Mecanica Solidos Beer-09

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MECHANICS OFMATERIALS

Third Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

CHAPTER

Deflection of Beams .

Lecture Notes:

J. Walt Oler

Texas Tech University

© 2002 The McGraw-Hill Companies, Inc. All rights reserved.



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Beer • Johnston • DeWolf

Deflection of Beams

Deformation of a Beam Under Transverse

Loading

Equation of the Elastic Curve

Direct Determination of the Elastic CurveFrom the Load Di...

Statically Indeterminate Beams

Sample Problem 9.8

Moment-Area Theorems

Application to Cantilever Beams and

Beams With Symmetric ...

Bending Moment Diagrams by Parts

Sample Problem 9.11

© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 2

.

Sample Problem 9.3

Method of Superposition

Sample Problem 9.7

Application of Superposition to Statically

Indeterminate ...

Application of Moment-Area Theorems toBeams With Unsymme...

Maximum Deflection

Use of Moment-Area Theorems With

Statically Indeterminate...



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Deformation of a Beam Under Transverse Loading

• Relationship between bending moment and

curvature for pure bending remains valid for

general transverse loadings.

EI

x M )(1=

ρ

• Cantilever beam subjected to concentrated

load at the free end,


EI Px−=

ρ 1

• Curvature varies linearly with x

• At the free end A, ∞== A A

ρ ρ

,01

• At the support B,PL

EI B

B

=≠ ρ ρ

,01



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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

moment is zero, i.e., at each end and at E .

M )(1=


• Beam is concave upwards where the bending

moment is positive and concave downwards

where it is negative.

• Maximum curvature occurs where the momentmagnitude is a maximum.

• An equation for the beam shape or elastic curve

is required to determine maximum deflection

and slope.





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Equation of the Elastic Curve

( ) 21

00

C xC dx x M dx y EI

x x

++= ∫∫

• Constants are determined from boundary

conditions

• Three cases for statically determinant beams,

– Simply supported beam


, == B A

y y

– Overhanging beam0,0 == B A y y

– Cantilever beam

0,0 == A A y θ

• More complicated loadings require multiple

integrals and application of requirement for

continuity of displacement and slope.



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Direct Determination of the Elastic Curve From the

Load Distribution• For a beam subjected to a distributed load,

( ) ( ) xw

dx

dV

dx

M d xV

dx

dM −===

2

2

• Equation for beam displacement becomes

yd M d −==

42


dxdx 42

( ) ( )

432221316

1 C xC xC xC

dx xwdxdxdx x y EI

++++

−= ∫∫∫∫

• Integrating four times yields

• Constants are determined from boundary

conditions.



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Statically Indeterminate Beams

• Consider beam with fixed support at A and rollersupport at B.

• From free-body diagram, note that there are four

unknown reaction components.

• Conditions for static equilibrium yield

000 =∑=∑=∑ A y x M F F

The beam is statically indeterminate.


( ) 21

00

C xC dx x M dx y EI

x x

++= ∫∫

• Also have the beam deflection equation,

which introduces two unknowns but provides

three additional equations from the boundary

conditions:

0,At00,0At ===== y L y x θ



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Sample Problem 9.1

si1029in7236814 64 ×==× E I W

SOLUTION:

• Develop an expression for M(x)

and derive differential equation for

elastic curve.

• Integrate differential equation twice


ft4ft15kips50 === a LP

For portion AB of the overhanging beam,

(a) derive the equation for the elastic curve,

(b) determine the maximum deflection,(c) evaluate ymax.

and apply boundary conditions toobtain elastic curve.

• Locate point of zero slope or point

of maximum deflection.

• Evaluate corresponding maximum

deflection.



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Sample Problem 9.1

SOLUTION:

• Develop an expression for M(x) and derive

differential equation for elastic curve.

- Reactions:

↑

+=↓= L

aP R

L

Pa R B A 1


- From the free-body diagram for section AD,

( ) L x x L

aP M <<−= 0

x L

aP

dx

yd EI −=

2

2

- The differential equation for the elastic

curve,



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Sample Problem 9.1

C y 0:0,0at 2 ===

• Integrate differential equation twice and apply

boundary conditions to obtain elastic curve.

213

12

6

1

2

1

C xC x L

aP y EI

C x L

aP

dx

dy EI

++−=

+−=

2


PaLC LC L LaP y L x

61

610:0,at 113 =+−===

x LP

dx EI −=

2

−=

32

6 L

x

L

x

EI

PaL y

PaLx x L

aP y EI

L

x

EI

PaL

dx

dyPaL x L

aPdx

dy EI

6

1

6

1

3166

1

2

1

3

2

2

+−=

−=+−=

Substituting,



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Sample Problem 9.1

• Locate point of zero slope or point

of maximum deflection.

−=32

x xPaL y

L L

x

L

x

EI

PaL

dx

dym

m 577.0

3

31

6

0

2

==

−==

• Evaluate corresponding maximum

deflection.


( )[ ]32

max 577.0577.06

−= EI

PaL y

EI

PaL y

60642.0

2

max =

( )( )( )

( )( )46

2

maxin723psi10296

in180in48kips500642.0

×= y

in238.0max =

y



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Sample Problem 9.3

SOLUTION:

• Develop the differential equation for

the elastic curve (will be functionally

dependent on the reaction at A).

• Integrate twice and apply boundary

conditions to solve for reaction at


,

reaction at A, derive the equation forthe elastic curve, and determine the

slope at A. (Note that the beam is

statically indeterminate to the first

degree)

and to obtain the elastic curve.

• Evaluate the slope at A.



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Sample Problem 9.3

• Consider moment acting at section D,

L

xw x R M

M

x

L

xw

x R

M

A

A

D

6

032

1

0

30

20

−=

=−

−

=∑


L

xw x R M

dx

yd EI A

6

30

2

2

−==

• The differential equation for the elastic

curve,



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Sample Problem 9.3

wd 32

• Integrate twice

21

5

03

1

402

12061

242

1

C xC L xw x R y EI

C L

xw x R EI

dx

dy EI

A

A

++−=

+−== θ

• Apply boundary conditions:


L

x

dx A

62

−==

01206

1:0,at

0242

1:0,at

:,a

21

403

1

302

2

=++−==

=+−==

===

C LC Lw

L R y L x

C Lw

L R L x

y x

A

Aθ

• Solve for reaction at A

030

1

3

1 40

3 =− Lw L R A ↑= Lw R A 010

1



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Sample Problem 9.3

x Lw

L

xw x Lw y EI

−−

= 3

0

503

0

120

1

12010

1

6

1

( ) x L x L x EIL

w y

43250 2120

−+−=

• Substitute for C1, C2, and RA in the

elastic curve equation,


( )42240 65120

L x L x EIL

w

dx

dy−+−==θ

EI

Lw A

120

30=θ

• Differentiate once to find the slope,

at x = 0,



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Method of Superposition


r nc p e o uperpos on:

• Deformations of beams subjected to

combinations of loadings may be

obtained as the linear combination of

the deformations from the individual

loadings

• Procedure is facilitated by tables of

solutions for common types of

loadings and supports.



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Sample Problem 9.7

For the beam and loading shown,

determine the slope and deflection at

point B.

SOLUTION:


.



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Sample Problem 9.7

Loading I

( ) EI

wL I B

6

3

−=θ ( ) EI

wL y I B

8

4

−=

Loading II

wL3

= wL

4

=


EI 48 EI 128

In beam segment CB, the bending moment is

zero and the elastic curve is a straight line.

( ) ( ) EI

wL II C II B

48

3

== θ θ

( ) EI

wL L

EI

wL

EI

wL y II B

384

7

248128

434

=

+=



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Sample Problem 9.7


( ) ( ) EI

wL

EI

wL II B I B B 486

33

+−=+= θ θ θ

( ) ( ) EI

wL

EI

wL y y y II B I B B

384

7

8

44

+−=+=

EI

wL B 48

7 3

=θ

EI

wL y B

384

41 4

=

Combine the two solutions,

T E



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Application of Superposition to Statically

Indeterminate Beams


• Method of superposition may be

applied to determine the reactions at

the supports of statically indeterminate

beams.

• Designate one of the reactions asredundant and eliminate or modify

the support.

• Determine the beam deformation

without the redundant support.

• Treat the redundant reaction as an

unknown load which, together with

the other loads, must producedeformations compatible with the

original supports.

T E



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Sample Problem 9.8

For the uniform beam and loading shown,

determine the reaction at each support and

the slope at end A.

SOLUTION:


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

T E



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Sample Problem 9.8

• Distributed Loading:

( )

EI

wL

L L L L L EI

w y w B

4

334

01132.0

3

2

3

22

3

2

24

−=

+

−

−=

• Redundant Reaction Loading:


( ) EI

L R L

L EIL

R

y B B

R B 01646.033

2

3 = =

• For compatibility with original supports, y B = 0

( ) ( ) EI

L R

EI

wL y y B

R Bw B

34

01646.001132.00 +−=+=

↑= wL R B 688.0

• From statics,

↑=↑= wL RwL R C A 0413.0271.0

T E



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Sample Problem 9.8


( ) EI

wL

EI

wLw A

33

04167.024

−=−=θ

( ) EI

wL L

L

L

EIL

wL

R A

322

03398.0336

0688.0=

−

=

θ

( ) ( ) EI

wL

EI

wL R Aw A A

33

03398.004167.0 +−=+= θ θ θ EI

wL A

3

00769.0−=θ

,

T

E



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Moment-Area Theorems

• Geometric properties of the elastic curve can

be used to determine deflection and slope.

Não épossívelapresentar aimagem.O computador pode não ter memóriasuficientepara abrir aimagemou aimagem podeter sido danificada.Reinicieo computador e,emseguida,abra o ficheiro novamente.Seo x vermelho continuar aaparecer,poderáter deeliminar aimagemeinseri-lanovamente.

• Consider a beam subjected to arbitrary loading,


• First Moment-Area Theorem:

area under (M/EI) diagram between

C and D.

Não épossívelapresentar aimagem.O computador podenão ter memóriasuficienteparaabrir aimagemou aimagempodeter sido danificada.Reinicieo computador e,emseguida,abrao ficheironovamente.Seo x vermelho continuar aaparecer,poderáter deeliminar aimagemeinseri-lanovamente.



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Application to Cantilever Beams and Beams With

Symmetric Loadings• Cantilever beam - Select tangent at A as the

reference.Não épossívelapresentar aimagem.O computador podenão ter memóriasuficienteparaabrir aimagemou aimagemp odeter sido danificada.Reinicieo computador e,emseguida, abrao ficheiro novamente.Seo x vermelho continuar aaparecer,poderáter deeliminar aimagemeinseri-la novamente.


• Simply supported, symmetrically loaded

beam - select tangent at C as the reference.

Não épossívelapresentar aimagem.O computador podenão ter memóriasuficienteparaabrir aimagemou aimagemp odeter sido danificada.Reinicieo computador e,emseguida, abrao ficheiro novamente.Seo x vermelho continuar aaparecer,poderáter deeliminar aimagemeinseri-la novamente.

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Bending Moment Diagrams by Parts

• Determination of the change of slope and the

tangential deviation is simplified if the effect of

each load is evaluated separately.

• Construct a separate ( M/EI ) diagram for each

load.

- The chan e of slo e θ is obtained b


adding the areas under the diagrams.- The tangential deviation, t D/C is obtained by

adding the first moments of the areas with

respect to a vertical axis through D.

• Bending moment diagram constructed from

individual loads is said to be drawn by parts.

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Sample Problem 9.11

For the prismatic beam shown, determine

SOLUTION:

• Determine the reactions at supports.

• Construct shear, bending moment and

( M/EI ) diagrams.


the slope and deflection at E .

reference, evaluate the slope andtangential deviations at E .

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Sample Problem 9.11

SOLUTION:

• Determine the reactions at supports.

wa R R D B ==

• Construct shear, bending moment and

M/EI dia rams.


( ) EI

waa

EI

wa A

EI

Lwa L

EI

wa A

623

1

422

32

2

221

−=

−=

−=

−=

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Beer Johnston DeWolf



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Sample Problem 9.11

• Slope at E:

EI

wa

EI

Lwa A A

C E C E C E

64

32

21 −−=+=

=+= θ θ θ θ

( )a L EI

wa E 23

12

2

+−=θ


−−

−−−=

−

+

+=

−=

EI

Lwa

EI

wa

EI

Lwa

EI

Lwa

L A

a A

La A

t t y C DC E E

168164

44

3

4

224223

121

( )a L

EI

wa y E +−= 2

8

3

• Deflection at E:

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Application of Moment-Area Theorems to Beams

With Unsymmetric Loadings• Define reference tangent at support A. Evaluate θ A

by determining the tangential deviation at B with

respect to A.Não épossívelapresentar aimagem.O computador podenão ter memóriasuficienteparaabrir aimagemou aimagem podeter sido danificada.Reinicieo computador e,emseguida,abra o ficheiro novamente.Se o x vermelho continuar aaparecer,poderá ter deeliminar aimagemeinseri-lanovamente.

• The slo e at other oints is found with res ect to


reference tangent. A D A D θ θ θ +=

• The deflection at D is found from the tangential

deviation at D.

T h i r

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Maximum Deflection

• Maximum deflection occurs at point K

where the tangent is horizontal.Não épossívelapresentar aimagem.O computador podenão ter memóriasuficientepara abrir aimagemou aimagempod eter sido danificada.Reinicieo computador e,emseguida, abrao ficheiro novamente.Seo x vermelho continuar aaparecer,poderáter deeliminar aimagemeinseri-lanovamente.


• Point K may be determined by measuring

an area under the ( M/EI ) diagram equal

to -θ A .

• Obtain ymax by computing the firstmoment with respect to the vertical axis

through A of the area between A and K .

T h i r

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Beer Johnston DeWolf

Use of Moment-Area Theorems With Statically

Indeterminate Beams• Reactions at supports of statically indeterminate

beams are found by designating a redundant

constraint and treating it as an unknown load which

satisfies a displacement compatibility requirement.

• The ( M/EI ) diagram is drawn by parts. The

resulting tangential deviations are superposed and


related by the compatibility requirement.

• With reactions determined, the slope and deflection

are found from the moment-area method.

Documents

Mecanica Solidos Beer-09