# Spring strain vs fractional extension

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.

• May 2, 2023 at 12:53

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, $$\epsilon_{xx}=\frac{u(L)}{L}=\frac{[L+u(L)]-L}{L}$$

• 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 $u(L)=x-L$. Sep 26, 2018 at 16:26
• 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. 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 $$\epsilon_{xx}\equiv\epsilon=\partial_s u=\frac{x-L}{L}$$