fourth edition mechanics of materialsyunus.hacettepe.edu.tr/~boray/9_1_beam_deflection... · 2008....

MECHANICS OFFourth Edition

MECHANICS OF MATERIALS

CHAPTER

MATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T DeWolf

Deflection of BeamsJohn T. DeWolf

Lecture Notes:J. Walt OlerTexas Tech University

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.


FourthEdition

Beer • Johnston • DeWolf

Deflection of Beams

Deformation of a Beam Under Transverse Loading

Sample Problem 9.8

Moment-Area TheoremsEquation of the Elastic Curve

Direct Determination of the Elastic Curve F th L d Di

Moment Area Theorems

Application to Cantilever Beams and Beams With Symmetric ...

From the Load Di...

Statically Indeterminate Beams

Sample Problem 9 1

Bending Moment Diagrams by Parts

Sample Problem 9.11Sample Problem 9.1

Sample Problem 9.3

Method of Superposition

Application of Moment-Area Theorems to Beams With Unsymme...

Maximum DeflectionMethod of Superposition

Sample Problem 9.7

Application of Superposition to Statically

Maximum Deflection

Use of Moment-Area Theorems With Statically Indeterminate...Application of Superposition to Statically

Indeterminate ...

© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 2


FourthEdition


Deformation of a Beam Under Transverse Loading• Relationship between bending moment and

curvature for pure bending remains valid for general trans erse loadingsgeneral transverse loadings.

EIxM )(1

=ρ

• Cantilever beam subjected to concentrated load at the free end,

EIPx

−=ρ1

• Curvature varies linearly with x

• At the free end A ∞== A ρ,01• At the free end A, A

A ρ

ρ,0

• At the support B, PLEI

B =≠ ρρ

,01


pp , PLBρ


FourthEdition


Deformation of a Beam Under Transverse Loading• Overhanging beam

• Reactions at A and C

• Bending moment diagram

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

EIxM )(1

=ρ EIρ

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

• Maximum curvature occurs where the moment magnitude is a maximum.

• An equation for the beam shape or elastic curvei i d t d t i i d fl ti


is required to determine maximum deflection and slope.


FourthEdition


Equation of the Elastic Curve

• From elementary calculus, simplified for beam parameters,p ,

2

2

232

2

1 yddxyd

≈⎤⎡

= 22321

dx

dxdy

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+

ρ

• Substituting and integrating,

( )21 xMydEIEI == ( )

( ) 1

2

CdxxMdyEIEI

xMdx

EIEI

x+=≈

==

∫θ

ρ

( )

( ) 21

10

CxCdxxMdxyEI

CdxxMdx

EIEI

xx++=

+

∫∫

∫θ


( ) 2100

CxCdxxMdxyEI ++= ∫∫


FourthEdition


Equation of the Elastic Curve• Constants are determined from boundary

conditions

( ) 2100

CxCdxxMdxyEIxx

++= ∫∫

• Three cases for statically determinant beams,

– Simply supported beam00 0,0 == BA yy

– Overhanging beam00 0,0 == BA yy

– Cantilever beam0,0 == AAy θ, AAy

• More complicated loadings require multiple integrals and application of requirement for


integrals and application of requirement for continuity of displacement and slope.


FourthEdition


Direct Determination of the Elastic Curve From the Load Distribution

• For a beam subjected to a distributed load,

( ) ( )xwdxdV

dxMdxV

dxdM

−=== 2

2

• Equation for beam displacement becomes

( )xwydEIMd−==

42( )xw

dxEI

dx 42

• Integrating four times yields( ) ( )

432

2213

161 CxCxCxC

dxxwdxdxdxxyEI

++++

−= ∫∫∫∫432216

• Constants are determined from boundary conditions.



FourthEdition


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

support at B.• From free body diagram note that there are four• From free-body diagram, note that there are four

unknown reaction components.• Conditions for static equilibrium yield

000 =∑=∑=∑ Ayx MFF

The beam is statically indeterminate.

( ) 21 CxCdxxMdxyEIxx

++= ∫∫

• Also have the beam deflection equation,

( ) 2100

CCy ∫∫which introduces two unknowns but provides three additional equations from the boundarythree additional equations from the boundary conditions:

0,At 00,0At ===== yLxyx θ


yy


FourthEdition


Sample Problem 9.1

SOLUTION:SOLUTION:

• Develop an expression for M(x) and derive differential equation for

psi1029in7236814 64 ×==× EIW

qelastic curve.

• Integrate differential equation twice

ft 4ft15kips50

p

=== aLP

For portion AB of the overhanging beam

and apply boundary conditions to obtain elastic curve.

For portion AB of the overhanging beam, (a) derive the equation for the elastic curve, (b) determine the maximum deflection, ( ) l

• Locate point of zero slope or point of maximum deflection.

E l t di i(c) evaluate ymax. • Evaluate corresponding maximum deflection.



FourthEdition


Sample Problem 9.1SOLUTION:

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

- Reactions:

↑⎟⎠⎞

⎜⎝⎛ +=↓=

LaPR

LPaR BA 1

- From the free-body diagram for section AD,

( )LxxaPM <<0( )LxxL

PM <<−= 0

- The differential equation for the elastic

xaPydEI =2

- The differential equation for the elastic curve,


xL

Pdx

EI −=2


FourthEdition


Sample Problem 9.1• Integrate differential equation twice and apply

boundary conditions to obtain elastic curve.

3

12

121

CCaPEI

CxLaP

dxdyEI +−=

Cyx 0:0,0at 2 ===

213

6CxCx

LPyEI ++−=

ayd 2

PaLCLCLLaPyLx

61

610:0,at 11

3 =+−===x

LaP

dxydEI −=2

S b tit ti

Lx

EIPaL

dxdyPaLx

LaP

dxdyEI 31

661

21 2

2

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=+−=

Substituting,

⎥⎤

⎢⎡ ⎞

⎜⎛

32 xxPaLPaLxx

LaPyEI

LEIdxLdx

61

61

662

3 +−=

⎥⎦⎢⎣ ⎠⎝


⎥⎥⎦⎢

⎢⎣ ⎠

⎞⎜⎝⎛−=

6 Lx

Lx

EIPaLyL 66


FourthEdition


Sample Problem 9.1• Locate point of zero slope or point

of maximum deflection.

LLxL

xEI

PaLdxdy

mm 577.0

331

60

2==

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−==

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

32

6 Lx

Lx

EIPaLy

• Evaluate corresponding maximum deflection.

⎥⎦⎢⎣ ⎠⎝6 LLEI

( )[ ]32

max 577.0577.06

−=EI

PaLy

EIPaLy6

0642.02

max =

( )( )( )2( )( )( )( )( )46

2max

in723psi10296in180in48kips500642.0

×=y


in238.0max =y


FourthEdition


Sample Problem 9.3

SOLUTION:

D l th diff ti l ti f• Develop the differential equation for the elastic curve (will be functionally dependent on the reaction at A).

For the uniform beam, determine the• Integrate twice and apply boundary

conditions to solve for reaction at AFor the uniform beam, determine the reaction at A, derive the equation for the elastic curve, and determine the slope at A (Note that the beam is

and to obtain the elastic curve.

• Evaluate the slope at Aslope at A. (Note that the beam is statically indeterminate to the first degree)

• Evaluate the slope at A.



FourthEdition


Sample Problem 9.3• Consider moment acting at section D,

M D 0=∑

MxLxwxRA

D

032

1 20 =−⎟

⎟⎠

⎞⎜⎜⎝

⎛−

∑

LxwxRM A 6

30−=

⎠⎝

• The differential equation for the elastic curve,

LxwxRM

dxydEI A 6

30

2

2−==



FourthEdition


Sample Problem 9.3• Integrate twice

14

021 CxwxREIdyEI A +−== θ

215

03

1

12061

242

CxCL

xwxRyEI

CL

xREIdx

EI

A

A

++−=

+θ

xwRMydEI3

02

1206 L

• Apply boundary conditions:000t C

LxRM

dxyEI A 6

02 −==

0242

1:0,at

0:0,0at

13

02

2

=+−==

===

CLwLRLx

Cyx

Aθ

01206

1:0,at

242

214

03 =++−== CLCLwLRyLx A

• Solve for reaction at A

0301

31 4

03 =− LwLRA ↑= LwRA 010

1


303 10


FourthEdition


Sample Problem 9.3

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

xLwL

xwxLwyEI ⎟⎠⎞

⎜⎝⎛−−⎟

⎠⎞

⎜⎝⎛= 3

05

030 120

112010

161

( )xLxLxEIL

wy 43250 2120

−+−=

• Differentiate once to find the slope,

( )42240 65120

LxLxEIL

wdxdy

−+−==θ

EILw

A 120

30=θat x = 0,



FourthEdition


Method of Superposition

P i i l f S itiPrinciple of Superposition:

• Deformations of beams subjected to combinations of loadings may be

• Procedure is facilitated by tables of solutions for common types ofcombinations of loadings may be

obtained as the linear combination of the deformations from the individual loadings

solutions for common types of loadings and supports.

loadings



FourthEdition


Sample Problem 9.7

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

SOLUTION:

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



FourthEdition


Sample Problem 9.7

Loading I

L3 L4( )

EIwL

IB 6

3−=θ ( )

EIwLy IB 8

4−=

Loading II

( ) wLIIC

3=θ ( ) wLy IIC

4=( )

EIIIC 48θ ( )

EIy IIC 128

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

( ) ( )EI

wLIICIIB 48

3== θθ( ) ( )

EIIICIIB 48

( )EI

wLLEI

wLEI

wLy IIB 3847

248128

434=

⎠⎞

⎜⎝⎛+=


EIEIEIII 384248128 ⎠⎝


FourthEdition


Sample Problem 9.7

Combine the two solutions,

( ) ( )EI

wLEI

wLIIBIBB 486

33+−=+= θθθ

EIwL

B 487 3

−=θEIEIIII 486

( ) ( ) wLwLyyy 7 44+=+=

EI48

wLy 41 4−=


( ) ( )EIEI

yyy IIBIBB 3848+−=+=

EIyB 384

−=


FourthEdition


Application of Superposition to Statically Indeterminate Beams

• Method of superposition may be applied to determine the reactions at the supports of statically indeterminate

• Determine the beam deformation without the redundant support.

beams.

• Designate one of the reactions as

• Treat the redundant reaction as an unknown load which, together with the other loads, must produce

redundant and eliminate or modify the support.

pdeformations compatible with the original supports.


fourth edition mechanics of materialsyunus.hacettepe.edu.tr/~boray/9_1_beam_deflection... · 2008....

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