- 721 9 28 Consider The Torsion Of A Rod With The Half Ring Cross Section As Shown Formulating The Problem In Polar Coord 1 (104.59 KiB) Viewed 54 times
721 9.28 Consider the torsion of a rod with the half-ring cross-section as shown. Formulating the problem in polar coordinates (see Exercise 9.6), the governing stress function equation becomes 320 1 20 1 020 ar2 + ört-2 -2μα Using Fourier methods to solve this problem, first show that we can expand the constant right- hand side in a Fourier since series to get –2ua = -2=8ua/[(2n +1)]sin(2n + 1)8. Based on this, use the series solution from $(r, 6) = m-oFn(r)sin(2n + 1) and show that the gov- erning equation leads to the ordinary differential equation dPF 1dF, (2n + 1) 8μα -Fn = dr2 r dr (2n + 1) Show that the solution to this equation to give by 8μα + Bir + Kyr?, where Kn= (2n + 1)(2n - 1)(2n + 3) Note that the stress function form already satisfies the zero boundary condition on 0 = 0 and 7. Finally apply the remaining boundary conditions on r1 and 12, to determine the constants A, and By and show that the stresses can be written as -2n-1 Fn=A,u2n+1 Tre 1 дф r ao [4,221 +B,r-21-2 + K,r](2n + 1)cos(2n+1) n=0 Toz o = - [4,(2n + 1), 2n – B,(2n + 1), -21–2 + 2K,r] sin(2n + 1) ar n=0
286 Chapter 9 Extension, torsion, and flexure of elastic cylinders 2