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′ abx 1   =    or [] X′ = [][] F X y′ cdy dx′ abdx 2   =    or [] dX′ = [][] F dX dy′ cddy 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 221 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 ′ cdy cx+ dy A  1  11 11 Deformation 1

 ′ ab  ax by x2  22x 22+ =    =   = []FX2 [] y ′ cdy 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′′ ab22ab11x aa21++ bc 21 ab 21 bd 21x   =     =    cdcd ca dc cb dd C y′′  22 11y  21++ 21 21 21y 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′′′ ab11ab22x aa12++ bc 12 ab 12 bd 12x   =     =    cdcd ca dc cb dd D y′′′  11 22y  12++ 12 12 12y 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’ 2 1 x = y’ 0 1 y

2 1 0 1 y (0,1)y’ (1,1) = 0 1 1 1

x x’ 2 1 1 2 = (1,0) (2,0) 0 1 0 0

Coordinate Transformation 2

y’ (0,2) x’ 1 0 x = y’ 0 2 y

y (0,1) 1 0 0 0 = 0 2 1 2 x x’ 1 0 1 1 (1,0) (1,0) = 0 2 0 0

Coordinate transformation 3 (transformation 1 followed by transformation 2)

x’’ 1 0 2 1 x = y’’ (1,2) y’’ 0 2 0 1 y

2 1 0 1 = y (0,1) 0 2 1 2

2 1 1 2 x x’’ = 0 2 0 0 (1,0) (2,0)

Coordinate transformation 4 (transformation 2 followed by transformation 1)

x’’ 2 1 1 0 x (2,2) = y’’ y’’ 0 1 0 2 y

2 2 0 2 y (0,1) = 0 2 1 2

x x’’ 2 2 1 2 = (1,0) (2,0) 0 2 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.. 000102.. 001x 1..0302 0 0100x   =     =    y′ 000.. 102000.. 101y 0..0000 1 0302y

Deformation 6 followed by deformation 5 gives deformation “7a”: x′ 102.. 001101.. 000x 1..0302 0 0101x   =     =    y′ 000.. 101000.. 102y 0..0000 1 0302y

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. Also see equation IV.C.5.

Stephen Martel 15-8 University of Hawaii GG303 Lecture 15 8/24/03 9

7 Example of IV.B.5: [ε] from superposed vs. sequenced deformations 102.. 001 002.. 001 101. 0  001. 0  F5 = J 5 = F6 = J 6 = u u  0101.   0001.   0102.   0002. 

a Linear superposition, assuming infinitesimal strain (approx.) »F5 = [1.02 0.01;0.00 1.01] »F6 = [1.01 0.00;0.00 1.02] F5 = F6 = 1.0200 0.0100 1.0100 0 0 1.0100 0 1.0200 1 T 1 T EFFI5 = [][]55− EFFI6 = [][]66− 2 [] 2 [] »E5 = 0.5*(F5'*F5-eye(2)) » E6 = 0.5*(F6'*F6-eye(2))

E5 = E6 = 0.0202 0.0051 0.0101 0 0.0051 0.0101 0 0.0202  1 T   1 T  ≈  J 55+ J  ≈  J 66+ J   2 []u []u   2 []u []u 

»E7 = E5 + E6 Linear superposition of strains E7 = (Infinitesimal approximation) 0.0302 0.0051 0.0051 0.0303

1 T  b Sequenced deformation (exact) EF≡ FI− []7 2[][]77 »F7 = F6*F5 See eq. IV.B.5 F7 = 1.0302 0.0101 0 1.0302

»E7 = 0.5*(F7'*F7-eye(2)) Convert def. gradients to strain E7 = Good match with approximation 0.0307 0.0052 0.0052 0.0307

Stephen Martel 15-9 University of Hawaii GG303 Lecture 15 8/24/03 10

8 Recap

The infinitesimal strain tensor can be used to find the change in the square of the length of a deformed line segment connecting two nearby points separated by distances dx and dy, 1 22 T ()ds''− () ds= [][][] dXε dX 2{} and, with the rotation tensor, to find the change in displacement of two points in a deformed medium that are initially separated by distances dx and dy: 1 1 []∆UdXdX= [][εω ]+ [][] 2 2

9 For infinitesimal strains the displacements are essentially the same no matter whether the pre- or post-deformation positions are used.

Stephen Martel 15-10 University of Hawaii