MAE4700/5700 Finite Element Analsis for Mechanical and Aerospace Design Cornell Universit, Fall 2009 Nicholas Zabaras Materials Process Design and Control Laborator Sible School of Mechanical and Aerospace Engineering 101 Rhodes Hall Cornell Universit Ithaca, NY 14853-3801 U N I V E R S I T Y 1
Kinematics in two-dimensions We start with the displacement vector at a given point in 2D. It is a vector with x- and the -components. We denote it using both matrix and vector notation: ux u = u ΔxΔ u = u i+ u j Let us consider a square (control volume) on the plane before and after deformation as shown: x Δ Can ou use this figure to define the strain components? Δx U N I V E R S I T Y 2
Extensional strain components The extensional strains ε xx and ε are defined as: ε ε xx u (, ) (, ) lim x x+ Δx ux x u = x Δ x 0 Δx x u ( x, + Δ) u ( x, ) u = lim Δ 0 Δ ε xx and ε represent the change in the lengths of the infinitesimal line segments in the x and directions, Δx and Δ, respectivel, divided b the original lengths of the line segments. U N I V E R S I T Y 3
Shear strain components The shear strain, γ x, measures the change in angle α 1 + α 2 between unit vectors in the x and directions (in radians) γ x u( x+δx, ) u( x, ) u ( x, +Δ) u ( x, ) u u = lim + lim + Δ x 0 Δx Δ 0 Δ x x x x We often use the engineering shear strain γ x introduced above, and the tensor shear strain component ε x = ½ γ x. U N I V E R S I T Y 4
Rotation In addition to axial elongations, the control volume also undergoes rotation. The rotation in 2D, denoted b ω x, is computed b 1 ( 2 1) 1 u ω x x α α = 2 2 u x If α 1 =α 2, the rotation is zero. For small deformations, the rotation ω x is small and does not affect the stress calculation. U N I V E R S I T Y 5
Strain matrix For our finite element calculations, we introduce the following notation for the strains: u x 0 x ε x x T u ux ε = ε ε γ = ε = = 0 u = γ x u u x + x x As shown above, the smmetric gradient matrix S is defined as: x x S 0 x S = 0 x U N I V E R S I T Y 6 S u
Traction vector and stress components The traction (force per unit area) on the plane with the normal vector n aligned along the x-axis is denoted b and its vector form is σ x = σxxi+ σx j Similarl, the traction with the outer normal unit vector aligned along the -axis is denoted as σ and its corresponding components are σ = σ xi + σ j σ x σ σ x and are called the stress vectors acting on the planes normal to x and directions, respectivel. The stress state in a 2D bod is described b two normal stresses σ xx and σ and shear stresses σ x and σ x. n Can ou show that σ x =σ x? (moment equilibrium) U N I V E R S I T Y 7
Stress components Positive stress components act in the positive direction on a positive face. The 1 st subscript on the stress corresponds to the direction normal to the plane and the 2 nd subscript denotes the direction of the force. The normal stresses are often written simpl as σ x and σ. In matrix form, we denote the stress components as σ xx T σ = σxx σ σx = σ σ or as x σ xx σ x τ = σx σ U N I V E R S I T Y 8
Traction in an arbitrar surface with normal n The stress vectors x and can be used to obtain the tractions on an surface of the bod with normal n. Thus the stresses provide information about tractions on an surface at a point. The force equilibrium of the triangular bod σ σ In matrix form shown requires that: tdγ σ xd σ dx = 0 tdγ σ xnxdγ σ ndγ= 0 t = σ xnx + σ n t = ( σxxi+ σx j) nx + ( σ xi+ σ j) n t = ( σxxnx + σ xn) i+ ( σ xnx + σ n) j x tx σ xx σ x nx, t τ n t = = σ x σ n U N I V E R S I T Y 9 t t τ t n
Stress equilibrium The forces acting on the bod are the traction vector t = txi+ t j along the boundar Γ and the bod force per unit volume b= b i+ b j. x Examples of the bod forces are gravit and magnetic forces, etc. Thermal stresses as we have seen can also been interpreted as bod forces. U N I V E R S I T Y 10
Stress equilibrium Consider the equilibrium of the infinitesimal region (on the plane) of unit thickness: Δx Δx Δ Δ σ x( x, ) Δ + σ x( x+, ) Δ σ ( x, ) Δ x+ σ ( x, + ) Δ x+ b( x, ) ΔxΔ = 0 2 2 2 2 Δx Δx Δ Δ σ x( x+, ) σ x( x, ) σ ( x, + ) σ ( x, ) 2 2 2 2 + + bx (, ) = 0 Δx Δ Δx, Δ 0 U N I V E R S I T Y 11
Stress equilibrium σ x σ + + bx (, ) = 0 x However: σ = σ i+ σ j x xx x σ = σ i+ σ j x Combining the above two equations ields the equilibrium equations for the x- and -directions: ( xx i x j ) ( x i σ + σ σ + σ j ) + + bx (, ) = 0 x σ x σ xx x x σ x + + bx = 0 σ + + b = 0 or iσ x + bx = 0 iσ + b = 0 U N I V E R S I T Y 12
Stress equilibrium σ x σ xx x x σ x + + bx = 0 σ + + b = 0 We can now re-write the equilibrium equation in matrix form: 0 σ xx x bx 0 σ + b = 0 0 x σ x b 0 T S σ T σ + b = S U N I V E R S I T Y 13 0
Constitutive equations Constitutive equation is the relation between stress and strain. Examples include elasticit, viscoelasticit, creep, plasticit, viscoplasticit, etc. Here, we discuss linear isotropic elasticit. Recall that in 1D, a linear elastic material is governed b the Hooke s law σ= Eε, where E is Young s modulus. In 2D, the linear relation between the stress and strain matrices can be written as σ = Dε where D is a 3x3 matrix. U N I V E R S I T Y 14
Constitutive equations σ = Dε This is the generalized Hooke s law. D is a smmetric, positive-definite matrix. In 2D, the matrix D depends on whether one assumes a plane stress or plane strain condition. These assumptions determine how the model is simplified from a 3D phsical bod to a 2D model. U N I V E R S I T Y 15
Plane strain conditions A plane strain model assumes that the bod is thick relative to the x -plane in which the model is constructed. Consequentl, the strain normal to the plane, ε z is zero and the shear strains that involve angles normal to the plane, γ xz and, γ z are assumed to vanish. When a bod is thick, significant stresses can develop on the z -faces, in particular the normal stress σ zz can be quite large. U N I V E R S I T Y 16
Plane stress A plane stress model assumes that the bod is thin relative to the dimensions in the x -plane. In that case, we assume that no loads are applied on the z - faces of the bod and that the stress normal to the x - plane, σ zz, vanishes. If a bod is thin, since the stress σ zz must vanish on the outside surfaces, there is no mechanism for developing a significant nonzero stress σ zz within the bod. U N I V E R S I T Y 17
Constitutive equations Here we assume an isotropic material, i.e. a material whose stress-strain law is independent of the coordinate sstem. For an isotropic material, D is the same regardless of the coordinate sstem. Of course this is not the best approximation for all problems! Plane stress: Plane strain: E D= 1 v 2 1 v 0 v 1 0 0 0 (1 v) 2 1 v v 0 E D = v 1 v 0 (1 + v)(1 2 v) 0 0 (1 2 v) 2 Young s modulus E and Poisson s ratio ν are the onl independent material properties for a linear isotropic elastic material U N I V E R S I T Y 18
Constitutive equations For an isotropic elastic material we have onl 2 independent material constants: E and ν (of course D can be written using an two other elastic material constants such as the shear modulus G=E/2(1+ν) and the bulk modulus K=E/3(1 ν). Note that for plane strain, as ν 0.5, D becomes infinite. A Poisson s ratio of 0.5 corresponds to an incompressible material. Modeling incompressible materials for plane strain (and 3D) problems requires special attention in finite element analsis (and special elements). U N I V E R S I T Y 19
Strong form of an elasticit problem T Equilibrium equation: σ + b = 0 or equivalentl: iσ x + b 0 S x = iσ + b = 0 Kinematics equation (strain-displacement relation): ε = Su Constitutive equation (stress-strain relation): σ = Dε Boundar conditions: The portion of the boundar where the traction is prescribed is denoted b Γ t and the portion of the boundar where the displacement is prescribed Γ u. The traction boundar condition is written as τ n= t on Γt or equivalentl: σ xin= tx, σ in= t on Γt The displacement boundar condition is written as u = u on Γ U N I V E R S I T Y 20 u
Strong form Essential boundar condition: The displacement boundar condition is the essential boundar condition satisfied b the displacement field. Natural boundar condition: The traction boundar condition is a natural boundar condition. The displacement and traction cannot both be prescribed on the same part of the boundar, thus Γ u Γ t =0. However, on an portion of the boundar, either the displacement or the traction must be prescribed, so Γ u Γ t = Γ U N I V E R S I T Y 21
Strong form for isotropic linear elasticit Find the displacement field u on Ω such that: iσ + b = 0, iσ + b = 0 on Ω x x where σ = D u with σ in= t, σ in= t on Γ u = u on Γ x x t u S U N I V E R S I T Y 22
Weak form for isotropic linear elasticit We pre-multipl the equilibrium equations in x and directions and the two natural boundar conditions b the corresponding weight functions and integrate over the corresponding domains as follows: iσ + b = 0, iσ + b = 0 on Ω σ x x Ω Ω x + bx wxdω= 0 wx U0, + b w dω= 0 w U. ( iσ ) ( iσ ) n= t, σ n= t on Γ x x t i i x( σ x x) w i n t dγ= 0 w U, w i n t dγ= 0 w U. ( σ ) U N I V E R S I T Y 23 Γ t Γ t 0 x 0 0
Weak form for isotropic linear elasticit Appl Green s theorem (integration b parts) to the first term in each of these equations and account for w = w = 0 on Γ Adding the two equations gives: t t Ω Ω + b w dω= 0 w U, ( iσ ) x x x x + b w dω= 0 w U. ( iσ ) w σ indγ w iσ dω+ b w dω= 0 w U, x x x x x x x Γ Ω Ω w σ indγ w iσ dω+ b w dω= 0 w U. Γ Ω Ω ( wxiσ x + wiσ ) dω= witdγ+ wibdω wx U0 Ω Γ Ω t U N I V E R S I T Y 24 0 0 x u 0 0
Weak form for isotropic linear elasticit The left hand side can be simplified b noticing that: ( wxiσ x + wiσ ) dω= witdγ+ wibdω wx U0 Ω Γ Ω wx w w x w wx iσ x + wiσ = σxx + σx + σx + σ = x x σ xx w w x w w x = + σ x x σ x U N I V E R S I T Y 25 t 0 x σ xx wx = 0 σ w = σ x x S T ( ) S w T σ
Weak form for isotropic linear elasticit ( wxiσ x + wiσ ) dω= witdγ+ wibdω wx U0 Ω Γ Ω w iσ + w iσ = w σ The matrix form of the weak form becomes: T t ( ) T x x S ( ) Sw σ dω= witdγ+ wibdω wx U0 Ω Γ Ω t U N I V E R S I T Y 26
Weak form for isotropic linear elasticit Substituting the kinematic ε = Su and constitutive σ = Dε, relations, the resulting weak form in 2D can be summarized as: Find u U on Ω such that T : ( ) S w σ dω= witdγ+ wibdω w U0 Ω Γ Ω 0 t { 1 } u where : U = u : u H, u = u on Γ and { 1 :, 0 } u U = w w H w= onγ Can ou think a virtual work interpretation of the above equation? U N I V E R S I T Y 27