W4 Beam Deflection-1- Rev2

8/10/2019 W4 Beam Deflection-1- Rev2

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MECHANICS OF

MATERIALS

Third Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

John T. DeWolf

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

2002 The McGraw-Hill Companies, Inc. All rights reserved.

9Deflection of Beams


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MECHANICS OF MATERIALSThird

Edition

Beer Johnston DeWolf

9 - 2

Deflection of Beams

Deformation of a Beam Under Transverse

LoadingEquation of the Elastic Curve

Example 9.01

Example 9.02

Direct Determination of the Elastic CurveFrom the Load Distribution

Example 9.04

Sample 9.1

Using singularity functions todetermine the slope and deflection

of a beam


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Edition


9 - 3

Deformation of a Beam Under Transverse Loading

Relationship between bending moment and

curvature for pure bending remains valid for

general transverse loadings.

EI

xM )(1

EI

Px

1

Curvature varies linearly withx

At the free endA, x=0, A

A

,01

At the supportB,x=L,PL

EIB

B

,01

A Cantilever beam AB of length L subjected

to concentrated load P at the free end A. We

have

M(x) = -Px. Then it become

x

MECHANICS OF MATERIA ST


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Edition


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Deformation of a Beam Under Transverse Loading Consider Overhanging beam AD support

concentrate load

Reactions atAand C (RA= 1kN an RC= 5 kN

Bending moment diagram

Curvature is zero at points where the bending

moment is zero, i.e., at each end and atE.

EIxM )(1

Between AE Beam is concave upwards where the

bending moment is positive and concave

downwards where it is negative at ED.

Maximum curvature occurs where the moment

magnitude is a maximum at the support C.

An equation for the beam shape or elastic curve

is required to determine maximum deflection

and slope.

MECHANICS OF MATERIALSTE


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Edition


9 - 5

Deformation of a Beam Under Transverse Loading

From the information obtained its curvature,

we get fairly good idea of the shape of the

deformed beam.

The analysis and design of the beam require

more precise information on the deflectionand the slope of the beam at various points.



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Type 1. Equation of the Elastic Curve

From elementary calculus, simplified for beam

parameters,

2

2

232

2

2

1

1

dx

yd

dxdy

dx

yd

Substituting and integrating,

2

2

1

0

1 2

0 0

1

( )

( )

x

x x

d y

EI EI M xdx

dyEI EI M x dx C

dx

EI y dx M x dx C x C

The slope dy/dx = is

very small the

square negligible.

From EIxM )(1

Where dy/dx = tan

= (x)



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Edition


9 - 7

Type 1. Equation of the Elastic Curve

2100

CxCdxxMdxyEI

xx

Constants are determined from boundary

conditions

Three cases for statically determinant beams,

Simply supported beam0, 0A By y

Overhanging beam

0,0 BA yy

Cantilever beam0,0 AAy

More complicated loadings require multiple

integrals and application of requirement for

continuity of displacement and slope.

deflection and slope.



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Edition


9 - 8

Example 9.01

The cantilever beamABis of uniform cross

section and carries a load P at its free end A.

Determine the equation of the elastic curve

and the deflection and slope at A..

Using the free-body diagram of the portionAC

of the beam where Cis located at a distancex

fromend A, we find

M= -PxEI d2y/dx2= - Px

Integrating in x, we obtain

EI dy/dx = -Px2+ C1We now observe that at fixed end B we have

x=L and = dy/dx = 0 and we getC1= PL

2

Then EI dy/dx = -Px2+ PL2

Integrating both member we write

EIy = -1/6 Px3

+ PL2

x + C2



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Edition


9 - 9

Example 9.01

But at B we have x=L, y=0 and we have

0 = -1/6 PL3+ PL3+C2C2= -1/3 PL

3

Carrying the value of C2

EIy = -1/6 Px3+ PL2x -1/3 PL3

Or

y = P/6EI (-x3

+ 3L2

x 2L3

)

The deflection and slope at A are obtained by

letting x=0. We find

yA= -PL3/3EI

and

A= (dy/dx)A= PL

2

/2EI



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Edition


9 - 10

Example 9.02 The simply supported prismatic beam AB

carries a uniformly distributed load wper

unit length. Determine the equation of theelastic curve and the maximum deflection of

the beam

M = wLx wx2

EI d2

y/dx2

= - wx2

+ wLx

Integrating twice in x, we have

EI dy/dx = -1/6 wx3+ wLx2+ C1

EIy = -1/24 wx4+ 1/12 wLx3+ C1x + C2

Observing that y=0 at both ends of the beam,

we first let x = 0 and y=0 and obtain C2=0.

We then make x=L and y=0 in the same

equation and write

0 = -1/24 wx4

+ 1/12wLx3

+ C1LC1= 1/24wL

3



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Edition


9 - 11

Example 9.02 Carrying the value of C1and C2back we

obtain the equation of the elastic curve:

EIy = -1/24 wx4+ 1/12wLx31/24wL3x

Or y = w/(24EI) (-x4+ 2Lx3L3x)

We check that the slope of the beam is zero

for x = L/2 and the elastic curve has a max at

the midpoint C of the beam

yC= w/(24EI)(-L4/16 + 2LL3/8L3L/2)

= - 5wL4/384EI

The maximum deflection, the maximum

absolute value of the deflection is

/y/max= 5wL4/384EI



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Edition


9 - 12

Sample Problem 9.1

1.2m4.5mkN200

GPa200mm10300101360 46

aLP

ExIW

For portionABof the overhanging beam,

(a) derive the equation for the elastic curve,(b) determine the maximum deflection,

(c) evaluateymax.

SOLUTION:

Develop an expression for M(x)

and derive differential equation for

elastic curve.

Integrate differential equation twiceand apply boundary conditions to

obtain elastic curve.

Locate point of zero slope or point

of maximum deflection.

Evaluate corresponding maximum

deflection.



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Edition


9 - 13

Sample Problem 9.1

SOLUTION:

Develop an expression for M(x) and derivedifferential equation for elastic curve.

- Reactions:

L

aPR

L

PaR BA 1

- From the free-body diagram for sectionAD,

LxxL

aPM 0

xL

aP

dx

ydEI

2

2

- The differential equation for the elastic

curve,

MECHANICS OF MATERIALST

E


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Edition


9 - 14

Sample Problem 9.1

PaLCLCLL

aPyLx

Cyx

6

1

6

10:0,at

0:0,0at

113

2

Integrate differential equation twice and apply

boundary conditions to obtain elastic curve.

213

12

6

1

2

1

CxCxL

aPyEI

CxL

aP

dx

dyEI

xL

aP

dx

ydEI

2

2

32

6 L

x

L

x

EI

PaLy

PaLxxL

aPyEI

Lx

EIPaL

dxdyPaLx

LaP

dxdyEI

6

1

6

1

3166

121

3

2

2

Substituting, a) equation ofthe elastic curve

MECHANICS OF MATERIALSTh

E


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Edition


9 - 15

Sample Problem 9.1

Locate point of zero slope or pointof maximum deflection in portion AB

32

6 L

x

L

x

EI

PaLy

LL

xL

x

EI

PaL

dx

dym

m 577.03

316

02

Evaluate corresponding maximum

deflection.

32

max 577.0577.06

EI

PaLy

EI

PaLy

6

0642.02

max

46

2

max

m103002006

4.5m1.2m200kN0642.0

xGPay

mm5.2maxy

Evalution of ymax


Ed


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Edition


9 - 16

2. Direct Determination of the Elastic Curve From

the Load Distribution

The equation of elastic curve can be obtained

EI

xM

dx

yd )(2

2

Differentiating both member with respect to x3

3 1 ( )d y dM V xdx EI dx EI

And differentiating again

4

4

1 ( )d y dV w x

dx EI dx EI

We conclude that when a prismatic beam

supports a distributed load w(x) its elastic curve

is governed by the forth order linear differential

equation

EI

xw

dx

yd )(4

4

Where dM/dx =V

and dV/dx = -w


Ed


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MECHANICS OF MATERIALShird

Edition


9 - 17

2. Direct Determination of the Elastic Curve From

the Load Distribution

xwdx

ydEI

dx

Md

4

4

2

2

13

3

)( CdxxwxVdx

ydEI

xwdxydEI

dxMd 4

4

2

2

Equation for beam displacement becomes

xwdx

ydEI

dx

Md

4

4

2

2

432

2213

161 CxCxCxC

dxxwdxdxdxxyEI

Integrating four times yields

Constants are determined from boundary

conditions.

212

2

)( CxCdxxwdxxMdx

ydEI

32212/1)( CxCxCdxxwdxdxxEIdx

dyEI


Ed


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dition


9 - 18

Example 9.04

13

3

CwxxVdx

ydEI

Since w = constant

wdx

ydEI

4

4

2122

2

2/1 CxCwxxMdx

ydEI

The boundary condition require that M=0 at both end

of the beam we first let x=0 and M=0 and obtain c2=0.

We then make x =L and M=0 and obtain C1= wL

Carrying the value of C1and C2EI d2y/dx2= - wx2+ wLx

EI dy/dx = -1/6 wx3+ wLx2 + C3EIy = -1/24 wx4+ 1/12 wLx3+ C3x + C4


Ed


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dition


9 - 19

Example 9.04

But the bondary conditions also require that y=0 at both

end of the beam. Let x=0 and y=0, we obtain C4=0, let

x=L and y=0 in the same equation

EIy = -1/24 wx4+ 1/12 wLx3+ C3x + C4

0 = -1/24 wL4+ 1/12 wLL3+ C3LC3= -1/24 wL

3

We obtain the equation of elastic curve

y = w/24EI (-x4+ 2Lx3L3x)

The value of the maximum deflection is obtained by

making x =L/2.

We have

/y/max = 5wL4/384EI


Ed


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dition


9 - 20


Ed


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dition


9 - 21


Ed


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dition


9 - 22


Ed


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dition


9 - 23


Ed

B J h t D W lf


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MECHANICS OF MATERIALSirddition


9 - 24


Ed

B J h t D W lf


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MECHANICS OF MATERIALSirdition


9 - 25

MECHANICS OF MATERIALSThi

Ed

B J h t D W lf


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9 - 26


Edi

B J h t D W lf


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9 - 27

Type 3. Using singularity functionLet us consider the beam and loading and draw the free

Body diagram of the beam. We write

V(x) = PP x - L > 0

We have

M(x) = P x - P < x - L >

EI d2y/dx2= P x - P < x - L >

and integrating in x

EI= EI dy/dx = 3/8 Px2 P 2+ C1

EIy = 1/8 Px31/6 P 3 + C1x + C2

The constants C1and C2can be determined from. Letx=0, y=0 and we have C2=0 in Eq.

Let x =L, y=0

0 = 1/8 PL31/6 P (3/4 L)3+ C1L

C1 = -7PL2/128

EIy = 1/8 Px3

1/6 P 3

- 7pL2

/128 x


Edi



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MECHANICS OF MATERIALSrdition


9 - 28

Type 3. Using singularity function

For the beam and loading shown, determine

(a) the slope at end A,

(b) the deflection at point B,(c) the deflection at end D


Edi



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MECHANICS OF MATERIALSrdtion


9 - 29

Type 3. Using singularity function


Edi



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9- 30

Method of Superposition

Principle of Superposition:

Deformations of beams subjected to

combinations of loadings may be

obtained as the linear combination of

the deformations from the individualloadings

Procedure is facilitated by tables of

solutions for common types of

loadings and supports.

MECHANICS OF MATERIALSThir

Edit Beer Johnston DeWolf


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9- 31

Sample Problem 9.7

For the beam and loading

shown, determine the slope and

deflection at pointB.

SOLUTION:

Superpose the deformations due toLoading IandLoading

II as shown.




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9- 32

Sample Problem 9.7

Loading I

EI

wLIB 6

3

EI

wLy IB 8

4

Loading II

EI

wLIIC 48

3

EI

wLy IIC 128

4

In beam segment CB, the bending

moment is zero and the elastic curve is a

straight line.

EI

wLIICIIB 48

3

EI

wLL

EI

wL

EI

wLy IIB 384

7

248128

434




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9- 33

Sample Problem 9.7

EI

wLEI

wLIIBIBB 486

33

EI

wL

EI

wLyyy IIBIBB 384

7

8

44

EIwLB

487

3

EI

wLyB

384

41 4

Combine the two solutions,




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9- 34

Application of Superposition to Statically

Indeterminate Beams

Method of superposition may be applied

to determine the reactions at the

supports of statically indeterminate

beams.

Designate one of the reactions as

redundant and eliminate or modify

the support.

Determine the beam deformation without

the redundant support.

Treat the redundant reaction as an

unknown load which, together with

the other loads, must produce

deformations compatible with the

original supports.




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9- 35

Sample Problem 9.8

For the uniform beam and loadingshown, determine the reaction at

each support and the slope at endA.

SOLUTION:

Release the redundant support at B, and find deformation.

Apply reaction atBas an unknown load to force zero displacement atB.




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MECHANICS OF MATERIALSrdion


9- 36

Sample Problem 9.8

Distributed Loading:

EI

wL

LLLLLEI

wy wB

4

334

01132.0

3

2

3

223

2

24

At pointB, Lx32

xLLxxEI

wy

wB334 2

24




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MECHANICS OF MATERIALSdion

9- 37

Sample Problem 9.8

Redundant Reaction Loading:

EIL

bPayax

3,At

22

EI

LR

L

LEIL

R

y

B

B

RB

3

22

01646.0

33

2

3

LbLa31

32 andFor


Editi Beer Johnston DeWolf


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9- 38

Sample Problem 9.8

From statics,

wLRwLR CA 0413.0271.0

For compatibility with original supports,yB= 0

EI

LR

EI

wLyy BRBwB

34

01646.001132.00

wLRB 688.0


Editi Beer Johnston DeWolf


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Sample Problem 9.8

EI

wL

EI

wLwA

33

04167.024

EIwLLLLEILwLEILbLPbRA32

2

22

03398.0336

0688.06

EI

wL

EI

wLRAwAA

33

03398.004167.0 EI

wLA

3

00769.0

Slope at endA,

Documents

W4 Beam Deflection-1- Rev2