Civil Engineering Reference
In-Depth Information
τ
A
B
τ
s
AC
τ
τ
s
BD
τ
D
C
τ
F
IGURE
7.12
Stresses on diagonal planes in element
A
A
B
B
45
°
approx
F
f
f
F
IGURE
7.13
Distortion due to shear in
element
D
C
Hence we see that on planes parallel to the diagonals of the element there are direct
stresses
σ
BD
(tensile) and
σ
AC
(compressive) both numerically equal to
τ
as shown
in Fig. 7.12. It follows from Section 7.8 that the direct strain in the direction BD is
given by
σ
BD
E
νσ
AC
E
τ
E
(1
ε
BD
=
+
=
+
ν
)
(7.18)
Note that the compressive stress
σ
AC
makes a positive contribution to the strain
ε
BD
.
In Section 7.5 we defined shear strain and saw that under pure shear, only a change
of shape is involved. Thus the element ABCD of Fig. 7.11(a) distorts into the shape
A
B
CD shown in Fig. 7.13. The shear strain
γ
produced by the shear stress
τ
is then
given by
B
B
BC
γ
=
φ
radians
=
(7.19)
since
φ
is a small angle. The increase in length of the diagonal DB to DB
is
approximately equal to FB
where BF is perpendicular to DB
. Thus
DB
−
FB
DB
DB
ε
DB
=
=
DB
Again, since
φ
is a small angle, B B
F
45
◦
so that
FB
=
BB
cos 45
◦
