MECHANICS OF CHAPTER 9MATERIALS - IIT...

MECHANICS OF MATERIALS

CHAPTER

9 Deflection of Beams


9- 2

Deformation of a Beam Under Transverse LoadingRelationship between bending moment and

curvature for pure bending remains valid for general transverse loadings.

EIxM )(1

Cantilever beam subjected to tip load,

EIPx

1

Curvature varies linearly with x

At free end A, AA

ρρ

,01

At support B, PLEI

BB

,01


9 - 3

Deformation of a Beam Under Transverse Loading• Overhanging beam

• Reactions at A and C

• Bending moment diagram

• Curvature is zero at points where bending moment is zero, i.e., at each end and at E.

EIxM )(1

• Beam is concave upwards where bending moment is positive and concave downwards where it is negative.

• Maximum curvature occurs where the moment magnitude is maximum.

• An equation for the beam shape, i.e., elastic curve, is required to determine maximum deflection and slope.


9 - 4

Equation of the Elastic Curve

• Thus,

2

2

232

2

2

1

1dx

yd

dxdy

dxyd

• Substituting and integrating,

21

1

2

21

CxCdxxMdxyEI

CdxxMdxdyEIEI

xMdx

ydEIEI

2/32

2

2

2

2

2

22

2

2

2

1

1

1

1tan1tantan

1

1

tan

dxdydx

yddxdy

dxyd

dxd

dxd

dxdy

dxd

dxd

dd

dxyd

dxd

dxdydx

ddsd

dxdy


9 - 5

Equation of the Elastic Curve

21 CxCdxxMdxyEI

• Constants are determined from boundary conditions

• Three cases for statically determinate beams,

– Simply supported beam0,0 BA yy

– Overhanging beam0,0 BA yy

– Cantilever beam0,0 AAy

• More complicated loadings require multiple integrals and application of requirement for continuity of displacement and slope.


9 - 6

Direct Determination of Elastic Curve from Load Distribution

• For a beam subjected to a distributed load,

xwdxdV

dxMdxV

dxdM

2

2

• Equation for beam displacement becomes

xwdx

ydEIdx

Md 4

4

2

2

432

2213

161 CxCxCxC

dxxwdxdxdxxyEI

• Integrating four times yields

• Constants are determined from boundary conditions.


9- 7

Example 1

For portion AB of the overhanging beam, (a) derive equation for the elastic curve, (b) find maximum deflection, (c) evaluate ymax.


9- 8

Example 1

xLaP

dxydEI 2

2

- Differential equation for the elastic curve,

SOLUTION:

Develop expression for M(x) and derive differential equation for elastic curve.

- Reactions:

LaPR

LPaR BA 1

- From FBD for section AD,

LxxLaPM 0


9- 9

Example 1

PaLCLCLLaPyLx

Cyx

61

610:0,at

0:0,0at

113

2

Integrate differential equation twice and apply boundary conditions to obtain elastic curve.

213

12

61

21

CxCxLaPyEI

CxLaP

dxdyEI

xLaP

dxydEI 2

2

32

6 Lx

Lx

EIPaLy

PaLxxLaPyEI

Lx

EIPaL

dxdyPaLx

LaP

dxdyEI

61

61

3166

121

3

22

Substituting,


9- 10

Example 1Locate point of zero slope i.e., point

of maximum deflection.

32

6 Lx

Lx

EIPaLy

LLxL

xEI

PaLdxdy

mm 577.0

331

60

2

Evaluate maximum deflection.

32

max 577.0577.06

EI

PaLy

EIPaLy

2

max 0642.0

469

23

max m10300Pa10200m5.4m2.1N102000642.0

y

mm2.5max y


9 - 11

Statically Indeterminate Beams• Consider beam fixed at A and roller support at B.

• There are four unknown reaction components.

• Conditions for static equilibrium are000 Ayx MFF

So beam statically indeterminate to degree one. Say RB is redundant.

21 CxCdxxMdxyEI • We also have beam deflection equation,

which introduces two unknowns but provides three additional equations from the boundary conditions (used to solve for C1 , C2 , RB ):

0,At 00,0At yLxyx


9 - 12

Example 2

For the uniform beam, find reaction at A, derive equation for elastic curve, and find slope at A.

Beam is statically indeterminate to one degree (i.e., one excess reaction which static equilibrium alone cannot solve for).

SOLUTION:

• Develop differential equation for elastic curve (will be functionally dependent on reaction at A).

• Integrate twice, apply boundary conditions, solve for reaction at A and obtain elastic curve.


9 - 13

Example 2• Consider moment acting at section D,

LxwxRM

MxLxwxR

M

A

A

D

6

032

1

0

30

20

LxwxRM

dxydEI A 6

30

2

2

• The differential equation for the elastic curve,


9 - 14

Example 2

LxwxRM

dxydEI A 6

30

2

2

• Integrate twice

215

03

14

02

12061

2421

CxCL

xwxRyEI

CLxwxREI

dxdyEI

A

A

• Apply boundary conditions:

01206

1:0,at

0242

1:0,at

0:0,0at

214

03

13

02

2

CLCLwLRyLx

CLwLRLx

Cyx

A

A

• Solve for reaction at A

0301

31 4

03 LwLRA LwRA 010

1


9 - 15

Example 2

xLwL

xwxLwyEI

3

05

030 120

112010

161

xLxLxEIL

wy 43250 2120

• Substitute for C1, C2, and RA in the elastic curve equation,

42240 65120

LxLxEIL

wdxdy

EILw

A 120

30

• Differentiate once to find the slope,

at x = 0,


9 - 16

Method of Superposition

Principle of Superposition:

• Deformations of beams subjected to combinations of loadings may be obtained as a linear combination of the deformations from the individual loadings

• Procedure is facilitated by tables of solutions for common types of loadings and supports.


9 - 17

Example 3

For the beam and loading shown, find slope and deflection at point B.

SOLUTION:

Superpose the deformations due to Loading I and Loading II as shown.


9 - 18

Example 3

Loading I

EI

wLIB 6

3

EIwLy IB 8

4

Loading II

EI

wLIIC 48

3

EIwLy IIC 128

4

In beam segment CB, bending moment is zero and the elastic curve is a straight line.

EI

wLIICIIB 48

3

EI

wLLEI

wLEI

wLy IIB 3847

248128

434


9 - 19

Example 3

EI

wLEI

wLIIBIBB 486

33

EI

wLEI

wLyyy IIBIBB 3847

8

44

EIwL

B 487 3

EIwLyB 384

41 4

Combine the two solutions,


9 - 20

Application of Superposition to Statically Indeterminate Beams

• Method of superposition can be applied to find reactions of statically indeterminate beams.

• Designate one of the reactions as the redundant and eliminate or modify the support.

• Note that you must ensure that redundant chosen does not make structure unstable

• Determine beam deformation without redundant reaction.

• Treat redundant reaction as an unknown load which, together with the other (i.e., applied) loads, must produce deformations compatible with the original supports.


9 - 21

Example 4

For the uniform beam and loading shown, find reaction at each support and slope at A.

SOLUTION:

• Release “redundant” support/reaction at B, and find deformation.• Apply reaction at B as an unknown load to ensure zero displacement at B.


9 - 22

Example 4• Distributed Load:

EIwL

LLLLLEI

wy wB

4

334

01132.0

32

322

32

24

• Redundant Reaction Load:

EI

LRLLEILRy BB

RB322

01646.033

23

• For compatibility with original supports, yB = 0

EI

LREI

wLyy BRBwB

3401646.001132.00

wLRB 688.0

• From statics, wLRwLR CA 0413.0271.0

xLLxxEI

wy w334 2

24

EILbPayax

3,At

22

a b


9 - 23

Example 4

EI

wLEI

wLwA

3304167.0

24

EI

wLLLLEIL

wLRA

322 03398.0

3360688.0

EI

wLEI

wLRAwAA

3303398.004167.0

EIwL

A3

00769.0

Slope at A,

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