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MANE 4240 & CIVL 4240 Introduction to Finite Elements. Prof. Suvranu De. Numerical Integration in 1D. Reading assignment: Lecture notes, Logan 10.4. Summary: Newton-Cotes Integration Schemes Gaussian quadrature. y. F. x. x. x=L. x=0. 2. 1. d 2x. d 1x. Axially loaded elastic bar.

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mane 4240 civl 4240 introduction to finite elements

MANE 4240 & CIVL 4240Introduction to Finite Elements

Prof. Suvranu De

Numerical Integration in 1D

slide2

Reading assignment:

Lecture notes, Logan 10.4

  • Summary:
  • Newton-Cotes Integration Schemes
  • Gaussian quadrature
slide3

y

F

x

x

x=L

x=0

2

1

d2x

d1x

Axially loaded elastic bar

A(x) = cross section at x

b(x) = body force distribution (force per unit length)

E(x) = Young’s modulus

x2

x1

For each element

Element stiffness matrix

where

slide4

Only for a linear finite element

Element nodal load vector

Question: How do we compute these integrals using a computer?

slide5

Any integral from x1 to x2 can be transformed to the following integral on (-1, 1)

Use the following change of variables

Goal: Obtain a good approximate value of this integral

1. Newton-Cotes Schemes (trapezoidal rule, Simpson’s rule, etc)

2. Gauss Integration Schemes

NOTE: Integration schemes in 1D are referred to as “quadrature rules”

slide6

g(x)

f(1)

f(-1)

f(x)

x

-1

1

Trapezoidal rule: Approximate the function f(x) by a straight line g(x) that passes through the end points and integrate the straight line

slide7

Requires the function f(x) to be evaluated at 2 points (-1, 1)

  • Constants and linear functions are exactly integrated
  • Not good for quadratic and higher order polynomials
  • How can I make this better?
slide8

f(1)

g(x)

f(x)

f(-1)

f(0)

x

-1

1

Simpson’s rule: Approximate the function f(x) by a parabola g(x) that passes through the end points and through f(0) and integrate the parabola

slide9

Requires the function f(x) to be evaluated at 3 points (-1,0, 1)

  • Constants, linear functions and parabolas are exactly integrated
  • Not good for cubic and higher order polynomials

How to generalize this formula?

slide10

Integration point

Weight

Notice that both the integration formulas had the general form

Trapezoidal rule:

M=2

Accurate for polynomial of degree at most 1 (=M-1)

Simpson’s rule:

M=3

Accurate for polynomial of degree at most 2 (=M-1)

slide11

Generalization of these two integration rules: Newton-Cotes

  • Divide the interval (-1,1) into M-1 equal intervals using M points
  • Pass a polynomial of degree M-1 through these M points (the value of this polynomial will be equal to the value of the function at these M-1 points)
  • Integrate this polynomial to obtain an approximate value of the integral

f(1)

f(x)

f(-1)

g(x)

x

-1

1

slide12

With ‘M’ points we may integrate a polynomial of degree ‘M-1’

exactly.

Is this the best we can do ?

With ‘M’ integration points and ‘M’ weights, I should be able to integrate a polynomial of degree 2M-1 exactly!!

Gauss integration rule

See table 10-1 (p 405) of Logan

slide13

Integration point

Weight

Gauss quadrature

How can we choose the integration points and weights to exactly integrate a polynomial of degree 2M-1?

Remember that now we do not know, a priori, the location of the integration points.

slide14

Example: M=1 (Midpoint qudrature)

How can we choose W1 and x1 so that we may integrate a

(2M-1=1) linear polynomial exactly?

But we want

slide15

Hence, we obtain the identity

For this to hold for arbitrary a0 and a1 we need to satisfy 2 conditions

i.e.,

slide16

For M=1

f(x)

f(0)

g(x)

x

-1

1

  • Midpoint quadrature rule:
  • Only one evaluation of f(x) is required at the midpoint of the interval.
  • Scheme is accurate for constants and linear polynomials (compare with Trapezoidal rule)
slide17

Example: M=2

How can we choose W1 ,W2x1 and x2 so that we may integrate a polynomial of degree (2M-1=4-1=3) exactly?

But we want

slide18

Hence, we obtain the 4 conditions to determine the 4 unknowns (W1 ,W2x1 and x2 )

Check that the following is the solution

slide19

For M=2

f(x)

*

x

*

-1

1

  • Only two evaluations of f(x) is required.
  • Scheme is accurate for polynomials of degree at most 3 (compare with Simpson’s rule)
slide20

Exercise: Derive the 6 conditions required to find the integration points and weights for a 3-point Gauss quadrature rule

Newton-Cotes

Gauss quadrature

1. ‘M’ integration points are necessary to exactly integrate a polynomial of degree ‘2M-1’

2. Less expensive

3. Exponential convergence, error proportional to

1. ‘M’ integration points are necessary to exactly integrate a polynomial of degree ‘M-1’

2. More expensive

slide21

Example

Exact integration

Integrate and check!

Newton-Cotes

To exactly integrate this I need a 4-point Newton-Cotes formula. Why?

Gauss

To exactly integrate this I need a 2-point Gauss formula. Why?

slide22

Gauss quadrature:

Exact answer!

slide23

Comparison of Gauss quadrature and Newton-Cotes for the integral

Newton-Cotes

Gauss quadrature

slide24

2

1

In FEM we ALWAYS use Gauss quadrature

Linear Element

Stiffness matrix

Nodal load vector

Usually a 2-point Gauss integration is used. Note that if A, E and b are complex functions of x, they will not be accurately integrated

slide25

1

3

Quadratic Element

2

Nodal shape functions

You should be able to derive these!

Stiffness matrix

Assuming E and A are constants

slide26

Need to exactly integrate quadratic terms.

Hence we need a 2-point Gauss quadrature scheme..Why?