40
2 Analysis of Stress
∴ σ y =
Axial load
cross − sectional area
=
180 × 10
3
30 × 25
= 240 N/mm
2
Now, by Cauchy’s formula, the stress components along x, y and z co-ordinates
are
T x = σ x l + τ xy m + τ xz n
T y = τ xy l + σ y m + τ yz n
T z = τ xz l + τ yz m + σ z n
(a)
And the normal stress acting on the plane whose normal has the direction cosines
l, m and n is,
σ = T x l + T y m + T z n
(b)
Case (i) For l = m =
1
√
2
and n = 0
Here,
σ x = 0, τ xy = 0, σ y = 240 N/mm
2
τ xz = 0, τ yz = 0, σ z = 0
Substituting the above in (a), we get
T x = 0, T y = σ y m =
240
√
2
, T z = 0
Substituting in (b), we get
σ = 0 +
240
√
2
1
√
2
+ 0 = 120 N/mm
2
Resultant stress on the plane is
T =
T 2
x + T 2
y + T 2
z
=
0 +
240
√
2
2
+ 0
T = 169.706 N/mm
2
2 Analysis of Stress
∴ σ y =
Axial load
cross − sectional area
=
180 × 10
3
30 × 25
= 240 N/mm
2
Now, by Cauchy’s formula, the stress components along x, y and z co-ordinates
are
T x = σ x l + τ xy m + τ xz n
T y = τ xy l + σ y m + τ yz n
T z = τ xz l + τ yz m + σ z n
(a)
And the normal stress acting on the plane whose normal has the direction cosines
l, m and n is,
σ = T x l + T y m + T z n
(b)
Case (i) For l = m =
1
√
2
and n = 0
Here,
σ x = 0, τ xy = 0, σ y = 240 N/mm
2
τ xz = 0, τ yz = 0, σ z = 0
Substituting the above in (a), we get
T x = 0, T y = σ y m =
240
√
2
, T z = 0
Substituting in (b), we get
σ = 0 +
240
√
2
1
√
2
+ 0 = 120 N/mm
2
Resultant stress on the plane is
T =
T 2
x + T 2
y + T 2
z
=
0 +
240
√
2
2
+ 0
T = 169.706 N/mm
2