9 Beam Deflection 2

8/9/2019 9 Beam Deflection 2

1/23

MECHANICS OF

MATERIALS

Third Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

John T. e!ol"

Le#ture Notes$

J. !alt Oler

Te%as Te#h &ni'ersit(

CHAPTER

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

9Deflection of Beams


2/23

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

MECHANICS OF MATERIALST h i r d Beer ) Johnston ) e!ol"

Deformation of a Beam Under Transverse Loading

• Relationship between bending moment and

curvature for pure bending remains valid for

general transverse loadings.

EI

x M )(1 = ρ

• Cantilever beam subected to concentrated

load at the free end!

EI

Px−=

ρ

1

• Curvature varies linearl" with x

• #t the free end A! ∞== A A

ρ ρ

!$1

• #t the support B! PL

EI B

B

=≠ ρ ρ

!$1

http://contents.ppt/


3/23


MECHANICS OF MATERIALST h i r d Beer ) Johnston ) e!ol"

Deformation of a Beam Under Transverse Loading• &verhanging beam

• Reactions at A and C

• 'ending moment diagram

• Curvature is ero at points where the bending

moment is ero! i.e.! at each end and at E .

EI x M )(1 = ρ

• 'eam is concave upwards where the bending

moment is positive and concave downwards

where it is negative.

• a*imum curvature occurs where the moment

magnitude is a ma*imum.

• #n e+uation for the beam shape or elastic curve

is re+uired to determine ma*imum deflection

and slope.T



4/23


MECHANICS OF MATERIALSTh i r d Beer ) Johnston ) e!ol"

Equation of the Elastic Curve

• rom elementar" calculus! simplified for beam

parameters!

2

2

2%2

2

2

1

1

dx

yd

dx

dy

dx

yd

≈

+

= ρ

• ubstituting and integrating!

( )

( )

( ) 21$$

1

$

2

21

C xC dx x M dx y EI

C dx x M dx

dy EI EI

x M dx

yd EI EI

x x

x

++=

+=≈

==

∫ ∫

∫ θ

ρ

MECHANICS OF MATERIA ST



5/23



Equation of the Elastic Curve

( ) 21$$

C xC dx x M dx y EI

x x

++= ∫ ∫

• Constants are determined from boundar"

conditions

• 0hree cases for staticall" determinant beams!

impl" supported beam$!$ == B A y y

&verhanging beam$!$ == B A y y

Cantilever beam$!$ == A A y θ

• ore complicated loadings re+uire multiple

integrals and application of re+uirement for

continuit" of displacement and slope.

MECHANICS OF MATERIALST



6/23



Direct Determination of the Elastic Curve From the

Load Distriution

• or a beam subected to a distributed load!

( ) ( ) xwdx

dV

dx

M d xV

dx

dM −===2

2

• 3+uation for beam displacement becomes

( ) xwdx

yd EI dx

M d −==,

,2

2

( ) ( )

,%2

221%

11 C xC xC xC

dx xwdxdxdx x y EI

++++

−= ∫ ∫ ∫ ∫ • 4ntegrating four times "ields

• Constants are determined from boundar"

conditions.




7/23© 2002 The McGraw-Hill Companies, Inc. All rights reserved.


!taticall" #ndeterminate Beams• Consider beam with fi*ed support at A and roller

support at B.

• rom free-bod" diagram! note that there are four

un6nown reaction components.

• Conditions for static e+uilibrium "ield

$$$ =∑=∑=∑ A y x M F F

0he beam is staticall" indeterminate.

( ) 21$$

C xC dx x M dx y EI

x x

++= ∫ ∫

• #lso have the beam deflection e+uation!

which introduces two un6nowns but providesthree additional e+uations from the boundar"

conditions7

$!#t$$!$#t ===== y L x y x θ






!am$le Prolem %&'

ft,ft1/6ips/$

psi1$29in52%81, ,

===×==×

a L P

E I W

or portion AB of the overhanging beam!

(a) derive the e+uation for the elastic

curve! (b) determine the ma*imum

deflection!

(c) evaluate ymax.

&:04&;7

•




!am$le Prolem %&'

&:04&;7

•




!am$le Prolem %&'

PaLC LC L L

a P y L x

C y x

1

1$7$!at

$7$!$at

11%

2

=+−===

===

• 4ntegrate differential e+uation twice and appl"

boundar" conditions to obtain elastic curve.

21%

12

1

2

1

C xC x L

a P y EI

C x L

a P

dx

dy EI

++−=

+−=

x L

a P

dx

yd EI −=

2

2

−=%2

L

x

L

x

EI

PaL y

PaLx x L

a P y EI

L x

EI PaL

dxdy PaL x

La P

dxdy EI

1

1

%1

121

%

22

+−=

−=+−=

ubstituting!






!am$le Prolem %&'

• ocate point of ero slope or point

of ma*imum deflection.

−=

%2

L

x

L

x

EI

PaL y

L L

x L

x

EI

PaL

dx

dym

m /55.$%

%1

$2

==

−==

• 3valuate corresponding ma*imum

deflection.

( )[ ]%2

ma* /55.$/55.$

−= EI

PaL y

EI

PaL y

$,2.$2

ma* =

( ) ( ) ( )

( )( ),2

ma*in52% psi1$29

in18$in,86ips/$$,2.$

×= y

in2%8.$ma* = y






!am$le Prolem %&(

or the uniform beam! determine the

reaction at A! derive the e+uation for

the elastic curve! and determine the

slope at A. (;ote that the beam is

staticall" indeterminate to the first

degree)

&:04&;7

•




!am$le Prolem %&(

• Consider moment acting at section D!

L

xw

x R M

M x

L

xw x R

M

A

A

D

$%2

1

$

%$

2$

−=

=−

−

=∑

L

xw x R M

dx

yd EI

A

%$

2

2

−==

• 0he differential e+uation for the elastic

curve!





MECHANICS OF MATERIALSThi r d Beer ) Johnston ) e!ol"

!am$le Prolem %&(

L

xw x R M

dx

yd EI A

%$

2

2−==

• 4ntegrate twice

21

/$%

1

,

$2

12$

1

2,2

1

C xC L

xw x R y EI

C L

xw x R EI

dx

dy EI

A

A

++−=

+−== θ

• #ppl" boundar" conditions7

$12$

17$!at

$2,2

17$!at

$7$!$at

21

,

$%

1

%$2

2

=++−==

=+−==

===

C LC Lw

L R y L x

C Lw

L R L x

C y x

A

Aθ

• olve for reaction at A

$%$

1

%

1 ,$

% =− Lw L R A ↑= Lw R A $1$

1

MECHANICS OF MATERIALST h





!am$le Prolem %&(

x Lw L

xw x Lw y EI

−−

= %$

/$%

$12$

1

12$1$

1

1

( ) x L x L x EIL

w y ,%2/$ 2

12$

−+−=

• ubstitute for C1! C2! and R # in the

elastic curve e+uation!

( ),22,$ /12$

L x L x

EIL

w

dx

dy−+−==θ

EI

Lw A

12$

%$=θ

•




)ethod of !u$er$osition

>rinciple of uperposition7

•



MECHANICS OF MATERIALShi r d Beer ) Johnston ) e!ol"

!am$le Prolem %&*

or the beam and loading shown!

determine the slope and deflection at

point B.

&:04&;7

uperpose the deformations due to L!adi"# I and L!adi"# II as shown.






!am$le Prolem %&*

L!adi"# I

( ) EI

wL I B

%−=θ ( )

EI

wL y I B 8

,−=

L!adi"# II

( ) EI

wL II C ,8

%=θ ( ) EI

wL y II C 128,=

4n beam segment C'! the bending moment is

ero and the elastic curve is a straight line.

( ) ( ) EI

wL II C II B ,8

%== θ θ

( ) EI

wL L

EI

wL

EI

wL y II B %8,

5

2,8128

,%,

=

+=






!am$le Prolem %&*

( ) ( ) EI

wL EI

wL II B I B B ,8

%% +−=+= θ θ θ

( ) ( ) EI

wL

EI

wL y y y II B I B B %8,

5

8

,,

+−=+=

EI wL

B,85 %=θ

EI

wL y B

%8,

,1 ,=

Combine the two solutions!






A$$lication of !u$er$osition to !taticall"

#ndeterminate Beams

• ethod of superposition ma" be

applied to determine the reactions at

the supports of staticall" indeterminate

beams.

•




!am$le Prolem %&+

or the uniform beam and loading shown!determine the reaction at each support and

the slope at end A.

&:04&;7• Release the ?redundant@ support at '! and find deformation.

• #ppl" reaction at B as an un6nown load to force ero displacement at B.






!am$le Prolem %&+

•


23/23


!am$le Prolem %&+

( ) EI

wL

EI

wLw A

%%

$,15.$2,

−=−=θ

( ) EI wL L L L

EILwL R A

%2

2 $%%98.$%%

$88.$ =

− =

θ

( ) ( ) EI

wL

EI

wL R Aw A A

%%

$%%98.$$,15.$ +−=+= θ θ θ EI

wL A

%

$$59.$−=θ

lope at end A!

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9 Beam Deflection 2