Linearized theory of elasticity

Transcription

1 Linearized theory of elasticity Arie Verhoeven CASA Seminar, May 24, 2006

2 Seminar: Continuum mechanics 1 Stress and stress principles Bart Nowak March 8 2 Strain and deformation Mark van Kraaij March 29 3 General principles Ali Etaati April 12 4 Constitutive equations Godwin Kakuba April 19 5 Fluid mechanics Peter in t Panhuis May 3 6 Linearized theory of elasticity Arie Verhoeven May 24 7 Complex representation Erwin Vondenhoff June 7 of the stress function 8 Principles of virtual work Andriy Hlod June 21 9 The boundary element method Zoran Ilievsky June 28

3 Outline 1 Introduction 2 Plane elasticity Rectangular coordinates Cylindrical and polar coordinates 3 Three-dimensional elasticity Direct solution of Navier s equation Examples 4 Summary

4 Outline 1 Introduction 2 Plane elasticity Rectangular coordinates Cylindrical and polar coordinates 3 Three-dimensional elasticity Direct solution of Navier s equation Examples 4 Summary

5 The Cauchy stress tensor For a linear elastic solid we have the identity: T Ji (X, t) = T Ji (x + u(x, t)). In terms of Cartesian components, the first Piola-Kirchhoff stress tensor T 0 Ji is related to the Cauchy stress T ri at the points x = X + u by T 0 Ji = ρ 0 ρ X J x r T ri. (1) The displacement-gradient components are small compared to unity. The equations of motion in the reference state are given by T 0 Ji X J + ρ 0 b 0i = ρ 0 d 2 x i dt 2. (2)

6 Field equations Field Equations of Linearized Isotropic Isothermal Elasticity T ji X j 3 Eqs. of Motion + ρb i = ρ 2 u i t 2 (3) 6 Hooke s Law Eqs. T ij = λe kk δ ij + 2µE ij (4) ( 6 Geometric Eqs. E ij = 1 ui 2 X j + u j (5) 15 eqs. for 6 stresses, 6 strains, 3 displacements X i ) The two Lamé elastic constants λ and µ, introduced by Lamé in 1852, are related to the more familiar shear modulus G, Young s modulus E, and Poisson s ratio ν as follows: µ = G = E 2(1 + ν) and λ = νe (1 + ν)(1 2ν).

7 Boundary conditions 1 Displacement boundary conditions, with the three components u i prescribed on the boundary. 2 Traction boundary conditions, with the three traction components t i = T ji n j prescribed at a boundary point. 3 Mixed boundary conditions include cases where 1 Displacement boundary conditions are prescribed on a part of the bounding surface, while traction boundary conditions are prescribed on the remainder, or 2 at each point of the boundary we choose local rectangular Cartesian axes X i and then prescribe: 1 ū 1 or t 1, but not both, 2 ū 2 or t 2, but not both, and 3 ū 3 or t 3, but not both.

8 Navier s displacement equations Equations of this form were given by Navier in a memoir of 1821, published in 1827, but they contained only one elastic constant because they were deduced from an inadequate molecular model. The two-constant version was given by Cauchy in Navier Equation (λ + µ) ( u) + µ 2 u + ρb = ρ 2 u t 2 (8) Traction Boundary Condition λ( u)ˆn + µ(u ˆn + u) ˆn = prescribed function (9) Elastostatics: 2 u t 2 = 0.

9 Outline 1 Introduction 2 Plane elasticity Rectangular coordinates Cylindrical and polar coordinates 3 Three-dimensional elasticity Direct solution of Navier s equation Examples 4 Summary

10 Outline 1 Introduction 2 Plane elasticity Rectangular coordinates Cylindrical and polar coordinates 3 Three-dimensional elasticity Direct solution of Navier s equation Examples 4 Summary

11 Plane elasticity (1) In plane deformation, the assumptions u z = 0 and u x and u y indepenent of z lead to only three independent strain components e x, e y, and e xy = 1 2 γ xy, which are independent of z, a state of plane strain parallel to the xy-plane. The isotropic Hooke s law reduces to with, in addition, where e = e x + e y, G = σ x = λe + 2Ge x σ y = λe + 2Ge y τ xy = 2Ge xy (3) σ z = λe = ν(σ x + σ y ), (4) E 2(1 + ν), and λ = Eν (1 + ν)(1 2ν). (5)

12 Plane elasticity (2) To these must be added two equations of motion σ x x τ xy x + τxy y + σy y and one compatibility equation + ρb x = ρ 2 u x t 2 + +ρb y = ρ 2 u y t 2 } (6) 2 e x y e y x 2 = 2 2 e xy x y, (7) if for small displacements we ignore the difference between the material coordinates X, Y and the spatial coordinates x, y.

13 Particular solution for body forces The linearity can also be used to construct the solution in two parts: σ ij = σij H + σij P, e ij = eij H + eij P. The particular solution σij P, ep ij satisfies the given equations with given body-force distributions but not the boundary conditions. The distribution σij H, eij H satisfies the homogeneous differential equations (with no body force) and suitably modified boundary conditions. When the body force is simply the weight, say b x = 0, b y = g, then a possible particular solution is where C is the value of σ P y at y = 0. σ P y = ρgy C, σ P x = τ P xy = 0, (8)

14 Airy stress function For plane strain with no body forces, the equilibrium equations are identically satisfied if the stresses are related to a scalar function φ(x, y), called Airy s stress function, by the equations σ x = 2 φ y 2, σ y = 2 φ x 2, τ xy = 2 φ x y. (9) The compatibility equation then becomes the biharmonic equation 2 1( 2 1)φ = 0 (10) or 4 1φ = 0, where 1 = is the 2D Laplace operator. x 2 y 2

15 Boundary conditions for Airy stress function Airy stress function useful for boundary conditions for tractions. Since t i = σ ji n j, we have t x = σ x n x + τ xy n y and t y = τ xy n x + σ y n y or t x = 2 φ dy y 2 ds + 2 φ dx x y ds = d ds t y = 2 φ x y dy 2 φ dx ds x 2 ds = d ds ( φ ( y φ x ) ). } (11) Hence, integrating along the boundary, we obtain φ x = C t yds + C 1 and φ y = C t xds + C 2. Now we can calculate dφ ds = φ dx x ds + φ dy y ds, dφ dn = φ dy x ds φ dx y ds and φ = dφ C ds ds + C 3 at the boundary.

16 Outline 1 Introduction 2 Plane elasticity Rectangular coordinates Cylindrical and polar coordinates 3 Three-dimensional elasticity Direct solution of Navier s equation Examples 4 Summary

17 Elasticity model in cylindrical coordinates In cylindrical coordinates, we have u = u r ê r + u θ ê θ + u z ê z. Then the displacement-gradient tensor u is derived by differentiating the variable unit vectors ê r, ê θ as well as the coefficients of the three unit vectors. Small-strain components in cylindrical coordinates are E rθ = 1 2 E rr = ur E θθ = 1 r u θ θ [ r ] 1 u r r θ + u θ r u θ r E zr = 1 ( uz 2 r + ur r E zz = uz z ( ) E θz = 1 uθ 2 z + 1 u z r θ (12) ). + ur z

18 Plane stress equations in polar coordinates Plane-strain components will be denoted for u = ue r + ve θ by e r = u r e θ = 1 v ( r θ + u r e rθ = 1 1 u 2 r θ + ) v r v r (13) Hooke s law in polar coordinates σ r = λe + 2Ge r σ θ = λe + 2Ge θ τ rθ = 2Ge rθ σ z = ν(σ r + σ θ. where e = e r + e θ (14)

19 Lamé solution for cylindrical tube (1) Consider a tube long in the z-direction, loaded by internal pressure p i and external pressure p o, with negligible body forces, and assume plane deformation with radial symmetry, independent of z and θ in the plane region a r b. The Navier equations then become, with u θ = u z = 0, u r = u, simply (λ + 2G) d dr [ 1 r d dr (ru) Two integrations with respect to r then give In polar coordinates we get ] = 0. (15) u = Ar + B r. (16) e r = u r r = A B r 2 e θ = u r r = A + B r 2 e rθ = 0 (17)

20 Lamé solution for cylindrical tube (2) Hooke s law gives σ r = 2Aλ + 2GA 2GB r 2 σ θ = 2Aλ + 2GA + 2GB r 2 τ rθ = 0 σ z = 4Aν(λ + G). We apply the boundary conditions at r = a and r = b. r = a p i = 2A(λ + G) 2GB } a 2 r = b p o = 2A(λ + G) 2GB b 2 whence } 2GB = (p i p o)a 2 b 2 b 2 a 2 2(λ + G)A = p i a 2 p ob 2 b 2 a 2 so that the stress distribution is p i p o a 2 b 2 b 2 a 2 r 2 σ r = p i a 2 p ob 2 b 2 a 2 σ θ = p i a 2 p ob 2 b 2 a 2 + p i p o a 2 b 2 b 2 a 2 r 2 σ z = 2ν(p i a 2 p ob 2 ) b 2 a 2 τ rθ = 0. (18) (19) (20) (21)

21 Outline 1 Introduction 2 Plane elasticity Rectangular coordinates Cylindrical and polar coordinates 3 Three-dimensional elasticity Direct solution of Navier s equation Examples 4 Summary

22 Outline 1 Introduction 2 Plane elasticity Rectangular coordinates Cylindrical and polar coordinates 3 Three-dimensional elasticity Direct solution of Navier s equation Examples 4 Summary

23 Direct solution of Navier equations (λ + G) e X k + G 2 u k + ρb k = ρ 2 u k t 2 where e = u k X k. Already in 1969, for elastostatics, with 2 u k = 0, the solution of such a t 2 set of three equations in three dimensions by finite-difference or finite-element methods is beginning to be a possibility. An advantage of direct solving the 3D problem instead of the less complex 2D problem is that strains can be obtained in terms of the first partial derivatives of the displacement field.

24 The Helmholtz representation Each continuously differentiable vector field u can be represented as For definiteness there is also the requirement u = φ + ψ. (22) ψ 0. Equations of motion in terms of the potentials: (λ + 2G)( 2 φ),k + Ge krs ( 2 ψ s ),r = ρ( 2 φ t 2 ),k + ρe krs ( 2 ψ t 2 ),r. Specific solutions also satisfy the wave equations. 2 φ = 1 c φ t 2 2 ψ k = 1 2 ψ k c2 2 t 2.

25 Papkovich-Neuber potentials When the Helmholtz representation is substituted into the elastostatic Navier Eq., we get 2 [α φ + ψ] = ρb G, where α = 2(1 ν) 1 2ν. The Papkovic-Neuber potentials are φ 0 and Φ, which is defined by Φ = α φ + ψ. Then we get Poisson s equations { 2 Φ = ρb G 2 φ 0 = ρb r G The solution can be found by Green s formula. In general, the potentials satisfy complicated boundary conditions.

26 Green s formula (1) Two sufficiently differentiable functions f and g in volume V bounded by S satisfy Green s Second Identity: [ f g n g f ] [ ds = f 2 g g 2 f ] dv. (23) n S Let P be an arbitrary field point and Q a variable source point. Then Green s formula expresses the value f P as follows: [ 1 f 4πf P = r 1 n f ( )] 1 1 ds 2 fdv. n r 1 S V r 1 V

27 Green s formula for the half-space In potential theory the Green s function G(P, Q) for a region is a symmetric function of the form G(P, Q) = 1 r 1 + g(p, Q) 2 g = 0. where r 1 = P Q. We obtain the formula for f in terms of Green s function: 4πf P = f G n ds G 2 fdv. S Let Q 2 be the image point of Q in the XY-plane and let r 2 = P Q 2. For the half-space, it is possible to determine G(P, Q): f (P) = 1 2π V G(P, Q) = 1 r 1 1 r 2. Z S f ds 1 G 2 fdv r 0 4π V

28 Outline 1 Introduction 2 Plane elasticity Rectangular coordinates Cylindrical and polar coordinates 3 Three-dimensional elasticity Direct solution of Navier s equation Examples 4 Summary

29 Normal traction problem Given on part S 1 of the boundary S (z = 0) of the half-space a distributed normal pressure of intensity q, the traction boundary conditions are T zz = q T zx = T zy = 0 on S 1. with zero tractions on the remainder of the boundary z = 0. We seek the Papkovich-Neubich potentials, assuming that φ 1 φ 2 0, such that u = φ 3 ê z 1 4(1 ν) (φ 0 + zφ 3 ) and 2 φ 0 = 2 φ 3 0. They can be found by using Green s formula for the half-space.

30 Boussinesq problem of concentrated normal force on boundary of half-space This problem is solved by taking limit S 1 O and q such that lim S1 O S 1 qds = P, which is a finite concentrated load at O in the positive z-direction.

31 Solution of Boussinesq problem We obtain φ 2 (x, y, z) = (1 ν)p πgr Z [φ 0(x, y, z)] = (1 ν)(1 2ν)P πgr φ 0 (x, y, z) = (1 ν)(1 2ν)P πg log(r + z) Now also the displacements and tractions can be computed.

32 Outline 1 Introduction 2 Plane elasticity Rectangular coordinates Cylindrical and polar coordinates 3 Three-dimensional elasticity Direct solution of Navier s equation Examples 4 Summary

33 Summary Navier s equations Plane elasticity Airy stress function Plane elasticity in polar coordinates Lamé solution for tube Helmholtz representation Papkovich-Neuber potentials Boussinesq problem

34 Literature. L.E. Malvern: Introduction to the mechanics of a continuous medium, Prentice-Hall, 1969, pp

35 Literature. L.E. Malvern: Introduction to the mechanics of a continuous medium, Prentice-Hall, 1969, pp Y.C. Fung: Foundations of solid mechanics, Englewood Cliffs, N.J.: Prentice-Hall, Inc., A.E. Green, W. Zerna: Theoretical elasticity, London: Oxford University Press, A.E.H. Love: Mathematical theory of elasticity, 4th ed. New York: Dover Publications, Inc., S. Timoshenko, J.N. Goodier: Theory of elasticity, 2nd ed. New York: McGraw-Hill Book Company, H.M. Westergaard: Theory of elasticity and plasticity, New York: Dover Publications, Inc., 1952.