48
2 Analysis of Stress
5 × 1 × 2 + 3 + 3 × 1 × 1 + F y = 0
∴ F y = −16
and
3 × 1 × 2 + 2 × (−1) + 6 × 1 × (−1) × 2 + 2 × 1 + (−1)
2
+ F z = 0
∴ F z = 5
The body force vector is given by
→
F = −11 ˆ
i − 16 ˆ
j + 5 ˆ
k
Example 2.12 The rectangular stress components at a point in a three-dimensional
stress system are as follows.
σ x = 20 N/mm
2
σ y = −40 N/mm
2
σ z = 80 N/mm
2
τ xy = 40 N/mm
2
τ yz = −60 N/mm
2
τ zx = 20 N/mm
2
Determine the principal stresses at the given point.
Solution The principal stresses are the roots of the cubic equation
σ
3
− I 1 σ
2
+ I 2 σ − I 3 = 0
The three-dimensional stresses can be expressed in the matrix form as below.
⎡
⎣
σ x τ xy τ xz
τ xy σ y τ yz
τ xz τ yz σ z
⎤
⎦ =
⎡
⎣
20 40 20
40 −40 −60
20 −60 80
⎤
⎦ N/mm
2
Here,
I 1 =
σ x + σ y + σ z
= (20 − 40 + 80)
= 60
I 2 = σ x σ y + σ y σ z + σ z σ x − τ
2
xy − τ
2
yz − τ
2
zx
=
20(−40) + (−40)(80) + 80(20) − (40)
2
− (−60)
2
− (20)
2
= −8000
I 3 = σ x σ y σ z − σ x τ
2
yz − σ y τ
2
zx − σ z τ
2
xy + 2τ xy τ yz τ xz