Wagner - Class 3 1

the strength of common materials is actually dictated not so much by bond strength but by

something else:

Defects

Wagner - Class 3 2

Griffith’s equation for the strength of materials

21

2

a

E

a = length of defect

= 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 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!

Wagner - Class 3 3

NANOSCALE Vs MICROSCALE

Griffith’s experiments with glass fibers

(1921)

FIBER DIAMETER (micron)

Strength of bulk glass: 170 MPa

Extrapolates to

11 GPa

1

2

3

TEN

SIL

E S

TR

EN

GTH

(G

Pa)

020 40 60 80 100 1200

Wagner - Class 3 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 0 would equal the theoretical strength (although Griffith’s model doesn’t predict this…)

Case 1 – Semi-circular defect at fiber surface

We use the classical analytical solution of Inglis:

0

Wagner - Class 3 5

0 1 2 3 4 5

0.5

1.0

1.5

2.0

2.5

3.0

3.5

2

2

2

2

0

xx

x

a3

x

a1

2

1

4

4

2

2

0

yy

x

a3

x

a2

2

1

X/a

INGLIS, 1913

Wagner - Class 3 6

A

If x = a (point A), then YYlocal = 30

If the local stress reaches the theoretical strength, then the applied stress (0) at failure is

0 = th/3

And assuming th ≈ E/10, we get, at failure:

0 E/30

A more realistic situation is that of a sharper crack:

0

0

Wagner - Class 3 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.

0

A021

r

aYYlocal

a

So again, if , and a = 1 micron, and r = 20 Å, then at fracture

And thus:

046 ltheoretica

ltheoreticalocal

460ltheoretica

0 E/460

Wagner - Class 3 8

Therefore, defects are indeed a major source of material weakness

• Defects are the major players for strength (and for other physical properties too…!)

• Griffith’s 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; Achille’s tendon;

• All the fibers may follow the same statistical strength distribution BUT not necessarily the same worst defect characteristics…!

Wagner - Class 3 9

The strength of fibers is statistical

2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.20

200

400

600

800

1000

1200

1400

1600 B O R O N F I L A M E N T S

Num

ber

of te

sts

Tensile strength (GPa)

Wagner - Class 3 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:

Wagner - Class 3 11

0 1 2 3 4 5

0.2

0.4

0.6

0.8

1.0

V0=1

PROB

ABILI

TY, F

(V)

VOLUME, V

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 criticaldefect – Thus: strength tends to be very high

At larger volume, F(V) climbs rapidly: A plateau is reached where size has no more effect.

Wagner - Class 3 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) = (/)

and therefore:

F(V) = 1 – exp[-V (/)] (the Weibull Distribution)

= scale parameter = shape parameter

Wagner - Class 3 13

0 1 2 3 4

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.5

exponential

1.5

2

=3WEIBULL DISTRIBUTION

(=1)

f(x)

X

Density function: As increases, the distribution is more narrow, and is proportional to the average of the distribution

Wagner - Class 3 14

The effect of size on strength:The ‘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() under a stress .

• Probability of survival of a link is 1 – F()• Probability of survival of a chain (= n links) is [1 – F()]n

• Probability of failure of the chain is

Fn() = 1 - [1 – F()]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 but lower The larger the specimen, the higher its P of failure!

Wagner - Class 3 15

1. INTRODUCTION – GENERAL PRINCIPLES AND BASIC CONCEPTS Composites in the real world; Classification of composites; scale effects; the role of interfacial area and adhesion; three simple models for a-priori materials selection; the role of defects; Stress and strain; thermodynamics of deformation and Hooke’s law; anisotropy and elastic constants; micromechanics models for elastic constants. Measuring the elastic constants {Lectures 1-3}

2. MATERIALS FOR COMPOSITES: FIBERS, MATRICES Types and physical properties of fibers; flexibility and compressive behavior; stochastic variability of strength; Limits of fiber performance; types and physical properties of matrices; combining the phases: residual thermal stresses; {Lectures 4-5}

3. THE PRINCIPLES OF FIBER REINFORCEMENTStress transfer; The model of Cox; The model of Kelly & Tyson; Other model; {Lectures

6-7}

4. INTERFACES IN COMPOSITES Basic issues, wetting and contact angles, interfacial adhesion, the fragmentation phenomenon, microRaman spectroscopy, transcrystalline interfaces, {Lectures 8-9}

5. FRACTURE PHYSICS OF COMPOSITES Griffith theory of fracture, current models for idealized composites, stress concentration, simple mechanics of materials, micromechanics of composite strength, composite toughness, measuring the strength and fracture toughness, indentation and nanoindentation testing {Lectures 10-12}

6. DESIGN EXAMPLE A composite flywheel {Lecture 13}

7. THE FUTURE:Composites based on nanoreinforcement, composites based on biology, ribbon- and platelet-reinforced materials, biomimetic concepts {Lectures 14-15}

Wagner - Class 3 16

Last topics in the ‘Basic Concepts’ Section

• Stress and strain – brief review of definitions• Thermodynamics of deformation and Hooke’s 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.

Wagner - Class 3 17

Stress (ctd)

• Stress = force/area• The state of stress at a point in a continuum can be

represented by 9 stress components ij (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:

Wagner - Class 3 18

• Stress is a tensor with 9 components (ij)

• First subscript (i) gives the normal to the plane on which stress acts; Second subscript defines the direction of the stress.

• The ii components are called normal stresses, the ij 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 ij = ji (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.

Wagner - Class 3 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: = 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 ij = ji (i≠j).

Wagner - Class 3 20

Thermodynamics of deformation – The origin of Hooke’s law

• We assume small deformations in a body; Those deformations occur slowly so that thermodynamic equilibrium can be assumed.

Wagner - Class 3 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 Hooke’s 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).

ikikdTdSdE

ikikd

Wagner - Class 3 22

By definition, the (Helmholtz) free energy of the body is = E-TS Thus:

ikikdSdTd

So that for an isothermal deformation process (T = constant), we have:

Tikik

Therefore, we need to know the free energy per unit volume, , as a function of ik

Wagner - Class 3 23

This is easily calculated: since we have small deformations, can be expanded in a Taylor series:

...3''2'0 ikikikik

where 0 is the free energy of the undeformed body, and the ’s are given as follows:

....

02

2'

0

ik

ik

ik

ik

Wagner - Class 3 24

By differentiating, we obtain:

...32 2'''

ikikik

And we know that this is equal to ik (for an isothermal process).

If there is no deformation, there are no internal stresses in the body, thus ik = 0 for ik = o, from which we obtain = 0.

Thus, no linear term in the expansion of in powers of ik:2'0 ik

by limiting the expansion to the second order: ~ 2

Wagner - Class 3 25

And we can therefore compute the stress tensor in terms of the strain tensor:

Tikik

or

ikik '2

This very simple expression provides a linear dependence between stress and strain: it is the basic form of Hooke’s law !

Also, remember the connection between Young’s modulus and the potential from the previous class?

....

02

2'

ik

ik

E

Wagner - Class 3 26

A common general form (valid for anisotropic bodies) of Hooke’s law is the following:

klijklij C

Where ij and kl 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 Hooke’s legacy

Wagner - Class 3 27

Elementary concepts of mechanics

• In 1676, Robert Hooke makes a discovery about springs:

Wagner - Class 3 28

“ut tensio, sic vis”(load ~ stretch)

UNDER TENSION:

Wagner - Class 3 29

Similarly, under bending:

load ~ deflection

Wagner - Class 3 30

Under shear…

…and torsion:

load ~ shear deformation

load ~ angular deformation

Wagner - Class 3 31

Thus, in all cases, Hooke observed:

The ratio

applied force/distortion

is a constant for the material.

This is (almost) Hooke’s Law

Wagner - Class 3 32

Hooke’s Law

"Modulus"*

/

/

"constant Spring"

Area

Lengthk

LengthDistortion

AreaForce

Strain

Stress

kDistortion

Force

This definition is valid whatever the mode of testing(tension, bending, torsion, shearing, hydrostatic compression, etc,

and a specific modulus is then defined)

Wagner - Class 3 33

The stress-strain curve

Wagner - Class 3 34

A general stress-strain curve

Elastic (fully reversible)

Plastic (irreversible)

Wagner - Class 3 35

Comparing various stress-strain curves

Wagner - Class 3 36

•We focus on Hooke’s law for various special cases of material symmetry

•There are 9 x 9 = 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.

Wagner - Class 3 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.

Wagner - Class 3 38

Class of Material Number of nonzero coefficients

Number of independent coefficients

3D

Triclinic 36 21

Monoclinic 20 13

Orthorombic [Orthotropic]

12 9

Transversely isotropic 12 5

Isotropic 12 2

2D

Triclinic 9 6

Monoclinic 9 6

Orthorombic [Orthotropic]

5 4

Transversely isotropic 5 4

Isotropic 5 2

Wagner - Class 3 39

(From J.F. Nye, ‘Physical Properties of Crystals’)

Form of the Cijkl matrix

Wagner - Class 3 40

Wagner - Class 3 41

Notations:

ij = Cijklkl

(i, j, k, l = 1,2,3)

ij = Sijklkl

Historical paradox:

The Cijkl are called the Stiffness components

The Sijkl are called the Compliance components

Contracted notations in the mechanics of composites:

Wagner - Class 3 42

Contracted notations in the mechanics of composites:

Wagner - Class 3 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

Wagner - Class 3 44

Wagner - Class 3 45

Transversely isotropic lamina (5 constants)

Wagner - Class 3 46

Isotropic lamina (2 constants)

Wagner - Class 3 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: 3 = 23 = 4 = 13 = 5 = 0

• Inserting this into the orthotropic stress-strain relation, we obtain (after some manipulations):

where (i,j = 1, 2, 6) (4 independent constants) 33

33

C

CCCQ jiijij

6

2

1

66

2212

1211

6

2

1

00

0

0

Q

QQ

QQ

Wagner - Class 3 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 Young’s modulus.

• Formal connections between mathematical and engineering constants are derived from ‘elementary experiments’. For example:

Wagner - Class 3 49

Elementary experiments

Wagner - Class 3 50

Remember: an orthotropic lamina (9 constants)

Wagner - Class 3 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:

2

22222 E

S

2

2212121 E

S

2

2232233 E

S

From which we obtain:

2

2112 ES

222

1

ES

2

2323 ES

Wagner - Class 3 52

All other elementary experiments provide similar links. Eventually we obtain what we wanted, the stress-strain relations in terms of engineering constants (E, , G):

Wagner - Class 3 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 more complicated.

Wagner - Class 3 54

Finally, the connection between the stiffness constants Cij and the compliance constants Sij are as follows:

Wagner - Class 3 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

12 = 13

• The 3 engineering constants in the isotropic case are related by

therefore, as necessary, only 2 constants are independent.

12

EG