mom chapter 10 new-edited

7/29/2019 MOM Chapter 10 New-edited

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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


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1. Reading Quiz

2. Applications

3. Equations of plane-strain transformation

4. Principal and maximum in-plane shear strain

5. Mohrs circle for plane strain6. Absolute maximum shear strain

7. Measurement of strains

8. Stress-strain relationship

9. Theories of failure

10. Concept Quiz

In-class Activities



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READING QUIZ

1) Which of the following statement is incorrectfor plane-strain?

a) z = yz = xz = 0 while the plane-strain has 3 components x, y

and xy

.

b) Always identical to the state of plane stress

c) Identical to the state of plane stress only when the Poissons

ratio is zero.

d) When the state of strain is represented by the principal strains,

no shear strain will act on the element.



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READING QUIZ (cont)

2) Which of the following failure criteria is notsuitable for brittle materials?

a) Maximum-normal-stress theory

b) Mohrs failure criterion

c) Tresca yield criterion

d) None of them



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APPLICATIONS



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EQUATIONS OF PLANE-STRAINTRANSFORMATION


In 3D, the general state of strain at a point is

represented by a combination of 3 components ofnormal strain x, y, z, and 3 components of shearstrain xy, yz, xz.

In plane-strain cases, z, xz and yz are zero.

The state of plane strain at a point is uniquelyrepresented by 3 components (x, y and xy) acting onan element that has a specific orientation at the point.


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EQUATIONS OF PLANE-STRAINTRANSFORMATION (cont)


Note:Plane-stress caseplane-strain case


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Positive normal strain x and y cause elongation

Positive shear strain xy 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 thenormal and shear strain components are to be

determined) will be defined using the angle . The angleis measured from the positive x- to positive x-axis. It ispositive if it follows the curl of the right-hand fingers.


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Normal and shear strains Consider the line segment dx

sin'

cos'

dydy

dxdx

2sin2

2cos22

2sin2

2cos22

cossinsincos'

cossincos'

'

'

22

'

xyyxyx

y

xyyxyx

x

xyxxx

xyyx

dx

x

dydydxx


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Similarly,

2sincossin

sincossin'

xyyx

xyyx dydydxdy

90cossincos

90sin90cos90sin

2

2

xyyx

xyyx

y


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2cos2

2sin22

sincoscossin

''

22

''

xyyxyx

xyyxyx


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EXAMPLE 1


A differential element of material at a point is subjected to astate of plane strain

which tends to distort the element as shown in Fig. 105a.Determine the equivalent strains acting on an element of thematerial oriented at the point, clockwise 30 from theoriginal position.

666 10200,10300,10500 xyyx


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EXAMPLE 1 (cont)


Since ispositive counter-clockwise,

Solutions

(Ans)10213

302sin2

10200302cos10

2

30050010

2

300500

2sin2

2cos22

6

'

666

'

x

xyyxyx

x

(Ans)10793

302cos2

10200302sin10

2

300500

2cos2

2sin22

6

''

66

''

yx

xyyxyx


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EXAMPLE 1 (cont)


By replacement,

Solutions

(Ans)104.13

602sin2

10200602cos10

2

30050010

2

300500

2sin2

2cos22

6

'

666

'

x

xyyxyx

y


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PRINCIPLE AND MAXIMUM IN-PLANE SHEARSTRAIN


Similar to the deviations for principal stresses and the

maximum in-plane shear stress, we have

And,

22

2,1222

2tan

xyyxyx

yx

xy

p

2,

222

2tan

22plane-inmax yx

avg

xyyx

xy

yx

S


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PRINCIPLE AND MAXIMUM IN-PLANE SHEARSTRAIN (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 interms 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 shearstrain and its associated average normal strain is 45from the element representing the principal strains.


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EXAMPLE 2


A differential element of material at a point is subjected to astate of plane strain defined by

which tends to distort the element as shown in Fig. 107a.Determine the maximum in-plane shear strain at the pointand the associated orientation of the element.

666 1080,10200,10350 xyyx


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EXAMPLE 2 (cont)


Looking at the orientation of the element,

For maximum in-plane shear strain,

Solutions

311and9.40

80

2003502tan

s

xy

yx

s

(Ans)10556222

6

planeinmax

22

planeinmax

xyyx


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MOHRS CIRCLE FOR PLANE STRAIN


A geometrical representation of Equations 10-5 and 10-

6; i.e.

Sign convention: is positive to the right, and /2 ispositive downwards.

22

22and

2

where

xyyxyx

avg R

22

''2

'2

Ryx

avgx


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MOHRS CIRCLE FOR PLANE STRAIN (cont)



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EXAMPLE 3


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

Determine the principal strains and the orientation of theelement.

666 10120,10150,10250 xyyx


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EXAMPLE 3 (cont)


From the coordinates of point E, we have

To orient the element, we can determine

the clockwise angle.

Solutions

6

6

planeinmax''

6planeinmax''

1050

10418

108.2082

avg

yx

yx

(Ans)7.36

35.82902

1

1

s

s


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ABSOLUTE MAXIMUM SHEAR STRAIN


State of strain in 3-dimensional space:

2

minmax

minmaxmaxabs

avg


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EXAMPLE 4


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

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

666 10150,10200,10400 xyyx


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EXAMPLE 4 (cont)


From the strain components, the centre of the circle is on the 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


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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 Mohrs circle is as follow,

Solutions

66min

66

max

1040910309100

1020910309100

(Ans)1061810409209 66minmaxplaneinmax

10409,0,102096

minint

6

max


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MEASUREMENT OF STRAINS BY STRAINROSETTES


Ways of arranging 3 electrical-resistance strain gauges

In general case (a):

ccxycycxc

bbxybybxb

aaxyayaxa

cossinsincos

cossinsincos

cossinsincos

22

22

22


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VARIABLE SOLUTIONS


Please click the appropriate icon for your computer to access the

variable solutions
http://localhost/var/www/apps/conversion/tmp/scratch_10/variable_solutions_10_5_stand_alone.exehttp://localhost/var/www/apps/conversion/tmp/scratch_10/show-variable_solutions_10_5.html


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MEASUREMENT OF STRAINS BY STRAINROSETTES (cont)


In 45 strain rosette [case (b)],

In 60 strain rosette [case (c)],

cabxy

cy

ax

2

cbxy

acby

ax

3

2

2221


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EXAMPLE 5


The state of strain at pointA on the bracket in Fig. 1017a ismeasured using the strain rosette shown in Fig. 1017b. Due

to the loadings, the readings from the gauges give

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

666 10264,10135,1060 cba


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EXAMPLE 5 (cont)


Measuring the angles counter-clockwise,

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

have

Using Mohrs circle, we have A(60(10-6), 60(10-6)) and center C (153(10-6),

0).

Solutions

120and60,0 cba

666 10149,10246,1060 zyx

(Ans)3.19

,109.33

,10272

101.119105.7460153

p2

6

1

6

1

6622

R


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STRESS-STRAIN RELATIONSHIP


Use the principle of superposition

Use Poissons ratio,

Use Hookes Law (as it applies in the uniaxial direction),

allongitudinlateral

E

yxzzzxyyzyxx vE

vE

vE

1

,1

,1


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STRESS-STRAIN RELATIONSHIP (cont)


Use Hookes Law for shear stress and shear strain

Note:

xzxzyzyzxyxyGGG

1

1

1

vE

G

12


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Dilatation (i.e. volumetric strain )zyxV

ve

zyx

E

ve

21


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For special case of hydrostatic loading,

Where the right-hand side is defined as bulk modulus R,

i.e.

vE

e

P

pzyx

213

vEk

213

EXAMPLE 6


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EXAMPLE 6


The copper bar in Fig. 1024 is subjected to a uniformloading 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 afterapplication of the load. Take 34.0,GPa120 cucu vE

EXAMPLE 6 ( t)


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EXAMPLE 6 (cont)


From the loading we have

The associated normal strains are determined from the generalized

Hookes law,

The new bar length, width, and thickness are therefore

Solutions

0,80,MPa500,MPa800 zxyx

000850.0,00643.0,00808.0 yxzzzxy

yzyx

xE

v

EE

v

EE

v

E

(Ans)mm98.1920000850.020'

(Ans)mm68.495000643.050'(Ans)mm4.30230000808.0300'

t

ba


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THEORIES OF FAILURE (for ductile material)


Maximum-shear-stress theory (orTresca yield criterion)

2

maxY


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THEORIES OF FAILURE (for ductile material) (cont)


For plane-stress cases:

signsoppositehave,

signssamehave,

2121

21

2

1

Y

Y

Y


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Maximum-distortion-energy theory (orVon Mises

criterion):

Applying Hookes Law yields

2

1u

233121232221 22

1 v

Eu


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For plane or biaxial-stress cases:

22

221

2

1 Y


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Maximum-normal-stress theory (for materials having

equal strength in tension and compression)

Maximum principle stress 1 in the material reaches alimiting value that is equal to the ultimate normal stress

the material can sustain when it is subjected to simpletension.


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ult2

ult1


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Mohrs 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 1 = 2 =

0, 3 = (ult)c Circle B represents tensile test results, 1 = (ult)t, 2 =3 = 0

Circle C represents puretorsion test results, reaching the ult.


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EXAMPLE 7


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EXAMPLE 7


The solid shaft has a radius of 0.5 cm and is made of steel

having a yield stress ofY= 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)


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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

MPa5.165kN/cm55.16

5.0

2

5.025.3

MPa195kN/cm10.195.0

15

2

4

2

J

Tc

A

P

xy

x

MPa6.286andMPa6.95

22

21

2

2

2,1

xy

yxyx

EXAMPLE 7 (cont)


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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

22

221

2

1

Y


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CONCEPT QUIZ

1) Which of the following statement is incorrect?

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

b) When Poissons ratio approaches 0.5, the bulk modulus tends toinfinity and the material behaves like incompressible.

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

d) Mohrs failure criterion is not suitable for brittle material havingdifferent strength in tension and compression.

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