Structure Analysis IStructure Analysis IChapter 8Chapter 8

Deflections

IntroductionIntroduction

• Calculation of deflections is an important part ofCalculation of deflections is an important part of structural analysis

• Excessive beam deflection can be seen as a mode of failure.– Extensive glass breakage in tall buildings can be attributed to excessive deflections

– Large deflections in buildings are unsightly (and unnerving) and can cause cracks in ceilings and wallsand can cause cracks in ceilings and walls.

– Deflections are limited to prevent undesirable vibrations

Beam DeflectionBeam Deflection

• Bending changes theBending changes the initially straight longitudinal axis of the beam into a curve that is called the 

flDeflection Curve or Elastic Curve

Beam DeflectionBeam Deflection

• To determine the deflection curve:To determine the deflection curve:– Draw shear and moment diagram for the beam

– Directly under the moment diagram draw a line for the beam and label all supports

– At the supports displacement is zero

h h i i h d fl i i– Where the moment is negative, the deflection curve is concave downward.

– Where the moment is positive the deflection curve isWhere the moment is positive the deflection curve is concave upward

– Where the two curve meet is the Inflection Point

Deflected Shape

Example 1Draw the deflected shape for each of the beams shown

Example 2Draw the deflected shape for each of the frames shown

Double Integration Method

Elastic‐Beam Theory

• Consider a differential element• Consider a differential element of a beam subjected to pure bending.g

• The radius of curvature ρ is measured from the center of curvature to the neutral axis

• Since the NA is unstretched, the dx=ρdθ

Elastic‐Beam TheoryElastic Beam Theory

• Applying Hooke’s law and the Flexure formula we• Applying Hooke s law and the Flexure formula, we obtain:

M1EI

=ρ

Elastic‐Beam Theory• The product EI is referred to as the flexural rigidity• The product EI is referred to as the flexural rigidity.• Since dx = ρdθ, then

M )(SlopedxEIMd =θ

In most calculus booksIn most calculus books

[ ]dxvd

= 3

22 /1

In most calculus books In most calculus books 

( )[ ]l titdxvdM

dxdv+22

23

2

)(//1ρ

( )[ ]solutionexact

dxdvEI+

=

2

23

2)(

/1

EIM

dxvd=2

2

The Double Integration MethodThe Double Integration MethodRelate Moments to Deflections

Md 2

EIM

dxvd=2

2

( )∫== dx

EIxM

ddvx

)()(θ

Do NotIntegration Constants∫ xEIdx )(

( )∫∫

xMUse Boundary Conditions to Evaluate Integration Constants( )

∫∫= 2

)()( dx

xEIxMxv

Constants

Assumptions and Limitations

Deflections caused by shearing action negligibly small compared to bending

Deflections are small compared to the cross‐sectional dimensions of the beam

All portions of the beam are acting in the elastic range

Beam is straight prior to the application of loads

ExamplesLx

y

PxPLM +−=x

P

PL

P

PxPLM +=

Mdx

ydEI =2

2

yd 2

@ x PxPLdx

ydEI +−=2

Integrating once 1

2

2cxPPLx

dxdyEI ++−=

2dx

@ x = 0 ( ) ( ) ( ) 020000 11

2

=⇒++−=⇒= ccPPLEIdxdy

32PLIntegrating twice 2

32

62cxPPLxEIy ++−=

@ x = 0 ( ) ( ) ( ) 00000 22

32 =⇒++−=⇒= ccPPLEIy

3( ) ( )

62 22y

62

32 xPPLxEIy +−=

@ x = L  y = ymax EIPL3

3

max =∆

62y

EIPLyPLLPLPLEIy3662

3

max

332

max −=⇒−=+−=

y

( )2xLWM −−=W

( )2

xLM =

Mdx

ydEI =2

2

2 WydL

xx

WL2

2WL

@ x ( )22 2xLW

dxydEI −−=WL2

Integrating once( )

1

3

32cxLW

dxdyEI +

−=

@ x = 0 ( ) ( )63

02

003

11

3 WLccLWEIdxdy

−=⇒+−

=⇒=

( )66

33 WLxLW

dxdyEI −−=∴

66dx

Integrating twice( ) 34

cxWLxLWEIy +−

=Integrating twice 2646cxEIy +−−=

@ x = 0 ( ) ( ) ( )24

064

06

004

22

34 WLccWLLWEIy =⇒+−−

−=⇒=

( )43

4 WLxWLxLWEIy +−−−=

( ) ( )24646 22y

( )24624

xxLEIy +=

Max occurs @ x = LMax. occurs @ x = L  

EIWLyWLWLLWEIy88246

4

max

444

max −=⇒−=+−=

EIWL8

4

max =∆EI8

Exampley x

L

x

WL WL2 2

22xWxxWLM −=

22 xWLyd222

xWxWLdx

ydEI −=

Integrating 1

32

3222cxWxWL

ddyEI +−=g g 13222dx

Since the beam is symmetric 02

@ ==dxdyLx

⎞⎜⎛⎞

⎜⎛

32 LL

( ) ⇒+⎠⎞

⎜⎝⎛

−⎠⎞

⎜⎝⎛

== 132

222

20

2@ c

LW

LWLEILx

24

3

1WLc −=

3

2464

332 WLxWxWL

dxdyEI −−=∴

Integrating2

343

cxWLxWxWLEIy +−−=g g2244634

cxEIy +

@ x = 0 y = 0 ( ) ( ) ( ) ( ) 2

343

0000 cWLWWLEI +−−=⇒ 02 =⇒ c( ) ( ) 2244634 2

xWLxWxWLEIy3

43 −−=∴ xxxEIy242412

=∴

Max occurs @ x = L /25 4WLEIy =Max. occurs @ x = L /2 384maxEIy −=

WL5 4

∆EI384max =∆

Exampley x P

L/2

x

P PL/2

f2 LPyd

xPMLx22

0for =<<2 2

20for

22

LxxPdx

ydEI <<=

Integrating 1

2

22cxP

ddyEI +=g g 122dx

Since the beam is symmetric 02

@ ==dxdyLx

⎞⎜⎛

2L

( ) ⇒+⎠⎞

⎜⎝⎛

== 122

20

2@ c

LPEILx

16

2

1PLc −=

2

164

22 PLxP

dxdyEI −=∴

Integrating2

23

cxPLxPEIy +−=g g21634

cxEIy +

@ x = 0 y = 0 ( ) ( ) ( ) 2

23

000 cPLPEI +−=⇒ 02 =⇒ c( ) ( ) 21634 2

xPLxPEIy2

3 −=∴ xxEIy1612

∴

Max occurs @ x = L /23PLEIy =Max. occurs @ x = L /2 

48maxEIy −=

PL3

∆EI48max =∆

Example

Example 5

Moment‐Area Theorems

Moment‐Area Theorems

Theorem 1: The change in slope between any two points onthe elastic curve equal to the area of the bending momentdiagram between these two points, divided by the product EI.

2

2

d v M dvdx EI dxd M M

θ

θ

= ⇒ =

⎛ ⎞

B

d M Md dxdx EI EI

M

θ θ ⎛ ⎞= ⇒ = ⎜ ⎟⎝ ⎠

B

B AA

M dxEI

θ = ∫

dt xdMθ=

⎛ ⎞

B B

Md dxEI

M M

θ ⎛ ⎞= ⎜ ⎟⎝ ⎠

B AA A

M Mt x dx x dxEI EI

= =∫ ∫

Moment‐Area Theorems

Theorem 2: The vertical distance of point A on a elasticcurve from the tangent drawn to the curve at B is equal tothe moment of the area under the M/EI diagram betweentwo points (A and B) about point A .

B

A BA

Mt x dxEI

= ∫A

B

A BA

Mt x dxEI

= ∫A

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

M/EI

‐30/EI‐20/EI

‐

( )20 1 10 22 1 2 2t ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − ⋅ ⋅ + ⋅ − ⋅ ⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟( )/

3

2 1 2 22 3

53.33 0.00741

C BtEI EI

kN m rad

= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − ⋅ =EI

Another Solution

Conjugate-Beam Method

Conjugate-Beam Method

2

2

dV d Mw wdx dx

= =

2

2

dx dxd M d v Mdx EI dx EIθ= =

dx EI dx EI

Integrating

V wdx M wdx dx

M M

⎡ ⎤= = ⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞

∫ ∫ ∫

∫ ∫ ∫M Mdx v dx dxEI EI

θ ⎡ ⎤⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∫ ∫ ∫

tspp

ort

m S

up-B

eam

ugat

e-Co

nju

C

Example 1Example 1Find the Max. deflection Take E=200Gpa, I=60(106)

EIVBB

5.562' −==θ

EIEIM

EI

BB5.14062)25(5.562

'−

===∆

Example 2Example 2Find the deflection at Point C

C

EIEIEIMCC

162)3(63)1(27'

−=−==∆

Example 3Example 3Find the deflection at Point D

3600∆

EI360

EI720 EI

M DD3600

' ==∆

Example 4Example 4Find the Rotation at A

10 ft

3.33θEIA

3.33=θ

Example 5Example 5

Copyright © 2009 Pearson Prentice Hall Inc.

Example 6Example 6

Moment Diagrams and Equations for Maximum DeflectionMaximum Deflection

Example 4Example 4 Find the Maximum deflection for the following structure based onThe previous diagrams p g