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EXAMPLE 1.4 Calculation of Principal Stresses
Suppose that the stresses at a point of the body have been computed so that the matrix
T is
151
30
27
T
=
30
190
10
(1)
27
10
79
Find the principal stresses and directions.
The characteristic equation for this matrix is
3
2
|
T
λ
I
| =− λ
+
420
λ
53 900
λ +
2 058 000
=
0
(2)
In factored form this equation becomes
70
)(λ
140
)(λ
210
) =
0
(3)
so that the eigenvalues are
λ
=
70
λ
=
140
λ
=
210
(4)
1
2
3
which are equal to the principal stresses
σ 1 ,
σ 2 , and
σ 3 .
The extreme values of the shear
stresses are found using Eq. (1.91).
The principal directions will be specified by determining a right-handed orthogonal triad
of unit vectors. A vector b is an eigenvector corresponding to the first eigenvalue
λ 1 if and
only if
Tb
= λ 1 b
(5)
or
151 b x
30 b y
27 b z
=
70 b x
30 b x
+
190 b y
+
10 b z
=
70 b y
(6)
27 b x
+
10 b y
+
79 b z
=
70 b z
or
81 b x
30 b y
27 b z =
0
30 b x +
120 b y +
10 b z =
0
(7)
27 b x +
10 b y +
9 b z =
0
The first and third equations are dependent. The first and second equations force the value
of b y to be zero, which may be verified by multiplying the first equation by 10, the second
by 27 and adding the resulting equations. This leaves one condition to be satisfied, namely
30 b x
+
10 b z
=
0or b z
=
3 b x
(8)
Thus, any vector
b
=
b x e x +
b y e y +
b z e z =
b e x +
3 b e z
(9)
with b a nonzero real number is an eigenvector corresponding to the eigenvalue
λ 1 . Usually,
b is set equal to one.
Similar considerations show that any vector
=
+
c
3 c e x
2 c e y
c e z
(10)
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