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Unformatted text preview: 1111117 1, Filamﬂr FXch/Lso as warm Jm +£><1looo< m NLJﬂ’t the mash Q What is We now. aka/mgr,
® 2. :34 44.6 Modal TszvoJluvc x1 @
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@ ® © rum/UH 7 Solution: If we number the triangles starting from the right angle
[[5.2.4];[5.6.2];[1.2.3];[1.4.2]]
the element wise conductivity matrix does not to be recomputed and it is still 1 .8000 0.9000 0.9000
0.9000 0.9000 0
0.9000 0 0.9000 as in exercise 47.
The heat load vector due to internal generation rate changes: nodes 3,6, get
contribution from just one element, nodes 1,4,5 get contribution from two elements, and node 2 gets contribution from four elements. The surface heat transfer matrix and load vector do not change and may be
reused. The solutIon Is Solution from T = exercise 48 32.61
32.0881 £391)
31.0920 [T] = 25'10
34.8659 _ 24.84
257471 Relative change of temperature at Node 2: 21.88
245402 >> (31 .092030.8094)l30.8094*100 ‘
21.8391 ans = 0.9173 ”2., Solvc +2! 41% Tun/UPLra/fmvt in ’l’ht. +n‘m3vJ—W/ dvlmdun
smallest Swvge/‘l' I)": ‘qu Flrr‘mgwm'ovx oh ’11“, w‘jl/ct. triﬂing +an %n=ﬂL(T‘Ta1
r— D .. 3
“3.0 w/m/oK/,e,=5~w/W3/°K/ la=9C, Q—Ioow/am Solution: Since the geometry, the material properties, the internal heat
generation rate, and the boundary conditions are all symmetric with respect
to three symmetry planes, the smallest subset of the mesh that one can
use is a single T3 element in the interior, and a single L2 element on the
boundary. Node Locations
[[0,0];[b12,0];[0,bl2*sin(30/180*pi)]]
Degree of freedom numbers: 1,2,3
Solution 17.8571
7.1429
36.6071 in degrees Celsius 3 Rbfmt EXeru'SL “r51 Ming ‘itvvis «mama/tan, rid—L Tim‘s Ira—Mme rth is SOML‘HMQS mum ”mamQ”. WWW“ ”HUM in‘ltﬁtmﬁvu WICHVIDA ’ﬁwm Ckwtrl'tr lo wombL
use MS demrt rug}. Solution: The integral [(5:2)] = [3(2) cV[N][N]T Az dS is evaluated at three quadrature points. At each ofthose three points
one of the basis functions assumes value 1, while the other two
functions are identically zero there. Therefore the capacity matrix is
evaluated as a diagonal matrix with entries equal to CVAZIS(E)
3. Element wise diagonal capacity matrices would be preferred by the
explicit Eulertime integration method.This method is only
conditionally stable, which means that the time step is typically very
short. Therefore to coverthe time interval of integration one would
need to take a lot of steps, and therefore making each step very
efficient is critical. Since the system matrix on the left is only the
capacity matrix for the explicit Euler method, the solution step can
be made very quickly. No benefit derives from using a diagonal capacity matrix for the
implicit methods discussed in Chapter 10. ...
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 Spring '08
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