Symmetric Matrix Eigenvalue Problems - Numerical Methods in Engineering with MATLAB - page 339

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