The Plane Stress Problem

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1 The Plane Stress Problem Martin Kronbichler Applied Scientific Computing (Tillämpad beräkningsvetenskap) February 2, 2010 Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

2 Outline Plane stress Discrete interpolation Triangular elements Rectangular elements The algebraic problem Solution, postprocessing Accuracy of FEM Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

Plane stress A 2D problem: plane stress Assumptions: thin specimen free to move normal to the plane (x 3 direction, i.e. direction of unit vector e 3 ) only loaded in the plane (x 1, x 2 direction) Image from http://en.

3 Plane stress A 2D problem: plane stress Assumptions: thin specimen free to move normal to the plane (x 3 direction, i.e. direction of unit vector e 3 ) only loaded in the plane (x 1, x 2 direction) Image from Thus no stress normal to the plane, but specimen typically bulges ([sv. bukta]) in the normal direction so displacements is typically not zero normal to the plane (u 3 0). Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

4 Plane stress Plane stress modeling No stress (force) normal to the plane σ e 3 = 0 third column of σ is zero, and, because of σ = σ T, also the third row σ 11 σ 12 0 ( ) σ = σ 12 σ 22 0 ˆσ 0 = T. 0 Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

5 Plane stress Plane stress modeling No stress (force) normal to the plane σ e 3 = 0 third column of σ is zero, and, because of σ = σ T, also the third row σ 11 σ 12 0 ( ) σ = σ 12 σ 22 0 ˆσ 0 = T. 0 Constitutive law: σ = λi tr ɛ + 2µɛ = E [ 1 + ν ( ) ˆɛ 0 ɛ = 0 T ɛ 33, ν 1 2ν I tr ɛ ɛ, where ɛ 33 = ν tr ˆɛ. 1 ν ] Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

6 Plane stress Plane stress modeling Express constitutive law in terms of upper 2 2 block, ˆσ = E [ ] ν I tr ˆɛ + ˆɛ or 1 + ν 1 ν ˆσ = ˆλI tr ˆɛ + 2µˆɛ, ˆλ Eν = (1 + ν)(1 ν) Conclusion: for plane stress, compute in 2D with modified constitutive law Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

7 Plane stress Plane strain Adjoint/dual problem to plane stress (observe: not the mathematical adjointness ) Assumptions: no displacements orthogonal to the plane (a thick specimen or a restricted movement) loads constant in direction orthogonal to the plane Assumptions ɛ 11 ɛ 12 0 ɛ = ɛ 12 ɛ 12 0 = ( ) ˆɛ 0 0 T. 0 Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

8 Finite Element Discretization for Plane Stress Problems Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

9 Finite Element (FE) discretiztation for 2D linear elasticity Consider the equilibrium equation ) ) ] [ˆλ tr (ˆɛ(v) tr (ˆɛ(u) + 2µˆɛ(v) : ˆɛ(u) dω = Ω or, in abbreviated form, Ω v f dω + v g ds, Γ 1 a(v, u) = l(v) for all admissible v Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

10 Finite Element (FE) discretiztation for 2D linear elasticity Consider the equilibrium equation ) ) ] [ˆλ tr (ˆɛ(v) tr (ˆɛ(u) + 2µˆɛ(v) : ˆɛ(u) dω = Ω or, in abbreviated form, Ω v f dω + v g ds, Γ 1 a(v, u) = l(v) for all admissible v FEM aims at satisfying this equation for a subset of admissible displacements work on a discretization associated with a triangulation the domain, consisting of nonoverlapping triangles (quadrilaterals also common) An example triangulation Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

11 Discrete interpolation Discrete interpolation Admissible displacements u h are continuous on Ω and polynomials on each triangle (or quadrilateral). An example of one component of an admissible displacement when u h is a linear function (first-order polynomial) at each triangle. Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

12 Discrete interpolation Discrete interpolation Admissible displacements u h are continuous on Ω and polynomials on each triangle (or quadrilateral). An example of one component of an admissible displacement when u h is a linear function (first-order polynomial) at each triangle. For piecewise-linear functions, the nodes are the vertices of the triangulations lowest-order linear Lagrange element or the three-node triangle Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

13 Discrete interpolation Each admissible displacement u h is expressed as a weighted sum of basis functions φ j (x): N u h = φ j (x)u j. j=1 Basis function for continuous, piecewise-linear functions The sum interpolates discrete displacement vectors u j located at nodes (interpolation points) A nodal basis function is a continuous function in the finite element space that is one at one node, and zero at all other nodes Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

14 Discrete interpolation Lagrange elements P p Interpolation properties of nodal basis functions ( ): Polynomials of degree p on each triangle (element) Continuous across each edge Linear: P 1 The three-node triangle u(x, y) = a 1 + a 2x + a 3y Quadratic: P 2 The six-node triangle u(x, y) = a 1 + a 2x + a 3y + a 4x 2 + a 5xy + a 6y 2 Cubic; P 3 The ten-node triangle Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

15 Discrete interpolation Example basis functions Functions within a single element: For quadratic Lagrangian elements, there are two distinct types of nodal basis functions: Associated with mesh vertex Associated with edge midpoints Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

16 Discrete interpolation Example basis functions II Basis functions on their support: The basis function φ i(x) when i corresponds to a corner node The basis function φ i(x) when i corresponds to an edge-midpoint node Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

17 Discrete interpolation Rectangular elements A common element type for 2D elasticity. The displacements at the nodes ( ) are interpolated by a product of one-dimensional polynomials in each coordinate direction The four-node rectangular element Q 1. Bilinear, 4 coefficients u(x, y) = a 1 + a 2 x + a 3 y + a 4 xy (obtained as product of φ(x) = s 1 +s 2 x and ψ(y) = t 1 + t 2 y) Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

18 Discrete interpolation Rectangular elements, II The nine-node rectangular element Q 2. Biquadratic, 9 coefficients u(x, y) = a 1 + a 2 x + a 3 y + a 4 xy + a 5 x 2 + a 6 y 2 + a 7 x 2 y + a 8 xy 2 + a 9 x 2 y 2 Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

19 Discrete interpolation The four-node Q 1 on general quadrilaterals: Linear on each edge Not in general of form q(x, y) = a 0 + a 1 x + a 2 y + a 3 xy! A quadrilateral is not the image of an affine map of a rectangle! Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

20 The algebraic problem The algebraic problem Problem (FE equation for linear elasticity and plane stress) Find displacement field u h, such that a(v h, u h ) = l(v h ) for all v h V h. (2) Form of an admissible displacement in the finite element subspace: N N u h (x) = φ j (x)u j = φ j (x)(u 1,j e 1 + u 2,j e 2 ) = 2N j=1 j=1 j=1 where, for j = 1,..., N, φ j (x) u j (3) u 2j 1 = u 1,j u 2j = u 2,j φ 2j 1 (x) = e 1 φ j (x) φ 2j (x) = e 2 φ j (x). Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

21 The algebraic problem Substitute (3) into FE equation and test with v h = φ k, k = 1,..., 2N. Hence, (2) is equivalent to requiring 2N j=1 a(φ k, φ j ) u j = l(φ k ) for k = 1,..., 2N, which is a linear system Au = b, where [ ( A kj = a(φ k, φ j ) = λ tr ɛ(φk ) ) tr ( ɛ(φ j ) ) + 2µɛ(φ k ) ɛ(φ j ) ] dω, Ω b k = l(φ k ) = φ k f dω + φ k g ds, Ω Γ 1 u = (u 1, u 2,..., u 2N ) T = (u 1,1, u 2,1, u 1,2, u 2,2,..., u 1,N, u 2,N ) T Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

22 The algebraic problem Practical implementation of elastic equations Assembly of matrix A by a loop over all cells; deal.ii open source finite element library based on C++, authors W. Bangerth, R. Hartmann, G. Kanschat, tutorial step-8 for (CellIterator cell = cell_begin; cell!= cell_end; ++cell) { phi.reinit (cell); cell_matrix = 0; for (int q_point = 0; q_point < n_q_points; ++q_point) for (int i = 0; i < dofs_per_cell; ++i) for (int j = 0; j < dofs_per_cell; ++j) cell_matrix(i,j) += (lambda * trace(phi[u].gradient(i,q_point)) * trace(phi[u].gradient(j,q_point)) + 2 * mu * phi[u].symmetric_gradient(i,q_point) * phi[u].symmetric_gradient(j,q_point) ) * phi.jxw(q_point); cell->get_dof_indices(dof_indices); global_matrix.add (dof_indices, cell_matrix); } Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

23 Solution, postprocessing Solution and postprocessing Solve linear system Au = b Direct: Gauss/Cholesky decomposition, COMSOL: UMFPACK (sparse direct solver) Iteratively: for larger size (> degrees of freedom in 2D, > in 3D) CG with preconditioning, multigrid Postprocessing: Translate components in u to displacements at certain points Get stresses and/or strains from derivatives of (3), using kinematic and constitutive relation Computer lab Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

24 Accuracy of FEM Accuracy of FEM On the accuracy of the discrete FEM solutions Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

25 Accuracy of FEM Error estimates Discretization errors measured in integral norms. If everything is nice, we have for triangular or rectangular elements ( u u h L = 2 (Ω) Ω ) u u h 2 1/2 dω Ch p+1. h: largest edge in mesh; p: element polynomial order; C depends on the (p + 1)-th derivatives of u. Estimate requires some niceness conditions, typically Nondegenerate mesh refinements: r outer /r inner bounded as h 0 (or that the minimum (or maximum) angle is bounded away from 0 (or 180 ) Smooth solutions. Accuracy typically reduced in vicinity of reentrant corners on the boundary (ω > 180 ) Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24 r inner r outer!

26 Accuracy of FEM Remarks Stresses are functions of derivatives of the displacement (constitutive & kinematic relation) Therefore: accuracy of stress approximations typically one order lower than the displacement approximations In particular: stresses are piecewise constant for the P 1 element (3-node triangle) poor stress approximation Rule of thumb: P 1 too inaccurate (P 2 much better), see computer lab, task 1.1. Adaptive grid refinement (e.g. around reentrant corners) and error estimates central aspects in CSM Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

27 Accuracy of FEM P 1 versus Q 1 Advantage Q 1 Q 1 (4-node rectangle) performs usually much better than P 1 (both have error term h 2!) Reason: the presence of the quadratic xy term for Q 1 (stresses not constant in element) Rectangular elements can give higher accuracy for the same number of degrees of freedom, particularly in 3D Disadvantage quadrilaterals Mesh generation more difficult and less automatic for rectangular meshes compared to triangles Martin Kronbichler (TDB) The Plane Stress Problem February 2, / 24

The Finite Element Method for Computational Structural Mechanics

The Finite Element Method for Computational Structural Mechanics Martin Kronbichler Applied Scientific Computing (Tillämpad beräkningsvetenskap) January 29, 2010 Martin Kronbichler (TDB) FEM for CSM January