8. strain transformation 1 apply the stress transformation methods derived in chapter 9 to similarly...

1

8. Strain Transformation

• Apply the stress transformation methods derived in Chapter 9 to similarly transform strain

• Discuss various ways of measuring strain

• Develop important material-property relationships; including generalized form of Hooke’s law

CHAPTER OBJECTIVES

2


CHAPTER OUTLINE

1. Plane Strain

2. General Equations of Plane-Strain Transformation

3. Strain Rosettes

4. Material-Property Relationships

3


10.1 PLANE STRAIN

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

• Strain components at a pt determined by using strain gauges, which is measured in specified directions.

• A plane-strained element is subjected to two components of normal strain (x, y) and one component of shear strain, xy.

4


10.1 PLANE STRAIN

• The deformations are shown graphically below.• Note that the normal strains are produced by

changes in length of the element in the x and y directions, while shear strain is produced by the relative rotation of two adjacent sides of the element.

5


10.1 PLANE STRAIN

• Note that plane stress does not always cause plane strain.

• In general, unless = 0, the Poisson effect will prevent the simultaneous occurrence of plane strain and plane stress.

• Since shear stress and shear strain not affected by Poisson’s ratio, condition of xz = yz = 0 requires xz = yz = 0.

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10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION

Sign Convention• To use the same convention as

defined in Chapter 2.2.• With reference to differential

element shown, normal strains x and y are positive if they cause elongation along the x and y axes

• Shear strain xy is positive if the interior angle AOB becomes smaller than 90.

7



Sign Convention• Similar to plane stress, when measuring the normal

and shear strains relative to the x’ and y’ axes, the angle will be positive provided it follows the curling of the right-hand fingers, counterclockwise.

Normal and shear strains• Before we develop the

strain-transformation eqn for determining x;, we must determine the elongation of a line segment dx’ that lies along the x’ axis and subjected to strain components.

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Normal and shear strains• Components of line dx and dx’ are elongated and

we add all elongations together.

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

cossincos' dydydxx xyyx

210cossinsincos 22' - xyyxx

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Normal and shear strains

• To get the transformation equation for x’y’, consider amount of rotation of each of the line segments dx’ and dy’ when subjected to strain components. Thus, sincossin' dydydxy xyyx

10



Normal and shear strains• Using Eqn 10-1 with = y’/x’,

• As shown, dy’ rotates by an amount .

310sincossin 2 - xyyx

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Normal and shear strains• Using identities sin ( + 90) = cos ,

cos ( + 90) = sin ,

• Thus we get

2

2

cossincos

90sin90cos90sin

xyyx

xyyx

410sincoscossin2 22

''

-

xyyx

yx

12



Normal and shear strains• Using trigonometric identities sin 2 = 2 sin cos,

cos2 = (1 + cos2 )/2 and sin2 + cos2 = 1, we rewrite Eqns 10-2 and 10-4 as

5102sin2

2cos22' -

xyyxyx

x

6-102cos2

2sin22

''

xyyxyx

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Normal and shear strains• If normal strain in the y direction is required, it can

be obtained from Eqn 10-5 by substituting ( + 90) for . The result is

6102sin2

2cos22' -

xyyxyx

y

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Principal strains• We can orientate an element at a pt such that the

element’s deformation is only represented by normal strains, with no shear strains.

• The material must be isotropic, and the axes along which the strains occur must coincide with the axes that define the principal axes.

• Thus from Eqns 9-4 and 9-5,

8102tan -yx

xyp

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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

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Maximum in-plane shear strain• Using Eqns 9-6, 9-7 and 9-8, we get

12102

-avgyx

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IMPORTANT• Due to Poisson effect, the state of plane strain is

not a state of plane stress, and vice versa.• A pt on a body is subjected to plane stress when

the surface of the body is stress-free.• Plane strain analysis may be used within the plane

of the stresses to analyze the results from the gauges. Remember though, there is normal strain that is perpendicular to the gauges.

• When the state of strain is represented by the principal strains, no shear strain will act on the element.

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IMPORTANT• The state of strain at the pt can also be represented

in terms of the maximum in-plane shear strain. In this case, an average normal strain will also act on the element.

• The element representing the maximum in-plane shear strain and its associated average normal strains is 45 from the element representing the principal strains.

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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 element oriented at the pt, clockwise 30 from the original position.

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EXAMPLE 10.1 (SOLN)

• Since is counterclockwise, then = –30, use strain-transformation Eqns 10-5 and 10-6,

6'

6

6

6

'

10213

302sin210200

302cos102

300500

102

300500

2sin2

2cos22

x

xyyxyxx

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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

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EXAMPLE 10.1 (SOLN)

• Strain in the y’ direction can be obtained from Eqn 10-7 with = –30. However, we can also obtain y’ using Eqn 10-5 with = 60 ( = –30 + 90), replacing x’ with y’

6'

6

6

6'

104.13

602sin210200

602cos102

300500

102

300500

y

y

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EXAMPLE 10.1 (SOLN)

• The results obtained tend to deform the element as shown below.

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EXAMPLE 10.2

A differential element of material at a pt is subjected to 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 at the pt and associated orientation of the element.

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EXAMPLE 10.2 (SOLN)

Orientation of the element

From Eqn 10-8, we have

Each of these angles is measured positive 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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EXAMPLE 10.2 (SOLN)

Principal strains

From Eqn 10-9,

6

26

1

66

6226

22

2,1

1035310203

109.277100.75

102

802

2003502

10200350

222

xyyxyx

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EXAMPLE 10.2 (SOLN)

Principal strains

We can determine which of these two strains deforms the element in the x’ direction by applying Eqn 10-5 with = –4.14. Thus

6'

6

66

'

10353

14.42sin2

1080

14.4cos102

20035010

2200350

2sin2

2cos22

x

xyyxyxx

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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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EXAMPLE 10.3

A differential element of material at a pt is subjected to 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-plane shear strain at the pt and associated orientation of the element.

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EXAMPLE 10.3 (SOLN)

Orientation of the element

From Eqn 10-10,

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

9.1309.40

,72.26118072.8172.812

1080

102003502tan

6

6

and

thatsoandThus,

s

s

xy

yxs

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EXAMPLE 10.3 (SOLN)

Maximum in-plane shear strainApplying Eqn 10-11,

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

6

622

22

10556

102

802

200350

222

plane-in

max

plane-in

max

xyyx

plane-in

max

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EXAMPLE 10.3 (SOLN)

Maximum in-plane shear strain

Thus tends to distort the element so that the right angle between dx’ and dy’ is decreased (positive sign convention).

6''

6

6

''

10556

9.402cos2

1080

9.402sin102

200350

2cos2

2sin22

yx

xyyxyx

plane-in

max

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EXAMPLE 10.3 (SOLN)

Maximum in-plane shear strain

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

These strains tend to cause the element to contract.

66 1075102

2003502

yx avg

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• We measure the normal strain in a tension-test specimen using an electrical-resistance strain gauge.

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

• Such strain gauges, arranged in a specific pattern are called strain rosettes.

• Note that only the strains in the plane of the gauges are measured by the strain rosette. That is ,the normal 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, y xy by solving the three 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

17103

2

2231

-cbxy

acby

ax

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EXAMPLE 10.8

State of strain at pt A on bracket is measured using the strain rosette shown. Due to the loadings, the readings 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)

Establish x axis as shown, measure the angles counterclockwise from the +x axis to center-lines of each gauge, we have a = 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 obtained directly from Eqn 10-17. Reference pt on Mohr’s circle is A [60(10-6), –74.5(10-6)] and center of circle, C is on the axis at avg = 153(10-6). From shaded triangle, radius is

EXAMPLE 10.8 (SOLN)

666 10149102461060 xyyx

6

622

102.119

105.7460153

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 an out-of-plane strain, in the z direction, although this value does not influence the calculated results.

EXAMPLE 10.8 (SOLN)

3.19

7.3860153

5.74tan2

108.33102.11910246

10272102.11910153

2

12

6662

6661

p

p

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10.6 MATERIAL-PROPERTY RELATIONSHIPS

Generalized Hooke’s law• Material at a pt subjected to a state of triaxial

stress, with associated strains.• We use principle of superposition, Poisson’s ratio

(lat = long), and Hooke’s law ( = E) to relate stresses to strains, in the uniaxial direction.

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

Ex

x '

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Generalized Hooke’s law

• With y applied, element contracts with a strain ‘’x in the x direction,

• Likewise, With z applied, a contraction is caused in the z direction,

Ey

x

''

Ez

x '''

43



Generalized Hooke’s law• By using the principle of superposition,

yxzz

zxyy

zyxx

E

E

E

1

18101

1

-

44



Generalized Hooke’s law

• If we apply a shear stress xy to the element, experimental observations show that it will deform only due to shear strain xy. Similarly for xz and xy, yz and yz. Thus, Hooke’s law for shear stress and shear strain is written as

1910111

-xzxzyzyzxyxy GGG

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Relationship involving E, , and G• We stated in chapter 3.7:

• Relate principal strain to shear stress,

• Note that since x = y = z = 0, then from Eqn 10-18, x = y = 0. Substitute into transformation Eqn 10-19,

2010

12-

E

G

21101max -

Exy

2max1xy

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Relationship involving E, , and G

• By Hooke’s law, xy = xy/G. So max = xy/2G.

• Substitute into Eqn 10-21 and rearrange to obtain Eqn 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 stress application, they become (1 + x)dx, (1 + y)dy, (1 + z)dz, respectively.


47



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 Hooke’s law, we write the dilatation in terms of applied stress.

dzdydxdzdydxV zyx 111

2210 -zyxdVV

e

231021

-zyxEe

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Dilatation and Bulk Modulus• When volume element of material is subjected to

uniform pressure p of a liquid, pressure is the same in all directions.

• As shear resistance of a liquid is zero, we can ignore shear stresses.

• Thus, an element of the body is subjected to principal stresses x = y = z = –p. Substituting into Eqn 10-23 and rearranging,

2410

213-

E

ep

49



Dilatation and Bulk Modulus• This ratio (p/e) is similar to the ratio of linear-elastic

stress to strain, thus terms on the RHS are called the volume modulus of elasticity or the bulk modulus. Having same units as stress with symbol k,

• For most metals, ≈ ⅓ so k ≈ E.

• From Eqn 10-25, theoretical maximum value of Poisson’s ratio is therefore = 0.5.

• When plastic yielding occurs, = 0.5 is used.

2510

213-

E

k

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IMPORTANT• When homogeneous and isotropic material is

subjected to a state of triaxial stress, the strain in one of the stress directions is influence by the strains produced by all stresses. This is the result of the Poisson effect, and results in the form of a generalized Hooke’s law.

• A shear stress applied to homogenous and isotropic material will only produce shear strain in the same plane.

• Material constants, E, G and are related mathematically.

51



IMPORTANT• Dilatation, or volumetric strain, is caused by only by

normal strain, not shear strain.• The bulk modulus is a measure of the stiffness of a

volume of material. This material property provides an upper limit to Poisson’s ratio of = 0.5, which remains at this value while plastic yielding occurs.

52


EXAMPLE 10.10

Copper bar is subjected to a uniform loading along its edges as shown. If it has a length a = 300 mm, width b = 50 mm, and thickness t = 20 mm before the load is applied, determine its new length, width, and thickness after application of the load. Take Ecu = 120 GPa, cu = 0.34.

53


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 Hooke’s law, Eqn 10-18;

00500800 zxyyx MPaMPa

00808.0500

10312034.0

103120800

MPaMPa

MPa

zvx

x EE

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Associated strains are determined from generalized Hooke’s law, Eqn 10-18;

00850.0500800

10312034.0

0

00643.080010312034.0

103120500

MPaMPa

MPaMPa

MPa

yxz

z

zxy

y

EE

EE

55



The new bar length, width, and thickness are

mmmmmm

mmmmmm

mmmmmm

98.1920000850.020'

68.495000643.050'

4.30230000808.0300'

t

b

a

56


EXAMPLE 10.11

If rectangular block shown is subjected to a uniform pressure of p = 20 kPa, determine the dilatation and change in length of each side. Take E = 600 kPa, = 0.45.

57



Dilatation

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

33 /01.0

203600

45.021

21

cmcm

kPakPa

zyxEe

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Change in length

Normal strain on each side can be determined from Hooke’s law, Eqn 10-18;

cm/cm

kPakPakPakPa

00333.0

202045.020600

1

1

zyxE

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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

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CHAPTER REVIEW

• When element of material is subjected to deformations that only occur in a single plane, then it undergoes plain strain.

• If the strain components x, y, and xy are known for a specified orientation of the element, then the strains acting for some other orientation of the element can be determined using the plane-strain transformation equations.

• Likewise, principal normal strains and maximum in-plane shear strain can be determined using transformation equations.

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CHAPTER REVIEW

• Hooke’s law can be expressed in 3 dimensions, where each strain is related to the 3 normal stress components using the material properties E, and , as seen in Eqns 10-18.

• If E and 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 a volume of material.

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