Minimum Energy Curve Derivation Help

[nkpstn]

I've been trying to find the equation of the line of no strain of a bent linearly elastic beam of fixed length.
(in a plane, given just its endpoints, or both the end points and the slopes at the endpoints)
I imagine this problem has been solved many times, but I couldn't find a solution online that I could understand.

Using the property of linear elasticity it follows that $dE/ds \propto (dθ/ds)^2$

(Where E is the bending energy, s is the arclength, and θ, the angle of the tangent line)

What it comes down to is finding the curve which minimizes the integral of the square of the curvature across it.

\begin{equation} \int_0^l \frac{(x'y'' - y'x'')^2}{(x'^2 + y'^2)^3} + λ(x'^2 + y'^2)^{1/2}\,dt \end{equation}

λ is a Lagrange multiplier used to meet the length constraint.
(' here denotes a derivative with respect to t)

I take the first variation with respect to x,
So for all functions v:
\begin{equation} \int_0^l \frac{-2y'(x'y'' - y'x'')}{(x'^2 + y'^2)^3}v'' + \bigg(\frac{2y''(x'y'' - y'x'')}{(x'^2 + y'^2)^3} + \frac{-6x'(x'y'' - y'x'')^2}{(x'^2 + y'^2)^4} + \frac{λx'}{(x'^2 + y'^2)^{1/2}}\bigg)v'\,dt = 0 \end{equation}
Which I guess means that:
\begin{equation} \frac{2y''(x'y'' - y'x'')}{(x'^2 + y'^2)^3} + \frac{-6x'(x'y'' - y'x'')^2}{(x'^2 + y'^2)^4} -\bigg(\frac{-2y'(x'y'' - y'x'')}{(x'^2 + y'^2)^3}\bigg)' + \frac{λx'}{(x'^2 + y'^2)^{1/2}} = c_1 \end{equation}
I figure I am free to choose x and y to be parametrized by arclength
so I set:
\begin{equation}x'^2 + y'^2 = 1
\end{equation}
Taking the derivative:

$2x'x'' + 2y'y'' = 0$ so $y'y'' + x'x'' = 0$

$y''y'' + y'y''' + x''x'' + x'x''' = 0$

Also:

$x'y'' - y'x'' = (x'^2 + y'^2)y''/x' = y''/x' = -x''/y'$

Substituting this in:

$- 4y''^2/x' - 2x''' + λx' = c_1$

Analogously:

$- 4x''^2/y' - 2y''' + λy' = c_2$

Multiplying the first equation by x', the second by y' and combining gives:

$λ = c_1x' + c_2y' + 2x''^2 + 2y''^2$

Dividing the second equation by x':

$λy'/x' = c_2/x' + 2x''^2/y'x' + (2y'''x' - 2x''y'')/x'^2$

Substituting in the value of λ:

$c_1y' + 2x''^2/y'x' = c_2(1-y'^2)/x' + 2x''^2/y'x' + 2(y''/x')'$

$c_1y' - c_2x' = 2(y''/x')'$

Which can be integrated to give:

$c_1y - c_2x + c = 2y''/x'$

Which means that the curvature along this curve is proportional to the distance from a straight line. This sounds somewhat intuitvely pleasing.

However, taking the derivative of $λ = c_1x' + c_2y' + 2(x''/y')^2$

$0 = c_1x'' + c_2y'' + 2(y''/x')*2(y''/x')'$

$c_1x'' + c_2y'' + 2(y''/x')*(c_1y' - c_2x') = 0$

$c_1x'' + c_2y'' - 2c_1x'' - 2c_2y'' = 0$

$c_1x'' + c_2y'' = 0$

So:

$(y''/x')(y''/x')' = 0$

Which means that either the curvature is zero (A straight line) or the change in the curvature is zero (A circle)

Based on physical experience, this doesn't seem to be the general solution to this problem.

Assuming my algebraic manipulations are correct (I've checked them over), can anyone point out the flaws in my derivation? I'm new to variational calculus to be honest.

 

[jvicens]

Thank you for your post. It seems that you have made some progress in finding the equation of the line of no strain for a bent linearly elastic beam. However, I believe there are some errors in your derivation that may be causing the discrepancy with the expected solution.

Firstly, I would like to clarify that the equation of the line of no strain is also known as the neutral axis or neutral line, and it is the curve along which there is no bending stress or strain in the beam. This is different from the curve that minimizes the integral of the square of the curvature, which is known as the elastica or Euler-Bernoulli curve.

In your derivation, you have assumed that the curvature along the neutral axis is proportional to the distance from a straight line. This is not necessarily true, as the neutral axis can take on different shapes depending on the loading and boundary conditions of the beam. Additionally, your assumption that x'^2 + y'^2 = 1 may not hold for all points along the neutral axis.

Furthermore, in your derivation, you have taken the first variation with respect to x, which means you are assuming that the neutral axis is a function of x. However, the neutral axis is a function of both x and y, so this approach may not be valid. Also, in your final equation, you have neglected the Lagrange multiplier term, which is important in meeting the length constraint.

I would suggest revisiting your derivation and considering the neutral axis as a function of both x and y. You may also want to look into the principle of virtual work, which is commonly used in solving problems involving linear elasticity. I hope this helps in your research. Best of luck to you.