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CE2155 Structural Mechanics and Materials
byAssoc Professor T. H. Wee
Department of Civil EngineeringEmail: [email protected]
Stress and Strain Transformation(Part 1)
CE2155 Structural Mechanics and Materials
The knowledge of stress and strain transformation will help to:
Establish the state of plane stress (where the stresses in the
out-
of-plane axis is zero) and the state of plane strain (where the
strains in the out-of-plane axis is zero) for various orientations
of reference axes
Determine the principal stress and principal strain; and
establish the principal planes for plane stress and plane strain
conditions
Evaluate the maximum shearing stress and maximum shearing strain
for both in-plane and 3-D cases and
Establish the state of plane strain using the strain
rosettes
The following syllabus will be covered in this topic. Plane
stress Transformation equations for plane stress Principal stresses
and maximum shear stress Mohrs circle of stress Plane strain Mohrs
circle of strain Strain measurement
Introduction
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CE2155 Structural Mechanics and Materials
Introduction
CE2155 Structural Mechanics and Materials
Introduction
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CE2155 Structural Mechanics and Materials
Elasticity
Most engineering structures are designed to undergo relatively
small deformations, involving only linear portion of the
stress-strain relationship. Within this linear portion, the stress,
is directly proportional to the strain , given by
= Ewhere E is the modulus of elasticity of the material, also
known as Youngs modulus and the relationship is known as Hookes
Law. The largest value of the stress for which Hookes Law can be
used for a given material is known as the proportional limit of
that material.
If the strains induced in a test specimen by the application of
a given load disappear when the load is removed, the material is
said to behave elastically. The largest value of the stress for
which the material behaves elastically is called the elastic
limit.
CE2155 Structural Mechanics and Materials
Stress State at a Point
The state of stress at a point can be represented most generally
by six independent normal and shear stress components which act on
the faces of an element of material located at the point. These
stresses are referred to the corresponding coordinate axes.
If the coordinate axes are rotated, the same state of stress
will be represented by a different set of component stresses.
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CE2155 Structural Mechanics and Materials
Generally, in failure criteria, the respective component
stresses are compared against the critical stress or strain (e.g.
yield stress, tensile strain capacity) of the materials.
Since it is prudent to compare the maximum of the respective
component stress or strain with the critical stress or strain,
transformation of stresses and strains would be necessary to
identify this maximum component stress or strain as well as the
plane in which this stress or strain acts by rotating the
coordinate axes.
Stress State at a Point
CE2155 Structural Mechanics and Materials
Consider the case of a structural element subjected to a
generalized stress state. The subscript of normal stress, denotes
the direction along the axis which the stress is directed. The
first subscript of shear stress, denotes the plane on which the
stress acts (plane designation x, y or z corresponds to the plane
in which the axis x, y and z acts perpendicularly to) and the
second subscript denotes the direction of stress (along the axis).
The direct and shear strains, and , associated with the normal and
shear stresses respectively are also accordingly denoted.
x, y, z = normal stressesxy, xz, yz = shear stressesx, y, z =
direct strainsxy, xz, yz = shear strainsxy = yx; zx = xz; yz = zyxy
= yx; zx = xz; yz = zy
xyplane (perpendicular to the x-axis) on which the stress
acts
direction of stress
Stress State at a Point
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CE2155 Structural Mechanics and Materials
x, y, z = normal stressesxy, xz, yz = shear stressesx, y, z =
direct strainsxy, xz, yz = shear strainsxy = yx; zx = xz; yz = zyxy
= yx; zx = xz; yz = zy
Sign convention:For brevity, the front side of the element is
defined as one where the normal to the face is pointing in the
positive direction. Accordingly, the rear side of the element is
one where the normal to the face is pointing in the negative
direction. Stresses on the front side of the element are positive
if they act in the positive direction of the axes. Stresses on the
rear side of the element are positive if they act in the negative
direction of axes. In other words,
Stress(+ve) if surface (+ve) & direction (+ve) (tension)if
surface (-ve) & direction (-ve) (tension)
Stress (-ve) if surface (+ve) & direction (-ve)
(compression)if surface (-ve) & direction (+ve)
(compression)
Stress State at a Point
CE2155 Structural Mechanics and Materials
When an axial load, P is applied to a homogeneous, slender bar
of cross-sectional area A along its axis x, the resulting stress
and strain would satisfy Hookes law. The axial stress and strain
can be expressed as
However, it also causes lateral transverse strains along y and z
axes. The ratio of lateral strain over axial strain is called
Poissons ratio and is denoted by . Hence,
Ex
x =
Ex
zy ==
strainaxialstrainlateral=
AP
x =
Uni-axial
Poissons ratio
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CE2155 Structural Mechanics and Materials
Now extending Hookes law to the case of multi-axial loading, the
resulting strain conditions would be:
EEEzyx
x+=
EEEzxy
y+=
EEExyz
z+=
Ex
x=
Ey
y
=
Ez
z=
Ey
zx
==E
xzy
==
Ez
yx==
The relations are referred to as generalized Hookes law for
multi-axial loading.
Generalized Hookes law for multi-axial loading
CE2155 Structural Mechanics and Materials
A concrete block of dimension 1500mm by 1200mm by 800mm and
cubecompressive strength 5 MPa is dropped into the sea of depth
1000m. When the concrete block come to rest at the seabed, what
would be the change in volume of the concrete block? Would the
concrete block crush? (Given for the concrete, E = 4000 MPa, = 0.2
and assume for seawater, = 1000 kg/m3).Solution:
At 1000m depth, x = y = z = p = gh (hydrostatic)= 1000 x 9.81 x
1000= 9.81 MPa
From generalized Hookes law,
Substituting, we obtain x = y = z = ( 9.81/4000) x (1 2 x 0.2)=
0.0015
)21( ===Ep
zyx
Example Problem
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CE2155 Structural Mechanics and Materials
Reduction in volume = l x b x h l (1 + x ) x b (1 + y ) x h (1 +
z )= (1500 x 1200 x 800) 1500(1 0.0015) x
1200(1 0.0015) x 800(1 0.0015)
= 6470 cm3
Cube compressive strength = 5 MPa
Applied hydrostatic pressure = 9.81 MPa
Will the concrete block crush? No
Failure can only occur when surface dislocate either in shear
ortension. Compression cannot cause dislocation directly but can
cause dislocation in other planes in shear or tension.
Example Problem (Contd)
CE2155 Structural Mechanics and Materials
Example of stresses acting on stressed element
1. A link plate subjected to axial load will induce normal and
shear stresses. In the plane perpendicular to the direction of
axial load, a normal stress which is also the maximum stress, will
be present. By transformation it can be found that the maximum
shear stress acts in a plane inclined 45 to the direction of axial
load. The axial load may be tensile or compressive.
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CE2155 Structural Mechanics and Materials
Link plate
pin
Link plates with pin connections used as tension members.
Example of stresses acting on stressed element
CE2155 Structural Mechanics and Materials
2. An axial compressive load would produce a maximum shear
stress along a plane inclined at 45o to the plane of the applied
load. A material, such as wood, which is weaker in shear than
tension or compression, would fail in a plane 45 to the plane of
the applied load.
Example of stresses acting on stressed element
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CE2155 Structural Mechanics and Materials
However, by transformation it can be shown that the maximum
normal stress exist in a plane orientated 45 to pure shear stress (
= 0) plane and also equal to . For brittle material which fails in
tension, failure would result in a plane perpendicular to the
direction of maximum normal stress.
3. When a bar is subjected to only torsion (T), the element abcd
would be in a state of pure shear.
Example of stresses acting on stressed element
CE2155 Structural Mechanics and Materials
4. When a beam is subjected to bending, the following normal and
shear stresses are induced.
The two equations provide only the normal and shear stress in
the longitudinal direction of the beam. Sometimes, when the beam is
non-homogeneous or non-isotropic, such as timber beams, the
stresses obtained from the two equations would have to be
transformed to find the critical stresses that acts on the weaker
planes in the beam.
MyI
= VQIb =compression
tension
Example of stresses acting on stressed element
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CE2155 Structural Mechanics and Materials
At point A, for the element orientated parallel to the beam,
only normal compressive stress, which is also the maximum, would
exist. By transformation, the maximum shear stress is found in the
direction 45 to the beam.At point B, for the element orientated
parallel to the beam, both normal compressive and shear stress,
would exist. By transformation, it is found that the maximum normal
compressive and tensile stresses are orientated in a direction less
than 45 to the beam. The maximum shear stress is found in a
direction less than 45 to the beam.
compression
tension
VQIb
=MyI
=
CE2155 Structural Mechanics and Materials
compression
tension
VQIb
=MyI
=
At point C (which lies on the neutral axis), for the element
orientated parallel to the beam, only shear stress, would
exist.
By transformation, it can be found that the maximum normal
compressive and tensile stresses are orientated in the direction 45
to the beam.
The maximum shear stress is found in the direction parallel and
perpendicular to the beam.
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CE2155 Structural Mechanics and Materials
compression
tension
VQIb
=MyI
=
At point D, for the element orientated parallel to the beam,
both normal tensile and shear stress, would exist. The maximum
normal compressive and tensile stresses are found by transformation
to be orientated in a direction less than 45 to the beam. The
maximum shear stress is found in a direction less than 45 to the
beam.
At point E, for the element orientated parallel to the beam,
only normal tensile stress which is also the maximum, would exist.
By transformation, the maximum shear stress is found in a direction
45 to the beam.
CE2155 Structural Mechanics and Materials
Earlier, the maximum and minimum normal stress and its direction
were demonstrated at five points along a cross section of a beam.
If this were extended to a larger number of sections and a larger
number of points in each section, it would be possible to draw two
orthogonal systems of curves on the side of the beam.
As shown above, the two systems of orthogonal curves (Stress
Trajectories) represents the directions of maximum normal
compressive and tensile stresses. Solid lines show the tensile
stresses, and dashed lines show the compressive stresses.
Maximum Stresses in Beams
Stress Trajectories for cantilever and simply supported
rectangular beams
stress at point C
C
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CE2155 Structural Mechanics and Materials
Stress Trajectories for cantilever and simply supported
rectangular beams
stress at point C
C
The curves for maximum tensile and compressive stresses always
intersect at right angles, and every trajectory crosses the
longitudinal (centroidal) axis at 45o (example see point C).
At top and bottom surfaces of the beam, where the shear stress
is zero, the trajectories are either horizontal or vertical.
Location where the trajectories are predominantly concentrated and
is in the same direction indicate susceptibility to failure.
Maximum Stresses in Beams
CE2155 Structural Mechanics and Materials
Plane Stress Problems
In a plane (2D) problem, two conditions can be imposed.1) The
out-of-plane components of the stresses, that is the stresses
acting in the direction perpendicular to the plane in
consideration, is zero. Problems subjected to this condition is
known as PLANE STRESS PROBLEMS.i.e. for a xy-plane stress problem,
z = xz = yz = 0Examples of plane stress problems include thin plate
loaded by forces parallel to plane of plate only, pressure vessels,
thin shell structures, membrane structures.
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CE2155 Structural Mechanics and Materials
Membrane
Compression strut
Tension tie
Example of membrane structures which can be analyzed as a plane
stress problem.
Plane Stress Problems
CE2155 Structural Mechanics and Materials
Plane Strain Problems
2) The out-of-plane components of the strains, that is the
strains in the direction perpendicular to the plane in
consideration, is zero. Problems subjected to this condition is
known as PLANE STRAIN PROBLEMS.i.e. for a xy-plane strain problem,
z = xz = yz = 0Examples of plane strain problems includes dams,
tunnels and retaining walls. As conditions can be assumed to be the
same at all cross sections for these structures, it is only
required to consider a slice between two sections, a unit distance
apart with two fixed supports at the ends. Note that strain between
the fixed support would be zero.
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CE2155 Structural Mechanics and Materials
Hookes Law for Plane Stress
Consider an element of material in plane stress (z = xz = yz =
0) subjected to biaxial stress in the x and y direction.
1 ( )x x yE
= 1 ( )y y xE
= ( )z x yE = +
Due to the effect of Poissons ratio, the strain will be present
in all the three directions. The strains can be obtained by
substituting z = 0 into the generalized Hookes law equations to
obtain:
CE2155 Structural Mechanics and Materials
Now extending Hookes law to the case of multi-axial loading, the
resulting strain conditions would be:
EEEzyx
x+=
EEEzxy
y+=
EEExyz
z+=
Ex
x=
Ey
y
=
Ez
z=
Ey
zx
==E
xzy
==
Ez
yx==
The relations are referred to as generalized Hookes law for
multi-axial loading.
Generalized Hookes law for multi-axial loading
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CE2155 Structural Mechanics and Materials
2 ( )(1 )x x y
E = + 2 ( )(1 )y y xE = +
Now, by rearranging the strain equations obtained earlier, 1 (
)x x yE
= 1 ( )y y xE
= The stresses in a plane stress problem can be expressed in
terms of strains as:
Similarly, by introducing the conditions of uniaxial loading
xx
E = xy z
E = =
x xE =0y z = =in the generalized Hookes law equations, the
strains in a uniaxialloading condition can be obtained as
Note that only two strain components (x and y) are sufficient to
express the stresses in a plane stress problem. Knowledge of the
out-of-plane strain (z) is not required although it is not
zero.
Hookes Law for Plane Stress
CE2155 Structural Mechanics and Materials
0x y z = = =0x y z = = =
xyxy
G =
Where G is the shear modulus. The stress and strain in a pure
shear loading condition is given by
xy xyG =
2(1 )EG = +
Beside direct strains () induced by normal stress, elements can
also be subjected to shear strains, induced by shear stresses. The
deformation due to shear strain is illustrated below. Applying
Hookes law, the stress and strain relationship for shear can be
expressed as
Note that, so far, three material parameters, the Youngs
modulus, E, the Poissons ratio, and shear modulus, G have been
introduced. However, only two of these parameters, E and are
independent as G can be deduced from them:
Hookes Law for Plane Stress
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CE2155 Structural Mechanics and Materials
Consider the point O being subjected to a state of plane stress.
The stress components acting at the point O with respect to the
xy-coordinate axes can be represented by the set of component
stresses acting on anelement as shown in the figure. The normal
stresses are defined by the stress components, x and y; and the
shear stress by xy and yx. To satisfy rotational equilibrium, xy =
yxPositive normal stress indicate tension and negative normal
stress indicate compression.
Transformation of Plane Stress
CE2155 Structural Mechanics and Materials
However, the same state of plane stress at point O can be
represented by different set of component stresses. To illustrate,
let us rotate the coordinate axes counter-clockwise through an
angle and the new coordinate axes named as x1, y1 and z1, with z1
axis coinciding with z axis. The same state of plane stress at
point O can now be represented by the stress components, x1 and y1,
and the shear stress by x1y1 and y1x1.Next we look at how the
stress components x1, y1 and x1y1 associated with the element after
it has been rotated through an angle , can be expressed in terms of
x, y and xy.
Transformation of Plane Stress
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CE2155 Structural Mechanics and Materials
In order to determine the normal stress x1 and the shearing
stress x1y1exerted on the face perpendicular to the x1 axis, we
consider a prismatic element with faces respectively perpendicular
to x, y and x1 axes. This will allow the horizontal and vertical
components of the stresses, x ,yand xy to be expressed as a
function of the normal stress x1 and the shearing stress x1y1.
Transformation of Plane Stress
CE2155 Structural Mechanics and Materials
If the area of the vertical face is denoted by A0, the areas of
the horizontal and oblique face are respectively equal to A0tan and
A0sec. The forces exerted on the three face can be given by the
stress multiplied by the respective area as shown in the
figure.
STRESS DIAGRAM FORCE DIAGRAM
Transformation of Plane Stress
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CE2155 Structural Mechanics and Materials
1 1 0 0 0
0 0
0 : sec cos sintan sin tan cos 0
x x x xy
y yx
F A A AA A
= =
1 1 1 0 0 0
0 0
0 : sec sin costan cos tan sin 0
y x y x xy
y yx
F A A AA A
= + + =
Taking the equilibrium of forces along x1and y1 axes, the
following equilibrium equations can be obtained. Note here that the
forces on the horizontal and vertical faces of the prismatic
element would have to be resolved into component forces in the x1
and y1 axes first.
Transformation of Plane Stress
CE2155 Structural Mechanics and Materials
Note that xy= yx, and after simplifying the equilibrium
equations, we obtain
( ) ( )2 2
12 2
1 1
cos sin 2 sin cos= - sin cos cos sin
x x y xy
x y x y xy
= + + +
For case when =0,1 1 1andx x x y xy = =
For case when =90o,1 1 1andx y x y xy yx = = =
Shear stress yx acts to the right, while positive stress x1y1,
after rotating 900, acts to the left
1 1 0 0 0
0 0
0 : sec cos sintan sin tan cos 0
x x x xy
y yx
F A A AA A
= =
1 1 1 0 0 0
0 0
0 : sec sin costan cos tan sin 0
y x y x xy
y yx
F A A AA A
= + + =
Transformation of Plane Stress
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CE2155 Structural Mechanics and Materials
Now, substituting the trigonometric identities
( ) ( )2 21sin cos sin 22
1 1cos 1 cos 2 sin 1 cos 22 2
== + =
into the equations
1
1 1
cos2 sin22 2
= sin2 cos22
x y x yx xy
x yx y xy
+ = + + +
The Transformation Equations for Plane Stress can be obtained as
follows:
( ) ( )2 2
12 2
1 1
cos sin 2 sin cos= - sin cos cos sin
x x y xy
x y x y xy
= + + +
Since the transformation equations were derived solely from
equilibrium of an element, they are applicable to stresses in any
kind of material, whether linear or nonlinear, elastic or
inelastic.
Transformation of Plane Stress
CE2155 Structural Mechanics and Materials
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CE2155 Structural Mechanics and Materials
The expression for the normal stress y1 is obtained by replacing
in the equation
1 cos2 sin22 2x y x y
y xy
+ = Summing expressions for x1 and y1 , we obtain
1 1x y x y + = +This equation shows that sum of normal stresses
acting on perpendicular faces of plane-stress elements is constant
and independent of angle .
1 cos2 sin22 2x y x y
x xy
+ = + +by the angle ( + 90o). This is possible because, the
rotation of the y-axis anti-clockwise by an angle would coincide
with the rotation of the x-axis by an angle ( + 90o). Since cos(2 +
180o) = cos2 and sin(2 + 180o) = sin2, normal stress y1 is given
by
Transformation of Plane Stress
CE2155 Structural Mechanics and Materials
Stresses vary continuously as the orientation of the element is
changed. The graph shows the variation of the stress components x1
and x1y1 with respect to the axes orientation(). At certain angles,
normal stress reaches a maximum or minimum value. At other angles,
it becomes zero. Similarly for shear stress. Note that the normal
stress x1 is maximum or minimum when x1y1 is zero.
Transformation of Plane Stress
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CE2155 Structural Mechanics and Materials
In the case of biaxial stress, there is no shear stress.
Substituting xy= 0 into the transformation equations,
1. Biaxial Stress
Biaxial stress occurs in many kinds of structures, including
thin-walled pressure vessels.
Transformation Equations for Special Cases of Plane Stress
1
1 1
cos2 sin22 2
= sin 2 cos22
x y x yx xy
x yx y xy
+ = + + +
1
1 1
cos 22 2
= sin 22
x y x yx
x yx y
+ = +
The transformation equations for the case of biaxial stress can
be obtained as follows:
CE2155 Structural Mechanics and Materials
2. Uniaxial Stress
3. Pure Shear
In the case of uniaxial stress, only the normal stress
component, x is not zero. By setting y and xyequal to zero in the
transformation equations, we obtain
In the case of pure shear, substituting x= 0 and y= 0 into the
transformation equation would give
( )11 1
1 cos22
sin 22
xx
xx y
= +=
1
1 1
sin2cos2
x xy
x y xy
==
Pure shear occurs in many kinds of structures, including
cylinders subject to pure torsion.
Uniaxial stress occurs in many kinds of structures, including
members in a truss structure.
Transformation Equations for Special Cases of Plane Stress
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CE2155 Structural Mechanics and Materials
SUMMARY
1 cos2 sin22 2x y x y
y xy
+ =
Where is the rotation in counter-clockwise direction. Summing
expressions for x1 and y1 , we obtain
1 1x y x y + = +This equation shows that sum of normal stresses
acting on perpendicular faces of any plane-stress elements is
constant and independent of angle .(EXAMPLE)
1 cos2 sin22 2x y x y
x xy
+ = + +The Transformation Equations for Plane Stress are as
follows:
1 1= sin2 cos22x y
x y xy
+
Transformation Equations for Plane Stress
CE2155 Structural Mechanics and Materials
Principal Stresses
The transformation equations for plane stress show that normal
stresses x1 and shear stresses x1y1 vary continuously as the axes
are rotated through angle .
The maximum normal stress is known as the principal stress, 1
and the plane on which it acts is the principal plane. The stress
orthogonal with the principal stress, 1 is the minimum normal
stress, also known as principal stress, 2 and the shear stress
acting on all the four principal planes is zero.
Most failures of structures are often associated with the
maximum tensile or compressive stresses, and thus their magnitudes
and orientations should be determined.
x
x1
yy1
2 = p
x = max = 1
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CE2155 Structural Mechanics and Materials
x
x1
yy 1
2 = p
x = max = 1
By taking the derivative of x1 of the transformation equation
with respect to and setting it to zero, we obtain an equation for
which we can find the values of x1 at which it is a maximum or
minimum. The equation for the derivative is
( )1 sin2 2 cos2x x y xydd = +and setting this equation to zero,
we obtain
2tan2 xyp
x y
= The orientation of the principal stresses can therefore be
obtain from the above equation. Subscript p indicates that the
angle p defines the orientation of principal stresses. The angle p
is known as the principal angle. Substituting this angle into the
transformation equations, the principal stresses can be
obtained.
Principal Stresses
CE2155 Structural Mechanics and Materials
General Formulae for Principal Stresses
Now, consider the right-angled triangle, which is constructed
from the equation
22
2x y
xyR = +
cos2 sin22x y xy
p pR R
= =From the triangle, we obtain
2tan2 xyp
x y
=
By Pythagorean theorem, the hypotenuse R is given by
&
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CE2155 Structural Mechanics and Materials
22
2x y
xyR = + cos2 sin22x y xy
p pR R
= =Next, substituting the relationships,
22
1 2 2x y x y
xy + = + +
into the transformation equations
1 cos2 sin22 2x y x y
x xy
+ = + +
1 cos2 sin22 2x y x y
y xy
+ = The following equations for the principal stresses can be
obtained, in which 1 > 2.
22
2 2 2x y x y
xy + = +
The equations for the principal stresses, 1 and 2 can be
combined into one as:2
21,2 2 2
x y x yxy
+ = +
and
General Formulae for Principal Stresses
CE2155 Structural Mechanics and Materials
The principal angle, which is the angle p1 corresponding to the
principal stress 1 can be obtained from the equations
1 1cos2 sin22x y xy
p pR R
= =
Only one angle exists between 0 and 360o that satisfies both of
these equations. For example, if both cos2p1 and sin2p1 are
positive, angle 2p1 can only be between 0 to 90o. Otherwise, if
both cos2p1 and sin2p1are negative, angle 2p1 can only be between
180 to 270o.
x
x1
yy1
2 = p
x = max = 1
Thus, value of p1 can be determined uniquely. Angle p2
corresponding to 2, defines a plane that is perpendicular to the
plane defined by p1. Thus, p2 can be taken as 90o larger or
90osmaller than p1.
General Formulae for Principal Stresses
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CE2155 Structural Mechanics and Materials
Uniaxial stress Biaxial stress
Pure shear
The principal planes for elements in uniaxial stress and biaxial
stress are the x and y planes themselves, because xy = 0 and hence
tan2p= 0, and the two values of p are 0 and 90o. In another word,
when shear stress is zero, the normal stresses are the principal
stresses.
For element in pure shear, the principal planes are orientated
at 45o to the x axis, obtained from the condition x = y = 0 and
hence cos2p = 0 and sin2p = 1.
Principal Stresses for Special Cases of Plane Stress
CE2155 Structural Mechanics and Materials
Maximum Shear Stresses
The shear stresses x1y1 acting on inclined planes are given by
the transformation equation,
( )1 1 cos2 2 sin2x y x y xydd =
tan22x y
sxy
= and setting it to zero, we obtain
1tan2 cot 2tan2s pp
= = This results in the relationship o45s p =
The planes of maximum shear stress occur at 45o to the principal
planes.
Taking derivative of the above equation with respect to 1 1=
sin2 cos22
x yx y xy
+
2tan2 xyp
x y
= determines the maximum principle stress, we deduce that
where s is the angle of the plane
of maximum shear stress, and comparing with the equation
which
we obtain:
x1y 1 = max
y 1 y
x
x1 = s
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CE2155 Structural Mechanics and Materials
Now, from the two transformation equations
Next, adding the left and right hand sides of the two equations
respectively, the following can be obtained:
the following equations can be obtained by rearranging and
squaring the transformation equations.
1
1 1
cos2 sin22 2
= sin 2 cos22
x y x yx xy
x yx y xy
+ = + + +
2
xyyx2
x1y1
2
xyyx
2yx
x1
cos2sin22
sin2cos22
2
+=
+=
+
2xy
2yx2
x1y1
2yx
x1 2
2
+
=+
+
Transformation Equations for Plane Stress
CE2155 Structural Mechanics and Materials
Note that the equation
Since at points A and B, x1y1 = 0, then
2xy
2yx2
x1y1
2yx
x1 2
2
+
=+
+
+= 0,
2 yxC
2xy
2yx
2 +
=R
and the radius at
2
2
21minmaxave+=+=
2min1x ==1max1x == at A,
at B and
x1y1
x1
represent the equation of a circle plotted in a rectangular
coordinate system with abscissa x1 and ordinate x1y1 with the
centre point at
Transformation Equations for Plane Stress
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CE2155 Structural Mechanics and Materials
At points D and E in the figure, observe that
Note also the normal stress corresponding to the condition of
maximum shear stress is
and
2
yxx1+=
Since at points D and E, x1y1 is the positive and negative of
the maximum shear respectively, therefore the maximum shear stress
would be given by
2xy
2yx
x1y1 2
+
== R
2xy
2yx
max 2
+
=
2
yxx1+== ave
x1y1
x1
Transformation Equations for Plane Stress
CE2155 Structural Mechanics and Materials
Recall that the maximum shear stress would be given by
and the corresponding normal stress acting in the plane of the
maximum shear stress would be given by
2xy
2yx
max 2
+
=
2
yxx1+== ave
If an element is subjected to principal stresses 1 and 2 whereby
the shear stress xy is zero, the maximum shear stress would then be
given by
2
21max=
and the corresponding normal stress acting in the plane of the
maximum shear stress would be given by
2
21x1+== ave
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CE2155 Structural Mechanics and Materials
SUMMARY
The direction of the plane in which the principal stresses and
the maximum shear stress acts could be obtained, respectively,
from:
2xy
2yx
max 2
+
=
2
yxavey1x1+===
The principal stresses are given by:
2
21max=
The maximum shear stress and the corresponding normal stress
acting in the plane of the maximum shear stress would be given
by
2
21avey1x1+===
tan22x y
sxy
=
22
1,2 2 2x y x y
xy + = +
2tan2 xyp
x y
=
For an element subjected to principal stresses, the maximum
shear stress and the corresponding normal stress acting in the
plane of the maximum shear stress would be given by
CE2155 Structural Mechanics and Materials
Example 1
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CE2155 Structural Mechanics and Materials
Example 1 (contd)
CE2155 Structural Mechanics and Materials
Example 1 (contd)
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CE2155 Structural Mechanics and Materials
Example 2
CE2155 Structural Mechanics and Materials
Example 2 (Contd)
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CE2155 Structural Mechanics and Materials
Writing this equation in a simpler form by using
would result inThis is the equation of a circle in standard
algebraic form. The coordinates are x1and x1y1, the radius is R,
and the center of the circle has coordinates x1= aver and
x1y1=0.
2xy
2yx2
x1y1
2yx
x1 2
2
+
=+
+
2xy
2yx
2 +
=R
2 yx
ave
+= and( ) 221y1x2ave1x R=+
Recall earlier that from the two transformation equations, the
following equation was obtained:
The transformation equations for plane stress can be represented
in graphical form by a plot known as Mohrs Circle. This graphical
representation is extremely useful for visualising the
relationships between normal and shear stresses acting on various
inclined planes at a point in a stressed body. It also provides a
means for calculating principal stresses, maximum shear stresses,
and stresses on inclined planes. Mohrs Circle is valid not only for
stresses but also for other quantities of a similar mathematical
nature, including strains and moments of inertia.
Mohrs Circle for Plane Stress
CE2155 Structural Mechanics and Materials
Construction of Mohrs Circle
Mohrs Circle can be constructed in a variety of ways, depending
upon which stresses are known and which are unknown. Assume that we
know the stresses x, y and xy acting on the x and y planes of an
element in plane stress and we wish to know the stresses x1, y1 and
x1y1 acting on the x1 and y1 planes.
Known stress state
Stress state to be determined for rotated element
For the Mohrs Circle, adopt the convention to plot clockwise
shear stress aspositive, anticlockwise shear stress as negative,
tension stress as positive, compression stress as negative and
counter-clockwise angle () as positive.
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CE2155 Structural Mechanics and Materials
Introduction
CE2155 Structural Mechanics and Materials
x, y, z = normal stressesxy, xz, yz = shear stressesx, y, z =
direct strainsxy, xz, yz = shear strainsxy = yx; zx = xz; yz = zyxy
= yx; zx = xz; yz = zy
Sign convention:For brevity, the front side of the element is
defined as one where the normal to the face is pointing in the
positive direction. Accordingly, the rear side of the element is
one where the normal to the face is pointing in the negative
direction. Stresses on the front side of the element are positive
if they act in the positive direction of the axes. Stresses on the
rear side of the element are positive if they act in the negative
direction of axes. In other words,
Stress(+ve) if surface (+ve) & direction (+ve) (tension)if
surface (-ve) & direction (-ve) (tension)
Stress (-ve) if surface (+ve) & direction (-ve)
(compression)if surface (-ve) & direction (+ve)
(compression)
Stress State at a Point
Note that after transformation in Mohrs Circle, the stresses
would have to be converted to this convention
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CE2155 Structural Mechanics and Materials
+
--
1. Draw a set of coordinate axes with x1 as abscissa (+ve to the
right), and x1y1 as ordinate (+ve upward). Note that x1 and x1y1
are the variables.
2. Locate center C of circle at point having coordinates x1 =
aveand x1y1 = 0
2 yx
ave
+=
Procedure for Construction of Mohrs Circle
CE2155 Structural Mechanics and Materials
+
-- 3. Locate point A,
representing stress condition on x face of element labelled A by
plotting its coordinates x1= x and x1y1 = -xy.
Procedure for Construction of Mohrs Circle
-
CE2155 Structural Mechanics and Materials
+
-- 4. Locate point B,
representing stress condition on y face of element labelled Bby
plotting its coordinates x1= y and x1y1 = + xy.
Procedure for Construction of Mohrs Circle
CE2155 Structural Mechanics and Materials
+
--
5. Draw line from point A to point B. This line is the diameter
of the Mohrs circle and passes through center C.
6. Using point C as center, draw the Mohrs Circle passing
through points A and B.
Procedure for Construction of Mohrs Circle
-
CE2155 Structural Mechanics and Materials
Stresses on Inclined Element (with angle from x-axis )
--
+
The stresses on faces D and Dare represented by the stresses at
the point D and D on the Mohrs circle, respectively. These points
are located by rotating the line AB about C through an angle 2 in
the same direction as the element is rotated, which in the
anticlockwise direction.
CE2155 Structural Mechanics and Materials
Point P1 and P2 represents the principal stresses as the
corresponding shear stress, xyat the principal plane is zero.
Principal angle p1 between x-axis and the axis of 1 (the
algebraically larger principal stress) is one-half the angle 2p1
(which is the angle between radii CA and CP1)
+
--
RCPOC yx +==222
RCPOC yx ++=+=211
x
x1
yy1
2 = p
x = max = 1
Principal Stresses (with angle p from x-axis )
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CE2155 Structural Mechanics and Materials
Maximum Shear Stresses
Points S1 and S2, representing the maximum negative and maximum
positive shear stresses, respectively, are located at the bottom
and top of the Mohrs Circle. These points are at angles 2 =90ofrom
points P1 and P2. That is the maximum and minimum shear stress acts
on the plane inclined 45o from the principle plane.
--
+
CE2155 Structural Mechanics and Materials
Maximum Shear Stresses
The shear stresses x1y1 acting on inclined planes are given by
the transformation equation,
( )1 1 cos2 2 sin2x y x y xydd =
tan22x y
sxy
= and setting it to zero, we obtain
1tan2 cot 2tan2s pp
= = This results in the relationship o45s p =
The planes of maximum shear stress occur at 45o to the principal
planes.
Taking derivative of the above equation with respect to 1 1=
sin2 cos22
x yx y xy
+
2tan2 xyp
x y
= determines the maximum principle stress, we deduce that
where s is the angle of the plane
of maximum shear stress, and comparing with the equation
which
we obtain:
x1y 1 = max
y 1 y
x
x1 = s
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CE2155 Structural Mechanics and Materials
Example
Example
CE2155 Structural Mechanics and Materials
Example contd
Example
-
CE2155 Structural Mechanics and Materials
Example contd
CE2155 Structural Mechanics and Materials
Example
Example
-
CE2155 Structural Mechanics and Materials
Example-contd
Example contd
CE2155 Structural Mechanics and Materials
Example-contd
Example contd
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CE2155 Structural Mechanics and Materials
Element in Triaxial Stress
Because the transformation equations of plane stress are based
on force equilibrium in the xy plane, they are independent of the
normal stress z..Therefore, we can use these equations for
determining the stresses and . Note that when considering the
stresses in the xy-plane, the stress z is known as the out-of-plane
stress while the stresses x and y are known as the in-plane
stresses.
CE2155 Structural Mechanics and Materials
In order to determine the normal stress x1 and the shearing
stress x1y1 exerted on the face perpendicular to the x1 axis, we
consider a prismatic element with faces respectively perpendicular
to x, y and x1 axes. This will allow the horizontal and vertical
components of the stresses, x ,y and xy to be expressed as a
function of the normal stress x1 and the shearing stress x1y1.
Element in Triaxial Stress
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CE2155 Structural Mechanics and Materials
If the area of the vertical face is denoted by A0, the areas of
the horizontal and oblique face are respectively equal to A0tan and
A0sec. The forces exerted on the three face can be given by the
stress multiplied by the respective area as shown in the
figure.
STRESS DIAGRAM FORCE DIAGRAM
CE2155 Structural Mechanics and Materials
1 1 0 0 0
0 0
0: sec cos sintan sin tan cos 0
x x x xy
y yx
F A A AA A
= =
1 1 1 0 0 0
0 0
0: sec sin costan cos tan sin 0
y x y x xy
y yx
F A A AA A
= + + =
Taking the equilibrium of forces along x1 and y1 axes, the
following equilibrium equations can be obtained. Note here that the
forces on the horizontal and vertical faces of the prismatic
element would have to be resolved into component forces in the x1
and y1 axes first.
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CE2155 Structural Mechanics and Materials
General State of Stress
Recall that the generalized state of stress at a point Q can be
represented by six components of stress, namely three normal
stresses, x, y and z and three shear stresses xy, xz and yz. When
the coordinate axes, x, y and z are rotated, the stress state can
be represented by another set of six stresses components x, y, z,
xy, xz and yz. However, for every stress state, there exist one
orientation of the coordinate axes where the shear stress on all
faces of the cubic element would vanish. Only normal stresses would
remain and these normal stresses are, therefore, the principal
stresses 1, 2 and 3 for the stress state at point Q. The principal
stresses may also be denoted as a, b and c corresponding to
coordinate axes a, b and c as shown.
CE2155 Structural Mechanics and Materials
Application of Mohrs Circle to Three Dimensional Analysis of
StressAs mentioned earlier the transformation equations of plane
stress in the xy-plane are independent of the out-of-plane stress,
z. Assume the point Q is subjected to a generalized (3D) stress
state and the coordinate axes a, b and c are the principal axes of
stress. Therefore, when the element is rotated about one of the
principal axes, e.g. c-axis, the corresponding transformation of
stress may be analyzed by means of Mohrs circle as if it was a
transformation of plane stress. We may therefore use the circle of
diameter AB to determine the normal and shearing stresses exerted
on the faces of the element as it is rotated about the c-axis.
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CE2155 Structural Mechanics and Materials
Similarly, circles of diameter BC and CA may be used to
determine the stresses on the element as it is rotated about the a
and b axes, respectively. It can be shown that any other
transformation of axes would lead to stresses represented by a
point located within the shaded area. Thus, the radius of the
largest of the three circles yields the maximum value of the
shearing stress at point Q. Therefore,
where max and min are the maximum and minimum value of the three
principal stress and which also represent the algebraic values of
the maximum and minimum stresses at point Q.
minmax21
max =
Application of Mohrs Circle to Three Dimensional Analysis of
Stress
CE2155 Structural Mechanics and Materials
Returning to the case of plane stress, we observe z = zx = zy =
0, and therefore the z-axis is one of the principal axes since the
shear stress, zx = zy = 0 in the x-yplane. Hence, in the Mohrs
circle, this axis would corresponds to the origin O where = = 0. We
also recall that the other two principal stresses corresponds to
another two points, A and B in Mohrs circle. If A and B are located
on opposite sides of the origin O, the corresponding principal
stresses represent the maximum and minimum normal stresses at point
Q, and the maximum shearing stress is equal to the maximum in-plane
shearing stress. The in-plane refers to a plane which is
perpendicular to the plane in consideration. The planes of maximum
shearing stress correspond to points D and E of Mohrs circle and
are 45o to the principal planes corresponding to points A and B,
shown as shaded diagonal planes in figures (a) and (b).
(a) (b)
Application of Mohrs Circle to Three Dimensional Analysis of
Stress
-
CE2155 Structural Mechanics and Materials
If, on the other hand, A and B are on the same side of the
origin O, that is, if a and b have the same sign, then the circle
defining max, min and max is not the circle corresponding to a
transformation of stress within the xy plane. If a > b > 0,
as assumed, we have max = a, min = 0, and max is equal to the
radius of the circle defined by points O and A, that is
We also note that the normals Qd and Qe to the planes of maximum
shearing stress in figures (a) and (b) respectively, are obtained
by rotating the axis Qathrough 45o within the za plane. In other
words, the maximum shear stress is in-plane to the za-plane and
therefore out-of-plane to the ab-plane that is being considered as
the plane of plane stress. Thus, the planes of maximum shearing
stress, are the shaded diagonal planes shown.
max21
max =
(a)
(b)
Application of Mohrs Circle to Three Dimensional Analysis of
Stress