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1 Appendix A Useful Formulae ( ) 34
2 Jeremić et al. A.. CHAPTER SUMMARY AND HIGHLIGHTS page: 35 of 536 A. Chapter Summary and Highlights A. Stress and Strain This section reviews small deformation stress and strain measures used in this report. A.. Stress In this work, the tensile stress is assumed positive, and in general we follow classical strength of materials (mechanics of materials) conventions for stress and strain. The stress tensor σ ij is defined as F j σ ij = lim A i 0 A i (A.) where F j is a traction (force) in the j direction and A i is an infinitesimal surface area with normal in i direction. Cauchy stress tensor has a total of nine components, six of which are independent (symmetry σ ij = σ ji ): σ xx σ xy σ xz σ = σ yx σ yy σ yz = σ x σ xy σ zx σ xy σ y σ yz (A.) σ zx σ zy σ zz σ zx σ yz σ z In small deformation theory, this stress is symmetric, that is, σ xy = σ yx, σ yz = σ zy, and σ zx = σ xz. There are only six independent components and sometimes the stress can be expressed in the vector form σ = {σ xx, σ yy, σ zz,σ xy, σ yz, σ zx } (A.3) The principle stresses σ, σ, and σ 3 (σ σ σ 3 ) are the eigenvalues of the symmetric tensor σ ij in Equation A. and can be obtained by solving the equation σ xx σ σ xy σ xz σ yx σ yy σ σ yz = 0 (A.4) σ zx σ zy σ zz σ or in alternative form σ 3 I σ I σ I 3 = 0 (A.5) The three first-type stress invariants are then I = σ ii = σ xx +σ yy +σ zz = σ +σ +σ 3 (A.6)
3 Jeremić et al. A.. STRESS AND STRAIN page: 36 of 536 I = σ ijσ ji = (σ xx σ yy +σ yy σ zz +σ zz σ xx )+(σxy +σyz +σzx) = (σ σ +σ σ 3 +σ 3 σ ) (A.7) I 3 = 3 σ ijσ jk σ ki = det(σ ij ) = σ xx σ yy σ zz +σ xy σ yz σ zx (σ xx σyz +σ yz σzx +σ zz σxy) = σ σ σ 3 (A.8) The stress σ ij can be decomposed into the hydrostatic stress σ m δ ij and deviatoric stress s ij as σ ij = σ m δ ij +s ij, with the definitions σ m = 3 I, s ij = σ ij 3 σ kkδ ij (A.9) where δ ij is the Kronecker operator such that δ ij = for i = j and δ ij = 0 for i j. Since both hydrostatic and deviatoric stresses are stress tensors, they have their own coordinateindependent stress invariants respectively. The three invariants of the hydrostatic stress are I = 3σ m = I, I = 3σ m = 3 I, I 3 = σ 3 m = 7 I3 (A.0) Since I, I and I 3 are all simple functions of I, the hydrostatic stress state can therefore be represented by only one variable I. The three eigenvalues of the deviatoric stresses s ij are called principal deviatoric stresses, with the order s s s 3. The three invariants of the deviatoric stress are J = s ii = 0 J = s ijs ji = 3 I +I = 6 [(σ σ ) +(σ σ 3 ) +(σ 3 σ ) ] = (s xx s yy +s yy s zz +s zz s xx )+(s xy +s yz +s zx) (A.) = (s +s +s 3) = (s s +s s 3 +s 3 s ) (A.) J 3 = 3 s ijs jk s ki = det(s ij ) = I I I + 7 I3 = I 3 3 I J 7 I3 = 7 (σ σ σ 3 )(σ σ 3 σ )(σ 3 σ σ ) = s xx s yy s zz +s xy s yz s zx (s xx s yz +s yy s zx +s zz s xy) = s s s 3 (A.3)
4 Jeremić et al. A.. STRESS AND STRAIN page: 37 of 536 The deviatoric stress state can therefore be represented by only two variables J and J 3. Combining hydrostatic and deviatoric stress, we can conclude that the stress state can be be represented by three variables I, J and J 3. Using the three invariants (I,J,J 3 ) or its equivalents instead of the nine components of σ ij is widely used in geomechanics. The stress state may also be described in three dimensional space (p,q,θ σ ), defined as p = 3 I q = 3J ( θ σ = 3 arccos 3 3 ) J 3 J 3 (A.4) (A.5) (A.6) where θ σij is the stress Lode s angle (0 θ σij π/3). A stress state with θ σ = 0 corresponds to the meridian of conventional triaxial compression (CTC), while θ σ = π/3 to the meridian of conventional triaxial extension (CTE). The relationship between (σ,σ,σ 3 ) and (p,q,θ σ ) is σ σ σ 3 cosθ σ = p+ 3 q cos(θ σ 3 π) cos(θ σ + 3 π) (A.7) The line of the principal stress space diagonal is called hydrostatic axis. Any plane perpendicular to the hydrostatic axis is an deviatoric plane, or π plane. The Haigh-Westergaard three dimensional stress coordinate system (ξ,ρ,θ σ ) Chen and Han (988a), is defined as ξ = I = 3p 3 ρ = J = 3 q (A.8) (A.9) The Haigh-Westergaard invariants have physical meanings. ξ is the distance of the deviatoric plane to the origin of the Haigh-Westergaard coordinates, and ρ is the distance of a stress point to the hydrostatic line and represents the magnitude of the deviatoric stress. The projections of the axes σ, σ and σ 3 on the deviatoric plane are assumed σ, σ and σ 3 respectively. (ρ,θ σ) is the polar coordinate system in the deviatoric plane with the σ the polar axis and θ σ the polar angle. The relationship between (σ,σ,σ 3 ) and (ξ,ρ,θ σ ) is σ σ σ 3 = cosθ σ ξ ρ cos(θ σ 3 π) cos(θ σ + 3 π) (A.0)
5 Jeremić et al. A.. STRESS AND STRAIN page: 38 of 536 A.. Strain Point P(x i ) and nearby point Q(x i +dx i ) displace due to applied loading to new positions P(x i +U i ) and Q(u i +( u i / x j )dx j ). We can define a displacement gradient tensor u i,j as u i,j = u i x j (A.) Matrix form of the displacement gradient can decomposed into the symmetric and antisymmetric parts u, u, u,3 u, u, u, u,3 = (u, +u, ) (u,3 +u 3, ) (u, +u, ) u, (u,3 +u 3, ) u 3, u 3, u 3,3 (u 3, +u,3 ) (u 3, +u,3 ) u 3,3 or 0 (u, u, ) (u,3 u 3, ) + (u, u, ) 0 (u,3 u 3, ) (u 3, u,3 ) (u 3, u,3 ) 0 (A.) u i,j = ǫ ij +w ij (A.3) where ǫ ij = (u i,j +u j,i ) w ij = (u i,j u j,i ) (A.4) (A.5) The symmetric part of the deformation gradient tensor, ǫ ij, is the small deformation strain tensor, while the antisymmetric part of the deformation gradient tensor, w ij, is the rotation motion tensor. The matrix form of the strain ǫ ij is ǫ = ǫ xx ǫ xy ǫ zx ǫ xy ǫ yy ǫ yz ǫ zx ǫ yz ǫ zz = ǫ x γ xy γ xy ǫ y γ zx γ zx γ yz γ yz ǫ z The engineering strain is usually expressed in the vector form ǫ = {ǫ x, ǫ y, ǫ z,γ xy, γ yz, γ zx } T Note that the engineering shear strain γ ij is the double of the corresponding strain component ǫ ij. Here the second and higher order derivative terms are neglected due to the small deformation assumption. (A.6) (A.7)
6 Jeremić et al. A.. STRESS AND STRAIN page: 39 of 536 Similar to the stress tensor, the strain tensor also has three principle strains ǫ i (ǫ ǫ ǫ 3 ), and three strain invariants I, I, and I 3, defined as I = ǫ ii = ǫ v = ǫ xx +ǫ yy +ǫ zz = ǫ +ǫ +ǫ 3 (A.8) I = ǫ ijǫ ji = (ǫ xx ǫ yy +ǫ yy ǫ zz +ǫ zz ǫ xx )+(ǫ xy +ǫ yz +ǫ zx) = (ǫ ǫ +ǫ ǫ 3 +ǫ 3 ǫ ) (A.9) I 3 = 3 ǫ ijǫ jk ǫ ki = det(ǫ ij ) = ǫ xx ǫ yy ǫ zz +ǫ xy ǫ yz ǫ zx (ǫ xx ǫ yz +ǫ yz ǫ zx +ǫ zz ǫ xy) = ǫ ǫ ǫ 3 (A.30) The first strain invariant is also called the volumetric strain ǫ v. The strain ǫ ij can be decomposed into the hydrostatic strain ǫ m δ ij and deviatoric strain e ij through ǫ ij = ǫ m δ ij +e ij where: ǫ m = 3 I, e ij = ǫ ij 3 ǫ kkδ ij (A.3) Since both hydrostatic and deviatoric strains are strain tensors, they have their own strain invariants respectively. The three invariants of the hydrostatic strain are I = 3ǫ m = I, I = 3ǫ m = 3 (I ), I 3 = ǫ 3 m = 7 (I ) 3 (A.3) The hydrostatic strain state can therefore be represented by only one variable I. The three eigenvalues of the deviatoric strains e ij are called principal deviatoric strains, with the order e e e 3. The three invariants of the deviatoric strain are J = e ii = 0 (A.33) J = e ije ji = 3 (I ) +I = 6 [(ǫ ǫ ) +(ǫ ǫ 3 ) +(ǫ 3 ǫ ) ] = (e xx e yy +e yy e zz +e zz e xx )+(e xy +e yz +e zx) = (e +e +e 3) = (e e +e e 3 +e 3 e ) (A.34)
7 Jeremić et al. A.3. DERIVATIVES OF STRESS INVARIANTS page: 30 of 536 J 3 = 3 e ije jk e ki = det(e ij ) = I I I + 7 (I ) 3 = I 3 3 I J 7 (I ) 3 = 7 (ǫ ǫ ǫ 3 )(ǫ ǫ 3 ǫ )(ǫ 3 ǫ ǫ ) = e xx e yy e zz +e xy e yz e zx (e xx e yz +e yy s zx +e zz e xy) = e e e 3 (A.35) The deviatoric strain state can therefore be represented by only two variables J and J 3. Combining the hydrostatic and deviatoric strain, we can conclude that the strain state can be be represented by three variables I, J and J 3. Strain state may also be represented with another three invariant (ǫ p,ǫ q,θ ǫ ), defined as ǫ p = I = ǫ v J ǫ q = 3 ( θ ǫ = 3 arccos 3 3 ) J 3 (J ) 3 (A.36) (A.37) (A.38) where θ ǫ is the strain Lode s angle and 0 θ ǫ π/3. The relationship between (ǫ,ǫ,ǫ 3 ) and (ǫ p,ǫ q,θ ǫ ) is ǫ ǫ ǫ 3 = 3 ǫ p + 3 ǫ q cosθ ǫ cos(θ ǫ 3 π) cos(θ ǫ + 3 π) A.3 Derivatives of Stress Invariants (A.39) In this part of the Appendix, we shall derive some useful formulae, that are rarely found in texts treating elasto plastic problems in mechanics of solid continua. First derivative of I with respect to stress tensor σ ij : I σ ij = σ kk σ ij = δ ij First derivative of J D with respect to stress tensor σ ij : if found at all.
8 Jeremić et al. A.3. DERIVATIVES OF STRESS INVARIANTS page: 3 of 536 J D = ( s mns nm ) = s mn s nm + s nm s mn = σ ij σ ij σ ij σ ij = s nm σ ij s mn = (σ nm 3 σ kkδ nm ) σ ij s mn = (δ ni δ jm 3 δ nmδ ki δ jk )s mn = = (δ ni δ jm 3 δ nmδ ij )s mn = δ ni δ jm s nm 3 δ nmδ ij s mn = s ij First derivative of J 3D with respect to stress tensor σ pq : J 3D = ( 3 s ijs jk s ki ) = σ pq σ pq 3 s ij s jk s ki s jk s ki + s ij s ki + s ij s jk = σ pq 3 σ pq 3 σ pq = s ij σ pq s jk s ki = (σ ij 3 σ kkδ ij ) σ pq s jk s ki = (δ ip δ qj 3 δ ijδ kp δ qk )s jk s ki = = δ ip δ qj s jk s ki 3 δ ijδ kp δ qk s jk s ki = s qk s kp 3 δ pqj D = t pq First derivative of s pq with respect to stress tensor σ mn, or second derivative of J D with respect to stress tensors σ pq and σ mn : s pq = ( σpq 3 δ ) pqσ kk σ mn σ mn = (( δmp δ nq 3 δ ) ) pqδ mn σmn = σ mn (δ mp δ nq 3 ) δ pqδ mn = = p pqmn Second derivative of J 3D with respect to stress tensors σ pq and σ mn : because δ nmδ ijs mn 0 since δijδ 3 kpδ qk s jk s ki = δ 3 kpδ qk s ik s ki = δqps 3 iks ki = δpqjd see also Chen and Han (988a) page 3
9 Jeremić et al. A.3. DERIVATIVES OF STRESS INVARIANTS page: 3 of 536 t pq = ( sqk s kp 3 δ ) pqj D = (s qks kp ) ( 3 δ ) pqj D = σ mn σ mn σ mn σ mn = (s qks kp ) σ mn 3 δ J D pq = s qk s kp s kp +s qk σ mn σ mn σ mn 3 δ pqs mn = = (δ qm δ nk 3 δ qkδ nm )s kp +s qk (δ km δ np 3 ) δ kpδ nm 3 δ pqs mn = = (δ qm s np 3 ) s qpδ nm + (s qm δ np 3 ) s qpδ nm 3 δ pqs mn = = s np δ qm +s qm δ np 3 s qpδ nm 3 δ pqs mn = w pqmn Multiplying stiffness tensor E ijkl with compliance tensor D klpq :
10 Jeremić et al. A.3. DERIVATIVES OF STRESS INVARIANTS page: 33 of 536 E ijkl D klpq = ( )( ) E +ν ν ν (+ν) E ν δ ijδ kl +δ ik δ jl +δ il δ jk +ν δ klδ pq +δ kp δ lq +δ kq δ lp = 4 (δ ikδ jl δ kp δ lq +δ il δ jk δ kp δ lq +δ ik δ jl δ kq δ lp +δ il δ jk δ kq δ lp ) + ν + ( ν) (δ ν ijδ kl δ kp δ lq +δ ij δ kl δ kq δ lp ) (+ν) (δ ikδ jl δ kl δ pq +δ il δ jk δ kl δ pq ) ( ν)(+ν) δ ijδ kl δ kl δ pq = (δ ipδ jq +δ iq δ jp ) + ν + ( ν) (δ ν ijδ kq δ kp +δ ij δ kp δ kq ) (+ν) (δ ilδ jl δ pq +δ il δ jl δ pq ) 3ν ( ν)(+ν) δ ijδ pq = (δ ipδ jq +δ iq δ jp ) + ν + ( ν) (δ ν ijδ pq +δ ij δ pq ) (+ν) (δ ijδ pq +δ ij δ pq ) ν 3ν ( ν)(+ν) δ ijδ pq = (δ ipδ jq +δ iq δ jp ) + ν + ( ν) δ ν ijδ pq (+ν) δ ijδ pq 3ν ( ν)(+ν) δ ijδ pq = (δ ipδ jq +δ iq δ jp ) + + ν(+ν) ν( ν) 3ν δ ij δ pq = ( ν)(+ν) (δ ipδ jq +δ iq δ jp ) = I sym ijpq
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