08032012-nice ppt strain transformation
Post on 06-Apr-2018
220 Views
Preview:
TRANSCRIPT
-
8/2/2019 08032012-Nice Ppt Strain Transformation
1/124
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
2/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
3/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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.
-
8/2/2019 08032012-Nice Ppt Strain Transformation
4/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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.
-
8/2/2019 08032012-Nice Ppt Strain Transformation
5/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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.
-
8/2/2019 08032012-Nice Ppt Strain Transformation
6/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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.
-
8/2/2019 08032012-Nice Ppt Strain Transformation
7/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
7
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
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.
-
8/2/2019 08032012-Nice Ppt Strain Transformation
8/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
8
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Normal and shear strains
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
9/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
9
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Normal and shear strains
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
10/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
10
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Normal and shear strains
Using Eqn 10-1 with = y/x,
As shown, dyrotates by an amount.
310sincossin 2 - xyyx
-
8/2/2019 08032012-Nice Ppt Strain Transformation
11/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
11
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
12/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
12
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Normal and shear strains
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
13/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
13
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Normal and shear strains
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
14/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
14
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
15/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
15
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
16/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
16
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Maximum in-plane shear strain
Using Eqns 9-6, 9-7 and 9-8, we get
12102
-avgyx
S i T f i
-
8/2/2019 08032012-Nice Ppt Strain Transformation
17/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
17
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
18/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
18
10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
19/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
20/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
21/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
22/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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
-
8/2/2019 08032012-Nice Ppt Strain Transformation
23/1242005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
23
EXAMPLE 10.1 (SOLN)
The results obtained tend to deform the element as
shown below.
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
24/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
24
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 atthe pt and associated orientation of the element.
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
25/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
26/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
27/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
28/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
28
EXAMPLE 10.2 (SOLN)
Principal strains
Hence x= 2. When subjected to the principal strains,the element is distorted as shown.
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
29/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
29
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-planeshear strain at the pt and associated orientation of theelement.
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
30/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
31/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
31
EXAMPLE 10.3 (SOLN)
Maximum in-plane shear strain
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
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
32/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
32
EXAMPLE 10.3 (SOLN)
Maximum in-plane shear strain
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
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
33/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
33
EXAMPLE 10.3 (SOLN)
Maximum in-plane shear strain
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
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
34/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
35/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
35
*10.3 MOHRS CIRCLE: PLANE STRAIN
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.
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
36/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
36
*10.3 MOHRS CIRCLE: PLANE STRAIN
Procedure for Analysis
Construction of the circle
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.
10 Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
37/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
37
*10.3 MOHRS CIRCLE: PLANE STRAIN
Procedure for Analysis
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.
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
38/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
38
*10.3 MOHRS CIRCLE: PLANE STRAIN
Procedure for Analysis
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.
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
39/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
39
*10.3 MOHRS CIRCLE: PLANE STRAIN
Procedure for Analysis
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
40/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
40
*10.3 MOHRS CIRCLE: PLANE STRAIN
Procedure for Analysis
Maximum in-plane shear strain
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.
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
41/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
41
*10.3 MOHRS CIRCLE: PLANE STRAIN
Procedure for Analysis
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.
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
42/124
2005 Pearson Education South Asia Pte Ltd
10. Strain Transformation
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.
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
43/124
2005 Pearson Education South Asia Pte Ltd
0 St a a s o at o
43
EXAMPLE 10.4 (SOLN)
Construction of the circle
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
44/124
2005 Pearson Education South Asia Pte Ltd 44
EXAMPLE 10.4 (SOLN)
Construction of the circle
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
45/124
2005 Pearson Education South Asia Pte Ltd 45
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
46/124
2005 Pearson Education South Asia Pte Ltd 46
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
47/124
2005 Pearson Education South Asia Pte Ltd 47
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.
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
48/124
2005 Pearson Education South Asia Pte Ltd 48
EXAMPLE 10.5 (SOLN)
Maximum in-plane shear strain
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
49/124
2005 Pearson Education South Asia Pte Ltd 49
EXAMPLE 10.5 (SOLN)
Maximum in-plane shear strain
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
50/124
2005 Pearson Education South Asia Pte Ltd 50
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.
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
51/124
2005 Pearson Education South Asia Pte Ltd 51
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
52/124
2005 Pearson Education South Asia Pte Ltd 52
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
53/124
2005 Pearson Education South Asia Pte Ltd 53
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
54/124
2005 Pearson Education South Asia Pte Ltd 54
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
55/124
2005 Pearson Education South Asia Pte Ltd 55
From the circles drawn,
we see that the absolutemaximum shear strain isdetermined from the circlehaving the larges radius.For this condition,
*10.4 ABSOLUTE MAXIMUM SHEAR STRAIN
15102
1410
minmax
minmax
-
and
-
avg
max
abs
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
56/124
2005 Pearson Education South Asia Pte Ltd 56
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.
*10.4 ABSOLUTE MAXIMUM SHEAR STRAIN
max'' maxmaxabs zx
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
57/124
2005 Pearson Education South Asia Pte Ltd 57
Plane strain
For material subjected toprincipal in-plane strains ofopposite signs,
*10.4 ABSOLUTE MAXIMUM SHEAR STRAIN
minmax
''max
yxmax
abs
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
58/124
2005 Pearson Education South Asia Pte Ltd 58
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.
*10.4 ABSOLUTE MAXIMUM SHEAR STRAIN
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
59/124
2005 Pearson Education South Asia Pte Ltd 59
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.
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
60/124
2005 Pearson Education South Asia Pte Ltd 60
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
61/124
2005 Pearson Education South Asia Pte Ltd 61
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
62/124
2005 Pearson Education South Asia Pte Ltd 62
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
63/124
2005 Pearson Education South Asia Pte Ltd 63
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
64/124
2005 Pearson Education South Asia Pte Ltd 64
10.5 STRAIN ROSETTES
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
65/124
2005 Pearson Education South Asia Pte Ltd 65
10.5 STRAIN ROSETTES
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
66/124
2005 Pearson Education South Asia Pte Ltd 66
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.
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
67/124
2005 Pearson Education South Asia Pte Ltd 67
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
68/124
2005 Pearson Education South Asia Pte Ltd 68
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
69/124
2005 Pearson Education South Asia Pte Ltd 69
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
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
70/124
2005 Pearson Education South Asia Pte Ltd 70
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
'
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
71/124
2005 Pearson Education South Asia Pte Ltd 71
10.6 MATERIAL-PROPERTY RELATIONSHIPS
Generalized Hookes law
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
'''
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
72/124
2005 Pearson Education South Asia Pte Ltd 72
10.6 MATERIAL-PROPERTY RELATIONSHIPS
Generalized Hookes law
By using the principle of superposition,
yxzz
zxyy
zyxx
E
E
E
1
18101
1
-
10. Strain Transformation
-
8/2/2019 08032012-Nice Ppt Strain Transformation
73/124
2005 Pearson Education South Asia Pte Ltd 73
10.6 MATERIAL-PROPERTY RELATIONSHIPS
Generalized Hookes law
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
10. Strain Transformation
0 6 MATERIAL PROPERTY RELATIONSHIPS
-
8/2/2019 08032012-Nice Ppt Strain Transformation
74/124
2005 Pearson Education South Asia Pte Ltd 74
10.6 MATERIAL-PROPERTY RELATIONSHIPS
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
10. Strain Transformation
10 6 MATERIAL PROPERTY RELATIONSHIPS
-
8/2/2019 08032012-Nice Ppt Strain Transformation
75/124
2005 Pearson Education South Asia Pte Ltd 75
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.
10.6 MATERIAL-PROPERTY RELATIONSHIPS
10. Strain Transformation
10 6 MATERIAL PROPERTY RELATIONSHIPS
-
8/2/2019 08032012-Nice Ppt Strain Transformation
76/124
2005 Pearson Education South Asia Pte Ltd 76
10.6 MATERIAL-PROPERTY RELATIONSHIPS
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
10. Strain Transformation
10 6 MATERIAL PROPERTY RELATIONSHIPS
-
8/2/2019 08032012-Nice Ppt Strain Transformation
77/124
2005 Pearson Education South Asia Pte Ltd 77
10.6 MATERIAL-PROPERTY RELATIONSHIPS
Dilatation and Bulk Modulus
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
10. Strain Transformation
10 6 MATERIAL PROPERTY RELATIONSHIPS
-
8/2/2019 08032012-Nice Ppt Strain Transformation
78/124
2005 Pearson Education South Asia Pte Ltd 78
10.6 MATERIAL-PROPERTY RELATIONSHIPS
Dilatation and Bulk Modulus
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
10. Strain Transformation
10 6 MATERIAL PROPERTY RELATIONSHIPS
-
8/2/2019 08032012-Nice Ppt Strain Transformation
79/124
2005 Pearson Education South Asia Pte Ltd 79
10.6 MATERIAL-PROPERTY RELATIONSHIPS
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.
10. Strain Transformation
10 6 MATERIAL PROPERTY RELATIONSHIPS
-
8/2/2019 08032012-Nice Ppt Strain Transformation
80/124
2005 Pearson Education South Asia Pte Ltd 80
10.6 MATERIAL-PROPERTY RELATIONSHIPS
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.
10. Strain Transformation
EXAMPLE 10 10
-
8/2/2019 08032012-Nice Ppt Strain Transformation
81/124
2005 Pearson Education South Asia Pte Ltd 81
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.
10. Strain Transformation
EXAMPLE 10 10 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
82/124
2005 Pearson Education South Asia Pte Ltd 82
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
10. Strain Transformation
EXAMPLE 10 10 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
83/124
2005 Pearson Education South Asia Pte Ltd 83
EXAMPLE 10.10 (SOLN)
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
10. Strain Transformation
EXAMPLE 10 10 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
84/124
2005 Pearson Education South Asia Pte Ltd 84
EXAMPLE 10.10 (SOLN)
The new bar length, width, and thickness are
mmmmmm
mmmmmm
mmmmmm
98.1920000850.020'
68.495000643.050'
4.30230000808.0300'
t
b
a
10. Strain Transformation
EXAMPLE 10 11
-
8/2/2019 08032012-Nice Ppt Strain Transformation
85/124
2005 Pearson Education South Asia Pte Ltd 85
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.
10. Strain Transformation
EXAMPLE 10 11 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
86/124
2005 Pearson Education South Asia Pte Ltd 86
EXAMPLE 10.11 (SOLN)
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
10. Strain Transformation
EXAMPLE 10 11 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
87/124
2005 Pearson Education South Asia Pte Ltd 87
EXAMPLE 10.11 (SOLN)
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
10. Strain Transformation
EXAMPLE 10 11 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
88/124
2005 Pearson Education South Asia Pte Ltd 88
EXAMPLE 10.11 (SOLN)
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. Strain Transformation
*10 7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
89/124
2005 Pearson Education South Asia Pte Ltd 89
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
10. Strain Transformation
*10 7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
90/124
2005 Pearson Education South Asia Pte Ltd 90
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.
10.7 THEORIES OF FAILURE
10. Strain Transformation
*10 7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
91/124
2005 Pearson Education South Asia Pte Ltd 91
10.7 THEORIES OF FAILURE
A. Ductile materials
1. Maximum-Shear-Stress Theory
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
10. Strain Transformation
*10 7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
92/124
2005 Pearson Education South Asia Pte Ltd 92
10.7 THEORIES OF FAILURE
A. Ductile materials
1. Maximum-Shear-Stress Theory
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
10. Strain Transformation
*10 7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
93/124
2005 Pearson Education South Asia Pte Ltd 93
10.7 THEORIES OF FAILURE
A. Ductile materials
1. Maximum-Shear-Stress Theory
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. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
94/124
2005 Pearson Education South Asia Pte Ltd 94
10.7 THEORIES OF FAILURE
A. Ductile materials
1. Maximum-Shear-Stress Theory
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
95/124
2005 Pearson Education South Asia Pte Ltd 95
10.7 THEORIES OF FAILURE
A. Ductile materials
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
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
96/124
2005 Pearson Education South Asia Pte Ltd 96
10.7 THEORIES OF FAILURE
A. Ductile materials
2. Maximum-Distortion-Energy Theory
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
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
97/124
2005 Pearson Education South Asia Pte Ltd 97
10.7 THEORIES OF FAILURE
A. Ductile materials
2. Maximum-Distortion-Energy Theory
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
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
98/124
2005 Pearson Education South Asia Pte Ltd 98
10.7 THEORIES OF FAILURE
A. Ductile materials
2. Maximum-Distortion-Energy Theory
Since maximum-distortion energy theory requiresud= (ud)Y, then for the case of plane or biaxial
stress, we have
3010222212
1 -Y
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
99/124
2005 Pearson Education South Asia Pte Ltd 99
0 O S O U
A. Ductile materials
2. Maximum-Distortion-Energy Theory
Comparing both theories, we get the followinggraph.
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
100/124
2005 Pearson Education South Asia Pte Ltd 100
A. Brittle materials
3. Maximum-Normal-Stress Theory
Figure shows how brittle materialsfail.
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
101/124
2005 Pearson Education South Asia Pte Ltd 101
A. Brittle materials
3. Maximum-Normal-Stress Theory
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
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
102/124
2005 Pearson Education South Asia Pte Ltd 102
A. Brittle materials
3. Maximum-Normal-Stress Theory
Experimentally, it was found to be in closeagreement with the behavior of brittle materials that
have stress-strain diagrams similar in both tensionand compression.
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
103/124
2005 Pearson Education South Asia Pte Ltd 103
A. Brittle materials
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.
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
104/124
2005 Pearson Education South Asia Pte Ltd 104
A. Brittle materials
4. Mohrs Failure Criterion
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.
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
105/124
2005 Pearson Education South Asia Pte Ltd 105
A. Brittle materials
4. Mohrs Failure Criterion
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.
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
106/124
2005 Pearson Education South Asia Pte Ltd 106
A. Brittle materials
4. Mohrs Failure Criterion
The Criterion can also be represented on a graphof principal stresses 1 and 2 (3 = 0).
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
107/124
2005 Pearson Education South Asia Pte Ltd 107
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.
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
108/124
2005 Pearson Education South Asia Pte Ltd 108
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.
10. Strain Transformation
*10.7 THEORIES OF FAILURE
-
8/2/2019 08032012-Nice Ppt Strain Transformation
109/124
2005 Pearson Education South Asia Pte Ltd 109
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.
10. Strain Transformation
*EXAMPLE 10.12
-
8/2/2019 08032012-Nice Ppt Strain Transformation
110/124
2005 Pearson Education South Asia Pte Ltd 110
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.
10. Strain Transformation
*EXAMPLE 10.12 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
111/124
2005 Pearson Education South Asia Pte Ltd 111
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.
10. Strain Transformation
*EXAMPLE 10.12 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
112/124
2005 Pearson Education South Asia Pte Ltd 112
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
10. Strain Transformation
*EXAMPLE 10.12 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
113/124
2005 Pearson Education South Asia Pte Ltd 113
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
10. Strain Transformation
*EXAMPLE 10.12 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
114/124
2005 Pearson Education South Asia Pte Ltd 114
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
10. Strain Transformation
*EXAMPLE 10.14
-
8/2/2019 08032012-Nice Ppt Strain Transformation
115/124
2005 Pearson Education South Asia Pte Ltd 115
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.
10. Strain Transformation
*EXAMPLE 10.14 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
116/124
2005 Pearson Education South Asia Pte Ltd 116
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
10. Strain Transformation
*EXAMPLE 10.14 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
117/124
2005 Pearson Education South Asia Pte Ltd 117
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
10. Strain Transformation
*EXAMPLE 10.14 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
118/124
2005 Pearson Education South Asia Pte Ltd 118
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
10. Strain Transformation
*EXAMPLE 10.14 (SOLN)
-
8/2/2019 08032012-Nice Ppt Strain Transformation
119/124
2005 Pearson Education South Asia Pte Ltd 119
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
10. Strain Transformation
CHAPTER REVIEW
-
8/2/2019 08032012-Nice Ppt Strain Transformation
120/124
2005 Pearson Education South Asia Pte Ltd 120
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.
10. Strain Transformation
CHAPTER REVIEW
-
8/2/2019 08032012-Nice Ppt Strain Transformation
121/124
2005 Pearson Education South Asia Pte Ltd 121
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.
10. Strain Transformation
CHAPTER REVIEW
-
8/2/2019 08032012-Nice Ppt Strain Transformation
122/124
2005 Pearson Education South Asia Pte Ltd 122
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.
10. Strain Transformation
CHAPTER REVIEW
-
8/2/2019 08032012-Nice Ppt Strain Transformation
123/124
2005 Pearson Education South Asia Pte Ltd 123
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.
10. Strain Transformation
CHAPTER REVIEW
-
8/2/2019 08032012-Nice Ppt Strain Transformation
124/124
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.
top related