Are a Centroid

8/12/2019 Are a Centroid

1/34

MECHANICS OFMATERIALS

Third Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

John T. DeWolf

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

2002 The McGraw-Hill Com anies, Inc. All ri hts reserved.

Deflection of Beams


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hird Beer Johnston DeWolf

9 - 2

Deflection of Beams

Deformation of a Beam Under TransverseLoading

Equation of the Elastic Curve

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

Statically Indeterminate Beams

Sample Problem 9.1

Sample Problem 9.3

Method of Superposition

Sample Problem 9.7Application of Superposition to Statically

Indeterminate ...

Sample Problem 9.8

Moment-Area Theorems

Application to Cantilever Beams and

Beams With Symmetric ...

Bending Moment Diagrams by Parts

Sample Problem 9.11

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

Maximum Deflection

Use of Moment-Area Theorems WithStatically Indeterminate...


3/34



9 - 3

Deformation of a Beam Under Transverse Loading

Relationship between bending moment andcurvature for pure bending remains valid for

general transverse loadings.

EI

x)(1=

Cantilever beam subjected to concentrated

load at the free end,

EIPx=

1

Curvature varies linearly withx

At the free endA, == A

,01

At the supportB, PL

EI

BB

=

,01


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9 - 4

Deformation of a Beam Under Transverse Loading

Overhanging beam

Reactions atA and C

Bending moment diagram

Curvature is zero at points where the bendingmoment is zero, i.e., at each end and atE.

EI

x)(1=

Beam is concave upwards where the bending

moment is positive and concave downwards

where it is negative.

Maximum curvature occurs where the moment

magnitude is a maximum.

An equation for the beam shape or elastic curve

is required to determine maximum deflectionand slope.


5/34 2002 The McGraw-Hill Com anies, Inc. All ri hts reserved.


9 - 5


From elementary calculus, simplified for beamparameters,

2

2

232

2

2

1

1dx

yd

dx

dydx

yd

+

=

Substituting and integrating,

( )

( )

( ) 2100

1

0

2

21

CxCdxxMdxyEI

CdxxMdx

dyEIEI

xMdx

ydEIEI

xx

x

++=

+=

==




9 - 6


( ) 21

00

CxCdxxMdxyEI

xx

++=

Constants are determined from boundaryconditions

Three cases for statically determinant beams,

Simply supported beam

0,0 ==

Byy

Overhanging beam0,0 == Byy

Cantilever beam0,0 ==y

More complicated loadings require multiple

integrals and application of requirement forcontinuity of displacement and slope.




9 - 7

Direct Determination of the Elastic Curve From the

Load Distribution

For a beam subjected to a distributed load,

( ) ( )xwdx

dV

dx

MdxV

dx

dM===

2

2

Equation for beam displacement becomes

( )xw

dx

ydEI

dx

Md==

4

4

2

2

( ) ( )

432

2213

161 CxCxCxC

dxxwdxdxdxxyEI

++++

=

Integrating four times yields

Constants are determined from boundary

conditions.

h




9 - 8

Statically Indeterminate Beams

Consider beam with fixed support atA and roller

support atB.

From free-body diagram, note that there are four

unknown reaction components.

Conditions for static equilibrium yield000 === AyxF

The beam is statically indeterminate.

( ) 2100

CxCdxxMdxyEI

xx

++=

Also have the beam deflection equation,

which introduces two unknowns but providesthree additional equations from the boundary

conditions:

0,At00,0At ===== yLxyx

h




9 - 9

Sample Problem 9.1

ft4ft15kips50

psi1029in7236814 64

===

==

aLP

EIW

For portionAB of the overhanging beam,

(a) derive the equation for the elastic curve,

(b) determine the maximum deflection,

(c) evaluateymax.

SOLUTION:

Develop an expression for M(x)

and derive differential equation forelastic curve.

Integrate differential equation twice

and apply boundary conditions toobtain elastic curve.

Locate point of zero slope or point

of maximum deflection.

Evaluate corresponding maximum

deflection.

h




9 - 10

Sample Problem 9.1

SOLUTION: Develop an expression for M(x) and derive

differential equation for elastic curve.

- Reactions:

+==L

aPR

L

PaR BA 1

- From the free-body diagram for sectionAD,

( )LxxL

aPM



ird Beer Johnston DeWolf

9 - 11

Sample Problem 9.1

PaLCLCLLaPyLx

Cyx

61

610:0,at

0:0,0at

113

2

=+===

===

Integrate differential equation twice and applyboundary conditions to obtain elastic curve.

213

12

6

1

2

1

CxCxL

aPyEI

CxL

aP

dx

dyEI

++=

+=

xL

a

Pdx

yd

EI =2

2

=

32

6 Lx

Lx

EIPaLy

PaLxx

L

aPyEI

Lx

EIPaL

dxdyPaLx

LaP

dxdyEI

6

1

6

1

3166

121

3

22

+=

=+=

Substituting,

h

f




9 - 12

Sample Problem 9.1

Locate point of zero slope or pointof maximum deflection.

=32

6 L

x

L

x

EI

PaLy

LL

xL

x

EI

PaL

dx

dym

m 577.03

316

02

==

==

Evaluate corresponding maximum

deflection.

( )[ ]32max 577.0577.06

=EI

PaLy

EI

PaLy

6

0642.02

max =

( )( )( )

( )( )462

maxin723psi10296

in180in48kips500642.0

=y

in238.0max =y

hi

B J h t D W lf




9 - 13

Sample Problem 9.3

For the uniform beam, determine the

reaction atA, derive the equation forthe elastic curve, and determine the

slope atA. (Note that the beam is

statically indeterminate to the first

degree)

SOLUTION:

Develop the differential equation for

the elastic curve (will be functionally

dependent on the reaction atA).

Integrate twice and apply boundary

conditions to solve for reaction atA

and to obtain the elastic curve.

Evaluate the slope atA.

hi

Beer Johnston DeWolf



rd Beer Johnston DeWolf

9 - 14

Sample Problem 9.3

Consider moment acting at sectionD,

L

xwxRM

Mx

L

xwxR

M

A

A

D

6

032

1

0

30

20

=

=

=

L

xwxRM

dx

ydEI

A 6

30

2

2

==

The differential equation for the elastic

curve,

hir Beer Johnston DeWolf


15/34



9 - 15

Sample Problem 9.3

L

xwxRM

dx

ydEI A

6

30

2

2

==

Integrate twice

21

503

1

402

1206

1

242

1

CxCL

xw

xRyEI

CL

xwxREI

dx

dyEI

A

A

++=

+==

Apply boundary conditions:

01206

1:0,at

0242

1:0,at

0:0,0at

21

4

03

1

302

2

=++==

=+==

===

CLCLw

LRyLx

CLw

LRLx

Cyx

A

A

Solve for reaction atA

0

30

1

3

1 40

3 = LwLRA = LwRA 010

1



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9 - 16

Sample Problem 9.3

xLwL

xwxLwyEI

= 30

503

0120

1

12010

1

6

1

( )xLxLxEIL

wy 43250 2

120+=

Substitute for C1, C2, and RA in theelastic curve equation,

( )42240 65120

LxLx

EIL

w

dx

dy+==

EI

LwA

120

30=

Differentiate once to find the slope,

atx = 0,



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9 - 17

Method of Superposition

Principle of Superposition:

Deformations of beams subjected to

combinations of loadings may be

obtained as the linear combination of

the deformations from the individualloadings

Procedure is facilitated by tables of

solutions for common types of

loadings and supports.



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d Beer Johnston DeWolf

9 - 18

Sample Problem 9.7

For the beam and loading shown,

determine the slope and deflection at

pointB.

SOLUTION:

Superpose the deformations due toLoading IandLoading II as shown.



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d

9 - 19

Sample Problem 9.7

Loading I

( )EI

wLIB 6

3

= ( )EI

wLy IB 8

4

=

Loading II

( )

EI

wLIIC

48

3

= ( )

EI

wLy IIC

128

4

=

In beam segment CB, the bending moment is

zero and the elastic curve is a straight line.

( ) ( )EI

wLIICIIB 48

3==

( )EI

wLL

EI

wL

EI

wLy IIB

384

7

248128

434

=

+=



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d

9 - 20

Sample Problem 9.7

Combine the two solutions,

EIwL

B487 3=( ) ( )

EIwL

EIwL

IIBIBB 48633 +=+=

EI

wL

yB 384

41 4

=( ) ( ) EIwL

EI

wL

yyy IIBIBB 384

7

8

44

+=+=


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22/34


d

9 - 22

Sample Problem 9.8

For the uniform beam and loading shown,

determine the reaction at each support and

the slope at endA.

SOLUTION:

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



23/34

2002 The McGraw-Hill Com anies, Inc. All ri hts reserved. 9 - 23

Sample Problem 9.8

Distributed Loading:

( )

EIwL

LLLLLEI

wy wB

4

334

01132.0

3

2

3

22

3

2

24

=

+

=

Redundant Reaction Loading:

( ) EILRL

LEIL

Ry BBRB

322

01646.033

2

3 =

=

For compatibility with original supports,yB = 0

( ) ( ) EILR

EI

wLyy BRBwB

34

01646.001132.00 +=+=

= wLRB 688.0

From statics,

== wLRwLR C 0413.0271.0



24/34


Sample Problem 9.8

Slope at endA,

( )EI

wL

EI

wLwA

33

04167.024

==

( )EIwLLLL

EILwLRA

32

2 03398.0336

0688.0 =

=

EI

wLA

3

00769.0=( ) ( )EI

wL

EI

wL

RA

wAA

33

03398.004167.0 +=+=



25/34



Geometric properties of the elastic curve canbe used to determine deflection and slope.

Consider a beam subjected to arbitrary loading,

First Moment-Area Theorem:

area under (M/EI) diagram between

CandD.



26/34



Second Moment-Area Theorem:

The tangential deviation of Cwith respect toD

is equal to the first moment with respect to avertical axis through Cof the area under the

(M/EI) diagram between CandD.

Tangents to the elastic curve atPandPintercepta segment of length dton the vertical through C.

= tangential deviation of C

with respect toD


A li i C il B d B Wi h


27/34


Application to Cantilever Beams and Beams With

Symmetric Loadings

Cantilever beam - Select tangent atA as the

reference.

Simply supported, symmetrically loaded

beam - select tangent at Cas the reference.


B di M t Di b P t


28/34


Bending Moment Diagrams by Parts

Determination of the change of slope and thetangential deviation is simplified if the effect of

each load is evaluated separately.

Construct a separate (M/EI) diagram for eachload.

- The change of slope, D/C, is obtained by

adding the areas under the diagrams.

- The tangential deviation, tD/Cis 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.


S l P bl 9 11


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Sample Problem 9.11

For the prismatic beam shown, determine

the slope and deflection atE.

SOLUTION:

Determine the reactions at supports.

Construct shear, bending moment and(M/EI) diagrams.

Taking the tangent at Cas the

reference, evaluate the slope andtangential deviations atE.


S l P bl 9 11


30/34


Sample Problem 9.11

SOLUTION:

Determine the reactions at supports.

waRRDB

==

Construct shear, bending moment and

(M/EI) diagrams.

( )EI

waa

EI

waA

EI

LwaL

EI

waA

623

1

422

32

2

22

1

=

=

=

=


S l P bl 9 11


31/34


Sample Problem 9.11

Slope at E:

EI

wa

EI

LwaAA

CECECE

64

32

21 =+=

=+=

( )aLEI

waE 23

12

2

+=

=

+

+=

=

EI

Lwa

EI

wa

EI

Lwa

EI

Lwa

LA

aA

LaA

tty CDCEE

168164

44

3

4

224223

121

( )aLEI

wayE += 28

3

Deflection at E:


A li ti f M t A Th t B


32/34


Application of Moment-Area Theorems to Beams

With Unsymmetric Loadings

Define reference tangent at supportA. Evaluate Aby determining the tangential deviation atB with

respect toA.

The slope at other points is found with respect to

reference tangent.

ADAD +=

The deflection atD is found from the tangential

deviation atD.


Maximum Deflection


33/34


Maximum Deflection

Maximum deflection occurs at pointKwhere the tangent is horizontal.

PointKmay be determined by measuring

an area under the (M/EI) diagram equal

to -A .

Obtainymaxby computing the first

moment with respect to the vertical axis

throughA of the area betweenA andK.


Use of Moment Area Theorems With Statically


34/34


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.

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