(^332) A Textbook of Engineering Mechanics
Now expanding the term inside the square root, by *Binomial theorem,
(^222)
2
1
1 1 ......
2
wx w x
H H
⎛⎞ ⎛⎞
+=+ +⎜⎟ ⎜⎟⎜⎟
⎝⎠ ⎝⎠
22
... Neglecting higher powers of 2
wx
H
⎛⎞
⎜⎟⎜⎟
⎝⎠
∴
22
2
1
1
2
wx
ds dx
H
⎡⎤⎛⎞
=+×⎢⎥⎜⎟⎜⎟
⎢⎥⎣⎦⎝⎠
and now integrating the above equation between the limits x = 0, and
2
l
x=
2 22
2
0
1
1
2
l
wx
sdx
H
⎡⎤⎛⎞
=+×⎢⎥⎜⎟⎜⎟
⎢⎥⎣⎦⎝⎠
∫
232 23
22
0
2232 24
l
wx l wl
sx
HH
⎡⎤⎛⎞⎛⎞
=+⎢⎥⎜⎟⎜⎟⎜⎟⎜⎟× =+ ×
⎢⎥⎣⎦⎝⎠⎝⎠
23
2 48 2
lwl
H
=+
A little consideration will show, that since the limits of integration were from 0 to l/2 (taking C
as origin) therefore the above equation gives the length of half of the cable,
∴ Total length of the cable,
23 23
(^22482224)
lwl wl
Ll
HH
⎛⎞
=+⎜⎟⎜⎟=+
⎝⎠
Now substituting the value of
2
8
wl
H
yc
= in the above equation,
∴
23 2
2 2
18
24 3
8
wl yc
Ll l
wl l
yc
=+ × =+
⎛⎞
⎜⎟⎜⎟
⎝⎠
Example 15.5. A light foot bridge is to be provided over a river crossing of 80 m span. Find
the length of each cable, if the central dip is 6 m.
Solution. Given : Span (l) = 80 m and central dip (yc) = 6 m
We know that length of each cable,
8 2 2
8(6)
80 81.2 m
3380
Ll yc
l
×
=+ = + =
×
Ans.
Example 15.6. A steel wire, of uniform section, is hung in the form of a parabola. Find the
maximum horizontal span, if the central dip is 1/12th of the span and the strees in steel wire is not to
exceed 120 N/mm^2. Take mass density of the steel as 7800 kg/m^3.
- As per Binomial theorem,
(1 ) 1 (–1)^2 (–1)(–2)^3 ......
12 123
+=++xnxn nn x+nn n x+
×××