Civil Engineering Reference
In-Depth Information
s
y
B
τ
xy
B
τ
xy
u
D
s
x
s
x
D
τ
xy
C
xy
A
C
A
τ
xy
p
p
xy
F
IGURE
14.14
Determination of
strains on an
inclined plane
2
2
s
y
(a)
(b)
To a first order of approximation
!
A
C
=
AC(1
+
ε
x
)
C
B
=
+
(14.15)
CB(1
ε
y
)
"
A
B
=
+
AB(1
ε
n
+
π/
2
)
where
ε
n
+
π/
2
is the direct strain in the direction AB. From the geometry of the triangle
A
B
C
in which angle B
C
A
=
π/
2
+
γ
xy
2(A
C
)(C
B
) cos
π
γ
xy
(A
B
)
2
(A
C
)
2
(C
B
)
2
=
+
−
2
+
or, substituting from Eq. (14.15)
(AB)
2
(1
ε
n
+
π/
2
)
2
(AC)
2
(1
ε
x
)
2
(CB)
2
(1
ε
y
)
2
+
=
+
+
+
+
2(AC)(CB)(1
+
ε
x
)(1
+
ε
y
) sin
γ
xy
Noting that (AB)
2
(CB)
2
and neglecting squares and higher powers of small
quantities, this equation may be rewritten
(AC)
2
=
+
2(AB)
2
ε
n
+
π/
2
=
2(AC)
2
ε
x
+
2(CB)
2
ε
y
+
2(AC)(CB)
γ
xy
Dividing through by 2(AB)
2
gives
ε
x
sin
2
θ
ε
y
cos
2
θ
ε
n
+
π/
2
=
+
+
sin
θ
cos
θγ
xy
(14.16)
The strain
ε
n
in the direction normal to the plane AB is found by replacing the angle
θ
in Eq. (14.16) by
θ
−
π/
2
.
Hence
γ
xy
2
ε
x
cos
2
θ
ε
y
sin
2
θ
ε
n
=
+
−
sin 2
θ
(14.17)
Now from triangle C
D
B
we have
2(C
D
)(D
B
) cos
π
γ
(C
B
)
2
(C
D
)
2
(D
B
)
2
=
+
−
2
−
(14.18)
