Simple examples illustrating the use of the deformation gradient tensor

Transcription

1 Simple examples illustrating the use of the deformation gradient tensor Nasser M. bbasi February, 2006 compiled on Wednesday January 0, 2018 at 08: M ontents 1 Introduction 1 2 Examples Square shape becomes longer with width fixed Square shape becomes both longer and wider square shape becomes wider and pulled at an angle Introduction This note illustrates using simple examples, how to evaluate the deformation gradient tensor F and derive its polar decomposition into a stretch and rotation tensors. Diagrams are used to help illustrate geometrically the effect of applying the stretch and the rotation tensors on a differential vector with the purpose of giving better insight into these operations. For simplicity, only 2D shapes are used. Starting by selecting some arbitrary differential vector in the un shape. The shape is then assumed to undergo a fixed form of deformation such that F is constant over the whole body (as opposed to being a field tensor where F would be a function of the position). Then the tensor F is computed and shown using diagrams how the differential vector in the un shape is mapped to the vector dr in the shape by successive application of the stretch tensor Ũ followed by a parallel translation operation, and followed by the application of the rotation tensor R. The point that is located at is labeled P in the un shape, and its image will be labeled P in the shape. The coordinates in the un shape will be upper case X 1, X 2 and in the shape will be lower case x 1, x 2. One observation found is that if the deformation is such that perpendicular lines in the un shape remain perpendicular to each others in the shape, then this implies that the rotation tensor R will come out to be the identity tensor. The first 2 examples below illustrate this case. In the third example the rotation tensor R is not the identity tensor because lines do not remain perpendicular to each others after deformation. 1

2 2 Examples 2.1 Square shape becomes longer with width fixed The following diagram is the un. (0,) (1,) (2,) (,) (1,2) (2,2) (,2) (1,1) (,1) Un X-coordinates system In this shape, the vector extends from the point (1, 1) to the point (2, 2). In this example, we assume a deformation whereby the shape is pulled upwards by some distance, causing the shape to become longer in the vertical direction and we assume the shape remain the same width. This is the simplest form of deformation. Let us assume for simplicity that the shape becomes times as long as before. 2

3 Traction force t (per unit area) c (0,) (1,) (2,) (,) b (1,2) (2,2) (,2) dr (0,) (1,) (2,) (,) a (1,1) (,1) (1,2) (1,1) (2,2) (,2) (,1) Un X-coordinates We observe the following. The lines,, have moved to new locations in the. For instance, the line started at (0, 1) and ended at (, 1) in the un shape coordinates. While the same line now labeled lower case a, starts from (0, ) and ends at (, ) in the shape using the un coordinates system. The first step in finding F is to determine the mapping between the X coordinates in the

4 un shape, and the x coordinates in the shape. In this example this mapping is constant over any region of the shape. We see immediately that since the width of the shape did not change, then x 1 X 1 and since the new shape is times as long as before then x 2 X 2 nd now we can calculate F. Since F [ x1 x 2 x 1 x 2 ] then given that x 1 1, x 1 0, x 2 0, x 2 we obtain the numerical value for F [ ] 1 0 F 0 We note here that F is the same for any region of the shape. deformation is uniform. Now we can find dr. This is because the Since from the un shape we see that dr F e 1 + e 2 Then [ ] [ ] dr 0 1 [ ] 1 hence, dr e 1 + e 2 Looking at the shape we see that this agrees with the expected shape of the dr vector. Now once F is found, we can determine the stretch tensor Ũ and the rotation tensor R. We will do this algebraically first, then verify the result geometrically. Since by definition F R Ũ Once F is known, we can find Ũ using the relation Ũ 2 F T F [ ] [ ] [ ]

5 Now we take the square root of the matrix Ũ 2 to find Ũ 1 and now that Ũ is known, we can find R To verify this result algebraically, we write Ũ [ ] R F Ũ 1 [ ] [ ] [ ] dr F R Ũ [ ] [ ] R 0 1 [ ] 1 R [ ] [ ] [ ] 1 dr e 1 + e 2 Which agrees with earlier result. To verify the result geometrically, we first apply the stretch tensor Ũ to, this results in a new differential vector which we call dr, then we slide dr without changing its slope (i.e. parallel translation) such that the vector dr starts at the point P in the, where the point P is the image of the point P in the un shape, and then we apply the rotation tensor R to dr to obtain dr. dr Ũ e 1 + e 2 [ ] [ ] To obtain the square root of a matrix, say matrix, follow these steps. 1. Determine the eigenvalues λ of the matrix. 2. For each eigenvalue λ n determine the correspending eigenvector V n. onstruct the Matrix N whose columns are the eigenvectors V n. i.e. the first column will be the vector V 1 etc onstruct matrix M with diagonal elements that contains the λ n. i.e. M (1, 1) λ 1, M (2, 2) λ 2, and so forth. (This is the Jordan form for real distinct eigenvalues) 5. Now N T MN In Matlab, the command expm() can be used to calculate sqrt of a matrix. [ ] 1 5

6 [ ] 1 0 Now we apply the rotation of R to dr, and since the rotation is a unit tensor, then this 0 1 operation will produce no effect. Traction force t (per unit area) Traction force t (per unit area) Traction force t (per unit area) (0,) (1,) (2,) (0,) (1,) (2,) (0,) (1,) (2,) (0,) Stretch tensor (1,) (1,2) P (1,1) Ũ (2,) (2,2) Un X-coordinates P (1,2) (1,2) (1,1) (2,) Parallel translation Of vector to point P, the image of point P R Rotation tensor Point P in the is the image of point P in the un configuratroin P (1,2) (1,2) (1,1) (2,) P (1,2) (1,2) dr (1,1) (2,) Steps taken when processing the deformation gradient tensor F 2.2 Square shape becomes both longer and wider. In this example we start with the same original shape as above, but we increase both the length and the width of the shape and not just its length. Let the length be times as long as the original length, and the width be 1.5 times as wide as the original width. 6

7 Traction force t (per unit area) c (0,) (1,) (2,) (,) b (1,2) (2,2) (,2) (0,) (1,) (1,2) (1,1) (2,) (2,2) Un X-coordinates (,) (,2) (,1) a dr (1,1) (,1) Traction force t (per unit area) s before, the first step in finding F is to determine the mapping between the X coordinates in the un shape, and the x coordinates in the shape. In this example, this mapping is constant over any region of the shape. We see that x 1 1.5X 1 and since the new shape is times as long as before then x 2 X 2 nd now we can calculate F. Since F [ x1 x 2 x 1 x 2 ] then given that x 1 1.5, x 1 0, x 2 0, x 2 we obtain numerical value for F [ ] F 0 7

8 Now let us find dr. From the un shape we see that dr F hence, e 1 + e 2 [ ] [ ] dr 0 1 [ ] 1.5 dr 1.5e 1 + e 2 Looking at the shape we see that this is indeed the case. Now once F is found, we can determine the stretch tensor Ũ and the rotation tensor R. We will do this algebraically first, then verify the result geometrically. Once F is known, we can find Ũ F R Ũ and now that Ũ is known, we can find R Ũ 2 F T F [ ] [ ] Ũ [ ] R F Ũ 1 [ ] [ ] [ ] [ ] To verify the result geometrically, we first apply the stretch Ũ to, this results in a new differential vector which we call dr, then we slide dr without changing its slope (i.e. parallel translation) such that the vector dr starts at the point P in the, where the point P is the image of the point P, and then we apply the rotation R to dr to obtain dr. dr Ũ [ ] [ ] 1 1 e 1 + e 2 [ ] 1 0 Now we apply the rotation of R to dr, and since the rotation is a unit tensor, then no 0 1 rotation will occur. 8 [ ] 1

9 Parallel translation Of vector to point P, the image of point P R Rotation tensor (0,) (1,) (2,) (0,) (1,) (2,) (0,) (1,) (2,) (,) Stretch tensor Ũ (1,2) (2,2) (1,2) (2,2) (1,2) (2,2) (,2) dr (0,) (1,) (2,) (1,1) (1,1) (1,1) (,1) (1,2) (1,1) (2,2) Un X-coordinates 2. square shape becomes wider and pulled at an angle. In this example, the same un shape shown in earlier examples will be to cause the rotation tensor to be something other than the identity tensor. We assume the following deformation (0,) (1,) (2,) (1,2) (1,1) (2,2) Un X-coordinates 9

10 The above deformation is constructed such that x 1 2X 1 x 2 X 1 + X 2 Now we can calculate F. Since F [ x1 x 2 x 1 x 2 ] then given that x 1 2, x 1 0, x 2 1, x 2 1 we obtain numerical value for F [ ] 2 0 F 1 1 Now we can find dr. From the un shape we see that dr F e 1 + e 2 [ ] [ ] dr [ ] 2 2 hence, dr 2e 1 + 2e 2 Looking at the shape we see that this is indeed the case. Now once F is found, we can determine the stretch tensor Ũ and the rotation tensor R. We will do this algebraically first, then verify the result geometrically. Once F is known, we can find Ũ F R Ũ Ũ 2 F T F [ ] [ ] [ ] Ũ [ ]

11 and now that Ũ is known, we can find R R F Ũ 1 [ ] [ ] R [ ] R To verify the result geometrically, we first apply the stretch tensor Ũ to, this results in a new differential vector which we call dr, then we slide dr without changing its slope (i.e. parallel translation) such that the vector dr starts at the point P in the, where the point P is the image of the point P, and then we apply the rotation tensor R to dr to obtain dr. dr Ũ [ e e 2 Now we apply the rotation to R to dr to obtain dr ] [ ] 1 1 [ ] dr R dr [ ] [ ] [ ] 2 2 2e 1 + 2e 2 which agrees with the result obtained above. The following diagram illustrates geometrically the action of R and Ũ. 11

12 pply stretch tensor Ũ dr Ũ P P Parallel translation pply rotation tensor R Of vector to point P, the image of point P dr R dr dr P P 12