20
2 Analysis of Stress
Table 2.1 Direction cosines
for transformed co-ordinates
X
Y
Z
X
l 1
m 1
n 1
Y
l 2
m 2
n 2
Z
l 3
m 3
n 3
(The notation corresponding to a complete set of direction cosines is shown in
Table 2.1).
The normal stress σ x is found by projecting T x , T y and T z in the X
direction and
adding:
σ x = T x l 1 + T y m 1 + T z n 1
(2.26)
Equations (2.24a), (2.24b), (2.24c) and (2.26) are combined to yield
σ x = σ x l
2
1 + σ y m
2
1 + σ z n
2
1 + 2
T xy l 1 m 1 + τ yz m 1 n 1 + τ xz l 1 n 1
(2.27)
Similarly by projecting T x , T y , T z in the y
and z
directions, we obtain,
respectively
τ x y = σ x l 1 l 2 + σ y m 1 m 2 + σ z n 1 n 2 + τ xy (l 1 m 2 + m 1 l 2 )
+ τ yz (m 1 n 2 + n 1 m 2 ) + τ xz (n 1 l 2 + l 1 n 2 )
(2.27a)
τ x z = σ x l 1 l 3 + σ y m 1 m 3 + σ z n 1 n 3 + τ xy (l 1 m 3 + m 1 l 3 )
+ τ yz (m 1 n 3 + n 1 m 3 ) + τ xz (n 1 l 3 + l 1 n 3 )
(2.27b)
Recalling that the stresses on three mutually perpendicular planes are required to
specify the stress at a point (one of these planes being the oblique plane in question),
the remaining components are found by considering those planes perpendicular to
the oblique plane. For one such plane n would now coincide with y
direction, and
expressions for the stresses σ y , τ y , τ y z would be derived. In a similar manner, the
stresses σ z , τ z x , τ z y are determined when n coincides with the z
direction. Owing to
the symmetry of stress tensor, only six of the nine stress components thus developed
are unique. The remaining stress components are as follows:
σ y = σ x l
2
2 + σ y m
2
2 + σ z n
2
2 + 2
τ xy l 2 m 2 + τ yz m 2 n 2 + τ xz l 2 n 2
(2.27c)
σ z = σ x l
2
3 + σ y m
2
3 + σ z n
2
3 + 2
τ xy l 3 m 3 + τ yz m 3 n 3 + τ xz l 3 n 3
(2.27d)
τ y z = σ x l 2 l 3 + σ y m 2 m 3 + σ z n 2 n 3 + τ xy (m 2 l 3 + l 2 m 3 )
+ τ yz (n 2 m 3 + m 2 n 3 ) + τ xz (l 2 n 3 + n 2 l 3 )
(2.27e)
2 Analysis of Stress
Table 2.1 Direction cosines
for transformed co-ordinates
X
Y
Z
X
l 1
m 1
n 1
Y
l 2
m 2
n 2
Z
l 3
m 3
n 3
(The notation corresponding to a complete set of direction cosines is shown in
Table 2.1).
The normal stress σ x is found by projecting T x , T y and T z in the X
direction and
adding:
σ x = T x l 1 + T y m 1 + T z n 1
(2.26)
Equations (2.24a), (2.24b), (2.24c) and (2.26) are combined to yield
σ x = σ x l
2
1 + σ y m
2
1 + σ z n
2
1 + 2
T xy l 1 m 1 + τ yz m 1 n 1 + τ xz l 1 n 1
(2.27)
Similarly by projecting T x , T y , T z in the y
and z
directions, we obtain,
respectively
τ x y = σ x l 1 l 2 + σ y m 1 m 2 + σ z n 1 n 2 + τ xy (l 1 m 2 + m 1 l 2 )
+ τ yz (m 1 n 2 + n 1 m 2 ) + τ xz (n 1 l 2 + l 1 n 2 )
(2.27a)
τ x z = σ x l 1 l 3 + σ y m 1 m 3 + σ z n 1 n 3 + τ xy (l 1 m 3 + m 1 l 3 )
+ τ yz (m 1 n 3 + n 1 m 3 ) + τ xz (n 1 l 3 + l 1 n 3 )
(2.27b)
Recalling that the stresses on three mutually perpendicular planes are required to
specify the stress at a point (one of these planes being the oblique plane in question),
the remaining components are found by considering those planes perpendicular to
the oblique plane. For one such plane n would now coincide with y
direction, and
expressions for the stresses σ y , τ y , τ y z would be derived. In a similar manner, the
stresses σ z , τ z x , τ z y are determined when n coincides with the z
direction. Owing to
the symmetry of stress tensor, only six of the nine stress components thus developed
are unique. The remaining stress components are as follows:
σ y = σ x l
2
2 + σ y m
2
2 + σ z n
2
2 + 2
τ xy l 2 m 2 + τ yz m 2 n 2 + τ xz l 2 n 2
(2.27c)
σ z = σ x l
2
3 + σ y m
2
3 + σ z n
2
3 + 2
τ xy l 3 m 3 + τ yz m 3 n 3 + τ xz l 3 n 3
(2.27d)
τ y z = σ x l 2 l 3 + σ y m 2 m 3 + σ z n 2 n 3 + τ xy (m 2 l 3 + l 2 m 3 )
+ τ yz (n 2 m 3 + m 2 n 3 ) + τ xz (l 2 n 3 + n 2 l 3 )
(2.27e)