----- Original Message ---- From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Tuesday, September 23, 2008 1:46:05 PM Subject: [math-fun] Thin strip of spring steel model I wonder if anyone is familiar with a simplified physics of uniform thin planar elastic strips, and wouldn't mind explaining it. (A.K.A. a leaf spring) In particular, suppose you have a thin strip of fixed length that for each endpoint is fixed to the plane at a fixed location, with a fixed tangent direction there. (There's also probably a spring constant to toss into the mix.) Is there a formula to determine the shape of the resulting curve based on the endpoint data? A differential equation it must satisfy? Or at least a way to calculate explicitly a quantity that is minimized by the shape, such as potential energy ? --Dan ________________________________ The strain energy per unit length is proportional to the square of the curvature. This is the same as for a thin fiber, untwisted and confined to a plane, and a curve that minimizes the energy is known as an "elastica". The intrinsic equation of an elastica is easily found. Let r = position vector along the curve t = unit tangent vector n = unit normal vector k = curvature s = arc length ' = differentiation with respect to s. We have r' = t, t' = k n, and from t.n = 0, n' = - k n. Then we can show that r' = t r'' = k n r'''' = (k'' - k^3) n - 3 k k' t. The total strain energy int(k^2, s = 0..L) = int(x''(s)^2 + y''(s)^2, s = 0..L) is to be minimized subject to the constraint of fixed total arc length. This constraint can be expresses as x'(s)^2 + y'(s)^2 = 1 at each point on the curve. This the variational problem has Lagragrian L = (x''(s)^2 +y''(s)^2) - m(s) (x'(s)^2 + y'(s)^1 - 1), where the Lagrange multiplier m must be a function of s since the constraint holds for each s.

From the calculus of variations, the minimizing curve satisfies

(dL/dx(s)) - (dL/dx'(s))' + (dL/dx''(s))'' = 0, and similarly for y. This becomes (m x')' + x'''' = 0, (m y')' + y'''' = 0, Which are the x and y componenets of the single vector equation (m r')' + r'''' = 0, or resolved along t and n, (m' - 3 k k') t + (m k + k'' - k^3) n = 0. The t component gives m' = (3/2) (k^2)' so that m(s) = (3/2) k^2 + A, where A is a constant. Then the n component gives k''(s) + (1/2) k(s)^3 + A k(s) = 0. Multiplying this by k' and integrating gives (k'(s))^2 + (1/4) k(s)^4 + A (k(s))^2 + B = 0. The solution k(s) involves elliptic functions, and introduces one more constant of integration, in the form k(s) = F(A, B, s + C). Integrating k to give t introduces only one more constant, the initial direction of t. Integration to r gives two more constants, the initial point, for a total of six constants of integration. These can be determined, e.g. by imposing initial and final points on the curve, and initial and final tangent directions. Three of the six degrees of freedom are rigid displacements in the plane, one is a scaling, and one is a displacement of the fineite length fiber along an infinite curve. The sixth degree of freedom is the shape of the curve. Gene