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

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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.