MATH45061: SOLUTION SHEET 1 II

Transcription

1 MATH456: SOLUTION SHEET II. The deformation gradient tensor has Cartesian components given by F IJ R I / r J ; and so F R e x, F R, F 3 R, r r r 3 F R r, F R r, F 3 R r 3, F 3 R 3 r, F 3 R 3 r, F 33 R 3 r 3. Writing the components in matrix form, we have e x F which has determinant given by det F e x + 5e x. The deformation is physically admissible provided that det F >, which is true provided that x does not extend to. For the two-dimensional lamina, x 3, so the deformed position is given by X e x, X x, X 3 x, which means that the corners map as follows,,,,;,, e,,;,,,,;,, e,,, and the deformed lamina moves out of the x x plane and is also stretched, see Figure x x.5 x.5.5 Deformed lamina Undeformed lamina Figure : Sketch of the undeformed and deformed laminæ. Any feedback to: Andrew.Hazel@manchester.ac.uk

2 . a. The easy to compute base vectors in component form in the global Cartesian basis are given by g r ξ ξ ξ ξ ξ ξ, g r ξ ξ ξ ξ ξ ξ ; G R χ χ χ χ χ χ, G R χ χ χ χ χ These are the natural choices for the covariant base vectors in the original and deformed positions. The other base vectors are unnatural in the sense that we are using a coordinate system associated with the other position and so we must make use of the mapping between the two. ξ ξ R r, ξ χ. so that G ξ ξ ξ ξ ξ, G ξ ξ ξ ξ ξ ; and leading to g χ χ χ χ r R χ b. The associated metric tensors are: g ij g g g g g g g g χ χ χ, g χ χ χ χ ξ ξ ξ ξ ξ ξ +ξ ξ, ; G G G ij G G 4ξ ξ 4ξ ξ G G G G 4ξ ξ 4+4ξ ξ, g g g ij g g χ χ 4 χ χ, g g g g 4 χ χ 4 χ χ G G G ij G G +χ χ χ χ ; G G G G χ χ χ χ and the differences are then 3ξ ξ 3ξ ξ A ij 3ξ ξ 3+3ξ ξ and A ij χ χ χ 3 4 χ χ 3 4 χ χ 3 4 χ χ.

3 3 The two tensors should be related by covariant transformation between the Eulerian and Lagrangian coordinates: A ij χk ξ i χ l ξ ja kl. Confirming this is a little painful algebraically as with many explicit tensor calculations, but perfectly straightforward. From the deformation χ ξ R χ χ r ξ ξ, and so χ ξ ξ, and χ ξ /χ /ξ. Thus coordinate transformation is given by χ χ ξ ξ χ χ ξ ξ ξ ξ /ξ ξ ; and we can write the transformation in the form of a matrix product A A ξ ξ T A A ξ ξ A A /ξ ξ A A /ξ ξ ξ /ξ ξ χ χ 3 4 χ χ ξ ξ ξ 3 4 χ χ 3 4 χ χ /ξ ξ. We now convert all components A ij into the Lagrangian coordinates ξ i, so that A A ξ /ξ ξ ξ ξ ξ ξ ξ A A ξ 3 ξ 3ξ ξ ξ ξ /ξ ξ ξ /ξ ξ 3 ξ 3 ξ + 3 /ξ 3ξ ξ 3ξ ξ ξ 3ξ ξ 3ξ ξ 3ξ ξ, + as expected phew!.,, 3. a. The deformation gradient is given by +t F t t which has det F + t + t >, when t, so the deformation is physically admissible for all t >. We have that the deformation is given by X +t x X t x X 3 t x 3

4 4 and in order to find the inverse transform we must simply invert the matrix to obtain x +t X x +t t+t X 3 x 3 t+t +t X 3 b. The Lagrangian velocity field is given by X X +t x t x t t X 3 t x 3 so v x, v x 3, v 3 x. The Eulerian velocity field is obtained by using the inverse transformation 3 in the expression for the velocity V +t +t t+t t+t +t +t t+t +t +t t+t X X X 3 so V X /+t, V tx +X 3 /+t, V 3 tx 3 X /+t. c. The function Φt,X I tx +X can be converted to Lagrangian coordinates by using the deformation φt,x I t+tx +x +tx 3. The material derivative is then simply φ t +tx +x 3 +t xi +t X + tx +X 3, +t after using the inverse deformation 3. d. The material derivative in Eulerian coordinates is DΦ Dt Φ t +V R Φ X + X +t, tx +X 3, tx 3 X +t +t X + tx +t + tx +X 3 +t +t +t X + tx +X 3 +t, which, of course, agrees with the answer in part c. x x x 3 X X X 3 t,,, 4. This proof does rely on some previous knowledge of vector calculus and differential equations so don t worry too much if you struggled.

5 5 The key piece of information is that the normal to a surface fχ i,t or any constant in fact is given by f, so ñ G i f,i, with length ñ G i f,i G j f,j G ij f,i f,j, where a comma,i denotes / χ i Thus, the covariant components of a unit normal to the surface are given by f,k n k. G ij f,i f,j We differentiate the defining expression of the surface with respect to time: f f t + χk f t χ, k where χ k / tisforagivenparticlewithinthesurface. Wesupposethattheparticles in the surface are moving with an arbitrary velocity field V χ i,t. Thus, for a given particle R t V X I t V I χi t X I χ i V I χi t χi X I VI V i, and so f f t + V k f χ k f t + V k n k G ij f,i f,j, which means that f t V k n k G ij f,i f,j ; in other words the normal speed of the particles must be the same as the speed of movement of the surface in order to remain within the surface. The material derivative of the surface function fχ i,t is Df Dt f t +Vi f χ i, where V i are the components of the underlying velocity of the continua. Using the above expression we have Df Dt V i V i n k G ij f,i f,j. If the particles are material then their velocity V must coincide with the material velocity V, which implies that Df/Dt from the above equation. For the sufficiency condition we can use the method of characteristics. We introduce a new variable s and let χ i s and ts, such that dχ i ds V i, and dt ds ;

6 6 so that df ds f dt t ds + f dχ i χ i ds f t +Vi f χ i Df Dt. We can find the solution of the equation by solving the two systems: dχ i ds V i, and dt ds. The second equation implies that s t because we are free to choose the arbitrary constant and then the remaining equation becomes dχ i dt V i, but this is the defining equation for the evolution of a material particle, which proves sufficiency. 5. a. We know that planes of constant x become planes of constant R. If R is a function of x then a change in x would cause a change in R, but this is a contradiction because varying x, but keeping x constant must not change R. A similar argument for x 3 demonstrates that R fx and similarly Θ gx and Z hx 3. b. If we let hx 3 λx 3, then we have the that deformed position is given by X RcosΘ fx cosgx R Y RsinΘ fx singx. Z Z λx 3 Thus, the covariant base vectors are G R f cosg f sing x fg sing fg cosg λ and the deformed metric tensor G ij has the matrix of components f G ij G i G j f g. λ c. If the deformation is of constant volume then G/g. The undeformed state is represented in Cartesians for which g and so G. Now G det G ij f f g λ, which means that G λff g. We now proceed to separate the variables λff /g.

7 7 The left-hand side is a function only of x, the right-hand side is a function of x and therefore the only way that they can be equal for all x and x is for both sides to be equal to the same constant, β, say. Hence, and integrating gives ff β/λ and g /β, f βx λ +c and g x β +c, where c and c are constants. It follows that βx fx λ +c and gx x β +c. For a specific problem the unknown constants will be fixed by the specific boundary conditions. 6. a. The deformation is given by R αr, which in Cartesian coordinates can be written as X αx X αx. X 3 αx 3 The deformation gradient tensor is given by α F r α. α The Green Lagrange strain tensor in Cartesian coordinates is e IJ F KIF KJ δ IJ, so the only non-zero components are the diagonal entries and we have that e IJ α b. As on Example Sheet I, let ξ r, ξ θ, ξ 3 φ and because R αr, we have that G i αg i. Hence, G ij α g ij and δ IJ. γ ij G ij g ij α g ij. From Example Sheet I, we know that g ij is diagonal with the entries g, g r, g 33 r sin θ, so γ α, γ α r, γ 33 α r sin θ. If you transform the undeformed coordinates back to Cartesian coordinates then g ij becomes δ IJ and we obtain the same result as in part a.

8 8 7. The equation can be written in the form G ij n j µg ij n j, X K ξ i X K ξ j nj µ x K ξ i x K ξ j nj X K ξ i X K ξ j nj µ x K ξ i x K ξ j nj. Now the transformation of contravariant components of the normal vector to from Cartesians is n j ξj x L n L, which means that the equation becomes X K ξ i X K ξ j ξ j x L n L µ x K ξ i x K ξ j ξ j x L n L, and so X K X K ξ iδ JLn L µ x K x K ξ δ JLn i L XK X K µ x K x K n J. ξ i The transformation / ξ i is assumed to be non-singular and the coordinates x I are independent, which means that the equation becomes XK X K µδ KI δ KJ n J, c IJ n J µδ IJ n J c IJ n J µn I, where c IJ X K / X K /. The principal stretches are therefore the square roots of the eigenvalues of the matrix c IJ, which for the deformation given in question 6. is α α α c F T F α α α. α α α Hence the eigenvalues are all α with eigenvectors given by,,,,, and,,. The principal stretches are then all equal to α, as we should expect from the nature of the deformation. 8. The deformation is given by X X X 3 γ a. The deformation gradient tensor is given by F γ x x x 3.

9 9 and so the Green Lagrange strain tensor is e FT F I γ γ γ γ γ.. b. The extreme values of the stretches are given by the principal stretches, which are the square-roots of the eigenvalues of the tensor c γ. γ +γ The eigenvalues are given by the roots of the characteristic equation µ[ µ+γ µ γ ] µ[+γ µ γ µ µγ +µ ] which has roots so µ[ µ+γ +µ ], µ,µ +γ ± +γ 4 µ,+ γ ±γ +γ ±γ 4+γ, + 4 γ. The associated directions eigenvectors are given by the solutions of the linear equations n n γ n µ n. γ +γ n 3 n 3 Thus, the eigenvalue associated with µ is simply e, µ, n,, T. Using the third linear equation for the other eigenvalues we have that + 4 γ γn ++γ n 3 + γ ±γ n 3 γn γ ±γ + 4 γ n 3 ; which means that if we choose n 3, then the associated eigenvalues are n, γ ± + T, 4 γ. The maximum stretch is given by the positive root and the minimum stretch by the negative root.

10 c. In general coordinates the angle between the deformed line elements is given by cosθ n i G ij m j ni G ij n j m i G ij m j; from the lecture notes. Transforming to Cartesian coordinates gives cosθ n I c IJ m J ni c IJ n J mi c IJ c J, and we choose n,, and m,, in the Cartesian coordinates, so cosθ c 3 c c 33 γ +γ. Hence the change in angle or shear between the two line elements is π γ cos. +γ As γ, cosθ and so there is no change in angle between the two line elements. As γ, cosθ and the change in angle tends to π/. 9. a. The entire bar is stretched by the amount λ in the x direction so the initial deformationisgivenby X λ x. TheseconddeformationisgivenbyX λ X and so the entire deformation is given by X λ λ x. b. The Green Lagrange strain tensor associated with the first deformation is simply e λ ; the incremental Green Lagrange strain tensor associated with the second deformation is ẽ λ, becauseweassumethatthelagrangiancoordinateisλ x. TheGreen Lagrange strain tensor associated with the entire deformation is ẽ Total λ λ. There is no simple relationship between the three tensors. c. The Hencky strain tensor associated with the first deformation is e H lnλ. Theincremental Hencky strain tensor associated with the second deformation is ẽ H lnλ.

11 The Hencky strain tensor associated with the total deformation is e HTotal lnλ λ. There is a simple relationship between the three tensors: e HTotal lnλ λ lnλ +lnλ e H +ẽ H. Thus, the sum of incremental Hencky strain tensors gives the total strain. For this reason it is sometimes called the true strain tensor and is often used in plasticity theory.. a. The deformation is given by X cosωt sinωt X sinωt cosωt X 3 +αt Thus the Lagrangian velocity is given by v ω sinωt ω cosωt v ωcosωt ωsinωt v 3 α The Eulerian velocity is found by making use of the inverse deformation x cosωt sinωt X x sinωt cosωt X x 3 +αt X 3 to write V V V 3 ω sinωt ω cosωt ωcosωt ωsinωt α ω ω α/+αt x x x 3 x x x 3.. cosωt sinωt sinωt cosωt +αt X X X 3 Hence, the Eulerian velocity gradient tensor is given by ω V I,J ω. α/+αt X X X 3 b. The rate of deformation tensor is the symmetric part of V I,J and the spin tensor is the antisymmetric part so ω D IJ W IJ ω. α/+αt The deformation consists of a rotation in the x -x plane with angular velocity ω and a stretch in the x 3 direction that increases with time. The rotation is associated only with the spin tensor and the stretch only with the rate of deformation tensor.

12 . The Eulerian rate of strain tensor is related to the Eulerian rate of deformation tensor by the formula Ė IJ D IJ E IK V K,J E JK V K,I, see the lecture notes. In one dimension we can only have and D V, which means that Ė D E V, E V, Ė V, E D E. Thus, if the rate of deformation tensor is zero, then the rate of strain tensor must also be zero. If the rate of deformation tensor is not zero then the rate of strain tensor is proportional to the rate of deformation, but will never equal it unless the strain is instantaneously zero.. a. The deformation is a stretch of α in the x direction followed by rigid-body rotation about the origin of the Cartesian coordinate system. b. The Eulerian rate of deformation tensor can be found by using the same techniques as in q. 9. The Eulerian velocity will be given by V αω sinωt ω cosωt cosωt/α sinωt/α X V αωcosωt ωsinωt sinωt cosωt X V 3 X 3 which means that ω ω V I,J W IJ X X X 3 ω ω where D IJ and there is no rate of deformation because the body is simply rotating rigidly after the initial instantaneous stretch. The spatial deformation gradient tensor is H IJ cosωt/α sinωt/α sinωt cosωt which means that the Eulerian strain tensor is then E IJ cosωt/α sinωt sinωt/α cosωt cos ωt α cosωtsinωtα cosωtsinωtα sin ωt α cosωt/α sinωt/α sinωt cosωt.

13 3 Note that the strain tensor is a function of t because the strained body is rotating rigidly about the origin. When t nπ/ω, the only non-zero component is E α, and when t nπ/ω, the only non-zero component is E α. The rate of strain tensor is given by which gives as non-zero entries Ė IJ D IJ E KI V K,J E JK V K,I, Ė E V, ωcosωtsinωtα, Ė Ė E V, E V, ωe E ω α sin ωt cos ωt, Thus we have Ė E V, ωcosωtsinωtα Ė IJ ωα sinωt cosωt cosωt sinωt which is only zero and therefore coincides with D IJ when α α ±. 3. For ease of notation, we drop the overbars in this solution so that in general coordinates ξ i we can write the i-th component of the cross product as From the definition of ω, we have that and so [ω v] i ǫ ijk ω j v k. ω j ǫ jmn W nm, [ω v] i ǫ ijk ǫ jmn W nm v k. Now using the relationship from Example Sheet I, q 5: and then ǫ ijk ǫ jmn g m k gn i g n k gm i, [ω v] i g m k gn i g n k gm i W nmv k W ik v k W ki v k W ik v k, using the antisymmetry property of the spin tensor. Thus The Eulerian acceleration is given by W v ω v. A V t +V R V.

14 4 For simplicity let s work in Cartesian coordinates in which case we can write A I V I t +V JV I,J. Adding and subtracting the term V J V J,I gives A I V I t +V I,J V J,I V J +V J,I V J, V I t +W IJV J + V JV J,I, which can be written in dyadic notation as A V t +W V + V V ; R and using the result above we finally obtain A V t +ω V + R V. 4. This is essentially a direct application of the Stokes theorem Γ V dr V d A ω d A. If we consider the limit as the area becomes very small then ω is approximately constant over the area and we can assume that the normal is also constant so that we have Γ ω nda ω n Γ/dA. Thus, the ratio of the circulation around a closed loop to the area enclosed by the loop is equal to the component of vorticity normal to the surface in the limit as the area tends to zero.