9 beam deflection

5/23/2018 9 Beam Deflection

1/34

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 reserved.

9Deflection of Beams


2/34


MECHANICS OF MATERIALSThird

Edition

Beer ) Johnston ) e!ol"

Deflection of Beams

Deformation of a Beam Under Transverse Loading

Equation of the Elastic Curve

Direct Determination of the Elastic Curve From the Load Di...

taticall! "ndeterminate Beams

am#le $ro%lem 9.&

am#le $ro%lem 9.'

(ethod of u#er#osition

am#le $ro%lem 9.)

*##lication of u#er#osition to taticall! "ndeterminate ...

am#le $ro%lem 9.+

(oment-*rea Theorems

*##lication to Cantilever Beams and

Beams ,ith !mmetric ...

Bending (oment Diagrams %! $arts

am#le $ro%lem 9.&&

*##lication of (oment-*rea Theorems to

Beams ,ith Uns!mme...

(aimum Deflection

Use of (oment-*rea Theorems ,ithtaticall! "ndeterminate...


3/34



Edition



/elationshi# %et0een %ending moment and

curvature for #ure %ending remains valid for

general transverse loadings.

I

!M 1&=

Cantilever %eam su%3ected to concentrated

load at the free end4

I

"!=

&

Curvature varies linearl! 0ith!

*t the free endA4 == AA

##

45&

*t the su##ort$4"%

I$

$

=

45&


4/34



Edition



7verhanging %eam

/eactions atAand C

Bending moment diagram

Curvature is 8ero at #oints 0here the %ending

moment is 8ero4 i.e.4 at each end and at.

I!M 1& =

Beam is concave u#0ards 0here the %ending

moment is #ositive and concave do0n0ards

0here it is negative.

(aimum curvature occurs 0here the moment

magnitude is a maimum.

*n equation for the %eam sha#e or elastic c&rve

is required to determine maimum deflection

and slo#e.

C CS O ST


5/34



Edition



From elementar! calculus4 sim#lified for %eam

#arameters4

2

2

2'2

2

2

&

&

d!

'd

d!

d'

d!

'd

+

=

u%stituting and integrating4

( )

( )

( ) 2&55

&

5

2

2&

C!Cd!!Md!'I

Cd!!Md!

d'II

!Md!

'dII

!!

!

++=

+=

==

MECHANICS OF MATERIALSTE


6/34



Edition



( ) 2&55

C!Cd!!Md!'I

!!

++=

Constants are determined from %oundar!

conditions

Three cases for staticall! determinant %eams4

; im#l! su##orted %eam545 == $A ''

; 7verhanging %eam545 == $A ''

; Cantilever %eam545 == AA'

(ore com#licated loadings require multi#le

integrals and a##lication of requirement for

continuit! of dis#lacement and slo#e.



7/34



Edition


Direct Determination of the Elastic Curve From the

Load Distriution

For a %eam su%3ected to a distri%uted load4

( ) ( )!wd!

d(

d!

Md!(

d!

dM===

2

2

Equation for %eam dis#lacement %ecomes

( )!wd!

'dId!

Md ==6

62

2

( ) ( )

6'2

22&'

&:& C!C!C!C

d!!wd!d!d!!'I

++++

= "ntegrating four times !ields

Constants are determined from %oundar!

conditions.



8/34



Edition


!taticall" #ndeterminate Beams

Consider %eam 0ith fied su##ort atA and roller

su##ort at$.

From free-%od! diagram4 note that there are four

un


9/34



Edition


!am$le Prolem %&'

ft6ft&


10/34



Edition


!am$le Prolem %&'

7LUT"7>=

Develo# an e#ression for (1 and derivedifferential equation for elastic curve.

- /eactions=

+==

%

a"

%

"a $A &

- From the free-%od! diagram for sectionA4

( )%!!%

a"M


11/34 2002 The McGraw-Hill Companies, Inc. All rights reserved.


Edition


!am$le Prolem %&'

"a%C%C%%

a"'%!

C'!

:

&

:

&5=54at

5=545at

&&'

2

=+===

===

"ntegrate differential equation t0ice and a##l!

%oundar! conditions to o%tain elastic curve.

2&'

&2

:

&

2

&

C!C!%

a"'I

C!%

a"

d!

d'I

++=

+=

!%

a"

d!

'dI =

2

2

='2

: %

!

%

!

I

"a%'

"a%!!%

a"'I

%!

I"a%

d!d'"a%!

%a"

d!d'I

:

&

:

&

'&::

&2&

'

22

+=

=+=

u%stituting4





Edition


!am$le Prolem %&'

Locate #oint of 8ero slo#e or #oint

of maimum deflection.

=

'2

: %

!

%

!

I

"a%'

%%

!%

!

I

"a%

d!

d'm

m )).5'

'&:

52

==

==

Evaluate corres#onding maimum

deflection.

( )[ ]'2

ma )).5)).5:

=I

"a%'

I

"a%'

:

5:62.52

ma=

( ) ( ) ( )

( )( )6:2

main)2'#si&529:

in&+5in6+




Edition


!am$le Prolem %&(

For the uniform %eam4 determine the

reaction atA4 derive the equation for

the elastic curve4 and determine the

slo#e atA. >ote that the %eam is

staticall! indeterminate to the first

degree1

7LUT"7>=

Develo# the differential equation for

the elastic curve 0ill %e functionall!

de#endent on the reaction atA1.

"ntegrate t0ice and a##l! %oundar!

conditions to solve for reaction atA

and to o%tain the elastic curve.

Evaluate the slo#e atA.

MECHANICS OF MATERIALST

E




Edition


!am$le Prolem %&(

Consider moment acting at section4

%

!w

!M

M!

%

!w!

M

A

A

:

5'2

&

5

'5

25

=

=

=

%

!w!M

d!

'dI

A :

'5

2

2

==

The differential equation for the elastic

curve4

MECHANICS OF MATERIALST

E




Edition


!am$le Prolem %&(

%

!w!M

d!

'dI A

:

'52

2==

"ntegrate t0ice

2&

5'

&

6

52

&25:

&

262

&

C!C%

!w!'I

C%

!w!I

d!

d'I

A

A

++=

+==

*##l! %oundar! conditions=

5&25:

&=54at

5262

&=54at

5=545at

2&

6

5'

&

'52

2

=++==

=+==

===

C%C%w

%'%!

C%w

%%!

C'!

A

A

olve for reaction atA

5'5

&

'

& 65

' = %w%A = %wA 5&5

&

MECHANICS OF MATERIALSTh

Ed




Edition


!am$le Prolem %&(

!%w%

!w!%w'I

= '5

5'

5&25

&

&25&5

&

:

&

( )!%!%!I%

w' 6'25 2

&25

+=

u%stitute for C&4 C24 and /* in the

elastic curve equation4

( )62265 :&25

%!%!

I%

w

d!

d'+==

I

%wA

&25

'5=

Differentiate once to find the slo#e4

at!? 54


Ed



MECHANICS OF MATERIALShird

Edition


)ethod of !u$er$osition

$rinci#le of u#er#osition=

Deformations of %eams su%3ected to

com%inations of loadings ma! %e

o%tained as the linear com%ination of

the deformations from the individualloadings

$rocedure is facilitated %! ta%les of

solutions for common t!#es of

loadings and su##orts.


Ed




dition


!am$le Prolem %&*

For the %eam and loading sho0n4

determine the slo#e and deflection at

#oint$.

7LUT"7>=

u#er#ose the deformations due to%oading Iand%oading II as sho0n.


Ed




dition


!am$le Prolem %&*

%oading I

( )I

w%I$ :

'= ( )

I

w%' I$ +

6=

%oading II

( )I

w%IIC 6+

'= ( )I

w%' IIC &2+6=

"n %eam segment CB4 the %ending moment is

8ero and the elastic curve is a straight line.

( ) ( )I

w%IICII$ 6+

'==

( )I

w%%

I

w%

I

w%' II$ '+6

)

26+&2+

6'6

=

+=


Ed




dition


!am$le Prolem %&*

( ) ( )I

w%Iw%

II$I$$6+:

''

+=+=

( ) ( )I

w%

I

w%''' II$I$$ '+6

)

+

66

+=+=

Iw%

$6+)

'

=

I

w%'$

'+6

6& 6=

Com%ine the t0o solutions4


Ed




dition


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

#ndeterminate Beams

(ethod of su#er#osition ma! %e

a##lied to determine the reactions at

the su##orts of staticall! indeterminate

%eams.

Designate one of the reactions as

redundant and eliminate or modif!

the su##ort.

Determine the %eam deformation

0ithout the redundant su##ort.

Treat the redundant reaction as an

un




dition


!am$le Prolem %&+

For the uniform %eam and loading sho0n4

determine the reaction at each su##ort and

the slo#e at endA.

7LUT"7>= /elease the @redundantA su##ort at B4 and find deformation.

*##l! reaction at$as an un




dition


!am$le Prolem %&+

Distri%uted Loading=

( )

I

w%

%%%%%Iw' w$

6

'

'6

5&&'2.5

'2

'22

'2

26

=

+=

/edundant /eaction Loading=

( )I

%%%

I%

' $$$

'22

5&:6:.5''

2

'=

=

For com#ati%ilit! 0ith original su##orts4'$? 5

( ) ( ) I

%

I

w%'' $$w$

'6

5&:6:.55&&'2.55 +=+=

= w%,$ :++.5

From statics4

== w%w% CA 56&'.52)&.5


Ed



MECHANICS OF MATERIALSirddition


!am$le Prolem %&+

( )I

w%

I

w%wA

''

56&:).526

==

( )Iw%%%%

I%w%A

'2

2 5''9+.5'':

5:++.5 =

=

( ) ( )I

w%

I

w%,AwAA

''

5''9+.556&:).5 +=+= I

w%A

'

55):9.5=

lo#e at endA4


Ed

B J h t ! l"



MECHANICS OF MATERIALSirdition


)oment,Area Theorems

eometric #ro#erties of the elastic curve can

%e used to determine deflection and slo#e.

Consider a %eam su%3ected to ar%itrar! loading4

)irst Moment-Area Theorem=

area under M/Idiagram %et0een

Cand.

=C-

MECHANICS OF MATERIALSThi

Ed

B J h t ! l"





)oment,Area Theorems

1econd Moment-Area Theorem=

The tangential deviation of C0ith res#ect to

is equal to the first moment 0ith res#ect to avertical ais through Cof the area under the

M/I1 diagram %et0een Cand.

Tangents to the elastic curve at"and"interce#t

a segment of length dton the vertical through C.

? tangential deviation of C

0ith res#ect to


Edi

B J h t ! l"





A$$lication to Cantilever Beams and Beams -ith

!"mmetric Loadings

Cantilever %eam - elect tangent atAas thereference.

im#l! su##orted4 s!mmetricall! loaded

%eam - select tangent at Cas the reference.


Edi

Beer Johnston e!ol"


28/34


MECHANICS OF MATERIALSrdition


Bending )oment Diagrams " Parts

Determination of the change of slo#e and the

tangential deviation is sim#lified if the effect ofeach load is evaluated se#aratel!.

Construct a se#arate M/I1 diagram for each

load.

- The change of slo#e4 /C4 is o%tained %!

adding the areas under the diagrams.

- The tangential deviation4 t/Cis o%tained %!

adding the first moments of the areas 0ith

res#ect to a vertical ais through D.

Bending moment diagram constructed from

individual loads is said to %e drawn +' parts.


Edi



29/34


MECHANICS OF MATERIALSrdtion


!am$le Prolem %&''

For the #rismatic %eam sho0n4 determinethe slo#e and deflection at.

7LUT"7>= Determine the reactions at su##orts.

Construct shear4 %ending moment and

M/I1 diagrams.

Ta


30/34




!am$le Prolem %&''

7LUT"7>=

Determine the reactions at su##orts.

wa $ ==

Construct shear4 %ending moment and

M/I1 diagrams.

( )I

waa

I

waA

I

%wa%

I

waA

:2'

&

622

'2

2

22

&

=

=

=

=

MECHANICS OF MATERIALSThir

Edit Beer ) Johnston ) e!ol"


31/34




!am$le Prolem %&''

lo#e at E=

I

wa

I

%waAA

CCC

:6

'2

2& =+=

=+=

( )a%I

wa 2'

&2

2

+=

=

+

+=

=

I

%wa

I

wa

I

%wa

I

%wa

%A

aA

%aA

tt' CC

&:+&:6

66

'

6

22622'

&2&

( )a%I

wa' += 2

+

'

Deflection at E=




32/34




A$$lication of )oment,Area Theorems to Beams

-ith Uns"mmetric Loadings

Define reference tangent at su##ortA. Evaluate A%! determining the tangential deviation at$0ith

res#ect toA.

The slo#e at other #oints is found 0ith res#ect toreference tangent.

A-A- +=

The deflection atis found from the tangential

deviation at.




33/34




)a.imum Deflection

(aimum deflection occurs at #oint3

0here the tangent is hori8ontal.

$oint3ma! %e determined %! measuring

an area under the M/I1 diagram equal

to -A.

7%tain'ma!%! com#uting the first

moment 0ith res#ect to the vertical ais

throughAof the area %et0eenAand3.




34/34


Beer Johnston e!ol"

Use of )oment,Area Theorems -ith !taticall"

#ndeterminate Beams

/eactions at su##orts of staticall! indeterminate%eams are found %! designating a redundant

constraint and treating it as an un

9 beam deflection

Documents