Civil Engineering Reference
In-Depth Information
Suppose that the tensile stress of 80N
/
mm
2
acts in the
x
direction. Then
σ
x
=
80N
/
mm
2
,
σ
y
=
45N
/
mm
2
. Substituting these values in Eqs (14.8) and
+
0 and
τ
xy
=
(14.9) in turn gives
2
80
2
80
2
+
1
100
.
2N
/
mm
2
σ
I
=
+
4
×
45
2
=
2
80
2
80
2
−
1
20
.
2N
/
mm
2
σ
II
=
+
4
×
45
2
=−
From Eq. (14.7)
2
×
45
80
tan 2
θ
=−
=−
1
.
125
from which
24
◦
11
θ
=−
(corresponding to
σ
I
)
24
◦
11
−
90
◦
=−
114
◦
11
.
=−
Also, the plane on which
σ
II
acts corresponds to
θ
The maximum shear stress is most easily found from Eq. (14.12) and is given by
100
.
2
−
(
−
20
.
2)
60
.
2N
/
mm
2
τ
max
=
=
2
The maximum shear stress acts on planes at 45
◦
to the principal planes. Thus
69
◦
11
and
θ
159
◦
11
give the planes of maximum shear stress.
=−
=−
θ
14.4 M
OHR'S
C
IRCLE OF
S
TRESS
The state of stress at a point in a structural member may be conveniently represented
graphically by
Mohr's circle of stress
. We have shown that the direct and shear stresses
on an inclined plane are given, in terms of known applied stresses, by
σ
x
cos
2
θ
σ
y
sin
2
θ
σ
n
=
+
−
τ
xy
sin 2
θ
(Eq. (14.5))
and
(
σ
x
−
σ
y
)
τ
=
sin 2
θ
+
τ
xy
cos 2
θ
(Eq. (14.6))
2
respectively. The positive directions of these stresses and the angle
θ
are shown in
Fig. 14.7. We now write Eq. (14.5) in the form
σ
y
2
(1
σ
x
2
(1
σ
n
=
+
cos 2
θ
)
+
−
cos 2
θ
)
−
τ
xy
sin 2
θ
or
1
2
(
σ
x
+
1
2
(
σ
x
−
σ
n
−
σ
y
)
=
σ
y
) cos 2
θ
−
τ
xy
sin 2
θ
(14.13)
