Engineering Tensors A BEH430 review session by Thomas Gervais [email protected] References: • Long, RR, Mechanics of Solids and Fluids, Prentice-Hall, 1960, pp 1-32 • Deen, WD, Analysis of transport phenomena, Oxford, 1998, p. 551-563 • Goodbody, AM, Cartesian tensors: With applications to Mechanics, Fluid Mechanics and Elasticity, John Wiley & Sons, 1982 Friday November 16, 2001 16h30 -Muddy Charles Scalars, Vectors & Tensors Scalar: Quantity that is invariant in itself (does not depend on any referential) Also known as a zeroth order tensor. Ex: mass (non relativistic referentials), Temperature, Energy, Concentration Vector: Quantity that possess both a direction and a magnitude located somewhere in space. It is a first vorder tensor. v =a1eˆ1 + a2eˆ2 + a3eˆ3 = a'1 eˆ'1 +a'2 eˆ'2 +a'3 eˆ'3 Note: A vector possessesv two invariants with respect to coordinate space: Its magnitude, v , and its direction. Ex: Force, Electric field, Flux Higher order tensors 2nd order tensor: An ordered set of nine numbers, each of which possessing 2 directions. An helpful analogy would be to imagine a vector whose three components each would be a vector. A tensor of order n can be reprensented by 3n numbers in tridimensional space. It is a matrix of dimension n. A 2nd possesses 3 invariants which are the coefficients of its characteristic polynomial arising from det T - l I = 0 . The trace and the determinant of the tensor are two of its 3 invariants. In general, a tensor of order n will have n+1 invariants. Physical applications of tensors: Discipline Phenomenon Quantity Example Physics: E-M Light transmission in Refraction index n Optically active anisotropic media polarizers Dynamics Angular momentum Moment of Inertia I Gyroscope Fluid mechanics Hindered transport in Effective Ca++ diffusion in porous media permeability K muscles Solid mechanics Tensile properties of Young’s modulus E Duct Tape, anisotropic solids Muscles Dyadic Product (Tensor Product) This the general form of a tensor product: Scalar A (0th order tensor) v v v v A = A e + A e + A e Vector 1 1 2 2 3 3 (1st order tensor) v v v v v v v v A = A e + A e + A e A = A e + A e + A e 2nd order 1 11 1 21 v2 31 3 3 13 1 23 2 33 3 v v v tensor A2 = A12e1 + A22e2 + A32e3 etc. 3 3 • By dividing the Aij in three vector components, A = A eˆ eˆ we get a 3rd order tensor, and so on... å å ij i j i=1 j =1 Notation z z’ v v 3 3 A A = å Aieˆi = å A'i eˆ'i i=1 i=1 y The projection of the vector on the axis, x, y, z can each be decomposed in the basis (x’, y’, z’). The relation y’ between each of the components is then given by: x’ x 3 = Ax = a11A'x + a21A' y + a31A'z Ai å aij A' j j =1 Ay = a12 A'x + a22 A' y + a32 A'z Az = a13A'x + a23A' y + a33A'z Ai = aij A' j (Einstein’s notation) é a a a ù 3 3 11 12 13 ê ú a = a eˆ eˆ = a a a å å ij i j ê 21 22 23ú i=1 j =1 ëê a31 a32 a33ûú Dot and Cross products Definition of dot product: v v 3 3 3 3 3 3 3 A× B = å Aieˆi × å B jeˆ j = å å AiB j eˆi × eˆ j = å å AiB j d ij = å AiBi i=1 j =1 i=1 j =1 i=1 j =1 i=1 nd Where dij is the Kronecker delta, a 2 order tensor. Does it hold in tensor notation? 3 Let’s test it using a change of coordinate: Ai = å aij A' j j =1 v v 3 3 3 3 3 3 3 3 A × B = å å aki A'k eˆk × å å alj B'l eˆl = å å å å akialj A'k B'l d kl i=1 k =1 j= 1 l=1 i=1 j =1 k =1 l=1 If what we said about the conservation of the magnitude of a vector from v v 3 3 one cartesian referential to another, then A× B = å Ai Bi = å A' k B' k i=1 k =1 For this to be true, we need: d ij = akialjd kl (Einstein notation) Dot and Cross product (cont.) d ij = akialjd kl expanded yields: 2 2 2 a11 + a12 + a13 = 1 a21a31 + a22 a32 + a23a33 = 0 2 2 2 a21 + a22 + a23 = 1 a31a11 + a32 a12 + a33a13 = 0 2 2 2 a31 + a32 + a33 = 1 a11a21 + a12a22 + a13a23 = 0 These are the normalization and the orthogonality conditions that any orthonormal base respects. Definition of cross product: v v 3 3 3 3 3 3 3 A× B = å Aieˆi ´ å B jeˆ j = å å AiB j eˆi ´ eˆ j = å å å AiBj e ijkeˆk i=1 j =1 i=1 j =1 i=1 j=1 k =1 Where e ijk is called the Levi-Civita density and is the cross product equivalent of the Kroneker delta, d ij , for the dot product. It is a third order tensor. 0, i=j, j=k, or i=k This is the famous e ijk 1, ijk= 123, 231, 312 “right hand rule” -1, ijk= 132, 321, 213 Gradients and Divergence v v 3 ¶f 3 We know: Ñ f = å eˆj and A = å Aieˆi ¶ j =1 x j v iv=1 Q: What about the gradient of a vector Ñ A ? A: The generalizationv of a vector provided by tensor analysis implies: v v 3 ¶ A 3 3 ¶A i A second order tensor! Ñ A = å eˆ j = å å eˆieˆj j =1 ¶ x j i=1 j=1 ¶ x j • The gradient of a tensor increases its order by one Q: What about the divergence of a 2nd order tensor? A: Using our definition of the dot product: v 3 3 ¶ 3 Tij ¶Tii Ñ × T = å å d ijeˆieˆ j = å eˆi A vector! i=1 j=1 ¶ x j i=1 ¶xi • The divergence of a tensor decreases its order by one Practical problem: How to set up a tensor from physical reasoning? Consider the following problem: We want to stretch a piece of anisotropic tissue and find the components of the Young’s Moduli v v E2 F F E1 q Apply a given displacement and compute the force as a function of the angle (assuming constant strain and a Poisson coefficient of “0”) v 2 y' v 2 2 2 u = u eˆ å i i F = å Fieˆ'i = å å Eijuieˆj i=1 E2 x' i=1 i=1 j =1 E1 y q é F1 ù é E11 E12 ù é u1 ù x ê ú = ê ú ê ú ë F2 û ë E21 E22 û ëu2 û Practical problem: How to set up a tensor from physical reasoning? (cont.) We need to compute the 4 matrix elements: • F1 ® u1 : E11 = Ex cosq • F1 ® u2 : E12 = Ex sin q é F1 ù é Ex cosq Ex sinq ù é u1 ù = ê ú • F ® u : E = - E sin q ê ú - q q ê ú 2 1 21 y ë F2 û ë E y sin Ey cos û ëu2 û • F2 ® u2 : E22 = E y cosq We are measuring the magnitude of the force required to impose a unity displacement in the x direction. 2 v 1.8 é F ù é1ù x' 2 2 1.6 = [Eij ] Þ F = F + F ê ú ê ú x ' y ' 1.4 Fy' ë0û ë û 1.2 Effective modulus v 1 0 0.5 1 1.5 F angle (rad) = E = E 2 cos2 q + E 2 sin 2 q u eff x y Normalized effective modulus as a function of the angle for E2 double of E1 Same problem: other approach using tensors v v 2 y' 2 2 2 F = F eˆ' = E' u' eˆ' u = å uieˆi å i i å å ij j j i=1 i=1 j=1 i=1 E2 x' E1 é E1 0 ù y E'ij = ê ú q ë 0 E2 û x 2 u'i eˆ'i = aiju jeˆ j å v 2 2 j =1 F = å å E'ij aiju jeˆ j é cosq sinq ù i=1 j =1 aij = ê ú ë - sin q cosq û Q: Are expressions equivalent, i.e. E'ij aij = Eij ? E 0 cosq sinq E cosq E sin q é 1 ù é ù é 1 1 ù A: Yes! E'ij aij = ê ú ê ú = ê ú ë 0 E2 û ë - sin q cosq û ë - E2 sinq E2 cosq û Fluid mechanics: The stress and rate of strain tensors v 3 vz' A) Stress tensor y' S = å Sieˆi v n i=1 z S x' 3 v = × ˆ ˆ y Si å n (Siei )e j j=1 x Stress vector v B) Rate of strain tensor Sij = s ij = n × (Sieˆi )eˆieˆ j v v v v v v(r,t) v(r + d r,t) By Taylor expansion to the first order: v v v v v v v v v v v t v = v(r,t) + d v d v = d r × Ñ v(r,t) = (Ñ v(r,t)) × d r v v v Ñ v(r,t)is a tensor (gradient of a vector)! v v v v & 1 t v v e = [(Ñ v) + (Ñ v)] t & 2 (Ñv ) = e + x v v v v & 1 t x = [(Ñ v) - (Ñ v)] Vorticity tensor 2 Conclusion • 2nd order tensors establish relation between two sets of vectors.