2-Lectures LEC 10

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Chapter 5

Deflection and Stiffness

A. Aziz Bazoune

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Outline

1. Spring Rates2. Deflection in Tension, Compression & Torsion

3. Deflection due to Bending

4. Strain Energy5. Castiglianos Theorem

6. Statically Indeterminate Problems

7. Compression MembersLong Columns with Central Loading

Intermediate Length Columns with Central Loading

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Spring Rates Tension Com ression and Torsion

Lec. 10

Deflection Due to Bending Superposition

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Rigid Body

A body is said to berigidif it exhibits no change in size orshape under the influence of forces or couples.

Introduction

A rigid body is a body of finite size and is such that thedistance between any two points within the body remainsconstant under the application of forces.

All real bodies deform under load, either elastically orplastically. Classification of a real body as a rigid is anidealization.

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Flexibility

Flexibilityis the ability of a body to distort by bending

that is, it is but one form of distortion- but the term isoften used interchangeably with distortion.

Elasticity

Elasticityis a property of a material that enables it toregain its original configuration after having beendeformed.

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A spring is a mechanical element that exerts a force when deformed,5-1 Spring Rates

( ) limy

F dFk y

y dy

= =

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Figure 5-1


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Linear Displacement Angular Displacement

Formula: Spring Constant Formula: Torsional Spring rate

Fk

y=

Tk

=

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Units:

[k]=N/m or [k]= lb/in

Units:

[k]=N.m/rad or [k]=lb.in/rad


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Example-1 (Problem 5-5)A bar in tension has a circular cross-sectionand includes a conical portion of lengthl ,as shown. Find the spring rate of the entirebar.

Solution

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5-3 Deflection due to Bending

Beams deflect great deal more than axially loaded members, and theproblem of bending probably occurs more often than any other loadingproblem in design.

Shafts, axles, cranks, levers, springs, brackets, and wheels must often

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and systems.


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The curvature of a beam subjected to bending momentMis given by

where is the radius of curvature. The curvature of a plane curve is

given by

1 M

E I=

2 2

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1

1

d y dx

dy dx=

+

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wherey is the deflection of the beam at any pointx along its length.

The slope of the beam at any point x is given by

dy

dx=


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Noting Eqs. (4-3) and (4-4) andsuccessively differentiating theprevious Eq. yields

3

3

2 4

2 4

dM d y

V EIdx dx

dV d M d yq EI

dx dx dx

= =

= = =

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It is convenient to displaythese relations in a group asfollows:


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Nomenclature: Refer to Fig. (5-2), a beam of length l =20 in is loaded by the uniform

load w =80 lbfper inch of beam length.

9/29/2007 Chapter 5 14Fig. (5-2)


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Example-2

0 x l For the beam in Fig.5-2, the bending moment equation, for , is

Determine the equations for slope and deflection of the beam, the slopesat the ends, and the maximum deflection.

0 x l

2

2 2

wl wM x x=

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Integrating Eq. (5-12) textbook as an indefinite integralgives

whereC1 is a constant of integration that is to be determined from theB.Cs.

2

2

M d yEI dx

=

2 31

4 6dy wl wEI Mdx x x Cdx

= = +


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0 x l

Integration of the previous Eq. gives

B.Cs for simply supported beam are :y(0)= y(L) =0.

Applyy(0) = 0 to the above Eq. gives

Applyy(L) = 0 to the above Eq. gives

3 41 2

12 24wl wEIy Mdx x x C x C= = + +

2 20 0 0 0 0C C= + + =

4 4 3wl wl wl

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Therefore,

1 112 24 24= =

( )

( )

4 3 3

3 2 3

224

4 624

wy x lx l x

EI

dy wx lx l

dx EI

= +

= = +


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0 x l

Comparing the previous Eq. with that given in Table A-9, beam 7, wesee complete agreement.

For the slope at the left end, substitutingx = 0 andx = l into theprevious Equation gives

and

At the mid-span, substitutingx=l/2 gives as expected.

3

0

,24x

wl

EI

=

=

3

24x l

wl

EI

=

=

0dy

dx=

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The maximum deflection occurs where . Substitutingx = l/2into the deflection Equation gives

Which again agrees with Table A-9-7.

4

max

5

384

wly

EI

=

0dydx

=


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Example-3

Determine the deflection at theleft end of the cantilever beamwith variable width as shown in

the Figure.

Solution

From the eometr of the beam

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it is clear that the width varieslinearly with the positionx.

Similar triangles give the relation

Therefore, the second moment ofarea of the beam may beexpressed as

b w bw x

L x L

= =


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Therefore, the second moment of area of the beam may be expressedas

where is the second moment of area at the support.

( ) 33 3

12 12 12

L

bx L hwh bh x xI I I

L L

= = = = =

3

12LI bh=

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The equation for the elastic curve y is obtained by successiveintegrations, that is

Successive integrations give

( )

2 2

2 2

L L

M d y d y M Px Px PLE

EI dx dx I I I x L I

= = = = =


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Apply B. Cs

1

2

1 2

2

L

L

dyEI PLx Cdx

PLEI y x C x C

= +

= + +

2

10 ,

dyx L C PL

dx= = =when

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Thus the equation for the elastic curve is

2 2

2 2L

PL x Ly Lx

EI

= +

3

20, ,

2PLy x L C= = = when


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At the left end of the beam where , the deflection is

It is also interesting to notice how the maximum flexural stress variesalong the length of the beam

0x=

( )3 3

3

60

2 L

PL PLy

I Ebh= =

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23

62

12

hPxMc PL

I bhbh x

L

= = =


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5-4 Beam Deflection Methods

2

2M d yEI dx

=

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1. Integration of moment equation (example 5-1)2. By use of superposition (Examples 5-2 & 5-3)3. By moment area method

4. Singularity functions (Ex. 5-5)5. Use of strain energy with Castiglianos theorem6. Numerical integration


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5-5 Finding Beam Deflections by

Superposition

Table A-9 provides some cases for results of beams subjected to simpleloads and boundary conditions.

Superposition resolves the effect of combined loading on a structure bydetermining the effect of each load separately and adding the results

.

In using the superposition principle, the followings are required:

1. Each effect is linearly related to the load that produces it.

2. A load does not create a condition that affects the results of anotherload.

3. The deformations resulting from any specific load are not largeenough to appreciably alter the geometric relations of the parts of thestructure.

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Example-4 (5-2 Textbook)

Find y as a function of x in the Figure Shown,

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======

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( )( )

( )

1 2

2 2 2

2 2

/ , /

6

26

AB

BC

R Fb L R Fa L

Fbxy x b lEIL

Fa l xy x a lx

EI

= =

= +

= +

Case 6 Table A-9 (Simple SupportsIntermediate Loads) Case 7 Table A-9 (Simple SupportsUniform Loads)

( )

( )

1 2

2 3 3

2 3 3

/ 2, / 2

224

2

24

AB

AB

R wl R wl

wxy lx x lEI

wxy lx x l

EI

= =

=

=


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( ) ( )( )

( ) ( )

1 2

2 2 2

2

2 3 3

2 3 32

/ , /

6

26

/ 2 / 2

2

24

224

AB

BC

wl wl

wxlx x l

E

R Fb L R Fa L

Fbxy x b l

EILFa l x

y x a lxE

I

wlx x l

EII

x

= + = +

= + +

= + +

Superposition:


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http://www.engapplets.vt.edu/statics/BeamView/BeamView.html

https://ecourses.ou.edu/cgi-bin/navigation.cgi?course=me

Useful links related to the subject

9/29/2007 Chapter 5 28

http://web.umr.edu/~mecmovie/index.html

http://www.engin.umich.edu/students/ELRC/me211/beamdef.html

2-Lectures LEC 10

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