2.23 Numerical Examples
41
But shear stress τ can be determined from the relation
T
2
= σ
2
+ τ
2
or
τ =
T 2 − σ 2
=
(169.706) 2 − (120) 2
τ = 120 N/mm
2
Case (ii) For l = m = n =
1
√
3
Again from (a),
T x = 0, T y = σ y m =
240
√
3
, T z = 0
Normal stress = σ = 0 +
240
√
3
1
√
3
+ 0 = 80.00 N/mm
2
Resultant stress on the plane is
T =
T 2
x + T 2
y + T 2
z
T =
0 +
240 √
3
2 + 0
τ = 113.13 N/mm
2
Shear stress = τ =
(138.56) 2 − (80) 2
τ = 113.13 N/mm
2
Example 2.6 A body is subjected to three-dimensional forces and the state of stress
at a point in it is represented as
⎡
⎣
200 200 200
200 −100 200
200 200 −100
⎤
⎦ MPa
Determine the normal stress, shearing stress and resultant stress on the octahedral
plane.
Solution For the octahedral plane, the direction cosines are
l = m = n =
1
√
3
Here,