2.23 Numerical Examples
45
⎡
⎣
129.3 −29.3 0
−29.3 −89.3 0
0
0 40
⎤
⎦ (KPa)
Example 2.9 The stress tensor at a point is given by the following array
⎡
⎣
50 −20 40
−20 20 10
40 10 30
⎤
⎦ (kPa)
Determine the stress vectors on the plane whose unit normal has direction cosines
1
√
2
,
1
2
,
1
2
Solution The stress vectors are given by
T x = σ x l + τ xy m + τ xz n
(a)
T y = τ xy l + σ y m + τ yz n
(b)
T z = τ xz l + τ yz m + σ z n
(c)
Substituting the stress components in (a), (b) and (c), we get
T x = 50
1
√
2
− 20
1
2
+ 40
1
2
= 45.35 kPa
T y = −20
1
√
2
+ 20
1
2
+ 10
1
2
= 0.858 kPa
T z = 40
1
√
2
+ 10
1
2
+ 30
1
2
= 48.28 kPa
Now, resultant stress is given by
T =
45.35 ˆ
i + 0.858 ˆ
j + 48.28 ˆ
k
kPa
Example 2.10 The stress tensor at a point is given by the following array
⎡
⎣
40 20 30
20 30 40
30 40 20
⎤
⎦ (kPa)
Calculate the deviator and spherical stress tensors.
Solution