Symmetric Matrix Eigenvalue Problems - Numerical Methods in Engineering with MATLAB - page 338

fori=j+1:n
L(i,j) = -dot(L(i,j:i-1), L(j:i-1,j)/L(i,i));

end

end
L(n,n) = 1/L(n,n); Linv = L;

EXAMPLE 9.1

40 MPa

30 MPa

3 0 MPa

80 MPa

60 MPa

The stress matrix (tensor)corresponding to the state of stress shown is

⎡
⎣

⎤
⎦

80 300
3040 0
0060

S = MPa

(each row of the matrix consists of the three stress components acting onacoordinate
plane). It can be shown that the eigenvalues of S are the principal stresses and the
eigenvectors are normal to the principal planes . (1) Determine the principalstresses
by diagonalizing S with a Jacobi rotation and (2)compute the eigenvectors.

Solution of Part (1) To e l iminate S 12 we must applyarotationinthe 1-2plane.With
k

= 1 and = 2Eq. (9.15) is

S 11 − S 22

2 S 12

80 40

2(30)

− 2

3

φ =− =− =−

Equation (9.16a) thenyields

φ −

sgn( ) 1

t =

1 =

+ (2

1 =−

0 . 53518

2

/ /

3) 2

+

| φ | + φ + 2 3

According to Eqs. (9.18), the changes in S due to the rotationare

S 11 = S 11 − tS 12 = 80 − ( − 0 . 53518) ( 30 ) = 96 . 055 MPa

S 22 =

S 22 + tS 12 = 40 + ( − 0 . 53518) (30) = 23 . 945 MPa

S 12 = S 21 = 0