2.23 Numerical Examples
49
= 20(−40)(80) − (20)(−60)
2
− (−40)(20)
2
− 80(40)
2
+ 2(40)(−60)(20)
= −344,000
Therefore, cubic equation becomes
σ
3
− 60σ
2
− 8000σ + 344,000 = 0
( a )
Solving the cubic equation for the principal stresses
σ 1 = 104.98 N/mm
2
σ 2 = −83.99 N/mm
2
σ 3 = 39.01 N/mm
2
Example 2.13 At a point in a given material, the three-dimensional state of stress is
given by
σ x = σ y = σ z = 10 N/mm
2
, τ xy = 20 N/mm
2 and τ yz = τ zx = 10 N/mm
2
Compute the principal planes if the corresponding principal stresses are
σ 1 = 37.3 N/mm
2
, σ 2 = −10 N/mm
2
, σ 3 = 2.7 N/mm
2
Solution The principal planes can be obtained by their direction Cosines l, m and n
associated with each of the three principal stresses, σ 1 , σ 2 and σ 3 .
(a) To find principal plane for stress σ 1
(10 − 37.3)
20
10
20
(10 − 37.3)
10
10
10
(10 − 37.3)
=
−27.3 20
10
20 −27.3 10
10
10 −27.3
Now,
A =
−27.3 10
10 −27.3
= 745.29 − 100
A = 645.29
B = −
20 10
10 −27.3
= −(−546 − 100)
B = 646
C =
20 −27.3
10 10