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Simple examples illustrating the use of the deformation gradient tensor

Nasser M. Abbasi

February 3, 2006 page compiled on July 2, 2015 at 5:14pm

Contents

1 Introduction
2 Examples
 2.1 Square shape becomes longer with width fixed
 2.2 Square shape becomes both longer and wider.
 2.3 square shape becomes wider and pulled at an angle.

1 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 dR  in the undeformed 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 dR  in the undeformed shape is mapped to the vector dr  in the deformed shape by successive application of the stretch tensor ˜U  followed by a parallel translation operation, and followed by the application of the rotation tensor R˜ .

The point that dR  is located at is labeled P  in the undeformed shape, and its image will be labeled       ′
     P in the deformed shape. The coordinates in the undeformed shape will be upper case X1, X2   and in the deformed shape will be lower case x1,x2   .

One observation found is that if the deformation is such that perpendicular lines in the undeformed shape remain perpendicular to each others in the deformed 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.

2 Examples

2.1 Square shape becomes longer with width fixed

The following diagram is the undeformed configuration.

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In this shape, the vector dR  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 3 times as long as before.

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We observe the following. The lines A,B,C have moved to new locations in the deformed configuration. For instance, the line A started at (0, 1)  and ended at (3,1)  in the undeformed shape coordinates. While the same line now labeled lower case a  , starts from (0, 3)  and ends at (3,3)  in the deformed shape using the undeformed coordinates system.

The first step in finding F˜  is to determine the mapping between the X  coordinates in the undeformed shape, and the x  coordinates in the deformed 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

x1 = X1

and since the new shape is 3 times as long as before then

x2 = 3X2

And now we can calculate ˜F.  Since

˜F =  [∂∂xX11-  ∂∂xX12]
      ∂x2-  ∂x2-
      ∂X1   ∂X2

then given that ∂x1-     ∂x1-    -∂x2-     ∂x2-
∂X1 = 1, ∂X2 = 0,∂X1 =  0,∂X2 = 3  we obtain the numerical value for ˜
F

    [     ]
˜F =  1  0
     0  3

We note here that F˜  is the same for any region of the deformed shape. This is because the deformation is uniform.

Now we can find dr  .

dr =  ˜F ⋅ dR

Since from the undeformed shape we see that

dR  = e1 + e2

Then

pict

hence,

dr  = e1 + 3e2

Looking at the deformed shape we see that this agrees with the expected shape of the deformed dr  vector.

Now once ˜F    is found, we can determine the stretch tensor U˜  and the rotation tensor R˜ .

We will do this algebraically first, then verify the result geometrically. Since by definition

˜F =  ˜R ⋅U˜

Once ˜F  is known, we can find U˜  using the relation

pict

Now we take the square root of the matrix ˜U2   to find U˜ 1

     [     ]
U˜ =   1  0
       0  3

and now that ˜
U  is known, we can find ˜
R

pict

To verify this result algebraically, we write

pict

Which agrees with earlier result.

To verify the result geometrically, we first apply the stretch tensor U˜  to dR  , 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 deformed configuration, where the point P′ is the image of the point P  in the undeformed shape, and then we apply the rotation tensor ˜R  to dr∗ to obtain dr .

Hence

pict

Now we apply the rotation of      [    ]
      1  0
˜R  =
      0  1 to dr ∗ , and since the rotation is a unit tensor, then this operation will produce no effect.

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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 3 times as long as the original length, and the width be 1.5 times as wide as the original width.

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As before, the first step in finding F˜  is to determine the mapping between the X  coordinates in the undeformed shape, and the x  coordinates in the deformed shape. In this example, this mapping is constant over any region of the shape. We see that

x1 = 1.5X1

and since the new shape is 3 times as long as before then

x2 = 3X2

And now we can calculate ˜
F.  Since

     [∂x1-  ∂x1]
˜F =   ∂X1   ∂X2
      ∂x2-  ∂x2-
      ∂X1   ∂X2

then given that ∂x1-      -∂x1-     ∂x2-     ∂x2-
∂X1 = 1.5,∂X2 =  0,∂X1 = 0, ∂X2 = 3  we obtain numerical value for ˜
F

˜F = [1.5  0]
      0   3

Now let us find dr  .

      ˜
dr =  F ⋅ dR

From the undeformed shape we see that

dR  = e1 + e2

Hence

pict

hence,

dr =  1.5e1 + 3e2

Looking at the deformed shape we see that this is indeed the case.

Now once ˜F    is found, we can determine the stretch tensor U˜  and the rotation tensor  ˜
R .

We will do this algebraically first, then verify the result geometrically.

˜F =  ˜R ⋅U˜

Once ˜
F  is known, we can find  ˜
U

pict

Hence

    [       ]
      1.5  0
˜U =
       0  3

and now that ˜U  is known, we can find ˜R

pict

To verify the result geometrically, we first apply the stretch ˜U  to dR  , 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 deformed configuration, where the point   ′
P is the image of the point P,  and then we apply the rotation ˜
R  to   ∗
dr to obtain dr .

Hence

pict

Now we apply the rotation of      [    ]
˜R =   1  0
      0  1 to dr ∗ , and since the rotation is a unit tensor, then no rotation will occur.

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2.3 square shape becomes wider and pulled at an angle.

In this example, the same undeformed shape shown in earlier examples will be deformed to cause the rotation tensor to be something other than the identity tensor. We assume the following deformation

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The above deformation is constructed such that

pict

Now we can calculate F˜.  Since

     [         ]
˜     ∂∂xX11-  ∂∂xX12-
F =   ∂x2-  ∂x2-
      ∂X1   ∂X2

then given that ∂x       ∂x       ∂x       ∂x
∂X11-= 2, ∂X12-= 0,∂X21-=  1,∂X22-= 1  we obtain numerical value for ˜F

    [     ]
˜    2  0
F =  1  1

Now we can find dr  .

dr =  ˜F ⋅ dR

From the undeformed shape we see that

dR  = e1 + e2

Hence

pict

hence,

dr = 2e1 + 2e2

Looking at the deformed shape we see that this is indeed the case. Now once F˜    is found, we can determine the stretch tensor U˜  and the rotation tensor ˜R  .

We will do this algebraically first, then verify the result geometrically.

˜    ˜   ˜
F =  R ⋅U

Once ˜
F  is known, we can find  ˜
U

pict

Hence

     [               ]
U˜ =  2.2136   0.3162
      0.3162   0.9487

and now that ˜U  is known, we can find ˜R

pict

To verify the result geometrically, we first apply the stretch tensor  ˜
U  to dR  , 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 deformed configuration, where the point P ′ is the image of the point P,  and then we apply the rotation tensor ˜R  to dr∗ to obtain dr .

Hence

pict

Now we apply the rotation to R˜  to dr ∗ to obtain dr

pict

which agrees with the result obtained above.

The following diagram illustrates geometrically the action of R˜  and ˜U.

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