30
2
3
10
3
⎡ x −1⎤
x
⎣
⎦ + 10 ⎡ x − 3⎤ + C x + C
⎣
⎦
EI yy uz = −28 + 30
1
2
3
3
3
3

36/
39
10
Boundary conditions (simply supports at points A and B) gives:
⎧u z = 0 @ x = 0 ⇒ C 2 = 0
⎪
⎨
⎪uz = 0 @ x = 5 ⇒ C1 = 100
⎩

38. Abscissa of maximum deflection



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39
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:
2
2
duz
2
= −28 x + 30 ⎡ x − 1⎤ + 10 ⎡ x − 3⎤ + 100
⎣
⎦
⎣
⎦
dx

39. Abscissa of maximum deflection

We can now solve the quadratic equation:
duz
2
2
= 0 ⇒ − 28 x + 30 ( x − 1) + 100 = 0
dx
−28 x 2 + 30x 2 − 60x + 30 + 100 = 0
2x 2 − 60x + 130 = 0
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39
60 ± 602 − 4 × 2 × 130
3,600 − 1,040
x=
= 15 ±
4
4
⎧
⎪
⎪ 27.65 → Root unacceptable
⎪
xmax = 15 ± 12.65 = ⎨
(outside the beam)
⎪
⎪2.35 → Root consistent with the
⎪
assumption 0 ≤ x ≤ 3
⎩

40. Maximum deflection

We can now evaluate the deflection at x=2.35 m:
uz ,max
3
3
⎧
⎫
⎡ xmax − 3⎤
xmax
3
1 ⎪
⎦ + 100 x ⎪
=
−28
+ 10 ⎡ xmax − 1⎤ + 10 ⎣
⎨
max ⎬
⎣
⎦
EI yy ⎪
3
3
⎪
⎩
⎭
3
⎧
⎫
3
⎡ 2.35 − 3⎤
3
1
2.35
⎪
⎦ + 235⎪
=
−28
+ 10 ⎡ 2.35 − 1⎤ + 10 ⎣
⎨
⎬
⎣
⎦
3
3
2.58 × 104 ⎪
⎪
⎩
⎭
−121.13+ 24.60 + 235
138.47
=
=
= 53.7 × 10−4 m = 5.4mm
2.58 × 104
2.58 × 104

39/
39
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.