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give the equations
2
a
1
a
2
+
(σ
2
a
1
a
3
−
λ
(σ
−
σ
)
−
σ
)
a
1
=
0
1
2
1
3
2
a
2
a
3
+
(σ
2
a
2
a
1
−
λ
(σ
−
σ
)
−
σ
)
=
a
2
0
2
3
2
1
2
a
3
a
1
+
(σ
2
a
3
a
2
−
λ
(σ
−
σ
)
−
σ
)
a
3
=
0
3
1
3
2
a
1
+
a
2
+
a
3
−
1
=
0
These equations have the following three solutions
2
1
√
2
λ
=
(σ
2
−
σ
3
)
a
1
=
0
a
2
=
a
3
=
2
2
1
√
2
λ
=
(σ
−
σ
)
3
1
a
2
=
0
a
3
=
a
1
=
(1.90)
2
2
1
√
2
λ
=
(σ
−
σ
)
1
2
a
3
=
0
a
1
=
a
2
=
2
The extreme values of the shear stresses are found by substituting the solutions of
Eq. (1.90) into Eq. (1.89). For example, for the third case with
a
1
=
1
√
2
a
2
=
and
a
3
=
0,
1
2
τ
=
(σ
−
σ
)
Eq. (1.89 ) gives
. Thus, in general the extreme values of the shear stresses
1
2
τ
=
|
σ
|
τ
=
|
σ
|
τ
=
|
σ
|
−
σ
−
σ
−
σ
2
3
3
1
1
2
(1.91)
2
2
2
It is seen that the planes on which shear stresses reach their extreme values make a 45
◦
angle with the principal directions. These planes are, in general, not free of normal stress.
EXAMPLE 1.3 Plane Stress Problems
For the plane stress problem of Section 1.3.1,
σ
=
τ
=
τ
=
0. The stress matrix of Eq.
z
xz
yz
(1.83) becomes
σ
τ
0
x
xy
T
=
τ
y
0
000
σ
(1)
xy
The principal stresses are obtained from Eq. (1.87) as
σ
2
=
σ
+
σ
−
σ
x
y
x
y
σ
±
+
τ
xy
(2)
max
,
min
2
2
These are expressions for the principal stresses in the
x, y
plane. The out-of-plane normal
stress
σ
min
can
vary. As a consequence there are several possibilities for choosing which principal stress
corresponds to
σ
z
=
0 is the third principal stress. The magnitudes and signs of
σ
max
and
σ
1
,
σ
2
,
or
σ
3
,
which usually are ordered from the algebraically largest
(σ
1
)
to the algebraically smallest
(σ
3
)
. The possibilities for the plane stress problem are
σ
=
σ
max
,
σ
=
σ
min
,
σ
=
0
1
2
3
σ
=
σ
max
,
σ
=
0
,
σ
=
σ
(3)
1
2
3
min
σ
=
0
,
σ
=
σ
max
,
σ
=
σ
1
2
3
min
The extreme values of the shear stress are given by Eq. (1.91), with the maximum shear
stress being the largest of these values.
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