12.005 Lecture Notes 14

12.005 Lecture Notes 14 Finite Strain The "paradox" of finite strain and preferred orientation – pure vs. simple shear Suppose: 1) Anisotropy develops as mineral grains grow such that they are preferentially oriented with the "easy slip" direction facilitating deformation. 2) New grain growth is determined by the state of stress. (What else would a mineral grain "know" about?) 3) The state of stress for pure shear and simple shear are identical – pure and simple shear stress differ only by a rotation of 45° in coordinate system. 4) In pure shear there is only one slip direction, whereas in pure shear, 2 systems could be active. Paradox: If the stress state is the same, how does the rock "know" to have 1 preferred orientation in simple shear, but not in pure shear? Figure 14.1 Up to now, we have considered infinitesimal strain, where eij = (∂ui/∂xj + ∂uj/∂xi)/2 <<1. For the kinds of deformations common in structural geology, as well as for the deforming telephone book example shown in class, this condition of "infinitesimal" deformation no longer holds. Life becomes complicated. In particular, it's important to know whether one is coming or going: Reference frame: Consider a medium undergoing motion & deformation, with the displacement a (vector) function of position. B x3 A a u b x2 x1 Figure 14.2 bj = f(ai) or bj = bj(ai) Since the mapping ai <—> bj is one-to-one, we could also say aj= aj(bi). It turns out (consider the telephone book example in class) that the description of strain depends on whether the reference frame is: Lagrangian ∫ fixed to particles (the "a" frame) ui = bi(aj) - ai Example: elasticity -- the particles return to where they started Eulerian ∫ fixed to space (the "b" frame) ui = bi -ai(bj) Example: fluid mechanics, geology -- no idea where the particles started or where they will end up; no stable frame fixed to the materials. Finite strain depends on whether the reference frame is fixed to the particle or to "space:" Lagrangian strain: Eij = (∂ui/∂aj + ∂uj/∂ai + ∂ul/∂ai ∂ul/∂aj)/2 Eulerian strain: eij = (∂ui/∂bj + ∂uj/∂bi - ∂ul/∂bi ∂ul/∂bj)/2 When strain is small, the 3rd term is small2, and both strains converge to the infinitesmal value. Physical interpretation: diagonal elements - e.g., elongation, given by ds = (1 + E1) ds0. x2 ds dso da1 db1 x1 Figure 14.3 For the Lagrangian system, E1 = (1 + 2E11)1/2 -1 off diagonal —>change in angles of coordinate axes: x2 x2 θ db' da' α = π/2 − θ db da x1 Figure 14.4 sin a = 2E12/[(1 + E11) (1 + E22)]1/2 These quantities go to the infinitesimal values when Eij << 1. x1 Because eij ≠ Eij, the principal values & principal directions of the Lagrangian and Eulerian strain tensors differ. We will discuss these later. Rotation Consider two particles initially separated by dai (note: not just a function of strain tensor) which undergo a displacement uj, and are then separated by dxi. Lagrangian: dx i = dai + ⎡ ∂ ui ∂u ⎤ daj = ⎢δ ij + i ⎥ da j ∂aj ∂ aj ⎦ ⎣ This can be written, including the finite strain tensor, ⎡ 1 ⎛ ∂u ∂u ∂u ∂u ⎞ 1 ⎛ ∂ u ∂u ∂ u ∂u ⎞ ⎤ dx i = ⎢δ ij + ⎜ i + j + s s ⎟ + ⎜ i − j − s s ⎟ ⎥ daj 2 ⎝ ∂ a j ∂ ai ∂ ai ∂a j ⎠ 2 ⎝ ∂a j ∂ai ∂ai ∂a j ⎠ ⎦ ⎣ [ ] = δ ij + Ε ij + Ω ij da j Ω ij — Lagrangian rotation tensor if E ij = 0 → rotation alone Similarly, for the Eulerian case: 1 ⎛ ∂u ∂ u ∂ u ∂u ⎞ ω ij = ⎜ i − j + s s ⎟ 2 ⎝ ∂ bj ∂ bi ∂ bi ∂ b j ⎠ [ ] dai = δ ij + eij + ω ij dx j for small deformation: Ω ij rotation needed to bring a fiber dai into position. For large deformation: no easy way to separate into pure rotation and pure shear—the two are intimately connected. THIS IS OF UTMOST IMPORTANCE IN PROBLEMS LIKE THE DEVELOPMENT OF ROCK FABRIC. ALSO—REGIONAL TECTONICS. Consider homogeneous strain Figure 14.5 Figure by MIT OCW. This holds if Eij = ⇒ ∂r ∂r 1 (δ lm l m − δ ij ) ≠ Eij ( Ri ) 2 ∂Ri ∂R j ∂l = constant ⇒ R & r linearly related   ∂j Define Ri → x, y ri → x ', y ' In general x ' = ax + by y ' = cx + dy ⎡ a 2 + c 2 − 1 ab + cd 0⎤ 1⎢ ⎥ 2 2 Eij = ⎢ ab + cd b + d − 1 0⎥ 2 ⎢ 0 0 0 ⎥⎦ ⎣ Examples x ' = kx y' = y Figure 14.6 Figure by MIT OCW. x ' = k1x y ' = k2 y Figure 14.7 Figure by MIT OCW. x ' = kx 1 y' = y k Figure 14.8 Figure by MIT OCW. x ' = x + 2sy y' = y Figure 14.9 Figure by MIT OCW. Figure 14.10 Figure by MIT OCW. Unstrained 1 x = 2 (dx '− by ') h 1 y = 2 (−cx '+ ay ') h Strained x ' = ax + by h2 = ad − bc y ' = cx + dy Principal Axes A, B Roots of R 4 − (a 2 + b2 + c 2 + d 2 )R 2 + h4 = 0 tan 2α = 2(ab + cd ) a + c 2 − b2 − d 2 tan 2α ' = 2 2(ac + bd ) a + b2 − c 2 − d 2 2 No Length Change tan δ = ±[(1 − B ) / ( A − 1)] 2 2 B( A2 − 1)1/ 2 A(1 − B 2 )1/ 2 relative to principal axis 1/ 2 tan δ ' = ± relative to reciprocal axis Maximum Shear b +d −a −c 2(ab + cd ) relative to coord. axis tan 2ϕ = 2 2 2 2 tan ϕ ' = ( B / A) relative to principal axis Principal axis of infinitesimal strain α +α' α0 = 2