Chapter 6 2D Elements Plate Elements

Institute of Structural Engineering Page 1 Chapter 6 2D Elements Plate Elements Method of Finite Elements I

Institute of Structural Engineering Page 2 Continuum Elements Plane Stress Plane Strain Toda s Lecture Etending (discretization) to higher Dimensions Structural Elements Plate Elements Method of Finite Elements I

Institute of Structural Engineering Page 3 Toda s Lecture (in more detail) In the previous lectures: The Galerkin method was presented The isoparametric concept was introduced The bar and beam element were presented Some numerical integration methods were presented In toda s lecture: Concepts from previous lectures are combined to formulate elements for the above cases Linear 3D elasticit equations are used as a starting point Plane stress/plane strain and plates are presented as special cases Method of Finite Elements I

Institute of Structural Engineering Page 4 3D elasticit problem Problem variables: n t Γ t Displacements: u u = u u z Stresses and strains: Γ u Γ 0 Ω ε ε ε ε z = ε ε εz εz εz ε z Method of Finite Elements I σ σ τ τ z = τ σ τ z τz τ z σ z

Institute of Structural Engineering Page 5 3D elasticit problem In FE analsis Voigt notation is tpicall used for strains and stresses: ε ε ε ε z = ε ε εz ε ε ε z z zz ε ε ε ε ε ε z ε z = = 2ε γ 2ε z γ z 2 εz γ z engineering strain σ Method of Finite Elements I σ τ τ z = τ σ τ z τ τ σ z z zz σ σ σ σ z = τ τ z τ z

Institute of Structural Engineering Page 6 3D elasticit problem n t Γ t Boundar conditions: Ω σn = 0 at Γ σn = t at Γ u= u at Γ u t 0 Γ u Γ 0 Method of Finite Elements I

Institute of Structural Engineering Page 7 3D elasticit equations Equilibrium equations: In vector form σ+ F=0 Component wise σ τ τz + + + F z = 0 τ σ τ z + + + F z = 0 τ τ z z σ z + + + Fz z = 0 Where FF is the applied bod force: F F = F F z Method of Finite Elements I

Institute of Structural Engineering Page 8 Strain definition: 3D elasticit equations In vector form s 1 ε= u= u+ u 2 Method of Finite Elements I T ( ) Component wise u 1 u u 1 u uz + + 2 2 z 1 u u u 1 u u z ε = + + 2 2 z 1 uz u 1 u u z uz + + 2 z 2 z z Using Voigt notation u u uz z ε = u u + u uz + z u u z + z

Institute of Structural Engineering Page 9 3D elasticit equations Constitutive equation, using Voigt notation can be convenientl written as: σ = Eε 1 ν ν ν 0 0 0 ν 1 ν ν 0 0 0 σ ε σ ν ν 1 ν 0 0 0 ε σ 1 2 z E ν 0 0 0 0 0 ε z = τ (1 )(1 2 ) 2 ν ν + γ 1 2ν τ z 0 0 0 0 0 γ z 2 τ z γ z 1 2ν 0 0 0 0 0 2 Method of Finite Elements I

Institute of Structural Engineering Page 10 3D elasticit equations The weak form of the problem can be obtained using the Galerkin method as: T T T ε ( w) Eε( u) dω= w FdΩ+ w t dγ Ω Ω Γ where w w = w w z is the weight function Method of Finite Elements I

Institute of Structural Engineering Page 11 Plane stress/strain In several cases of practical interest: The third dimension of the problem is: Either ver small Or ver large but includes no variation in the problem parameters The problem equations can be simplified resulting in a 2D problem Method of Finite Elements I

Institute of Structural Engineering Page 12 Plane stress assumptions First let s consider a structure where: The length in one dimension is much smaller than the other two Loads are applied onl within a plane The in-plane stresses, strains and displacements are constant in the third dimension Normal and shear stresses in the third dimension are negligible Such a structure is said to be in a state of plane stress

Institute of Structural Engineering Page 13 Plane stress equations Assuming that stresses along the third dimension are zero: Assumptions: Equilibrium equations: σ σ τ σ z = σ = σ = τ = τ z (, ) (, ) (, ) = τ z = 0 σ τ τ + σ + + + F F = 0 = 0 Strain definition Constitutive equation u u ε z = ε z u = ε =, εz = 0 γ εz = 0 u u + ε 2 z σ 1 ν 0 ε E σ = ν 1 0 ε 1 ν σ 1 ν γ 0 0 2 How is this derived? Method of Finite Elements I

Institute of Structural Engineering Page 14 Eamples of plane stress problems Thin Plate With Central Hole Circular Plate Under Edge Loadings

Institute of Structural Engineering Page 15 Plane strain assumptions Net we consider a structure where: The length in one dimension is much larger than the other two Loads are applied onl within a plane Loads are constant in the third dimension Displacements and strains along the third dimension are negligible Such a structure is said to be in a state of plane strain

Institute of Structural Engineering Page 16 Plane strain equations Net we assume that strains along the third dimension are zero: Assumptions: Equilibrium equations: u = u(, ) u = u(, ) u z = 0 σ τ τ + σ + + + F F = 0 = 0 Strain definition Constitutive equation u ε εz = 0 u ε = ε =, γ z = 0 γ γ z = 0 u u + σ 1 ν ν 0 ε E σ = ν 1 ν 0 ε (1 ν)(1 2 ν) σ + 1 2ν γ 0 0 2 Method of Finite Elements I

Institute of Structural Engineering Page 17 Eamples of plane strain P z z Long Clinders Under Uniform Loading Semi-Infinite Regions Under Uniform Loadings

Institute of Structural Engineering Page 18 Weak form of the problem The weak form of the problem is similar to the one used for the 3D problem: T T T ε ( w) Eε( u) dω= w FdΩ+ w t dγ Ω Ω Γ where: stresses, strains, displacements and constitutive matrices correspond to the plain stress/strain case integration is carried out over an area instead of a volume Method of Finite Elements I

Institute of Structural Engineering Page 19 Discretization Several options are available for discretizing the plane stress/strain problem One of the most common choices is isoparametric Lagrange elements In the following we will briefl review some basic properties of isoparametric elements Some specific, and widel used, elements will be presented for plane stress/strain analsis Method of Finite Elements I

Institute of Structural Engineering Page 20 Using the isoparametric concept, geometr can be discretized as: n Isoparametric formulation = h(,) rs; = h(,) rs i i i i i= 1 i= 1 n and displacements as: n u = h(,) r su ; u = h(,) r su i i i i i= 1 i= 1 n where i, i, ui, ui are nodal values of the spatial coordinates and displacement components

Institute of Structural Engineering Page 21 Isoparametric quadrilaterals For a 4 noded linear isoparametric quadrilateral (q4), coordinates r and s are defined based on the following transformation: 4(, ) 4 4 3(, ) 3 3 4 ( 1,1) 3 (1,1) s r 2(, ) 2 2 1( 1, 1) Method of Finite Elements I 1 ( 1, 1) 2 (1, 1)

Institute of Structural Engineering Page 22 Isoparametric quadrilaterals Then the displacement and geometr approimations specialize to: 4 4 = h(,) rs; = h(,) rs i i i i i= 1 i= 1 4 4 u = h(,) r su ; u = h(,) r su i i i i i= 1 i= 1 with 1 1 h1 = (1 r)(1 s); h2 = (1 + r)(1 s); 4 4 1 1 h3 = (1 + r)(1 + s); h4 = (1 r)(1 + s); 4 4

Institute of Structural Engineering Page 23 Isoparametric quadrilaterals h1 h2 h3 h4

Institute of Structural Engineering Page 24 Isoparametric quadrilaterals The shape functions can also be written in matri form as: u1 u 1 u 2 u h1 0 h2 0 h3 0 h4 0 u2 = u 0 h1 0 h2 0 h3 0 h u 4 3 u N u 3 u4 u 4 u 1 1 2 h1 0 h2 0 h3 0 h4 0 2 = 0 h1 0 h2 0 h3 0 h 4 3 N 3 4 4 u= Nu = N

Institute of Structural Engineering Page 25 Shape function derivatives Derivatives with respect to the spatial coordinates can be obtained as: r r r = s s s r J = J = Γ r r 1 Γ= J In addition, the infinitesimal surface element can be transformed as: dω= d d = det( J) dr ds

Institute of Structural Engineering Page 26 Shape function derivatives Using the above transformations, strain components can be obtained, for instance: ε u hi(,) rs hi(,) rs hi(,) rs with hi, (,) rs = = Γ 11 + Γ 12 r s 4 = = hi, (,) rs ui i= 1 The above can be written in matri form as: ε u u u u 2 = h1, 0 h2, 0 h3, 0 h4, 0 u3 u u u 1 1 2 3 4 4

Institute of Structural Engineering Page 27 Shape function derivatives All strain components can be similarl computed and gathered in a matri using Voigt notation: u1 u 1 u 2 h1, 0 h2, 0 h3, 0 h4, 0 u2 ε = 0 h1, 0 h2, 0 h3, 0 h4, u3 h1, h1, h2, h2, h3, h3, h4, h 4, u 3 B u4 u 4 u ε= Bu

Institute of Structural Engineering Page 28 Stiffness matri and load vectors Net the strains can be substituted in the weak form to obtain the stiffness matri as: K T = B EBdΩ Ω In the above, the domain of integration is defined with respect to the sstem and the shape functions with respect to r and s. Therefore a change of variables has to be performed as follows: 1 1 K T T = B EB dω= B EB det( J ) drds Ω 1 1

Institute of Structural Engineering Page 29 Stiffness matri and load vectors In a similar manner load vectors can be obtained, for instance due to a bod force: 1 1 T f NFdet( J) drds F = 1 1 For surface tractions, integration has to be carried out along the sides of the quadrilateral, e.g.: 1 T ft = N tdet( Js1) dr where det(jj ssss ) is the Jacobean determinant 1 of the side where the load is applied

Institute of Structural Engineering Page 30 Stiffness matri and load vectors B reviewing the derived epressions: 1 1 T K B EBdet( J) drds = 1 1 1 1 T f NFdet( J) drds F = 1 1 1 T ft = N tdet( Js1) dr 1 It can be observed that: integration over a general quadrilateral was reduced to integration over a square the epressions to be integrated are now significantl more complicated due to the use of the chain rule analtical integration might not be possible

Institute of Structural Engineering Page 31 Stiffness matri and load vectors To overcome this difficult, numerical integration is tpicall emploed, thus reducing integrals to sums: 1 1 T T K = B EBdet( J) drds = ww i jb ( ri, sj) EB( ri, sj)det[ J( ri, sj)] 1 1 i j where wi, ri, si are the weights and coordinates of the Gauss points used Notice that the Jacobian is also evaluated at the different Gauss points since it is not constant in general!

Institute of Structural Engineering Page 32 Numerical integration Gauss point weights and coordinates are tpicall precomputed can be found in tables. For instance for a linear quadrilateral: 1 4 3 2 s 4 3 1 2 r In 1D: w w 1 = 1, r =, 3 1 = 1, r = 3 1 1 2 2 The above can be combined to obtain the 2D coordinates and weights

Institute of Structural Engineering Page 33 Other isoparametric elements The procedure presented above can be performed for different kinds of elements just b changing the shape functions and Gauss points used, for instance: 8 noded quadratic quadrilateral (q8) Shape functions 1 h1 = (1 r )(1 s ) 1 r s 4 ( ) 1 4 8 Method of Finite Elements I 7 5 3 6 2 s 4 7 3 8 6 1 5 2 r 1 h2 = (1 + r )(1 s ) 4 1 + r s 1 h3 = (1 + r )(1 + s ) 4 1 + r+ s 1 h4 = (1 r )(1 + s ) 4 1 r+ s h h h 5 6 7 h = 8 = = = 2 ( 1 r )( 1 s) 2 2 ( 1+ r)( 1 s ) 2 2 ( 1 r )( 1+ s) 2 2 ( 1 r)( 1 s ) 2 ( ) ( ) ( )

Institute of Structural Engineering Page 34 Other isoparametric elements The procedure presented above can be performed for different kinds of elements just b changing the shape functions and Gauss points used, for instance: 8 noded quadratic quadrilateral (q8) Gauss points In 1D: 1 7 8 9 4 5 6 2 3 s 7 8 9 4 5 6 1 2 3 r w w w = 5 / 9, r = 0.6 = 8 / 9, r = 0 1 1 2 2 = 5 / 9, r = 0.6, 3 3 The above can be combined to obtain the 2D coordinates and weights

Institute of Structural Engineering Page 35 Other isoparametric elements The procedure presented above can be performed for different kinds of elements just b changing the shape functions and Gauss points used, for instance: Linear or constant strain triangle (CST) Shape functions 3 3 h h 1 2 = r = s h = r s 3 1 s 1 2 1 r 2

Institute of Structural Engineering Page 36 Other isoparametric elements The procedure presented above can be performed for different kinds of elements just b changing the shape functions and Gauss points used, for instance: Linear or constant strain triangle (CST) Gauss points w = 1, r = 1/3, s = 1/3 1 1 1 3 3 s 1 2 1 r 2

Institute of Structural Engineering Page 37 The plate problem We net consider structures where: One dimension (thickness) is sufficientl smaller that the other two dimensions The geometr is planar Loads are applied both in plane and normal to the plane The problem can again be reduced to a 2D problem

Institute of Structural Engineering Page 38 The plate problem Depending on the loading, the problem can be reduced to two separate problems: In plane loading Normal loading Plane stress (or membrane) problem Plate bending problem

Institute of Structural Engineering Page 39 Plate theories Two alternative theories eist for the bending of plates which are in complete analog to beam theories: Beam theor Bernoulli-Euler Kinematic assumptions Plane sections that are normal to the ais of the beam remain plane and normal to the ais. Valid for Beams with small height to length ratio Plate theor Kirchhoff-Love Kinematic assumptions Straight lines that are normal to the mid-surface remain straight and normal to the mid-surface. Valid for Thin plates

Institute of Structural Engineering Page 40 Plate theories Two alternative theories eist for the bending of plates which are in complete analog to beam theories: Beam theor Timoshenko Kinematic assumptions Plane sections that are normal to the ais of the beam remain plane but not necessaril normal to the ais. Valid for Beams with an height to length ratio Plate theor Mindlin-Reissner Kinematic assumptions Straight lines that are normal to the mid-surface remain straight but not necessaril normal to the mid-surface. Valid for Thin and thick plates

Institute of Structural Engineering Page 41 Mindlin-Reissner plate theor In the following we will focus on the Mindlin-Reissner theor, since most elements used in practice are based on that theor. The basic equations for this theor can be derived from the 3D elasticit equations b emploing the following assumptions: Deflections don t var along the thickness direction In plane displacements, and as a result strains and stresses, var linearl along the thickness Normal stresses along the thickness direction vanish Method of Finite Elements I

Institute of Structural Engineering Page 42 Mindlin-Reissner plate theor The kinematic assumptions can be written as: u = zθ (, ); u = zθ (, ); u = u(, ) z z where θ are the rotations of the normal to the mid-plane along, θ the z and z planes z It should be noted that in this theor rotations are not defined as the derivatives of the deflections Instead the are independent variables B A z A θ B γ z

Institute of Structural Engineering Page 43 Mindlin-Reissner plate theor The kinematic assumptions can be written as: u = zθ (, ); u = zθ (, ); u = u(, ) z z Substituting in the strain definition and separating the in-plane from the transverse shear components: ε u θ u θ ε = = z γ u u θ θ + + uz u uz θ γ + z z = = γ z u u z uz θ + z

Institute of Structural Engineering Page 44 Mindlin-Reissner plate theor Using plane stress constitutive relations, the stresses can be obtained: θ σ 1 ν 0 E θ σ = z ν 1 0 2 1 ν τ 1 ν 0 0 θ 2 θ + uz θ τ z E = τ z 2(1 + ν ) uz θ In plane stresses var linearl along the thickness Transverse shear stresses are constant along the thickness

Institute of Structural Engineering Page 45 Mindlin-Reissner plate theor B separating integration along the thickness and on the area, the weak form can be written as: 1 ν 0 ε γ da dz da dz w p da 0 0 2 t/2 t/2 E E z ε 1 0 2 ε γ ν ε γ z γ z z z 1 ν + = 2(1 + ν) γ t/2 A t/2 A z A 1 ν γ where: t ε, ε, γ, γ z, γ z w z pz A is the plate thickness are strain components computed using the weight function is the weight function components in the z direction is the normal load applied to the plate is the area of the plate

Institute of Structural Engineering Page 46 Mindlin-Reissner plate theor Once integration along the thickness has been carried out, the weak form can be re-written as: T κ Eκ + T b γ Eγ s = A A A with: da da w p da z z κ θ θ θ θ + = E b 1 ν 0 3 Et = 1 0 2 ν 12( 1 ν ) 1 ν 0 0 2 γ uz θ = uz θ E s Ekt 1 0 = 2(1 + ν ) 0 1 k is a corrective factor to account for variation of shear stresses, usuall k = 5/6

Institute of Structural Engineering Page 47 Mindlin-Reissner plate theor Moments and shear forces are obtained from the curvatures and transverse shear stresses through the constitutive equations: M θ M 1 ν 0 3 Et θ = Ebκ, M = ν 1 0 2 12( 1 ν ) M 1 ν 0 0 θ 2 θ + uz Q θ Et 1 0 Q= Esγ, = Q 2(1 ν ) 0 1 + uz θ

Institute of Structural Engineering Page 48 Isoparametric plate elements Since rotations and deflections are independent, the are also discretized independentl: n n n u = h(,) rs u; θ = h(,) rsθ ; θ = h(,) rsθ z i zi i i i i i= 1 i= 1 i= 1 One of the most commonl used elements for this kind of analsis is the isoparametric quadrilateral previousl presented The shape functions, derivative definitions, jacobian determinant and Gauss points used for this element are identical to the ones presented for the plane stress element

Institute of Structural Engineering Page 49 Isoparametric plate elements Element geometr, and degree of freedom definitions: u z4 θ u z3 θ 4 3 θ θ 4 3 z u z1 θ u 1 z2 θ 2 θ θ 1 2

Institute of Structural Engineering Page 50 Isoparametric plate elements Curvatures and shear strains can then be obtained as: uz1 θ 1 θ 1 uz2 θ 2 0 h1, 0 0 h2, 0 0 h3, 0 0 h4, 0 θ 2 κ = 0 0 h1, 0 0 h1, 0 0 h1, 0 0 h1, uz3 0 h1, h1, 0 h2, h2, 0 h3, h3, 0 h4, h 4, θ 3 B b θ 3 u z4 θ 4 θ 4 u κ B u = b

Institute of Structural Engineering Page 51 Isoparametric plate elements Curvatures and shear strains can then be obtained as: uz1 θ 1 θ 1 uz2 θ 2 h1, h1 0 h2, h2 0 h3, h3 0 h4, h4 0 θ 2 γ = h1, 0 h1 h2, 0 h2 h3, 0 h3 h4, 0 h 4 uz3 B s θ 3 θ 3 u z4 θ 4 θ 4 u γ B u = s

Institute of Structural Engineering Page 52 Isoparametric plate elements The element stiffness matri can be obtained as: 1 1 1 1 T T b b b drds s s s 1 1 1 1 K= BEB det( J) + BEBdet( J) drds 1 1 T f = N p det( J) drds 1 1 z Numerical integration can be carried out as for the plane stress/strain elements.

Institute of Structural Engineering Page 53 Isoparametric plate elements The element presented: Is simple to formulate and implement Offers good performance for relativel large thickness to length ratios (>1/10) Suffers from locking problems for smaller ratios Method of Finite Elements I

Institute of Structural Engineering Page 54 The shear locking problem Let s consider the presented element in a state of pure bending: Element deformation Actual deformation (for a thin plate) Transverse shear strain develops No transverse shear θ θ 1 2 θ θ 1 2 As the thickness decreases, the spurious energ produced b this shear deformation dominates the total energ leading to unrealisticall small deflections!

Institute of Structural Engineering Page 55 The shear locking problem Several remedies have been proposed for the shear locking problem such as: Alternative numerical integration schemes Alternative interpolation for the shear strain components Specialized variational formulations B emploing one of the above alternatives, the element can also be used for thin plates! Method of Finite Elements I