CONTINUUM MECHANICS for ENGINEERS

1
Obviously, n from the second of these equations, and from the other
2 0
two, ( 1)
4 ( 1)
n so that, from the normalizing condition, nini = 1, we see
3 =− n1
3
2
( 1)
9
( 1)
3
( 1)
4
that ( n1 ) = which gives n1 =± and n3 = m . The fact that the first
25
5
5
and third equations result in the same relationship is the reason the normalizing
condition must be used.
Next for σ (2) = 50, Eq 3.6-3 gives
2 2
which are satisfied only when n1= n3=
0 . Then from the normalizing
( 2)
condition, nini = 1, n2 =± 1.
Finally, for σ (3) = 75, Eq 3.6-3 gives
as well as
( ) =
( ) ( )
1 3
2 2
7n + 24n = 0
( ) ( )
1 3
2 2
24n − 7n = 0
( ) ( )
( ) ( )
1
3
3 3
− 18n + 24n = 0
3
− 25n = 0
Here, from the second equation , and from either of the other two
equations , so that from nini = 1 we have and .
From these values of , we now construct the transformation matrix
[aij] in accordance with the table of Figure 3.10b, keeping in mind that to
assure a right-handed system of principal axes we must have .
Thus, the transformation matrix has the general form
n 3
2 0
( 3)
( 3)
( 3)
4
( 3)
3
4n3= 3n1
n1 =± n3 =±
5
5
n
( q)
i
ˆ( 3) 1 2
n nˆ( )
nˆ(
)
= ×
[ aij] =
( )
2
( ) ( )
1 3
3 3
24n − 32n = 0
( ) =
⎡ 3 4 ⎤
⎢±
0 m
5 5
⎥
⎢
⎥
⎢
⎥
⎢ 0 ± 1 0⎥
⎢
⎥
⎢ 4 3⎥
⎢±
0 ± ⎥
⎣ 5 5⎦