Elastic energy of a bent steel rod

[cedricyu803]

Homework Statement

[Math. for Physicists, M. Stone Problem 1.4]

Assume that a rod of length L is only slightly bent into the yz plane and lies close to the z axis, show that the elastic energy can be approximated as
$$U[y]= \int_{0}^{L} \frac{1}{2}YI(y'')^2 dz$$

Homework Equations

It is given that the elastic energy per unit length of a bent rod , $$u=\frac{1}{2}YI/R^{2}$$
R is the radius of curvature due to the bending, Y is the Young's modulus of the steel and I is the moment of inertia of the rod's cross section about an axis through its centroid and
perpendicular to the plane in which the rod is bent.

The Attempt at a Solution

I don't quite get the picture.
Does it mean that each infinitesimal piece is a segment of a circle R with a different center? Or should I consider the whole bent rod as a segment of a circle of radius R?

But still the infinitesimal rod length should be $$\sqrt{1+(y')^2} dz$$, so how can I get $$y''^2$$?

Thank you very much!

The Attempt at a Solution

 

[TSny]

Think of the bent rod as lying along a curve y(z). Maybe you can use the mathematical expression for the radius of curvature of a curve. http://en.wikipedia.org/wiki/Radius_of_curvature_(mathematics)