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The Plane Stress Problem

Martin Kronbichler

Applied Scientific Computing (Tillämpad beräkningsvetenskap)

February 2, 2010

Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 1 / 24

Outline

Plane stress

FEM in 2DDiscrete interpolation

Triangular elementsRectangular elements

The algebraic problemSolution, postprocessing

Accuracy of FEM


Plane stress

A 2D problem: plane stress

Assumptions:I thin specimenI free to move normal to the plane

(x3 direction, i.e. direction ofunit vector e3)

I only loaded in the plane (x1, x2

direction)

Image from http://en.wikipedia.org.

Thus no stress normal to the plane, but specimen typically “bulges” ([sv.bukta]) in the normal direction so displacements is typically not zeronormal to the plane (u3 6= 0).


Plane stress

Plane stress modeling

No stress (force) normal to the plane ⇒ σ · e3 = 0⇒ third column of σ is zero, and, because of σ = σT, also the third row

σ =

σ11 σ12 0σ12 σ22 00 0 0

=(σ 00T 0

).

Constitutive law:

σ = λI tr ε+ 2µε =E

1 + ν

[ν

1− 2νI tr ε+

22ε

]⇒ ε =

(ε 00T ε33,

), where ε33 = − ν

1− νtr ε.


Plane stress

Plane stress modeling

Express constitutive law in terms of upper 2× 2 block,

σ =E

1 + ν

[ν

1− νI tr ε+ ε

]or

σ = λI tr ε+ 2µε, λ =Eν

(1 + ν)(1− ν)

Conclusion: for plane stress, compute in 2D with modified constitutivelaw


Plane stress

Plane strainAdjoint/dual problem to plane stress (observe: not the mathematical“adjointness”)

Assumptions:I no displacements orthogonal to the

plane (a thick specimen or a restrictedmovement)

I loads constant in direction orthogonalto the plane

Assumptions ⇒

ε =

ε11 ε12 0ε12 ε12 00 0 0

=(ε 00T 0

).


FEM in 2D

Finite Element Discretization for PlaneStress Problems


FEM in 2D

Finite Element (FE) discretiztation for 2D linear elasticity

Consider the equilibrium equation∫Ω

[λ tr(ε(v)

)tr(ε(u)

)+ 2µε(v) : ε(u)

]dΩ =

∫Ωv · f dΩ +

∫Γ1

v · g ds,

or, in abbreviated form,

a(v,u) = l(v) for all admissible v

FEM aims at satisfying this equation for asubset of admissible displacements

→ work on a discretization associated witha triangulation the domain, consisting ofnonoverlapping triangles (quadrilaterals alsocommon) An example triangulation


FEM in 2D Discrete interpolation

Discrete interpolation

Admissible displacements uh arecontinuous on Ω and polynomials on eachtriangle (or quadrilateral).

An example of one component of anadmissible displacement when uh is a linearfunction (first-order polynomial) at eachtriangle.

For piecewise-linear functions, the nodes are the vertices of thetriangulations → lowest-order linear Lagrange element or thethree-node triangle



Each admissible displacement uh isexpressed as a weighted sum ofbasis functions φj(x):

uh =N∑

j=1

φj(x)uj .Basis function for continuous,

piecewise-linear functions

I The sum interpolates discrete displacement vectors uj located atnodes (interpolation points)

I A nodal basis function is a continuous function in the finite elementspace that is

I one at one node, andI zero at all other nodes



Lagrange elements Pp

Interpolation properties of nodal basis functions (•):I Polynomials of degree p on each triangle (element)I Continuous across each edge

Linear: P1

The three-node triangleu(x, y) = a1 + a2x+ a3y

Quadratic: P2

The six-node triangleu(x, y) = a1 + a2x+

a3y + a4x2 + a5xy + a6y

2

Cubic; P3

The ten-node triangle



Example basis functions

Functions within a single element:

For quadratic Lagrangian elements, there are two distinct types of nodalbasis functions:

Associated with mesh vertex Associated with edge midpoints



Example basis functions II

Basis functions on their support:

The basis function φi(x) when i correspondsto a corner node

The basis function φi(x) when icorresponds to an edge-midpoint node



Rectangular elements

A common element type for 2D elasticity.The displacements at the nodes (•) are interpolated by a product ofone-dimensional polynomials in each coordinate direction

The four-node rectangular element Q1.Bilinear, 4 coefficients

u(x, y) = a1 + a2x+ a3y + a4xy

(obtained as product of φ(x) = s1 +s2x and ψ(y) =t1 + t2y)



Rectangular elements, II

The nine-node rectangular element Q2.Biquadratic, 9 coefficients

u(x, y) = a1 + a2x+ a3y + a4xy + a5x2 + a6y

2

+ a7x2y + a8xy

2 + a9x2y2



The four-node Q1 on general quadrilaterals:I Linear on each edgeI Not in general of formq(x, y) = a0 + a1x+ a2y + a3xy!

I A quadrilateral is not the image of an affinemap of a rectangle!


FEM in 2D The algebraic problem

The algebraic problem

Problem (FE equation for linear elasticity and plane stress)Find displacement field uh, such that

a(vh,uh) = l(vh) for all vh ∈ V h. (2)

Form of an admissible displacement in the finite element subspace:

uh(x) =N∑

j=1

φj(x)uj =N∑

j=1

φj(x)(u1,je1 + u2,je2) =2N∑j=1

φj(x)uj (3)

where, for j = 1, . . . , N ,

u2j−1 = u1,j

u2j = u2,j

φ2j−1(x) = e1φj(x)

φ2j(x) = e2φj(x).



Substitute (3) into FE equation and test with vh = φk, k = 1, . . . , 2N .Hence, (2) is equivalent to requiring

2N∑j=1

a(φk,φj)uj = l(φk) for k = 1, . . . , 2N,

which is a linear system Au = b , where

Akj = a(φk,φj) =∫

Ω

[λ tr(ε(φk)

)tr(ε(φj)

)+ 2µε(φk) · ε(φj)

]dΩ ,

bk = l(φk) =∫

Ωφk · f dΩ +

∫Γ1

φk · g ds,

u = (u1, u2, . . . , u2N )T = (u1,1, u2,1, u1,2, u2,2, . . . , u1,N , u2,N )T



Practical implementation of elastic equations

Assembly of matrix A by a loop over all cells; deal.II open source finite element librarybased on C++, authors W. Bangerth, R. Hartmann, G. Kanschat,http://www.dealii.org, 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);


FEM in 2D Solution, postprocessing

Solution and postprocessing

I Solve linear system Au = bI Direct: Gauss/Cholesky decomposition, COMSOL: UMFPACK (sparse

direct solver)I Iteratively: for larger size (> 500 000 degrees of freedom in 2D,> 100 000 in 3D) → CG with preconditioning, multigrid

I Postprocessing:I Translate components in u to displacements at certain pointsI Get stresses and/or strains from derivatives of (3), using kinematic and

constitutive relation

−→ Computer lab


Accuracy of FEM

Accuracy of FEM

On the accuracy of the discrete FEMsolutions


Accuracy of FEM

Error estimatesDiscretization errors measured in integral norms. If everything is “nice”, wehave for triangular or rectangular elements∥∥∥u− uh

∥∥∥L2(Ω)

=(∫

Ω

∣∣∣u− uh∣∣∣2 dΩ

)1/2

≤ Chp+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

I Nondegenerate mesh refinements:router/rinner bounded as h→ 0 (or thatthe minimum (or maximum) angle isbounded away from 0 (or 180)

I Smooth solutions. Accuracy typicallyreduced in vicinity of reentrant cornerson the boundary (ω > 180)

routerrinner

!


Accuracy of FEM

Remarks

I Stresses are functions of derivatives of the displacement (constitutive& kinematic relation)

I Therefore: accuracy of stress approximations typically one orderlower than the displacement approximations

I In particular: stresses are piecewise constant for the P1 element(3-node triangle) → poor stress approximation

I Rule of thumb: P1 too inaccurate (P2 much better), see computerlab, task 1.1.

I Adaptive grid refinement (e.g. around reentrant corners) and errorestimates central aspects in CSM


Accuracy of FEM

P1 versus Q1

I Advantage Q1

Q1 (4-node rectangle) performs usually much better than P1 (bothhave error term h2!)Reason: the presence of the quadratic xy term for Q1 (stresses notconstant in element)⇒ Rectangular elements can give higher accuracy for the samenumber of degrees of freedom, particularly in 3D

I Disadvantage quadrilateralsMesh generation more difficult and less automatic for rectangularmeshes compared to triangles


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