Structure I Lecture20

8/20/2019 Structure I Lecture20

1/18

6.06.0 ELASTIC DEFLECTION OF BEAMS ELASTIC DEFLECTION OF BEAMS

6.1 Introduction

6.2 Double-Integration Method

6.3 Examples

6.4 Moment Area Method

6. Examples


2/18

Introductionx! !

"

Elastic cur#e

$he de%lection is measured %rom the original neutral axis to the neutral axis o% the

de%ormed beam.

$he displacement " is de%ined as the de%lection o% the beam.

It ma" be necessar" to determine the de%lection " %or e#er" #alue o% x along the

beam. $his relation ma" be &ritten in the %orm o% an e'uation &hich is %re'uentl"

called the e'uation o% the de%lection cur#e (or elastic cur#e) o% the beam

Importance o% *eam De%lections

A designer should be able to determine de%lections+ i.e.

In building codes "max ,beam/300

Anal"ing staticall" indeterminate beams in#ol#e the use o% #arious de%ormation relationships.


3/18

Methods o% Determining *eam De%lections

a) Double-Integration Method

b) Moment-Area Method

c) Elastic Energ" Methods

d) Method o% singularit" %unctions


4/18

Double-Integration Method$he de%lection cur#e o% the bent beam is M

dx

yd EI =

2

2

In order to obtain "+ abo#e e'uation needs to be integrated t&ice.

"

ρ

adius o%

cur#ature"

x

)(Curvature1

κ ==ρ

⇒ρ

= EI

M EI M

An expression %or the cur#ature at an" point along the cur#e representing the

de%ormed beam is readil" a#ailable %rom di%%erential calculus. $he exact %ormula

%or the cur#ature is

2

32

2

2

1

+

=κ

dx

dy

dx

yd

smallisdx

dy2

2

dx

yd =κ ∴ M

dx

yd EI =

2

2


5/18

The Integration Procedure

Integrating once "ields to slope d"/dx at an" point in the beam.

Integrating t&ice "ields to de%lection " %or an" #alue o% x.

$he bending moment M must be expressed as a %unction o% the coordinate xbe%ore the integration

Di%%erential e'uation is 2nd order+ the solution must contain t&o constants o%

integration. $he" must be e#aluated at no&n de%lection and slope points(i.e. at a simple support de%lection is ero+ at a built in support both slope

and de%lection are ero)


6/18

Sign Convention

!ositi#e *ending egati#e *ending

Assumptions and Limitations

De%lections caused b" shearing action negligibl" small compared to bending

De%lections are small compared to the cross-sectional dimensions o% the beam

All portions o% the beam are acting in the elastic range

*eam is straight prior to the application o% loads


7/18

Examples x

x

"

!

!

!

Px PL M +−=

M dx

yd EI =2

2

5 x Px PLdx

yd EI +−=

2

2

Integrating once 1

2

2c

x P PLx

dx

dy EI ++−=

5 x 0 ( ) ( ) ( ) 0

2

0000 11

2

=⇒++−=⇒= cc P PL EI dx

dy

Integrating t&ice2

32

62c

x P

PLx EIy ++−=

5 x 0 ( ) ( ) ( )

06

00200

22

32

=⇒++−=⇒= cc P PL

EI y

62

32 x P

PLx EIy +−=

5 x " "max

EI

PL y

PL L P

L PL EIy

3662

3

max

332

max −=⇒−=+−=

EI

PL

3

3

max =∆


8/18

x x

"

( ) 22

x LW

M −−=

M dx

yd EI =2

2

5 x ( )2

2

2

2 x L

W

dx

yd EI −−=

Integrating once ( ) 13

32c x LW

dxdy EI +−=

5 x 0 ( ) ( )

63

0

200

3

11

3WL

cc LW

EI dx

dy−=⇒+

−=⇒=

W per unit length

2

2WL

( )66

33 WL

x LW

dx

dy EI −−=∴


9/18

( )

24624

434 WL

xWL

x LW

EIy +−−−=

Max. occurs 5 x

EI

WL y

WLWL LW EIy

88246

4

max

444

max −=⇒−=+−=

EI

WL

8

4

max =∆

Integrating t&ice ( )

2

34

646c x

WL x LW EIy +−

−−=

5 x 0 ( ) ( ) ( )24

064

0

600

4

22

34 WLccWL LW EI y =⇒+−−−=⇒=


10/18

Example

x

" x

2

WL

2

WL

22

xWx x

WL M −=

22

2

2

2 xW x

WL

dx

yd EI −=

Integrating 1

32

3222c

xW xWL

dx

dy EI +−=

7ince the beam is s"mmetric 02

@ ==dx

dy L x

( ) ⇒+

−

== 1

32

3

2

22

2

20

2@ c

LW

LWL

EI L

x24

3

1

WLc −=

2464

332 WL x

W x

WL

dx

dy EI −−=∴


11/18

Integrating2

343

244634c x

WL xW xWL EIy +−−=

5 x 0 " 0 ( ) ( ) ( ) ( ) 2343

0244

0

63

0

40 cWLW WL EI +−−=⇒ 02 =⇒ c

xWL

xW

xWL

EIy

242412

343 −−=∴

Max. occurs 5 x /2384

5 4

max

WL EIy −=

EI WL

3845

4

max =∆


12/18

Example"

20for

22

2 L x x

P

dx

yd EI


13/18

Integrating2

23

1634c x

PL x P EIy +−=

5 x 0 " 0 ( ) ( ) ( ) 223

0163

0

40 c PL P EI +−=⇒ 02 =⇒ c

x PL

x P

EIy

1612

23 −=∴

Max. occurs 5 x /248

3

max

PL EIy −=

EI PL48

3

max =∆


14/18

Moment-Area MethodFirst Moment –Area Theorem

$ a n g e n t a t A∆

The first moment are theorem states that8 $he angle bet&een the

tangents at A and * is e'ual to the

area o% the bending moment diagram

bet&een these t&o points+ di#ided b"

the product EI.

∫ =θ B

A

dx EI

M

$he second moment area theorem states that8 $he #ertical distance o% point * on a de%lection

cur#e %rom the tangent dra&n to the cur#e at A is e'ual to the moment &ith respect to the #ertical

through * o% the area o% the bending diagram bet&een A and *+ di#ided b" the product EI.

dx EI

Mx B

A∫ =∆

A*

ρ

$ a n g e n

t a t *

θ

dθ

dθ

x dx

ds

θ=ρ⇒θρ=

ρ=

d

dsd ds

EI M dxwithdsreplacesdeflectiolateralsmallisitds

EI

M d =θ

dx EI

M d dx

EI

M d

B

A

∫ ∫ =θ=θ=θ !ivewill!ite!rati ∫ =∆⇒=θ B

A

dx EI

Mxdx

EI

Mx xd

M


15/18

The Moment Area Procedure

1. $he reactions o% the beam are determined

2. An approximate de%lection cur#e is dra&n. $his cur#e must be consistent &ith

the no&n conditions at the supports+ such as ero slope or ero de%lection

3. $he bending moment diagram is dra&n %or the beam. 9onstruct M/EI diagram

4. 9on#enient points A and * are selected and a tangent is dra&n to the assumed

de%lection cur#e at one o% these points+ sa" A

. $he de%lection o% point * %rom the tangent at A is then calculated b" the secondmoment area theorem

Comparison of Moment Area and Double Integration Methods

I% the de%lection o% onl" a single point o% a beam is desired+ the moment-area

method is usuall" more con#enient than the double integration method.

I% the e'uation o% the de%lection cur#e o% the entire beam is desired the

double integration method is pre%erable.

Assumptions and Limitations

7ame assumptions as Double Integration Method holds.


16/18

Examples

!

!

!

A

*

∆

?

$angent at A

$angent at *

θ

!M

( )

33

2

2

3 PL L

PL L

EI −=

−=∆ EI

PL

3

3

−=∆

( ) PL L

EI −=θ2 EI

PL

2

2

−=θ


17/18

2

2

WL

$angent A

A W per unit length

*

∆

!

2

2WL

x

LWL

A23

1 2=

L x 4

3

=

84

3

23

4

2 WL L L

W L EI −= −=∆ EI

WL

8

4

−=∆


18/18

Example

a!

a!

! !

aa L

−2

!a

$angent A

A ∆ :

a Paa

aa L

a L

Pa EI 3

2

2242+

+−

−=∆

322

32448a

P a La La L Pa +

−−+=

−=−= 3

3332

432468 L

a La PL Pa PaL

−=∆

3

33 43

24 L

a

L

a

EI

PL

Structure I Lecture20

Documents

Transcript of Structure I Lecture20