Complex Representation in Two-Dimensional Theory of Elasticity
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1 Complex Representation in Two-Dimensional Theory of Elasticity E. Vondenhoff 7-june-2006
2 Literature: Muskhelishvili : Some Basic Problems of the Mathematical Theory of Elasticity, Chapter 5 Prof. ir. C. de Pater: Collegediktaat Elasticiteitstheory, Technische Hogeschool Twente (afdeling der werktuigbouwkunde)
3 This presentation consists of the following parts: 1. Basic equations in theory of elasticity 2. The Airy stress function in 2D problems 3. Cauchy Riemann equations 4. Determination of the displacements from the stress function 5. Representation of the Airy stress function by two complex analytic functions 6. Complex representation of displacements and stresses 7. Example 8. Results
4 1. Basic equations in theory of elasticity Solutions of problems in elasticity have to satisfy Conservation laws Compatibility conditions Constitutive equations Variables: u: displacement vector E: deformation tensor T : stress tensor
5 Conservation of momentum: ( j T ij,j ) + F i = 0 Relation between deformation and displacement: ε ij = 1 2 (u i,j + u j,i ) or E = 1 2 ( u + ut )
6 Constitutive equations (Hooke s law) T ij = λδ ij ( k ε kk ) + 2µε ij or T = λ(tre)i + 2µE where λ and µ are the Lamé parameters.
7 Combining Balance laws and constitutive equations and assuming that there are no body forces we get ( ) µ u i + (λ + µ) u k,ki = 0. These are called the equations of Navier-Cauchy. Differentiating with respect to x i and summing over i we get (tre) = 0 k
8 2. The Airy stress function in 2D problems We assume that there exists a function U = U(x, y) which is called the Airy stress function such that T 11 = 2 U y 2 T 22 = 2 U x 2 T 12 = 2 U x y The balance equations are automatically satisfied: T i1 x + T i2 y = 0
9 From (trt ) = 0 we get U = 0 The function U is called a biharmonic function.
10 3. Cauchy-Riemann equations Let ϕ : C C be a complex function ϕ(z) = p(x, y) + iq(x, y). ϕ is differentiable at z C then there exists a C C such that for small h C ϕ(z + h) = ϕ(z) + Ch + O( h 2 ). Write C = D + ie and h = ξ + iη
11 We get ϕ(z + h) = p(x + ξ, y + η) + iq(x + ξ, y + η) = p(x, y) + iq(x, y) + ξ p x + η p q + iξ y x ϕ(z) + Ch = ϕ(z) + Dξ Eη + i(eξ + Dη). Differentiable functions satisfy the Cauchy-Riemann equations p y p x = D = q y = E = q x + iη q y + O( h 2 ),
12 4. Determination of displacements from the stress function Let u and v denote the components of the displacement vector u in the twodimensional case. From the constitutive equations we get λtre + 2µ u x = 2 U y 2 λtre + 2µ v y = 2 U x 2 ( v µ x + u ) = 2 U y x y
13 We can easily derive 2µ u x = 2 U y 2 λ 2(λ + µ) U because 2µ v y = 2 U x 2 λ 2(λ + µ) U U = trt = 2(λ + µ)tre.
14 Introduce the function P := U. This P is harmonic: P = U = 0. We get 2µ u x = 2 U x 2 + λ + 2µ 2(λ + µ) P 2µ v y = 2 U y 2 + λ + 2µ 2(λ + µ) P
15 Define Q as the harmonic conjugate of P : P x = Q y, P y = Q x Define f : G C C by f(x + iy) = P (x, y) + iq(x, y) f : G C is analytic. This means that for all a G there exist A k s such that in a neighborhood of a f(z) = A k (z a) k. k=0
16 Define ϕ by and define p and q ϕ(z) := 1 4 z 0 f(ζ)dζ. p(z) := Re ϕ(z), q(z) := Im ϕ(z). We have ϕ (z) = p x + i q x = 1 (P + iq). 4 By the Cauchy-Riemann equations we get P = 4 p x = 4 q y Q = 4 p y = 4 q x
17 We can write After integration we get 2µ u x = U 2(λ + 2µ) p 2 + x2 λ + µ x 2µ v y = U 2(λ + 2µ) q 2 + y2 λ + µ y. 2µu = U x 2µv = U y 2(λ + 2µ) + λ + µ p + f 1(y) 2(λ + 2µ) + λ + µ q + f 2(x).
18 From the Cauchy Riemann equations and ( v µ x + u ) = 2 U y x y we get The general form must be f 1(y) + f 2(x) = 0. f 1 (y) = 2µ( εy + α) f 2 (x) = 2µ(εx + β). These terms f 1 and f 2, which only give rigid body displacement, are often omitted.
19 After omitting f 1 and f 2 2µu = U x 2µv = U y 2(λ + 2µ) + λ + µ p + 2(λ + 2µ) λ + µ q.
20 5. Representation of the biharmonic function by two complex analytic functions Theorem: (U px qy) = 0 where ϕ(z) = p(z) + iq(z) as introduced earlier. Write for a harmonic function p 1. U = px + qy + p 1 Let q 1 be the harmonic conjugate of p 1 and define the analytic function χ = χ(z) by χ(z) = p 1 (z) + iq 1 (z).
21 One may write or U = 1 2 U = Re( zϕ(z) + χ(z)) ( ) zϕ(z) + zϕ(z) + χ(z) + χ(z). Every U of this form is biharmonic if ϕ and χ are analytic.
22 The method of deduction by Goursat. Consider the biharmonic equation U = 0. Introduce new variables ξ = x + iy = z and η = x iy = z. Then the preceding equation takes the form whence it follows directly that 4 U ξ 2 η 2 = 0, U = ϕ 1 (z) + ϕ 2 (z) + zχ 1 (z) + zχ 2 (z), where ϕ 1, ϕ 2, χ 1 and χ 2 are "arbitrary" functions. If U is real then it is easily seen that one must put ϕ 2 (z) = ϕ 1 (z) and χ 2 (z) = χ 1 (z).
23 It is easily found that and 2 U x = ϕ(z) + zϕ (z) + ϕ(z) + zϕ (z) + χ (z) + χ (z) 2 U ( ) y = i ϕ(z) + zϕ (z) + ϕ(z) zϕ (z) + χ (z) χ (z). It turns out to be convenient to work with where We have U x + i U y = ϕ(z) + zϕ (z) + ψ(z) ψ = χ. U = 4 Re ϕ (z).
24 6. Complex representation of displacements and stresses We already found the following relation between the displacement vector and U: ( ) U 2µ(u + iv) = x + i U 2(λ + 2µ) + y λ + µ ϕ(z). A more convenient formula is 2µ(u + iv) = κϕ(z) zϕ (z) ψ(z), where κ = λ + 3µ λ + µ.
25 In order to find a representation of the stress components by means of the functions ϕ and ψ we will find an expression for the forces acting on an arc AB lying in the plane Oxy. t : tangential vector n: normal vector.
26 The force (T xn ds, T yn ds) acting on an element ds of the arc AB, will be understood to be the force exerted on the side of the positive normal: T xn = T xx (n, e x ) + T xy (n, e y ) = 2 U y (n, e x) 2 U 2 x y (n, e y), T yn = T xy (n, e x ) + T yy (n, e y ) = 2 U x y (n, e x) + 2 U x (n, e y). 2
27 We have (n, e x ) = (t, e y ) = dy ds, (n, e y ) = (t, e x ) = dx ds. Introducing these values into the preceding formulae, one finds ( ) U T xn = d ds T yn = d ds y ( ) U. x
28 We get T xn + it yn = d ( ) U ds y i U x = i d ( ) U ds x + i U y = i d ( ) ϕ(z) + zϕ ds (z) + ψ(z)
29 Let ds have the direction of Oy, then T xn = T xx and T yn = T xy and T xx + it xy = ϕ (z) + ϕ (z) zϕ (z) ψ (z). Let ds have the direction of Ox, then T xn = T xy and T yn = T yy and T yy it xy = ϕ (z) + ϕ (z) + zϕ (z) + ψ (z).
30 7. Example Use the Airy stress function to calculate displacements.
31 Use a stress function of the form U(x, y) = αxy + βxy 3, such that T 11 = 6βxy, T 22 = 0 and T 12 = α + 3βy 2. The top side and the bottom side are stress free and C C T xy dy = P.
32 We get the stress function the stress tensor T = U(x, y) = 3P 4C xy P 4C 3xy3, ( P 4C 3 xy 3P 4C + 3P 3P 4C + 3P 4C 3 y 2 4C 3 y 2 0 and the deformations can be calculated from ( ) ( ) 1 ( ) ε11 λ + µ λ T11 = ε 22 λ λ + µ T 22 ) and ε 12 = T 12 2µ.
33 After introducing some other boundary conditions, the displacement can be calculated.
34 8. Results Introduction of the Airy stress function Determination of displacements from the stress function Representation of the stress function in terms of two complex analytic functions Complex representation of displacements and stresses
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