Civil Engineering Reference
In-Depth Information
and
sin 2
θ
2
π
2
τ
xy
(
σ
x
−
+
=
σ
y
)
2
4
τ
xy
+
cos 2
θ
2
−
(
σ
x
−
σ
y
)
π
+
=
(
σ
x
−
σ
y
)
2
4
τ
xy
+
Rewriting Eq. (14.5) as
σ
x
2
(1
σ
y
2
(1
σ
n
=
+
cos 2
θ
)
+
−
cos 2
θ
)
−
τ
xy
sin 2
θ
and substituting for
{
sin 2
θ
, cos 2
θ
}
and
{
sin 2(
θ
+
π/
2), cos 2(
θ
+
π/
2)
}
in turn gives
(
σ
x
−
σ
x
+
σ
y
1
2
σ
I
=
+
σ
y
)
2
+
4
τ
xy
(14.8)
2
(
σ
x
−
σ
y
)
2
σ
x
+
σ
y
1
2
σ
II
=
−
+
4
τ
xy
(14.9)
2
where
σ
I
is the
maximum
or
major principal stress
and
σ
II
is the
minimum
or
minor
principal stress
;
σ
I
is algebraically the greatest direct stress at the point while
σ
II
is
algebraically the least. Note that when
σ
II
is compressive, i.e. negative, it is possible
for
σ
II
to be numerically greater than
σ
I
.
From Eq. (14.6)
d
τ
d
θ
=
(
σ
x
−
σ
y
) cos 2
θ
−
2
τ
xy
sin 2
θ
=
0
giving
(
σ
x
−
σ
y
)
2
τ
xy
tan 2
θ
=
(14.10)
It follows that
(
σ
x
−
σ
y
)
sin 2
θ
=
(
σ
x
−
σ
y
)
2
+
4
τ
xy
2
τ
xy
cos 2
θ
=
(
σ
x
−
σ
y
)
2
4
τ
xy
+
sin 2
θ
2
(
σ
x
−
π
σ
y
)
+
=−
(
σ
x
−
σ
y
)
2
4
τ
xy
+
cos 2
θ
2
2
τ
xy
(
σ
x
−
σ
y
)
2
π
+
=−
+
4
τ
xy
Substituting these values in Eq. (14.6) gives
(
σ
x
−
1
2
σ
y
)
2
4
τ
xy
τ
max,min
=±
+
(14.11)
