Lecture # 10 - Aamir Razaq Elastic Compliance and
Stiffness constant
• Stress and strain: What are they and why are they
• Elastic behavior: When loads are small, how much
deformation occurs? What materials deform least?
• Plastic behavior: At what point do dislocations
cause permanent deformation? What materials are
most resistant to permanent deformation?
• Toughness and ductility: What are they and how
do we measure them?
• Ceramic Materials: What special provisions/tests
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Normal &amp; Shear components of stress
Normal stress is perpendicular to the cross section,  (sigma).
Shear stress is parallel to the cross section,  (tau).
y
y
3D case
xy
yx
yz
zx
z
Second subscript
indicates the positive
direction of the shear
stress
xy
zy
z
First subscript indicates
the axis that is
perpendicular to the face
xz
x
x
Due to equilibrium condition;
xy = yx
zx = xz
zy = yz
Normal &amp; Shear components of stress
Two Dimensional Case
y
yx
xy
x
x
xy
yx
y
Normal Stress Due to Axial Load
A positive sign is used to indicate a
tensile stress (tension), a negative
sign to indicate a compressive stress
(compression)
Normal Stress Due to Bending Load
Stress distribution
Maximum stress at the surface
M change in position
Where I is area moment of inertia
Shear Stress Due to Torque (twisting)
Torsional stress is caused by twisting a member
Stress distribution
Maximum shear stress at the surface
Where J is polar area moment of inertia
Torsional Stress- examples
Structural member
Power transmission
Mixer
Combined Stresses - examples
Bicycle pedal arm and
lug wrench, bending
and torsion stresses
Trailer hitch, bending
and axial stresses
Billboards and traffic
signs, bending, axial
and torsion stresses
Power transmission,
bending and torsion stresses
Understanding stress in terms of its components
 Stress is a Second Order Tensor
 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 stress:
2D  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
 xx
Plane
Direction
2D
  xx

  yx
 xy 
 yy 
  xx  xy 


  yx  yy 
  11  12 




22 
 21
Let us consider a body in the presence of external agents (constraints and forces) bringing
about stresses in the body. A unit region in the body (assumed having constant stresses) is
analyzed. (body forces are ignored)
Note: the directions of
the stresses shown are
arbitrary (the stresses in
general could be
compression/tension and
shear could be opposite
in sign)
• Elastic Compliance and Stiffness Constants
• The generalized Hooke’s law is an
assumption, which is reasonably accurate
for many material subjected to small strain,
for a given temperature, time and location.
• A mathematical expression of the stress-strain
relation for the elastic deformation of materials
was first suggested by Robert Hooke
F is the applied force
u is the deformation of the elastic body
k is the material dependent spring constant
Anisotropy and Isotropy
• In a single crystal, the physical and mechanical
properties often differ with orientation.
•
An anisotropic material is a material which
does not behave the same way in all
directions. Take wood for example. Wood
is very strong along the grain. Against the
grain, however, it will easily break.
The opposite of an anisotropic material is
an isotropic material. Most metals (steel,
aluminum) are isotropic materials. They
respond the same way in all directions.
Figure 7-1 Anisotropic materials: (a) rolled material, (b)
wood, (c) glass-fiber cloth in an epoxy matrix, and (d) a
crystal with cubic unit cell.
• Real materials are never perfectly isotropic. In
some cases (e.g. composite materials) the
differences in properties for different directions
are so large that one can not assume isotropic
behavior - Anisotropic.
• There is need to discuss Hooke’s Law for
anisotropic cases in general. This can then be
reduced to isotropic cases - material property
(e.g., elastic constant) is the same in all
directions.
• In the general 3-D case, there are six components of
stress and a corresponding six components of strain.
• In highly anisotropic materials, any one component of
stress can cause strain in all six components.
• For the generalized case, Hooke’s law may be
expressed as:
where,
 i  Cij j
 i  Sij j
C  Stiffness (or Elastic constant )
S  Compliance
• Both Sijkl and Cijkl are fourth-rank tensor quantities.
• Expansion of either Eqs. will produce nine (9)
equations, each with nine (9) terms, leading to 81
constants in all.
• It is important to note that both ij and  ij are
symmetric tensors.
• Symmetric tensor
Means that the offdiagonal components are equal. For example, in
case of stress:
 13   31 ,  12   21
 23   32
We can therefore write:
  11  12  13    11  12  13 

 

  21  22  23     12  22  23 
       
 31 32 33   13 23 33 
Similarly, the strain tensor can be written as:
 11 12 13   11 12 13 

 

  21  22  23    12  22  23 
       
 31 32 33   13 23 33 
Elastic Deformation
1. Initial
bonds
stretch
initial

F
Elastic means reversible.
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Plastic Deformation (Metals)
1. Initial
F
Plastic means permanent.
linear
elastic
linear
elastic
plastic

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Typical stress-strain
behavior for a metal
showing elastic and
plastic deformations,
the proportional limit P
and the yield strength
σy, as determined
using the 0.002 strain
offset method (where there
is noticeable plastic deformation).
to plastic transition.
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Plastic Deformation
(permanent)
• From an atomic perspective, plastic
deformation corresponds to the breaking of
bonds with original atom neighbors and
then reforming bonds with new neighbors.
• After removal of the stress, the large
number of atoms that have relocated, do
• Yield strength is a measure of resistance
to plastic deformation.
25
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• Localized deformation of a ductile material during a
tensile test produces a necked region.
• The image shows necked region in a fractured sample
Permanent Deformation
• Permanent deformation for metals is
accomplished by means of a process called
slip, which involves the motion of
dislocations.
• Most structures are designed to ensure that
only elastic deformation results when stress
is applied.
• A structure that has plastically deformed, or
experienced a permanent change in shape,
may not be capable of functioning as
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intended.
Yield Strength, y
tensile stress, 
y
engineering strain, 
p = 0.002
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Stress-Strain Diagram
ultimate
tensile
strength
3
yield
strength
Strain
Hardening
 UTS
y
necking
Fracture
5
2
Elastic region
slope =Young’s (elastic) modulus
yield strength
Plastic region
ultimate tensile strength
strain hardening
fracture
Plastic
Region
Elastic
Region
σ Eε
σ
E
ε
1
E
σy
ε 2  ε1
4
Strain (  ) (DL/Lo)
Stress-Strain Diagram (cont)
• Elastic Region (Point 1 –2)
after the material is unloaded( like a rubber band).
- The stress is linearly proportional to the strain in
this region.
σ Eε
σ
or
σ
E
ε
: Stress(psi)
E : Elastic modulus (Young’s Modulus) (psi)
ε : Strain (in/in)
- Point 2 : Yield Strength : a point where permanent
deformation occurs. ( If it is passed, the material will
Stress-Strain Diagram (cont)
• Strain Hardening
- If the material is loaded again from Point 4, the
curve will follow back to Point 3 with the same
Elastic Modulus (slope).
- The material now has a higher yield strength of
Point 4.
- Raising the yield strength by permanently straining
the material is called Strain Hardening.
Stress-Strain Diagram (cont)
• Tensile Strength (Point 3)
- The largest value of stress on the diagram is called
Tensile Strength(TS) or Ultimate Tensile Strength
(UTS)
- It is the maximum stress which the material can
support without breaking.
• Fracture (Point 5)
- If the material is stretched beyond Point 3, the stress
decreases as necking and non-uniform deformation
occur.
- Fracture will finally occur at Point 5.
The stress-strain curve for an aluminum alloy.
• Stress-strain
behavior
found for
some steels
with yield
point
phenomenon.
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Toughness is
the ability to
absorb
energy up to
fracture (energy
Toughness
Lower toughness: ceramics
Higher toughness: metals
per unit volume of
material).
A “tough”
material has
strength and
ductility.
Approximated
by the area
under the
stress-strain
curve.
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Toughness
• Energy to break a unit volume of material
• Approximate by the area under the stress-strain
curve.
Engineering
tensile
stress, 
smaller toughness (ceramics)
larger toughness
(metals, PMCs)
smaller toughnessunreinforced
polymers
Engineering tensile strain, 
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Linear Elastic Properties
• Hooke's Law:
=E
• Poisson's ratio:
n  x/y
metals: n ~ 0.33
ceramics: n ~0.25
polymers: n ~0.40
Modulus of Elasticity, E:
(Young's modulus)
Units:
E: [GPa] or [psi]
n: dimensionless
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Elastic Strain Energy
• If the sample
obeys Hooke's
Law, and is below
the elastic limit,
the Elastic Strain
Energy can be
calculated by the
formula:
• E = &frac12;Fx
Example 2:
Young’s Modulus - Aluminum Alloy
From the data in Example 1, calculate the modulus of
elasticity of the aluminum alloy.
Assignment
E  29 10 6 psi
D  1.07 in. d  0.618 in.
Determine the deformation of
the steel rod shown under the