Graphics Programs Reference
In-Depth Information
Hence the diagonalized stress matrix is
96
.
055
00
S =
0 23
.
945
0
0
0 60
where the diagonalterms are the principalstresses.
Solution of Part (2) To c ompute the eigenvectors, westart with Eqs. (9.17) and (9.19),
which yield
1
1
c
=
1
=
1
53518) 2 =
0
.
88168
+
t 2
+
(
0
.
s
=
tc
=
(
0
.
53518)(0
.
88168)
=−
0
.
47186
s
0
.
47186
τ =
=
88168 =−
0
.
25077
1
+
c
1
+
0
.
Weobtain the changes in the transformationmatrix P fromEqs. (9.20).Because P is
initialized to the identitymatrix ( P ii
=
1 and P i j
=
0, i
=
j ) the first equation gives us
P 11 =
P 11
s ( P 12 + τ
P 11 )
=
1
(
0
.
47186) [0
+
(
0
.
25077) (1)]
=
0
.
88167
P 21 =
P 21
s ( P 22 + τ
P 21 )
=
0
(
0
.
47186)[1
+
(
0
.
25077) (0)]
=
0
.
47186
Similarly, the second equation of Eqs. (9.20) yields
P 12 =−
P 22 =
0
.
47186
0
.
88167
The third row and column of P are not affectedbythe transformation. Thus
0
.
88167
0
.
47186 0
P =
0
.
47186
0
.
88167 0
0
0 1
The columnsof P are the eigenvectorsof S .
EXAMPLE 9.2
L
L
2 L
i 2
i 3
i 1
3 C
i 1
C
C
i 2
i 3
Search WWH ::

Custom Search