General Michell Solution in Polar Coordinates

Transcription

1 General Michell Solution in Polar Coorinates A. Narayana Rey [Reference: The solution is aopte from the book on Introuction to Continuum Mechanics by W. M. Lai, D. Rubin, an E. Krempl] Let the solution for biharmonic equation is separable, i.e., the Airy stress function φr, θ = frgθ. 1 is the solution of the following partial ifferential equation. 4 φ = r 1 r r 1 φ r θ r 1 r r 1 φ = 0. r θ We assume that the function gθ can be represente by the Fourier series. Therefore, the function gθ can be written as gθ = p n cosnθ q n sinnθ. 3 q=0 It can be observe that every term is in the form fr cosnθ or fr sinnθ. Our task now is to get the function form of fr if the solution were to be in the form of fr cosnθ or fr sinnθ. Solution involving the terms cos nθ: Let the solution φr, θ = fr cosnθ then φ θ = cosnθ fr = n fr cosnθ = n φ. 4 θ Substituting this relation in above partial ifferential equation, we get r 1 r r 1 φ r θ r 1 r r n r φ = 0. 5 φ r r 1 r r n 1 φ r r r r 1 r r n 1 φ r r θ r 1 r r n r φ = 0. [ 4 φ r 4 r 1 φ r r 1 n r r r r ] r n r φ 3 ] n r [ 1 3 φ r r 1 φ 3 r r 1 r 3 r n r 3 r n r φ 4 1 [ φ r 1 ] r r n r φ = 0.

2 Expaning all terms, we get 4 φ r 3 φ 4 r r 1 n 3 r φ r 1 n r 3 r n 4 n r 4 φ = 0. 6 It can be note that instea of cosine if we assume sine function, i.e., φr, θ = fr cosnθ, then also yiels same ifferential equation. substituting φr, θ = fr cosnθ, we obtain the following orinary ifferential equation. 4 f r 3 f 4 r r 1 n f 3 r r 1 n f r 3 r n 4 n f = 0. 7 r 4 We now use change of variables to get simple form of ifferential equation. Let r = e t. Then we have r t = et = r, an t r = 1 r. 8 We now express f r, f r, 3 f r, an 4 f f in terms of 3 r4 t, f t, 3 f t, an 4 f. Using change 3 t4 of variable f t = f r r t. Substituting Eq. 8, we get r f r = f t. 9 Taking both sies ifferentiation with respect to r, we get Substituting Eq. 8, we get Substituting Eq. 9, we get r f r r = r r r f r = t = f r t f t t r. r f r r f r = f t. r f r = f t f t. 10 Taking ifferentiation on both sies of equation with respect to r, we get r 3 f r 3 r f = r 3 f r 3 r f r = t Substituting Eq. 10, we get r = f r t f t f t f t t r = r 3 3 f r f 3 r r = 3 f t f 3 t Since t r = 1 r r 3 3 f r 3 = 3 f t 3 3 f t f t. 11.

3 Taking ifferentiation with respect to r on both sies of equation, we get r 3 4 f r 4 3r 3 f = r 3 4 f r 4 3r 3 f r 3 = t r = 3 f 3 r t f 3 3 t f t 3 f t f t 3 3 t f t r = r 4 4 f r 3 f 4 3r3 r = 4 f 3 t f 4 33 t f 3 t Substituting Eq. 11, we get Since t r = 1 r r 4 4 f r 4 = 4 f t 4 63 f t 3 11 f t 6f t. 1 The substitution of Eqs. 9, 10, 11, an 1 in Eq. 7 yiels The characteristic equation The equation can written as 4 f t f 4 43 t 4 3 n f f 4n t t n 4 n f = m 4 4m 4 n m 4n m n 4 n = m n 4m m 1 4n m 1 = 0 = m n 4m 1m n = 0 = m n m n 4m 4 = 0 = m n m n = 0. Thus, roots of characteristic equation can be written as m = ±n, an m = ± n. 15 Clearly, the roots are not repeating if n. For repeating roots we nee to fin some other inepenent solutions. If n then Substituting r = e t, we get The general solution for n, we have φr, θ = ft = c n1 e nt c n e nt c n3 e nt c n4 e nt. 16 fr = c n1 r n c n r n c n3 r n c n4 r n. 17 cn1 r n c n r n c n3 r n c n4 r n cosnθ. 18 n= We now fin the solution to n = 0 an also for n = 1. Problem 1. Fin the general solution to the following orinary ifferential equation g t 4g 4g = 0. t 3

4 Proof. The characteristic equation for the given ifferential equation can be written as m 4m 4 = 0. The roots of the equation are m =,. This shows that the roots are equal. We have a solution gt = c 1 e t. Since the roots are repetitive, we can obtain another inepenent solution by Thus, we have the general solution e m 1t m 1 = te m 1t. gt = c 1 c te t. Therefore, if roots are repetitive one can obtain other inepenent solutions by ifferentiating with root as variable. Casei n = 0: Clearly it can be observe from Eq. 15 that the roots m = 0, 0,, if n = 0. Therefore, a solution for the ifferential equation c 1 r 0 c r or c 1 c r. We now apply the same technique that was use in above problem to get the general solution. The inepenent solutions from roots information are r n cosnθ an r n cosnθ. The other inepenent solutions n rn cosnθ = [r n log r cosnθ r n θ sinnθ] = log r. 19 [ r n cosnθ ] = [ r n log r cosnθ r n θ sinnθ ] n = r log r. 0 The general solution for n = 0 φr, θ = c 1 c r c 3 log r c 4 r log r. 1 Caseii n = 1: Clearly it can be observe from Eq. 15 that the roots m = 1, 1, 1, 3 if n = 1. The root one is repeate twice. Therefore, we can get the following inepenent solutions similar to previous case. n rn cosnθ = [r n log r cosnθ r n θ sinnθ] [ r n cosnθ ] n The general solution for n = 1 = r log r cosθ rθ sinθ. = [ r n log r cosnθ r n θ sinnθ ] = r log r cosθ rθ sinθ. 3 φr, θ = c 5 r cosθ c 6r cosθ c 7 r 3 cosθ c 8 r log r cosθ 9 rθ sinθ. 4 4

5 Solution involving the terms sin nθ: Following similar steps that of previous solution proceure, we obtain the following solution for case n. φr, θ = n1 r n n r n n3 r n n4 r n sinnθ. 5 n= We present solution for the cases n = 0 an n = 1 as there are repeate roots. Casei n = 0: Clearly it can be observe from Eq. 15 that the roots m = 0, 0,, if n = 0. Therefore, a solution for the ifferential equation c 1 r 0 c r or c 1 c r. Similar to the previous case, the inepenent solutions from roots information are r n sinnθ an r n sinnθ. The other inepenent solutions n rn sinnθ = [r n log r sinnθ r n θ cosnθ] = θ. 6 [ r n sinnθ ] = [ r n log r sinnθ r n θ cosnθ ] n = r θ. 7 The general solution for the case n = 0 can be written as φr, θ = 1 sin0θ r sin0θ 3 θ 4 r θ = 3 θ 4 r θ. 8 Caseii n = 1: Clearly it can be observe from Eq. 15 that the roots m = 1, 1, 1, 3 if n = 1. The root one is repeate twice. Therefore, we can get the following inepenent solutions similar to previous case. n rn sinnθ = [r n log r sinnθ r n θ cosnθ] [ r n sinnθ ] n The general solution for n = 1 = r log r sinθ rθ cosθ. 9 = [ r n log r sinnθ r n θ cosnθ ] = r log r sinθ rθ cosθ. 30 φr, θ = 5 r sinθ 6r sinθ 7 r 3 sinθ 8 r log r sinθ c 9 rθ cosθ. 31 Combining the solutions shown in Eqs. 1, 4, 8, 31, 18, an 5, we get the following general Michell solution to the biharmonic equation. φr, θ = c 1 c r c 3 log r c 4 r log r 3 θ 4 r θ c5 r c 6r c 7 r 3 c 8 r log r c 9 rθ cosθ 5 r 6r 7 r 3 8 r log r 9 rθ sinθ cn1 r n c n r n c n3 r n c n4 r n cosnθ n= n= n1 r n n r n n3 r n n4 r n sinnθ 3 5