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From equilibrium principles:
xy
=
yx
,
xz
=
zx
,
zy
=
yz
3D Stress Components
Normal Stresses
Shear Stresses
Normal stress () : the subscript identifies the face on which thestress acts. Tension is positive and compression is negative.
Shear stress () : it has two subscripts. The first subscript
denotes the face on which the stress acts. The second subscript
denotes the direction on that face. A shear stress is positive if itacts on a positive face and positive direction or if it acts in a
negative face and negative direction.
The most general state of stress at a point may
be represented by 6 components
zyx
xzyzxy
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==
zyzxz
zyyxy
zxyxx
ij
For static equilibrium xy
=
yx
,
xz
=
zx
,
zy
=
yz
resulting insix independent
scalar quantities. These six scalars can be arranged in a 3x3 matrix, giving us astress
tensor.
The sign convention for the stress elements is that a
positive force on a positive face or a negative force
on a negative face is positive. All others are negative.
The stress state is a second order tensor since it is a quantity associated with two
directions (two subscripts direction of the surface normal and direction of the stress).
Same state of stress is represented by a different set of components if axes are rotated.There is a special set of components (when axes are rotated) where all the shear
components are zero (principal stresses).
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A property of a symmetric tensor is that there exists an orthogonal set of axes 1,2 and
3 (called principal axes) with respect to which the tensor elements are all zero except
for those in the diagonal.
==
zyzxz
zyyxy
zxyxx
ij
==
3
2
1
'
00
00
00
'
ij
Eigen values
In matrix notation the transformation is known as theEigen-values.
The principal stresses are the new-axescoordinate system. The angles between the
old-axesand the new-axesare known as theEigen-vectors.
principal stress
Cosine of angle
between X and the
principal stress
Cosine of angle
between Y and the
principal stress
Cosine of angle
between Z and the
principal stress
1 k1 l1 m1
2 k2 l2 m2
3 k3 l3 m3
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State of stress in which two faces of the cubic element are free of stress. For the
illustrated example, the state of stress is defined by
.0and,, xy === zyzxzyx
Plane Stress
Sign Conventions for Shear Stress and Strain
The Shear Stress will be considered
positive when a pair of shear stress
acting on opposite sides of the
element produce a counterclockwise(ccw) torque (couple).
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A shear strain in an element is positive when the angle between two positive faces(or two negative faces) is reduced, and is negative if the angle is increased.
000
0
0
yyxy
yxxx
000
0
0
1111
1111
yyyx
xyxx
000
00
00
2
1
1
22
1
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Stresses on Inclined SectionsKnowing the normal and shear stresses acting in the element denoted by thexy axes,
we will calculate the normal and shear stresses acting in the element denoted by the
axisx1y1.
x1y1
x
y
y
x
x1
x1y1
yxxy
Equilibrium of forces:
Acting inx1
Cos
SinA
Cos
SinASinAA
AOYXOYOXYOX
OX +++= cossincos
cos1
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EliminatingAo ,sec = 1/cos and
xy=yx
cossin2sincos 221 XYYXX ++=
Acting in y1
x1y1Aosec = xAosin + xyAocos + yAotancos yxAotansin
EliminatingAo ,sec = 1/cos and xy=yx
cossin2cossin 221 XYYXY +=
22
11 sincoscossincossin ++= xyyxyx
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Case 1: Uniaxial stress
Special Cases
=
+=
===
2
2
2
21
00
1,1
1
xy
Sin
Cos
xyx
xx
yxy
Case 2 : Pure Shear
2
2
0
1,1
1
Cos
Sin
xyyx
xyx
yx
=
=
==
2cos2sin2
2sin2cos22
11
1
xy
yx
yx
xy
yxyx
x
+
=
+
++
=
Case 3: Biaxial stress
22
222
0
1,1
1
Sin
Cos
yx
yx
yxyx
x
xy
=
++
=
=
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An element in plane stress is subjected to stresses x=16000psi, y=6000psi, and
xy=yx= 4000psi (as shown in figure below). Determine the stresses acting on an
element inclined at an angle =45o (counterclockwise - ccw).
Solution: We will use the following transformation equations:
2cos2sin
2
2sin2cos22
2sin2cos22
11
1
1
xy
yx
yx
xyyxyxy
xy
yxyx
x
+
=
+
=
+++=
( )600016000 ++
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Numerical substitution
( )
( )
( ) ( )
( ) ( )( ) ( ) psi
psi
psi
psi
psi
psi
yx
y
x
xy
yx
yx
50000400015000
7000140000500011000
15000140000500011000
4000
50002
600016000
2
110002
600016000
2
11
1
1
=+===
=++=
=
=
=
=+
=+
( ) ( )( ) ( ) 0902
1902
45
0
0
0
==
==
+=
CosCos
SinSin
(ccw)For x-axis
( ) ( )( ) ( ) 02702
12702
9045
0
0
00
====
++=
CosCos
SinSin
(ccw)For y-axis
Note:11 yxyx
+=+
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A plane stress condition exists at a point on the surface of a loaded structure such as
shown below. Determine the stresses acting on an element that is oriented at a
clockwise (cw) angle of 15o with respect to the original element, the principalstresses, the maximum shear stress and the angle of inclination for the principal
stressesSolution: We will use the following transformation equations:
2cos2sin2
2sin2cos22
11
1
xy
yx
yx
xyyxyx
x
+
=
+++=
( )
( )
( ) ( )( ) ( )( )
( )( ) ( )( ) MPa
MPaMPa
MPa
MPa
yx
x
xy
yx
yx
31866.0195.029
6.325.019866.0291719
292
1246
2
172
1246
2
11
1
=+=
=++==
=
=
=+
=+
( ) ( )( ) ( ) 866.0302
5.0302
15
0
0
0
==
==
=
CosCos
SinSin
(cw)For x-axis
MPay
yxyx
4.11
11
=
+=+
A rectangular plate of dimensions 3 0 in x 5 0 in is formed by welding two
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Determine the normal stress sw acting perpendicular to the line or the weld and the shear stress
tw acting parallel to the weld. (Assume sw is positive when it acts in tension and tw is positivewhen it acts counterclockwise against the weld).
Solution:
psi
psipsi
psipsipsi
y
yxx
xy
yxyx
25
375375
04252
1752
1
111
===
==
=+
00 926129630
5
3tan
..
For x-axis
==
=
psiy
yxyx
251
11
=
+=+
-375psi
= 30.96o
375psi25psi
x
y
Stresses acting on the weld
w
w
25psi
375psi = 30.96o
A rectangular plate of dimensions 3.0 in x 5.0 in is formed by welding two
triangular plates (see figure). The plate is subjected to a tensile stress of 600psi
in the long direction and a compressive stress of 250psi in the short direction.
w
= -25psi and
w
= 375psi
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Principal Stresses and Maximum Shear Stresses
The sum of the normal stresses acting on perpendicular faces of plane stress
elements is constant and independent of the angle .
YXYX +=+ 11As we change the angle there will bemaximum and minimum normal and
shear stresses that are needed for design purposes.
The maximum and minimum normal
stresses are known as the principal
stresses. These stresses are found bytaking the derivative of x1 with respect
to and setting equal to zero.
=
=+=
+
++
=
2
2tan
02cos22sin)(
2sin2cos22
1
1
yx
xy
P
xyyxx
xy
yxyx
x
cossin2sincos 221 XYYXX ++=
cossin2cossin 221 XYYXY +=
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The subscriptp indicates that the anglep defines the orientation of the principal
planes. The angle p has two values that differ by90o. They are known as the principal
angles. For one of these angles x1 is amaximumprincipal stress and for the other a
minimum. The principal stresses occur in mutually perpendicular planes.
( )2
2tanyx
xy
P
=
2
)(2cos2sin
RR
yx
P
xy
P
==
P
1
stressmaximumfor the
2sin2cos22
=
+
++
= xyyxyx
x
( )
( )
( ) ( )22
12
2
2
2
1
2
2
1
22
2But
2
1
2
11
22222
xyyxyx
xyyx
xy
yxyx
xy
yxyxxy
xy
yxyxyx
R
R
RRRR
+
++=+
=
+
++
=
+
+
+=
+
+
+=
P i i l St
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The plus sign gives the algebraically
larger principal stress and the minus
sign the algebraically smaller principal
stress.
This are the in-plane principal
stresses. Thethird stress is zero in
plane stress conditions
( )
( ) R
R
Averagexy
yxyx
Averagexy
yxyx
=+
+=
+=+
+
+=
2
2
2
2
2
1
22
22
Principal Stresses
The location of the angle for the maximum shear stress is obtained by taking thederivative of x1y1 with respect to and setting it equal to zero.
2cos2sin
211 xy
yx
yx +
=
( )
xy
yx
S
xyyx
yx
22tan
02sin22cos)(11
=
==
Maximum Shear Stresses
045d)(
2i2
yxxy
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( )
( )
( )
( )o
P
o
P
P
o
P
o
P
P
S
S
P
Pxy
yx
S
902cos
902sin
290cos
290sin
2sin
2cos
2cos
2sin
2cot2tan
1
2
)(2tan
=
==
==
=
0
P1s111 45and2
)(2sin2cos ===
RR
yx
s
xy
s
The planes for maximum shearstress occurs at45o to the principal
planes. The plane of the maximum
positive shear stress max is defined
by the angle S1 for which thefollowing equations apply:
The corresponding maximum shear is given
by the equation
Another expression for the maximum shear
stress
The normal stresses associated with the
maximum shear stress are equal to
( ) Rxyyx
MAX =+
=
2
2
2
( )
2
21 =
MAX
Therefore,2s-2p=-90o or
s= p +/- 45o
2
yx
AVER
+
=
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Equations of a Circle
2sin2cos2
2sin2cos22
1
1
xy
yx
AVERx
xy
yxyx
x
+
=
+
++
=General equation
Consider
Equation (1)
Equation (2)
2cos2sin211 xy
yx
yx +
=
Equation (1) + Equation (2)
( )2
2
1 2sin2cos2
+= xy
yx
AVERx
( )2
2
11 2cos2sin2
+
=
xy
yx
yx
( )22
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( ) ( )2112
1 2cos2sin2
2sin2cos2
+
+
+
=+
xy
yx
xy
yx
yxAVERx
( ) ( )
( ) ( )
( ) 222
222
22
222
22
2
2cos2sin2
22cos2sin2
2cos2sin2
2cos2sin2
22sin2cos2
2sin2cos2
RSUM xyyx
xy
yx
xy
yx
xy
yx
xyyx
xyyx
xyyx
=+
=
+
=
+
++
=
+
( ) ( )22
11
2
1 RyxAVERx =+
Mohr Circle
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x
y
P
C
( )22
2 xy
yxR
+
=
+
2
yx
The radius of the Mohr circle is the
magnitudeR.
Mohr Circle
( )22
2 xy
yxR
+
=
The center of the Mohr circle is the
magnitude
2
yx
AVER
+=
( ) ( ) ( )22
211
21
2 xyyxyxAVERx
+
=+State of Stresses
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Alternative sign conversion for shear
stresses:
(a)clockwise shear stress,
(b)counterclockwise shear stress, and
(c) axes for Mohrs circle.
Note that clockwise shear stresses are
plotted upward and counterclockwise
shear stresses are plotted downward.
a) We can plot the normal stress x1 positive to the right and the shear stress
x1y1 positive downwards, i.e. the angle2 will be positive when
counterclockwise or
b) We can plot the normal stress x1positive to the right and the shear stress x1y1positive upwards, i.e. the angle2 will be positive when clockwise.
Both forms are mathematically correct. We use (a)
Forms of Mohrs Circle
T f f M h i l
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Two forms of Mohrs circle:
x1y1 is positive downward and the angle2 is
positive counterclockwise, and
x1y1 is positive upward and the angle2 is positive
clockwise. (Note: The first form is used here)
Construction of Mohrs circle for plane stress.
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Construction of Mohr s circle for plane stress.
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At a point on the surface of a pressurized cylinder, the
material is subjected to biaxial stresses x = 90MPa and y =
20MPa as shown in the element below.
Using the Mohr circle, determine the stresses acting on an
element inclined at an angle = 30o (Sketch a properly
oriented element).
Because the shear stress is zero,
these are the principal stresses.
Construction of the Mohrs circle
(x = 90MPa, y = 20MPa and xy = 0MPa)
( )MPa
yx
Average 552
2090
2=
+=
+=
The center of the circle is
The radius of the circle is( ) ( ) MPaR xy
yx350
2
2090
2
2
2
2
2
=+
=+
=
Stresses on an element inclined at = 30o
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Stresses on an element inclined at = 30
By inspection of the circle, the coordinates of pointD are
( )MPaCosCosR
MPaSinSinR
MPaCosCosR
Averagey
oo
yx
Averagex
5.3760355560
3.30603560
5.7260355560
0
1
11
0
1
===
===
=+=+=
An element in plane stress at the surface of a large
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machine is subjected to stresses x = 15000psi, y =
5000psi and xy = 4000psi.
Using the Mohrs circle determine the following:a) The stresses acting on an element inclined at an
angle = 40o
b) The principal stresses and
c) The maximum shear stresses.
Construction of Mohrs circle:
Center of the circle (Point C):
Radius of the circle:
Point A, representing the stresses on thex face of the element ( = 0o
) has thecoordinates x1 = 15000psi and x1y1 = 4000psi
Point B, representing the stresses on they face of the element ( = 90o) has the
coordinates y1 = 5000psi and y1x1 = - 4000psiThe circle is now drawn through points A and B with center C and radius R
( )psi
yx
Average 100002
500015000
2=
+=
+=
( ) ( ) psiR xyyx
640340002
500015000
2
2
2
2
2
=+
=+
=
Stresses on an element inclined at
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By inspection the angle ACP1 for
the principal stresses (point P1) is :
Then, the angle P1CD is 80o
38.66o = 41.34o
( ) oACPACP 66.385000
4000tan 11 ==
= 40o
These are given by the coordinates of
pointD which is at an angle2 = 80o
from point A
Knowing this angle, we can calculate the coordinates of point D (by inspection)
( )
psiCosCosR
psiSinSinR
psiCosCosR
ooAveragey
oo
yx
oo
Averagex
519034.4164031000034.41
423034.41640334.41
1481034.4164031000034.41
1
11
1
===
===
=+=+=
And of course, the sum of the normal stresses is
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Principal Stresses
The principal stresses are represented by
points P1 and P2 on Mohrs circle.
The angle it was found to be 2 = 38.66o
or = 19.3o
psiR
psiR
Average
Average
3597640310000
16403640310000
2
1
===
=+=+=
14810psi + 5190psi = 15000psi + 5000psi
Maximum Shear Stresses
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These are represented by point S1 and S2 in Mohrs circle. Algebraically the
maximum shear stress is given by theradius of the circle.
The angle ACS1 from point A to point S1 is 2 S1 = 51.34o. This angle is negativebecause is measured clockwise on the circle. Then the corresponding S1 value is 25.7o.
0400015000 0016403
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Psi
000
050004000
0400015000
Psi
000
035970
00164033-D stress state
Transform to
In matrix notation the transformation is known as the Eigen-values.
The principal stresses are the new-axes coordinate system. The angles betweenthe old-axes and the new-axes are known as the Eigen-vectors.
principal stress
Cosine of angle
between X and theprincipal stress
Cosine of angle
between Y and theprincipal stress
Cosine of angle
between Z and theprincipal stress
16403.1242 0.94362832 0.331006939 0
3596.876 -0.33101 0.943628 0
0 0 0 1
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At a point on the surface of a generator shaft the stresses are
x = -50MPa, y = 10MPa and xy = - 40MPa as shown in the
figure. Using Mohrs circle determine the following:(a)Stresses acting on an element inclined at an angle = 45o,(b)The principal stresses and
(c)The maximum shear stresses
Construction of Mohrs circle
Center of the circle (Point C):
Radius of the circle:.
Point A, representing the stresses on thex face of the element ( = 0o) has the
coordinates x1 = -50MPa and x1y1 = - 40MPaPoint B, representing the stresses on they face of the element ( = 90o) has thecoordinates y1 = 10MPa and y1x1 = 40MPaThe circle is now drawn through points A and B with center C and radius R.
( ) ( )MPa
yx
Average 202
1050
2=
+=
+=
( ) ( ) ( ) ( ) MPaR xyyx 504021050
22
2
2
2
=+
=+
=
Stresses on an element
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inclined at = 45o
These stresses are given by
the coordinates of point D(2 = 90o of point A). Tocalculate its magnitude we
need to determine the angles
ACP2 and P2CD.
Then, the coordinates of point D are
And of course, the sum of the normal stresses is -50MPa+10MPa = -60MPa +20MPa
( )
( )( ) MPaCosCosR
MPaSinSinR
MPaCosCosR
oo
Averagey
oo
yx
oo
Averagex
2087.36502087.36
3087.365087.36
6087.36502087.36
1
11
1
=+==
===
=+=+=
tan ACP2=40/30=4/3
ACP2=53.13o
P2CD = 90o 53.13o = 36.87o
Principal Stresses
Th d b i P dMaximum Shear Stresses
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They are represented by points P1 and
P2 on Mohrs circle.
The angle ACP1 is 2P1 = 180o + 53.13o
= 233.13o or P1 = 116.6o
The angle ACP2 is 2P2 = 53.13o or P2= 26.6o
These are represented by point S1 and S2in Mohrs circle.
The angle ACS1 is 2S1 = 90o + 53.13o =143.13o or = 71.6o .The magnitude of the maximum shear
stress is 50MPa and the normal stresses
corresponding to point S1 is the average
stress -20MPa.
MPaR
MPaR
Average
Average
705020
305020
2
1
====+=+=
04050 0030
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MPa
000
01040
04050
MPa
000
0700
00303-D stress state Transform to
In matrix notation the transformation is known as theEigen-values.
The principal stresses are the new-axes coordinate system. The angles betweenthe old-axes and the new-axes are known as theEigen-vectors.
principal stress
Cosine of angle
between X and theprincipal stress
Cosine of angle
between Y and theprincipal stress
Cosine of angle
between Z and theprincipal stress
30 -0.44721359 0.894427193 0
-70 0.894427 0.447214 0
0 0 0 1
Hookes Law for Plane Stress
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The stress transformations equations were derived solely from equilibrium
conditions and they are material independent.
Here the material properties will be considered (strain) taking into account the
following:
a)The material is uniform throughout the body (homogeneous)
b)The material has the same properties in all directions (isotropic)c)The material follows Hookes law (linearly elastic material)
Hookes law: Linear relationship between stress and strain
For uniaxial stress:
(E = modulus of elasticity or Youngs modulus)
Poissons ratio:
For pure shear : (G = Shear modulus of elasticity)
E=allongitudin
transverse
straina
strainlateral
==
xial
G=
Element of material in plane stress (z
= 0).
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Consider the normal strains x
,
y
,
z
in plane
stress.
All are shown with positive elongation.
The strains can be expressed in terms of the
stresses by superimposing the effect of the
individual stresses.
For instance the strain x
in the x direction:a)Due to the stress
x
is equal to x
/E .
b)Due to the stress y
is equal to y
/E .
The resulting strain is: EE
EE
EE
yx
z
yxy
yxx
=
+=
=
The shear stress causes a distortion of the element
h th t h f b h b
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such that each z face becomes a rhombus.
GXYXY
=
The normal stressesx
and y
have no
effect on the shear strain xy
or rearranging the equations:
G
XYXY
=
( ) ( ) ( ) ( ) XYyx
Y
yxx G
EEEE
=+
+
=
+
= XY2222 11
11
EEEEEE
yxz
yxy
yxx
=+==
These equations are known collectively as Hookes Law for plane stress
These equations contain three material constants (E, G and
) but only two are
independent because of the relationship:
)1(2 += EG
Special cases of Hookes law
z
= 0)
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EEx
zx
x
xyy
===
==
y
00
Uniaxial stress :
G
xy
xyzyx
yx
====
==
0
0
Pure Shear :
Biaxial stress :
-
- yxyx
y
yxx
xy
=+==
=
z
0
When a solid object undergoes strains, both its
dimensions and its volume will change.Consider an object of dimensions a, b, c. The
original volume is Vo
= abc and its final volume is
V
1
= a + a
x
) b + b
y
) c + c
z
)
V
1
= abc 1+
x
) 1+
y
) 1+
z
)
Volume Change
Upon expanding the terms:
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V
1
= V
o
1 +
x
+
y
+
z
+
x
y
+
x
z
+
y
z
+
x
y
z
)
For small strains:
V
1
= V
o
1 +
x
+
y
+
z
)
The volume change is
V = V
1
V
o
= V
o
x
+
y
+
z
)
The unit volume change e, also known as dilatation is defines as:e = V / V
o
=
x
+
y
+
z
Positive strains are considered for elongations and negative strains for
shortening, i.e. positive values of efor an increase in volume.
EEEEEE
yxz
yxy
yxx
=+==
( )( )EV
Ve yx 21+== For uniaxial stress y = 0 ( )
Ee x 21=
We can notice that the maximum possible value of Poissons ratio is 0.5, because a
larger value means that the volume decreases when the material is in tension(contrary to physical behavior).
STRAIN ENERGY DENSITY IN PLANE STRESS
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The strain energy density u is the strain energy stored in a unit volume of the
material.Because the normal and shear strains occur independently, we can add the
strain energy of these two elements to obtain the total energy.
Work done = Force x distance
( )( )( )
( )( ) ( )y
x
bac
abc
2direction-yin thedoneWork
2direction-xin thedoneWork
y
x
=
=
The sum of the energies due to normal stresses:
( )yyxxabcU += 2
Then the strain energy density (strain per unit volume) ( )yyxxu +=
2
11
Similarly, the strain energy density associated with the
shear strain:
By combining the strain energy densities for the normaland shear strains:
xyxyu 2
12=
( )xyxyyyxxu ++=21
The strain energy density in terms
f t l GEEEu XYYXYX
222
222
++=
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of stress alone: GEEE 222
The strain energy density in termsof strain alone: ( )( )
2222 2
212
XYYXYX GEu ++=
For the special case of uniaxial stress: For the special case of pure shear:
2
or2
000
22
xx
xyxyxyy
Euu
==
====
2
or2
000
22
xyxy
xyyx
Gu
Gu
==
====
An element of the material subjected to normal
stresses x
,
y
and z
acting in three mutually
perpendicular directions is said to be in a stateof triaxial stress. Since there is no shear in x, y
or z faces then the stresses x
, y
and z
are
the principal stresses in the material.
TRIAXIAL STRESS
If an inclined plane parallel to thez-axis is cut through the element, the only stress
of the inclined face are the normal stress and the shear stress both of which act
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of the inclined face are the normal stressand the shear stress,both of which act
parallel to thexyplane.
The same general conclusion hold for normal and shear stresses
acting on inclined planes cut through the element parallel to the
x andy axes.
Maximum Shear Stress For a material in triaxial stress, the maximum shear stressesoccur on elements oriented at angles of45o to thex,y andz
axes.
( )
( )
( ) 2
2
2
ZX
YMAX
ZY
XMAX
YX
ZMAX
=
=
=
Because these stresses are independent of the z, we can usethe transformation equations of plane stress, as well as the
Mohrs circle for plane stress, when determining the stresses
and in triaxial stress.
for the inclined plane // z-axis
for the inclined plane // x-axis
for the inclined plane // y-axis
The absolute maximum of the shear stress is the numerically largest of the above.
The stresses acting on elements oriented at
various angles to the x y and z axes can be
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various angles to thex,y andz axes can be
visualized with the aid of the Mohrs
circle.
In this case x
>
y
>
z
Hookes Law for Triaxial Stress
If H k l i b d i i ibl b i h
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If Hookes law is obeyed, it is possible to obtain the
relationship between normal stresses and normal strains
using the same procedure as for plane stress.
EEE
EEE
zyxz
zyxy
zyxx
+=
+=
=
Or
( )( )( )[ ]
( )( ) ( )[ ]
( )( ) ( )[ ]zyxz
zyxy
zyxx
E
E
E
+++
=
+++
=
+++=
1211
1211
1211
They are known as the
Hookes law for triaxialstress.
Unit Volume ChangeThe unit volume change (or dilatation) for an
element in triaxial stress is obtained in the
same manner as for plane stress.
( ) ( )ZYXO
ZYX
O
EV
Ve
VVe
++==
++==
21
thenapply,lawssHooke'If
Strain Energy Density
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The strain energy density for an
element in triaxial stress is
obtained by the same method
used for plane stress.
( ) ( )zyzxyxzyx
zzyyxx
EEu
u
++++=
++=
-2
1
:stressesofIn terms
222
222
( )( )( )( ) ( )[ ]21
2112
222
zyzxyxzyx
Eu
+++++
+=
In terms of the strains:
Spherical StressA special case of triaxial stress, called spherical stress, occurs
whenever all three normal stresses are equal:0 === zyx
The Mohrs circle is reduced to a single point. Any plane cut through the element
will be free of shear stress and will be subjected to the same normal stressso andit is a principal plane.
The normal strains in spherical stress are also
the same in all directions, provided the material
is isotropic and if Hookes law applies:
( )
( )
E
O
OO
2133e
changevolumeThe21
O
==
=
K= bulk or volume modulus of elasticity
E
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Element in spherical stress.
( ) eK
EK 0
213
=
=
If = 1/3 thenK = E
If = 0 thenK = E/3
If = 1/2 thenK= infinite (rigid material having no
change in volume)These formulas also apply to and element in uniform compression,
for example rock deep within the earth or material submerged in
water (hydrostatic stress).
PLAIN STRAIN
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Strains are measured by strain gages.
A material is said to be in a state of plain strain if the only deformations are those
in thexyplane, i.e. it has only three strain components x, y and xy.
Plain stress is analogous to plane stress, but under ordinary conditions they do
not occur simultaneously
Exception when x = -y and when = 0
Strain components x, y, and xy in the
xyplane (plane strain).
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Comparison of plane stress and plane strain.
Th f i i f l lid h i
APPLICATION OF THE TRANSFORMATION EQUATIONS
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The transformation equations for plane stress are valid even when z is not zero,
because z
does not enter the equations of equilibrium.Therefore,the
transformations equations for plane stress can also be used for stresses in plane
strain.
Similarly, the strain transformation equations that will be derived for the case of
plain strain in thexyplane are valid even when z is not zero, because the strain zdoes not affect the geometric relationship used for the derivation.Therefore,the
transformations equations for plane strain can also be used for strains in plane
stress.
Transformation Equations for Plain Strain
We will assume that the strain x, y and xyassociated with thexyplane are known.
We need to determine the normal and shear strains(x1 and x1y1) associated with thex1y1 axis. y1 can
be obtained from the equation of x1by substituting
+90 for .
For an element of sizedx,dy
In thex direction:
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the strain xproduces an elongation xdx.
The diagonal increases in length by xdx cos.
In they direction:
the strain yproduces an elongation ydy.
The diagonal increases in length by ydy sin .
The shear strain xy in the planexyproduces a
distortion of the element such that the angle at thelower left corner decreases by an amount equal to
the shear strain. Consequently, the upper face
moves to the right by an amount xydy. This
deformation results in an increase in the length ofthe diagonal equal to: xydy cos
The total increase of the diagonal is the sum of the preceding three expressions,
thus: ( ) ( ) ( ) CosdySindyCosdxd xyyx ++=
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( ) ( ) ( ) yy xyyx
But ( ) ( ) ( )
Sinds
dyCos
ds
dx
Cosds
dySinds
dyCosds
dx
ds
dxyyxx
==
++=
=
1
CosSinSinCos xyyxx ++=22
1
Shear Strain x1y1 associated withx1y1 axes.
This strain is equal to the decrease in angle
between lines in the material that were initially
along thex1 andy1 axes.
Oa and Ob were the lines initially along thex1 and
y1 axis respectively. The deformation caused by
the strains x, y and xy caused the Oa and Ob lines
to rotate and angle and from thex1 andy1 axis
respectively. The shear strain x1y1
is the decrease
in angle between the two lines that originally were
at right angles, therefore, x1y1=+.
The angle can be found from the deformations
produced by the strains x,y and xy . The strains
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p y x y xyx andxyproduce a clockwise rotation, while the
strain y produces a counterclockwise rotation.
Let us denote the angle of rotation produced by
x , y and xy as 1 , 2 and 3 respectively.
dsdySin
ds
dyCos
ds
dx
Sin
xy
y
x
=
=
=
1
2
1 Sinds
dy
Cosds
dx
==
( ) 2321 SinCosSin xyyx =+=
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PRINCIPAL STRAINS
Th l f th i i l t i ixy
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The angle for the principal strains is :
The value for the principal strains are ( ) yxxy
yx
xy
P
=
==
2
22tan
( )
( ) 222
22
1
222
222
+
+=
+
+
+=
xyyxyx
xyyxyx
Maximum Shear
The maximum shear strains in
thexyplane are associated withaxes at45o to the directions of
the principal strains
( )21
22
or222
=
+
+= Max
xyyxMax
For isotropic materials, at a given point in an stressed body, the principal strainsand principal stresses occur in the same directions.
MOHRS CIRCLE FOR PLANE STRAIN
It is constructed in the same mannerx
x
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as the Mohrs circle for plane stress
with the following similarities:
2
2
11
1
11
1
yx
x
xy
y
yx
x
xy
y
Strain Measurements
An electrical-resistancestrain gage is a device for measuring normal strains () on
the surface of a stressed object
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the surface of a stressed object.The gages are small (less than inch) made of wires that are bonded to the surface of the object. Each
gage that is stretched or shortened when the object is strained at the point, changes its electrical
resistance. This change in resistance is converted into a measurement of strain.
From three measurements it is possible to calculate the strains in any direction. A
group of three gages arranged in a particular pattern is called astrain rosette.
Because the rosette is mounted in the surface of the body, where the material is inplane stress, we can use the transformation equations for plane strain to calculate the
strains in various directions.
45 strain rosette, and element oriented at
an angle
to thexy axes.
General Equations
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Other Strain Rosette
An element of material in plane strain undergoes the following strains:x=340x10-6
; = 110x10-6 ; = 180x10-6 Determine the following quantities:
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;y = 110x106 ; xy = 180x10
6 . Determine the following quantities:
(a) the strains of an element oriented at an angle = 30o
;(b) the principal strains and
(c) the maximum shear strains.
(a) Element oriented at an angle = 30o (2 = 60o)
6
1
6
1
1
10360
10602
18060
2
110340
2
110340
2sin2
2cos22
=
+
++
=
+
++
=
x
x
xyyxyx
x
SinCos
90
360110340
1
1
11
=
+=+
+=+
y
y
yxyx
225=Average
6611
11
105510602
18060
2
110340
2
2cos2
2sin22
=
+
=
+=
CosSinyx
xyyxyx
(b)Principal Strains and Angle of Rotation
( ) 22
+ 78260
552tan =
== xy
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( )
802
180
2
110340225
3702
180
2
110340225
222
22
2
22
1
2,1
=
+
=
=
+
+=
+
+= xyyxyx
019
7826.0110340
2tan
=
=
=
=
P
yx
P
(c) In-Plane Maximum Shear Strain
180110-3402222
xyyxM
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( )
29080370
145
2
180
2
110340
222
21 ===
=
+
=
+
+=
Max
xyyxMax
225=Average
(d) Out-of-Plane Maximum Shear Strain
( ) 370037031 ===Max
cossin22
sincos 221
++= XYYXX
Transformation Equations
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( ) 2211 sincos2
cossincossin
2
++= XYYXYX
21
YXX
cossin22
cossin 221
+= XYYXY
[ ]
[ ]
=
=
22
22
11
1
1
1
11
1
1
YX
Y
X
XY
Y
X
XY
Y
X
YX
Y
X
T
T
[ ]
( )
=
22
22
22
sincoscossincossin
cossin2cossin
cossin2sincos
T
( )
=
2
sincoscossincossin
cossin2cossincossin2sincos
2
22
22
22
11
1
1
XY
Y
X
YX
Y
X
30cos30sin230sin30cos22
1 XX
For =30 degrees
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( )
=
=
=
8.55
6.883.361
90
110340
5.0438.0438.0
876.075.025.0876.025.075.0
2
230sin30cos30cos30sin30cos30sin30cos30sin230cos30sin
2
11
1
1
22
22
11
1
YX
Y
X
XY
Y
YX
Y
zxyxxx
112
1
2
1
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[ ] TensorStrain
zzyzxz
zyyyxy _
2
1
2
12
1
2
1
=
=
[ ]
=
=
000
0790
00371
_
000
01102
180
02
180340
2
1
2
121
21
2
1
2
1
ValuesEigen
zzyzxz
zyyyxy
zxyxxx
Example
A45o strain rosette (rectangular rosette) consists of three electrical-resistance strain gages,
arranged to measure strains in two perpendicular directions and also at a45o angle (as shown
b l ) Th i b d d h f f h b f i i l d d G A B d
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below). The rosette is bonded to the surface of the structure before it is loaded. GagesA,B and
Cmeasure the normal strainsa, b and c in the directions of the lines Oa, Ob and Oc,respectively.
Explain how to obtain the strainsx1, y1 and x1y1, associated with an element oriented at an
angle
to thexy axes.
cossin22
sincos22
++= xyyxa
Angles with respect to x-axis: (a) is zero ; (b) is
45 degrees CCW ; (c) 90 degrees CCW
0cos0sin220sin0cos
22
++=
xy
yxa
( )45cos45sin22
45sin45cos 22
++= xyyxb
( )90cos90sin22
90sin90cos 22
++= xyyxb
cabxy
by
ax
=
==
2
The following results are obtained from a 600 strain gauge
rosette:
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Strain in direction of strain gauge A = 750;
Strain in direction of SG B, 600 to A = 350;
Strain in direction of SG C, 1200 to A = 100 .
Determine the principal strains and their directions.
750== xa
( )
( ) ( ) ( )433.0275.025.0
60cos60sin22
60sin60cos 22
++=
++=
xy
yxb
xy
yxb
( )
( ) ( ) ( )433.02
75.025.0
120cos120sin22
120sin120cos 22
++=
++=
xy
yxb
xy
yxb
289
50
=
=
xy
y
[ ]
=
= 050289
02
289750112
1
2
1zxyxxx
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[ ]
=
=
100
0981.01945.0
01945.0981.0
_
000
0210
00779
_
000
0502
2
1
2
1 22
VectorsEigen
ValuesEigen
zzyzxz
zyyyxy
ArcCos(angle)=0.981Angle =11.2degress
(A) Using the transformation equations define the maximum and minimum
principal strains, maximum shearing strain and principal angles given
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p p , g p p g g
X = 3500 ; Y = 700 and XY = -1050
(B) Repeat using the Mohrs circle.
[ ]
=
100
0984.0178.0
0178.0984.0
_
000
08.6040
002.3595
_
000
07002
1050
02
10503500
VectorsEigen
ValuesEigen
ArcCos(angle)=0.984
Angle =10.28degress
(c) In-Plane Maximum Shear Strain
( ) 4.29908.6042.359521 ===Max
(d) Out-of-Plane Maximum Shear Strain
( ) 2.359502.359531 ===Max
The state of stress at a point in a structural member is determined to be as shown.
Knowing that for this material E=210GPa and
=0.3, use the Mohrs circle to
determine: (1) the principal stresses; (2) the in-plane maximum shear stress; (3) the
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determine: (1) the principal stresses; (2) the in plane maximum shear stress; (3) the
absolute maximum shear stress; (4) principal angles; (5) the strains and the principalstrains; (6) the maximum shear strain; (7) the principal angles.
14MPa
56MPa
11.2MPa
x
y ( ) RAveragexy
yxyx =+
+= 2
2
2,122
( )
( )( )
MPa
MPaR
Average 352
1456
8.232.112
1456 22
=
+=
=+
=
MPaMPa8.588.23352.118.2335
2
1
===+=
1. Principal Stresses
[ ] MPaTensorStress
==000
0142.11
02.1156
_ MPaValuesEigen
8.5800
02.110
000
_
MPaMPa
MPa
8.582.11
0
3
2
1
==
=
2. In-Plane Maximum Shear Stress MPaRMax 8.23==
3 The Absolute Maximum Shear Stress (Out of plane)
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3. The Absolute Maximum Shear Stress (Out of plane)
( )MPaMax 4.29
2
8.580
231 =
=
=
4. Angle between the x-axis and the
Principal Stresses
( ) ( ) ( )( )
( )
deg07.282
533.021
2.112tan
21456
2.11
2
2tan)_(
=
=
=
=
==
P
P
yx
xy
PPlaneIn
12425.09701.0
09701.02425.0
000
_VectorsEigen
For 1=0ArcCos(angle)=0.0
Angle =90degress
For 2=-11.2MPaArcCos(angle)=0.2425
Angle =76degress
For 3=-58.8MPaArcCos(angle)=0.9701
Angle =14degress
5. Strains and Principal Strains
EEEzyx
x
= ( ) ( ) 6.246206.266
210000
143.0
210000
56=+=
=x
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EEE
zyxz
zyxy
+=
+=
G
XYXY
=
( ) ( )
( ) ( ) ( )
1002080210000
143.0
210000
563.0
4.136.6680210000
14210000
563.0
=+=
=
==+=
z
y
13980770
2.11===
G
xy
xy
( ) GPaE
G 77.803.012
210
)1(2 =+=+=
[ ]
=
=
0.26400
08.300
00100
_
1000004.132139
02
1396.246
2
1
2
12
1
2
12
1
2
1
ValuesEigen
zzyzxz
zyyyxy
zxyxxx
( )
RAverage
xyyxyx
=
+
+=
2,1
22
2,1222
( )
4.1472
139
2
4.136.246
6.1162
4.136.246
22
=
+
=
=+
=
R
Average
264
8.30
2
1
=
=
6. Maximum Shear Strain and Absolute Maximum Shear Strain
( )
4.1472
139
2
4.136.246_
2
22
=
+
==Max Rplanein
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8.294=Max
( ) ( )
364
1822
264100
2__
2
31
=
=
=
=
Max
Max planeofout
( ) ( ) 534.04.136.246139
2
22tan ===== yx
xy
yx
xy
P
7. In-plane and Out-of-plane angles
( )
deg07.282
533.02tan
=
=
P
P
12425.09701.009701.02425.0
000
_VectorsEigen
For 1=100
ArcCos(angle)=0.0Angle =90degress
For 2=30.8
ArcCos(angle)=0.2425Angle =76degress
For 3=-264
ArcCos(angle)=0.9701Angle =14degress