Finite Element Analysis. Barna Szabó
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Название: Finite Element Analysis

Автор: Barna Szabó

Издательство: John Wiley & Sons Limited

Жанр: Физика

Серия:

isbn: 9781119426462

isbn:

СКАЧАТЬ left-parenthesis k right-parenthesis Baseline upper N Subscript i Baseline right-parenthesis d xi equals sigma-summation Underscript i equals 1 Overscript p Subscript k plus 1 Baseline Endscripts b Subscript i Superscript left-parenthesis k right-parenthesis Baseline r Subscript i Superscript left-parenthesis k right-parenthesis"/>

      which is computed from the given data and the shape functions.

      Example 1.5 Let us assume that f left-parenthesis x right-parenthesis is a linear function on Ik. In this case f left-parenthesis x right-parenthesis can be written as

f left-parenthesis x right-parenthesis equals StartFraction 1 minus xi Over 2 EndFraction f left-parenthesis x Subscript k Baseline right-parenthesis plus StartFraction 1 plus xi Over 2 EndFraction f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis equals f left-parenthesis x Subscript k Baseline right-parenthesis upper N 1 left-parenthesis xi right-parenthesis plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis upper N 2 left-parenthesis xi right-parenthesis period

      Using the Legendre shape functions we have:

StartLayout 1st Row 1st Column r 1 Superscript left-parenthesis k right-parenthesis Baseline equals 2nd Column f left-parenthesis x Subscript k Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 1 squared d xi plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 1 upper N 2 d xi equals StartFraction script l Subscript k Baseline Over 6 EndFraction left-parenthesis 2 f left-parenthesis x Subscript k Baseline right-parenthesis plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis right-parenthesis 2nd Row 1st Column r 2 Superscript left-parenthesis k right-parenthesis Baseline equals 2nd Column f left-parenthesis x Subscript k Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 1 upper N 2 d xi plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 2 squared d xi equals StartFraction script l Subscript k Baseline Over 6 EndFraction left-parenthesis f left-parenthesis x Subscript k Baseline right-parenthesis plus 2 f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis right-parenthesis 3rd Row 1st Column r 3 Superscript left-parenthesis k right-parenthesis Baseline equals 2nd Column f left-parenthesis x Subscript k Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 1 upper N 3 d xi plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis StartFraction script l Subscript k Baseline Over 2 EndFraction integral Subscript negative 1 Superscript plus 1 Baseline upper N 2 upper N 3 d xi 4th Row 1st Column equals 2nd Column minus StartFraction script l Subscript k Baseline Over 6 EndFraction StartRoot three halves EndRoot left-parenthesis f left-parenthesis x Subscript k Baseline right-parenthesis plus f left-parenthesis x Subscript k plus 1 Baseline right-parenthesis right-parenthesis period EndLayout

      Exercise 1.11 Assume that f left-parenthesis x right-parenthesis is a linear function on Ik. Using the Legendre shape functions compute r 4 Superscript left-parenthesis k right-parenthesis and show that r Subscript i Superscript left-parenthesis k right-parenthesis Baseline equals 0 for i greater-than 4. Hint: Make use of eq. (1.55).

      Exercise 1.12 Let

f left-parenthesis x right-parenthesis equals f Subscript k Baseline sine StartFraction x minus x Subscript k Baseline Over script l Subscript k Baseline EndFraction pi comma x element-of upper I Subscript k Baseline

      where fk is a constant. Compute r 5 Superscript left-parenthesis k right-parenthesis numerically in terms of fk and ℓk using 3, 4 and 5 Gauss points. See Appendix E. Use the Legendre basis functions.

      Exercise 1.13 Assume that f left-parenthesis x right-parenthesis is a linear function on I. Using the Lagrange shape functions for p equals 2, compute r 1 Superscript left-parenthesis k right-parenthesis.

      1.3.5 Assembly

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