Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

The forces on the x, y, and z faces are shown as three components, each of whose magnitude
is the product of the area by the appropriate stress. If we use l, m, and n to represent cos α,
cos β, and cos γ , respectively, then the force equilibrium equations in the x, y, and z directions
are as follows:
F = S dA = σ dA ⋅ l + τ dA⋅ m + τ dA ⋅ n
x x x yx zx
F = S dA = σ dA ⋅ m + τ dA⋅ n + τ dA ⋅ l
y y y zy xy
F = S dA = σ dA ⋅ n + τ dA⋅ l + τ dA ⋅ m
z z z xz yz
From these equations, we obtain the three orthogonal components of the resultant stress:
533
gENERAL STATE OF STRESS
AT A POINT
The terms:
l = cos α
m = cos β
n = cos γ
are called direction cosines.
S = σ ⋅ l + τ ⋅ m + τ ⋅ n
x x yx zx
S = τ ⋅ l + σ ⋅ m + τ ⋅ n
y xy y zy
S = τ ⋅ l + τ ⋅ m + σ ⋅ n
z xz yz z
(a)
The normal component σ n of the resultant stress S equals S x ∙ l + S y ∙ m + S z ∙ n; therefore,
from Equation (a), it follows that the normal stress on any oblique plane through the point is
2 2 2
σ = σ l + σ m + σ n + 2τ lm + 2τ mn + 2τ
nl (12.25)
n x y z xy yz zx
2 2 2
The shear stress τ nt on the oblique plane can be obtained from the relation S = σn + τnt.
Thus, for any given problem, the values of S and σ n will be obtained from Equations (a)
and (12.25).
magnitude and orientation of principal Stresses
A principal plane was previously defined as a plane on which the shear stress τ nt is zero.
The normal stress σ n on such a plane was defined as a principal stress σ p . If the oblique
plane of Figure 12.18 is a principal plane, then S = σ p and, in addition, S x = σ p l, S y = σ p m,
and S z = σ p n. Substituting these components into Equation (a) then produces the following
homogeneous linear equations in terms of the direction cosines l, m, and n:
( σ − σ ) l + τ m + τ n = 0
x p yx zx
( σ − σ ) m + τ n + τ l = 0
y p zy xy
( σ − σ ) n + τ l + τ m = 0
z p xz yz
(b)
This set of equations has a nontrivial solution only if the determinant of the coefficients of
l, m, and n is equal to zero. Thus,
( σ − σ ) τ τ
x p yx zx
τ ( σ − σ ) τ
xy y p zy
τ τ ( σ − σ )
xz yz z p
= 0
(12.26)
Expansion of the determinant yields the stress cubic equation for determining the principal
stresses:
3 2
σ − I σ + I σ − I = 0
(12.27)
p 1 p 2 p 3