This problem is from the section on deformations obbeying the linear relation between reaction force and deformation length. This one requires knowledge of Young modulus which allows to express reaction force as a function of relative deformation. In one-dimensional case the law looks as follows:
$$ \sigma = E\varepsilon $$where $\sigma=F/A$ is the reaction tension (${\rm N/m^2}$) of the material in response to the relative deformation $\varepsilon=\Delta L / L$.
This problem is taken from the book Savchenko et al and it took me some time to understand to my satisfaction, therefore it is written up here.
Problem statement
Determine the minimum amount of work required to bend a rod of length $l$ with square cross section ($a \times a$) into a ring. Young modulus of the material is $E$ and $l \gg a$.
Solution
The approach to take here was hinted to me by the previous problem which asked for the expression of the energy density for a deformed material. It is relatively easy to derive using the above expression for the deformation tension that the energy density per unit volume ($\omega$) can be expressed as follows:
$$ \omega = \frac{dW}{dV} = \frac{E\varepsilon^2}{2} $$So the problem now is reduced to finding the ring configuration that would minimize the energy (work) due to deformation of the rod:
$$ W = \int\omega dV \Rightarrow W_{\rm min} = \min_{V_{\rm ring}}\left( \int \omega dV \right) $$In the bent state we can divide the ring into multiple strips of width $dr$ with the smallest having the radius $R$. We assume that the length of the longest (outer) strip would be $R+a$, i.e. here we neglect radial deformations and assume that the width of the rod as well as its height won't change. Then the deformation energy $dW$ stored in each strip would be:
$$ dW = \omega dV = \frac{E\varepsilon^2}{2}dV =\frac{E\varepsilon^2}{2}a\cdot 2\pi r dr $$Therefore the total energy of the deformed rod would be a sum of energy of each of the strips:
$$ W = \int\limits_{R}^{R+a}\frac{E\varepsilon^2}{2}a\cdot 2\pi r dr $$Then noting that for each strip the relative deformation $\varepsilon$ is $(2 \pi r -l )/l $, we get: $$ W(R) = \int\limits_{R}^{R+a}\frac{E(2 \pi r - l)^2}{2l^2}a\cdot 2\pi r dr $$
Now we need to find the radius of the shortest strip $R$ (which would also define the radii of all the longer strips) that minimizes $W(R)$, i.e.:
$$ W_{\min} = \min_{R} W(R) = \min_{R}\left( \int\limits_{R}^{R+a}\frac{E(2 \pi r - l)^2}{2l^2}a\cdot 2\pi r dr \right) $$Intuitively, I figured that the minimizing configuration would include the undeformed strip (i.e. the one in bent state having the same length $l$ as before). To minimize the deviations of the other strips, the undeformed strip would have to be in the middle, so that leaves us with $R = \frac{l}{2\pi} - \frac{a}{2}$ and once we have this the solution is straightforward. But I still would like to find the minimizing $R$ in a more rigorous way.
If one accepts the above arguments then (If not please see another way of determining $R$ in the note section at the end of the post) the expression for $W_{\min}$ becomes:
$$ W_{\min}=W\left(R=\frac{l}{2\pi} - \frac{a}{2}\right)=\int\limits_{\frac{l}{2\pi} - \frac{a}{2}}^{\frac{l}{2\pi} + \frac{a}{2}}\frac{E(2 \pi r - l)^2}{2l^2}a\cdot 2\pi r dr $$Making a substitution $y = 2 \pi r$: $$ W_{\min} = \frac{Ea}{4\pi l^2}\int\limits_{l - \pi a}^{l + \pi a}(y - l)^2\cdot y dy $$
Then simplifying: $$ W_{\min} = \frac{Ea}{4\pi l^2}\left(\int\limits_{\:l - \pi a}^{l + \pi a}(y - l)^3 dy + l\cdot\int\limits_{l - \pi a}^{l + \pi a}(y - l)^2 dy \right) $$
And another substitution for convenience $z=y-l$: $$ W_{\min} = \frac{Ea}{4\pi l^2}\left(\int\limits_{\: - \pi a}^{\pi a}z^3 dz + l\cdot\int\limits_{- \pi a}^{\pi a}z^2 dz \right) $$
Noticing that the first integral above is from an odd function over a symmetric interval with respect to 0, therefore it is 0 (or you could simply integrate to convince yourself) and we get: $$ W_{\min} = \frac{Ea}{4\pi l^2}l\cdot\int\limits_{- \pi a}^{\pi a}z^2 dz = \frac{Ea}{4\pi l^2}l\cdot \frac{2 \pi^3 a^3}{3} $$
Then simplifying we get the final expression for the minimum amount of work needed to bend the rod:
$$ \boxed{W_{\min} = \frac{E\pi^2 a^4}{6 l}} $$Note on finding the minimum radius of the minimizing configuration of bent strips
The general expression for the deformation energy (work) of the bent rod is:
$$ \min_{R} W(R) = \min_{R}\left( \int\limits_{R}^{R+a}\frac{E(2 \pi r - l)^2}{2l^2}a\cdot 2\pi r dr \right) $$We need to find the $R_0$ for which $W(R)$ is minimized. The minimizing configuration should contain the undeformed strip and therefore we can limit our search for $R_0$ to the closed interval $\left[ \frac{l}{2\pi} - a, \frac{l}{2\pi} \right ]$.
Reorganizing the terms
$$ W(R) = \min_{R}\left( \frac{Ea}{8\pi l^2} \int\limits_{2\pi R - l}^{2\pi R+ 2\pi a - l}z^2\cdot (z + l) dz \right) $$Let's change the independent variable from $R$ to $Z = 2\pi R - l$ (where $Z \in [-2\pi a\,,\: 0]$), for convenience:
$$ W(Z) = \min_{Z}\left( \frac{Ea}{8\pi l^2} \int\limits_{Z}^{Z+ 2\pi a}z^2\cdot (z + l) dz \right) $$Equivalently, we need to determine $Z$, which will minimize the following function:
$$ I(Z) = \int\limits_{Z}^{Z+ 2\pi a}z^2\cdot (z + l) dz $$To find the extrema of $I(Z)$ we solve for roots of $\frac{dI(Z)}{dZ}=0$.
By differentiation of the integral with variable limits we get the following expression for the derivative:
$$ \frac{dI(Z)}{dZ}=(Z+2\pi a)^2 \cdot (Z+2\pi a + l) - Z^2 \cdot (Z + l)=0 $$Simplifying we get the following equation for the extrema of $I(Z)$:
$$ Z^2 2 \pi a + 4\pi a Z^2 + 4\pi a Z (2\pi a + l) + 4\pi^2 a^2 Z + 4\pi^2 a^2 (2\pi a + l) = 0 $$Simplifying further, we get the following quadratic equation with respect to $Z$:
$$ 3Z^2 + 2 Z (3\pi a + l) + 2\pi a (2\pi a + l) = 0 $$The exact solution is then:
$$ Z_{1,2} = \frac{-(3\pi a + l) \pm \sqrt{l^2 - 3 \pi^2a^2}}{3} $$then using the condition $l \gg a$ we neglect second order term $3\pi^2(a/l)^2$:
$$ \left [ \begin{array}{l} Z_1 = -\pi a \\ Z_2 = -\pi a - 2l/3 \\ \end{array}\right. $$Since $\frac{dI(Z)}{dZ}$ is quadratic, and $Z_2 < Z_1$, it is easy to see that
$$ \left\{ \begin{array}{l} \frac{dI(Z)}{dZ} \geq 0\,,\; Z\leq Z_2 \\ \frac{dI(Z)}{dZ} \leq 0\,,\; Z_2 < Z \leq Z_1 \\ \frac{dI(Z)}{dZ} > 0\,,\; Z > Z_1 \\ \end{array}\right. $$Therefore we can conclude that $Z_1$ is a local mininmum, and $Z_2$ is a local maxinmum of $I(Z)$. Hence $Z_1 = -\pi a$ is a candidate for a minimum of $I(Z), Z\in [-2 \pi a\,,\: 0]$. To make sure that $Z_1$ is the minimum we need to check that $I(Z_1) < I(-2 \pi a)$.
Let's compute directly these values:
$$ I(Z_1) = \int\limits_{-\pi a}^{\pi a}z^2\cdot (z + l) dz = \frac{2}{3}l \pi^3 a^3 $$$$ I(-2 \pi a) = \int\limits_{-2\pi a}^{0}z^2\cdot (z + l) dz = 4 \pi^4 a^4 + \frac{8}{3}l \pi^3 a^3 $$From the two above expressions it is obvious that $Z=-\pi a$ minimizes $I(Z), Z \in [-2\pi\,,\: 0]$.
Finally, we can find the $R_0$ which minimizes energy of deformation:
$$ Z_0 = -\pi a = 2 \pi R_0 - l \Rightarrow \boxed{R_0 = \frac{l}{2\pi} - \frac{a}{2}} $$Which is the same as we guessed before.
