GG303 Lecture 15 8/24/03 1 FINITE STRAIN AND INFINITESIMAL STRAIN I Main Topics (on infinitesimal strain) A The finite strain tensor [E] B Deformation paths for finite strain C Infinitesimal strain and the infinitesimal strain tensor ε II The finite strain tensor [E] A Used to find the changes in the squares of lengths of line segments in a deformed body. B Definition of [E] in terms of the deformation gradient tensor [F] Recall the coordinate transformation equations: x′ abx 1 = or [] X′ = [][] F X y′ cdy dx′ abdx 2 = or [] dX′ = [][] F dX dy′ cddy dx T If = []dX, then [ dx dy ]= [] dX ; transposing a matrix is switching dy its rows and columns 222 dx TT 3 ()ds= () dx+ () dy= [] dx dy = [][]dX dX= [][][] dX I dX , dy 10 where I = is the identity matrix. 01 222 dx′ T 4 ()ds′ = () dx′ + () dy′ = [] dx′′ dy = []dX′ [] dX′ dy′ Now dX’ can be expressed as [F][dX] (see eq. II.B.2). Making this substitution into eq. (4) and proceeding with the algebra 2 T TT 5 ()ds′ = [][][] F dX[][][] F dX= [][] dX F[][] F dX 22 TT T 6 ()ds′ − () ds′ = [][] dX F[][] F dX− [][][] dX I dX 22 TT 7 ()ds′ − () ds′ = [][] dX[] F[] F− I[] dX 1 221 TT T 8 ()ds′ − () ds′ = [][] dX F[] F− I[] dX≡ [][][] dX E dX 2{}2 [] 1 T 9 []EF≡ [][]FI− = finite strain tensor 2[] Stephen Martel 15-1 University of Hawaii GG303 Lecture 15 8/24/03 2 IIIDeformation paths Consider two different finite strains described by the following two coordinate transformation equations: ′ ab ax by x1 11x 11+ = = = []FX1 [] y ′ cdy cx+ dy A 1 11 11 Deformation 1 ′ ab ax by x2 22x 22+ = = = []FX2 [] y ′ cdy cx+ dy B 2 22 22 Deformation 2 Now consider deformation 3, where deformation1 is acted upon (followed) by deformation 2 (i.e., deformation gradient matrix F1 first acts on [X], and then F2 acts on [F1][X]) x′′ ab22ab11x aa21++ bc 21 ab 21 bd 21x = = cdcd ca dc cb dd C y′′ 22 11y 21++ 21 21 21y Deformation 3 Next consider deformation 4, where deformation 2 is acted upon (followed) by deformation 1 (i.e., deformation gradient matrix F2 first acts on [X], and then F1 acts on [F2][X]. x′′′ ab11ab22x aa12++ bc 12 ab 12 bd 12x = = cdcd ca dc cb dd D y′′′ 11 22y 12++ 12 12 12y Deformation 4 EA comparison of the net deformation gradient matrices in C and D shows they generally are different. Hence, the net deformation in a sequence of finite strains depends on the order of the deformations. (If the b and c terms [the off-diagonal terms] are small, then the deformations are similar) Stephen Martel 15-2 University of Hawaii GG303 Lecture 15 8/24/03 3 Coordinate Transformation 1 x’ 21 x = y’ 01 y 21 0 1 y (0,1)y’ (1,1) = 01 1 1 x x’ 21 1 2 = (1,0) (2,0) 01 0 0 Coordinate Transformation 2 y’ (0,2) x’ 10 x = y’ 02 y y (0,1) 10 0 0 = 02 1 2 x x’ 10 1 1 (1,0) (1,0) = 02 0 0 Coordinate transformation 3 (transformation 1 followed by transformation 2) x’’ 10 21 x = y’’ (1,2) y’’ 02 01 y 21 0 1 = y (0,1) 02 1 2 21 1 2 x x’’ = 02 0 0 (1,0) (2,0) Coordinate transformation 4 (transformation 2 followed by transformation 1) x’’ 21 10 x (2,2) = y’’ y’’ 01 02 y 22 0 2 y (0,1) = 02 1 2 x x’’ 22 1 2 = (1,0) (2,0) 02 0 0 Stephen Martel 15-3 University of Hawaii GG303 Lecture 15 8/24/03 4 IV Infinitesimal strain and the infinitesimal strain tensor [ε] A What is infinitesimal strain? Deformation where the displacement derivatives are small relative to one (i.e., the terms in the corresponding displacement gradient matrix J are very small), so that the products of the derivatives []u are very small and can be ignored. B Why consider infinitesimal strain if it is an approximation? 1 Many important geologic deformations are small (and largely elastic) over short time frames (e.g., fracture earthquake deformation, volcano deformation). 2 The terms of the infinitesimal strain tensor [ε] have clear geometric meaning (clearer than those for finite strain) 3 Infinitesimal strain is much more amenable to sophisticated mathematical treatment than finite strain (e.g., elasticity theory). 4 The net deformation for infinitesimal strain is independent of the deformation sequence. 5 Example 102.. 001 101. 0 002.. 001 001. 0 F5 = F6 = J 5 = J 6 = u u 0101. 0102. 0001. 0002. Deformation 5 followed by deformation 6 gives deformation 7: x′ 101.. 000102.. 001x 1..0302 0 0100x = = y′ 000.. 102000.. 101y 0..0000 1 0302y Deformation 6 followed by deformation 5 gives deformation “7a”: x′ 102.. 001101.. 000x 1..0302 0 0101x = = y′ 000.. 101000.. 102y 0..0000 1 0302y The net deformation is essentially the same in the two cases. Stephen Martel 15-4 University of Hawaii GG303 Lecture 15 8/24/03 5 C The infinitesimal strain tensor (Taylor series derivation) Consider the displacement of two neighboring points, where the distance from point 0 to point 1 initially is given by dx and dy. Point 0 is displaced by an amount u0, and we wish to find u1. We use a truncated Taylor series; it is linear in dx and dy (dx and dy are only raised to the first power). ∂u ∂u 1 u 10= u +++x dx x dy ... x x ∂x ∂y ∂u ∂u y y 2 u 10= u +++dx dy ... y y ∂x ∂y These can be rearranged into a matrix format: ∂u ∂u x x 1 0 u u ∂x ∂y dx 3 x x UJ0 dX 1 = 0 + ∂u ∂u = + [] u u y y dy []u y y ∂x ∂y Now split J into two matrices: the symmetric infinitesimal strain []u T matrix [ε], and the anti-symmetric rotation matrix [ω] by using J , []u ef T eg J = , J = , []u gh []u fh ee++ f g 0 fg− J + J T = , J − J T = []u u gfhh++ []u u gf− 0 1 1 1 1 4 J =+J J + J TT− J =+J J T + J − J T = []εω+ [] []u 2 []u u 2 []u u 2 []u u 2 []u u Now the displacement expression can be expanded using [ε] and [ω] ∂u ∂u ∂u ∂u ∂u ∂u x x x y x y + + 0 −− 1 ∂x ∂x ∂y ∂x 1 ∂y ∂x 5 []ε = , []ω = 2 ∂u ∂u ∂u ∂u 2 ∂u ∂u y x y y y x + + − − 0 ∂x ∂y ∂y ∂y ∂x ∂y Equations (3) and (5) show that the deformation can be decomposed into a translation, a strain, and a rotation. D Geometric interpretation of the infinitesimal strain components Stephen Martel 15-5 University of Hawaii GG303 Lecture 15 8/24/03 6 Infinitesimal Deformation of Line Elements AB and AD (modified from Chou and Pagano, 1967) ∂u x dy y ∂y ∂u C’ y dy B’ ∂y Positive angles Ψ2 measured counterclockwise π/2 − Ψ B C dy D’ Ψ1 ∂u dy A’ y dx x u ∂ uy A dx ∂u D x dx ∂x y dx x xuux Line length changes (dx + ∂ u x dx) - dx εxx = A’D’-AD ¯ ∂x = ∂ux AD dx ∂x (dy + ∂ u y dy) - dy εyy = A’B’-AB ¯ ∂y = ∂uy AB dy ∂y Angle changes ∂u ∂u ∂u tanΨ1 = x dy = x ¯ Ψ1 if x is small ∂y ∂y ∂y dy ∂u y dx ∂x ∂u ∂u tanΨ2 = = y ¯ Ψ2 if y is small dx ∂x ∂x Ψ = Ψ1 − Ψ2= change in right angle (the minus sign accounts for Ψ2 being negative) γ = tan Ψ εxy = εyx = (1/2) tan Ψ γ = engineering shear strain ε = tensor shear strain Stephen Martel 15-6 University of Hawaii GG303 Lecture 15 8/24/03 7 Infinitesimal Strains a dux Normal strain exx Deformed exx = ∂ux/∂x Undeformed dx b Normal strain e du yy y eyy = ∂uy/∂y dy dy dx c Shear strains exy, eyx † dux exy = eyx = 1/2(change in right angle) Ψ2 dy = 1/2(Ψ1 - Ψ2)* Ψ1 duy But for small angles, Ψ = tanΨ dx exy = eyx = 1/2(∂ux/∂y+∂uy/∂x) d (This is pure shear; there is no rotation) du x Rotations ωyx, ωxy * dy ωxy = ωyx = (change in right angle) Ψ2 = 1/2(Ψ1 + Ψ2)* Ψ1 duy dx But for small angles, Ψ = tanΨ Note that "simple shear strain" involves a ωxy = -ωyx = 1/2(∂uy/∂x-∂ux/∂y) shear and a rotation! Here Ψ1 is zero and is negative. Ψ2 † The shear strain exy = eyx is half the dux shear strain γ Ψ2 * Positive angles are measured about dy the z-axis using a right hand rule. In (b) the angle Ψ2 is clockwise (negative), but dux is positive. In (d) Ψ is counter-clockwise, and du < 0. dx 2 x Stephen Martel 15-7 University of Hawaii GG303 Lecture 15 8/24/03 8 E Relationship between [ε] and [E] From eq. II.B.9, [E] is defined in terms of deformation gradients: 1 T 1 []EF≡ [][]FI− = finite strain tensor 2[] The tensor [E] also can be solved for in terms of displacement gradients because FJ =+u I . 1 T 2 []EJIJII= []u + []u + − 2[] T ∂∂uxxu ∂∂uxxu 1 dx dy 10 dx dy 10 10 3 []E = + + − 2 ∂∂uyyu 01 ∂∂uyyu 01 01 dx dy dx dy ∂u ∂uy ∂∂uxxu x +1 +1 1 dx dy 10 4 []E = dx dx − 2 ∂u ∂uy ∂∂uyyu 01 x +1 +1 dy dy dx dy ∂∂u u ∂∂uyy u ∂∂u u ∂∂uyy u xx+11+ + − 1xx+ 1 + + 1 1 dx dx dx dx dx dy dx dy 5 []E = 2 ∂∂u u ∂∂∂uyy u ∂∂u u ∂∂uyy u xx +1 + +1111 xx ++ + − dy dx dy dx dy dy dy dy If the displacement gradients are small relative to 1, then the products of the displacements are very small relative to 1, and in infinitesimal strain theory they can be dropped, yielding [ε]: ∂∂∂u u u ∂uy xxx+ + 1 dx dx dy dx 1 T 6 ε ≈ = J + J [] []u []u 2 ∂ux ∂∂∂uyyy u u 2 + + dy dx dy dy This suggests that for multiple deformations, infinitesimal strains might be obtained by matrix addition (i.e., linear superposition ) rather than by matrix multiplication; the former is simpler.