If I roll an elastic plate into a cylinder, does it shrink? Suppose I start with a rectangular elastic (to keep things simple, zero Poisson's ratio) sheet of length $2\pi R$, thickness $h$, and (immaterial) width $W$. I roll it up into a cylinder of radius $R$, and allow it to relax. By symmetry, if it deforms at all, it will do so by uniformly scaling of the cross section to a cylinder of radius $R+\Delta R$.
I'm trying to study the behavior of the cylinder using Kirchoff-Love plate theory, but have gotten myself confused. If I use Cauchy strain, I get that the strain is proportional to
$$\frac{z+\Delta R}{R}$$
where $z\in (-h/2, h/2)$ is the normal coordinate. I then get that the energy density of the cylinder depends on $(\Delta R)^2$ but not $\Delta R$, and therefore setting $\Delta R=0$ minimizes energy -- the cylinder neither grows nor shrinks.
However, if I instead use Green strain, the deformation energy has a nonzero linear term in $\Delta R$, and I get that the cylinder shrinks ever so slightly. I find this very counter-intuitive.
Why is using Cauchy strain inappropriate for approaching this problem? If I perform this experiment with a real sheet, will I really observe (very slight) shrinking? Is there intuition for why this occurs?
 A: I suppose that when you say "Cauchy strain" you mean the infinitesimal strain tensor. In general, I would not use this strain measure since it is valid for
$$\frac{\partial u_i}{\partial X_j} \ll 1\, .$$
Now, let's consider the problem that you are proposing. The configuration is given by
\begin{align}
&x = (R - Y) \sin\left(X/R\right)\, ,\\
&y = R - (R - Y) \cos\left(X/R\right)\, ,\\
&z = Z\, ,
\end{align}
where $R$ is the radius of curvature of the deformation, $X, Y, Z$ the material coordinates, and $x, y, z$ the spatial ones.
For this problem the deformation gradient is
$$F = \left[\begin{matrix}\frac{\left(R - Y\right) \cos{\left(\frac{X}{R} \right)}}{R} & - \sin{\left(\frac{X}{R} \right)} & 0\\- \frac{\left(- R + Y\right) \sin{\left(\frac{X}{R} \right)}}{R} & \cos{\left(\frac{X}{R} \right)} & 0\\0 & 0 & 1\end{matrix}\right]\, ,$$
and the jacobian is
$$J = 1 - \frac{Y}{R}\, .$$
Notice that this is the change of volume of the body. Thus, it does shrink, at least for points above $Y=0$.
Now, let's move on to different strain measures. If we use the Lagrange strain we have the following
$$E = \frac{1}{2}\left(F^T F - I\right) = \left[\begin{matrix}
- \frac{Y}{R} + \frac{Y^{2}}{2 R^{2}} & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0\end{matrix}\right]\, ,$$
if we want to compute the extension, we could use
$$E_1 = \sqrt{1 + 2E_{11}} - 1\approx - \frac{Y}{R}\, ,$$
that would be positive/negative depending on the sign of $Y$.
Finally, let's consider the infinitesimal strain tensor.
$$\epsilon_{ij} = \frac{1}{2}\left(\frac{\partial u_i}{\partial X_j} + \frac{\partial u_j}{\partial X_i}\right)\, ,$$
with
$$\begin{align}
u = x - X\, ,\\
v = y - Y\, ,\\
z = z - Z\, .
\end{align}$$
Thus,
$$\epsilon = \left[\begin{matrix}\cos{\left(\frac{X}{R} \right)} - 1 - \frac{Y \cos{\left(\frac{X}{R} \right)}}{R} & - \frac{Y \sin{\left(\frac{X}{R} \right)}}{2 R} & 0\\- \frac{Y \sin{\left(\frac{X}{R} \right)}}{2 R} & \cos{\left(\frac{X}{R} \right)} - 1 & 0\\0 & 0 & 0\end{matrix}\right]\, ,$$
and considering the first term in the Taylor series
$$\epsilon \approx \left[\begin{matrix}- \frac{Y}{R} & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{matrix}\right]\, .$$
So, we could say that it can shrink or extend depending on $Y$. The extension depends on $Y/R$ linearly.
