Civil Engineering Reference
In-Depth Information
radius of curvature
R
, we see, by direct comparison with Eq. (13.2) that
d
2
ζ
d
x
2
1
R
=
(13.12)
or, substituting from Eq. (13.11)
d
2
u
d
x
2
d
2
v
d
x
2
sin
α
R
cos
α
R
=
=
(13.13)
We observe from the derivation of Eq. (9.31) that
M
y
I
z
−
M
z
I
zy
I
z
I
y
−
E
sin
α
R
=
I
zy
E
cos
α
R
M
z
I
y
−
M
y
I
zy
I
z
I
y
−
=
I
zy
Therefore, from Eq. (13.13)
d
2
u
d
x
2
M
y
I
z
−
M
z
I
zy
E
(
I
z
I
y
−
=
(13.14)
I
zy
)
d
2
v
d
x
2
M
z
I
y
−
M
y
I
zy
E
(
I
z
I
y
−
=
(13.15)
I
zy
)
E
XAMPLE
13.15
Determine the horizontal and vertical components of the deflec-
tion of the free end of the cantilever shown in Fig. 13.18. The second moments of area
of its unsymmetrical section are
I
z
,
I
y
and
I
zy
.
y
D
L
G
F
z
F
IGURE
13.18
Deflection of a cantilever of
unsymmetrical cross section carrying a
concentrated load at its free end
(Ex. 13.15)
x
W
The bending moments at any section of the beam due to the applied load
W
are
M
z
=−
W
(
L
−
x
),
M
y
=
0
Then Eq. (13.14) reduces to
d
2
u
d
x
2
W
(
L
x
)
I
zy
E
(
I
z
I
y
−
−
=
(i)
I
zy
)
