08032012-Nice Ppt Strain Transformation

8/2/2019 08032012-Nice Ppt Strain Transformation

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2005 Pearson Education South Asia Pte Ltd

10. Strain Transformation

1

Apply the stresstransformation methodsderived in Chapter 9 tosimilarly transform strain

Discuss various ways ofmeasuring strain

Develop importantmaterial-property

relationships; including generalized form ofHookes law

Discuss and use theories to predict the failure of a

material

CHAPTER OBJECTIVES


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2

CHAPTER OUTLINE

1. Plane Strain

2. General Equations of Plane-StrainTransformation

3. *Mohrs Circle: Plane Strain

4. *Absolute Maximum Shear Strain

5. Strain Rosettes

6. Material-Property Relationships

7. *Theories of Failure


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3

10.1 PLANE STRAIN

As explained in Chapter 2.2, general state of strainin a body is represented by a combination of 3components of normal strain (x, y, z), and 3components of shear strain (xy, xz, yz).

Strain components at a pt determined by usingstrain gauges, which is measured in specifieddirections.

A plane-strained element is subjected to twocomponents of normal strain (x, y) and onecomponent of shear strain, xy.


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4

10.1 PLANE STRAIN

The deformations are shown graphically below.

Note that the normal strains are produced bychanges in length of the element in thex andydirections, while shear strain is produced by the

relative rotation of two adjacent sides of theelement.


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5

10.1 PLANE STRAIN

Note that plane stress does not always cause planestrain.

In general, unless = 0, the Poisson effect willprevent the simultaneous occurrence of plane strain

and plane stress. Since shear stress and shear

strain not affected by Poissons

ratio, condition of xz = yz = 0

requires xz = yz = 0.


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6

10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION

Sign Convention

To use the same convention asdefined in Chapter 2.2.

With reference to differential

element shown, normal strainsxz and yz are positive if theycause elongation along thexandy axes

Shear strain xy is positive if the interior angle AOBbecomes smaller than 90.


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7


Sign Convention

Similar to plane stress, when measuring the normaland shear strains relative to thexandyaxes, theangle will be positive provided it follows thecurling of the right-hand fingers, counterclockwise.

Normal and shear strains

Before we develop thestrain-transformation eqn fordetermining

x;, we must determine

the elongation of a line segment dxthat lies along thexaxis andsubjected to strain components.




8



Components of line dx and dxare elongated andwe add all elongations together.

From Eqn 2.2, the normal strain along the line dxisx=x/dx. Using Eqn 10-1,

cossincos' dydydxx xyyx

210cossinsincos 22' - xyyxx




9



To get the transformation equation for xy, consideramount of rotation of each of the line segments dxand dywhen subjected to strain components.

Thus, sincossin' dydydxy xyyx




10



Using Eqn 10-1 with = y/x,

As shown, dyrotates by an amount.

310sincossin 2 - xyyx




11



Using identities sin (+ 90) = cos ,cos (+ 90) = sin ,

Thus we get

2

2

cossincos

90sin90cos90sin

xyyx

xyyx

410sincoscossin2 22''

-

xyyx

yx




12



Using trigonometric identities sin 2=2 sincos,cos2= (1 + cos2 )/2 and sin2+ cos2= 1, werewrite Eqns 10-2 and 10-4 as

5102sin2

2cos22

' -

xyyxyx

x

6-102cos2

2sin22

''

xyyxyx




13



If normal strain in they direction is required, it canbe obtained from Eqn 10-5 by substituting (+ 90)for . The result is

6102sin2

2cos22

' -

xyyxyx

y




14


Principal strains

We can orientate an element at a pt such that theelements deformation is only represented by

normal strains, with no shear strains.

The material must be isotropic, and the axes alongwhich the strains occur must coincide with the axesthat define the principal axes.

Thus from Eqns 9-4 and 9-5,

8102tan -yx

xyp




15


Principal strains

Maximum in-plane shear strain

Using Eqns 9-6, 9-7 and 9-8, we get

910222

22

2,1 -

xyyxyx

1110222

22plane-in

max

-

xyyx

10102tan -

xy

yxs




16



Using Eqns 9-6, 9-7 and 9-8, we get

12102

-avgyx

S i T f i




17


IMPORTANT

Due to Poisson effect, the state of plane strain is nota state of plane stress, and vice versa.

A pt on a body is subjected to plane stress whenthe surface of the body is stress-free.

Plane strain analysis may be used within the planeof the stresses to analyze the results from thegauges. Remember though, there is normal strainthat is perpendicular to the gauges.

When the state of strain is represented by theprincipal strains, no shear strain will act on theelement.

10 S i T f i




18


IMPORTANT

The state of strain at the pt can also be representedin terms of the maximum in-plane shear strain. Inthis case, an average normal strain will also act on

the element. The element representing the maximum in-plane

shear strain and its associated average normalstrains is 45 from the element representing the

principal strains.

10 St i T f ti




19

EXAMPLE 10.1

A differential element of material at a pt is subjected to

a state of plane strain x = 500(10-6), y = 300(10-6),which tends to distort the element as shown.Determine the equivalent strains acting on an elementoriented at the pt, clockwise 30 from the originalposition.

10 St i T f ti




20

EXAMPLE 10.1 (SOLN)

Since is counterclockwise, then =30, use

strain-transformation Eqns 10-5 and 10-6,

6

'

6

6

6

'

10213

302sin2

10200

302cos102

300500

102

300500

2sin2

2cos22

x

xyyxyxx

10 St i T f ti




21

EXAMPLE 10.1 (SOLN)

Since is counterclockwise, then =30, use

strain-transformation Eqns 10-5 and 10-6,

6''

6

''

10793

302cos210200

302sin2

300500

2cos2

2sin22

yx

xyyxyx

10 St i T f ti




22

EXAMPLE 10.1 (SOLN)

Strain in theydirection can be obtained from Eqn

10-7 with =30. However, we can also obtain yusing Eqn 10-5 with = 60 (=30 + 90),replacing xwith y

6

'

6

6

6'

104.13

602sin2

10200

602cos102

300500

102

300500

y

y

10 St i T f ti




23

EXAMPLE 10.1 (SOLN)

The results obtained tend to deform the element as

shown below.

10 Strain Transformation


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24

EXAMPLE 10.2


a state of plane strain defined by x =350(10-6),y = 200(10

-6), xy = 80(10-6), which tends to distort the

element as shown. Determine the principal strains atthe pt and associated orientation of the element.



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25

EXAMPLE 10.2 (SOLN)

Orientation of the element

From Eqn 10-8, we have

Each of these angles is measuredpositive counterclockwise, from the

x axis to the outward normals on

each face of the element.

9.8514.4

,17218028.828.82

)10(200350

)10(802tan

6

6

and

thatsoandThus

p

p

yx

xyp



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26

EXAMPLE 10.2 (SOLN)

Principal strains

From Eqn 10-9,

6261

66

6226

22

2,1

1035310203

109.277100.75

102

80

2

200350

2

10200350

222

xyyxyx



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27

EXAMPLE 10.2 (SOLN)

Principal strains

We can determine which of these two strains deformsthe element in thexdirection by applying Eqn 10-5with =4.14. Thus

6'

6

66

'

10353

14.42sin2

1080

14.4cos102

20035010

2

200350

2sin2

2cos22

x

xyyxyxx



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28

EXAMPLE 10.2 (SOLN)

Principal strains

Hence x= 2. When subjected to the principal strains,the element is distorted as shown.



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29

EXAMPLE 10.3


a state of plane strain defined by x =350(10-6),y = 200(10

-6), xy = 80(10-6), which tends to distort the

element as shown. Determine the maximum in-planeshear strain at the pt and associated orientation of theelement.



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30

EXAMPLE 10.3 (SOLN)

Orientation of the element

From Eqn 10-10,

Note that this orientation is 45 from that shown inExample 10.2 as expected.

9.1309.40

,72.26118072.8172.8121080

102003502tan

6

6

and

thatsoandThus,

s

s

xy

yxs



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31

EXAMPLE 10.3 (SOLN)


Applying Eqn 10-11,

The proper sign of can be obtained by applyingEqn 10-6 with s = 40.9.

6

622

22

10556

102

80

2

200350

222

plane-in

max

plane-in

max

xyyx

plane-in

max



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32

EXAMPLE 10.3 (SOLN)


Thus tends to distort the element so that theright angle between dxand dyis decreased (positive

sign convention).

6

''

6

6

''

10556

9.402cos2

1080

9.402sin10

2

200350

2cos2

2sin22

yx

xyyxyx

plane-in

max



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33

EXAMPLE 10.3 (SOLN)


There are associated average normal strains imposedon the element determined from Eqn 10-12:

These strains tend tocause the element to contract.

66

1075102

200350

2

yx

avg



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34

*10.3 MOHRS CIRCLE: PLANE STRAIN

Advantage of using Mohrs circle for plane strain

transformation is we get to see graphically how thenormal and shear strain components at a pt varyfrom one orientation of the element to the next.

Eliminate parameter in Eqns 10-5 and 10-6 andrewrite as

22

avg

22

2avg

222

where

13-102

xyyxyx

xyx

R

R



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35


Procedure for Analysis

Construction of the circle

Establish a coordinate system such that theabscissa represents the normal strain , with

positive to the right, and the ordinate represents halfthe value of the shear strain, /2, with positivedownward.

Using positive sign convention for x, y, and xy,

determine the center of the circle C, which is locatedon the axis at a distance avg = (x + v)/2 from theorigin.



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36




Plot the reference ptA having coordinates (x,xy/2).This pt represents the case for which thexaxis

coincides with thex axis. Hence = 0. Connect ptA with center C

of the circle and from theshaded triangle determine

the radiusR of the circle.

Sketch the circle.



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37



Principal strains Principal strains 1 and 2 are

determined from the circle asthe coordinates of ptsB and

D (= 0).

Determine the orientation of theplane on which 1 acts bycalculating 2p1, using trigonometry.This angle is measuredcounterclockwise from the radialreference lines CA to CB.



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38



Principal strains

Remember that the rotation of p1,must be in this same direction,

from the elements reference axisx to thex axis. When 1 and 2 are indicated as being positive as

shown earlier, the element shown here will elongatein thexandydirections as shown by the dashed

outline.



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39



Maximum in-plane shear strain Average normal strain and half the

maximum in-plane shear strainare determined from the circleas the coordinates of ptsEand F.

Orientation of the plane on whichand avg act can be

determined from the circle bycalculating 2s1 using trigonometry.This angle is measured clockwisefrom the radial reference linesCA to CF.

planein

max



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40




Remember that the rotation ofps1, must be in this same

direction, from the elementsreference axisx to thex axis.

Strains on arbitrary plane

Normal and shear strain components x

and xy

fora plane specified at an angle , can be obtainedfrom the circle using trigonometry to determine thecoordinates of pt P.



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41



Strains on arbitrary plane To locate P, the known angle of

thexaxis is measured on thecircle as 2. This measurement is

made from the radial reference line CA to the radialreference line CA to CP. Remember thatmeasurements for 2on the circle must be in thesame direction as for thexaxis.

If value of y is required, it can be determined bycalculating the coordinate of pt Q. The line CQ lies180 away from CP and thus represents a rotation of90 of thexaxis.



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42

EXAMPLE 10.4

State of plane strain at a pt represented by the

components x = 250(10-6), y =150(10-6), andxy = 120(10

-6). Determine the principal strainsand the orientation of the element.



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0 St a a s o at o

43

EXAMPLE 10.4 (SOLN)


The and /2 axes areestablished as shown. Notethat the positive /2 axis must

be directed downward so thatcounterclockwise rotations ofthe element correspond tocounterclockwise rotation

around the circle, and viceversa. Center of the circle is located on the axis at

66 1050102

150250

avg



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2005 Pearson Education South Asia Pte Ltd 44

EXAMPLE 10.4 (SOLN)


Since xy/2 = 60(10-6), the

reference ptA (= 0) hascoordinates [250(10-6), 60(10-6)].

From shaded triangle, radiusof circle is CA:

6622 108.208106050250 R



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EXAMPLE 10.4 (SOLN)

Principal strains

The coordinates of ptsB andDrepresent the principal strains.They are

The direction of the positive principal strain 1 is

defined by the counterclockwise 2p1, measured fromthe radial reference lines CA to CB.

662

66

1

10159108.20850

10259108.20850



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EXAMPLE 10.4 (SOLN)

Principal strains

We have

Hence, the side dxof theelement is orientedcounterclockwise 8.35.This also defines the direction of

1

.

The deformation of the element is also shown.

35.8

50250602tan

1

1

p

p



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EXAMPLE 10.5

State of plane strain at a pt represented by the

components x = 250(10-6), y =150(10-6), andxy = 120(10

-6). Determine the maximum in-planeshear strains and orientation of the element.



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EXAMPLE 10.5 (SOLN)


Half the maximum in-plane shear strain and averagenormal strain are represented by the coordinates ofptsEand Fon the circle. From coordinates of ptE

6

6''

6''

1050

10418

108.208

avg

plane-in

max

plane-inmax

2

yx

yx



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EXAMPLE 10.5 (SOLN)


To orientate the element, determine the clockwiseangle 2s1 from the circle,

Since shear strain defined from ptEon the circle has a positive value andaverage normal strain is also positive,corresponding positive shear stressand positive average normal stressdeform the element into dashedshape as shown.

6.36

35.82902

1

1

s

s



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EXAMPLE 10.6

State of plane strain at a pt represented by an

element having the components x =300(10-6),y =100(10

-6), and xy = 100(10-6). Determine the

state of strain on an element oriented 20 clockwisefrom this reported position.



51/124


EXAMPLE 10.6 (SOLN)

Construction of circle

The and /2 axes areestablished as shown.Center of circle is on

the axis at

Coordinates of reference ptA is [300(10-6), 50(10-6)].Radius CA determined from shaded triangle,

6622 108.1111050200300 R

66 10200102

100300

avg



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EXAMPLE 10.6 (SOLN)

Strains on inclined elements

As we orient element 20 clockwise, first establish aradial line CP, 2(20) = 40 clockwise, measured fromCA (= 0). Coordinates of pt P (x, xy/2) are

obtained from the geometry of the circle.

Thus

6

''

6''

66'

100.52

1043.13sin8.1112

103091043.13cos8.111200

yx

yx

x

57.26

200300

50tan 1



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EXAMPLE 10.6 (SOLN)

Strains on inclined elements

Normal strain ycan be determined from the coordinate of pt Q on the circle. Why?

As a result of these strains, theelement deforms relative to the

x,yaxes as shown.

66' 103.911043.13cos8.111200 y



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Assume three principal strains

(max, int, min) cause elongationsalong thex, yandzaxes as shown.

Use Mohrs circle to determine

maximum in-plane shear strain forthex-y,x-zandy-zplanes.

*10.4 ABSOLUTE MAXIMUM SHEAR STRAIN



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From the circles drawn,

we see that the absolutemaximum shear strain isdetermined from the circlehaving the larges radius.For this condition,


15102

1410

minmax

minmax

-

and

-

avg

max

abs



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Plane strain

When material subjected toprincipal in-plane strains of thesame sign, the largest circle has

a radius ofR = (xz)max/2.

This value represents theabsolute maximum shearstrain for the material. It islarger than the maximum

in-plane shear strain.


max'' maxmaxabs zx



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Plane strain

For material subjected toprincipal in-plane strains ofopposite signs,


minmax

''max

yxmax

abs



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IMPORTANT

General 3-D state of strain at a pt can berepresented by an element oriented so that onlythree principal strains act on it.

From this orientation, the orientation of the element

representing the absolute maximum shear straincan be obtained by rotating the element 45 aboutthe axis defining the direction of int.

The absolute maximum shear strain will be larger

that the maximum in-plane shear strain wheneverthe in-plane principal strains have the same sign,the absolute maximum shear strain will act out ofthe plane.




59/124


EXAMPLE 10.7

Plane of strain at a pt is represented by the strain

components x = 400(10-6), y = 200(10-6),xy = 150(10

-6). Determine the maximum in-planeshear strain and the absolute maximum shear strain.



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EXAMPLE 10.7 (SOLN)

Maximum in-plane strain

Using Mohrs circle method, center of circle is on the

-axis at

Since xy/2 = 75(10-6), reference pt has coordinatesA[400(10-6), 75(10-6)]. Radius of circle is

66 10100102

200400

avg

6622 103091075100400 R



61/124


EXAMPLE 10.7 (SOLN)

Maximum in-plane strain

Computing in-plane principal strains,we have

From the circle, maximum in-plane shear strain is

6666

1040910309100

1020910309100

min

max

66minmax 1061810409209 plane-in

max



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EXAMPLE 10.7 (SOLN)

Absolute maximum shear strain

From the results, we have max = 209(10-6),int = 409(10

-6). The 3 Mohrs circles plotted for

element orientations about each of thex,yandzaxes are shown. We see that the principal in-planestrains have opposite signs, and maximum in-planeshear strain is also the absolutemaximum shear strain

610618 plane-in

abs



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We measure the normal strain in a tension-test

specimen using an electrical-resistance straingauge.

For general loading on a body, the normal strainsat a pt are measured using a cluster of 3 electrical-resistance strain gauges.

Such strain gauges, arranged in a specific patternare called strain rosettes.

Note that only the strains in the plane of the gaugesare measured by the strain rosette. That is ,thenormal strain on the surface is not measured.

10.5 STRAIN ROSETTES



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Apply strain transformation

Eqn 10-2 to each gauge:

We determine the values of x, yxy by solving thethree equations simultaneously.

1610cossinsincos

cossinsincos

cossinsincos

22

22

22

-ccxycycxc

bbxybybxb

aaxyayaxa



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For rosettes arranged in the 45

pattern, Eqn 10-16 becomes

For rosettes arranged in the 60 pattern,Eqn 10-16 becomes

cabxy

cy

ax

2

1710

3

2

223

1

-cbxy

acby

ax



66/124


EXAMPLE 10.8

State of strain at ptA on bracket is measured using

the strain rosette shown. Due to the loadings, thereadings from the gauges give a = 60(10

-6),b= 135(10

-6), and c = 264(10-6). Determine the

in-plane principal strains at the pt and the directions

in which they act.



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EXAMPLE 10.8 (SOLN)

Establishx axis as shown, measure the

angles counterclockwise from the +x axisto center-lines of each gauge, we havea = 0, b = 60, and c = 120Substitute into Eqn 10-16,

)3(433.075.025.0

120cos120sin120sin120cos10264

)2(433.075.025.0

60cos60sin60sin60cos10135

)1(0cos0sin0sin0cos1060

226

226

226

xyyx

xyyx

xyyx

xyyx

xxyyx



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Solving Eqns (1), (2) and (3) simultaneously, we get

The in-plane principal strains can also be obtaineddirectly from Eqn 10-17. Reference pt on Mohrs circle

isA [60(10-6),74.5(10-6)] and center of circle, Cis onthe axis at avg = 153(10

-6).From shaded triangle, radius is

EXAMPLE 10.8 (SOLN)

666 10149102461060 xyyx

6622

102.119105.7460153

R

R



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The in-plane principal strains are thus

Deformed element is shown dashed.Due to Poisson effect, element also subjected to anout-of-plane strain, in thez direction, although thisvalue does not influence the calculated results.

EXAMPLE 10.8 (SOLN)

3.19

7.3860153

5.74

tan2

108.33102.11910246

10272102.11910153

2

1

2

6662

6661

p

p



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10.6 MATERIAL-PROPERTY RELATIONSHIPS

Generalized Hookes law

Material at a pt subjected to a state of triaxialstress, with associated strains.

We use principle of superposition, Poissons ratio

(lat =long), and Hookes law (=E) to relatestresses to strains, in the uniaxial direction.

With x applied, element elongates in thexdirection and strain is this direction is

E

xx

'



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With y applied, element contracts with a strain xin thex direction,

Likewise, With z applied, a contraction is causedin thez direction,

E

yx

''

E

zx

'''



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By using the principle of superposition,

yxzz

zxyy

zyxx

E

E

E

1

18101

1

-



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If we apply a shear stress xy to the element,experimental observations show that it will deformonly due to shear strain xy. Similarly for xz and xy,

yz andyz. Thus, Hookes law for shear stress andshear strain is written as

1910111

-xzxzyzyzxyxyGGG


0 6 MATERIAL PROPERTY RELATIONSHIPS


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Relationship involvingE, , and G

We stated in chapter 3.7:

Relate principal strain to shear stress,

Note that since x =y =z = 0, then from Eqn10-18, x =y = 0. Substitute into transformationEqn 10-19,

2010

12-

EG

21101max -

E

xy

2max1

xy




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Relationship involvingE, , and G

By Hookes law, xy = xy/G. So max =xy/2G.

Substitute into Eqn 10-21 and rearrange to obtainEqn 10-20.

Dilatation and Bulk Modulus Consider a volume element subjected to principal

stresses x, y, z.

Sides of element are dx, dy and dz, and after stressapplication, they become (1 + x)dx, (1 + y)dy,(1 + z)dz, respectively.





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Dilatation and Bulk Modulus

Change in volume of element is

Change in volume per unit volume is the

volumetric strain or dilatation e.

Using generalized Hookes law, we write thedilatation in terms of applied stress.

dzdydxdzdydxV zyx 111

2210 -zyxdV

Ve

231021 -zyxE

e




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When volume element of material is subjected touniform pressurep of a liquid, pressure is the samein all directions.

As shear resistance of a liquid is zero, we canignore shear stresses.

Thus, an element of the body is subjected toprincipal stresses x = y = z =p. Substituting

into Eqn 10-23 and rearranging,

2410

213-

E

e

p




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This ratio (p/e) is similar to the ratio of linear-elasticstress to strain, thus terms on the RHS are calledthe volume modulus of elasticity or the bulkmodulus. Having same units as stress withsymbol k,

For most metals, sok E.

From Eqn 10-25, theoretical maximum value ofPoissons ratio is therefore= 0.5.

When plastic yielding occurs, = 0.5 is used.

2510

213-

Ek




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IMPORTANT

When homogeneous and isotropic material issubjected to a state of triaxial stress, the strain inone of the stress directions is influence by thestrains produced by all stresses. This is the result

of the Poisson effect, and results in the form of ageneralized Hookes law.

A shear stress applied to homogenous andisotropic material will only produce shear strain in

the same plane. Material constants,E, G and are related

mathematically.




80/124



IMPORTANT

Dilatation, or volumetric strain, is caused by only bynormal strain, not shear strain.

The bulk modulus is a measure of the stiffness of avolume of material. This material property providesan upper limit to Poissons ratio of= 0.5, whichremains at this value while plastic yielding occurs.


EXAMPLE 10 10


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EXAMPLE 10.10

Copper bar is subjected to a uniform loading along its

edges as shown. If it has a length a = 300 mm, widthb = 50 mm, and thickness t= 20 mm before the loadis applied, determine its new length, width, andthickness after application of the load.

TakeEcu = 120 GPa, cu = 0.34.


EXAMPLE 10 10 (SOLN)


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EXAMPLE 10.10 (SOLN)

By inspection, bar is subjected to a state of plane

stress. From loading, we have

Associated strains are determined from generalized

Hookes law, Eqn 10-18;

00500800 zxyyx MPaMPa

00808.050010312034.0

103120

800

MPaMPa

MPa

zvx

xEE




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Associated strains are determined from generalized

Hookes law, Eqn 10-18;

00850.0500800

103120

34.00

00643.080010312034.0

103120500

MPaMPa

MPaMPa

MPa

yxz

z

zxy

y

EE

EE




84/124



The new bar length, width, and thickness are

mmmmmm

mmmmmm

mmmmmm

98.1920000850.020'

68.495000643.050'

4.30230000808.0300'

t

b

a


EXAMPLE 10 11


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EXAMPLE 10.11

If rectangular block shown is subjected to a uniform

pressure ofp = 20 kPa, determine the dilatation andchange in length of each side.TakeE= 600 kPa, = 0.45.




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Dilatation

The dilatation can be determined using Eqn 10-23with x = y = z =20 kPa. We have

33

/01.0

203600

45.021

21

cmcm

kPakPa

zyxE

e




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Change in length

Normal strain on each side can be determined fromHookes law, Eqn 10-18;

cm/cm

kPakPakPakPa

00333.0

202045.020600

1

1

zyxE




88/124



Change in length

Thus, the change in length of each side is

The negative signs indicate that each dimension is

decreased.

cmcm

cmcm

cmcm

0100.0300333.0

00667.0200333.0

0133.0400333.0

c

b

a


*10 7 THEORIES OF FAILURE


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When engineers design for a material, there is a

need to set an upper limit on the state of stress thatdefines the materials failure.

For ductile material, failure is initiated by yielding.

For brittle material, failure is specified by fracture. However, criteria for the above failure modes is not

easy to define under a biaxial or triaxial stress.

Thus, four theories are introduced to obtain the

principal stresses at critical states of stress.

10.7 THEORIES OF FAILURE




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A. Ductile materials

1. Maximum-Shear-Stress Theory

Most common cause of yielding ofductile material (e.g., steel) is slipping.

Slipping occurs along the contactplanes of randomly-ordered crystalsthat make up the material.

Edges of planes of slipping as they appear on the

surface of the strip are referred to as Lders lines.

The lines indicate the slip planes in the strip, whichoccur at approximately 45 with the axis of the

strip.





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The lines indicate the slip planes inthe strip, which occur at approximately

45 with the axis of the strip. Consider an element, determine maximum shear

stress from Mohrs circle,

Thus, in 1868, Henri Trescaproposed the maximum-shear-stresstheory or Tresca yield criterion.

26102max

-Y




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If the two in-plane principalstresses have the same sign,

failure will occur out of the plane:

If in-plane principal stresses are of opposite signs,failure occurs in the plane:

2max

max

abs

2

minmax

max

abs




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Thus, we express the maximum-shear-stresstheory for plane stress for any two in-plane principal

stresses for 1 and 2 by the following criteria:

signs.oppositehave

27-10signs.samehave

signs.samehave

2121

212

211

,}

,}

,}

Y

Y

Y


*10.7 THEORIES OF FAILURE


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2. Maximum-Distortion-Energy Theory

Energy per unit volume of material is called thestrain-energy density.

Material subjected to a uniaxial stress , thestrain-energy density is written as

28102

1-u

3322112

1

2

1

2

1 u




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For linear-elastic behavior, applying Hookes lawinto above eqn:

Maximum-distortion-energy theory is defined as the

yielding of a ductile material occurs when thedistortion energy per unit volume of the materialequals or exceeds the distortion energy per unitvolume of the same material when subjected to

yielding in a simple tension test.

2910

221

233121

23

22

21 -

Eu




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To obtain distortion energy per unit volume,

In the case of plane stress,

For uniaxial tension test, 1 = Y, 2 = 3 = 0

2132322216

1

E

ud

2221213

1

E

ud

23

1YYd E

u




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Since maximum-distortion energy theory requiresud= (ud)Y, then for the case of plane or biaxial

stress, we have

3010222212

1 -Y




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0 O S O U



Comparing both theories, we get the followinggraph.




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A. Brittle materials

3. Maximum-Normal-Stress Theory

Figure shows how brittle materialsfail.




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The maximum-normal-stress theorystates that a brittle material will fail

when the maximum principal stress1 in the material reaches a limiting value that isequal to the ultimate normal stress the material cansustain when subjected to simple tension.

For the material subjected to plane stress

31102

1

-ult

ult




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Experimentally, it was found to be in closeagreement with the behavior of brittle materials that

have stress-strain diagrams similar in both tensionand compression.




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4. Mohrs Failure Criterion

Use for brittle materials where the tension andcompression properties are different.

Three tests need to be performed on material todetermine the criterion.




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Carry out a uniaxial tensile test to determine theultimate tensile stress (ult)t

Carry out a uniaxial compressive test to determinethe ultimate compressive stress (ult)c

Carry out a torsion test to determine the ultimateshear stress ult.

Results are plotted in Mohr circles.




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Circle A represents the stress condition 1 = 2 = 0,3 =(ult)c

Circle B represents the stress condition 1 = (ult)t,2 = 3 = 0

Circle C represents thepure-shear-stress conditioncaused by ult.




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The Criterion can also be represented on a graphof principal stresses 1 and 2 (3 = 0).




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IMPORTANT

If material is ductile, failure is specified by theinitiation of yielding, whereas if it is brittle, it isspecified by fracture.

Ductile failure can be defined when slipping occursbetween the crystals that compose the material.

This slipping is due to shear stress and themaximum-shear-stress theory is based on this

idea. Strain energy is stored in a material when

subjected to normal stress.




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IMPORTANT

The maximum-distortion-energy theory depends onthe strain energy that distorts the material, and notthe part that increases its volume.

The fracture of a brittle material is caused by themaximum tensile stress in the material, and not thecompressive stress.

This is the basis of the maximum-normal-stress

theory, and it is applicable if the stress-straindiagram is similar in tension and compression.




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IMPORTANT

If a brittle material has a stress-strain diagram thatis different in tension and compression, thenMohrs failure criterion may be used to predict

failure.

Due to material imperfections, tensile fracture of abrittle material is difficult to predict, and so theoriesof failure for brittle materials should be used with

caution.


*EXAMPLE 10.12


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Steel pipe has inner diameter of 60 mm and outer

diameter of 80 mm. If it is subjected to a torsionalmoment of 8 kNm and a bending moment of3.5 kNm, determine if these loadings cause failure asdefined by the maximum-distortion-energy theory.

Yield stress for the steel found from a tension test isY = 250 MPa.


*EXAMPLE 10.12 (SOLN)


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Investigate a pt on pipe that is subjected to a state of

maximum critical stress.Torsional and bending moments are uniformthroughout the pipes length.

At arbitrary section a-a, loadings produce the stressdistributions shown.




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By inspection, pts A and B subjected to same state of

critical stress. Stress at A,

MPa9.101

m03.0m04.04m04.0mN3500

MPa4.116m03.0m04.02

m04.0mN8000

44A

44

IMc

J

TcA




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Mohrs circle for this state of stress has center located

at

The radius is calculated from the

shaded triangle to beR = 127.1and the in-plane principalstresses are

MPa9.502

9.1010avg

MPa0.1781.1279.50

MPa2.761.1279.50

2

1




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Using Eqn 10-30, we have

Since criterion is met, material within the pipe will notyield (fail) according to the maximum-distortion-energy theory.

OK!500,62100,51

?0.1780.1782.762.76Is222

22221

21

Y

Y


*EXAMPLE 10.14


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Solid shaft has a radius of 0.5 cm and made of steel

having yield stress of Y = 360 MPa. Determine if theloadings cause the shaft to fail according to themaximum-shear-stress theory and the maximum-distortion-energy theory.




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State of stress in shaft caused by axial force and

torque. Since maximum shear stress caused bytorque occurs in material at outer surface, we have

MPa5.165kN/cm55.16

cm5.02

cm5.0cmkN25.3

MPa191kN/cm10.19cm5.0

kN15

2

4

2

2

xy

xy

x

J

Tc

A

P




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Stress components acting on an element of material

at pt A. Rather than use Mohrs circle, principalstresses are obtained using stress-transformationeqns 9-5:

MPa6.286

MPa6.95

1.1915.95

5.1652

0191

2

0191

22

2

1

22

22

2,1

xyyxyx




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Maximum-shear-stress theory

Since principal stresses have opposite signs,absolute maximum shear stress occur in the plane,apply Eqn 10-27,

Thus, shear failure occurs by maximum-shear-stress

theory.

Fail!3602.382

?3606.2866.95Is

21

Y




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Maximum-distortion-energy theory

Applying Eqn 10-30, we have

However, using the maximum-distortion-energytheory, failure will not occur. Why?

OK!600,1299.677,118

?3606.2866.2866.956.95Is222

2221

21

Y


CHAPTER REVIEW


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When element of material is subjected to

deformations that only occur in a single plane, thenit undergoes plain strain.

If the strain components x, y, and xy are knownfor a specified orientation of the element, then thestrains acting for some other orientation of theelement can be determined using the plane-straintransformation equations.

Likewise, principal normal strains and maximumin-plane shear strain can be determined usingtransformation equations.


CHAPTER REVIEW


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Strain transformation problems can be solved

in a semi-graphical manner using Mohrs circle. Establish the and /2 axes, then compute

center of circle [(x +y)/2, 0] and controlling pt[, /2], before plotting the circle.

Radius of circle extends between these two ptsand is determined from trigonometry.

Absolute maximum shear strain equals the

maximum in-plane shear strain provided thein-plane principal strains are of opposite signs.


CHAPTER REVIEW


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If the in-plane principal strains are of same signs,

then absolute maximum shear strain will occur outof plane and is determined from max = max/2.

Hookes law can be expressed in 3 dimensions,

where each strain is related to the 3 normal stress

components using the material propertiesE, and ,as seen in Eqns 10-18.

IfEand are known, then G can be determined

using G =E/[2(1 + ]. Dilatation is a measure of volumetric strain, and the

bulk modulus is used to measure the stiffness of avolume of material.


CHAPTER REVIEW


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Provided the principal stresses for a material

are known, then a theory of failure can be usedas a basis for design.

Ductile materials fail in shear, and here themaximum-shear-stress theory or the maximum-

distortion-energy theory can be used to predictfailure.

Both theories make comparison to the yield

stress of a specimen subjected to uniaxialstress.


CHAPTER REVIEW


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Brittle materials fail in tension, and so the

maximum-normal-stress theory or Mohrsfailure criterion can be used to predict failure.

Comparisons are made with the ultimate tensilestress developed in a specimen.

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