Lecture 5 Strain Transformation

Chapter Objectives

Navigate between rectilinear co-ordinate system for strain components

Determine principal strains and maximum in-plane shear strain Determine the absolute maximum shear strain in 2D and 3D cases Know ways of measuring strains Define stress-strain relationship Predict failure of material

Copyright © 2011 Pearson Education South Asia Pte Ltd

1. Reading Quiz

2. Applications

3. Equations of plane-strain transformation

4. Principal and maximum in-plane shear strain

5. Mohr’s circle for plane strain

6. Absolute maximum shear strain

7. Measurement of strains

8. Stress-strain relationship

9. Theories of failure

10. Concept Quiz

In-class Activities


APPLICATIONS





EQUATIONS OF PLANE-STRAIN TRANSFORMATION


• In 3D, the general state of strain at a point is represented by a combination of 3 components of normal strain σx, σy, σz, and 3 components of shear strain γxy, γyz, γxz.

• In plane-strain cases, σz, γxz and γyz are zero.

• The state of plane strain at a point is uniquely represented by 3 components (σx, σy and γxy) acting on an element that has a specific orientation at the point.

EQUATIONS OF PLANE-STRAIN TRANSFORMATION (cont)


Note: Plane-stress case ≠ plane-strain case



• Positive normal strain σx and σy cause elongation• Positive shear strain γxy causes small angle AOB• Both the x-y and x’-y’ system follow the right-hand rule• The orientation of an inclined plane (on which the

normal and shear strain components are to be determined) will be defined using the angle θ. The angle is measured from the positive x- to positive x’-axis. It is positive if it follows the curl of the right-hand fingers.



• Normal and shear strains– Consider the line segment dx’

sin'

cos'

dydy

dxdx

2sin2

2cos22

2sin2

2cos22

cossinsincos'

cossincos'

'

'

22'

xyyxyxy

xyyxyxx

xyxxx

xyyx

dx

x

dydydxx



• Similarly,

2sincossin

sincossin'

xyyx

xyyx dydydxdy

90cossincos

90sin90cos90sin2

2

xyyx

xyyx

y



2cos2

2sin22

sincoscossin

''

22''

xyyxyx

xyyxyx

EXAMPLE 1


A differential element of material at a point is subjected to a state of plane strain which tends to distort the element as shown in Fig. 10–5a. Determine the equivalent strains acting on an element of the material oriented at the point, clockwise 30° from the original position.

666 10200 , 10300 , 10500 xyyx

EXAMPLE 1 (cont)


• Since θ is positive counter-clockwise,

Solutions

(Ans) 10213

302sin2

10200302cos10

2

30050010

2

300500

2sin2

2cos22

6'

666

'

x

xyyxyxx

(Ans) 10793

302cos2

10200302sin10

2

300500

2cos2

2sin22

6''

66

''

yx

xyyxyx

EXAMPLE 1 (cont)


• By replacement,

Solutions

(Ans) 104.13

602sin2

10200602cos10

2

30050010

2

300500

2sin2

2cos22

6'

666

'

x

xyyxyxy

PRINCIPLE AND MAXIMUM IN-PLANE SHEAR STRAIN


• Similar to the deviations for principal stresses and the maximum in-plane shear stress, we have

• And,

22

2,1 222 2tan

xyyxyx

yx

xyp

2 ,

222

2tan

22

plane-inmax yxavg

xyyx

xy

yxS

PRINCIPLE AND MAXIMUM IN-PLANE SHEAR STRAIN (cont)


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

• The state of strain at a point 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 strain is 45° from the element representing the principal strains.

EXAMPLE 2


A differential element of material at a point is subjected to a state of plane strain defined by which tends to distort the element as shown in Fig. 10–7a. Determine the maximum in-plane shear strain at the point and the associated orientation of the element.

666 1080 , 10200 , 10350 xyyx

EXAMPLE 2 (cont)


• Looking at the orientation of the element,

• For maximum in-plane shear strain,

Solutions

311 and 9.40

80

2003502tan

s

xy

yxs

(Ans) 10556

2226

planein max

22

planein max

xyyx

MOHR’S CIRCLE FOR PLANE STRAIN


• A geometrical representation of Equations 10-5 and 10-6; i.e.

• Sign convention: ε is positive to the right, and γ/2 is positive downwards.

22

22 and

2

where

xyyxyx

avg R

2

2

''2' 2

Ryxavgx

MOHR’S CIRCLE FOR PLANE STRAIN (cont)


EXAMPLE 3


The state of plane strain at a point is represented by the components:

Determine the principal strains and the orientation of the element.

666 10120 , 10150 , 10250 xyyx

EXAMPLE 3 (cont)


• From the coordinates of point E, we have

• To orient the element, we can determine the clockwise angle.

Solutions

6

6

planein max ''

6planein max ''

1050

10418

108.2082

avg

yx

yx

(Ans) 7.36

35.82902

1

1

s

s

ABSOLUTE MAXIMUM SHEAR STRAIN


• State of strain in 3-dimensional space:

2minmax

minmaxmax abs

avg

EXAMPLE 4


The state of plane strain at a point is represented by the components:

Determine the maximum in-plane shear strain and the absolute maximum shear strain.

666 10150 , 10200 , 10400 xyyx

EXAMPLE 4 (cont)


• From the strain components, the centre of the circle is on the ε axis at

• Since , the reference point has coordinates

• Thus the radius of the circle is

Solutions

66 10100102

200400

avg

610752

xy

66 1075,10400 A

9622 103091075100400

R

EXAMPLE 4 (cont)


• Computing the in-plane principal strains, we have

• From the circle, the maximum in-plane shear strain is

• From the above results, we have

• Thus the Mohr’s circle is as follow,

Solutions

66

min

66max

1040910309100

1020910309100

(Ans) 1061810409209 66minmaxplanein max

10409 , 0 , 10209 6minint

6max

MEASUREMENT OF STRAINS BY STRAIN ROSETTES


• Ways of arranging 3 electrical-resistance strain gauges

• In general case (a):

ccxycycxc

bbxybybxb

aaxyayaxa

cossinsincos

cossinsincos

cossinsincos

22

22

22

VARIABLE SOLUTIONS


Please click the appropriate icon for your computer to access the variable solutions

MEASUREMENT OF STRAINS BY STRAIN ROSETTES (cont)


• In 45° strain rosette [case (b)],

• • In 60° strain rosette [case (c)],

cabxy

cy

ax

2

cbxy

acby

ax

3

2

222

1

EXAMPLE 5


The state of strain at point A on the bracket in Fig. 10–17a is measured using the strain rosette shown in Fig. 10–17b. Due to the loadings, the readings from the gauges give

Determine the in-plane principal strains at the point and the directions in which they act.

666 10264 , 10135 , 1060 cba

EXAMPLE 5 (cont)


• Measuring the angles counter-clockwise,

• By substituting the values into the 3 strain-transformation equations, we have

• Using Mohr’s circle, we have A(60(10-6), 60(10-6)) and center C (153(10-6), 0).

Solutions

120 and 60 ,0 cba

666 10149 , 10246 , 1060 zyx

(Ans) 3.19

, 109.33

, 10272

101.119105.7460153

p2

61

61

6622

R

STRESS-STRAIN RELATIONSHIP


• Use the principle of superposition

• Use Poisson’s ratio,

• Use Hooke’s Law (as it applies in the uniaxial direction),

allongitudinlateral

E

yxzzzxyyzyxx vE

vE

vE

1

, 1

, 1

STRESS-STRAIN RELATIONSHIP (cont)


• Use Hooke’s Law for shear stress and shear strain

• Note:

xzxzyzyzxyxy GGG 1

1

1

vE

G

12



• Dilatation (i.e. volumetric strain )zyxV

ve

zyxE

ve

21



• For special case of “hydrostatic” loading,

• Where the right-hand side is defined as bulk modulus R, i.e.

vE

e

P

pzyx

213

vE

k213

EXAMPLE 6


The copper bar in Fig. 10–24 is subjected to a uniform loading along its edges as shown. If it has a length a = 300 mm, b = 500 mm, and t = 20 mm before the load is applied, determine its new length, width, and thickness after application of the load. Take 34.0 , GPa 120 cucu vE

EXAMPLE 6 (cont)


• From the loading we have

• The associated normal strains are determined from the generalized Hooke’s law,

• The new bar length, width, and thickness are therefore

Solutions

0 , 80 , MPa 500 , MPa 800 zxyx

000850.0 , 00643.0 , 00808.0 yxz

zzxy

yzyx

x E

v

EE

v

EE

v

E

(Ans) mm 98.1920000850.020'

(Ans) mm 68.495000643.050'

(Ans) mm 4.30230000808.0300'

t

b

a

THEORIES OF FAILURE (for ductile material)


• Maximum-shear-stress theory (or Tresca yield criterion)

2maxY

THEORIES OF FAILURE (for ductile material) (cont)


• For plane-stress cases:

signs opposite have ,

signs same have ,

2121

21

2

1

Y

Y

Y



• Maximum-distortion-energy theory (or Von Mises criterion):

• Applying Hooke’s Law yields

2

1u

23312123

22

21 2

2

1 vE

u



• For plane or biaxial-stress cases:

22221

21 Y



• Maximum-normal-stress theory (for materials having equal strength in tension and compression)

• Maximum principle stress σ1 in the material reaches a limiting value that is equal to the ultimate normal stress the material can sustain when it is subjected to simple tension.




ult2

ult1



• Mohr’s failure criterion (for materials having different strength in tension and compression

• Perform 3 tests on the material to obtain the failure envelope

• Circle A represents compression test results σ1 = σ2 = 0, σ3 = – (σult)c

• Circle B represents tensile test results, σ1 = (σult)t, σ2 = σ3 = 0

• Circle C represents pure torsion test results, reaching the τult.




EXAMPLE 7


The solid shaft has a radius of 0.5 cm and is made of steel having a yield stress of σY = 360 MPa. Determine if the loadings cause the shaft to fail according to the maximum-shear-stress theory and the maximum-distortion-energy theory.

EXAMPLE 7 (cont)


• Since maximum shear stress caused by the torque, we have

• Principal stresses can also be obtained using the stress-transformation equations,

Solutions

MPa 5.165kN/cm 55.16

5.02

5.025.3

MPa 195kN/cm 10.195.0

15

2

4

2

J

Tc

A

P

xy

x

MPa 6.286 and MPa 6.95

22

21

2

2

2,1

σσ

xyyxyx

EXAMPLE 7 (cont)


• Since the principal stresses have opposite signs, the absolute maximum shear stress will occur in the plane,

• Thus, shear failure of the material will occur according to this theory.

• Using maximum-distortion-energy theory,

• Using this theory, failure will not occur.

Solutions

3602.382

3606.2866.95

21

Y

1296009.118677

3606.2866.2866.956.95

222

22221

21

Y

CONCEPT QUIZ

1) Which of the following statement is incorrect?

a) Dilatation is caused only by normal strain, not shear strain.

b) When Poisson’s ratio approaches 0.5, the bulk modulus tends to infinity and the material behaves like incompressible.

c) Von Mises failure criterion is not suitable for ductile material.

d) Mohr’s failure criterion is not suitable for brittle material having different strength in tension and compression.


Documents

Lecture 5 Strain Transformation