GG303 Lecture 5 0609 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 & y = a b & x& c d y or [ X ] = [ F] [ X ] d x & 2 d y = a b & dx& c d dy or [ d X ] = [ F ][ dx ] If dx dy & = [ dx ] then [ dx dy ] = [ dx ]T ; transposing a matrix is switching its rows and columns dy& = [ dx]t dx where I = 0 0 & is the identity matrix. 2 = d x 2 d y 2 = [ d x d y ] d x & d y = d X 3 ds 2 = dx 2 dy 2 = [ dx dy] dx 4 d s [ ] = [ dx ] T [ I] dx [ ] T [ d X ] [ ] 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 = [[ F] [ dx] ] T [[ F ][ dx ] = [ dx ] T [ F ] T [ F] [ dx] 5 d s 6 d s 2 ds 2 = [ dx] T [ F] T [ F] [ dx] [ dx] T [ I] [ dx] [ ] dx 7 d s 2 ds 2 = [ dx] T [ F] T [ F] I [ ] 8 2 d s 2 ds & dx [ ] dx { 2 } = [ ] T [ F] T [ F] I 2 [ ] * dx [ ]T E [ ][ dx] 9 E [ ] 2 [ [ ] I] = finite strain tensor & [ F ]T F Stephen Martel Fall 2009: Conrad 5- University of Hawaii
GG303 Lecture 5 0609 2 III Deformation paths Consider two different finite strains described by the following two coordinate transformation equations: A x & y = a b & x& c d y = a x b y & c x d y = F [ ] X [ ] Deformation B C x & 2 y = a 2 b 2& x& c 2 2 d 2 y = a 2 x b 2 y & c 2 x d 2 y = F 2 [ ] X [ ] Deformation 2 Now consider deformation 3 where deformation is acted upon followed by deformation 2 i.e. deformation gradient matrix F first acts on [X] and then F2 acts on [F][X] x & y = a 2 b 2 & a b & x& c 2 d 2 c d y = a 2 a b 2 c a 2 b b 2 d & x& c 2 a d 2 c c 2 b d 2 d y Deformation 3 Next consider deformation 4 where deformation 2 is acted upon followed by deformation i.e. deformation gradient matrix F2 first acts on [X] and then F acts on [F2][X]. D x & y = a b & a 2 b 2 & x& c d c 2 d 2 y = a a 2 b c 2 a b 2 b d 2 & x& c a 2 d c 2 c b 2 d d 2 y Deformation 4 E A 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 Fall 2009: Conrad 5-2 University of Hawaii
GG303 Lecture 5 0609 3 Stephen Martel Fall 2009: Conrad 5-3 University of Hawaii
GG303 Lecture 5 0609 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 u [ ] are very small so that the products of the derivatives are very small and can be ignored. B Why consider infinitesimal strain if it is an approximation? 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.02 0.0 F5 = 0.0& F6 =.0 0 0.02& 0.02 0.0 J u 5 = 0 0.0& J u 6 = 0.0 0 0 0.02& Deformation 5 followed by deformation 6 gives deformation 7: x &.0 0.00&.02 0.0& x&.0302 0.000& x& = = y 0.00.02 0.00.0 y 0.0000.0302 y Deformation 6 followed by deformation 5 gives deformation 7a : x &.02 0.0&.0 0.00& x&.0302 0.00& x& = = y 0.00.0 0.00.02 y 0.0000.0302 y The net deformation is essentially the same in the two cases. Stephen Martel Fall 2009: Conrad 5-4 University of Hawaii
GG303 Lecture 5 0609 5 C The infinitesimal strain tensor Taylor series derivation Consider the displacement of two neighboring points where the distance from point 0 to point initially is given by dx and dy. Point 0 is displaced by an amount u 0 and we wish to find u. We use a truncated Taylor series; it is linear in dx and dy dx and dy are only raised to the first power. u = ux 0 u x x x dx u x y u u 2 u y = uy 0 y y x dx y dy... dy... These can be rearranged into a matrix format: u 3 x u = u x 0 u x u x x y dx y & u 0 u u y & y y dy& = U0 & J u x y & Now split J u [ ] dx [ ] [ ] into two matrices: the symmetric infinitesimal strain matrix [ ] T [ ] = e f [ ] T = e g [ε] and the anti-symmetric rotation matrix [ω] by using J u J u g h& J u J T [ u ] = e e f g g f h h & 4 [ J u ] = 2 J u J u [ ] 2 J T T J u u [ ] = 2 J u J u T J u f [ J u J u T ] = [ ] [ 2 J u J u T ] = h& 0 f g& g f 0 [ ] [ ] Now the displacement expression can be expanded using [ε] and [ω] * u x 5 [ ] = x u x u u & x y x y - * & x 0 u u x y 2 u y & x u [ 0] = y - & x u u x y y & y y 2 u y & y x u x 0 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 Fall 2009: Conrad 5-5 University of Hawaii
GG303 Lecture 5 0609 6 Stephen Martel Fall 2009: Conrad 5-6 University of Hawaii
GG303 Lecture 5 0609 7 Stephen Martel Fall 2009: Conrad 5-7 University of Hawaii
GG303 Lecture 5 0609 8 E Relationship between [ε] and [E] From eq. II.B.9 [E] is defined in terms of deformation gradients: E [ ] 2 [ [ ] I] & [ F ]T F = finite strain tensor The tensor [E] also can be solved for in terms of displacement gradients because F = J u I. 2 E 3 E 4 E 5 E [ ] = 2 [ ] = 2 [ ] = 2 [ ] = 2 [ [ ] & I] [ J u I]T J u I T &u x &u x * * &u x &u x * * * dx dy &u 0 * dx dy y 0 &u 0 * y 0-0 * 0 dx dy dx dy &u x dx *&u x dx dx &u x * * dy &u x dy dy &u - 0 * y dx dy 0 * &u x dx &u x dx dx dx &u x &u x dy dx dy dx &u x dx &u x dy dx dy - &u x &u x dy dy dy dy. If the displacement gradients are small relative to then the products of the displacements are very small relative to and in infinitesimal strain theory they can be dropped yielding [ε]: * u x & 6 [ ] & dx u x & u x & dx dy u y &- dx 2 u x & dy u y & u y & dx dy u = * y & [ 2 J u] [ J u ] T -. 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 Fall 2009: Conrad 5-8 University of Hawaii
GG303 Lecture 5 0609 9 7 Example of IV.B.5: [ε] from superposed vs. sequenced deformations.02 0.0 0.02 0.0 F5 = J 5 = F6 =.0 0 J 6 = 0.0 0 0.0& u 0 0.0& 0.02& u 0 0.02& a Linear superposition assuming infinitesimal strain approx.»f5 = [.02 0.0;0.00.0]»F6 = [.0 0.00;0.00.02] F5 =.0200 0.000 0.000 E5 = [ 2 F 5]T F 5 F6 = [ [ ] I] E6 = 2 F 6»E5 = 0.5*F5*F5-eye2.000 0 0.0200 [ [ [ ]T F 6] I]» E6 = 0.5*F6*F6-eye2 E5 = 0.0202 0.005 0.005 0.00 [ 2 J u 5 ] [ J u 5] T &* E6 = 0.00 0 0 0.0202 [ 2 J u 6 ] [ J u 6] T &*»E7 = E5 E6 Linear superposition of strains E7 = 0.0302 0.005 0.005 0.0303 Infinitesimal approximation [ ] 2 b Sequenced deformation exact E 7 F & * 7»F7 = F6*F5 See eq. IV.B.5 F7 =.0302 0.00 0.0302 [ ] T [ F 7 ] I»E7 = 0.5*F7*F7-eye2 Convert def. gradients to strain - E7 = 0.0307 0.0052 0.0052 0.0307 Good match with approximation Stephen Martel Fall 2009: Conrad 5-9 University of Hawaii
GG303 Lecture 5 0609 0 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 { 2 ds 2 ds 2 } = [ dx] T [][ dx] 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: [U] = [ 2 ][ dx] [ 2 ][ dx ] 9 For infinitesimal strains the displacements are essentially the same no matter whether the pre- or post-deformation positions are used. Stephen Martel Fall 2009: Conrad 5-0 University of Hawaii