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

Table 12.1 Direction cosines for planes of maximum and minimum Shear Stress
Minimum
Maximum
1 2 3 4 5 6
l ±1 0 0 ± 1/2 ± 1/2 0
m 0 ±1 0 ± 1/2 0 ± 1/2
n 0 0 ±1 0 ± 1/2 ± 1/2
y
z
x
the largest shear stress of all surfaces whose normal is perpendicular to the y axis. Next,
consider surfaces whose normal is perpendicular to the x axis; that is, the direction cosine
has the value l = 0. From Equation (j), m =± 1/2 , and from Equation (d), n =± 1/2. The
normal to this surface makes an angle of 45° with both the y and z axes. This surface has
the largest shear stress of all surfaces whose normal is perpendicular to the x axis. Repeating
the preceding procedure by eliminating l and m in turn from Equation (h) yields other
values for the direction cosines that make the shear stresses maximum or minimum. All of
the possible combinations are listed in Table 12.1. In the last row of the table, the planes
corresponding to the direction cosines in the column above are shown shaded. Note that in
each case only one of the two possible planes is shown.
The first three columns of Table 12.1 give the direction cosines for planes of minimum
shear stress. Since we are here considering the special case in which the given stresses σ x ,
σ y , and σ z are principal stresses, columns 1, 2, and 3 are simply the principal planes for
which the shear stress must be zero. Hence, the minimum shear stress is τ nt = 0.
To determine the magnitude of the maximum shear stress, values of direction cosines
from Table 12.1 are substituted into Equation (12.29), replacing σ x , σ y , and σ z with σ p1 ,
σ p2 , and σ p3 , respectively. Direction cosines from column 4 of Table 12.1 give the following
expression for the maximum shear stress:
2
1 1 ⎛ 1 1 ⎞ σ p − σ
τmax = σ 2 2
p1
+ σ p2
+ 0 − σ + σ
⎝
⎜ p1 p2⎠
⎟ =
2 2
2 2 2
Similarly, direction cosines from columns 5 and 6 give
τ
max
σ
=
− σ
p 1 p 3
and
p 2 p 3
τ max =
2
σ
− σ
2
1 p2
The shear stress of largest magnitude from these three possible results is τ abs max ; hence, the
absolute maximum shear stress can be expressed as
τ
absmax
σ
=
max
−
2
σ
min
(12.30)
which confirms Equation (12.18). The maximum shear stress acts on the plane that bisects
the angle between the maximum and minimum principal stresses.
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