Civil Engineering Reference
In-Depth Information
and
(
ε
x
−
ε
x
+
ε
y
1
2
ε
II
=
−
ε
y
)
2
+
γ
xy
(14.22)
2
Since the shear strain
γ
is zero on these planes it follows that the shear stress must also
be zero and we deduce fromSection 14.3 that the directions of the principal strains and
principal stresses coincide. The related planes are then determined from Eq. (14.7) or
from
γ
xy
ε
x
−
ε
y
tan 2
θ
=−
(14.23)
In addition the maximum shear strain at the point is given by
γ
2
(
ε
x
−
1
2
max
=
ε
y
)
2
+
γ
xy
(14.24)
or
γ
2
ε
I
−
ε
II
max
=
(14.25)
2
(cf. Eqs (14.11) and (14.12)).
14.8 M
OHR'S
C
IRCLE OF
S
TRAIN
The argument of Section 14.7 may be applied to Mohr's circle of stress described in
Section 14.4. A circle of strain, analogous to that shown in Fig. 14.11, may be drawn
when
σ
x
,
σ
y
, etc., are replaced by
ε
x
,
ε
y
, etc., as specified in Section 14.7. The horizontal
extremities of the circle represent the principal strains, the radius of the circle half the
maximum shear strain, and so on.
E
XAMPLE
14.5
Astructuralmember is loaded in such away that at a particular point
in the member a two-dimensional stress system exists consisting of
σ
x
=+
60N
/
mm
2
,
40N
/
mm
2
and
τ
xy
=
50N
/
mm
2
.
σ
y
=−
(a) Calculate the direct strain in the
x
and
y
directions and the shear strain,
γ
xy
, at the
point.
(b) Calculate the principal strains at the point and determine the position of the
principal planes.
(c) Verify your answer using a graphical method. Take
E
200 000N/mm
2
=
and
Poisson's ratio,
ν
=
0
.
3
.
(a) From Section 7.8
1
200 000
(60
10
−
6
ε
x
=
+
0
.
3
×
40)
=
360
×
1
200 000
(
10
−
6
ε
y
=
−
40
−
0
.
3
×
60)
=−
290
×
