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0.979(2)
0.490(2)
0.245(2)
0.122(2) kekA =h 0.156
0.157
0.157
0.157
0.157
0.157 Table 1.3.2: Errors in L2 and strain energy for the piecewiselinear nite element solution
of Example 1.3.1. (Numbers in parenthesis indicate a power of 10.) Problems
1. The integral involved in obtaining the mass matrix according to (1.3.13) may, of
course, be done symbolically. It may also be evaluated numerically by Simpson's
rule which is exact in this case since the integrand is a quadratic polynomial. Recall,
that Simpson's rule is Z The mass matrix is 0 h F(x)dx h F(0) + 4F(h=2) + F(h)]:
6
Z xj j1
Mj =
j 1 j ]dx:
j
x;
Using (1.3.4), determine Mj by Simpson's rule to verify the result (1.3.13). The
; ; j 1 use of Simpson's rule may be simpler than symbolic integration for this example
since the trial functions are zero or unity at the ends of an element and one half at
its center.
2. Consider the solution of the linear system AX = F (1.3.24a) 20 Introduction
where F and X are N dimensional vector...
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 Spring '14
 JosephE.Flaherty

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