Strain and its Transformation

We can’t get very far in the understaning of machines and structures without thinking about their deformation. For solids, we usually describe the deformation in terms of strain. For a fluid, we usually speak of strain rate.

If we start with a bar that is 1 m long and stretch it by 1 ${\mu}$ m, it has an average longitudinal strain of 1 ${\mu}$ m/m or simply ${10^{-6}}$ . For most cases of interest, the strain isn’t constant over the volume of a body or fluid, so we define the strain on infinitesimal volumes and think of it as varying throughout the volume.

1. Strain in Cartesian Coordinates

Consider the infinitesimal rectangular element sketched below. We can think of marking the points ${P}$ , ${Q}$ , and ${R}$ in a continuum and watching them move to ${P_1}$ , ${Q_1}$ , and ${R_1}$ as the material deforms. The longitudinal strain in the ${x}$ direction is ${(P_1 R_1-PR)/PR}$ , the longitudinal strain in the ${y}$ direction is ${(P_1 Q_1-PQ)/PQ}$ and the shear strain is ${\phi_1 + \phi_2}$ .

Suppose that the vertex ${P}$ of the element is located at the location ${(x,y,z)}$ within the continuum, and that the continuum undergoes small deformations ${u_x}$ , ${u_y}$ , and ${u_z}$ —all functions of ${x}$ , ${y}$ , and ${z}$ . Because we will be taking the limit as ${dx}$ , ${dy}$ , and ${dz}$ go to zero, we can write the deformations using the first term of a Taylor series. For example, if the ${y}$ displacement of point ${P}$ is ${u_y(x,y,z)}$ , the ${y}$ displacment at point ${R}$ is expressed as

$\displaystyle u_y(x+dx,y,z) \approx u_y(x,y,z) + \left[ \frac{\partial u_y}{\partial x} (x,y,z) \right] dx$

and we needn’t bother with the higher-order terms.

Under this deformation, the length of the line segment ${PR}$ from ${(x,y,z)}$ to ${(x+dx,y,z)}$ changes by the difference between the ${x}$ displacements at each of its ends. This change in length divided by the original length ${dx}$ gives the longitudinal strain ${\epsilon_x (x,y,z)}$ in the ${x}$ direction:

$\displaystyle \epsilon_x(x,y,z) = \frac{1}{dx} \left[ u_x(x+dx,y,z) - u_x(x,y,z) \right] \ \ \ \ \ (1)$

In the limit as ${dx}$ goes to zero, the right-hand side of this expression becomes simply ${\partial u_x / \partial x}$ . Similar expressions can be obtained for the other edges of the element, so that in summary we have the following expressions for the longitudinal strains in the ${x}$ , ${y}$ , and ${z}$ directions:

$\displaystyle \epsilon_x = \frac{\partial u_x}{\partial x} \; \; \; \epsilon_y = \frac{\partial u_y}{\partial y} \; \; \; \epsilon_z = \frac{\partial u_z}{\partial z} \ \ \ \ \ (2)$

The shear strains are defined as the change in right angles within the continuum. Consider again the element of material shown in the figure above. The change in angle is ${\phi_1+\phi_2}$

$\displaystyle \begin{array}{rcl} \gamma_{xy}(x,y,z) & = & \dfrac{ u_x(x,y+dy,z)-u_x(x,y,z) }{ dy } \\ & & + \dfrac{ u_y(x+dx,y,z) - u_y(x,y,z) }{ dx } \end{array}$

Thus we have ${\gamma_{xy} = \partial u_x / \partial y + \partial u_y \partial x}$ . Similar expressions are obtained by considering deformation in the ${xz}$ and ${yz}$ planes, so that the shear strains are written as

$\displaystyle \begin{array}{rcl} \gamma_{xy} & = & \dfrac{\partial u_y}{\partial x} + \dfrac{\partial u_x}{\partial y} \\ \gamma_{xz} & = & \dfrac{\partial u_x}{\partial z} + \dfrac{\partial u_z}{\partial x} \\ \gamma_{yz} & = & \dfrac{\partial u_y}{\partial z} + \dfrac{\partial u_z}{\partial y} \end{array}$

The reader should be aware that occasionally a definition of shear strains ${1/2}$ as large as these is employed. Such strains are usually denoted for example as ${\epsilon_{xy} = \gamma_{xy} / 2}$ . When the distinction between these two strains is made, ${\epsilon_{xy}}$ is usually called “strain,” and ${\gamma_{xy}}$ is called “engineering strain.” But in most engineering work (especially for isotropic materials) we use ${\gamma_{xy}}$ and just call it “strain.”

The following simple examples should help give a feel for the basics of strain in the ${xyz}$ coordinate system. The reader is suggested to sketch each field and see how it corresponds to the algebra.

Example 1 A displacement ${u_x = cx}$ , ${u_y = u_z = 0}$ is imposed on a continuum. All of the strains are zero execpt ${\epsilon_x}$ which has the value of ${c}$ . (Note that ${c}$ is unitless, which is correct for strain.) This case is called “uniaxial strain.”

Example 2 The displacement is ${u_x = cy}$ , ${u_y = cx}$ , ${u_z = 0}$ . All of the strains are zero except ${\gamma_{xy}}$ which has the value of ${2c}$ . This is called “pure shear.”

Example 3 The displacement is ${u_x = -cy}$ , ${u_y = cx}$ , ${u_z = 0}$ . The strain is zero. This is a counter-clockwise rigid-body rotation.

2. Strain in Other Directions

It is often important to compute or express the strain along some directions other then the ${xyz}$ axes. A complete development of this topic is fairly long, but for this article we give a part of the story. Consider again a line ${PQ}$ that deforms to line ${P_1 Q_1}$ as a continuum deforms. And suppose now that ${PQ}$ is at an angle ${\theta}$ to the ${x}$ axis as sketched below.

What is the longitudinal strain in the direction of ${PQ}$ ?

Line ${PQ}$ stretches in the ${x}$ direction by

$\displaystyle \delta_x = u_x(x+dx,y+dy)-u_x(x,y) = \dfrac{\partial u_x}{\partial x} dx + \dfrac{\partial u_x}{\partial y} dy \ \ \ \ \ (3)$

and in the ${y}$ direction by

$\displaystyle \delta_y = u_y(x+dx,y+dy)-u_y(x,y) = \dfrac{\partial u_y}{\partial x} dx + \dfrac{\partial u_y}{\partial y} dy \ \ \ \ \ (4)$

These are relative displacements, so they can be though of as a vector composed of ${\delta_x \vec{e}_x + \delta_y \vec{e}_y}$ , where ${\vec{e}_x}$ and ${\vec{e}_y}$ are the unit vectors in the ${x}$ and ${y}$ directions. (Note that strains cannot be treated as vectors, as we shall see.)

To get the longitudinal strain along the line, our next step is to project the stretching displacement along the line ${PQ}$ . We can write a unit vector along the line ${PQ}$ (pointing in the ${x'}$ direction in the figure) as

$\displaystyle \begin{array}{rcl} \vec{e}_{x'} & = & \cos\theta \vec{e}_x + \sin\theta \vec{e}_y \\[2ex] & = & \dfrac{dx}{PQ} \vec{e}_x + \dfrac{dy}{PQ} \vec{e}_y \end{array}$

Taking the dot product of the vector of relative displacements with ${\vec{e}_{x'}}$ and dividing by the length ${PQ}$ , we obtain an expression for the strain ${\epsilon_{x'}}$ along line ${PQ}$ :

$\displaystyle \begin{array}{rcl} \epsilon_{x'} & = & \dfrac{ \delta_x ( dx/PQ ) + \delta_y (dy/PQ) }{ PQ } \\[4ex] & = & \cos^2 \theta \epsilon_x + \sin^2 \theta \epsilon_y + \cos\theta \sin\theta \gamma_{xy} \end{array}$

Example 4 The displacement is ${u_x = cy}$ , ${u_y = cx}$ , ${u_z = 0}$ . What is the strain along a direction rotated (a) ${45^\circ}$ and (b) ${135^\circ}$ from the ${x}$ axis?

We saw in Example~2 that this deformation field yields ${\epsilon_x= \epsilon_y = 0}$ and ${\gamma_{xy} = 2c}$ . Along the ${45^\circ}$ line, ${\cos\theta = \sin\theta = 1/\sqrt{2}}$ , so ${\epsilon_{x'} = c}$ . Along the ${135^\circ}$ line, ${\sin\theta = -\cos\theta = 1/\sqrt{2}}$ , so ${\epsilon_{x'} = -c}$ . If you start with a sketch of a square and sketch the way it would deform into a diamond like shape, you should be able to verify these answers directly.

As you can see from this example, strains don’t transform like vectors. They are tensors, and need to be handled differently than more familiar vector quantities like displacements and forces. Look for a future article on tensor manipulation.

Mechanics and Machines

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notes on mechanics, machine design, and analysis