Understanding Stress

1
Understanding Stress Strain
In these brief set of slides we try to get a
grasp of tensile, compressive and shear stresses
No strain, no gain!
Sorry, this will not help with worldly stresses!!!
2
What will you learn in this chapter?

• Why stress strain? Why not work with loads and
  elongations? (Which are more easy to comprehend
  as compared to stress strain).
• How do forces lead to stresses and strains?Can
  stress exist without strain and can strain exist
  without stress?
• How to physically understand stress?
• Understanding stress and strain tensors in terms
  of their components.
• Hydrostatic and deviatoric components of stress
  and strain.
• Planes of maximum shear stress (and their role in
  plasticity).
• Understanding surface stress.
• Residual stress. Microstructural origins of
  residual stress.

Please revise common types of loading and
deformation induced by them first (click here)
3
Why Stress?

• Consider the experiment below in which 3 rods of
  the same material are extended by a force P.

Why Strain?
Case-2
Same area
Increase L
Case-1
Increase A
Case-3
Same length

• Note that increasing the length (L2 gt L1)
  increases the elongation (?2 gt ?1), while
  increasing the area (A2 gt A1) decreases the
  elongation (?3 gt ?1). The load-elongation (P-?)
  curve for each of these bodies will be different.
  Clearly load is not a good parameter to
  characterize what the material experiences.
  This load is borne by a larger cross-sectional
  area in case-3 which decreases the elongation.
• If we define stress (s or ?) as load/area (P/A),
  then larger the load bearing area, lower will be
  the stress in the material. Stress in case-1 2
  are the same. Stress in case-3 is lower and hence
  the elongation is smaller (as compared to
  case-1). In this 1D example- i.e. elongation
  is along 1D.

Continued
4

• In case-2 the elongation is more (?2 gt ?1). As
  there is more material along the force.This
  implies that elongation is not a good measure of
  how the elongation feels to the material. If
  we define strain (e or ?) as change in length by
  original length (Lfinal ? Linitial)/Linitial
  ?L/L0 , then we will observe that strain in
  case-1 and case-2 is the same (i.e. ?1/L1
  ?2/L2).
• If we plot the load-elongation (P-?) curves for
  the samples (cases), we will have 3 plots (for
  linear elastic materials these will be straight
  lines).
• The slope of these curves is a structure/geometry
  dependent property called stiffness.
• If on the other hand we plot stress-strain (s-e
  or ?-?) curves, we will get one master curve and
  the slope of this curve will be an inherent
  material property the Youngs (or elastic)
  modulus.
• Stress is the internal response of a material to
  externally applied loads.

• Note
• Both stress and strain are defined at each point
  in the material. In the stated example, they are
  average stress and strain.
• In general stress and strain are second order
  tensors (with 9 components in 3D).

Replot as ?-? curve
Actually this is the definition of Engineering
strain (e). But, if the strain is small e ?
(i.e. is approximately equal numerically to the
true strain).
5
Stress and Strain

• In normal life we are accustomed to loads/forces
  and displacements. These are most appropriate
  variables when one talks about point masses,
  rigid bodies or is sitting outside a body.
• Inside a body (typically a deformable body with
  mass and extent), one can locate other
  (appropriate) field variables to describe the
  state of the system. E.g. inside a gas kept in a
  cylinder, instead of tracking the velocities of
  the molecules, we come up with a field variable
  called pressure, which describes the momentum
  transferred by these molecules per unit area per
  unit time (pressure is a time averaged
  macroscopic quantity).
• To understand the above point let us consider the
  pulling of a body in tension (figure to right).
  Assume there is a weak plane (AA), the two sides
  of which can slip past one another. We note as in
  the graphic that the inclined plane shears even
  though we applied tensile forces. That is the
  plane feels shear stresses (?).
• Hence, when we apply only tensile forces to the
  body (in the simple example considered), certain
  field develops within the body, which depending
  on the orientation of the plane in the body (or a
  unit volume being considered) can undergo shear
  and/or dilatation.
• This field is the stress field and is a second
  order tensor with 9 components in general in 3D
  (4 in 2D).

A
A
?
It may so happen that some planes do not feel
any shear stresses (like the horizontal and
vertical planes in in figures)
6

• Stress in 1D is defined as Stress Force/Area.
  This implies that in 1D stress is a scalar.
  Clearly, this is valid in 1D only, where a even a
  tensor looks like a scalar!!
• Similar to the stress field (which we noted to be
  the force dependent term within the body), we
  can define a strain field, which is a
  displacement dependent term. Strain is also a
  2nd order tensor with 9 components in general in
  3D. Strain in 1D is Strain change in
  length/original length (usually for small
  strains).

Some general constraint

• In summaryExternal forces and constraints give
  rise to a stress field within a body. Depending
  on the orientation of a unit element (cube in the
  figure), the cube may stretch along one or two
  directions and/or may shear.

Unit cube
We do not apply stresses
Stress develop inside the body
We apply forces
Some general loading
Variables in solid mechanics
Stress
Force
Variables in rigid Body Mechanics
Related variables inside the material
Strain
Displacement
7
Funda Check
What can happen to a unit volume inside a body on
the application of external loads/forces/constrain
ts?
Contraction/dilation
Volume change
What can happen to a unit volume in a material
when we apply forces/constraints to the outside
of the body
Or a combination of these
Shear
Shape change
Rigid body translation/rotation
Position/Orientation change
As such no distortion to the volume element (but
affects neighbouring elements)
8
Funda Check
Does stress cause strain or does strain cause
stress? Can we have stress-free strain and
strain-free stress? How can strain and stress
arise in a material?

• First point stresses can exist without strains
  (heating a body between rigid walls) and strains
  can exist without stresses (heating a
  unconstrained/free-standing body).
• What we are asking here is which came first
  (something like the proverbial chicken and egg
  problem!).
• Both situations are possible (at least from a
  perspective of easy understanding).
• If we load a body and this leads to stress inside
  the body? this will lead to strains in a
  deformable body. I.e. stress gives rise to
  strain. Load ? Stress ? Strain.
• Now if a cubic phase transforms to another cubic
  phase with a larger lattice parameter (i.e. the
  transformation involves volume expansion), we can
  assume two situations1) the transforming
  material is small and the whole volume transforms
  (Fig.1a)2) the transforming volume is small, but
  now embedded in a matrix (Fig.1b).
• In case (1) above there are no stresses. In case
  (2) above the surrounding matrix will try to
  constrain the expansion, leading to stresses. The
  primary causative agent in case (2) is strains
  (due to phase transformation), which further
  causes stresses. Phase transformation? Strain ?
  Stress.

• It is important to note that in most cases it is
  strain which is measured experimentally and
  converted to stress via stress-strain relations
  involving material properties.

Fig.1b Strains with stresses
Fig.1a Strains but no stresses(dilatation
during phase transformation)
9
Funda Check
How can strain and stress arise in a material?

• Some of the origins of stress and strain we have
  already seen in the previous slide. We consider
  here a few more.
• Basically, stress and strain can arise because
  of (i) external loads/constraints/effects, (ii)
  internal loads/constrains/effects (iii) other
  stimuli (via cross-coupling coefficients).
• Among external factors, heating (leading to an
  increased T) is an important one. Phase
  transformation, which in itself can be caused by
  T, P, etc., is an important internal
  effect.
• Other stimuli can also lead to stress/strain.
  Paramagnetic to ferromagnetic ordering (say on
  cooling), can lead to strain in materials with a
  strong spin-lattice coupling. The strain induced
  by magnetization is called magnetostriction and
  arises due to cross-coupling between magnetic
  and strain parameters.
• In inverse piezo-electric effect, application of
  an electric fields leads to strain/stress.

External loads/constrains/effects
Basic origins of stress and strain
Internal loads/constrains/effects
Other stimuli
Magnetic fields, electric fields,...
Via cross-coupling coefficients
10
Scalar, Vector and Tensor Quantities

• To describe a property at a point inside the
  material we may require to specify just ? a
  number (magnitude of that property like
  temperature or density), ? a magnitude and
  direction (i.e. 3 numbers in 3D like for electric
  field or pyroelectric coefficient) or ? even more
  numbers. The number of values required is given
  by 3n in 3D (2n in 2D) where n is called the
  rank of the tensor.
• To understand why tensors are required let us
  consider a force F (with a ?F) applied along
  the x-axis and ask the question what will
  happen to the body?.
• Clearly the information available is insufficient
  to answer unequivocally. If the forces are
  applied as in Fig.1b the body will elongate,
  while if applied as in Fig.1c the body will
  shear.
• Hence, we need to specify the plane on which the
  force is acting. In case in Fig.1b the F force
  is acting on the x-plane along the x direction?
  this is written as Fxx. In Fig.1c the force F is
  acting along the y-plane along the x direction?
  written as Fyx.
• This implies (in this example of mechanical
  deformation) two directions are required to
  specify the force (and hence determine its
  effect) (i) the direction of the plane normal
  and (ii) the direction of the force.
• A combination of these forces Fxx Fxy could
  also be acting on the body.

Fyx
F
Note. For tension/compression F on x face is
positive and similarly, ?F force on ?x plane is
also positive. For shear shear causing clockwise
rotation is positive.
Fig.1c
F
?F
Fig.1a
Fig.1b
F
?F
Fxx
?F
For now we will assume that other forces are
present to give us force and moment balance (i.e.
equilibrium condition). ? Also note that force
(F) is a vector and we are trying to understand
its effect as a prelude to tensors.
11
Q A
Give examples of axial and polar vectors.

• Polar vectors Force, Electric field,
  Polarization.
• Axial vectors Mechanical moment, Angular
  momentum, Curl of a polar vector.

Some basic points...

• A scalar does not require a coordinate system to
  define and hence is independent of the coordinate
  axes chosen. We require just one number at each
  point for a scalar.
• A vector can be represented by its components
  along a set of coordinate axes. We require three
  numbers in 3D at each point to specify a vector.
  A change in the coordinate axes system (change
  in the angle between the basis vectors,
  translation/rotation of the basis vectors) will
  change the components along the axes but, will
  the vector itself will remain unchanged.
• The transformation of a vector (rotation,
  inversion, mirror, etc.) can be carried out using
  a transformation matrix.
• A second rank tensor can also can be represented
  by its components along a set of coordinate axes.
  It can be visualized as a combination of two
  directions. We require nine numbers in 3D at
  each point to specify a second rank tensor.A
  change in the coordinate axes system will change
  the components along the axes but, will the
  tensor itself will remain unchanged.
• The transformation of a second rank tensor can be
  carried out using two transformation matrices.

We will repeat some of these concepts soon.
12
Tensors of various ranks

• Tensors can be used to describe (i) fields
  (field tensors) or (ii) properties (property
  tensors).
• These tensors can belong to various ranks zeroth
  rank, first rank, second rank, etc.
• E.g. Temperature field is a scalar field, where
  each point in space is described by one number
  the T at that point (T(x,y,z)). Scalar fields
  are tensor fields of rank-0. On the other hand
  some fields require more numbers to be specified
  at each point in space.
• Electric field (polar vector) and Magnetic field
  (axial vector) require three numbers (in 3D) to
  be specified at each point. These 3 numbers are
  the components along the coordinate axes and give
  the direction and magnitude of the vector. Such a
  field is a vector field. Vectors are tensors of
  rank-1.
• Some other fields require more numbers to be
  specified. E.g. to describe the state of stress
  at a point, we need 9 numbers (in 3D) in general
  (stress being a symmetric tensor we actually need
  only 6 numbers). Stress is a tensor of rank-2.

Q A
Why do we need coordinate transformations?

• We have already seen (Matrix representation of
  symmetry operators) that symmetry operators can
  be written as transformation matrices which
  involve operations like rotation (about a
  crystallographic axis), inversion (about an
  inversion centre? usually the centre of the unit
  cell) and mirror (about a plane? usually passing
  through lattice points are exactly between
  lattice points).

Note only one number needed at each point. If
there is a to point to point variation of
temperature a set of T have to be specified for
each (x,y,z)? which gives rise to the temperature
field.
13
Funda Check
What is the order of a tensor? How can we
understand force and stress from this perspective?

• The order of a tensor basically tells us the
  number of directions involved in describing the
  quantity.
• Two types of tensors may be distinguished ?
  field tensors (like stress) and ? property
  tensors (like electrical conductivity).
• The direction may be visualized as a vector. The
  direction itself may be prescribed under
  Cartesian, polar or other coordinate systems. The
  number of basis vectors required to specify the
  direction will depend on the dimension the
  direction lives in. In 2D we need two basis
  vectors and in 3D three basis vectors.
• A scalar (zeroth order tensor) has no directions
  involved. E.g. density (?).
• A vector (first order tensor) requires one
  direction to be specified. E.g. electric field
  vector (E), magnetic field vector (H).
• A second order tensor requires 2 directions to be
  specified. E.g. stress (?ij), strain (?ij),
  thermal expansion coefficient (?ij) .
• Taking this forward, a nth order tensor requires
  n directions for its specification. Examples of
  higher order tensors include piezoelectric
  coefficient (dijk) and elastic constant (Eijkl).

Order of the Tensor Field Tensor Property Tensor Property Tensor
Equation
0 Energy (E) Density (?) -
1 Electric Field (Ei) Pyroelectric Coefficient (pi) Pi pi ?T
2 Stress Coefficient of thermal expansion (?ij) ?ij aij?T
3 - Piezoelectric Coefficient (dijk) Pi dijk?jk
14
Examples of some vector and tensor quantities
Let us start by considering some important
quantities
Quantity Type Acts
Force (Fi) (Polar) vector At a point mass
Torque Pseudo Vector (Axial Vector) About an axis
Traction (Ti) (Polar) Vector On a surface element
Stress (?ij) Second order Tensor Acts on a volume element

• Traction vector is the internal force vector on
  a cross-section divided by the cross-sections
  area. Traction has units of stress (e.g. MPa) but
  is a vector and not the stress tensor. In a
  continuous body the tractions on the opposite
  internal surface cancel each other.
• Traction and stress may vary with position,
  orientation and time i.e., are field quantities
  with spatial and temporal variations (next
  slide).
• Polar vectors reflect in a mirror, axial vectors
  do not reflect.

With the exception of surface stress which acts
only on the surface. External force is also
called surface tractions.

• The scalar component of the normal stress is
  given by

15
Funda Check
What is the difference between Traction and
Stress?

• Sometimes Traction is also called as the stress
  vector (thus adding to the confusion!).
• Traction is a vector (1st order tensor), while
  stress is a tensor (2nd order tensor).
• Traction is a measure of the intensity of the
  force and is defined as the force per unit area
  (on a cut plane inside a body). Traction can act
  in any direction, i.e. need not be parallel to
  the normal.
• On opposite surfaces created by the cut plane (at
  a point), the traction vector is equal in
  magnitude (but opposite in direction). I.e. at P
  and P (in the Fig.1) the traction vectors are T
  ?T.
• The components of the Traction vector when
  divided by the unit area gives us the components
  of the stress tensor (In Fig.2 a special case is
  considered for simplicity, wherein the area
  normal is along z and ?A is the unit area
  (shaded yellow)).
• When the cutting procedure is carried out along
  the 3 orthogonal planes and we compute the
  stresses acting, we get the 9 components of the
  stress tensor.

Note the traction vector has been moved to the
corner of the area (actually it should be at P)
Fig.1
Fig.2
Infinitesimal area (i.e. in the limit ?A?0)
16
Stress

• Stress is a second order tensor and best
  understood in terms of its effect on a unit body
  (cube in 3D and square in 2D), in terms of its
  components.
• Stresses can be Compressive, Tensile or Shear (in
  terms of specific components).
• We may apply forces/constraints and stresses will
  develop within the material (including the
  surface) ? we apply forces (or constraints) and
  not stresses.
• The source of stress could be an external agent
  (forces etc.) or could be internal (dislocations,
  coherent precipitates etc.) ? i.e. stresses can
  exist in a body in the absence of external agents
  (residual stress).
• The effect of stress at a particular point in the
  material is not dependent on how the stress came
  about (i.e. could be external or internal
  factors) ? just the components of stress matter
  in determining the response of the material.
• We can have stress without strain and strain
  without stress (ideal circumstances)? Strain
  without stress ? heat a unconstrained body (it
  will expand and no stresses will develop)?
  Stress without Strain ? heat a body constrained
  between rigid walls (it will not be able to
  expand but stresses will develop).
• In 3D if two of the three principal stresses are
  equal it is called cylindrical state of stress
  (?1 ?2 ? ?3) and if all the principal stresses
  are equal, it is referred to hydrostatic state of
  stress (?1 ?2 ?3).
• The cause behind the strains can be ? stress ?
  electric field ? temperature change, etc.

Will learn about this soon.
17
We can apply forces and not stresses- stresses
develop within the body
Note
Shear
E.g.
Only shear tends to change the shape of a body
without changing its volume
Actually this kind of simple shear is disallowed
to due lack of moment balance

In anisotropic crystals it may do more (may even
shear the crystal)!
Anisotropic crystals
Note we apply shear force and shear stresses
develop in the interior of the material
18
Funda Check
How do I understand the sign of stress (if
compressive or tensile)?

• We will try understand this in 4 pages Pages
  1-4.
• In tensile stress material points want to come
  towards each other as they have been stretched
  from their equilibrium positions.
• In compressive stress the reverse is true?
  material points want to go away from one another
  (as they have been compressed as compared to
  their equilibrium positions).

Physical Understanding of Stress
Any of these may be used depending on the
situation
Method A
Method B
Effect on points, lines, surfaces and volumes in
the body
Effect on release of constraint
This visualization may or may not be easy in many
situations
Page-1
19
Method A
Tensile Stress
Compressive Stress
Forces on the external surface of a body
Uniaxial compressive stress tends to reduce the
length of the body (shorten the body)
Uniaxial tensile stress tends to elongate the body
Page-2
20
Method B
Let us get a physical feel for TENSILE STRESS
2
1
Pull a body of length L0 to new length L1 and
hold it at this length
3
4
Introduce a cut (crack) in the body
The crack will open up due to the tensile stress
That is when the constraint is removed points in
the body move towards each other I.e. under
tensile stress the points in a body tend to move
towards one another (while the crack faces move
apart) This is because we have increased the
interatomic distance over the equilibrium value.
!!
Page-3
21
Alternately if the external constraint is removed
points in the body move towards each other I.e.
under tensile stress the points in the body tend
to move towards one another
The reverse will happen under compressive
stress That is when the constraint is removed
points in the body move away from each other I.e.
under compressive stress the points in the body
tend to move away one another
Page-4
22
Funda Check
How can tensile and shear stresses arise inside a
material?

• Stresses of a particular type can arise inside a
  material by (Case-1) Shear or tensile loading,
  (Case-2) geometry of loading, (Case-3)
  orientation of planes within the material.

(Case-1) Shear loading leading to shear stresses
(note focus on the sponge- else it will look
like case-2)
Shear stress on the bolt here
(Case-3) Loading is tensile but inclined planes
feel shear.
(Case-2) Geometry of loading leading to shear
stresses (Loads applied are purely tensile).
23
The Stress ( Strain) Tensors

• Tensors which measure crystal properties (e.g.
  magnetic susceptibility) have a definite
  orientation within a crystal and its components
  are dictated by the crystal symmetry. These are
  Material Property Tensors or Material Property
  Tensors.
• The stress and strain tensors can have any
  orientation within a crystal and can even be
  defined for amorphous (or isotropic) materials.
• The stress tensor develops the material in
  response to forces.
• The stress and strain tensors are Field Tensors.
  On the other hand, the elastic constant is a
  Material Property Tensor (4th order, Eijkl).
• (Say) when forces are applied to a body, stress
  and strain tensor fields develop within the body.
• In a isotropic materials the direction of
  principal stresses coincides with the directions
  of principal strains.

The order of a tensor basically tells us the
number of directions involved in describing the
quantity. A scalar (zeroth order tensor) has no
directions involved. A vector (first order
tensor) requires one direction to be specified. A
second order tensor like stress, requires 2
directions to be specified. Taking this forward,
a nth order tensor requires n directions for
its specification.
24
Understanding stress in terms of its components

• Stress is a Second Order Tensor. (It is a
  symmetric tensor ?ij ?ji in usual materials).
• It is easier to understand stress in terms of its
  components and the effect of the components in
  causing deformations to a unit body within the
  material.
• These components can be treated as vectors.
• Components of a stress2D (plane stress) ? 4
  components 2 ? (tensile) and 2 ? (shear) 3D ?
  9 components 3 ? (tensile) and 6 ? (shear)
• ? written with subscripts not equal implies ?
  (shear stress). E.g. ?xy ? ?xy.
• First index refers to the plane and the second to
  the direction.
• Close to 2D state of stress (plane stress) can
  occur in very thin bodies and 2D state of strain
  (plane strain) very thick bodies. In plane stress
  components of stress with z coordinates are
  zero.
• Shear stresses are responsible for plastic
  deformation in metallic materials (by slip).

x-plane, x-direction Also written as ?x
Direction
x-plane, y-direction
As stress is a symmetric tensor in normal
materials
Plane
?
?
25

• Let us consider a body in the presence of
  external agents (constraints and forces), which
  causes stresses in the body.
• Stresses are defined at a point and may vary from
  point to point but, we consider a sample
  rectangle in 2D or a sample cube in 3D. This
  representation helps in visualizing the effect of
  stress on unit element. The sample region
  (rectangle or cube) is shown to be of finite
  extent for illustration purposes, but should be
  of infinitesimal extent.
• A unit region in the body (assumed having
  constant stresses) is analyzed. (body forces are
  ignored)

2D
Note the directions of the stresses shown are
arbitrary (the stresses in general could be
compression/tension and shear could be opposite
in sign)
Note ?xx ?x
26
Understanding how stress develops inside a
material based on the load applied
2D

• The normal stresses (?x ?y) tend to elongate
  the body (the square in the figure below) ? this
  will give rise to volume changes.
• The shear stress (?xy ?yx) will tend to change
  the shape of the body ? without changing its
  volume.
• Depending on the orientation of the unit volume
  considered, the stresses acting on its faces will
  change.
• A good feel for the same can be got by looking at
  stress in 2D (plane stress, with 3 independent
  components).
• We have already noted that even if we apply
  tensile/compressive forces, shear stresses can
  develop on inclined planes.
• Stress on one axes set (x, y) can be mapped to
  stress on another axes (x, y) set by the
  formulae as below.
• There will always be one unique axis set (x,
  y), wherein the shear stresses are zero. The
  corresponding planes are the principal planes and
  the principal normal stresses are labeled ?1 and
  ?2 (More about this soon).

27

• Points to be noted (some of these will be
  illustrated via figures in coming slides)
• Planes of maximum/minimum normal stress (?)
  correspond to zero shear stress (?xy 0) ? known
  as the principal planes. The corresponding
  stresses are the principal stresses (labeled ?1
  and ?2).
• There exist planes where shear stress is zero.
  These planes also correspond to extremum in
  normal stresses. Planes of extremum shear stress
  are 45? from planes of zero shear stress (which
  correspond to the principal planes).
• The period of the functions is 180? (as above
  equations are functions of Sine and Cosine of 2?)
  ? the maxima of the functions is separated from
  the minima by 90?. This is expected e.g. the
  stress in x (?xx) is expected to be same as
  stress in x (?xx).
• Extremum in shear stress occurs midway in angle
  between extrema in normal stress.
• Shear stress is symmetric, i.e. ?xy ?yx.
  Minimum value of shear stress (Maximum value
  of shear stress).

Principal stresses
Maximum shear stress
28
Now we will consider special cases of importance
Case-1

• The simplest case can be loading in uniaxial
  tension.
• For x and y as in the figure below only the
  vertical and horizontal planes feel no shear
  stress (every other plane feels shear stress).
  This is in spite of the fact that we applied only
  a tensile force.
• Shear stress is maximum at 45?. For ?xx 100MPa,
  ?max 50 MPa.
• Rotation of 90? implies that x goes to y and y
  goes to x (which is same as x).
• The principal stress is the resultant of what we
  applied ? Px (i.e. ?1 100 MPa).

100 MPa
The above stress state can be thought of arising
from a loading as below
Normal stresses reaches extremum when shear
stress is zero
Note that every inclined plane feels shear stress
29
Case-2

• If we push along one direction (say y) and pull
  along another direction (say x), with equal
  magnitude.(Biaxial push-pull).
• For x and y as in the figure below the vertical
  and horizontal planes feel no shear stress. The
  are the principal planes and the principal stress
  are (trivially) ?1 100MPa, ?2 100MPa
• Shear stress is maximum at 45? (at this angle
  both normal stresses are zero).For ?xx 100MPa
  ?yy 100MPa, ?max 100 MPa ? the shear
  stress equals the normal stresses in magnitude
  (even though we did not apply shear forces) This
  implies this push-pull configuration gives rise
  to a higher value of shear stress. This aspect
  can be physically visualized as well.
• All stress functions (? ?) are identical and
  only phase shifted from each other.

100 MPa
Load applied
Body
100 MPa
The above stress state can be thought of arising
from a loading as in the figure to right
This loading is equivalent of applying shear
stress at planes inclined at 45?.
This loading is equivalent of applying shear
stress at planes inclined at 45?.
Note that (?x ?y) 0 for all ?
30
Case-3

• If we pull along one direction (say x) and push
  along another direction (say y) with lesser
  force.
• For x and y as in the figure below the vertical
  and horizontal planes feel no shear stress.
• Shear stress is maximum at 45?. For ?xx 100MPa
  ?yy 100MPa, ?max 75 MPa.
• There are no planes where both normal stresses
  are zero.

50 MPa
100 MPa
31
Case-4

• Equi-biaxial tension (2D hydrostatic state of
  stress).
• All planes feel equal normal stress.
• There is no shear stress on any plane.
• Usual materials (metallic) will not plastically
  deform (by slip) under this state of stress (in
  plane? i.e. the planes inclined in the third
  dimension may experience shear stress, which can
  lead to plastic deformation by slip).

100 MPa
All planes feel equal normal stress
100 MPa
All planes feel zero shear stress
32
Case-5

• Pure shear Only shear forces applied. (Can be
  considered a case for pure shear).
• This leads to a stress state identical to case-2,
  but with phase shift of 45?.
• Though we applied only shear forces, normal
  stresses develop in all planes except the planes
  where shear stresses are maximum.

Load applied
Body
Note that (?x ?y) 0 for all ?
How is pure shear different from simple shear
(click here to know more)
33
Hydrostatic and Deviatoric Components of Stress

• (In metallic materials) Hydrostatic components of
  stress can cause elastic volume changes and not
  plastic deformation.
• Yield stress (of metals) is not dependent on the
  hydrostatic stress. However, fracture stress (?f)
  is strongly affected by hydrostatic stress.
• We understand the concept of hydrostatic and
  deviatoric stress in 2D first.
• Hydrostatic stress is the average of the two
  normal stresses.

34
For only normal loads applied on a rectangular
body (equal/zero), what is the increasing order
in which there is a propensity to cause plastic
deformation? (Considering only in-plane stresses
and plasticity by slip).
Funda Check
Best for plastic deformation
Worst for plastic deformation
Fig.A
Fig.C
100 MPa
Fig.B
100 MPa
100 MPa
100 MPa
100 MPa
same as pure shear

• Note if we add ve ?yy to uniaxial tension this
  is bad for plastic deformation.
• Similarly in 3D triaxial (tensile) state of
  stress is bad for plastic deformation.
• Hence, triaxial state of stress suppresses
  plastic deformation and promotes fracture.

• This is 2D hydrostatic state of stress.
• Note slip can still take place on the planes
  inclined in the 3rd dimension

100 MPa

• This reiterates the important point that we
  already know that, hydrostatic states of stress
  tend to cause volume changes, while shear stress
  tend to cause shape changes.

35
Mohrs circle representation of stress

• A nice geometrical way of understanding stress is
  the Mohrs circle representation of stress.
• In plane stress condition (2D) the stresses can
  be written as

This can be rearranged as
From this we can get
This is of the form
Which is the equation of a circle with
36
Features of the Mohrs circle of stress

• The axes (Coordinates) are . The
  centre of the circle is always on the x-axis.
• Angle ? in physical element is represented by
  2? on Mohr's circle. So 45? on the physical
  element is 90? in Mohrs circle.
• Shear stress causing clockwise rotation in the
  physical element is plotted as a positive number
  (above the horizontal axis).
• Any point on the Mohr's circle gives the
  magnitude and direction of normal and shear
  stresses on any plane in the physical element.
  The inside of the Mohrs circle has no physical
  meaning (only the circumference).
• Traversing along the x-axis gives the two
  priciple stresses ?1 ?2.
• Points G, H in Fig.1 correspond to planes G and H
  in the physical element (Fig.2).

Fig.1
Fig.2
37
Mohrs circle for various specific cases

• Considering specific cases can help us understand
  the utility of the Mohrs circle.

Uniaxial tension
Case-1

• Planes A B are principal planes.
• ?2 0.
• r (?1/2)

Uniaxial compression

• Planes A B are principal planes.
• ?1 0.
• ?2 is negative.

Case-2
38
Equi-biaxial tension
Case-3

• The circle collapses to a point.(rMohrs cicle
  0).
• ?2 ?2.
• ?xy is zero.
• Since there is no shear stress, plastic
  deformation by slip cannot occur (in-plane).

Equi Tension-compression
Case-4

• Planes A B are principal planes.
• ?2 0.
• The Mohrs circle will look exactly identical for
  the case below of pure shear? just that planes C
  D are 45?. (Considered in case-5)

39
Pure Shear
Case-5

• The principal planes are at 45? to the C D
  planes. C D are the principal shear planes.

40
Generalized Plane Stress
41
3D state of stress

• In general, a point in a body may exist in a 3D
  state of stress, wherein the 3 principal stresses
  (?1, ?2, ?3) are not equal. The list of
  possibilities in this context are? 3 unequal
  principal stresses (?1, ?2, ?3) ? Triaxial state
  of stress? 2 our of the 3 principal stresses are
  equal (say ?1, ?2 ?3) ? Cylindrical state of
  stress? All 3 principal stresses are equal (say
  ?1 ?2 ?3) ? Hydrostatic/spherical state of
  stress? One of the 3 principal stresses is zero
  (say ?1, ?2, ?3 0) ? Biaxial/2D state of
  stress? One of the 3 principal stresses is zero
  the remaining two are equal to each other (say
  ?1 ?2, ?3 0) ? 2D hydrostatic state of
  stress? Two of the 3 principal stresses is zero
  (say ?1, ?2 ?3 0) ? Uniaxial state of stress.
• We can start with the state of stress on an unit
  cube and observe the state of stress as the
  orientation of the cube is changed (by rotation
  in 3D) or we can look at an inclined plane with
  direction cosines l (Cos?), m (Cos?), n
  (Cos?). This is akin to the square we used in 2D
  and rotate it about the z-axis.

42
Planes which experience maximum shear stress/no
shear stress

• Plastic deformation by slip is caused by shear
  stress (at the atomic level). Hence, we would
  like to identify planes of maximum shear stress.
• For uniaxial tension, biaxial hydrostatic
  tension, triaxial hydrostatic tension, etc., we
  try to identify planes experiencing maximum shear
  stress.

Biaxial hydrostatic tension
Same in magnitude
?1
But yielding can take place due to planes
inclined in the third dimension which fell shear
stresses
These planes feel maximum shear stress
Uniaxial tension
These planes feel no shear stress
(0kl) type
?1
These planes feel maximum shear
stress
?1
These planes feel no shear stress
(hk0) type
43
Triaxial hydrostatic tension
Push-pull normal stresses
?1 ?3
No plane feels any shear stress
These planes feel shear stress
?1 ?2
These planes feel maximum shear stress
twice the other planes (above)
44
Surface Stress
45
Surface Stress

• Surface is associated with surface energy (see
  topic on Surface Energy and Surface Tension).
• Hence a body wants to minimizes its surface area.
  In the process surface atoms want to move towards
  each other.
• The surface of a body (say a liquid) is under
  tensile stress (usual surfaces are under tensile
  stress, under some circumstances (e.g. polar
  surfaces) can be under surface compression).
• As the molecules of water want to come towards
  one another (to minimize surface area) the stress
  has to be tensile.
• This can also be understood by releasing a
  constraint as in coming slides (as before).

46
Consider a soap film held between fixed sliders
At a section AB in the film the surface tension
forces balance the reaction of the slider
If a constraint is removed then the film will
tend to shrink as the points want to move towards
each other ? the surface is under tension
47
Residual Stress
48
Residual stress
What is residual stress and how can it arise in
a material (/component)?

• The stress present in a material/component in the
  absence of external loading/forces or constraints
  (i.e. in a free-standing body) is called residual
  stress.
• Residual stress can be in the macro-scale or
  micro-scale and can be deleterious or beneficial
  depending on the context (diagram below).
• Residual stress may have multiple origins as in
  the diagrams (next slide).
• We have already noted that residual stress is an
  important part of the definition of
  microstructure (it can have profound impact on
  properties).

Macro-scale
Macro-strain
Based on scale
Residual Stress
Corresponding strains will be
Micro-scale
Micro-strain

• Residual stress can be beneficial () or
  detrimental ()
• E.g. ? ? Stress corrosion cracking ?
  Residual Surface Stress (e.g. in toughened glass)

49
Unlike a void or a crack, a dislocation is
naturally associated with residual stresses. A
crack or a void only amplify far field stresses.
Microstructure
Residual Stress

Phases
Defects

• Vacancies
• Dislocations
• Voids
• Cracks

Defects

• Vacancies
• Dislocations
• Twins
• Stacking Faults
• Grain Boundaries
• Voids
• Cracks

Phase Transformation reactions

• Mismatch in coefficient of thermal expansion

Thermal origin
Physical properties

• Thermal
• Magnetic
• Ferroelectric

Residual Stress
Origins/Related to
Geometrical entities
50
Residual stresses due to an edge dislocation in a
cylindrical crystal
Plot of ?x stress contours
Due to phase transformation
Residual stresses due to an coherent precipitate
Due to a dislocation (a crystallographic defect)
51

• Often one gets a feeling that residual stress is
  only harmful for a material, as it can cause
  warpage of the component- this is far from true.
• Residual stress can both be beneficial and
  deleterious to a material, depending on the
  context.
• Stress corrosion cracking leading to an
  accelerated corrosion in the presence of internal
  stresses in the component, is an example of the
  negative effect of residual stresses.
• But, there are good numbers of examples as well
  to illustrate the beneficial effect of residual
  stress such as in transformation toughened
  zirconia (TTZ). In this system the crack tip
  stresses (which are amplified over and above the
  far field mean applied stress) lead to the
  transformation of cubic zirconia to tetragonal
  zirconia. The increase in volume associated with
  this transformation imposes a compressive stress
  on the crack which retards its propagation. This
  dynamic effect leads to an increased toughness in
  the material.
• Another example would be the surface compressive
  stress introduced in glass to toughen it (Surface
  of molten glass solidified by cold air, followed
  by solidification of the bulk ? the contraction
  of the bulk while solidification, introduces
  residual compressive stresses on the surface ?
  fracture strength can be increased 2-3 times).

52
Funda Check
What is the difference between simple and pure
shear?

• Usually we apply simple shear forces on a body.
  Though this is called simple shear it is clear
  that with just two forces the body will not be in
  equilibrium (moment balance is not satisfied).
  This implies that there has to be additional
  hidden forces (as shown in Fig.1b). These
  forces ensure moment balance. To understand this
  let us consider a block on a table being sheared
  by force T. Friction provides the opposite
  force on bottom surface (?T).
• At the material level, pure shear can be
  considered as simple shear rotation of ?/2 (for
  small shear).

Fig.1
b
c
a
Note the bottom
For small deformations
Usually we apply simple shear forces on a material
Simple Shear
The way the diagram is drawn the body is not in
equilibrium!
Pure shear of ?/2 Simple shear of ? ACW
rotation of ?/2
Shear
OR
Pure Shear
Simple shear of ? Pure shear of ?/2 CW
rotation of ?/2
53
Visualizing residual stress/strain
How to see stress?

• Stress typically gives rise to strain, which can
  then be measured/visualized.
• Using high-resolution lattice fringe imaging
  (HRLFI) we can measure the local lattice
  parameter and hence determine the local strain.
• XRD can give a global picture, wherein
  macro-strain gives rise to peak shifting and
  micro-strain (arising from dislocations, coherent
  precipitates, etc. give rise to peak broadening).
• One nice way to visualize strain in transparent
  materials is via imaging using polarized light.
• In the pictures below glass has been heated in a
  furnace between 600-650?C and then cooled in air
  (either by leaving the sample outside or blowing
  air on both the sides). The samples are viewed
  by (a) placing it front of a TFT screen, (b)
  polarizer setup. The colour/shaded contours arise
  due to residual strain in the glass, which arose
  due to the processing.

Also found in understanding_stress_and_strain.pp
t
Experiments and Images by Mr. Shankar Lahura
(CGBS SWF)
(b) Viewed via a polarizer setup
(a) Placed in front of a TFT screen
54
More images showing strain (using polarized light
microscopy and seen through an analyzer)
Packaging material seen in normal light
Packaging material seen in polarized light
55
More images showing strain (using polarized light
microscopy and seen through an analyzer)
After loading the specimen
Crack tip stresses
Unloaded condition
Packing material with cracks
Finite Element Method (FEM) simulated stress
contours at a crack-tip.
Unloaded condition (polarizer rotated by 90?)
56
More images showing strain (using polarized light
microscopy and seen through an analyzer)
Finite Element Method (FEM) simulated stress
contours for a crystallite in an amorphous
matrix.
Molten Aluminium poured ( solidified) in a hole
in a sheet of perspex (transparent polymer)
Note. ? The similarity in the images. In both
cases there is a misfit between the matrix and
the inclusion. ? The lengthscales (one is in mm
and the other is in nm)!
Cu10Zr7 crystal in am amorphous matrix
(composition of (Cu64Zr36)96Al4)
Diameter of the metal part (black) is 5 mm
Aluminium
Transparent plastic
Various rotations of the polarizer
The metal being opaque, no colors are seen there.
Diameter of the crystal is 10 nm
Plot of hydrostatic stress contours (values in
GPa)
Experiments and Images by Mr. Shankar Lahura
(CGBS SWF)