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

February 3, 2006   Compiled on May 23, 2020 at 2:17am

### 1 Introduction

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

The point that $$\mathbf{dR}$$ is located at is labeled $$P$$ in the undeformed shape, and its image will be labeled $$P^{\prime }$$ in the deformed shape. The coordinates in the undeformed shape will be upper case $$X_{1},X_{2}$$ and in the deformed shape will be lower case $$x_{1},x_{2}$$.

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 $$\mathbf{\tilde{R}}$$ will come out to be the identity tensor. The ﬁrst 2 examples below illustrate this case. In the third example the rotation tensor $$\mathbf{\tilde{R}}$$ 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. In this shape, the vector $$\mathbf{dR}$$ extends from the point $$\left ( 1,1\right )$$ to the point $$\left ( 2,2\right )$$. 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 $$\left ( 0,1\right )$$ and ended at $$\left ( 3,1\right )$$ in the undeformed shape coordinates. While the same line now labeled lower case $$a$$, starts from $$\left ( 0,3\right )$$ and ends at $$\left ( 3,3\right )$$ in the deformed shape using the undeformed coordinates system.

The ﬁrst step in ﬁnding $$\mathbf{\tilde{F}}$$ 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 $x_{1}=X_{1}$ and since the new shape is 3 times as long as before then $x_{2}=3X_{2}$ And now we can calculate $$\mathbf{\tilde{F}.}$$ Since $\mathbf{\tilde{F}=}\begin{bmatrix} \frac{\partial x_{1}}{\partial X_{1}} & \frac{\partial x_{1}}{\partial X_{2}}\\ \frac{\partial x_{2}}{\partial X_{1}} & \frac{\partial x_{2}}{\partial X_{2}}\end{bmatrix}$ then given that $$\frac{\partial x_{1}}{\partial X_{1}}=1,\frac{\partial x_{1}}{\partial X_{2}}=0,\frac{\partial x_{2}}{\partial X_{1}}=0,\frac{\partial x_{2}}{\partial X_{2}}=3$$ we obtain the numerical value for $$\mathbf{\tilde{F}}$$ $\mathbf{\tilde{F}=}\begin{bmatrix} 1 & 0\\ 0 & 3 \end{bmatrix}$ We note here that $$\mathbf{\tilde{F}}$$ is the same for any region of the deformed shape. This is because the deformation is uniform.

Now we can ﬁnd $$\mathbf{dr}$$.

$\mathbf{dr}=\mathbf{\tilde{F}}\cdot \mathbf{dR}$ Since from the undeformed shape we see that $\mathbf{dR}=\mathbf{e}_{1}+\mathbf{e}_{2}$ Then \begin{align*} \mathbf{dr} & =\begin{bmatrix} 1 & 0\\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix} \\ & =\begin{bmatrix} 1\\ 3 \end{bmatrix} \end{align*}

Hence$\mathbf{dr}=\mathbf{e}_{1}+3\mathbf{e}_{2}$ Looking at the deformed shape we see that this agrees with the expected shape of the deformed $$\mathbf{dr}$$ vector.

Now once $$\mathbf{\tilde{F}}$$ is found, we can determine the stretch tensor $$\mathbf{\tilde{U}}$$ and the rotation tensor $$\mathbf{\tilde{R}}$$. We will do this algebraically ﬁrst, then verify the result geometrically. Since by deﬁnition$\mathbf{\tilde{F}=\tilde{R}\cdot \tilde{U}}$ Once $$\mathbf{\tilde{F}}$$ is known, we can ﬁnd $$\mathbf{\tilde{U}}$$ using the relation\begin{align*} \mathbf{\tilde{U}}^{2} & \mathbf{=}\mathbf{\tilde{F}}^{T}\cdot \mathbf{\tilde{F}}\\ & =\begin{bmatrix} 1 & 0\\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & 3 \end{bmatrix} =\begin{bmatrix} 1 & 0\\ 0 & 9 \end{bmatrix} \end{align*}

Now we take the square root of the matrix $$\mathbf{\tilde{U}}^{2}$$ to ﬁnd $$\mathbf{\tilde{U}}$$1 $\mathbf{\tilde{U}=}\begin{bmatrix} 1 & 0\\ 0 & 3 \end{bmatrix}$ and now that $$\mathbf{\tilde{U}}$$ is known, we can ﬁnd $$\mathbf{\tilde{R}}$$ \begin{align*} \mathbf{\tilde{R}} & \mathbf{=}\mathbf{\tilde{F}\cdot \tilde{U}}^{-1}\\ & =\begin{bmatrix} 1 & 0\\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & \frac{1}{3}\end{bmatrix} =\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{align*}

To verify this result algebraically, we write\begin{align*} \mathbf{dr} & =\mathbf{\tilde{F}}\cdot \mathbf{dR}\\ & =\mathbf{\tilde{R}\cdot \tilde{U}}\cdot \mathbf{dR}\\ & =\mathbf{\tilde{R}\cdot }\begin{bmatrix} 1 & 0\\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix} \\ & =\mathbf{\tilde{R}\cdot }\begin{bmatrix} 1\\ 3 \end{bmatrix} \\ & =\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1\\ 3 \end{bmatrix} \\ & =\begin{bmatrix} 1\\ 3 \end{bmatrix} \\ \mathbf{dr} & =\mathbf{e}_{1}+3\mathbf{e}_{2} \end{align*}

Which agrees with earlier result.

To verify the result geometrically, we ﬁrst apply the stretch tensor $$\mathbf{\tilde{U}}$$ to $$\mathbf{dR}$$, this results in a new diﬀerential vector which we call $$\mathbf{dr}^{\ast }$$, then we slide $$\mathbf{dr}^{\ast }$$ without changing its slope (i.e. parallel translation) such that the vector $$\mathbf{dr}^{\ast }$$ starts at the point $$P^{\prime }$$ in the deformed conﬁguration, where the point $$P^{\prime }$$ is the image of the point $$P$$ in the undeformed shape, and then we apply the rotation tensor $$\mathbf{\tilde{R}}$$ to $$\mathbf{dr}^{\ast }$$ to obtain $$\mathbf{dr}$$.

Hence \begin{align*} \mathbf{dr}^{\ast } & =\mathbf{\tilde{U}\cdot dR=}\begin{bmatrix} 1 & 0\\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix} =\begin{bmatrix} 1\\ 3 \end{bmatrix} \\ & =\mathbf{e}_{1}+\mathbf{3e}_{2} \end{align*}

Now we apply the rotation of $$\mathbf{\tilde{R}=}\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$ to $$\mathbf{dr}^{\ast }$$, 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 $$\mathbf{\tilde{F}}$$ 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 $x_{1}=1.5X_{1}$ and since the new shape is 3 times as long as before then $x_{2}=3X_{2}$ And now we can calculate $$\mathbf{\tilde{F}.}$$ Since $\mathbf{\tilde{F}=}\begin{bmatrix} \frac{\partial x_{1}}{\partial X_{1}} & \frac{\partial x_{1}}{\partial X_{2}}\\ \frac{\partial x_{2}}{\partial X_{1}} & \frac{\partial x_{2}}{\partial X_{2}}\end{bmatrix}$ then given that $$\frac{\partial x_{1}}{\partial X_{1}}=1.5,\frac{\partial x_{1}}{\partial X_{2}}=0,\frac{\partial x_{2}}{\partial X_{1}}=0,\frac{\partial x_{2}}{\partial X_{2}}=3$$ we obtain numerical value for $$\mathbf{\tilde{F}}$$ $\mathbf{\tilde{F}=}\begin{bmatrix} 1.5 & 0\\ 0 & 3 \end{bmatrix}$ Now let us ﬁnd $$\mathbf{dr}$$.$\mathbf{dr}=\mathbf{\tilde{F}}\cdot \mathbf{dR}$ From the undeformed shape we see that $\mathbf{dR}=\mathbf{e}_{1}+\mathbf{e}_{2}$ Hence \begin{align*} \mathbf{dr} & =\begin{bmatrix} 1.5 & 0\\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix} \\ & =\begin{bmatrix} 1.5\\ 3 \end{bmatrix} \end{align*}

hence, $\mathbf{dr}=1.5\mathbf{e}_{1}+3\mathbf{e}_{2}$ Looking at the deformed shape we see that this is indeed the case.

Now once $$\mathbf{\tilde{F}}$$ is found, we can determine the stretch tensor $$\mathbf{\tilde{U}}$$ and the rotation tensor $$\mathbf{\tilde{R}}$$.

We will do this algebraically ﬁrst, then verify the result geometrically. $\mathbf{\tilde{F}=\tilde{R}\cdot \tilde{U}}$ Once $$\mathbf{\tilde{F}}$$ is known, we can ﬁnd $$\mathbf{\tilde{U}}$$\begin{align*} \mathbf{\tilde{U}}^{2} & \mathbf{=}\mathbf{\tilde{F}}^{T}\cdot \mathbf{\tilde{F}}\\ & =\begin{bmatrix} 1.5 & 0\\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1.5 & 0\\ 0 & 3 \end{bmatrix} =\begin{bmatrix} 2.25 & 0\\ 0 & 9 \end{bmatrix} \end{align*}

Hence $\mathbf{\tilde{U}=}\begin{bmatrix} 1.5 & 0\\ 0 & 3 \end{bmatrix}$ and now that $$\mathbf{\tilde{U}}$$ is known, we can ﬁnd $$\mathbf{\tilde{R}}$$ \begin{align*} \mathbf{\tilde{R}} & \mathbf{=}\mathbf{\tilde{F}\cdot \tilde{U}}^{-1}\\ & =\begin{bmatrix} 1.5 & 0\\ 0 & 3 \end{bmatrix}\begin{bmatrix} \frac{2}{3} & 0\\ 0 & \frac{1}{3}\end{bmatrix} =\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{align*}

To verify the result geometrically, we ﬁrst apply the stretch $$\mathbf{\tilde{U}}$$ to $$\mathbf{dR}$$, this results in a new diﬀerential vector which we call $$\mathbf{dr}^{\ast }$$, then we slide $$\mathbf{dr}^{\ast }$$ without changing its slope (i.e. parallel translation) such that the vector $$\mathbf{dr}^{\ast }$$ starts at the point $$P^{\prime }$$ in the deformed conﬁguration, where the point $$P^{\prime }$$ is the image of the point $$P,$$ and then we apply the rotation $$\mathbf{\tilde{R}}$$ to $$\mathbf{dr}^{\ast }$$ to obtain $$\mathbf{dr}$$. Hence \begin{align*} \mathbf{dr}^{\ast } & =\mathbf{\tilde{U}\cdot dR=}\begin{bmatrix} 1 & 0\\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix} =\begin{bmatrix} 1\\ 3 \end{bmatrix} \\ & =\mathbf{e}_{1}+\mathbf{3e}_{2} \end{align*}

Now we apply the rotation of $$\mathbf{\tilde{R}=}\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$ to $$\mathbf{dr}^{\ast }$$, 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 \begin{align*} x_{1} & =2X_{1}\\ x_{2} & =X_{1}+X_{2} \end{align*}

Now we can calculate $$\mathbf{\tilde{F}.}$$ Since $\mathbf{\tilde{F}=}\begin{bmatrix} \frac{\partial x_{1}}{\partial X_{1}} & \frac{\partial x_{1}}{\partial X_{2}}\\ \frac{\partial x_{2}}{\partial X_{1}} & \frac{\partial x_{2}}{\partial X_{2}}\end{bmatrix}$ then given that $$\frac{\partial x_{1}}{\partial X_{1}}=2,\frac{\partial x_{1}}{\partial X_{2}}=0,\frac{\partial x_{2}}{\partial X_{1}}=1,\frac{\partial x_{2}}{\partial X_{2}}=1$$ we obtain numerical value for $$\mathbf{\tilde{F}}$$ $\mathbf{\tilde{F}=}\begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix}$ Now we can ﬁnd $$\mathbf{dr}$$.

$\mathbf{dr}=\mathbf{\tilde{F}}\cdot \mathbf{dR}$ From the undeformed shape we see that $\mathbf{dR}=\mathbf{e}_{1}+\mathbf{e}_{2}$ Hence \begin{align*} \mathbf{dr} & =\begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix} \\ & =\begin{bmatrix} 2\\ 2 \end{bmatrix} \end{align*}

Therefore$\mathbf{dr}=2\mathbf{e}_{1}+2\mathbf{e}_{2}$ Looking at the deformed shape we see that this is indeed the case. Now once $$\mathbf{\tilde{F}}$$ is found, we can determine the stretch tensor $$\mathbf{\tilde{U}}$$ and the rotation tensor $$\mathbf{\tilde{R}}$$.

We will do this algebraically ﬁrst, then verify the result geometrically. $\mathbf{\tilde{F}=\tilde{R}\cdot \tilde{U}}$ Once $$\mathbf{\tilde{F}}$$ is known, we can ﬁnd $$\mathbf{\tilde{U}}$$\begin{align*} \mathbf{\tilde{U}}^{2} & \mathbf{=}\mathbf{\tilde{F}}^{T}\cdot \mathbf{\tilde{F}}\\ & =\begin{bmatrix} 2 & 1\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix} =\begin{bmatrix} 5 & 1\\ 1 & 1 \end{bmatrix} \end{align*}

Hence $\mathbf{\tilde{U}=}\begin{bmatrix} 2.2136 & 0.3162\\ 0.3162 & 0.9487 \end{bmatrix}$ and now that $$\mathbf{\tilde{U}}$$ is known, we can ﬁnd $$\mathbf{\tilde{R}}$$ \begin{align*} \mathbf{\tilde{R}} & \mathbf{=}\mathbf{\tilde{F}\cdot \tilde{U}}^{-1}\\ \mathbf{\tilde{R}} & =\begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix}\begin{bmatrix} 0.4743 & -0.1581\\ -0.1581 & 1.1068 \end{bmatrix} \\ \mathbf{\tilde{R}} & =\begin{bmatrix} 0.9487 & -0.3162\\ 0.3162 & 0.9487 \end{bmatrix} \end{align*}

To verify the result geometrically, we ﬁrst apply the stretch tensor $$\mathbf{\tilde{U}}$$ to $$\mathbf{dR}$$, this results in a new diﬀerential vector which we call $$\mathbf{dr}^{\ast }$$, then we slide $$\mathbf{dr}^{\ast }$$ without changing its slope (i.e. parallel translation) such that the vector $$\mathbf{dr}^{\ast }$$ starts at the point $$P^{\prime }$$ in the deformed conﬁguration, where the point $$P^{\prime }$$ is the image of the point $$P,$$ and then we apply the rotation tensor $$\mathbf{\tilde{R}}$$ to $$\mathbf{dr}^{\ast }$$ to obtain $$\mathbf{dr}$$.

Hence \begin{align*} \mathbf{dr}^{\ast } & =\mathbf{\tilde{U}\cdot dR=}\begin{bmatrix} 2.2136 & 0.3162\\ 0.3162 & 0.9487 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix} =\begin{bmatrix} 2.5298\\ 1.2649 \end{bmatrix} \\ & =2.5298\ \mathbf{e}_{1}+1.2649\ \mathbf{e}_{2} \end{align*}

Now we apply the rotation to $$\mathbf{\tilde{R}}$$ to $$\mathbf{dr}^{\ast }$$ to obtain $$\mathbf{dr}$$\begin{align*} \mathbf{dr} & =\mathbf{\tilde{R}\cdot dr}^{\ast }\\ & =\begin{bmatrix} 0.9487 & -0.3162\\ 0.3162 & 0.9487 \end{bmatrix}\begin{bmatrix} 2.5298\\ 1.2649 \end{bmatrix} \\ & =\begin{bmatrix} 2\\ 2 \end{bmatrix} \\ & =2\mathbf{e}_{1}+2\mathbf{e}_{2} \end{align*}

which agrees with the result obtained above.

The following diagram illustrates geometrically the action of $$\mathbf{\tilde{R}}$$ and $$\mathbf{\tilde{U}.}$$ 