# Deformation tensor

A tensor describing the locations of the points of a body after deformation with respect to their location before deformation. It is a symmetric tensor of the second rank,

$$u_{ik}=\frac12\left(\frac{\partial u_i}{\partial x_k}+\frac{\partial u_k}{\partial x_i}+\frac{\partial u_l}{\partial x_i}\frac{\partial u_l}{\partial x_k}\right),\label{*}\tag{*}$$

where $x_i$ are the Cartesian rectangular coordinates of a point in the body prior to deformation and $u_i$ are the coordinates of the displacement vector $\mathbf u$. In the theory of elasticity the deformation tensor is decomposed into two constituent tensors:

$$u_{ik}=u_{ik}'+u_{ik}''.$$

The tensor $u_{ik}'$ describes a spatial deformation and is known as the spherical deformation tensor:

$$u_{ik}'=\frac13\delta_{ik}u_{ll}.$$

The tensor $u_{ik}''$ describes solely the change in form, and the sum of its diagonal elements is equal to zero:

$$u_{ik}''=u_{ik}-\frac13\delta_{ik}u_{ll}.$$

The tensor $u_{ik}''$ is known as the deviator of the deformation tensor.

In the case of a small deformation, second-order magnitudes are neglected, and the deformation tensor \eqref{*} is defined by the expression:

$$u_{ik}=\frac12\left(\frac{\partial u_i}{\partial x_k}+\frac{\partial u_k}{\partial x_i}\right).$$

In spherical coordinates $r,\theta,\phi$ the linearized deformation tensor \eqref{*} assumes the form:

$$u_{rr}=\frac{\partial u_r}{\partial r},\quad u_{\theta\theta}=\frac1r\frac{\partial u_\theta}{\partial\theta}+\frac{u_r}{r},$$

$$u_{\phi\phi}=\frac{1}{r\sin\theta}\frac{\partial u_\phi}{\partial\phi}+\frac{u_\theta}{r}\operatorname{cotan}\theta+\frac{u_r}{r},$$

$$2u_{\theta\phi}=\frac1r\left(\frac{\partial u_\phi}{\partial\theta}-u_\phi\operatorname{cotan}\theta\right)+\frac{1}{r\sin\theta}\frac{\partial u_\theta}{\partial\phi},$$

$$2u_{r\theta}=\frac{\partial u_\theta}{\partial r}-\frac{u_\theta}{r}+\frac1r\frac{\partial u_r}{\partial\theta},$$

$$2u_{\phi r}=\frac{1}{r\sin\theta}\frac{\partial u_r}{\partial\phi}+\frac{\partial u_\phi}{\partial r}-\frac{u_\phi}{r}.$$

In cylindrical coordinates $r,\phi,z$ it has the form

$$u_{rr}=\frac{\partial u_r}{\partial r},\quad u_{\phi\phi}=\frac1r\frac{\partial u_\phi}{\partial\phi}+\frac{u_r}{r},\quad u_{zz}=\frac{\partial u_z}{\partial z},$$

$$2u_{\phi z}=\frac1r\frac{\partial u_z}{\partial\phi}+\frac{\partial u_\phi}{\partial z},\quad2u_{rz}=\frac{\partial u_r}{\partial z}+\frac{\partial u_z}{\partial r},$$

$$2u_{r\phi}=\frac{\partial u_\phi}{\partial r}-\frac{u_\phi}{r}+\frac1r\frac{\partial u_r}{\partial\phi}.$$

#### References

[1] | L.D. Landau, E.M. Lifshits, "Theory of elasticity" , Pergamon (1959) (Translated from Russian) |

[2] | , [Soviet] Physical Encyclopedic Dictionary , 1 , Moscow (1960) pp. 553 (In Russian) |

#### Comments

Strictly speaking the $u_{ik}$ of \eqref{*} do not form a tensor in the mathematical sense; they do not transform as tensors. The "tensor" $u_{ik}$ is also known as the strain tensor or the rate-of-strain tensor.

Let $dl=(dx_1^2+dx_2^2+dx_3^3)^{1/2}$ and $dl'=(dx_1'^2+dx_2'^2+dx_3'^2)^{1/2}$ denote line elements (infinitesimal distances) before and after deformation. Then $dl'^2-dl^2=2\sum u_{ik}dx_idx_k$, so that $u_{ik}$ describes the change in an element of length as the body is deformed.

#### References

[a1] | A.C. Eringen, "Mechanics of continua" , Pergamon (1967) |

**How to Cite This Entry:**

Deformation tensor.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Deformation_tensor&oldid=44710