10-Linear Strain Triangle and Other Types of 2d Elements

Linear Strain Triangle and other types of 2D elements

ByS. Ziaei Rad

Linear Strain Triangle (LST or T6)

This element is also called quadratic triangular element.

Quadratic Triangular Element

Linear Strain Triangle

There are six nodes on this element: three corner nodes andthree midside nodes. Each node has two degrees of freedom(DOF) as before. The displacements (u, v) are assumed to bequadratic functions of (x, y),

where bi (i = 1, 2, ..., 12) are constants. From these, the strains are found to be,


which are linear functions. Thus, we have the linear straintriangle(LST), which provides better results than the CST.

In the natural coordinate system we defined earlier, the sixshape functions for the LST element are,

in which . Each of these six shape functionsrepresents a quadratic form on the element as shown in thefigure.


Shape Function N1 for LSTDisplacements can be written as,

The element stiffness matrix is still given bybut here is quadratic in x and y. In general, the integral has to be computed numerically.

Linear Quadrilateral Element (Q4)

Linear Quadrilateral Element

Linear Quadrilateral Element (Q4)

There are four nodes at the corners of the quadrilateral shape. In the natural coordinate system , the four shape functions are,

Note that at any point inside the element, asexpected. The displacement field is given by

which are bilinear functions over the element.

Isoparametric ElementIf we use the same parameters (shape functions) to express Geometry, we are using an isoparametric formulation.

=

4

4

3

3

2

2

1

1

4321

4321

00000000

vuvuvuvu

NNNNNNNN

vu

44332211

44332211

yNyNyNyNyxNxNxNxNx

+++=+++=

=

4

4

3

3

2

2

1

1

4321

4321

00000000

yxyxyxyx

NNNNNNNN

yx

Isoparametric Element

=

yfxf

yx

yx

f

f

or [ ]

=

yfxf

Jf

f

[ ]

=

f

f

J

yfxf

1

=

=2221

1211][JJJJ

yx

yx

J

=

+

=

v

v

u

u

JJJJJJ

JJ

J

xv

yuyvxu

12221121

1121

1222

0000

det1}{

A1

Isoparametric Element

++++

++++

=

4

4

3

3

2

2

1

1

)1(0)1(0)1(0)1(0)1(0)1(0)1(0)1(0

0)1(0)1(0)1(0)1(0)1(0)1(0)1(0)1(

41

vuvuvuvu

v

v

u

u

A2

}]{[}]{2][1[}{ dBdAA ==

=1

1

1

1

det]][[][][ ddJBEBtK Te

Quadratic Quadrilateral Element (Q8)

This is the most widely used element for 2-D problems dueto its high accuracy in analysis and flexibility in modeling.

Quadratic Quadrilateral Element


),(),( iii GFN =),( iF Give a value of zero along the sides of the element

That the given node does not contact

Select such that when multiply by Fi, it will produceA value of unity at node i and a value of zero at other neighboring nodes.

),( iG

Example: Consider N3)1)(1(),(3 ++=F 3213 ),( cccG ++=

1)1,1()1,1()1,1(; 0)1,0(; 0)0,1( 33333 ==== GFNGG


1)(41)1,1(00)1,0(00)0,1(

3213

313

213

=++==+==+=

cccNccGccG

4/1 4/1

4/1

3

2

1

===

ccc

)1)(1)(1(41

3 +++= N


There are eight nodes for this element, four corners nodesand four midside nodes. In the natural coordinate system ,the eight shape functions are,


Again, we have at any point inside the element.

The displacement field is given by

which are quadratic functions over the element. Strains and stresses over a quadratic quadrilateral element are linear functions, which are better representations.Notes:Q4 and T3 are usually used together in a mesh with linear elements.Q8 and T6 are usually applied in a mesh composed ofquadratic elements.Quadratic elements are preferred for stress analysis,because of their high accuracy and the flexibility inmodeling complex geometry, such as curved boundaries.

Example 3.2A square plate with a hole at the center and under pressurein one direction.

The dimension of the plate is 10 in. x 10 in., thickness is0.1 in. and radius of the hole is 1 in. Assume E = 10x10E6 psi,v= 0.3 and p = 100 psi. Find the maximum stress in the plate.

Example 3.2FE Analysis:From the knowledge of stress concentrations, we shouldexpect the maximum stresses occur at points A and B on theedge of the hole. Value of this stress should be around 3p (=300 psi) which is the exact solution for an infinitely large platewith a hole.

We use the ANSYS FEA software to do the modeling(meshing) and analysis, using quadratic triangular (T6 or LST)linear quadrilateral (Q4) and quadratic quadrilateral (Q8)elements. Linear triangles (CST or T3) is NOT available inANSYS.The stress calculations are listed in the following table,along with the number of elements and DOF used, forcomparison.

Example 3.2

Discussions: Check the deformed shape of the plate Check convergence (use a finer mesh, if possible) Less elements (~ 100) should be enough to achieve the same accuracy with a better or smarter mesh Well redo this example in next chapter employing thesymmetry conditions.

Example 3.2

FEA Mesh (Q8, 493 elements)FEA Stress Plot (Q8, 493 elements)

Transformation of LoadsConcentrated load (point forces), surface traction (pressureloads) and body force (weight) are the main types of loadsapplied to a structure. Both traction and body forces need to beconverted to nodal forces in the FEA, since they cannot beapplied to the FE model directly. The conversions of theseloads are based on the same idea (the equivalent-work concept)which we have used for the cases of bar and beam elements.

Traction on a Q4 element

Transformation of LoadsSuppose, for example, we have a linearly varying traction qon a Q4 element edge, as shown in the figure. The traction isnormal to the boundary. Using the local (tangential) coordinates, we can write the work done by the traction q as,

where t is the thickness, L the side length and the componentof displacement normal to the edge AB.

For the Q4 element (linear displacement field), we have

The traction q(s), which is also linear, is given in a similar way,

Transformation of Loads

Thus, we have,

and the equivalent nodal force vector is,


Note, for constant q, we have,

For quadratic elements (either triangular or quadrilateral),the traction is converted to forces at three nodes along the edge,instead of two nodes.Traction tangent to the boundary, as well as body forces,are converted to nodal forces in a similar way.


1 @ 1 @ 0

====

Lss dLds

2=)1(

2+= Ls

L1 2

q1 q2

s

34

21

21

)1(21)1(

21

)1(21)1(

21

qqq

vvv

++=

++=


[ ]

dL

qq

vvtWq 2)1(

21)1(

21

)1(21

)1(21

2

11

121

+

+

=

=

2

1

2

1

2112

6 qqtL

ff

L1 2

q1q1

s

4 3

[ ]

dL

qq

vvtWq 2)1(21

)1(21

1

11

121

+

=

=

11

21

2

1 tLqff

Transformation of Loads1- Point Load considered in a usual manner by having a Structural node at the point.2- Traction Force As it was seen in the previous exampleFirst the force and the deflection along the side express byUse of shape functions and then numerical integration willBe used to calculate the equivalent nodal forces.3- Body Force A Body force which is a distributed forcePer unit volume, contribute to the global force vector F. Assume as constant within each element.Tyx ffF ] [}{ =

=V e

eTT fdFdVu Where the 8*1 element body force is given by

= y

xTe

e

ff

dJdNtf1

1

1

1

det

Stress CalculationThe stress in an element is determined by the followingrelation,

where B is the strain-nodal displacement matrix and d is thenodal displacement vector which is known for each elementonce the global FE equation has been solved.

Stresses can be evaluated at any point inside the element(such as the center) or at the nodes. Contour plots are usuallyused in FEA software packages (during post-process) for usersto visually inspect the stress results.

The von Mises StressThe von Mises stress is the effective or equivalent stress for2-D and 3-D stress analysis. For a ductile material, the stresslevel is considered to be safe, if

where is the von Mises stress and the yield stress of thematerial. This is a generalization of the 1-D (experimental)result to 2-D and 3-D situations.

The von Mises stress is defined by

in which and are the three principle stresses at theconsidered point in a structure.

The von Mises StressFor 2-D problems, the two principle stresses in the planeare determined by

Thus, we can also express the von Mises stress in terms ofthe stress components in the xy coordinate system. For planestress conditions, we have,

Averaged Stresses:Stresses are usually averaged at nodes in FEA softwarepackages to provide more accurate stress values. This optionshould be turned off at nodes between two materials or othergeometry discontinuity locations where stress discontinuity doesexist.

Discussions1) Know the behaviors of each type of elements:T3 and Q4: linear displacement, constant strain and stress;T6 and Q8: quadratic displacement, linear strain and stress.2) Choose the right type of elements for a given problem:When in doubt, use higher order elements or a finer mesh.3) Avoid elements with large aspect ratios and corner angles:

where Lmax and Lmin are the largest and smallest characteristiclengths of an element, respectively.

Elements with Bad Shapes Elements with Nice Shapes

Discussions4) Connect the elements properly:Dont leave unintended gaps or free elements in FE models.

Improper connections (gaps along AB and CD)

Linear Strain Triangle and other types of 2D elementsLinear Strain Triangle (LST or T6)Linear Strain TriangleLinear Strain TriangleLinear Strain TriangleLinear Quadrilateral Element (Q4)Linear Quadrilateral Element (Q4)Isoparametric ElementIsoparametric ElementIsoparametric ElementQuadratic Quadrilateral Element (Q8)Quadratic Quadrilateral Element (Q8)Quadratic Quadrilateral Element (Q8)Quadratic Quadrilateral Element (Q8)Quadratic Quadrilateral Element (Q8)Example 3.2Example 3.2Example 3.2Example 3.2Transformation of LoadsTransformation of LoadsTransformation of LoadsTransformation of LoadsTransformation of LoadsTransformation of LoadsTransformation of LoadsStress CalculationThe von Mises StressThe von Mises StressDiscussionsDiscussions

10-Linear Strain Triangle and Other Types of 2d Elements

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