2.23 Numerical Examples
43
l
2
+ m
2
+ n
2
= 1
∴
1
4
2
+
1
2
2
+ n
2
= 1
∴ n =
√
11
4
(a) Using Cauchy’s formula,
T x = −800
1
4
+ 400
1
2
+ 500
√
11
4
= 414.60 kPa
T y = 400
1
4
+ 1200
1
2
− 600
√
11
4
= 202.51 kPa
T z = 500
1
4
− 600
1
2
− 400
√
11
4
= −506.66 kPa
(b) Normal stress,
σ = T x l + T y m + T z n
= 414.60
1
4
+ 202.51
1
2
− 506.66
√
11
4
σ = −215.20 kPa
Resultant stress on the plane = T =
(41460) 2 + (20251) 2 + (50666) 2
= 685.28 MPa
Shear stress on the plane = τ =
(685.28) 2 − (−215.20) 2
= 650.61 kPa
Example 2.8 Given the state of stress at a point as below
⎡
⎣
100 80 0
90 −60 0
0 0 40
⎤
⎦ kPa
Considering another set of co-ordinate axes, x
y
z
in which z
coincides with z
and x
is rotated by 30° anticlockwise from x-axis, determine the stress components
in the new co-ordinates system.
Solution The direction cosines for the transformation are given by
43
l
2
+ m
2
+ n
2
= 1
∴
1
4
2
+
1
2
2
+ n
2
= 1
∴ n =
√
11
4
(a) Using Cauchy’s formula,
T x = −800
1
4
+ 400
1
2
+ 500
√
11
4
= 414.60 kPa
T y = 400
1
4
+ 1200
1
2
− 600
√
11
4
= 202.51 kPa
T z = 500
1
4
− 600
1
2
− 400
√
11
4
= −506.66 kPa
(b) Normal stress,
σ = T x l + T y m + T z n
= 414.60
1
4
+ 202.51
1
2
− 506.66
√
11
4
σ = −215.20 kPa
Resultant stress on the plane = T =
(41460) 2 + (20251) 2 + (50666) 2
= 685.28 MPa
Shear stress on the plane = τ =
(685.28) 2 − (−215.20) 2
= 650.61 kPa
Example 2.8 Given the state of stress at a point as below
⎡
⎣
100 80 0
90 −60 0
0 0 40
⎤
⎦ kPa
Considering another set of co-ordinate axes, x
y
z
in which z
coincides with z
and x
is rotated by 30° anticlockwise from x-axis, determine the stress components
in the new co-ordinates system.
Solution The direction cosines for the transformation are given by