1

Gauss Quadrature

■ The above integral may be evaluated analytically with the help of a table of integrals or numerically.

■ Gauss quadrature is a means for numerical integration, which evaluates an integral as the sum of a finite number of terms:

where φi is the value of φ(ξ) at ξ=ξi . ξi is called a Gauss point.Wi is the weight of the function value at that Gauss point.

■ Hence, the integral is approximated by the weighted sum of function values at specially selected locations.

in

iiWdI φξξφ ∑≈∫=

=− 1

1

1 )(

2

Order of Gauss Quadrature ■ n, the number of terms in the summation, is the order of the Gauss quadrature.

3

Example: Polynomial Integrand■ Prob. 4.10a:

3/2 )( :1

1

32 =∫ +=−

ξξξ dIExact

0)0(2 :1 11 ==≈= φφWIn

3/2)3/3(22)()( :2 222211 ===+−=+≈= aaaWWIn φφφφ

321332211 95

98

95 :3 φφφφφφ ++=++≈= WWWIn

( ) 3/296.0109/10)(95)0(

98)(

95 22 ===++−=⇒ bbbI φφφ

■ Hence, Gauss rules of order 2 and 3 give the exact value.

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Example: Non-polynomial Integrand ■ If φ =φ(ξ) is a polynomial, n-point (nth order) Gauss quadrature yields the exact integral if φ is of degree 2n-1 or less. (A cubic polynomial is exactly integrated by a two-point rule, three-point rule, etc.)

■ If the integrand is not a polynomial, Gauss quadrature yields an approximate result, with better accuracy for higher n.

■ Prob. 4.10b: 1

1

: cos1.5 (4 / 3)sin1.5 1.330Exact I dξ ξ−

= = =∫

error % 50.4 ; 20cos2)0(2 :1 ==≈= φIn

error % 2.6- ; 296.15.1cos2)()( :2 ==+−≈= aaaIn φφ

error % 0.05 ; 331.10cos985.1cos

910

95

98

95 :3 321 =+=++≈= bIn φφφ

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Gauss Quadrature In Two dimensions ■ Integration is over a quadrilateral.

Wi Wj is the product of one-dimensional weights. Usually, m=n.

■ If m=n=1, ■ Below are Gauss points for four-point and nine-point rules.

)0,0(44 1 φφ =≈I

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Four- and Nine-Point Rules ■ Integrals for four-point and nine-point rules are

■ The 8x8 stiffness matrix of an isoparametric quadrilateral can thus be obtained by using Gauss quadrature. Because of symmetry, 36 entries of the matrix need to be evaluated. The number of Gauss points to be used should neither be too many nor too few.

4321 φφφφ +++≈I

586429731 8164)(

8140)(

8125

φφφφφφφφφ ++++++++≈I

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2-D Example ■ Use one- and four-point rules to evaluate the integral

ηξηξ ddI

211

1

1

1∫ ∫ +

+=

− −

■ Solution:197.23ln2

22 :Exact

1

1==∫ +

=− η

ηdI

error % 9 ; 202014)0,0(4 :rulepoint One =

++

=≈− φI

33 whereerror, % 0.7-

; 182.221

21

21

21 :rulepoint Four

=

=+++

−++

+−+

−−≈−

b

bb

bb

bb

bbI

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Stress Calculation

■ Stresses calculated from

are most accurate at Gauss points that were used for integrationof the stiffness matrix.

■ Usually an order two Gauss rule (four points) are used to integrate k of 4- and 8-node plane elements.

■ Eight points are commonly used for 3-D elements.

■ Strains and stresses are calculated at those points. Stresses at nodes and other locations are obtained by extrapolation or interpolation from Gauss point values.

EBd=σ

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Symmetry

■ Symmetry types:

◆ Reflective (mirror) symmetry◆ Skew symmetry◆ Axial symmetry◆ Cyclic symmetry

■ Exploitation of symmetry reduces the FE model size and the computational cost.

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Reflective (Mirror) Symmetry ■ Symmetry of geometry, loads, support conditions, AND elastic properties w.r.t. a plane.

■ The structure shown has reflective symmetry about yz and xz planes.

■ Analysis of one quarter is adequate. Px/2 and Py/2 are applied as loads.

■ Supports are introduced on the symmetry planes.

■ At these supports, motion is not allowed normal to the symmetry plane.

(v=0 @ y=0 and u=0 @ x=0)

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Reflective Symmetry (cont.)

■ In the previous structure if Px=Py , the planes x=y and x=-y are also planes of symmetry.

■Then, only one octant of the structure need be analyzed.

■ Motion normal to the plane x=y is prevented.

■ Motion normal to the plane y=0 is also prevented.

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Reflective Symmetry and Antisymmetry in Beams

■ Everything is symmetric in beam (a) so only half need be analyzed with rotation about the z-axis prevented at x=0.

■ The problem in (b) is antisymmetric. Again half of the structure is adequate with transverse displacement prevented at x=0 but rotation allowed.

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Summary of Correct Support Conditions ■ Support conditions to be applied to the boundary nodes in a plane of reflective symmetry:

◆ No translation normal to the plane of symmetry◆ No rotation about the axes that define the symmetry plane

■ Support conditions to be applied to the boundary nodes in an antisymmetric problem:

◆ No translation parallel to the plane of antisymmetry◆ No rotation about the axis that is normal to the plane of

antisymmetry

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Use of Symmetry Concepts When Loading is not Symmetric

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Skew Symmetry (Inversion Symmetry)

■ In (a), a half revolution of the structure and loads about the z-axis results in self-coincidence.■ In (b), the load on the revolved half needs to be reversed also to give self-coincidence.

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Axial Symmetry and Cyclic Symmetry

■ Axial symmetry is present when a solid is generated by rotation of a plane shape about an axis in the plane. (A cylindrical tank, dish antenna, ...)

■ Cyclic symmetry is present when a structure that is not axially symmetric exhibits a rotational repetition of geometry, materialproperties, supports and loads.

■ Periodic structures also have repetition of geometry, etc. as in some long slender structures such as long-body transport aircraft.

◆ Analysis of one repetitive portion is adequate for a periodic structure or when cyclic symmetry is present

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Example on Cyclic Symmetry

■ Boundary CD is a repetition of AB ⇒ dof along these two boundaries in (b) must match exactly (in number, type, constraints, etc.) (Displacements must match in respective n and s directions.)