How do I show that the strain tensor

$$\epsilon_{ij}=\partial_i u_j + \partial_j u_i$$

in the case of a one-dimensional spring is reduced to

$$\epsilon= \frac{x-L}{L},$$

where L is the initial length of the spring. I can see that the only component that survives in this case is $\epsilon_{xx}=\partial_x u_x$, but I don't see how $L$ appears at the denominator.


2 Answers 2


For a homogeneous deformation of the spring, $$u=\epsilon_{xx}x$$where $\epsilon_{xx}$ is a constant. So, at x=L, $$u(L)=\epsilon_{xx}L$$Therefore, $$\epsilon_{xx}=\frac{u(L)}{L}=\frac{[L+u(L)]-L}{L}$$

  • $\begingroup$ I am still a bit confused, because I am trying to fit this into the formal definition of $u$, which is $u(s)=X'(s)-X(s)$, where $X$ is the position of the point $s$ in the reference configuration, with $s$ the arc length. If I use this definition, $X'(L)=x,\ X(L)=L$, so $u(L)=x-L$. $\endgroup$ Sep 26, 2018 at 16:26
  • $\begingroup$ So $x=L+u(L)$. You and I are using the symbol x to represent two different things. My x is the same as your X. $\endgroup$ Sep 26, 2018 at 16:38

In the reference configuration, the spring is a straight line:

$$X(s)=s,\ s\in[0,L]$$

its length being $$\ell_0=\intop_0^L |\dot X(s)| ds=L.$$

When the spring has stretched to have length $x$, its expression will be

$$X'(s)=\lambda s$$

The length has now changed, and by requiring that it is indeed $x$ we can find $\lambda$:

$$\ell(x)=\intop_0^L \lambda ds = x\ \ \Longrightarrow \lambda = x/L$$

Therefore $$X'(s) = \frac{x}{L}\ s.$$

The displacement is defined as $u(s)=X'(s)-X(s)$, which in this case becomes:

$$u(s)=\left(\frac{x}{L}-1\right)\ s$$

The strain is $$\epsilon_{xx}\equiv\epsilon=\partial_s u=\frac{x-L}{L}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.