the  strength  of common materials is actually dictated not so much by  bond  strength but by something else:  Defects
Griffith’s equation for the strength of materials <ul><li>Thus, going from the macroscale to the atomic scale (via the nan...
NANOSCALE Vs MICROSCALE   Griffith’s experiments with glass fibers (1921) FIBER DIAMETER (micron) Strength of bulk glass: ...
Materials’ strength is critically sensitive to defects   <ul><li>Example : surface cracks </li></ul><ul><li>What is the ‘w...
INGLIS, 1913
A If x = a (point A), then   YYlocal  = 3  0   If the local stress reaches the theoretical strength, then the applied st...
<ul><li>Case 2 – Sharp (elliptical) defect at fiber surface </li></ul><ul><li>Inglis’ result in this case is, at point A: ...
Therefore, defects are indeed  a major source of material weakness <ul><li>Defects are the major players for strength (and...
The strength of fibers is statistical
Probabilistic argument <ul><li>Freudenthal [ A.M. Freudenthal, in H. Liebowitz, ed., Fracture, Vol. 2, Academic Press, New...
Probability of occurrence of a critical defect (F(V)= Probability of  failure) against size for a given defect concentrati...
However : no real physics in the previous equation.  How do we draw stress into the picture?   <ul><li>Weibull:  The origi...
Density function: As    increases, the distribution is more narrow, and    is proportional to the average of the distrib...
The effect of size on strength: The ‘Weakest-Link’ model for a fiber <ul><li>Assume that a fiber is viewed as a chain of ‘...
<ul><li>1. INTRODUCTION – GENERAL PRINCIPLES AND BASIC CONCEPTS  </li></ul><ul><li>Composites in the real world; Classific...
Last topics in the ‘Basic Concepts’ Section <ul><li>Stress and strain – brief review of definitions </li></ul><ul><li>Ther...
Stress (ctd) <ul><li>Stress = force/area </li></ul><ul><li>The state of stress at a point in a continuum can be represente...
<ul><li>Stress is a tensor with 9 components (  ij ) </li></ul><ul><li>First subscript  (i) gives the normal to the plane...
Strain <ul><li>A body subject to a state of stress will develop strains. There are several definitions of strain, the most...
<ul><li>Thermodynamics of deformation – The origin of Hooke’s law </li></ul><ul><li>We assume small deformations in a body...
<ul><li>Internal stresses are set up within the body, due to deformation. </li></ul><ul><li>OBJECTIVE : To find a relation...
By definition, the (Helmholtz) free energy of the body is   = E-TS  Thus: So that for an  isothermal  deformation proces...
This is easily calculated: since we have small deformations,   can be expanded in a Taylor series: where   0  is the fr...
By differentiating, we obtain: And we know that this is equal to   ik  (for an isothermal process). If there is no deform...
And we can therefore compute the stress tensor in terms of the strain tensor: or This very simple expression provides a li...
A common general form (valid for anisotropic bodies) of Hooke’s law is the following: Where   ij  and   kl  are 2d rank ...
Elementary concepts of mechanics <ul><li>In 1676, Robert Hooke makes a discovery about springs: </li></ul>
“ ut tensio, sic vis” (load ~ stretch) UNDER TENSION:
Similarly, under bending: load ~ deflection
Under shear… … and torsion: load ~ shear deformation load ~ angular  deformation
Thus, in all cases, Hooke observed: The ratio  applied force/distortion is a constant for the material.  This is (almost) ...
Hooke’s Law This definition is valid whatever the mode of testing (tension, bending, torsion, shearing, hydrostatic compre...
The stress-strain curve
A general stress-strain curve Elastic (fully reversible) Plastic  (irreversible)
Comparing various stress-strain curves
<ul><li>We focus on Hooke’s law for various special cases of material symmetry </li></ul><ul><li>There are 9 x 9 = 81 comp...
<ul><li>The number of elastic constants can be further reduced as it depends on the  crystalline class : </li></ul><ul><li...
2 5 Isotropic 4 5 Transversely isotropic 4 5 Orthorombic [Orthotropic] 6 9 Monoclinic 6 9 Triclinic 2D 2 12 Isotropic 5 12...
(From J.F. Nye, ‘Physical Properties of Crystals’) Form of the C ijkl  matrix
 
Notations:  ij  = C ijkl  kl (i, j, k, l = 1,2,3)  ij  = S ijkl  kl Historical paradox: The C ijkl  are called t...
Contracted notations in the mechanics of composites:
Orthotropic lamina (9 constants) <ul><li>Observations: </li></ul><ul><li>There are no interactions between normal stresses...
 
Transversely  isotropic lamina (5 constants)
Isotropic lamina (2 constants)
Orthotropic material under plane stress <ul><li>In many structural applications, composite materials are used in the form ...
How can we derive relations between mathematical  and engineering constants ? <ul><li>Stress-strain relations presented be...
Elementary experiments
Remember: an orthotropic lamina (9 constants)
Example <ul><li>Uniaxial tensile stress in (say) transverse direction (2) causes a strain in the ‘2’ direction:  </li></ul...
All other elementary experiments provide similar links. Eventually we obtain what we wanted, the stress-strain relations i...
<ul><li>From the symmetry of the compliance matrix S ij , we conclude that: </li></ul>In general, we conclude that the rel...
Finally, the connection between the stiffness constants C ij  and the compliance constants S ij  are as follows:
Last remarks <ul><li>In the case of a  transversally  isotropic material with the 2-3 plane as plane of isotropy, we have:...
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Class3

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Class3

  1. 1. the strength of common materials is actually dictated not so much by bond strength but by something else: Defects
  2. 2. Griffith’s equation for the strength of materials <ul><li>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). </li></ul><ul><li>Note that the Griffith model predicts that defects have no effect on the modulus, only on strength </li></ul><ul><li>But note: the model also predicts that defects of zero length lead to infinitely strong materials, an obvious impossibility! </li></ul>a = length of defect  = surface energy
  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 TENSILE STRENGTH (GPa) 0 20 40 60 80 100 120 0
  4. 4. Materials’ strength is critically sensitive to defects <ul><li>Example : surface cracks </li></ul><ul><li>What is the ‘weakening effect’ due to a defect at the surface of a fiber ? </li></ul><ul><li>Physically: Without any defect, the measured (applied) strength  0 would equal the theoretical strength (although Griffith’s model doesn’t predict this…) </li></ul><ul><li>Case 1 – Semi-circular defect at fiber surface </li></ul><ul><li>We use the classical analytical solution of Inglis: </li></ul> 0  
  5. 5. INGLIS, 1913
  6. 6. A If x = a (point A), then  YYlocal = 3  0 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
  7. 7. <ul><li>Case 2 – Sharp (elliptical) defect at fiber surface </li></ul><ul><li>Inglis’ result in this case is, at point A: </li></ul><ul><li>a = crack length </li></ul><ul><li>r = radius of curvature at A. </li></ul> 0 A   a So again, if , and a = 1 micron , and r = 20 Å , then at fracture And thus:  0  E/460
  8. 8. Therefore, defects are indeed a major source of material weakness <ul><li>Defects are the major players for strength (and for other physical properties too…!) </li></ul><ul><li>Griffith’s experiments and model are the historical basis of the ‘fracture mechanics’ approach </li></ul><ul><li>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; </li></ul><ul><li>All the fibers may follow the same statistical strength distribution BUT not necessarily the same worst defect characteristics…! </li></ul>
  9. 9. The strength of fibers is statistical
  10. 10. Probabilistic argument <ul><li>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: </li></ul><ul><li>F(V) = 1 – exp[-(V/V 0 )] </li></ul><ul><li>where V 0 is the mean volume occupied by a defect (thus: 1/V 0 is the mean cc of defects). </li></ul><ul><li>Plot: </li></ul>
  11. 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. 12. However : no real physics in the previous equation. How do we draw stress into the picture? <ul><li>Weibull: The original density of defects in the material (1/V 0 ) increases as the applied stress increases according to some physical (power) law </li></ul><ul><li>(1/V 0 ) = (  /  )  </li></ul><ul><li>and therefore: </li></ul><ul><li>F(V) = 1 – exp[-V (  /  )  ] (the Weibull Distribution) </li></ul><ul><li> = scale parameter </li></ul><ul><li> = shape parameter </li></ul>
  13. 13. Density function: As  increases, the distribution is more narrow, and  is proportional to the average of the distribution
  14. 14. The effect of size on strength: The ‘Weakest-Link’ model for a fiber <ul><li>Assume that a fiber is viewed as a chain of ‘links’ or ‘units’ having each the same probability of failure F(  ) under a stress  . </li></ul><ul><li>Probability of survival of a link is 1 – F(  ) </li></ul><ul><li>Probability of survival of a chain (= n links) is [1 – F(  )] n </li></ul><ul><li>Probability of failure of the chain is </li></ul><ul><li>F n (  ) = 1 - [1 – F(  )] n </li></ul><ul><li>Do this : insert a Weibull distribution for F (thus for a link) and observe that Weibull is again obtained for F n (the fiber), with the same  but lower  The larger the specimen, the higher its P of failure! </li></ul>
  15. 15. <ul><li>1. INTRODUCTION – GENERAL PRINCIPLES AND BASIC CONCEPTS </li></ul><ul><li>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} </li></ul><ul><li>2. MATERIALS FOR COMPOSITES: FIBERS, MATRICES </li></ul><ul><li>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} </li></ul><ul><li>3. THE PRINCIPLES OF FIBER REINFORCEMENT </li></ul><ul><ul><li>Stress transfer; The model of Cox; The model of Kelly & Tyson; Other model; {Lectures 6-7} </li></ul></ul><ul><li>4. INTERFACES IN COMPOSITES </li></ul><ul><ul><li>Basic issues, wetting and contact angles, interfacial adhesion, the fragmentation </li></ul></ul><ul><ul><li>phenomenon, microRaman spectroscopy, transcrystalline interfaces, {Lectures 8-9} </li></ul></ul><ul><li>5. FRACTURE PHYSICS OF COMPOSITES </li></ul><ul><li>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} </li></ul><ul><li>6. DESIGN EXAMPLE </li></ul><ul><li>A composite flywheel {Lecture 13} </li></ul><ul><li>7. THE FUTURE: </li></ul><ul><li>Composites based on nanoreinforcement, composites based on biology, ribbon- and platelet-reinforced materials, biomimetic concepts {Lectures 14-15} </li></ul>
  16. 16. Last topics in the ‘Basic Concepts’ Section <ul><li>Stress and strain – brief review of definitions </li></ul><ul><li>Thermodynamics of deformation and Hooke’s law </li></ul><ul><li>Anisotropy and elastic constants – Relevance to composite materials </li></ul><ul><li>Stress </li></ul><ul><li>(old concept – Hooke in the 1680s; Cauchy & Poisson in the 1820s) </li></ul><ul><li>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 . </li></ul>
  17. 17. Stress (ctd) <ul><li>Stress = force/area </li></ul><ul><li>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: </li></ul>
  18. 18. <ul><li>Stress is a tensor with 9 components (  ij ) </li></ul><ul><li>First subscript (i) gives the normal to the plane on which stress acts; Second subscript defines the direction of the stress. </li></ul><ul><li>The  ii components are called normal stresses, the  ij components are called shear stresses. </li></ul><ul><li>Tensile stresses are positive, compressive stresses are negative. </li></ul><ul><li>It can be shown from force equilibrium considerations, that the shear stress components are related by  ij =  ji (i≠j). </li></ul><ul><li>Therefore, we have only 6 independent components of stress. </li></ul><ul><li>Knowledge of all components allow us to define the stress acting on any plane within the body. </li></ul>
  19. 19. Strain <ul><li>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. </li></ul><ul><li>Strain is dimensionless. </li></ul><ul><li>Like stress, strain is a tensor with 9 components, 6 of them only being independent because  ij =  ji (i≠j). </li></ul>
  20. 20. <ul><li>Thermodynamics of deformation – The origin of Hooke’s law </li></ul><ul><li>We assume small deformations in a body; Those deformations occur slowly so that thermodynamic equilibrium can be assumed. </li></ul>
  21. 21. <ul><li>Internal stresses are set up within the body, due to deformation. </li></ul><ul><li>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.) </li></ul><ul><li>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), </li></ul><ul><li>Thus, we have: </li></ul><ul><li>(per unit volume). </li></ul>
  22. 22. By definition, the (Helmholtz) free energy of the body is  = 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,  , as a function of  ik
  23. 23. This is easily calculated: since we have small deformations,  can be expanded in a Taylor series: where  0 is the free energy of the undeformed body, and the  ’s are given as follows:
  24. 24. By differentiating, we obtain: 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 : by limiting the expansion to the second order:  ~  2
  25. 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 Hooke’s law ! Also, remember the connection between Young’s modulus and the potential from the previous class?
  26. 26. A common general form (valid for anisotropic bodies) of Hooke’s law is the following: Where  ij and  kl are 2d rank tensors and C ijkl is a 4 th rank tensor with 3x3x3x3 = 81 components [or 9 stress components x 9 strain components = 81]. The C ijkl are called the elastic constants. Historical parenthesis: Robert Hooke’s legacy
  27. 27. Elementary concepts of mechanics <ul><li>In 1676, Robert Hooke makes a discovery about springs: </li></ul>
  28. 28. “ ut tensio, sic vis” (load ~ stretch) UNDER TENSION:
  29. 29. Similarly, under bending: load ~ deflection
  30. 30. Under shear… … and torsion: load ~ shear deformation load ~ angular deformation
  31. 31. Thus, in all cases, Hooke observed: The ratio applied force/distortion is a constant for the material. This is (almost) Hooke’s Law
  32. 32. Hooke’s Law This definition is valid whatever the mode of testing (tension, bending, torsion, shearing, hydrostatic compression, etc, and a specific modulus is then defined)
  33. 33. The stress-strain curve
  34. 34. A general stress-strain curve Elastic (fully reversible) Plastic (irreversible)
  35. 35. Comparing various stress-strain curves
  36. 36. <ul><li>We focus on Hooke’s law for various special cases of material symmetry </li></ul><ul><li>There are 9 x 9 = 81 components of C ijkl but we know that only 6 x 6 = 36 of these are independent. </li></ul><ul><li>It can also be shown that provided that a strain free energy function exists, the number of distinct elastic constants reduces to 21 because C ijkl = C klij . </li></ul>
  37. 37. <ul><li>The number of elastic constants can be further reduced as it depends on the crystalline class : </li></ul><ul><li>Generally anisotropic materials ( triclinic ) possess 21 independent elastic constants. </li></ul><ul><li>Monoclinic systems (one plane of elastic symmetry) have 13 non-zero independent moduli. </li></ul><ul><li>Orthorombic crystals (3 planes of symmetry perpendicular to each other) have 9 moduli ( remember polyethylene ?). They are termed ‘ orthotropic ’ in the composite materials community. </li></ul>
  38. 38. 2 5 Isotropic 4 5 Transversely isotropic 4 5 Orthorombic [Orthotropic] 6 9 Monoclinic 6 9 Triclinic 2D 2 12 Isotropic 5 12 Transversely isotropic 9 12 Orthorombic [Orthotropic] 13 20 Monoclinic 21 36 Triclinic 3D Number of independent coefficients Number of nonzero coefficients Class of Material
  39. 39. (From J.F. Nye, ‘Physical Properties of Crystals’) Form of the C ijkl matrix
  40. 41. Notations:  ij = C ijkl  kl (i, j, k, l = 1,2,3)  ij = S ijkl  kl Historical paradox: The C ijkl are called the Stiffness components The S ijkl are called the Compliance components Contracted notations in the mechanics of composites:
  41. 42. Contracted notations in the mechanics of composites:
  42. 43. Orthotropic lamina (9 constants) <ul><li>Observations: </li></ul><ul><li>There are no interactions between normal stresses and shear strains </li></ul><ul><li>There are no interactions between shear stresses and normal strains </li></ul>
  43. 45. Transversely isotropic lamina (5 constants)
  44. 46. Isotropic lamina (2 constants)
  45. 47. Orthotropic material under plane stress <ul><li>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 </li></ul><ul><li>Inserting this into the orthotropic stress-strain relation, we obtain (after some manipulations): </li></ul><ul><li>where (i,j = 1, 2, 6) (4 independent constants) </li></ul>
  46. 48. How can we derive relations between mathematical and engineering constants ? <ul><li>Stress-strain relations presented before have more physical/intuitive meaning when expressed in terms of familiar engineering constants such as the Young’s modulus. </li></ul><ul><li>Formal connections between mathematical and engineering constants are derived from ‘elementary experiments’. For example: </li></ul>
  47. 49. Elementary experiments
  48. 50. Remember: an orthotropic lamina (9 constants)
  49. 51. Example <ul><li>Uniaxial tensile stress in (say) transverse direction (2) causes a strain in the ‘2’ direction: </li></ul><ul><li>but also in the ‘1’ and ‘3’ directions: </li></ul>From which we obtain:
  50. 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):
  51. 53. <ul><li>From the symmetry of the compliance matrix S ij , we conclude that: </li></ul>In general, we conclude that the relations between the compliances S ij and the engineering constants are simple. It can be shown that the relations between the stiffnesses C ij and the engineering constants are more complicated.
  52. 54. Finally, the connection between the stiffness constants C ij and the compliance constants S ij are as follows:
  53. 55. Last remarks <ul><li>In the case of a transversally isotropic material with the 2-3 plane as plane of isotropy, we have: </li></ul><ul><li>E 2 = E 3 </li></ul><ul><li>G 12 = G 13 </li></ul><ul><li> 12 =  13 </li></ul><ul><li>The 3 engineering constants in the isotropic case are related by </li></ul><ul><li>therefore, as necessary, only 2 constants are independent. </li></ul>

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