Civil Engineering Reference
In-Depth Information
Mohr's circle of strain is as shown in Fig. 14.17; the shear strains
γ
a
,
γ
b
and
γ
c
do not
feature in the discussion and are therefore ignored. From Fig. 14.17
1
2
(
ε
a
+
ε
c
)
OC
=
1
2
(
ε
a
−
CN
=
ε
a
−
OC
=
ε
c
)
1
2
(
ε
a
+
QN
=
CM
=
ε
b
−
OC
=
ε
b
−
ε
c
)
The radius of the circle is CQ and
CN
2
QN
2
=
+
CQ
Hence
1
2
(
ε
a
−
ε
c
)
2
ε
b
−
ε
c
)
2
1
2
(
ε
a
+
CQ
=
+
which simplifies to
(
ε
a
−
1
√
2
ε
b
)
2
ε
b
)
2
CQ
=
+
(
ε
c
−
Therefore
ε
I
, which is given by
ε
I
=
+
OC
radius of the circle
is
(
ε
a
−
1
2
(
ε
a
+
1
√
2
ε
I
=
+
ε
b
)
2
+
(
ε
c
−
ε
b
)
2
ε
c
)
(14.34)
Also
ε
II
=
OC
−
radius of the circle
i.e.
(
ε
a
−
1
2
(
ε
a
+
1
√
2
ε
II
=
ε
c
)
−
ε
b
)
2
+
(
ε
c
−
ε
b
)
2
(14.35)
Finally the angle
θ
is given by
QN
CN
=
ε
b
−
(1
/
2)(
ε
a
+
ε
c
)
tan 2
θ
=
(1
/
2)(
ε
a
−
ε
c
)
i.e.
2
ε
b
−
ε
a
−
ε
c
tan 2
θ
=
(14.36)
ε
a
−
ε
c
A similar approach can be adopted for a 60
◦
rosette.
