UCSD NANO106 - 08 - Principal Directions and Representation Quadrics
•
4 likes•4,150 views
NANO106 is UCSD Department of NanoEngineering's core course on crystallography of materials taught by Prof Shyue Ping Ong. For more information, visit the course wiki at http://nano106.wikispaces.com.
Recommended
More Related Content
What's hot
What's hot (20)
Viewers also liked
Viewers also liked (20)
Similar to UCSD NANO106 - 08 - Principal Directions and Representation Quadrics
Similar to UCSD NANO106 - 08 - Principal Directions and Representation Quadrics (20)
More from University of California, San Diego
More from University of California, San Diego (6)
Recently uploaded
Recently uploaded (20)
UCSD NANO106 - 08 - Principal Directions and Representation Quadrics
- 1. Principal Directions and Representation Quadrics Shyue Ping Ong Department of NanoEngineering University of California, San Diego
- 2. Summary of rank 2 tensors for all crystal systems ¡ Note that rank 2 tensors take these forms only when the measurement axes are based on the crystal axes as per the analysis. In a general direction, a rank 2 tensor has 9 non-zero components (though considerably fewer independent components, as we shall see later) NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 2 σij = σ11 σ12 σ13 σ21 σ22 σ23 σ31 σ32 σ33 ! " # # ## $ % & & && Triclinic σij = σ11 0 σ13 0 σ22 0 σ31 0 σ33 ! " # # ## $ % & & && Monoclinic σij = σ11 0 0 0 σ22 0 0 0 σ33 ! " # # ## $ % & & && Orthorhombic σij = σ11 0 0 0 σ11 0 0 0 σ33 ! " # # ## $ % & & && Tetragonal, Trigonal, Hexagonal Cubic σij = σ11 0 0 0 σ11 0 0 0 σ11 ! " # # ## $ % & & &&
- 3. Principal Directions ¡ A general rank 2 tensor is simply a matrix with 9 non-zero elements. ¡ However, in a certain basis, a rank 2 tensor will take the form ¡ These basis vectors are known as the principal axes. Note that these axes are not always orthogonal, unlike our assumption thus far. ¡ When a “force” is applied in the principal directions, the response is always parallel to the applied “force”. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 3 σij = σ11 0 0 0 σ22 0 0 0 σ33 ! " # # ## $ % & & &&
- 4. Finding the Principal Axes ¡ Finding the principal axes is essentially an eigenvalue problem. You are trying to find vectors where: ¡ Where we have used T to denote a general tensor and λ is the eigenvalue. Rewriting the above equation, we have ¡ To obtain useful solutions, we look for values satisfying the characteristic equation: NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 4 Tv = λv (T − λI)v = 0 det(T − λI) = 0
- 5. Example ¡ Consider the following tensor with the form: NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 5 σij = 5 1 1 1 3 1 1 1 3 ! " # # # $ % & & & Blackboard
- 6. Magnitude of Tensor in a Given Direction ¡ Consider the electrical conductivity: ¡ A typical experimental measurement may involve applying the electric field (voltage / unit length) in a particular direction (e.g., [111]) and then measuring the current density (charge / time / area) in the same direction. Note that j is in general not parallel to E, and the value of the tensor is not given by ¡ Instead, we define the value of the tensor along that direction as the component of j in the direction of E, divided by the magnitude of E, i.e. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 6 ji =σikEk j E j111 E
- 7. Magnitude of Tensor in a Given Direction NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 7 Consider j=σE The experiment measures the component of j in the direction of E. This is given by the dot product of j with the unit vector in the direction of E. Hence, the tensor σ in the direction of E is given by σ[E] = j⋅E E ⋅ 1 E = σE⋅E E2 = σikEkEi E2 Now, if E = E(l1, l2, l3), where li are the direction cosines for the vector E. Therefore, σ[E] =σiklilk The magnitude of a tensor in a given direction is specified by the direction cosines of that direction.
- 8. Example ¡ Let’s say we are interested in the value of a tensor along the [100] direction. We have ¡ Hence, the diagonal elements of the tensor is always the value of the tensor along the three orthogonal axis directions. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 8 Blackboard
- 9. The representation quadric ¡ The representation quadric is a useful geometric representation of a rank 2 tensor property. ¡ Consider the following equation: ¡ This equation can be plotted as a 3-dimensional surface that completely describes the tensor T. ¡ To make this easier to handle, let us consider the form of the tensor in the principal axes, in which case the off- diagonal elements of T are zero. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 9 Tij xi xj =1 T11x1 2 +T22 x2 2 +T33x3 2 +(T12 +T21)⋅ x1x2 +(T13 +T31)⋅ x1x3 +(T23 +T32 )⋅ x2 x3 =1
- 10. The representation quadric ¡In the principal directions, the representation quadric equations becomes: ¡Which can be rewritten as: NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 10 T11x1 2 +T22 x2 2 +T33x3 2 =1 x1 2 1/T11 + x2 2 1/T22 + x3 2 1/T33 =1
- 11. Forms of the representation quadric NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 11 Ellipsoid, T11,T22,T33 > 0
- 12. Forms of the representation quadric NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 12 Hyperboloid of 1 sheet, T11 < 0, T22,T33 > 0
- 13. Forms of the representation quadric NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 13 Hyperboloid of 2 sheets, T11,T22 < 0, T33 > 0
- 14. Interpreting the representation surface ¡ Recall that the representation surface is given by: ¡ Let us now write , where li is the ith direction cosine for x and r is the radius. We get ¡ Recall that we earlier showed that the value of a tensor in a given direction is given by ! Hence, using the representation surface, we can geometrically obtain the value of a tensor in a direction simply by measuring the radius! NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 14 Tij xi xj =1 xi = rli Tijrlirlj =1 Tijlilj = 1 r2 Tijlilj
- 15. The radius normal property ¡ Besides just the magnitude, we can also geometrically determine the direction of the response. We will do this analysis again in the principal axes to simplify the math. Consider the following: NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 15 p = Tq or pi = Tijqj Let qj = qlj Consider a point on the representation surface such that OQ is parallel to q. This point is given by r(l1, l2, l3). The tangent plane to this point is given by (substitute r(l1, l2, l3) into the quadric eqn).: T11x1rl1 +T22 x2rl2 +T33x3rl3 =1 The normal to this plane is therefore r(T11l1,T22l2,T33l3), which is parallel to p=q(T11l1,T22l2,T33l3)!
- 16. Example ¡ We will use an ellipsoid quadric surface (all principal values >0, which is the most common form for property tensors). The analysis will be done in 2D (assume qz component is 0) to facilitate visualization. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 16 R Let's say the generalized force q is in the direction indicated by the blue arrow, with magnitude q. Using the representation quadric, we can determine that the magnitude of the tensor in the direction of q is given by T//q = 1 R2 The magnitude of the response p in the direction of q is therefore p//q = q R2 The direction of the actual response is given by the purple arrow.
- 17. Example ¡Consider the following tensor: ¡Draw the projection of the representation quadric in the X1-X2 plane. ¡Determine from the representation quadric the value of the tensor in the [110] direction and the direction of the response. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 17 σij = 5 0 1 0 3 1 1 1 3 ! " # # # $ % & & &
- 18. Stress & strain ¡ Before embarking on the journey beyond rank 2 tensors, we will now go in-depth into stress (force / unit area) and strain (change in length / unit length). Both of these are rank-2 tensors, but they are not property tensors. Typically, they are known as field tensors to distinguish them from property tensors. ¡ Unlike property tensors, crystal symmetry restrictions do not apply to stress and strain. ¡ We will subsequently discuss higher order tensors that relate stress and strain to other measurements. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 18
- 19. Stress ¡ The stress tensor describes the force acting on a specimen. ¡ First subscript refers to the direction of the force, the second to the normal to the face on which the force acts. To prevent translational motion, each force is balanced by an equal and opposite force on the reverse side of the specimen. ¡ Xii are tensile stresses in which both the force and the normal are along Zi, and Xij are shear stresses in which a force along Zi acts on a face normal to Zj. ¡ Static equilibrium requirement (balanced torques) => Stress tensor must be symmetric with Xij = Xji. ¡ Stress state is specified by six independent components: three tensile stresses X11, X22, and X33, and three shear components X12, X13, and X23. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 19 σij = σ11 σ12 σ13 σ12 σ22 σ23 σ13 σ23 σ33 ! " # # ## $ % & & &&
- 20. Stress Quadric ¡ As stress is a rank-2 tensor, it can be represented as a quadric. ¡ Everything we have derived about the representation quadric for rank-2 tensors applies to the stress quadric as well: ¡ The square of the radius to a point is the reciprocal of the value of stress in that direction, given by . Note that for the stress tensor, the direction cosines is based on the normal to the surface of the generalized “displacement”. ¡ The radius normal property applies as well. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 20 σij xi xj =1 σijlilj li
- 21. Forms of Stress Tensor NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 21 σij = σ11 0 0 0 0 0 0 0 0 ! " # # ## $ % & & && Uniaxial tension/compression σij = σ11 0 0 0 σ22 0 0 0 0 ! " # # ## $ % & & && Biaxial stress σij = σ11 0 0 0 σ22 0 0 0 σ33 ! " # # ## $ % & & && Triaxial stress σij = −p 0 0 0 −p 0 0 0 −p " # $ $ $$ % & ' ' '' Hydrostatic pressure σij = σ 0 0 0 −σ 0 0 0 0 " # $ $ $ % & ' ' ' Pure shear σij = 0 !σ 0 !σ 0 0 0 0 0 " # $ $ $ % & ' ' '
- 22. Strain ¡ Strain is the fractional change in shape of a specimen. ¡ Like stress, strain is represented by a symmetric rank 2 tensor. ¡ The symmetry of the strain tensor arises from the elimination of pure rotations that do not involve any deformation of the object. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 22 εij = δui δxj = ε11 ε12 ε13 ε12 ε22 ε23 ε13 ε23 ε33 ! " # # ## $ % & & &&
- 23. Displacement tensor ¡ To illustrate the symmetry of the strain tensor, let us consider a generalized fractional displacement tensor denoted by e (note that this is different from the symbol used for strain!). e is in general not symmetric. ¡ For any non-symmetric tensor, we can always write it in terms of the sum of a symmetric tensor ε and an anti- symmetric tensor w as follows: NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 23 eij =εij + wij where εij = 1 2 (eij +eji ) and wij = 1 2 (eij −eji )
- 24. Rigid body rotations ¡ For a rigid body rotation, we have a displacement vector u that is always perpendicular to the vector r everywhere. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 24 u r Blackboard
- 25. Definition of strain ¡Rigid body rotations correspond to anti-symmetric displacement tensors. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 25 eij =εij + wij rotationActual strain Graphical example using pure shear strain
- 26. Dilation ¡ What is the fractional volume change of a strained specimen? ¡ For simplicity, we will again consider a diagonalized strain tensor and work in the principal axes. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 26 Blackboard
- 27. Forms of the strain tensor NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 27 εij = ε11 0 0 0 0 0 0 0 0 ! " # # ## $ % & & && Tensile or compressive strain εij = ε11 0 0 0 ε22 0 0 0 0 ! " # # ## $ % & & && Plane strain εij = ε 0 0 0 −ε 0 0 0 0 " # $ $ $ % & ' ' ' Shear strain (special case of plane strain!) Note that pure shear has no dilation, in line with our intuition εij = 0 !ε 0 !ε 0 0 0 0 0 " # $ $ $ % & ' ' '
- 28. Voigt notation ¡ Method of representing symmetric tensors by reducing its order. ¡ Instead of rank 2 tensors (matrices), stress and strain can be represented as a 6-element vectors. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 28 σij = σ11 σ12 σ13 σ12 σ22 σ23 σ13 σ23 σ33 ! " # # ## $ % & & && → σ1 σ2 σ3 σ4 σ5 σ6 ! " # # # # # # # # $ % & & & & & & & & = σ11 σ22 σ33 σ23 σ13 σ12 ! " # # # # # # # # $ % & & & & & & & & , εij = ε11 ε12 ε13 ε12 ε22 ε23 ε13 ε23 ε33 ! " # # ## $ % & & && → ε1 ε2 ε3 γ4 γ5 γ6 ! " # # # # # # # # $ % & & & & & & & & = ε11 ε22 ε33 2ε23 2ε13 2ε12 ! " # # # # # # # # $ % & & & & & & & &
- 29. Voigt notation, contd. ¡ Note that the off-diagonal strain elements has a factor of 2 in the vector notation. γij = 2εij are known as the engineering shear strains. ¡ The benefit of this representation is that the following scalar invariance is preserved: NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 29 E =σij ⋅εij = σ1 σ2 σ3 σ4 σ5 σ6 " # $ $ $ $ $ $ $ $ % & ' ' ' ' ' ' ' ' ⋅ ε1 ε2 ε3 γ4 γ5 γ6 " # $ $ $ $ $ $ $ $ % & ' ' ' ' ' ' ' '
- 30. Voigt notation for higher order tensors ¡The same principle can be (and in fact is frequently) applied to higher order tensors. ¡For example, the rank 3 piezoelectric tensor is represented as a 6x3 matrix (or 3x6 in the case of the converse piezoelectric effect), while the elastic and compliance moduli tensor are represented as 6x6 matrices. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 30
- 31. Electric polarization ¡ The electric polarization is the vector field that expresses the density of permanent or induced electric dipole moments in a dielectric material. When a dielectric is placed in an external electric field, its molecules gain electric dipole moment and the dielectric is said to be polarized. NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 31 +-
- 32. Electric polarization NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 8 32 +- d Electric dipole moment p = qd where d is the displacement vector Polarization P = p V (average dipole moment per unit volume) ¡ Electric polarization is a rank 1 field tensor.