14
2 Analysis of Stress
Upon substitution of stress resultants from Eq. (2.9), Eq. (2.10) become
σ x = σ x cos
2
θ + σ y sin
2
θ + 2τ xy sin θ cos θ
τ x y = τ xy
cos
2
θ − sin
2
θ
+
σ y − σ x
sin θ cos θ
(2.12)
The stress σ y is obtained by substituting
θ +
π
2
for θ in the expression for σ x .
By means of trigonometric identities
cos
2
θ =
1
2
(1 + cos 2θ), sin θ cos θ =
1
2
sin 2θ,
sin
2
θ =
1
2
(1 − cos 2θ)
(2.13)
The transformation equations for stresses are now written in the following form:
σ x =
1
2
σ x + σ y
+
1
2
σ x − σ y
cos 2θ + τ xy sin 2θ
(2.13a)
σ y =
1
2
σ x + σ y
−
1
2
σ x − σ y
cos 2θ − τ xy sin 2θ
(2.13b)
τ x y = −
1
2
σ x − σ y
sin 2θ + τ xy cos 2θ
(2.13c)
2.9 Principal Stresses in Two Dimensions
To ascertain the orientation of x
y
corresponding to maximum or minimum σ x , the
necessary condition,
dσ x
dθ
= 0, is applied to Eq. (2.12a), yielding
−
σ x − σ y
sin 2θ + 2τ xy cos 2θ = 0
(2.14)
Therefore,
tan θ =
2τ xy
σ x − σ y
(2.15)
As 2θ = tan (π +2θ ), two directions, mutually perpendicular, are found to satisfy
Eq. (2.14). These are the principal directions, along which the principal or maximum
and minimum normal stresses act.
When Eq. (2.13c) is compared with Eq. (2.14), it becomes clear that τ x y = 0
on a principal plane. A principal plane is thus a plane of zero shear. The principal
stresses are determined by substituting Eq. (2.15) into Eq. (2.13a)