42
2 Analysis of Stress
σ x = 200 MPa
σ y = −100 MPa
σ y = −100 MPa
τ xy = τ yz = τ zx = 200 MPa
Substituting the above in Cauchy’s formula, we get
T x = 200
1
√
3
+ 200
1
√
3
+ 200
1
√
3
= 346.41 MPa
T y = 200
1
√
3
− 100
1
√
3
+ 200
1
√
3
= 173.20 MPa
T z = 200
1
√
3
+ 200
1
√
3
− 100
1
√
3
= 173.20 MPa
Normal stress on the plane is given by
σ = T x l + T y m + T z n
= 346.41
1
√
3
+ 173.20
1
√
3
+ 173.20
1
√
3
σ = 400 MPa
Resultant stress = T =
T 2
x + T 2
y + T 2
z
=
(346.41) 2 + (173.20) 2 + (173.20) 2
T = 424.26 MPa
Also, tangential stress = τ =
(424.26) 2 − (400) 2
= 141.41 MPa
Also,
tangential stress = τ =
(424.26) 2 − (400) 2
= 141.41 MPa
Example 2.7 The state of stress at a point is given as follows:
σ x = −800 kPa, σ y = 1200 kPa, σ z = −400 kPa
τ xy = 400 kPa, τ yz = −600 kPa, τ zx = 500 kPa
Determine (a) the stresses on a plane whose normal has direction cosines l =
1
4
, m =
1
2
and (b) the normal and shearing stresses on that plane.
Solution We have the relation,