Gaussian Quadrature in FEM

Project report on

Application of Gaussain quadrature inFEM

Submitted byKhaja Mohiuddin(2010A2PS353H)

Under supervision ofDr. Chandu Parimi

Department of Civil EngineeringBITS-Pilani Hyderabad Campus

November 2012

ACKNOWLEDGEMENT

I would like to express my sincere thanks and gratitude to Dr. Chandu Parimi forhis continuous and unfailing support, guidance and help, which have beeninvaluable during the course of this project. His knowledge, insight and constantmotivation at each step of the project has been instrumental in its completion.

S.M.Khaja Mohiuddin(2010A2PS353H)

Table of contents

1. Introduction 042. Gaussian quadrature 053. Triangular co-ordinates 064. Shape functions 095. Gaussian quadrature for triangles 116. Sample code 127. Rectangular elements 158. sample code 179. Tetrahedral elements 1810. Gaussian quadrature for tetrahedral elements 2011. sample code 21

References 22

Introduction

The Finite Element Method (FEM) is a numerical technique to findapproximate solutions of partial differential equations.It was originated from theneed of solving complex elasticity and structural analysis problems in Civil andother engineering branches. FEM allows for detailed visualization and indicatesthe distribution of stresses and strains inside the body of a structure.Thispowerful design tool has significantly improved both the standard of engineeringdesigns and the methodology of the design process in many industrialapplications.

The basic finite element problem can be divided into following steps1. Discretisation of the continuum: The continuum is divided into a numberof elements by imaginary lines or surfaces.2. Identification of variables: The elements are assumed to be connected attheir intersecting points referred to as nodal points. At each node, unknowndisplacements are to be prescribed.3. Choice of approximating functions: Displacement function is the startingpoint of the mathematical analysis. This represents the variation of thedisplacement within the element.The displacement function may beapproximated in the form a linear function or a higherorder function.Aconvenient way to express it is by polynomial expressions. The shape orgeometry of the element may also be approximated.4. Formation of the element stiffness matrix: After continuum is discretisedwith desired element shapes, the individual element stiffness matrix isformulated.The geometry of the element is defined inreference to the global frame. Coordinate transformation must be done forelements where it is necessary.5. Formation of overall stiffness matrix: After the element stiffness matricesin global coordinates are formed, they are assembled to form the overallstiffness matrix. The assembly is done through the nodes which are common toadjacent elements. The overall stiffness matrix is symmetric and banded.6. Formation of the element loading matrix: The loading forms an essentialparameter in any structural engineering problem. The loading inside an elementis transferred at the nodal points and consistent element matrix is formed.7. Formation of the overall loading matrix: Like the overall stiffness matrix,the element loading matrices are assembled to form the overall loading matrix.This matrix has one column per loading case and it is either a column vector ora rectangular matrix depending on the number of loading cases.8. Incorporation of boundary conditions: The boundary restraint conditionsare to be imposed in the stiffness matrix. There are various techniques availableto satisfy the boundary conditions.9. Solution of simultaneous equations: The unknown nodal displacementsare calculated by the multiplication of force vector with the inverse of stiffnessmatrix.10. Calculation of stresses or stress-resultants: Nodal displacements areutilized for the calculation of stresses or stress-resultants.

The basic finite element problem can be divided into following steps1. Discretisation of the continuum: The continuum is divided into a numberof elements by imaginary lines or surfaces.2. Identification of variables: The elements are assumed to be connected attheir intersecting points referred to as nodal points. At each node, unknowndisplacements are to be prescribed.3. Choice of approximating functions: Displacement function is the startingpoint of the mathematical analysis. This represents the variation of thedisplacement within the element.The displacement function may beapproximated in the form a linear function or a higherorder function.Aconvenient way to express it is by polynomial expressions. The shape orgeometry of the element may also be approximated.4. Formation of the element stiffness matrix: After continuum is discretisedwith desired element shapes, the individual element stiffness matrix isformulated.The geometry of the element is defined inreference to the global frame. Coordinate transformation must be done forelements where it is necessary.5. Formation of overall stiffness matrix: After the element stiffness matricesin global coordinates are formed, they are assembled to form the overallstiffness matrix. The assembly is done through the nodes which are common toadjacent elements. The overall stiffness matrix is symmetric and banded.6. Formation of the element loading matrix: The loading forms an essentialparameter in any structural engineering problem. The loading inside an elementis transferred at the nodal points and consistent element matrix is formed.7. Formation of the overall loading matrix: Like the overall stiffness matrix,the element loading matrices are assembled to form the overall loading matrix.This matrix has one column per loading case and it is either a column vector ora rectangular matrix depending on the number of loading cases.8. Incorporation of boundary conditions: The boundary restraint conditionsare to be imposed in the stiffness matrix. There are various techniques availableto satisfy the boundary conditions.9. Solution of simultaneous equations: The unknown nodal displacementsare calculated by the multiplication of force vector with the inverse of stiffnessmatrix.10. Calculation of stresses or stress-resultants: Nodal displacements areutilized for the calculation of stresses or stress-resultants.

The integrations, we generally encounter in finite element methods, arequite complicated and it is not possible to find a solution to those problems. Exactevaluation of the integral associated to the element matrices and the loadingvector is not always possible because of the algebraic complexity of thecoefficient of the different equation (i.e., the stiffness influence coefficients,elasticity matrix, loading functions etc.) so we use numerical integratio. In thefinite element analysis, we face the problem of evaluating the following types ofintegrations

[k] = zyxBT( ) D( ) B( )

d

d

d

here B is strain displacement matrix and D is constituent relationship matrix

to find elemental stiffness matrix [k].Approximate solutions to such problems arepossible using certain numerical techniques. Several numerical techniques areavailable, in mathematics for solving definite integration problems,including,mid-point rule, trapezoidal-rule, Simpson’s 1/3rd rule, Simpson’s 3/8th rule andGauss Quadrature formula. Among these, Gauss Quadrature technique is mostuseful one for solving problems in finite element method.

Gauss Quadrature

Let us start the discussion with one dimensional gaussain quadrature.The n pointgaussain quadrature can be represented as follows for a definite integral from -1 to1.

1−

1

xf x( )

d

1

n

i

wif xi( )( )=

=

where xi are quadrature points and wi are weights of those points. The locationand weight coefficients of Gauss points are calculated by Legendre polynomials.Hence this method is also sometimes referred as Gauss-Legendre Quadraturemethod.The summation of these values at nsampling points gives the exact solution of a polynomial integrand of an order upto 2n-1. For example, considering sampling at two Gauss points we can get exactsolution for a polynomial of an order (2×2-1) or 3. The use of more number ofGauss points has no effect on accuracy of results but takes more computationtime.

The following table gives the values of wi and xi for n = 1 to 6 and theintegral should be evaluated from -1 to 1.

Number of

points, nGauss Point Location, x1 Weight, w1

1 0.0 2.0

2 ±0.5773502692 1.0

3 0.0 0.8888888889 (=8/9)

±0.7745966692 0.5555555556 (=5/9)

4 ±0.3399810436 0.6521451549

±0.861363116 0.3478548451

5 0.0 0.5688888889

±0.5384693101 0.4786286705

±0.9061798459 0.2369268851

6 ±0.2386191861 0.4679139346

±0.6612093865 0.3607615730

±0.9324695142 0.1713244924

For two dimensional integration problems the above mentioned method can beextended by first evaluating the inner integral, keeping η constant, and thenevaluating the outer integral.

1−

1

y1−

1

xf x y,( )

d

d

1

n

i 1

n

j

wiwjf xi yj,( )( )=

=

=

similarly for 3-d integration.

For two dimensional integration problems the above mentioned method can beextended by first evaluating the inner integral, keeping η constant, and thenevaluating the outer integral.

1−

1

y1−

1

xf x y,( )

d

d

1

n

i 1

n

j

wiwjf xi yj,( )( )=

=

=

similarly for 3-d integration.

As we have seen now the integration for a function can be done easily bygaussain quadrature if the integrals are from -1 to 1, but once the integral limitschange applying quadrature rules become difficult.We can change the limits for a one dimensional problem directly as follows

a

b

xf x( )

db a−( )

2=

1

n

i

wifb a−( )zi a b+( )+

2

=It is not that easy for a 2-d integral as it may have variable limits.To simplify theapplication of quadrature rules to 2-d and 3-d integrals we use natural co-ordinatesystem.

Natural Co-ordinate system

Natural coordinate system is basically a local coordinate system which allows thespecification of a point within the element by a set of dimensionless numberswhose magnitude never exceeds unity.This coordinate system is found to be very effective in formulating the elementproperties in finite element formulation. This system is defined in such that themagnitude at nodal points will have unity or zero or a convenient set of fractions.It also facilitates the integration to calculate element stiffness.

Triangular elements:The natural coordinate system for a triangular element is generally called

as triangular coordinate system. The coordinate of any point P inside the triangleis x,y in Cartesian coordinate system.Here the co-ordinates ,L1,L2,L3 can beused to define the location of points in terms of natural co-ordinates.The point pcan be defined by the following set of natural co-ordinates

L1 = A1/A ;L2 = A2/A ;L3 = A3/A ;

where A1,A2,A3,A are the area of the triangles P23,P13,P12,123It can be seen that A = A1+A2+A3;

so L1+L2+L3 = 1;Therefore, the natural coordinate of three nodes will be: node 1 (1,0,0); node 2(0,1,0); and node 3 (0,0,1).

Shape functions

Shape Function using Area CoordinatesNow as we have a understanding about triangular co-ordinate system we

can move to shape functions.Shape functions are functions which interpolate thepoints in the geometry in terms of their nodes.So if N1,N2,N3 are shape functionsthe we can write

x = N1 x1 +N2 x2 +N3 x3

y = N1 y1 +N2 y2 +N3 y3

Its is important to know that area co-ordinates are dependent and L1+L2+L3 = 1and also that L1 = 1 while L2,L3=0 at node 1 and L2 = 1 while L1,L3=0 at node 2and L3 = 1 while L2,L1 =0.So we can write any point p in vector notations as alinear combination of L1,L2,L3

P = a1L1 + a2L2 + a3L3;and considering nodes

x1

x2

x3

1

0

0

0

1

0

0

0

1

=a1

a2

a3

thusa1

a2

a3

1

0

0

0

1

0

0

0

1

=x1

x2

x3

so

P L1 L2 L3( )=1

0

0

0

1

0

0

0

1

x1

x2

x3

so we can say see thatx = L1 x1 +L2 x2 +L3 x3y = L1 y1 +L2 y2 +L3 y3

thus the shape functions for a 3 noded triangle are L1,L2,L3 itself .

Six noded triangleThe below figure shows a triangular element with six nodes. The additional threenodes (4, 5, and 6) are situated at the midpoints of the sides of the element.

Using the above field variable function, one can reach the following expressionusing interpolation function and the nodal values.

x = N1 x1 +N2 x2 +N3 x3 +N4 x4 +N5 x5 +N6 x6y = N1 y1 +N2 y2 +N3 y3+N5 y4 +N5 y5 +N6 y6

Here, the every shape function must be such that its value will be unity ifevaluated at its related node and zero if evaluated at any of the other fivenodes. Now the interpolation functions can be constructed with the helpof area coordinates from the above diagram. For example, the interpolationfunction N should be unity at node 1 and zero at all other five nodes.According to the above diagram, the value of L1 is 1 at node 1 and ½ at node 4and 6. Again, L will be 0 at nodes 2, 3 and 5. To satisfy all these conditions,one can propose following expression

N1(x,y) = N1(L1,L2,L3) = L1(L1-1/2)Evaluating the above expression, the value of N is becoming ½ at node 1though it must become unity. Therefore, the above expression is slightlymodified satisfying all the conditions and will be as follows:

N1(L1,L2,L3) = 2*L1(L1-1/2)=L1(2 L1 - 1)the above equation assures the required conditions at all the six nodes and is aquadratic function, as L is a linear function of x and y. The remaining fiveinterpolation functions can also be obtained in similar fashion applying therequired nodal conditions. Thus, the shape function for the six node triangleelement can be written as given below.

N1 = L1(2 L1 - 1)N2 = L2(2 L2 - 1)N3 = L2(2 L3 - 1)N4 = 4L1L2N5 = 4L2L3N6 = 4L3L1

Using the above field variable function, one can reach the following expressionusing interpolation function and the nodal values.

x = N1 x1 +N2 x2 +N3 x3 +N4 x4 +N5 x5 +N6 x6y = N1 y1 +N2 y2 +N3 y3+N5 y4 +N5 y5 +N6 y6

Here, the every shape function must be such that its value will be unity ifevaluated at its related node and zero if evaluated at any of the other fivenodes. Now the interpolation functions can be constructed with the helpof area coordinates from the above diagram. For example, the interpolationfunction N should be unity at node 1 and zero at all other five nodes.According to the above diagram, the value of L1 is 1 at node 1 and ½ at node 4and 6. Again, L will be 0 at nodes 2, 3 and 5. To satisfy all these conditions,one can propose following expression

N1(x,y) = N1(L1,L2,L3) = L1(L1-1/2)Evaluating the above expression, the value of N is becoming ½ at node 1though it must become unity. Therefore, the above expression is slightlymodified satisfying all the conditions and will be as follows:

N1(L1,L2,L3) = 2*L1(L1-1/2)=L1(2 L1 - 1)the above equation assures the required conditions at all the six nodes and is aquadratic function, as L is a linear function of x and y. The remaining fiveinterpolation functions can also be obtained in similar fashion applying therequired nodal conditions. Thus, the shape function for the six node triangleelement can be written as given below.

N1 = L1(2 L1 - 1)N2 = L2(2 L2 - 1)N3 = L2(2 L3 - 1)N4 = 4L1L2N5 = 4L2L3N6 = 4L3L1

Gauss Quadrature for Triangular Elements

As we have changed the co-ordinate system in triangular elements totriangular co-ordinates the quadrature also changes as follows

I yxF L1 L2, L3,( )

d

d=1

n

i

wif L1i L2i, L3i,( )( )=

= J

Where,J is jacobian to transform dx,dy and is equal to area for triangular systemL terms are the triangular area coordinates and the w terms are the weightsassociated withthose coordinates. The locations of integration points are shown infigure

For a linear triangle(ie. one sampling point)w1 = 1 L1 = 1/3 L2 = 1/3 L3 = 1/3

For a quadratic triangle (ie. n=3)w1 = 1/3 L1 = 1/2 L2 = 1/2 L3 = 0w1 = 1/3 L1 = 0 L2 = 1/2 L3 = 1/2w1 = 1/3 L1 = 1/2 L2 = 0 L3 = 1/2

For a cubic triangle (ie.n=7)w1 = 27/60 L1 = 1/3 L2 = 1/3 L3 = 1/3w1 = 8/60 L1 = 1/2 L2 = 1/2 L3 = 0w1 = 8/60 L1 = 0 L2 = 1/2 L3 = 1/2w1 = 8/60 L1 = 1/2 L2 = 0 L3 = 1/2w1 = 3/60 L1 = 1 L2 = 1 L3 = 0w1 = 3/60 L1 = 0 L2 = 1 L3 = 1w1 = 3/60 L1 = 1 L2 = 0 L3 = 1

Code for a 3 node triangle

In the code below

x1,y1,x2,y2,x3,y3 are x,y co -ordinates for a 3 noded trianglen Number of sampling pointsf Function to be integrated

function out = shape(x1,y1,x2,y2,x3,y3,n,f);if(n==1) mata = [1 1 1;x1 x2 x3;y1 y2 y3]; q = mata*[1/3;1/3;1/3]; area = det(mata)/2;

if(area<0) area = area*(-1);

end notarea = f(q(2,1),q(3,1)); out = area*notarea;endif(n==3) mata = [1 1 1;x1 x2 x3;y1 y2 y3]; area = det(mata)/2;

if(area<0) area = area*(-1);

end q1 = mata*[0.5;0.5;0]; q2 = mata*[0;0.5;0.5]; q3 = mata*[0.5;0;0.5]; notarea =(1/3*f(q1(2,1),q1(3,1)))+(1/3*f(q2(2,1),q2(3,1)))+(1/3*f(q3(2,1),q3(3,1))); out = area*notarea;endif(n==7) mata = [1 1 1;x1 x2 x3;y1 y2 y3]; area = det(mata)/2;

if(area<0) area = area*(-1);

end q1 = mata*[1/3;1/3;1/3]; q2 = mata*[0.5;0.5;0]; q3 = mata*[0;0.5;0.5]; q4 = mata*[0.5;0;0.5]; q5 = mata*[1;0;0]; q6 = mata*[0;1;0]; q7 = mata*[0;0;1]; notarea =(27/60*f(q1(2,1),q1(3,1)))+(8/60*f(q2(2,1),q2(3,1)))+(8/60*f(q3(2,1),q3(3,1)))+(8/60*f(q4(2,1),q4(3,1)))+(3/60*f(q5(2,1),q5(3,1)))+(3/60*f(q6(2,1),q6(3,1)))+(3/60*f(q7(2,1),q7(3,1))); out = area*notarea;end

function out = shape(x1,y1,x2,y2,x3,y3,n,f);if(n==1) mata = [1 1 1;x1 x2 x3;y1 y2 y3]; q = mata*[1/3;1/3;1/3]; area = det(mata)/2;

if(area<0) area = area*(-1);

end notarea = f(q(2,1),q(3,1)); out = area*notarea;endif(n==3) mata = [1 1 1;x1 x2 x3;y1 y2 y3]; area = det(mata)/2;

if(area<0) area = area*(-1);

end q1 = mata*[0.5;0.5;0]; q2 = mata*[0;0.5;0.5]; q3 = mata*[0.5;0;0.5]; notarea =(1/3*f(q1(2,1),q1(3,1)))+(1/3*f(q2(2,1),q2(3,1)))+(1/3*f(q3(2,1),q3(3,1))); out = area*notarea;endif(n==7) mata = [1 1 1;x1 x2 x3;y1 y2 y3]; area = det(mata)/2;

if(area<0) area = area*(-1);

end q1 = mata*[1/3;1/3;1/3]; q2 = mata*[0.5;0.5;0]; q3 = mata*[0;0.5;0.5]; q4 = mata*[0.5;0;0.5]; q5 = mata*[1;0;0]; q6 = mata*[0;1;0]; q7 = mata*[0;0;1]; notarea =(27/60*f(q1(2,1),q1(3,1)))+(8/60*f(q2(2,1),q2(3,1)))+(8/60*f(q3(2,1),q3(3,1)))+(8/60*f(q4(2,1),q4(3,1)))+(3/60*f(q5(2,1),q5(3,1)))+(3/60*f(q6(2,1),q6(3,1)))+(3/60*f(q7(2,1),q7(3,1))); out = area*notarea;end

Code for a 6-noded triangle

In the code below

x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6 are x,y co -ordinates for a 6noded triangle the order of co-ordinate input is important first 3 points arecorner points given in counter clockwise direction and next 3 points are midpoints given in counter clockwise direction.n Number of sampling pointsf Function to be integrated

function out = shape6(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,n,f);if(n==1) mata = [1 1 1 1 1 1;x1 x2 x3 x4 x5 x6;y1 y2 y3 y4 y5 y6]; [n1,n2,n3,n4,n5,n6] = ns(1/3,1/3,1/3); q = mata*[n1;n2;n3;n4;n5;n6] matb = [1 1 1;x1 x2 x3;y1 y2 y3]; area = det(matb)/2;l1=1/3;l2=1/3;l3=1/3;

if(area<0) area = area*(-1);

end notarea = f(q(2,1),q(3,1)); out = notarea*area;endif(n==3) mata = [1 1 1 1 1 1;x1 x2 x3 x4 x5 x6;y1 y2 y3 y4 y5 y6]; matb = [1 1 1;x1 x2 x3;y1 y2 y3]; area = det(matb)/2;

if(area<0) area = area*(-1);

end [n1,n2,n3,n4,n5,n6] = ns(0.5,0.5,0); q1 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0,0.5,0.5); q2 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0.5,0,0.5); q3 = mata*[n1;n2;n3;n4;n5;n6]; notarea =(1/3*f(q1(2,1),q1(3,1)))+(1/3*f(q2(2,1),q2(3,1)))+(1/3*f(q3(2,1),q3(3,1))); out = area*notarea;endif(n==7) mata = [1 1 1 1 1 1;x1 x2 x3 x4 x5 x6;y1 y2 y3 y4 y5 y6]; matb = [1 1 1;x1 x2 x3;y1 y2 y3]; area = det(matb)/2;

if(area<0) area = area*(-1);

end [n1,n2,n3,n4,n5,n6] = ns(1/3,1/3,1/3); q1 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0.5,0.5,0); q2 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0,0.5,0.5); q3 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0.5,0,0.5); q4 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(1,0,0); q5 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0,1,0); q6 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0,0,1); q7 = mata*[n1;n2;n3;n4;n5;n6]; notarea =(27/60*f(q1(2,1),q1(3,1)))+(8/60*f(q2(2,1),q2(3,1)))+(8/60*f(q3(2,1),q3(3,1)))+(8/60*f(q4(2,1),q4(3,1)))+(3/60*f(q5(2,1),q5(3,1)))+(3/60*f(q6(2,1),q6(3,1)))+(3/60*f(q7(2,1),q7(3,1))); out = area*notarea;endendfunction [n1,n2,n3,n4,n5,n6]= ns(l1,l2,l3)n1 = l1*(2*l1-1);n2 = l2*(2*l2-1);n3 = l3*(2*l3-1);n4 = 4*l1*l2;n5 = 4*l2*l3;n6 = 4*l3*l1;end

function out = shape6(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,n,f);if(n==1) mata = [1 1 1 1 1 1;x1 x2 x3 x4 x5 x6;y1 y2 y3 y4 y5 y6]; [n1,n2,n3,n4,n5,n6] = ns(1/3,1/3,1/3); q = mata*[n1;n2;n3;n4;n5;n6] matb = [1 1 1;x1 x2 x3;y1 y2 y3]; area = det(matb)/2;l1=1/3;l2=1/3;l3=1/3;

if(area<0) area = area*(-1);

end notarea = f(q(2,1),q(3,1)); out = notarea*area;endif(n==3) mata = [1 1 1 1 1 1;x1 x2 x3 x4 x5 x6;y1 y2 y3 y4 y5 y6]; matb = [1 1 1;x1 x2 x3;y1 y2 y3]; area = det(matb)/2;

if(area<0) area = area*(-1);

end [n1,n2,n3,n4,n5,n6] = ns(0.5,0.5,0); q1 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0,0.5,0.5); q2 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0.5,0,0.5); q3 = mata*[n1;n2;n3;n4;n5;n6]; notarea =(1/3*f(q1(2,1),q1(3,1)))+(1/3*f(q2(2,1),q2(3,1)))+(1/3*f(q3(2,1),q3(3,1))); out = area*notarea;endif(n==7) mata = [1 1 1 1 1 1;x1 x2 x3 x4 x5 x6;y1 y2 y3 y4 y5 y6]; matb = [1 1 1;x1 x2 x3;y1 y2 y3]; area = det(matb)/2;

if(area<0) area = area*(-1);

end [n1,n2,n3,n4,n5,n6] = ns(1/3,1/3,1/3); q1 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0.5,0.5,0); q2 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0,0.5,0.5); q3 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0.5,0,0.5); q4 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(1,0,0); q5 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0,1,0); q6 = mata*[n1;n2;n3;n4;n5;n6]; [n1,n2,n3,n4,n5,n6] = ns(0,0,1); q7 = mata*[n1;n2;n3;n4;n5;n6]; notarea =(27/60*f(q1(2,1),q1(3,1)))+(8/60*f(q2(2,1),q2(3,1)))+(8/60*f(q3(2,1),q3(3,1)))+(8/60*f(q4(2,1),q4(3,1)))+(3/60*f(q5(2,1),q5(3,1)))+(3/60*f(q6(2,1),q6(3,1)))+(3/60*f(q7(2,1),q7(3,1))); out = area*notarea;endendfunction [n1,n2,n3,n4,n5,n6]= ns(l1,l2,l3)n1 = l1*(2*l1-1);n2 = l2*(2*l2-1);n3 = l3*(2*l3-1);n4 = 4*l1*l2;n5 = 4*l2*l3;n6 = 4*l3*l1;end

Rectangular elements

It is easy to define the new co-ordinate system for rectangular elements. Justa change of origin will do the job. Let the co-ordinates in new system be η andξ.

Now we define the interpolation function of the element in natural coordinatesystem and can be found as:

Ni

N1

N2

N3

N4

1 ξ−( )1 η−( )

4

1 ξ+( )1 η−( )

4

1 ξ+( )1 η+( )

4

1 ξ−( )1 η+( )

4

=:=

These can found by using simple lagrange interpolation theorem

So now x and y can be written as

x N1 x1 N2 x2+ N3 x3+ N4 x4+=y N1 y1 N2 y2+ N3 y3+ N4 y4+=

We have got x,y in terms of nodes now we should convert dxdy to dηdξ thiscan be done by the introduction of jacobian term defined as follows

So I can be evaluated using gaussain quadrature as follows

I1−

1

ξ1−

1

ηϕ η ξ,( ) J

d

d=1

n

i 1

n

j

wiwj( ) ϕ η ξ,( ) J =

=

=

Sample code for a four noded rectangular element

Here x1,y1,x2,y2,x3,y3,x4,y4 are co ordinates of the givenrectangular element in anti-clockwise directionn Number of pointsf Function to be integrated

function out = shapequad(x1,y1,x2,y2,x3,y3,x4,y4,n,f);[r,cnst] = lgwt(n,-1,1);integretion = 0;for i=1:n jx=0; l1= r(i,1);

for j=1:n l2 = r(j,1); [n1,n2,n3,n4,jaco]=ns(l1,l2,x1,y1,x2,y2,x3,y3,x4,y4); x=n1*x1+n2*x2+n3*x3+n4*x4; y=n1*y1+n2*y2+n3*y3+n4*y4; q = jaco*f(x,y); jx = jx + (cnst(j,1)*q);

end integretion = integretion + (cnst(i,1)*jx);endout = integretion;endfunction [n1,n2,n3,n4,j] = ns(l1,l2,x1,y1,x2,y2,x3,y3,x4,y4)n1 = 0.25*(1-l1)*(1-l2);n2 = 0.25*(1+l1)*(1-l2);n3 = 0.25*(1+l1)*(1+l2);n4 = 0.25*(1-l1)*(1+l2);op=[-0.25*(1-l1) 0.25*(1-l1)0.25*(1+l1) -0.25*(1+l1);-0.25*(1-l2) -0.25*(1+l2) 0.25*(1+l2)0.25*(1-l2)];oq = [x1 y1;x2 y2;x3 y3;x4 y4];j = op*oq;j = det(j);end

Tetrahedral elements

There are two basic families of three-dimensional elements similar to two -dimensional case.Extension of triangular elements will produce tetrahedrons inthree dimensions. Similarly, rectangular parallelepipeds are generated on theextension of rectangular elements. Figure shows few commonly used solidelements for finite element analysis.

Tetrahedral elements

The simplest element of the tetrahedral family is a four node tetrahedron asshown in figure. The node numbering has been followed in sequential manner,i.e, in this case anti-clockwise direction. Similar to the area coordinates, theconcept of volume coordinates has been introduced here. The coordinates of thenodes are defined both in Cartesian and volume coordinates. Point P(x, y, and z)as shown in below figure is an arbitrary point in the tetrahedron.

The linear shape function for this element can be expressed as,

The linear shape function for this element can be expressed as,

NT = [ L1 L2 L3 L4]

Here L1,L2,L3,L4 are natural co ordinates defined as follows

LiVi

V=

Where Vi is the volume of the sub element which is bound by point P and face iand V is the total volume of the element. For example Li may be interpreted asthe ratio of the volume of the sub element P234 to the total volume of theelement 1234. The volume of the element V is given by the determinant of thenodal coordinates as follows:

V1

6

1

x1

y1

z1

1

x2

y2

z2

1

x3

y3

z3

1

x4

y4

z4

:=

The relationship between the Cartesian and natural coordinates of point P maybe expressed as

1

x

y

z

L1

L2

L3

L4

1

x1

y1

z1

1

x2

y2

z2

1

x3

y3

z3

1

x4

y4

z4

:=

so L1 L2 L3 L4 can found by inversing the matrix

Gaussain quadrature for tetrahedral elementsThe Gauss Quadrature for triangles can be effectively extended to includetetrahedron elements in terms of tetrahedron volume coordinates.

The Gauss Quadrature for triangles can be effectively extended to includetetrahedron elements in terms of tetrahedron volume coordinates.

I zyxF L1 L2, L3, L4,( )

d

d

d=1

n

i

wif L1i L2i, L3i, L4i,( )( )=

=

J

Where, L terms are the volume coordinates and the wi terms are the weightsassociated with those coordinates. The locations of Gauss points are shown infigure below

The sampling points and their associated weights are described below:

For sampling point = 1 (Linear tetrahedron)

w1 = 1 L11=L21=L31=L41=1/4

For sampling point = 4 (Quadratic tetrahedron)

w1 = 1/4 L11=0.5854102 L21=L31=L41=0.1381966w2 = 1/4 L22=0.5854102 L12=L32=L42=0.1381966w3 = 1/4 L33=0.5854102 L13=L23=L43=0.1381966w4 = 1/4 L44=0.5854102 L24=L34=L14=0.1381966

For sampling points = 5 (Cubic tetrahedron)

w1 = -4/5 L11=L21=L31=L41=1/4w2 = 9/20 L12=1/3 L22=L32=L42=1/6w3 = 9/20 L24=1/3 L14=L34=L44=1/6w4 = 9/20 L33=1/3 L13=L23=L43=1/6w5 = 9/20 L44=1/3 L24=L34=L14=1/6

Sample code for a tetrahedron

function out = shapetetra(x1,y1,z1,x2,y2,z2,x3,y3,z3,x4,y4,z4,n,f);if(f==1)mata = [1 1 1 1;x1 x2 x3 x4;y1 y2 y3 y4;z1 z2 z3 z4];volume = det(mata)/6;if(volume<0)volume = volume*(-1);endout = volume;elseif(n==1) mata = [1 1 1 1;x1 x2 x3 x4;y1 y2 y3 y4;z1 z2 z3 z4]; q = mata*[1/4;1/4;1/4;1/4]; volume = det(mata)/6;

if(volume<0) volume = volume*(-1);

end notvolume = f(q(2,1),q(3,1),q(4,1)); out = volume*notvolume;endif(n==4) mata = [1 1 1 1;x1 x2 x3 x4;y1 y2 y3 y4;z1 z2 z3 z4]; volume = det(mata)/6;

if(volume<0) volume = volume*(-1);

end q1 = mata*[0.5854102;0.1381966;0.1381966;0.1381966]; q2 = mata*[0.1381966;0.5854102;0.1381966;0.1381966]; q3 = mata*[0.1381966;0.1381966;0.5854102;0.1381966]; q4 = mata*[0.1381966;0.1381966;0.1381966;0.5854102]; notvolume =(1/4*f(q1(2,1),q1(3,1),q1(4,1)))+(1/4*f(q2(2,1),q2(3,1),q2(4,1)))+(1/4*f(q3(2,1),q3(3,1),q3(4,1)))+(1/4*f(q4(2,1),q4(3,1),q4(4,1))); out = volume*notvolume;endif(n==5) mata = [1 1 1 1;x1 x2 x3 x4;y1 y2 y3 y4;z1 z2 z3 z4]; volume = det(mata)/6;

if(volume<0) volume = volume*(-1);

end q1 = mata*[1/4;1/4;1/4;1/4]; q2 = mata*[1/3;1/6;1/6;1/6]; q3 = mata*[1/6;1/3;1/6;1/6]; q4 = mata*[1/6;1/6;1/3;1/6]; q5 = mata*[1/6;1/6;1/6;1/3]; notvolume =((-4/5)*f(q1(2,1),q1(3,1),q1(4,1)))+((9/20)*f(q2(2,1),q2(3,1),q2(4,1)))+((9/20)*f(q3(2,1),q3(3,1),q3(4,1)))+(9/20*f(q4(2,1),q4(3,1),q4(4,1)))+(9/20*f(q5(2,1),q5(3,1),q5(4,1))); out = volume*notvolume;endend

function out = shapetetra(x1,y1,z1,x2,y2,z2,x3,y3,z3,x4,y4,z4,n,f);if(f==1)mata = [1 1 1 1;x1 x2 x3 x4;y1 y2 y3 y4;z1 z2 z3 z4];volume = det(mata)/6;if(volume<0)volume = volume*(-1);endout = volume;elseif(n==1) mata = [1 1 1 1;x1 x2 x3 x4;y1 y2 y3 y4;z1 z2 z3 z4]; q = mata*[1/4;1/4;1/4;1/4]; volume = det(mata)/6;

if(volume<0) volume = volume*(-1);

end notvolume = f(q(2,1),q(3,1),q(4,1)); out = volume*notvolume;endif(n==4) mata = [1 1 1 1;x1 x2 x3 x4;y1 y2 y3 y4;z1 z2 z3 z4]; volume = det(mata)/6;

if(volume<0) volume = volume*(-1);

end q1 = mata*[0.5854102;0.1381966;0.1381966;0.1381966]; q2 = mata*[0.1381966;0.5854102;0.1381966;0.1381966]; q3 = mata*[0.1381966;0.1381966;0.5854102;0.1381966]; q4 = mata*[0.1381966;0.1381966;0.1381966;0.5854102]; notvolume =(1/4*f(q1(2,1),q1(3,1),q1(4,1)))+(1/4*f(q2(2,1),q2(3,1),q2(4,1)))+(1/4*f(q3(2,1),q3(3,1),q3(4,1)))+(1/4*f(q4(2,1),q4(3,1),q4(4,1))); out = volume*notvolume;endif(n==5) mata = [1 1 1 1;x1 x2 x3 x4;y1 y2 y3 y4;z1 z2 z3 z4]; volume = det(mata)/6;

if(volume<0) volume = volume*(-1);

end q1 = mata*[1/4;1/4;1/4;1/4]; q2 = mata*[1/3;1/6;1/6;1/6]; q3 = mata*[1/6;1/3;1/6;1/6]; q4 = mata*[1/6;1/6;1/3;1/6]; q5 = mata*[1/6;1/6;1/6;1/3]; notvolume =((-4/5)*f(q1(2,1),q1(3,1),q1(4,1)))+((9/20)*f(q2(2,1),q2(3,1),q2(4,1)))+((9/20)*f(q3(2,1),q3(3,1),q3(4,1)))+(9/20*f(q4(2,1),q4(3,1),q4(4,1)))+(9/20*f(q5(2,1),q5(3,1),q5(4,1))); out = volume*notvolume;endend

References:WikipediaGoogleNptel video lecturesIntroduction to Finite Element Methods course at University

of colorado at Boulder

References:WikipediaGoogleNptel video lecturesIntroduction to Finite Element Methods course at University

of colorado at Boulder

Number of

points, nGauss Point Location, x1 Weight, w1

1 0.0 2.0

2 ±0.5773502692 1.0

3 0.0 0.8888888889 (=8/9)

±0.7745966692 0.5555555556 (=5/9)

4 ±0.3399810436 0.6521451549

±0.861363116 0.3478548451

5 0.0 0.5688888889

±0.5384693101 0.4786286705

±0.9061798459 0.2369268851

6 ±0.2386191861 0.4679139346

±0.6612093865 0.3607615730

±0.9324695142 0.1713244924

Where Vi is the volume of the sub element which is bound by point P and face iand V is the total volume of the element. For example Li may be interpreted asthe ratio of the volume of the sub element P234 to the total volume of theelement 1234. The volume of the element V is given by the determinant of thenodal coordinates as follows:

The Gauss Quadrature for triangles can be effectively extended to includetetrahedron elements in terms of tetrahedron volume coordinates.

The Gauss Quadrature for triangles can be effectively extended to includetetrahedron elements in terms of tetrahedron volume coordinates.

I zyxF L1 L2, L3, L4,( )

d

d

d=1

n

i

wif L1i L2i, L3i, L4i,( )( )=

=

J

Where, L terms are the volume coordinates and the wi terms are the weightsassociated with those coordinates. The locations of Gauss points are shown infigure below