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

536
STRESS TRANSFORMATIONS
third principal stress is intermediate in value and has no particular significance. The preceding
discussion demonstrates that the set of principal stresses includes the maximum and
minimum normal stresses at the point.
S z
σ
p3
S y
σ
p2
S
σ
( Sx, Sy, Sz
)
τnt
σ p1 > σ p2 > σp3
n
σ
p1
S x
magnitude and orientation of maximum Shear Stress
Continuing with the special case in which the given stresses σ x , σ y , and σ z are
principal stresses, we can develop equations for the maximum shear stress at
the point under consideration. The resultant stress S on the oblique plane is
given by the equation
S = S + S + S
2 2 2 2
x y z
Substituting values for S x , S y , and S z from Equation (a), with zero shear stresses,
yields
2 2 2 2 2 2 2
S = σ l + σ m + σ n
(f)
x y z
FIGURE 12.19
Also, from Equation (12.25),
σ = ( σ l + σ m + σ n )
2 2 2 2 2
n x y z
(g)
Since S 2 = σn 2 + τnt
2 , we obtain, from Equations (f) and (g), the following equation for the
shear stress τ nt on the oblique plane:
2 2 2 2 2 2 2 2 2 2
τ = σ l + σ m + σ n − ( σ l + σ m + σ n ) (12.29)
nt x y z x y z
The planes on which maximum and minimum shear stresses occur can be found from
Equation (12.29) by differentiating with respect to the direction cosines l, m, and n. One of
the direction cosines in Equation (12.29) (e.g., n) can be eliminated by solving Equation (d)
for n 2 and substituting into Equation (12.29). Thus,
τ = {( σ − σ ) l + ( σ − σ ) m + σ
2 2 2 2 2 2 2
nt x z y z z
− [( σ − σ ) l + ( σ − σ ) m + σ ] }
2 2 2 1/2
x z y z z
(h)
Taking the partial derivatives of Equation (h), first with respect to l and then with respect to
m and setting the partial derivatives equal to zero, we obtain the following equations for the
direction cosines associated with planes having maximum and minimum shear stress:
l ⎡1
σ − σ − σ − σ − σ − σ
⎤
⎣⎢ 2 ( ) ( ) l 2 ( ) m 2
x z x z y z
⎦⎥ = 0
(i)
m
⎡1
l m
2 ( ) ( ) 2
y z x z ( y z ) 2
σ − σ − σ − σ − σ − σ
⎤
⎣⎢
⎦⎥ = 0
(j)
One solution of these equations is, obviously, l = m = 0. Then, from Equation (d), n = ±1.
Solutions different from zero are also possible for this set of equations. For instance, consider
surfaces in which the direction cosine has the value m = 0. From Equation (i),
l =± 1/2, and from Equation (d), n =± 1/2. Thus, the normal to this surface makes an
angle of 45° with both the x and z axes, and is perpendicular to the y axis. This surface has