Defining sign of stress tensor Kittel’s Fig. 15 may be confusing about sign of T xx (which he calls X x ) Stress tensor component T xx is defined as the x-component of force, transmitted in –x direction. The compressive force shown (labeled X x ) is in the –x direction (note that the x axis points left!) and is transmitted in the –x direction, so the sign of T xx is negative. Tension would be positive.
Elasticity (Chapter 3) – only 6 of 9 elements are independent. Replace 3x3 matrix by a 6-component column vector: Lecture 21 Oct. 8, 2010 PH 481/581 Recall Young’s modulus Y: Stress = elasticity constant x strain In general, stress and strain are 3x3 matrices, so the most general relation would be Where M has 4 indices (3x3x3x3 =81 elements) Fortunately, not all elements of T and are independent: these matrices are symmetric: ←L+ L → ← L → F F
Elasticity (continued) T is symmetric matrix – only 6 of 9 elements are independent. Replace 3x3 matrix By 6-element column vector: How about strain tensor? Again replace the 3x3 symmetric matrix By a 6-element column vector: Note factor of 2, only in off- diagonal elements. (Later, makes energy formula simpler.) We will use and to represent the composite indices 1..6, and and to represent x,y,z.
Elasticity (continued) Now we can write the most general linear relationship between stress and strain as a matrix equation e = S T: Recalling that the 1..6 indices are short for (xx),...(xy), this is
Elasticity (continued) In the case of cubic symmetry, the axes are equivalent, so that S (xx)(xx) = S (yy)(yy), i.e. S 11 = S 22. Similarly, S (xy)(xy) = S (yz)(yz), so S 66 = S 44. Also, S 13 = S all the off-diagonal elements in the upper left 3x3 matrix are equal. Also, anything like S (xx)(xy) that has a single y index must vanish due to the y↔ -y symmetry. This leads to We call S (in e = S T) the compliance matrix. We denote the inverse matrix (“stiffness matrix”) by C = S -1, so that T = C e. In a system with no symmetry, all 36 components of C and S are independent. Cubic symmetry only The stiffness matrix C has the same form -- a cubic material has 3 independent stiffness coefficients C 11, C 12, and C 44.
Elasticity (continued) Important special case: uniaxial stress, T xx >0, other T = 0. Then This experiment defines Young’s modulus Y = (F/A)/( L/L) = T xx / xx = 1/S 11, as well as Poisson’s ratio P = - yy / xx = -S 12 /S 11. Note that yy < 0 when stress is applied along x only, so S 12 < 0. if cubic symmetry ←L+ L → ← L → F F Elastic Energy: energy/volume U is (C can be chosen to be symmetric)
Elastic Waves (Chapter 3) Lecture 23 Oct. 13, 2010 PH 481/581 Write equation of motion for a volume element from position r, now at displaced position r + R(t): becomes F F But displacement R is related to the strain tensor, which is related to the stress tensor T. Expressing everything in terms of R, Origin r R So F=ma becomes Like a wave equation.
Elastic Waves (continued) Derived a wave equation for displacement R: Solutions: try a plane wave Writing part of the =x component explicitly, This is Eq. (57a) in Kittel. Try k = (K, 0, 0) & R along x: Soand the wave speed is so
Elastic Waves (Chapter 3) Lecture 24 Oct. 13, 2010 PH 481/581 Review: becomes in terms of displacement R – wave equation. Insert a plane wave Try k along x, i.e. (100) & R along 100: v 2 = C 11 & k along (100) & R along 010: v 2 = C 44 & k along (110) & R along 110: v 2 = ½(C 11 + C C 44 ) & k along (110) & R along 001: v 2 = C 44 & k along (110) & R along (1-10): v 2 = ½(C 11 - C 12 ) Condition for isotropy: C 11 - C 12 = 2C 44 Kittel gives table – W is close to isotropic Lamé constants: T = 2 + tr 1; derive C 11 =2 C 12 = C 44 =