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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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 2 / 24
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).
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 3 / 24
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.
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 4 / 24
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.
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 4 / 24
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 5 / 24
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
).
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 6 / 24
FEM in 2D
Finite Element Discretization for PlaneStress Problems
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 7 / 24
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 8 / 24
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 8 / 24
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 9 / 24
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 9 / 24
FEM in 2D Discrete interpolation
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 10 / 24
FEM in 2D Discrete interpolation
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 11 / 24
FEM in 2D Discrete interpolation
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 12 / 24
FEM in 2D Discrete interpolation
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 13 / 24
FEM in 2D Discrete interpolation
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)
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 14 / 24
FEM in 2D Discrete interpolation
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 15 / 24
FEM in 2D Discrete interpolation
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!
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 16 / 24
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).
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 17 / 24
FEM in 2D The algebraic problem
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 18 / 24
FEM in 2D The algebraic problem
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);
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 19 / 24
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 20 / 24
Accuracy of FEM
Accuracy of FEM
On the accuracy of the discrete FEMsolutions
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 21 / 24
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
!
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 22 / 24
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 23 / 24
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
Martin Kronbichler (TDB) The Plane Stress Problem February 2, 2010 24 / 24
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