Griffiths equation for the strength of materials

1
the strength of common materials is actually
dictated not so much by bond strength but by
something else
Defects
2
Griffiths equation for the strength of materials

• a length of defect
• g surface energy

• Thus, going from the macroscale to the atomic
  scale (via the nanoscale), defects progressively
  become smaller and/or are eliminated, which is
  why the strength increases (see equation).
• Note that the Griffith model predicts that
  defects have (almost) no effect on the modulus,
  only on strength
• But note the model also predicts that defects of
  zero length lead to infinitely strong materials,
  an obvious impossibility!

3
(No Transcript)
4
Materials strength is critically sensitive to
defects

• Example surface cracks
• What is the weakening effect due to a defect
  at the surface of a fiber ?
• Physically Without any defect, the measured
  (applied) strength s0 would equal the theoretical
  strength (although Griffiths model doesnt
  predict this)
• Case 1 Semi-circular defect at fiber surface
• We use the classical analytical solution of
  Inglis

s0
s0
5
INGLIS, 1913
6
If x a (point A), then sYYlocal 3s0 If the
local stress reaches the theoretical strength,
then the applied stress (s0) at failure is s0
sth/3 Or, assuming sth E/10, we get, at
failure s0 ? E/30 A more realistic situation
is that of a sharper crack
s0
A
s0
7

• Case 2 Sharp (elliptical) defect at fiber
  surface
• Inglis result in this case is, at point A
• a crack length
• r radius of curvature at A.

s0
A
a
s0
So again, if , and a
1 micron, and r 20 Å, then at fracture And
thus
s0 ? E/460
8
Therefore, defects are indeed a major source of
material weakness

• Defects are the major players for strength (and
  for other physical properties too!)
• Griffiths experiments and model are the
  historical basis of the fracture mechanics
  approach
• There is also a probabilistic approach to
  strength why do we need it at all? Because there
  is a whole population of defects present at the
  surface of fibers and within the bulk too, with
  varying degrees of severity. And because fibers
  come in bundle form, which consist of hundreds or
  thousands of fibers in parallel. Examples carbon
  fiber bundle bamboo Achilles tendon
• All the fibers may follow the same statistical
  strength distribution BUT not necessarily the
  same worst defect characteristics!

9
The strength of fibers is statistical
10
Probabilistic argument

• Freudenthal A.M. Freudenthal, in H. Liebowitz,
  ed., Fracture, Vol. 2, Academic Press, New York,
  592 (1968) proposed a link between the
  probability of occurrence of a critical defect,
  F(V), in a solid of (dimensionless) volume V, the
  concentration of defects, and the size (length,
  area, volume) of a solid
• F(V) 1 exp-(V/V0)
• where V0 is the mean volume occupied by a defect
  (thus 1/V0 is the mean cc of defects).
• Plot

11
Probability of occurrence of a critical defect
(F(V) Probability of failure) against size for
a given defect concentration
At very small volume, low P of occurrence of a
critical defect Thus strength tends to be very
high At larger volume, F(V) climbs rapidly A
plateau is reached where size has no more effect.

12
However no real physics in the previous
equation. How do we draw stress into the
picture?

• Weibull The original density of defects in the
  material (1/V0) increases as the applied stress
  increases according to some physical (power) law
• (1/V0) (s/a)b
• and therefore
• F(V) 1 exp-V (s/a)b (the Weibull
  Distribution)
• a scale parameter
• b shape parameter

13
Density function As b increases, the
distribution is more narrow, and a is
proportional to the average of the distribution
14
The effect of size on strengthThe
Weakest-Link model for a fiber

• Assume that a fiber is viewed as a chain of
  links or units having each the same
  probability of failure F(s) under a stress s.
• Probability of survival of a link is 1 F(s)
• Probability of survival of a chain ( n links) is
  1 F(s)n
• Probability of failure of the chain is
• Fn(s) 1 - 1 F(s)n
• Do this insert a Weibull distribution for F
  (thus for a link) and observe that Weibull is
  again obtained for Fn (the fiber), with the same
  b but lower a The larger the specimen, the
  higher its P of failure!

15

• INTRODUCTION GENERAL PRINCIPLES AND BASIC
  CONCEPTS Lectures 1-3
• Composites in the real world
• Classification of composites
• Three simple models for a-priori materials
  selection the role of defects
• Scale effects
• Stress and strain
• Thermodynamics of deformation and Hookes law
  Anisotropy and elastic constants
• Micromechanics models for elastic constants
  Measuring the elastic constants.

16
Last topics in the Basic Concepts Section

• Stress and strain brief review of definitions
• Thermodynamics of deformation and Hookes law
• Anisotropy and elastic constants Relevance to
  composite materials
• Stress
• (old concept Hooke in the 1680s Cauchy
  Poisson in the 1820s)
• Continuum view of materials no molecules (so
  that field quantities such as displacement,
  stress, etc can be defined as continuous
  functions of space and time), and homogeneity.

17
Stress (ctd)

• Stress force/area
• The state of stress at a point in a continuum can
  be represented by 9 stress components sij (i,j
  1,2,3) acting on the sides of an elemental cube
  with sides parallel to the axes 1,2,3 of a
  reference coordinate system

18
Stress is a tensor with 9 components (sij)

• First subscript (i) gives the normal to the plane
  on which stress acts Second subscript defines
  the direction of the stress.
• The sii components are called normal stresses,
  the sij components are called shear stresses.
• Tensile stresses are positive, compressive
  stresses are negative.
• It can be shown from force equilibrium
  considerations, that the shear stress components
  are related by sij sji (i?j).
• Therefore, we have only 6 independent components
  of stress.
• Knowledge of all components allow us to define
  the stress acting on any plane within the body.

19
Strain

• A body subject to a state of stress will develop
  strains. There are several definitions of strain,
  the most usual (used in linear elasticity) is the
  engineering strain e dl/l, where l is the
  initial length.
• Strain is dimensionless.
• Like stress, strain is a tensor with 9
  components, 6 of them only being independent
  because eij eji (i?j).

20

• Thermodynamics of deformation The origin of
  Hookes law
• We assume small deformations in a body Those
  deformations occur slowly so that thermodynamic
  equilibrium can be assumed.

21

• Internal stresses are set up within the body, due
  to deformation.
• OBJECTIVE To find a relation between the applied
  deformation (or strains) and induced stresses in
  the body. (In other words, to derive Hookes law
  from basic principles.)
• Thermodynamics an infinitesimal increment of the
  total (internal) energy per unit volume dE is
  equal to the sum of (1) the amount of heat TdS (T
  temperature, S entropy) acquired by the unit
  volume considered and (2) the work done by the
  internal stresses due to the deformation (per
  unit volume),
• Thus, we have
• (per unit volume).

22
By definition, the (Helmholtz) free energy of the
body is f E-TS Thus
So that for an isothermal deformation process (T
constant), we have
Therefore, we need to know the free energy per
unit volume, f, as a function of eik
23
This is easily calculated since we have small
deformations, f can be expanded in a Taylor
series
where f0 is the free energy of the undeformed
body, and the xs are given as follows
24
By differentiating, we obtain
And we know that this is equal to sik (for an
isothermal process). If there is no deformation,
there are no internal stresses in the body, thus
sik 0 for eik o, from which we obtain x
0. Thus, no linear term in the expansion of f in
powers of eik
by limiting the expansion to the second order f
e2
25
And we can therefore compute the stress tensor in
terms of the strain tensor
or
This very simple expression provides a linear
dependence between stress and strain it is the
basic form of Hookes law, as obtained from
purely thermodynamic considerations ! Also,
remember the connection between Youngs modulus
and the potential from the previous class?
26
A common general form (valid for anisotropic
bodies) of Hookes law is the following
Where sij and ekl are 2d rank tensors and Cijkl
is a 4th rank tensor with 3x3x3x3 81 components
or 9 stress components x 9 strain components
81. The Cijkl are called the elastic constants.
Historical parenthesis Robert Hookes legacy
27
Elementary concepts of mechanics

• In 1676, Robert Hooke makes a discovery about
  springs

28
ut tensio, sic vis(stretch force)
UNDER TENSION
29
Similarly, under bending
load deflection
30
Under shear
load shear deformation
and torsion
load angular deformation
31
Thus, in all cases, Hooke observed
The ratio applied force/distortion is constant
for a given material and specimen geometry.
This is (almost) Hookes Law
32
Hookes Law
This definition is now a Material Constant. It
is valid whatever the mode of testing (tension,
bending, torsion, shearing, hydrostatic
compression, etc, and a specific modulus is then
defined)
33
The stress-strain curve
34
A general stress-strain curve
Plastic (irreversible)
Elastic (fully reversible)
35
Comparing various stress-strain curves
36

• We focus on Hookes law for various special cases
  of material symmetry
• There are 9 x 9 (or 3 x3 x3 x 3) 81 components
  of Cijkl but we know that only 6 x 6 36 of
  these are independent.
• It can also be shown that provided that a strain
  free energy function exists, the number of
  distinct elastic constants reduces to 21 because
  Cijkl Cklij.

37

• The number of elastic constants can be further
  reduced as it depends on the crystalline class
• Generally anisotropic materials (triclinic)
  possess 21 independent elastic constants.
• Monoclinic systems (one plane of elastic
  symmetry) have 13 non-zero independent moduli.
• Orthorombic crystals (3 planes of symmetry
  perpendicular to each other) have 9 moduli
  (remember polyethylene?). They are termed
  orthotropic in the composite materials
  community.

38
(No Transcript)
39
Form of the Cijkl matrix
(From J.F. Nye, Physical Properties of Crystals)
40
(No Transcript)
41
Notations sij Cijklekl (i, j, k, l
1,2,3) eij Sijklskl
Historical paradox The Cijkl are called the
Stiffness components The Sijkl are called the
Compliance components
Contracted notations in the mechanics of
composites
42
Contracted notations in the mechanics of
composites
43
Orthotropic lamina (9 constants)

• Observations
• There are no interactions between normal stresses
  and shear strains
• There are no interactions between shear stresses
  and normal strains

44
(No Transcript)
45
Transversely isotropic lamina (5 constants)
46
Isotropic lamina (2 constants)
47
Orthotropic material under plane stress

• In many structural applications, composite
  materials are used in the form of thin laminates
  loaded within the plane of the laminate. This is
  a plane stress situation in which all stress
  components in the out-of-plane direction (say,
  the 3-direction) are zero s3 t23 t4 t13
  t5 0
• Inserting this into the orthotropic stress-strain
  relation, we obtain (after some manipulations,
  please check this)
• where (i,j 1,
  2, 6) (4 independent constants)

48
How can we derive relations between mathematical
and engineering constants ?

• Stress-strain relations presented before have
  more physical/intuitive meaning when expressed in
  terms of familiar engineering constants such as
  the Youngs modulus.
• Formal connections between mathematical and
  engineering constants are derived from
  elementary experiments. For example

49
Elementary experiments
50
Remember an orthotropic lamina (9 constants)
51
Example

• Uniaxial tensile stress in (say) transverse
  direction (2) causes a strain in the 2
  direction
• but also in the 1 and 3 directions

(by definition of the Poisson ratios nij
-ej/ei). Thus, we obtain
52
All other elementary experiments provide similar
links. Eventually we obtain what we wanted, the
stress-strain relations in terms of engineering
constants (E, n, G)
53

• From the symmetry of the compliance matrix Sij,
  we conclude that

In general, we conclude that the relations
between the compliances Sij and the engineering
constants are simple. It can be shown that the
relations between the stiffnesses Cij and the
engineering constants are a little more
complicated.
54
Finally, the connection between the stiffness
constants Cij and the compliance constants Sij
are as follows
55
Last remarks

• In the case of a transversally isotropic material
  with the 2-3 plane as plane of isotropy, we have
• E2 E3
• G12 G13
• n12 n13
• The 3 engineering constants in the isotropic case
  are related by
• therefore, as necessary, only 2 constants are
  independent.