Symmetry and Properties of Crystals (MSE638) Stress and Strain Tensor

Transcription

1 Symmetry and Properties of Crystals (MSE638) Stress and Strain Tensor Somnath Bhowmick Materials Science and Engineering, IIT Kanpur April 6, 2018

2 Tensile test and Hooke s Law Upto certain strain (0.75), ratio of stress and strain is a constant and the slope of the line is called the modulus of elasticity or Young s modulus. 2 / 23

3 How to define stress? How many components do we need to specify mass, energy, temperature, specific heat? One, because they are scalars. How many components do we need to specify force, displacement, temperature gradient, electric field? Three, because they are vectors. How many components do we need to specify stress (force per unit area)? Nine, because it is a second rank tensor. Stress tensor σ ij i direction of force j direction of plane normal σ ii normal stress σ ij shear stress (τ ij ) Equal and opposite force on reverse face prevents translational motion 3 / 23

4 What is a tensor? In a 3D space, a tensor of rank n has 3 n components Scalar is a rank 0 tensor (no subscript, 1 component) Vector is a rank 1 tensor (one subscript, 3 components) Example: Force F = (F 1, F 2, F 3 ), Area A = (A 1, A 2, A 3 ) Stress is a rank 2 tensor (two subscripts, 9 components) σ 11 σ 12 σ 13 σ 11 τ 12 τ 13 σ 11 τ 12 τ 13 σ 21 σ 22 σ 23 = τ 21 σ 22 τ 23 = τ 12 σ 22 τ 23 σ 31 σ 32 σ 33 τ 31 τ 32 σ 33 τ 13 τ 23 σ 33 Stress tensor is symmetric: ensures static equilibrium (net torque 0) Relation between force, area and stress: F i = j σ ija j = σ ij A j Einstein summation convention: sum over repeated indices i free index and j dummy index Important: two first rank tensors are related by a second rank tensor 4 / 23

5 Plane stress σ₂₂ τ₁₂ x₂ τ₂₁ σ₁₁ x₁ σ₁₁ τ₂₁ τ₁₂ Let the stress tensor be: σ = σ₂₂ [ σ11 = 8 ] τ 12 = 3 τ 21 = 3 σ 22 = 4 5 / 23

6 Eigenvalues and eigenvectors Find the eigenvalues and eigenvectors of the matrix [ ] 8 3 σ = 3 4 Eigenvalues are determined by solving det(σ λi) = 0 Eigenvalues are λ 1 = 9.6 and λ 2 = 2.4 Eigenvectors are determined by solving (σ λ i I)ˆx i = 0 [ ] λ 1 = 9.6, ˆx = λ 2 = 2.4, ˆx 2 = 0.47 ] [ ˆx 1 ˆx 2 = 0 they can be used as reference vectors New coordinate system rotated w.r.t. old ˆx 1 ˆx 1 = 28.36, ˆx 1 ˆx 2 = etc. How does the stress tensor look like in rotated coordinate system? 6 / 23

7 Principal stress σ₂₂ σ₂ τ₁₂ σ₁₁ τ₂₁ x'₂ τ₁₂ x₂ σ₁ x'₁ σ₂₂ x₁ σ₂ τ₂₁ σ₁ σ₁₁ Principal stress: σ = [ σ1 = 9.6 τ = 0 ] τ = 0 σ 2 = 2.4 Rotating coordinate system is useful - let s do it systematically 7 / 23

8 Eigenvectors as reference axes Vector ˆx, when multiplied by Matrix σ, is rotated to ˆx Eigenvectors are special such that ˆx and ˆx overlap Symmetric matrix:use eigenvectors ˆx 1 and ˆx 2 as reference vectors In this new coordinate system [ ] [ ] [ ] [ ] , Note that, coordinate axes does not rotate further if we multiply by [ ] Multiplication with reference vectors yields same vector times 9.6/2.4 Components of matrix change upon coordinate transform [ ] [ ] / 23

9 Recap We know tensile test and Hooke s law (stress strain) Stress is a tensor of rank 2 (same to be shown for strain later) Stress tensor is symmetric - only 6 independent components Diagonalize the stress tensor- find eigenvalues and eigenvectors Choose the eigenvectors as reference axes- rotate coordinate system How does the components of stress tensor look like after rotation? Claim: diagonal matix with eigenvalues as diagonal elements Let s prove this 9 / 23

10 Rotation of reference axes x₁ x'₁ x₃ x'₃ x 1 x 2 x 3 x 1 a 11 a 12 a 13 x'₂ x 2 a 21 a 22 a 23 x₂ x 3 a 31 a 32 a 33 Table: Angle between new and old set of axes, a ij = cos(x i x j) Change coordinate system from (ˆx 1, ˆx 2, ˆx 3 ) to ( ˆx 1, ˆx 2, ˆx 3) Orthogonal coordinate system only 3 direction cosines independent How does the coordinate system transform? Forward transform: x i = a ijx j and inverse transform: x i = a ji x j [a ij ] 1 = [a ji ] = [a ij ] T Thus, [a ij ] is an orghogonal matrix 10 / 23

11 How does the components of a vector change? x'₃ v₁ x₁ x'₁ v₃ x₃ v v₂ x'₂ x₂ x 1 x 2 x 3 x 1 a 11 a 12 a 13 x 2 a 21 a 22 a 23 x 3 a 31 a 32 a 33 Table: Angle between new and old set of axes, a ij = cos(x i x j) Change coordinate system from (ˆx 1, ˆx 2, ˆx 3 ) to ( ˆx 1, ˆx 2, ˆx 3) Component of a vector changes from (v 1, v 2, v 3 ) to (v 1, v 2, v 3 ) Length does not change v1 2 + v2 2 + v2 3 = v v v 2 3 How does the components transform? Treat components v 1, v 2, v 3 as vectors and write in ( ˆx 1, ˆx 2, ˆx 3) Forward transform: v i = a ijv j and inverse transform: v i = a ji v j [a ij ] 1 = [a ji ] = [a ij ] T Thus, [a ij ] is an orghogonal matrix 11 / 23

12 How does the components of a rank 2 tensor change? Let p be a rank 1 tensor and forward transform: p i = a ikp k Let q be a rank 1 tensor and reverse transform: q l = a jl q j Remember: two first rank tensors are related by a second rank tensor Let σ be a rank 2 tensor and p k = σ kl q l Now, p i = a ikp k = a ik σ kl q l = a ik a jl σ kl q j Forward transform: σ ij = a ika jl σ kl Reverse transform: σ ij = a ki a lj σ kl List of transformations for tensors of different rank: Rank Forward Reverse 0 φ = φ φ = φ 1 p i = a ijp j p i = a ji p j 2 σ ij = a ika jl σ kl σ ij = a ki a lj σ kl 3 σ ijk = a ila jm a kn σ lmn σ ijk = a li a mj a nk σ lmn 4 σ ijkl = a ima jn a ko a lp σ mnop σ ijkl = a mi a nj a ok a pl σ mnop 12 / 23

13 Apply to stress tensor x'₂ x₂ θ x'₁ x₁ x 1 x 2 x 3 x 1 a 11 a 12 a 13 x 2 a 21 a 22 a 23 x 3 a 31 a 32 a 33 Assume x 3 = z axis Table: a ij = cos(x i x j) a i3 = a 3i = 0, a 33 = 1, a 11 = a 22 = cos θ, a 12 = sin θ, a 21 = sin θ Transform rank 2 stress tensor: σ ij = a ika jl σ kl σ 11 = 1 2 (σ 11 + σ 22 ) 1 2 (σ 22 σ 11 ) cos 2θ + τ 12 sin 2θ σ 22 = 1 2 (σ 11 + σ 22 ) (σ 22 σ 11 ) cos 2θ + τ 12 sin 2θ τ 12 = 1 2 (σ 22 σ 11 ) sin 2θ + τ 12 cos 2θ Put θ = 28.36, σ 11 = 8, σ 22 = 4, τ 12 = 3 σ 11 = σ 1 = 9.6, σ 22 = σ 2 = 2.4, τ 12 = 0 (principal stress!!) Rotate coordinate system & align along eigenvectors [ ] [ ] Imp: trace of stress tensor same before and after transformation 13 / 23

14 Mohr circle τ P 2θ O σ₂ A R (σ₂₂,τ₁₂) C B Q σ σ₁ σ 11 = 1 2 (σ 1 +σ 2 ) 1 2 (σ 2 σ 1 ) cos 2θ σ 22 = 1 2 (σ 1 +σ 2 )+ 1 2 (σ 2 σ 1 ) cos 2θ τ 12 = 1 2 (σ 2 σ 1 ) sin 2θ Invariant: σ 1 + σ 2 = σ 11 + σ (σ₁+σ₂) T (σ₁₁,-τ₁₂) Max, min normal stress: σ 1, σ 2 Max, min shear stress: 1 2 (σ 2 σ 1 ), 0 OC = OP + P Q 2 = σ (σ 1 σ 2 ) = 1 2 (σ 1 + σ 2 ) = 1 2 (σ 11 + σ 22 ) CP = OC OP = 1 2 (σ 1 + σ 2 ) σ 2 = 1 2 (σ 1 σ 2 ) (radius r) CA = CB = 1 2 (σ 1 11 σ 22 ) r = 4 (σ 11 σ 22 ) 2 + τ12 2 = 1 2 (σ 1 σ 2 ) tan 2θ p = 2τ 12 σ 11 σ 22 σ 1 = OC + CQ = 1 2 (σ 11 + σ 22 ) + r = σ max σ 2 = OC CP = 1 2 (σ 11 + σ 22 ) r = σ min τ max = 1 2 (σ 1 2 σ 1 ) = r = 4 (σ 11 σ 22 ) 2 + τ12 2 θ s = θ p + 45 (orientation corresponding maximum shear stress) 14 / 23

15 Representing stress tensor square and column matrix 1=11, 2=22, 3=33, 4=23, 5=13, 6=12 σ 1 = σ 11 σ 11 τ 12 τ 13 σ 2 = σ 22 τ 12 σ 22 τ 23 = σ 3 = σ 33 τ 13 τ 23 σ 33 σ 4 = τ 23 σ 5 = τ 13 σ 6 = τ 12 Stress tensor transforms like: σ ij = a ika jl σ kl and σ ij = a ki a lj σ kl How to write it when stress is represented as a column matrix? σ 1 α 11 α 12 α 13 α 14 α 15 α 16 σ 1 σ 2 α 21 α 22 α 23 α 24 α 25 α 26 σ 2 σ 3 σ 4 = α 31 α 32 α 33 α 34 α 35 α 36 α 41 α 42 α 43 α 44 α 45 α 46 σ 3 σ 4 σ 5 α 51 α 52 α 53 α 54 α 55 α 56 σ 5 σ 6 α 61 α 62 α 63 α 64 α 65 α 66 σ 6 Einstein conventation: σ i = α ijσ j 15 / 23

16 Transformation matrices for stress tensor Elements of α: α mn = α ijkl = a ik a jl + (1 δ kl )a il a jk Elements of α 1 : αmn 1 = α 1 ijkl = a kia lj + (1 δ kl )a kj a li 16 / 23

17 Recap We know tensile test and Hooke s law (stress strain) Stress is a tensor of rank 2 (same to be shown for strain later) Stress tensor is symmetric - only 6 independent components Diagonalize the stress tensor- find eigenvalues and eigenvectors Choose the eigenvectors as reference axes- rotate coordinate system How does the components of stress tensor look like after rotation? Diagonal matix with eigenvalues as diagonal elements (proved) Eigenvalues of stress tensor are known as the principal stresses Mohr circle is an elegant way to represent the stress tensor 17 / 23

18 Strain tensor x₂ x'₂ r r' Consider deformation of a square Consider ˆx 1 x 1 OB=OA+AB=(1+e 11 )OA BC tan θoa = e 21 OA x'₁ C O θ A B x 1 = (1 + e 11)ˆx 1 + e 21ˆx 2 x 2 = e 12ˆx 1 + (1 + e 22 )ˆx 2 r(x 1, x 2 ) moves to r after deformation x₁ r = x 1 x 1 + x 2 x 2 (uniform deformation) Displacement U = r r = x 1 ( x 1 ˆx 1) + x 2 ( x 2 ˆx 2) U( r) = (e 11 x 1 + e 12 x 2 ) ˆx }{{} 1 + (e 21 x 1 + e 22 x 2 ) ˆx }{{} 2 u i = e ij x j u 1 ( r) u 2 ( r) Taylor series expansion: u 1 (0 + x 1, 0 + x 2 ) u 1 (0, 0) + u 1 x 1 x 1 + u 1 x 2 x 2 Comparing: e 11 = u 1 x 1, e 12 = u 1 x 2, e 22 = u 2 x 2, e 21 = u 2 x 1 In general: e ij = u i x j 18 / 23

19 Strain tensor e ij = 1 2 (e ij + e ji ) (e ij e ji ) e ij = ε ij + ω ij ( ) Strain tensor: ε ij = 1 ui 2 x j + u j x i (symmetric) ( ) Rotation tensor: ω ij = 1 ui 2 x j u j x i (anti-symmetric) Normal strain: ε 11 = u 1 x 1, ε 22 = u 2 x 2, ε 33 = u 3 x 3 ( ) Shear strain: γ 12 = γ 21 = 1 u1 2 x 2 + u 2 x 1 ( ) Shear strain: γ 13 = γ 31 = 1 u1 2 x 3 + u 3 x ( 1 ) Shear strain: γ 23 = γ 32 = 1 u2 2 x 3 + u 3 x 2 Note that, x 1 x 2 e 12 + e 21 = u 1 x 2 + u 2 x 1 Thus, shear strain is related to change of angle Normal strain is related to change of volume Combination of two - both angle and volume changes 19 / 23

20 Strain tensor Strain tensor: ε 11 γ 12 γ 13 γ 12 ε 22 γ 23 = γ 13 γ 23 ε 33 u 1 1 ( x 1 ) 2 1 u1 2 x 2 + u 2 x ( u1 x 3 + u 3 x 1 ) 1 2 ( ) ( u1 x 2 + u 2 1 u1 x 1 2 u 2 1 x 2 2 ( ) u2 x 3 + u 3 x 2 ) x 3 + u 3 x 1 ) x 2 ( u2 x 3 + u 3 u 3 x 3 Column representation (1=11, 2=22, 3=33, 4=23, 5=13, 6=12): ε 1 = ε 11 ε 11 γ 12 γ 13 ε 2 = ε 22 γ 12 ε 22 γ 23 = ε 3 = ε 33 γ 13 γ 23 ε 33 ε 4 = γ 23 ε 5 = γ 13 ε 6 = γ 12 Similar to principal stress, we can define principal strain Shear strain=0, normal strain= max and min in principal direction 20 / 23

21 Elastic constant Hooke s law: for small deformations, strain stress ε ij = S ijkl σ kl ε i = S ij σ j, S = Elastic compliance constant ε 1 S 11 S 12 S 13 S 14 S 15 S 16 σ 1 ε 2 S 21 S 22 S 23 S 24 S 25 S 26 σ 2 ε 3 ε 4 = S 31 S 32 S 33 S 34 S 35 S 36 σ 3 S 41 S 42 S 43 S 44 S 45 S 46 σ 4 ε 5 S 51 S 52 S 53 S 54 S 55 S 56 σ 5 ε 6 S 61 S 62 S 63 S 64 S 65 S 66 σ 6 σ ij = C ijkl ε kl σ i = C ij ε j, C = Elastic stiffness constant σ 1 C 11 C 12 C 13 C 14 C 15 C 16 ε 1 σ 2 C 21 C 22 C 23 C 24 C 25 C 26 ε 2 σ 3 σ 4 = C 31 C 32 C 33 C 34 C 35 C 36 ε 3 C 41 C 42 C 43 C 44 C 45 C 46 ε 4 σ 5 C 51 C 52 C 53 C 54 C 55 C 56 ε 5 σ 6 C 61 C 62 C 63 C 64 C 65 C 66 ε 6 Elastic constants are fourth rank tensors 21 / 23

22 Elastic energy density Elastic energy density (similar to a stretched spring): E = 1 2 C ij ε i ε j Stress components found by taking derivative of E w.r.t. associated strain component (force is derivative of potnetial energy) σ i = E ε i σ 1 = C 11 ε ( C 1j + C j1 )ε j j=2 Same stress-strain relationship expressed in another form Comparing with previous form: C ij = 1 2 ( C ij + C ji ) Thus, elastic constant tensor is symmetric 36 elastic constants reduced to / 23

23 Summary 23 / 23