Information Technology Reference
In-Depth Information
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
−
(λ
−
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)
Search WWH ::
Custom Search