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MEEG 202 Strengthof Materials

26-06-2012

Beam Deflections Using SingularityFunctions


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Admin

Preview Example Problems

BEFORE Class

Homework #4 Due Thursday Next class Thursday 1-4

Please be on Time


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Lesson Objectives

1. Understand the meaning of a singularityfunctions.

2. Be able to integrate singularity functions.

3. Be able to calculate the deflection of a


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Surprise QuizName:_____________________________ Roll #____________________

1) The drawing below shows a simply supported beam with a center load. Thecross section is triangular.

a) At what point in the cross section do you expect the transverse shearstress (due to the shear load) to be the greatest?

b) What is its value? Sketch:

x

A

m2

C

m1

kN50

B

mm90

mm90

bhIxx

3

36

1

2) What was the subject of the pre-class example problem for today?

Ib

SQ

I

My


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Beam Deflection by

Integration SummaryF

x

Deflection at any point is y

yd 4

xfy

dx

dy

dxyd

EIM

dx

yd

EI

S

dxEI

2

2

3

3


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Review Beam Deflection by Integration

ProcessF

xDeflection at any point is y

yd 4

xF S

MA

Fl

FlFxM

2) Put M(x) into the integral: Mdxdxdy

EI dxFlFxdxdy

EI

3) Integrate to get slope function 21d

C

A

1) Write Moment Function as a function of x

xfy

dx

dy

dxyd

EIM

dx

yd

EI

S

dxEI

2

2

3

3

1

2xx

dx

4) Integrate again to get deflection

21

23

1

2

2

1

6

1

2

1

CxCFlxFxEIy

dxCFlxFxEIy

5) Apply slope and deflection boundary

conditions to get C1 and C2.

00

00

y

At y=0

0

0

2

1

C

CTherefore:


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Singularity Functions

Know as Macaulays Method for

finding beam deflections

Unit Step Function

x

1

a

0axxf

1

0

0

0

ax

ax

What does the Integral of

ax

ax

0

ax Look like?

dxax 0


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Unit Ramp Function

1

1

axxf

1

10

axdxax

1

axxf

xa

axax

1

What is the Integral of

ax

1ax ?

1

2

2

1 axdxax

22

02

22

2

axax

ax

ax

ax


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Singularity FunctionsWhat is the Derivative of the Unit Step Function?

x

0

axxf

1

a

1 axxf

1

01

1

ax

axaxax

01

axdxax

x

1 axxf

a

This is the Unit Impulse (Concentrated Force Function)


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x

1 axxf

a

What is the Derivative of the Unit Impulse Function?

The derivative of the Unit impulse is the unit doublet

(or concentrated moment function)?

2

axxf

2

2

0

ax

ax

ax

ax

12

axdxaxx

2 axxf

a


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Singularity Function Summary

2

2

0

ax

ax

ax

ax

12 axdxaxx

2 axxf

a

1

0

1

1

ax

ax

ax

ax

01 axdxax

1 axxf

x

1

a

0

axxf

1

0

0

0

ax

ax

ax

ax

10

axdxax

xa

x

1

a

1

axxf axax

ax

1

1

0

ax

ax

1

2

2

1 axdxax


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Singularity Function

Process1) Write the load function w(x) in

terms of singularity functions.

2) Integrate again to get S(x)

3) Integrate twice to get M(x)

ydS

dx

yd

EI

3

3

4

4

n egra e aga n o ge x p us an

integration constant

5) One more integration gets you

EIy(x) with another integration

constant

6) Use boundary conditions to find

integration constants.

xfydx

dy

dx

yd

EI

M

x

2

2

Fi d f ti th t d ib th d fl ti f thExample 10 Beam Deflection Using Singularity Functions

Sketch: F


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Find a function that describes the deflection of the

beam shown at right as a function of x.

Solution:

Problem Type:

Find:

Given: The figure of the simply supported beam

at right.

y(x)

Beam Deflections

First find the reactions.

l

FaR

FalR

M

C

C

A

0

0

Sketch:

x

A

l

C

a

F

x

A

l

C

a

F

B

B

AR CR

l

FbR

FblR

M

A

A

C

0

0

b

Now write an equation for the loading in terms of singularity functions.

Next, Integrate w(x) to get S(x), the shear loading function

No constant of integration is needed

since this completely describes the

shear.

111 lxl

FaaxFx

l

Fbxw

000 lxl

FbaxFx

l

FaxS

Next, Integrate S(x) to get M(x), the bending moment function

No constant of integration is needed

again since this completely describes

the moment function.

111 lxl

FaaxFx

l

FbxM

Now integrate to get EI(x)


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Note that the first term in the singularity function will always be evaluated as x2 since x is

always greater than 0 and the last term will always be zero since xl. We can then simplify

the last two equations.

g g ( )

1

222

222Clx

l

Faax

Fx

l

FbxEI

Now we need to have a constant

of integration to make sure any

physical boundary conditions are

met.

Lastly, integrate one more time to get EIy(x)

21333

666 CxClxl

Fa

ax

F

xl

Fb

xEIy We need 1 more constant ofintegration.

1

22

22Cax

Fx

l

FbxEI

2133

CxCaxF

xFb

xEIy

0

00

ly

y

Now apply Boundary Conditions

And

0

0000

006

06

0

66

2

2

21

33

21

33

C

C

CCaF

l

FbEI

CxCaxFxl

FbxEIy

221

1

33

1

33

1

33

21

33

6

660

660

066

0

66

bll

FbC

lCbF

ll

Fb

lCalFll

Fb

ClCalF

ll

FbEI

CxCaxF

xl

FbxEIy

alb

sub


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To do for next time

Next Class on Thursday 1-4

We will practice beam deflection

problems

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