Displacement

When forces are applied to a body, it deforms to some extent. This deformation can be described in terms of the movement of material particles, a concept known as displacement.

In classical mechanics, a particle is treated as a point mass that can move along the x-, y-, and z-axes, without possessing rotational degrees of freedom.

If the position of a particle in the initial configuration¹ is denoted by X, and in the current configuration by x, the displacement vector is defined as
\[ \mathbf{u} = \mathbf{x} - \mathbf{X}. \] This vector field has three components along the x₁, x₂, and x₃ (or x, y, and z) axes by u₁, u₂, and u₃ (or u, v, and w).

In general, displacement field is not uniform; different points in the body may experience different displacements. Thus, it is, in general, a function of position and time t. There are two common ways to express this dependency:

• As a function of the initial (reference) position. In this case, we often denote the displacement by the upper case U:
  \(\mathbf{U} = \mathbf{U}(\mathbf{X}, t)\)
• As a function of the current (deformed) position:
  \(\mathbf{u} = \mathbf{u}(\mathbf{x}, t)\)

In metals with small elastic deformations, when evaluating functions of coordinates, the distinction between the coordinate systems X and x can often be ignored.

In this chapter, we focus on small strain theory, where both the displacement and its gradient are assumed to be small. For typical elastic behavior in metals — where strain is less than 0.2% — this assumption is well suited.

Rotation and Strain

Displacement can occur without deformation. For example, a body may undergo rigid body translation or rigid body rotation, during which the relative positions of particles remain unchanged. For example, consider the beam shown in the figure below. The overhanging segments undergo pure rotation without being subjected to internal forces or moments. These segments exhibit displacement, but not deformation.In such cases, although displacement exists, there is no internal deformation and hence no internal forces (or stress).

What gives rise to internal stresses is the relative displacements between neighboring particles. The relative displacement between two points increases with their initial separation. Therefore, to describe deformation, we should normalize the relative motion of two points by their initial distance.

For example, consider two points A and B whose connecting line is parallel with the x-axis, with a distance of Δx between them. Suppose point A undergoes a displacement \((u, v)\), while point B undergoes a displacement \((u + \Delta u, v + \Delta v)\).

The components of the relative displacement per unit initial distance between the two points are:
\[ \begin{aligned} &\frac{u_B-u_A}{|AB|}=\frac{(u + \Delta u) - u}{\Delta x} = \frac{\Delta u}{\Delta x},\\ &\frac{v_B-v_A}{|AB|}=\frac{(v + \Delta v) - v}{\Delta x} = \frac{\Delta v}{\Delta x} \end{aligned} \]
As \(\Delta x \to 0\), these expressions approach the partial derivatives:
\[ \frac{\partial u}{\partial x} \quad \text{and} \quad \frac{\partial v}{\partial x}. \] This shows that, to describe deformation, we need to evaluate the spatial derivatives (or gradient) of the displacement components.

The displacement gradient tensor is given by:
\[ \begin{bmatrix} u\\ v\\ w \end{bmatrix} \begin{bmatrix} \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y}& \dfrac{\partial }{\partial z} \end{bmatrix} = \begin{bmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} & \dfrac{\partial u}{\partial z} \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} & \dfrac{\partial v}{\partial z} \\ \dfrac{\partial w}{\partial x} & \dfrac{\partial w}{\partial y} & \dfrac{\partial w}{\partial z} \end{bmatrix} \]

The displacement gradient alone is not sufficient to describe deformation, because relative motion between points in a body can also result from rigid body rotation, which does not produce stress. Although rotation may be important in certain problems, when our goal is to focus on stress‑induced deformation, we must subtract out the effects of rotation.

To eliminate the rotations, we consider an element whose sides are parallel to the x- and y-axis (see the following figure). For small rotations, the rotation of a line initially parallel to the x-axis is
\[ \omega_1 \approx \tan \omega_1 \approx \frac{\Delta v}{\Delta x}, \] and the rotation of a line initially parallel to the y-axis is
\[ \omega_2 \approx \tan \omega_2 \approx -\frac{\Delta u}{\Delta y}. \]

The average rotation of the element in the xy-plane is then defined as
\[ \omega_{yx} = \frac{1}{2} \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right). \]

In general, the rotation is described by \[ \begin{bmatrix} 0 & \omega_{xy} & \omega_{xz}\\ \omega_{yx} & 0 & \omega_{yz}\\ \omega_{zx} & \omega_{zy} & 0 \end{bmatrix}= \begin{bmatrix} 0 & \frac{1}{2}\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right) & \frac{1}{2}\left(\frac{\partial u}{\partial z}-\frac{\partial w}{\partial x}\right)\\ \frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial x}\right) & 0 & \frac{1}{2}\left(\frac{\partial v}{\partial z}-\frac{\partial w}{\partial y}\right)\\ \frac{1}{2}\left(\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}\right) & \frac{1}{2}\left(\frac{\partial w}{\partial y}-\frac{\partial v}{\partial z}\right) & 0 \end{bmatrix} \] \[ \omega_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i}\right) \] Notice that this is an antisymmetric matrix.

Subtracting the rotation from the displacement gradient gives strain, which describes the local changes in the shape of the body

\[ \bbox[5px,border:1px #f2f2f2;background-color:#f2f2f2]{\begin{bmatrix} \epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz}\\ \epsilon_{yx} & \epsilon_{yy} & \epsilon_{yz}\\ \epsilon_{zx} & \epsilon_{zy} & \epsilon_{zz} \end{bmatrix}= \begin{bmatrix} \dfrac{\partial u}{\partial x} & \dfrac{1}{2}\left(\dfrac{\partial u}{\partial y}+\dfrac{\partial v}{\partial x}\right) & \dfrac{1}{2}\left(\dfrac{\partial u}{\partial z}+\dfrac{\partial w}{\partial x}\right)\\ \dfrac{1}{2}\left(\dfrac{\partial v}{\partial x}-\dfrac{\partial u}{\partial x}\right) & \dfrac{\partial v}{\partial y} & \dfrac{1}{2}\left(\dfrac{\partial v}{\partial z}+\dfrac{\partial w}{\partial y}\right)\\ \dfrac{1}{2}\left(\dfrac{\partial w}{\partial x}+\dfrac{\partial u}{\partial z}\right) & \dfrac{1}{2}\left(\dfrac{\partial w}{\partial y}+\dfrac{\partial v}{\partial z}\right) & \dfrac{\partial w}{\partial z} \end{bmatrix}.} \] Therefore, the components of the strain are given by \[ \bbox[5px,border:1px #f2f2f2;background-color:#f2f2f2]{\epsilon_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right).} \] The components on the main diagonal are called normal strains, and the rest of components are called shear strain.

• Notice that the matrix corresponding to strain is symmetric.
• Notice that \[ \frac{\partial u_i}{\partial x_j}=\epsilon_{ij}+\omega_{ij}. \]
• In engineering applications, it is common to use a simplified notation for normal strain components: when both indices are the same, only one subscript is used. For example, ϵ_xx​ is often written as ϵ_x​, and ϵ_yy​ as ϵ_y​. \[ \epsilon_x=\frac{\partial u}{\partial x},\quad \epsilon_y=\frac{\partial v}{\partial y}, \quad \epsilon_z=\frac{\partial w}{\partial z}. \]
• In many engineering problems, we often use twice the shear strain ϵ_ij​ and denote it by Ɣ_ij: \[ \begin{aligned} \gamma_{xy}&=2\epsilon_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x},\\ \gamma_{xz}&=2\epsilon_{xz}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x},\\ \gamma_{yz}&=2\epsilon_{yz}=\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y}. \end{aligned} \]

Geometric Interpretation

Consider an element in 2D that is deformed. For simplicity, let’s move the corner of the element in the deformed state to the corner of the element before deformation. The normal strains are the change in length per unit initial length, while the shear strain is half the decrease in the angle shown below.

Sign of Strain Components

The normal strain is considered positive when the length in the corresponding direction increases. For example, ϵ_xx is positive when the element elongates along the x-axis.

The shear strain ϵ_xy is considered positive if the angle between the +x and +y axes (or between the –x and –y axes) decreases, and the angle between the +x and –y axes (or between the –x and +y axes) increases (see the following figure). In other words, the change in angles produced by a positive shear stress is defined as a positive shear strain, and vice versa.

Units of Strain

Notice that strain is dimensionless, so it has no units.

References

1. Malvern, L. E. (1969). Introduction to the mechanics of a continuous medium. Prentice-Hall.
2. McClintock, F. A., & Argon, A. S. (1966). Mechanical behavior of materials. Addison-Wesley.