22
2 Analysis of Stress
Thus,
∂σ x
∂l
= 0,
∂σ x
∂m
= 0
(2.28a)
Differentiating Eq. (2.28), in terms of the quantities in Eqs. (2.24a), (2.24b) and
(2.24c), we obtain
T x + T z
∂n
∂l
= 0,
T y + T z
∂n
∂m
= 0,
(2.28b)
From n
2
= 1 − l
2
− m
2 , we have
∂n
∂l
= −
l
n
and
∂n
∂m
= −
m
n
Introducing the above into Eq. (2.28b), the following relationship between the
components of T and n is determined
T x
l
=
T y
m
=
T z
n
(2.28c)
These proportionalities indicate that the stress resultant must be parallel to the unit
normal and therefore contains no shear component. Therefore, from Eqs. (2.24a),
(2.24b) and (2.24c), we can write as below denoting the principal stress by σ P
T x = σ P l T y = σ P m T z = σ P n
(2.28d)
These expressions together with Eqs. (2.24a), (2.24b) and (2.24c) lead to
(σ x − σ P )l + τ xy m + τ xz n = 0
τ xy l +
σ y − σ P
m + τ yz n = 0
τ xy l + τ yz m + (σ z − σ P )n = 0
(2.29)
A non-trivial solution for the direction cosines requires that the characteristic
determinant should vanish.
⎡
⎣
(σ x − σ P )
τ xy
τ xz
τ xy
(σ y − σ P )
τ yz
τ xz
τ yz
(σ z − σ P )
⎤
⎦ = 0
(2.30)
Expanding (2.30) leads to
σ
3
P − I 1 σ
2
P + I 2 σ P − I 3 = 0
(2.31)