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Displacement and Strain: Strain Measures

In this section, we give various definitions and interpretations of strain measures. We will start the section with the definition of uniaxial strains:

Uniaxial Strain Measures:

Definitions:

Assume that a bar of length L_0 is uniformly stretched to a length L. The quantity \lambda=\frac{L}{L_0} is termed the stretch ratio. There are various uniaxial strain measures that we can use to define how the length of the bar changes:
The Engineering Strain is defined as:

    \[ \varepsilon_{eng}=\frac{L-L_0}{L_0}=\lambda -1 \]

The True (Logarithmic) Strain is defined as:

    \[ \varepsilon_{true}=\ln\frac{L}{L_0}=\ln\lambda \]

The Green (Lagrangian) Strain is defined as:

    \[ \varepsilon_{Green}=\frac{L^2-L_0^2}{2L_0^2}=\frac{1}{2}(\lambda^2-1) \]

The stretch ratio has the limits 0<\lambda<\infty, which limits the strain measures as follows:

    \[ -1<\varepsilon_{eng}<\infty\qquad -\infty<\varepsilon_{true}<\infty\qquad -0.5<\varepsilon_{Green}<\infty \]

The above three strain measures are related as follows:

    \[ \varepsilon_{true}=\ln {(\varepsilon_{eng}+1)}\qquad \varepsilon_{eng}=\mathrm{e}^{\varepsilon_{true}}-1 \]

    \[ \varepsilon_{Green}=\varepsilon_{eng}+\frac{\varepsilon_{eng}^2}{2}\qquad \varepsilon_{eng}=-1+\sqrt{1+2\varepsilon_{Green}} \]

There are advantages to each of the above strain measures. The engineering strain \varepsilon_{eng} is the most intuitive measure as it is simply the ratio of the change in length over the original length. The true strain \varepsilon_{true} is additive which led to the terminology: “True”! For example, if a bar is stretch from L_0 to L_1 and then to L_2. Then, the true strain measured from L_0 to L_2 is also equal to the sum of the true strain from L_0 to L_1 plus the true strain from L_1 to L_2.

    \[ \varepsilon_{true}=\ln\frac{L_2}{L_0}=\ln\frac{L_2L_1}{L_1L_0}=\ln\frac{L_1}{L_0}+\ln\frac{L_2}{L_1} \]

The Green strain, on the other hand, can easily be computed in a three dimensional object with nonlinear deformation.

There is a very subtle difference between the engineering and the true strain measures when viewed as integrals. The true strain is the integral of the function 1/L while the engineering strain is the integral of the function 1/L_0:

    \[ \varepsilon_{true}=\int_{L_0}^{L_1} \frac{\mathrm{d}l}{L}\qquad \varepsilon_{eng}=\int_{L_0}^{L_1} \frac{\mathrm{d}l}{L_0} \]

The following tool draws the variation of the above three strain measures when the stretch ratio varies between 0 and 2. By changing the slider you can also pick a value for the stretch ratio between 0 and 2 and the tool will display the corresponding values of the strain measures. Notice that the three strain measures agree when the stretch ratio is close to 1, i.e., for small strains, the three measures agree.

 

Nonuniform Axial Strain:

In the above discussion, we assumed that the bar deforms uniformly. I.e., the extension or contraction is the same at every point on the bar. However, a more general case is when the bar deforms non-uniformly. In this case, we can assume that the original position of the bar is given by a function X=\xi where \xi varies between 0 and L_0. The new position is given by x=f(X)=f(\xi). We can think of the local stretch as a the limit:

    \[ \lambda=\lim_{\Delta X\rightarrow 0}{\frac{\Delta x}{\Delta X}}=\frac{\partial f}{\partial X} \]

In this case, the three strain measures would be given by:

    \[ \varepsilon_{eng}=\frac{\partial f}{\partial X} -1 \]

    \[ \varepsilon_{true}=\ln\left(\frac{\partial f}{\partial X}\right) \]

    \[ \varepsilon_{Green}=\frac{1}{2}\left(\left(\frac{\partial f}{\partial X}\right)^2-1\right) \]

Example:

Assume a bar of length L_0=1 with the ends situated at X=0 and X=1. Assuming that the new position x=X+0.2X^2, then, the stretch is different at every point and is given by:

    \[ \lambda=\frac{\partial f}{\partial X}=1+0.4X \]

I.e., the stretch is zero at the end X=0, while is the highest at the end X=1. In this case, the strain is not uniformly distributed, but rather is given by the functions:

    \[ \varepsilon_{eng}=0.4X \]

    \[ \varepsilon_{true}=\ln\left(1+0.4X \right) \]

    \[ \varepsilon_{Green}=\frac{1}{2}\left(\left(1+0.4X\right)^2-1\right) \]

The following tool draws the sketches of the reference and deformed configurations. The vertical lines in the reference configuration are equidistant but they are not after deformation. Change the density of the vertical lines and notice that after deformation, the distances between the vertical lines on the right side of the bar are more than the distances on the left side.

The following tool draws the distribution of the strain as a function of the original position X. As predicted, the strain on the left side of the bar after deformation, is higher than the strain on the right side. You can change the slider to pick a specific value for X for which the tool will calculate and display the value of the different measures of strain.

Three Dimensional Strain Measures:

In general, strain measures give a measure of how the lengths or angles change when we compare vectors in the undeformed configuration with vectors in the deformed configuration. If a tangent vector in the reference configuration is denoted dX, while its deformed image is denoted dx, then we can calculate the various strain measures along the direction dx by the simple computations:

    \[ \varepsilon_{eng \mbox{ along }dX}=\frac{\| dx \| - \| dX \|}{\| dX \|} \]

    \[ \varepsilon_{true \mbox{ along }dX}=\ln\frac{\| dx \|}{\| dX \|} \]

    \[ \varepsilon_{Green \mbox{ along }dX}=\frac{\| dx \|^2 - \| dX \|^2}{2\| dX \|^2} \]

where we can use the relationship between the dot product and the norm when convenient:

    \[ \|dx\|^2=dx\cdot dx\qquad \|dX\|^2=dX\cdot dX \]

However, a more convenient and traditional way to measure strain is to define strain tensors by which we can calculate strain along certain directions. We will discuss the two most common strain tensors in the literature:

Small (Infinitesimal) Strain Tensor:

The small of infinitesimal strain tensor \varepsilon is defined as the symmetric part of the displacement gradient \nabla u:

    \[ \varepsilon= \frac{\nabla u+\nabla u^T}{2} \]

Which has the following component form:

    \[ \varepsilon= \left(\begin{array}{ccc} \frac{\partial u_1}{\partial X_1} & \frac{1}{2}\left(\frac{\partial u_1}{\partial X_2}+\frac{\partial u_2}{\partial X_1}\right) & \frac{1}{2}\left(\frac{\partial u_1}{\partial X_3}+\frac{\partial u_3}{\partial X_1}\right)\\ \frac{1}{2}\left(\frac{\partial u_1}{\partial X_2}+\frac{\partial u_2}{\partial X_1}\right)& \frac{\partial u_2}{\partial X_2} & \frac{1}{2}\left(\frac{\partial u_2}{\partial X_3}+\frac{\partial u_3}{\partial X_2}\right)\\ \frac{1}{2}\left(\frac{\partial u_1}{\partial X_3}+\frac{\partial u_3}{\partial X_1}\right)& \frac{1}{2}\left(\frac{\partial u_2}{\partial X_3}+\frac{\partial u_3}{\partial X_2}\right)& \frac{\partial u_3}{\partial X_3} \end{array}\right) \]

which can be written in a simple form as follows \forall i,j\leq 3:

    \[ \varepsilon_{ij}= \frac{1}{2}\left(\frac{\partial u_i}{\partial X_j}+\frac{\partial u_j}{\partial X_i}\right) \]

In the case of small deformations, the small strain tensor can be used to compute the engineering longitudinal and shear strains as shown below.

Calculating Longitudinal Strains along General Vectors:

In the case of small deformations, the engineering longitudinal strain along a tangent vector dX in the reference configuration can be calculated using the relationship:

    \[ \varepsilon_{eng \mbox{ along }dX}=\frac{dX \cdot \varepsilon dX }{\| dX \|^2} \]

To show the above relationship, we are going to use the relationship between the deformed tangent vector dx and its original vector dX as shown in the displacement gradient tensor section:

    \[ dx=dX + \varepsilon dX + W_{inf} dX \]

The length of dx can be estimated as follows:

    \[ \|dx\|=\sqrt{dx\cdot dx}=\sqrt{(dX + \varepsilon dX + W_{inf} dX)\cdot (dX + \varepsilon dX + W_{inf} dX)} \]

Since W_{inf} is a skewsymmetric tensor, we have: dX\cdot W_{inf}dX=0. For small deformations, the following terms can be neglected W_{inf}dX\cdot W_{inf}dX, \varepsilon dX\cdot W_{inf}dX and \varepsilon dX\cdot \varepsilon dX which simplifies the above relationship to:

    \[ \|dx\|\simeq\sqrt{dX\cdot dX + 2dX\cdot\varepsilon dX } \simeq \|dX\|\sqrt{1 + \frac{2dX\cdot\varepsilon dX}{dX\cdot dX} } \]

Since the term \frac{2dX\cdot\varepsilon dX}{dX\cdot dX} <<1, the above relationship can be further simplified as follows:

    \[ \|dx\| \simeq \|dX\| + \frac{dX\cdot\varepsilon dX}{\|dX\|} \]

Thus, for small deformations, the engineering strain along the reference configuration tangent vector dX can be calculated as follows:

    \[ \varepsilon_{eng \mbox{ along }dX}= \frac{\| dx \| - \| dX \|}{\| dX \|} \simeq \frac{dX \cdot \varepsilon dX }{\| dX \|^2} \]

Calculating Longitudinal Strains Along the Basis Vectors:

The strain along the basis vectors e_1, e_2 and e_3 in the reference configuration can be calculated using the above relationship as follows \forall i\leq 3:

    \[ \varepsilon_{eng \mbox{ along }e_i}\simeq \frac{e_i \cdot \varepsilon e_i }{\| e_i \|^2} \]

Therefore,

    \[ \varepsilon_{eng \mbox{ along }e_1}\simeq \frac{\partial u_1}{\partial X_1}\qquad \varepsilon_{eng \mbox{ along }e_2}\simeq \frac{\partial u_2}{\partial X_2}\qquad \varepsilon_{eng \mbox{ along }e_3}\simeq \frac{\partial u_3}{\partial X_3}\qquad \]

Therefore, the diagonal components of the strain matrix give the value of the longitudinal strains along the basis vectors of the reference configuration.

Calculating Angle Change between General Vectors:

Given two vectors dX and dY separated by an angle \Theta in the reference configuration that deform into the two vectors dx and dy separated by an angle \theta in the deformed configuration, then half the change in the dot produce between the two vectors before and after deformation can be calculated as:

    \[ \frac{1}{2}\frac{dx\cdot dy - dX\cdot dY}{\|dX\|\|dY\|}=\frac{1}{2}\frac{\|dx\|\|dy\|\cos\theta - \|dX\|\|dX\|\cos\Theta}{\|dX\|\|dY\|} \]

Using the relationship between the deformed tangent vectors dx, dy and their original vectors dX, dY as shown in the displacement gradient tensor section we can write dx and dy as functions of dX and dY:

    \[ dx\cdot dy=(dX + \varepsilon dX + W_{inf} dX)\cdot(dY + \varepsilon dY + W_{inf} dY) \]

Since \varepsilon is symmetric and W_{inf} is skewsymmetric and for small deformations, the dot product can be approximated as follows:

    \[ dx\cdot dy\simeq dX\cdot dY + 2dX\cdot\varepsilon dY \]

Then half the change in the dot produce between the two vectors before and after deformation can be calculated using \varepsilon as follows:

    \[ \frac{1}{2}\frac{dx\cdot dy - dX\cdot dY}{\|dX\|\|dY\|}\simeq \frac{dX\cdot\varepsilon dY}{\|dX\|\|dY\|} \]

When the deformations are small such that \|dx\|\simeq\|dX\| and \|dy\|\simeq\|dY\| then half the difference in the cosines of the angles before and after deformation can be calculated as follows:

    \[ \frac{\cos\theta-\cos\Theta}{2}\simeq \frac{dX\cdot\varepsilon dY}{\|dX\|\|dY\|} \]

If the original vectors dX and dY are orthogonal to each other, then \cos\Theta=0. In that case, the above relationship gives half the engineering shear strain of planes parallel to dX and perpendicular to dY.

Calculating Engineering Shear Strain in the Planes of the Basis Vectors:

Using the above relationship, the engineering shear strains in the planes of the basis vectors e_i and e_j with i\neq j can be calculated as follows:

    \[ \gamma_{ij}=2\frac{e_i\cdot\varepsilon e_j}{\|e_i\|\|e_j\|}=2\varepsilon_{ij}=\frac{\partial u_i}{\partial X_j}+\frac{\partial u_j}{\partial X_i} \]

i.e., the off diagonal components give the engineering shear strains in the planes e_1–e_2, e_1–e_3 and e_2–e_3

    \[ \gamma_{12}=2\varepsilon_{12}=\frac{\partial u_1}{\partial X_2}+\frac{\partial u_2}{\partial X_1}\qquad \gamma_{13}=2\varepsilon_{13}=\frac{\partial u_1}{\partial X_3}+\frac{\partial u_3}{\partial X_1}\qquad \gamma_{23}=2\varepsilon_{23}=\frac{\partial u_2}{\partial X_3}+\frac{\partial u_3}{\partial X_2}\qquad \]

Example:

In the following example, a two dimensional square centred around the origin is deformed using the displacement relationship:

    \[ x_1=a_{11}X_1+a_{12}X_2 \]

    \[ x_2=a_{12}X_2+a_{22}X_2 \]

The following tool lets you vary the values of a_{11} and a_{22} between -0.8 and 1.2 and the values of a_{12} and a_{21} between -0.2 and 0.2. The tool then draws the square before deformation and then after deformation. You can also select two unit vectors dX and dY by varying their respective angles of inclination with the horizontal axis \Theta_1 and \Theta_2. The two vectors are drawn before and after deformation. dX is drawn in blue and dY is drawn in red. The tool calculates the deformation gradient F, the gradient of displacement tensor \nabla u, the small strain matrix \varepsilon and the infinitesimal rotation matrix W_{inf}. The longitudinal strains along dX and dY are calculated underneath the figures in addition to \cos\theta where \theta is the angle between dx and dy which are the respective images of dX and dY. Vary the values of a_{ij} to see their effect on the deformation and the components of the strain matrix \varepsilon. Also, check your strain computations against the results shown for various choices of dX and dY. Note that the computation of \cos\theta is approximate because it assumes small deformation and that

    \[ \|dx\|=\|dX\| \qquad\mbox{and}\qquad \|dy\|=\|dY\| \]

 

 

Symmetry of the Strain Tensor:

The symmetry of the infinitesimal strain matrix implies that there is a coordinate system in which the strain matrix is diagonal. The eigenvalues of the strain matrix are called the principal strains. If an orthonormal coordinate system aligned with the eigenvectors is chosen for the basis set, then the strain matrix is diagonal (no shear strains). In other words, in this coordinate system, no shear strains exist on the planes perpendicular to the basis vectors.

Example:

The example shown above is repeated. In the first row, the tool draws the original shape and the deformed shape showing the deformation of the original coordinate system. In the second row, the tool draws the original shape of a small square aligned with the eigenvectors of the strain matrix which are drawn in blue and red. Notice that the square deforms into a rectangle (i.e., no shear strain!). The new rectangle might be slightly rotated as the deformation in this case is decomposed into an infinitesimal stretch described by \varepsilon and an infinitesimal rotation described by W_{inf}. You can vary the components a_{ij} to investigate their effect on the deformed shape. Which components control the stretch and which components control the rotation?

 

Calculating Volumetric Strains:

For small deformations, the diagonal components of the strain matrix can be used to calculate the volumetric strain as follows:

    \[ \frac{V-V_0}{V_0}=\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}=Trace(\varepsilon) \]

where V and V_0 are the deformed and original volumes of a small element inside the deformed object.

Behaviour under Pure Rotations:

A major disadvantage of the infinitesimal strain matrix is that it predicts strains for bodies that undergo large rotations even when the actual strain is zero. For example, consider a two dimensional displacement described by the rotation matrix:

    \[ Q= \left(\begin{array}{cc} \cos{\theta} & \sin{\theta}\\ -\sin{\theta} & \cos{\theta} \end{array}\right) \]

i.e.,

    \[ x=QX \]

Then, the displacement vector at every point is described by:

    \[ u=x-X=(Q-I)X \]

The gradient of displacement matrix is given by:

    \[ \nabla u=Q-I \]

The infinitesimal strain matrix is then given by:

    \[ \varepsilon =\left(\frac{(Q-I)^T+(Q-I)}{2}\right)=\frac{1}{2}(Q+Q^T-2I) =\left( \begin{array}{cc} \cos{\theta}-1 & 0\\ 0 & \cos{\theta}-1 \end{array} \right)\neq 0 \]

For small values of \theta, \varepsilon\simeq 0. However, at larger values of rotation, the strain matrix will predict high strains even though the object rotating is not being stretched. The value \cos{\theta}-1 for the compressive strain in the direction of the basis vectors e_1 and e_2 should not come as a surprise. If a square of unit length rotates with an angle \theta, \cos{\theta}-1 is actually the length of the projections of the square’s sides onto the coordinate axis!

 

Green Strain Tensor:

The Green strain tensor is \varepsilon_{Green} is defined as follows:

    \[ \varepsilon_{Green}= \frac{1}{2}(F^TF-I) \]

By utilizing the relationship \nabla u = F-I, we can replace F to obtain the relationship:

    \[ \varepsilon_{Green}= \frac{1}{2}((\nabla u + I)^T(\nabla u + I)-I)=\frac{1}{2}\left(\nabla u^T+\nabla u + \nabla u^T\nabla u\right) \]

In component form, the Green strain tensor components can be written as:

    \[ \varepsilon_{ij}=\frac{1}{2}\left( \frac{\partial u_j}{\partial X_i} + \frac{\partial u_i}{\partial X_j} + \sum\limits_{k=1}^3 {\frac{\partial u_k}{\partial X_j}\frac{\partial u_k}{\partial X_i}} \right) \]

Similar to the small strain tensor, the Green strain tensor is symmetric, i.e., there is a coordinate system in which the strain matrix has diagonal components.

Calculating Longitudinal Green Strains along General Vectors:

The Green strain tensor can be used to calculate the uniaxial (longitudinal) Green strain along general vectors. Let dX be a vector in the reference configuration and dx be its deformed image, then the Green strain along dX can be calculated as follows:

    \[ \varepsilon_{\mbox{Green along }dX}=\frac{1}{2}\frac{dx\cdot dx - dX\cdot dX}{\|dX\|^2} \]

Utilizing the relationship dx=FdX we get:

    \[ \varepsilon_{\mbox{Green along }dX}=\frac{1}{2}\frac{FdX \cdot FdX - dX\cdot dX}{\|dX\|^2}=\frac{1}{2}\frac{dX\cdot F^TFdX - dX\cdot dX}{\|dX\|^2}=\frac{dX\cdot \frac{1}{2}(F^TF-I)dX }{\|dX\|^2} \]

Therefore:

    \[ \varepsilon_{\mbox{Green along }dX}=\frac{dX\cdot \varepsilon_{Green} dX }{\|dX\|^2} \]

Calculating Change of the Dot Product between General Vectors:

Similarly, given two vectors dX and dY before deformation and their respective images dx and dy after deformation, half the difference between the dot product in the deformed configuration and the dot product in the reference configuration can be calculated using the Green strain tensor as follows:

    \[ \frac{1}{2}\frac{dx\cdot dy - dX\cdot dY}{\|dX\|\|dY\|}=\frac{dX\cdot \varepsilon_{Green} dY }{\|dX\|\|dY\|} \]

Behaviour under Pure Rotations:

One of the major advantages of the Green strain tensor is that it predicts zero strain in the cases of pure rotation. Assuming that a deformation is described by a rotation matrix Q, i.e.: x=QX. Then, the Green strain tensor is calculated as follows:

    \[ \varepsilon_{Green}= \frac{1}{2}(F^TF-I)=\frac{1}{2}(Q^TQ-I)=\frac{1}{2}(I-I)=0 \]

i.e., under pure rotations, the Green strain matrix has zero components!

Eigenvectors:

The eigenvectors of the Green strain matrix stay perpendicular to each other after deformation. This can be shown as follows. Let v_1 and v_2 be two distinct eigenvectors of \varepsilon_{Green}. Since \varepsilon_{Green} is symmetric, then v_1\cdot v_2=0. After deformation we have:

    \[ Fv_1 \cdot F v_2 = v_1\cdot F^TF v_2 = v_1\cdot \left(2\varepsilon_{Green}+I\right)v_2=v_1\cdot 2\varepsilon_{Green}v_2+ v_1\cdot v_2 \]

However, since v_2 is an eigenvector of \varepsilon_{Green}, then there exists a real number \lambda_2 such that 2\varepsilon_{Green}v_2=\lambda_2v_2. Therefore:

    \[ Fv_1 \cdot F v_2 = 2v_1\cdot \lambda_2v_2+ v_1\cdot v_2=(2\lambda_2+1)(v_1\cdot v_2) \]

Since v_1\cdot v_2=0 we have:

    \[ Fv_1 \cdot F v_2 =(2\lambda_2+1)(v_1\cdot v_2) = 0 \]

This is not necessarily true for the eigenvectors of the small strain matrix. You can find a counter example for the small strain matrix using the following tool!

Example:

The above example is repeated but with higher ranges of the components a_{ij} to account for larger deformations. In the first row, the square and its deformed shape are shown. In the second row, a square aligned with the eigenvectors of the small strain matrix is drawn along with its image. In the third row, a square aligned with the eigenvectors of the Green strain matrix is drawn along with its image. Vary the components a_{ij} and try to answer the following:

  • At what values of a_{ij} do the Green strain matrix and the small strain matrix have similar or different values?
  • What happens to the eigenvectors of the small strain matrix at higher rotations? Are they still perpendicular to each other after deformation?
  • What happens to the eigenvectors of the Green strain matrix at higher rotations? Are they still perpendicular to each other after deformation?

Examples and Problems:

Example 1:

Calculate the infinitesimal and Green strain matrices for the following position function:

    \[ x=\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)=\left(\begin{matrix}1.2 & 0.2 & 0.2\\0.2& 1.3 & 0.1\\0.9&0.5&1\end{matrix}\right)\left(\begin{array}{c}X_1\\X_2\\X_3\end{array}\right) \]

Also, find the displacement function, the uniaxial small and Green strains along the direction of the vector u=\left(\begin{array}{c}1\\1\\1\end{array}\right).

Solution:

The displacement function is:

    \[ u=x-X=\left(\begin{matrix}0.2 & 0.2 & 0.2\\0.2& 0.3 & 0.1\\0.9&0.5&0\end{matrix}\right)\left(\begin{array}{c}X_1\\X_2\\X_3\end{array}\right) \]

The small strain tensor is:

    \[ \varepsilon_{small}=\frac{1}{2}\left(\nabla u + \nabla u^T\right)=\left(\begin{matrix}0.2 & 0.2 & 0.55\\0.2& 0.3 & 0.3\\0.55&0.3&0\end{matrix}\right) \]

The Green strain tensor is:

    \[ \varepsilon_{Green}=\frac{1}{2}\left(\nabla u + \nabla u^T+\nabla u^T\nabla u\right)=\left(\begin{matrix}0.645 & 0.475 & 0.58\\0.475& 0.49 & 0.335\\0.58&0.335&0.025\end{matrix}\right) \]

The deformation is very large as shown by applying this deformation to a unit cube (see figure below), so the strain measures are different.
Strain1

The uniaxial small and Green strain along the vector u can be obtained as follows:

    \[ \varepsilon_{\mbox{small along }u}=\frac{u\cdot \varepsilon_{small}u}{\|u\|^2}=0.87\qquad \varepsilon_{\mbox{Green along }u}=\frac{u\cdot \varepsilon_{Green}u}{\|u\|^2}=1.31 \]

View Mathematica Code
F={{1.2,0.2,0.2},{0.2,1.3,0.1},{0.9,0.5,1}};
X={X1,X2,X3};
x=F.X;
u=x-X;
Gradu=Table[D[u[[i]],X[[j]]],{i,1,3},{j,1,3}]  
einfinitesimal=1/2*(Gradu+Transpose[Gradu]);
%//MatrixForm  
egreen=1/2*(Gradu+Transpose[Gradu]+Transpose[Gradu].Gradu);
%//MatrixForm  
u={1,1,1};
n=u/Norm[u];
n.einfinitesimal.n  
n.egreen.n  
Box1=Cuboid[{0,0,0},{1,1,1}];
Box2=GeometricTransformation[Box1,F];
Graphics3D[{EdgeForm[Thickness[0.01]],Box1,EdgeForm[Thickness[0.005]],Box2},Axes->True,AxesOrigin->{0,0,0},BaseStyle->Directive[Bold,15],AxesLabel->{e1,e2,e3}]
 

Example 2:

A cube of unit dimensions undergoes a rigid body rotation in \mathbb{R}^3 of 20^\circ, then 20^\circ, and then 30^\circ around the basis vectors e_3, e_2, and e_1 respectively. Use Mathematica to visualize the box. Compare the two strain measures \varepsilon_{small} and \varepsilon_{Green}.

Solution:

The deformation gradient is given by:

    \[ F=Q_{e_1}Q_{e_2}Q_{e_3}=\left(\begin{matrix}0.883&-0.3214 & 0.3420\\0.4569&0.7553 & -0.4698\\-0.1073 & 0.5712 & 0.8138\end{matrix}\right) \]

Note that Q_{e_3} is on the right since it is applied first.
The Green strain is identically zero while the small strain tensor predicts the following strains:

    \[ \varepsilon_{small}=\left(\begin{matrix}-0.117&0.068& 0.117\\0.068&-0.245 & 0.050\\0.117 & 0.051& -0.186\end{matrix}\right) \]

Strain2
View Mathematica Code

Qx=RotationMatrix[thx,{1,0,0}];
Qy=RotationMatrix[thy,{0,1,0}];
Qz=RotationMatrix[thz,{0,0,1}];
thx=30Degree  
thy=20Degree  
thz=20Degree  
F=Qx.Qy.Qz;
X={X1,X2,X3};
x=F.X;
u=x-X;
Gradu=Table[D[u[[i]],X[[j]]],{i,1,3},{j,1,3}];
einfinitesimal=1/2*(Gradu+Transpose[Gradu]);
einfinitesimal=FullSimplify[einfinitesimal];
%//MatrixForm  
egreen=1/2*(Gradu+Transpose[Gradu]+Transpose[Gradu].Gradu);
egreen=FullSimplify[egreen];
%//MatrixForm  
Box1=Cuboid[{0,0,0},{1,1,1}];
Box2=GeometricTransformation[Box1,F];
Graphics3D[{Box1,Box2},Axes->True,AxesOrigin->{0,0,},AxesLabel->{e1,e2,e3}]
 

Example 3:

A cube of unit length undergoes a simple shearing motion of an angle \theta in the direction of e_1 and perpendicular to e_2. Find the infinitesimal and Green strain measures that describe this motion. Comment on the difference between the two strain measures.

Solution:

The position function is given by:

    \[ x=\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)=\left(\begin{array}{c}X_1+\tan\theta X_2\\X_2\\X_3\end{array}\right) \]

The deformation gradient is given by:

    \[ F=\left(\begin{matrix}1&\tan\theta & 0\\0&1&0\\0&0&1\end{matrix}\right) \]

The small strain and Green strain tensors are given by:

    \[ \varepsilon_{small}=\left(\begin{matrix}0&\frac{\tan\theta}{2} & 0\\\frac{\tan\theta}{2}&0&0\\0&0&0\end{matrix}\right)\qquad \varepsilon_{Green}=\left(\begin{matrix}0&\frac{\tan\theta}{2} & 0\\\frac{\tan\theta}{2}&\frac{\tan^2\theta}{2}&0\\0&0&0\end{matrix}\right) \]

Both measures predict the same shear strains. For small deformations, both measures predict zero normal strains. As the \theta increases, the vertical lines start developing large axial strains which the \varepsilon_{Green} can predict while \varepsilon_{small} cannot. The undeformed and deformed shapes of the cube are shown below for \theta=30^\circ.
Strain3

View Mathematica Code
F={{1,Tan[theta],0},{0,1,0},{0,0,1}};
X={X1,X2,X3};
x=F.X;
u=x-X;
Gradu=Table[D[u[[i]],X[[j]]],{i,1,3},{j,1,3}]  
einfinitesimal=1/2*(Gradu+Transpose[Gradu]);
einfinitesimal=FullSimplify[einfinitesimal];
%//MatrixForm
egreen= 1/2*(Gradu+Transpose[Gradu]+Transpose[Gradu].Gradu);
egreen=FullSimplify[egreen];
%//MatrixForm  
Box1=Cuboid[{0,0,0},{1,1,1}];
Box2=GeometricTransformation[Box1,F];
theta=30Degree;
Graphics3D[{Box1,Box2},Axes->True,AxesOrigin->{0,0,0},AxesLabel->{e1,e2,e3}]

 

Example 4:

The following small strain matrix represents a two dimensional state of strain

    \[ \varepsilon_{small}=\left(\begin{matrix}0.01 & 0.012 \\ 0.012 & 0\end{matrix}\right) \]

Find the principal strains and their directions. If a coordinate system is aligned with the principal directions, find the components of the strain matrix in this new coordinate system.

Solution:

The principal strains \varepsilon_1 and \varepsilon_2 and their directions ev_1, and ev_2 be obtained using Mathematica by finding the eigenvalues and eigenvectors of \varepsilon_{small}. The following are the eiganvalues:

    \[ \varepsilon_1=0.018\qquad \varepsilon_2=-0.008 \]

The following are the normalized eigenvectors.

    \[ ev_1=\left(\begin{array}{c}0.83205\\0.5547\end{array}\right)\qquad ev_2=\left(\begin{array}{c}-0.5547\\0.83205\end{array}\right) \]

The transformation matrix between the original coordinate system and the new coordinate system is:

    \[ Q=\left(\begin{matrix}0.83205 & 0.5547 \\ -0.5547 & 0.83205 \end{matrix}\right) \]

The strain matrix in the new coordinate system has the form:

    \[ \varepsilon'=Q\varepsilon Q^T=\left(\begin{matrix}0.018 & 0 \\ 0 & -0.008 \end{matrix}\right) \]

View Mathematica Code
eps={{0.01,0.012},{0.012,0}};
s=Eigensystem[eps]  
Q={-s[[2,1]],-s[[2,2]]}  
epsdash=Chop[Q.eps.Transpose[Q]]

 

Example 5:

A state of uniform small strain is described by the following small strain matrix:

    \[ \varepsilon_{small}=\left(\begin{matrix}-0.01 & 0.002 & -0.023 \\ 0.002 & 0.05 & 0\\ -0.023 & 0 & 0.02\end{matrix}\right) \]

Determine the following:

  • The longitudinal strain along the direction of the vector dX=\left(\begin{array}{c}1\\2\\1\end{array}\right).
  • The change in the cosines of the angles between the vectors dX=\left(\begin{array}{c}1\\0\\1\end{array}\right) and dY=\left(\begin{array}{c}1\\1\\1\end{array}\right) and comment on whether the angle between the vectors decreased or increased after deformation.
  • The shear strains of planes parallel to the vector dX=\left(\begin{array}{c}-1\\1\\0\end{array}\right) and perpendicular to the vector dY=\left(\begin{array}{c}1\\1\\0\end{array}\right).
Solution:

The strains along dX=\left(\begin{array}{c}1\\2\\1\end{array}\right) can be calculated as follows:

    \[ \varepsilon_{small \mbox{ along }dX}=\frac{dX\cdot\varepsilon_{small}dX}{\|dX\|^2}=0.0287 \]

The change in the cosines of the angles between the vectors dX=\left(\begin{array}{c}1\\0\\1\end{array}\right) and dY=\left(\begin{array}{c}1\\1\\1\end{array}\right) can be calculated as follows:

    \[ \frac{\cos\theta-\cos\Theta}{2}\simeq \frac{dX\cdot\varepsilon dY}{\|dX\|\|dY\|}=-0.0139 \]

The angle \Theta between the vectors dX and dY before deformation can be calculated using the dot product:

    \[ \cos\Theta = \frac{dX\cdot dY}{\|dX\|\|dY\|}=\sqrt{\frac{2}{3}}\Rightarrow \Theta=0.6155rad=35.26^\circ \]

Therefore, the angle \theta after deformation between the vectors is larger because:

    \[ \cos\theta-\sqrt{\frac{2}{3}}=-0.0248\Rightarrow \theta=0.662rad=37.93^\circ \]

Since the vectors dX=\left(\begin{array}{c}-1\\1\\0\end{array}\right) and dY=\left(\begin{array}{c}1\\1\\0\end{array}\right) are perpendicular, the shear strains of planes parallel to the vector dX and perpendicular to the vector dY can be obtained as follows:

    \[ \mbox{Shear strains of planes parallel to }dX\mbox{ and perpendicular to }dY=\frac{dX\cdot\varepsilon dY}{\|dX\|\|dY\|}=0.03rad \]

View Mathematica Code
Eps={{-0.01,0.002,-0.023},{0.002,0.05,0},{-0.023,0,0.02}};
dX={1,2,1};
Norm[dX]  
Eps.dX  
dX.Eps.dX/Norm[dX]^2  
dX={1,0,1};
dY={1,1,1};
anglechange=dX.Eps.dY/Norm[dX]/Norm[dY]  
THETA=N[ArcCos[dX.dY/Norm[dX]/Norm[dY]]]  
THETA/Degree  
theta=ArcCos[2*anglechange+Cos[THETA]]  
theta/Degree  
dX={-1,1,0};
dY={1,1,0};
ShearSTrain=dX.Eps.dY/Norm[dX]/Norm[dY]
 

Example 6:

The longitudinal engineering strains on the surface of a test specimen were measured using a strain gauge rosette to be 0.005, 0.002 and –0.001 along the three directions: a, b, and c respectively, where a=\{\cos {30^\circ}, \sin {30^\circ\}}, b=\{0,1\}, and c=\{-\cos{30^\circ},\sin{30^\circ}\}. If the material is assumed to be in a small strain state, find the principal strains and their directions on the surface of the test specimen.

Solution:

The state of strain on the surface of the material can be represented by the symmetric small strain matrix:

    \[ \varepsilon_{small}=\left(\begin{matrix}\varepsilon_{11} & \varepsilon_{12} \\ \varepsilon_{12} & \varepsilon_{22}\end{matrix}\right) \]

Since the longitudinal strain is known along three directions, three equations can be written to find the three unknown components of the strain matrix as follows:

    \[ \frac{a\cdot \varepsilon_{small}a}{\|a\|^2}=0.005 \qquad \frac{b\cdot \varepsilon_{small}b}{\|b\|^2}=0.002 \qquad \frac{c\cdot \varepsilon_{small}c}{\|c\|^2}=-0.001 \]

The norm of each of a, b, and c is 1. Therefore, the three equations are:

    \[ \begin{split} \frac{1}{4}\left(3\varepsilon_{11}+2\sqrt{3}\varepsilon_{12}+\varepsilon_{22}\right) & = 0.005 \\ \varepsilon_{22}& = 0.002\\ \frac{1}{4}\left(3\varepsilon_{11}-2\sqrt{3}\varepsilon_{12}+\varepsilon_{22}\right) & = -0.001 \end{split} \]

Solving the above three equations yields:

    \[ \varepsilon_{small}=\left(\begin{matrix}\varepsilon_{11} & \varepsilon_{12} \\ \varepsilon_{12} & \varepsilon_{22}\end{matrix}\right)=\left(\begin{matrix}0.002 & 0.003464 \\ 0.003464 & 0.002\end{matrix}\right) \]

The principal strains and their directions are the eigenvalues and eigenvectors of the small strain matrix. The principal strains are:

    \[ \varepsilon_{1}=0.0055\qquad \varepsilon_2=-0.00146 \]

The principal directions are:

    \[ ev_1=\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{array}\right)\qquad ev_2=\left(\begin{array}{c}\frac{-1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{array}\right) \]

View Mathematica Code
eps=Table[Subscript[e,i,j],{i,1,2},{j,1,2}]
a={Cos[30Degree],Sin[30Degree]};
b={0,1};
c={-Cos[30Degree],Sin[30Degree]};
FullSimplify[a.eps.a/Norm[a]^2]  
FullSimplify[b.eps.b/Norm[b]^2]  
FullSimplify[c.eps.c/Norm[c]^2]  
a=Solve[{a.eps.a/Norm[a]^2==0.005,b.eps.b/Norm[b]^2==0.002,c.eps.c/Norm[c]^2==-0.001,eps[[1,2]]==eps[[2,1]]},Flatten[eps]]  
eps=eps/.a[[1]];
Eigensystem[eps]//MatrixForm
 

Example 7:

In the above examples, the deformation was described by a uniform strain matrix (i.e., a strain matrix that is not function in position). In this example, a nonuniform strain, i.e., the strain matrix is function of position is explored. Assume that the two dimensional displacement function of a 2units by 2units plate has the following form:

    \[ u=\left(\begin{array}{c}u_1\\u_2\end{array}\right)=\left(\begin{array}{c}0.2X_1\\0.09X_2X_1\end{array}\right) \]

  • Find the deformed configuration function x=f(X).
  • Find \varepsilon_{small} and \varepsilon_{Green}.
  • Draw the vector plot of the displacement function on the plate.
  • Draw the contour plots of {\varepsilon_{Green}}_{11}, {\varepsilon_{Green}}_{12}, {\varepsilon_{small}}_{11}, and {\varepsilon_{small}}_{12} on the plate.
Solution:

The position function of the deformed configuration is given by:

    \[ x=X+u=\left(\begin{array}{c}1.2X_1\\X_2+0.09X_2X_1\end{array}\right) \]

The deformation gradient is given by:

    \[ F=\left(\begin{matrix}1.2 & 0 \\0.09X_2 &1+ 0.09X_1\end{matrix}\right) \]

The displacement gradient is given by:

    \[ \nabla u=F-I=\left(\begin{matrix}0.2 & 0 \\0.09X_2 & 0.09X_1\end{matrix}\right) \]

Therefore, the small strain matrix is given by:

    \[ \varepsilon_{small}=\left(\begin{matrix}0.2 & 0.045X_2 \\0.045X_2 & 0.09X_1\end{matrix}\right) \]

The Green strain matrix is given by:

    \[ \varepsilon_{Green}=\left(\begin{matrix}0.22+0.00405X_2^2 & (0.045+0.00405X_1)X_2 \\(0.045+0.00405X_1)X_2 & (0.09+0.00405X_1)X_1\end{matrix}\right) \]

The required plots are shown below.
strain5111
Strain52
View Mathematica Code

Clear[x1,x2,X1,X2]  
X={X1,X2} ;
u={0.2X1,0.09X2*X1} ;
x=u+X;
Gradu=Table[D[u[[i]],X[[j]]],{i,1,2},{j,1,2}]  
einfinitesimal=1/2*(Gradu+Transpose[Gradu]);
egreen=1/2*(Gradu+Transpose[Gradu]+Transpose[Gradu].Gradu);
einfinitesimal//MatrixForm  
FullSimplify[egreen]//MatrixForm  
VectorPlot[u,{X1,0,2},{X2,0,2},BaseStyle->Directive[Bold,15],AspectRatio->Automatic,PlotLabel->"u"]  
ContourPlot[einfinitesimal[[1,1]],{X1,0,2},{X2,0,2} ,BaseStyle->Directive[Bold,15],AspectRatio->Automatic,ContourLabels->All,PlotLabel->"einf11"]  
ContourPlot[egreen[[1,1]],{X1,0,2},{X2,0,2} ,BaseStyle->Directive[Bold,15],AspectRatio->Automatic,ContourLabels->All,PlotLabel->"egreen11"]  
ContourPlot[einfinitesimal[[1,2]],{X1,0,2},{X2,0,2} ,BaseStyle->Directive[Bold,15],AspectRatio->Automatic,ContourLabels->All,PlotLabel->"einf12"]  
ContourPlot[egreen[[1,2]],{X1,0,2},{X2,0,2} ,BaseStyle->Directive[Bold,15],AspectRatio->Automatic,ContourLabels->All,PlotLabel->"egreen12"]  
ContourPlot[einfinitesimal[[2,2]],{X1,0,2},{X2,0,2} ,BaseStyle->Directive[Bold,15],AspectRatio->Automatic,ContourLabels->All,PlotLabel->"einf22"]  
ContourPlot[egreen[[2,2]],{X1,0,2},{X2,0,2} ,BaseStyle->Directive[Bold,15],AspectRatio->Automatic,ContourLabels->All,PlotLabel->"egreen22"]  

Problems:

  1. Which of the following functions describing motion is/are not physically possible and why?
    • x_1=X_1, x_2=X_2, x_3=-X_3.
    • x_1=X_1, x_2=X_2, x_3=X_2.
    • x_1=X_1+X_2, x_2=X_2, x_3=X_3.
    • x_1=X_3, x_2=X_1, x_3=X_2.
  2. Calculate the limits of \alpha and \beta for the following position functions to be physically possible:
    • x_1=\alpha X_1+X_2, x_2=2X_2,x_3=5X_3.
    • x_1=\beta (X_1+X_2), x_2=X_2+X_3,x_3=X_1+X_3.
    • x_1=\beta (X_1+X_2), x_2=X_2+X_3,x_3=X_3.
  3. Let a cube of unit length be represented by the set of vectors \Omega_0\subset\mathbb{R}^3 such that three of the cube’s sides are aligned with the three orthonormal basis set vectors e_1, e_2, and e_3 and one of the cube vertices lies on the origin of the coordinate system. Let \Omega\subset \mathbb{R}^3 be the deformed configuration such that f:X\in\Omega_0\mapsto x\in \Omega is defined as x=f(X):

        \[\begin{split} x_1&=1.1X_1+0.02X_1^2+0.01X_2+0.03X_3\\ x_2&=0.001X_1+0.9X_2+0.003X_3\\ x_3&=0.001X_1+0.005X_2+0.009X_3^2+0.9X_3 \end{split} \]

    Find the following:

    • The displacement function u.
    • The new position of the 8 vertices of the cube.
    • The deformed curves of the three edges of the cube that are aligned with the basis vectors e_1, e_2, and e_3.
    • The infinitesimal strain matrix and the volumetric strain as a function of the position inside the cube.
    • The Green strain matrix as a function of the position inside the cube.
    • Evaluate the infinitesimal strain matrix at two of the eight cube vertices.
    • Evaluate the Green strain matrix at two of the eight cube vertices.
  4. The reference and deformed configurations of an object exhibiting a two dimensional motion (u_3=0) is shown in the figure below.

    Find the following:

    • The two dimensional position function that can describe this shown motion.
    • The associated two dimensional displacement function.
    • The associated small strain tensor.
    • The associated Green Lagrange strain tensor.
  5. A cube undergoes a deformation such that the infinitesimal strain is described by the matrix:

        \[ \varepsilon_{small}=\left(\begin{matrix}0.015&0.005&0\\0.005&-0.002&0\\0&0&0\end{matrix}\right) \]

    Find the following:

    • The principal strains and their directions.
    • The longitudinal strains along the direction of the vector:

          \[ dX=\left(\begin{array}{c}1\\1\\1\end{array}\right) \]

    • The approximate change in the angles between the vectors dX and

          \[ dY=\left(\begin{array}{c}-1\\1\\1\end{array}\right) \]

    • The coordinate system transformation in which the strain matrix is diagonal. Express the strain matrix in the new coordinate system.
  6. The longitudinal engineering strains on the surface of a test specimen were measured using a strain rosette to be 0.005, 0.002, and –0.001 along the three directions: a, b, and c\in\mathbb{R}^2 where a is along the direction of the basis vector e_1, b is oriented 60^\circ anticlockwise from a, and c is oriented 60^\circ anticlockwise from b. If the material is assumed to be in a small strain state, find the principal strains and their directions on the surface of the test specimen.
  7. The longitudinal engineering strains on the surface of a test specimen were measured using a strain rosette to be 0.006, -0.001, and 0.002 along the three directions: a, b, and c\in\mathbb{R}^2 where a is along the direction of the basis vector e_1, b is oriented 30^\circ anticlockwise from a, and c is oriented 60^\circ anticlockwise from b. If the material is assumed to be in a small strain state, find the principal strains and their directions on the surface of the test specimen.
  8. The measured strains for the three axes of a strain gauge rosette are: \varepsilon_1=-0.0004, \varepsilon_2=-0.000085, and \varepsilon_3=0.00025. If \varepsilon_1 coincides with the basis vector e_1 while \varepsilon_2 and \varepsilon_3 coincide with vectors that are anticlockwise 60^\circ and 120^\circ respectively from e_1. Find the components of the strain tensor.
  9. Assume that a two dimensional rectangular plate with dimensions 5 units along the basis vector e_1 and 1 unit along the basis vector e_2 is situated such that the origin of the coordinate system is at the midpoint of the left side of the plate. If the displacement function of the plate is described by:

        \[ \begin{split} u_1&=(-0.005X_1^2+0.004X_1)X_2\\ u_2&=-0.00016X_1^3+0.0012X_1^2\\ u_3&=0 \end{split} \]

    find the following:

    • The position function x=f(X).
    • The infinitesimal strain matrix as a function of the position.
    • The Green strain matrix as a function of the position.
    • Draw the vector plot of the displacement function.
    • Draw the contour plots of {\varepsilon_{small}}_{11}, {\varepsilon_{Green}}_{11}, {\varepsilon_{small}}_{12}, and {\varepsilon_{Green}}_{12}.
  10. Repeat the previous question if the displacement function is given by:

        \[ \begin{split} u_1&=(-0.001X_1^2+0.003X_1)X_2\\ u_2&=(-0.007X_2^2+0.003X_1^2)X_1\\ u_3&=0 \end{split} \]

  11. The general equation for the coordinates of a point x on an ellipsoidal surface whose axes are oriented along the Cartesian coordinate system is:

        \[ \left(\frac{x_1}{a}\right)^2+\left(\frac{x_2}{b}\right)^2+\left(\frac{x_3}{c}\right)^2=1 \]

    where a, b, and c are the three semi-diameters of the ellipsoidal surface. When a=b=c=r, the surface is a sphere with radius r. Assuming that the motion of an object is described by the homogeneous deformation gradient F\in\mathbb{M}^3 such that the diagonal components of F are the only non-zero components.

    • Show that a sphere of radius r deforms into an ellipsoid surface.
    • Find the relationship between the semi-diameters of the ellipsoid and the components F_{ij} of the deformation gradient.
    • Find the relationship between the semi-diameters of the ellipsoid and the components \varepsilon_{ij} of the infinitesimal strain tensor.
  12. Three unit square elements in \mathbb{R}^2 deform as shown in the figure below.StrainE1
    For each case:

    • Write expressions for the displacement components u_1 and u_2 as functions of X_1, X_2, and the variables shown for each square.
    • Determine the components of the infinitesimal strain tensor and the infinitesimal rotation tensor.
    • Determine the components of the Green strain tensor.

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