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## Simple examples illustrating the use of the deformation gradient tensor

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

### 1 Introduction

This note illustrates using simple examples, how to evaluate the deformation gradient tensor 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 undeformed shape. The shape is then assumed to undergo a ﬁxed form of deformation such that is constant over the whole body (as opposed to being a ﬁeld tensor where would be a function of the position). Then the tensor is computed and shown using diagrams how the diﬀerential vector in the undeformed shape is mapped to the vector in the deformed shape by successive application of the stretch tensor followed by a parallel translation operation, and followed by the application of the rotation tensor .

The point that is located at is labeled in the undeformed shape, and its image will be labeled in the deformed shape. The coordinates in the undeformed shape will be upper case and in the deformed shape will be lower case .

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 will come out to be the identity tensor. The ﬁrst 2 examples below illustrate this case. In the third example the rotation tensor 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.

In this shape, the vector extends from the point to the point . 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.

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 and ended at in the undeformed shape coordinates. While the same line now labeled lower case , starts from and ends at in the deformed shape using the undeformed coordinates system.

The ﬁrst step in ﬁnding is to determine the mapping between the coordinates in the undeformed shape, and the 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

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

And now we can calculate Since

then given that we obtain the numerical value for

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

Now we can ﬁnd .

Since from the undeformed shape we see that

Then

hence,

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

Now once is found, we can determine the stretch tensor and the rotation tensor .

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

Once is known, we can ﬁnd using the relation

Now we take the square root of the matrix to ﬁnd 1

and now that is known, we can ﬁnd

To verify this result algebraically, we write

Which agrees with earlier result.

To verify the result geometrically, we ﬁrst apply the stretch tensor to , this results in a new diﬀerential vector which we call , then we slide without changing its slope (i.e. parallel translation) such that the vector starts at the point in the deformed conﬁguration, where the point is the image of the point in the undeformed shape, and then we apply the rotation tensor to to obtain .

Hence

Now we apply the rotation of to , and since the rotation is a unit tensor, then this operation will produce no eﬀect.

#### 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.

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

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

And now we can calculate Since

then given that we obtain numerical value for

Now let us ﬁnd .

From the undeformed shape we see that

Hence

hence,

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

Now once is found, we can determine the stretch tensor and the rotation tensor .

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

Once is known, we can ﬁnd

Hence

and now that is known, we can ﬁnd

To verify the result geometrically, we ﬁrst apply the stretch to , this results in a new diﬀerential vector which we call , then we slide without changing its slope (i.e. parallel translation) such that the vector starts at the point in the deformed conﬁguration, where the point is the image of the point and then we apply the rotation to to obtain .

Hence

Now we apply the rotation of to , and since the rotation is a unit tensor, then no rotation will occur.

#### 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

The above deformation is constructed such that

Now we can calculate Since

then given that we obtain numerical value for

Now we can ﬁnd .

From the undeformed shape we see that

Hence

hence,

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

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

Once is known, we can ﬁnd

Hence

and now that is known, we can ﬁnd

To verify the result geometrically, we ﬁrst apply the stretch tensor to , this results in a new diﬀerential vector which we call , then we slide without changing its slope (i.e. parallel translation) such that the vector starts at the point in the deformed conﬁguration, where the point is the image of the point and then we apply the rotation tensor to to obtain .

Hence

Now we apply the rotation to to to obtain

which agrees with the result obtained above.

The following diagram illustrates geometrically the action of and