3.17 Numerical Examples
85
Hence, Eqs. (ii) and (iii) are identical, if
0.03251
r 2
=
3
4
i.e., r =
4×0.03251
3
= 0.2082
and
cos 3θ
4
=
0.00112
r 3
or cos 3θ =
4×0.00112
(0.2082)
3 = 0.496 ∼ = 0.5
∴ 3θ = 60
0 or θ 1 =
60
3
= 20
0
θ 2 = 100
0
θ 3 = 140
0
∴ ε 1 = r 1 cos θ 1 +
J 1
3
= 0.2082 cos 20
◦
+
0.1
3
ε 1 = 0.228
ε 2 = r 2 cos θ 2 +
J 1
3
= 0.2082 cos 100
◦
+
0.1
3
= −0.0031
ε 3 = r 3 cos θ 3 +
J 1
3
= 0.2082 cos 140
◦
+
0.1
3
= −0.126
To find principal directions
(a) Principal direction for ε 1
⎡
⎣
(0.1 − ε 1 )
0.15
−0.04
0.15 (−0.05 − ε 1 )
0.05
−0.04
0.05
(0.05 − ε 1 )
⎤
⎦
=
⎡
⎣
(0.1 − 0.228)
0.15
−0.04
0.15
(−0.05 − 0.228)
0.05
−0.04
0.05
(0.05 − 0.228)
⎤
⎦
=
⎡
⎣
−0.128 0.15 −0.04
0.15 −0.278 0.05
−0.04 0.05 −0.178
⎤
⎦
Now, A 1 =
−0.278 0.05
0.05 −0.178
= (−0.278) (−0.178) − (0.05) (0.05)