SOLUTION
cos u
max
cos 38.66
°
*7–108.
The cable is subjected to a uniform loading of
w = 200 lb
>
ft. Determine the maximum and minimum
tension in the cable.
100 ft
20 ft
y
x
AB
200 lb/ft
7–109.
If the pipe has a mass per unit length of 1500 kg m,
determine the maximum tension developed in the cable.
SOLUTION
we can write
Applying the boundary condition at x=0, results in c1=0.
Applying the boundary condition y=0 at x =0 results in c2 =0. Thus,
Applying the boundary condition y =3 m at x=15 m, we have
Substituting this result into Eq. (1), we have
The maximum tension occurs at either points at Aor Bwhere the cable has the
greatest angle with the horizontal. Here,
Thus,
T
max =F
H
cos umax
=551.8(103)
cos 21.80° =594.32(103) N =594 kN
umax =tan–1 ady
dx
`
15 m b= tan–1 [0.02667(15)] =21.80°
dy
dx =0.02667x
F
H=551.81(103) N3 =7.3575(103)
F
H
(15)2
y=7.3575(103)
F
H
x2
dy
dx =0
=1
F
Ha14.715(103)
2x2+c1x+c2b
y=1
F
HLaLw0dxbdx
>
30 m
3 m
AB
7–110.
If the pipe has a mass per unit length of 1500 kg m,
determine the minimum tension developed in the cable.
SOLUTION
we can write
Applying the boundary condition at x=0, results in c1=0.
Applying the boundary condition y=0 at x =0 results in c2 =0. Thus,
Applying the boundary condition y =3 m at x=15 m, we have
Substituting this result into Eq. (1), we have
The minimum tension occurs at the lowest point of the cable, where .Thus,
T
min =F
H=551.81(103) N =552 kN
u =0°
dy
dx =0.02667x
F
H=551.81(103) N3 =7.3575(103)
F
H
(15)2
y=7.3575(103)
F
H
x2
dy
dx =0
=1
F
Ha14.715(103)
2x2+c1x+c2b
y=1
F
HLaLw0dxbdx
>
30 m
3 m
AB
*7–112.
SOLUTION
Boundary Conditions:
at , then from Eq.[1]
at , then from Eq.[2]
Thus,
[3]
[4]
, at then from Eq.[3]
at and the maximum tension occurs when .From
Eq.[4]
Thus,
The maximum tension in the cable is
10 =
18.125
h(0.6)
1
20.0256h2+1
Tmax =FH
cos umax
cos umax =1
20.0256h2+1
tan umax =dy
dx 2x–12.5m
=w0
18.125
hw0
x=0.0128h(12.5) =0.160h
u=umax
x=12.5 mu=umax
h=w0
2FH
(12.52)FH=78.125
hw0
x=12.5m,y=h
dy
dx =w0
FH
x
y=w0
2FH
x2
0=1
FH
(C1)C1=0x=0
dy
dx =0
0=1
FH
(C2)C2=0x=0y=0
dx =1
FH
T
h
e ca
bl
e w
ill
b
rea
k
w
h
en t
h
e max
i
mum tens
i
on reac
h
es
Determine the minimum sag hif it supports
the uniform distributed load of w=600 N>m.
T
max =10 kN.
h
25 m
600N
/m
7–113.
T
h
e ca
bl
e
i
s su
bj
ecte
d
to t
h
e para
b
o
li
c
l
oa
di
ng
where xis in ft. Determine the
equation
which defines the cable shape AB and
the maximum tension in the cable
.
y=f1x2
w
=15011–1x>50222lb>ft,
SOLUTION
At
,
At
,
y=1
FHa75x2–x2
200 b
C2=0y=0x=0
C1=0
dy
dx =0x=0
dy
dx =150x
FH
–1
50FH
x3+C1
FH
y=1
FH
(75x2–x4
200 +C1x+C2)
y=1
FHL[150(x–x3
3(50)2)+C1]dx
y=1
FHL(Lw(x)dx)dx
100ft
20 ft
y
x
AB
150lb/ft
7–114.
The power transmission cable weighs . If the
resultant horizontal force on tower is required to be
zero, determine the sag hof cable .
SOLUTION
Applying the boundary condition of cable AB, at
Solving by trial and error yields
Since the resultant horizontal force at B is required to be zero,
Applying the boundary condition of cable BC
Ans. =4.44 ft
h=11266.62
10 ccosh
B
10(–100)
11266.62
R
–1s
y=h at x=-100 ft to Eq. (1), we obtain
(F
H)BC =(F
H)AB =11266.62 lb.
(F
H)AB =11266.63 lb
10 =(F
H)AB
10
B
cosh
¢
10(150)
(F
H)AB
≤
–1
R
x=150 ft,y=10 ft
y=F
H
10
B
cosh
¢
10
F
H
x
≤
–1
R
ft
y=F
H
w0
B
cosh
¢
w0
F
H
x
≤
–1
R
BC
BD
10 lb
>
ft
ABhC
D
300 ft
10 ft
200 ft
7–115.
The power transmission cable weighs . If ,
determine the resultant horizontal and vertical forces the
cables exert on tower .
SOLUTION
Applying the boundary condition of cable AB, at
Solving by trial and error yields
Applying the boundary condition of cable BC,at to Eq. (2),
we have
Solving by trial and error yields
Thus, the resultant horizontal force at Bis
Ans.
Using Eq. (1), and
Thus, the vertical force of cables AB and BC acting
on point Bare
The resultant vertical force at Bis therefore
(F
v)R=(F
v)AB +(F
v)BC =1504.44 +1006.64
(F
v)BC =(F
H)BC tan (uB)BC =5016.58(0.20066) =1006.64 lb
(F
v)AB =(F
H)AB tan (uB)AB =11266.63(0.13353) =1504.44 lb
sin h
B
10(–100)
5016.58
R
=0.20066.
tan (uB)BC =
tan (uB)AB = sin h
B
10(150)
11266.63
R
=0.13353
(F
H)R=(F
H)AB –(F
H)BC =11266.63 –5016.58 =6250 lb =6.25 kip
(F
H)BC =5016.58 lb
10 =(F
H)BC
10
B
cosh
¢
10(100)
(F
H)BC
≤
–1
R
x=-100 fty=10 ft
(F
H)AB =11266.63 lb
10 =(F
H)AB
10
B
cos h
¢
10(150)
(F
H)AB
≤
–1
R
x=150 ft,y=10 ft
y=F
H
10
B
cos h
¢
10
F
H
x
≤
–1
R
ft
y=F
H
w0
B
cos h
¢
w0
F
H
x
≤
–1
R
BD
h=10 ft10 lb
>
ft
ABhC
D
300 ft
10 ft
200 ft
738
*7–116.
SOLUTION
Performing the integration yields
(1)
From Eq. 7–14
(2)
Boundary Conditions:
at From Eq. (2)
Then, Eq. (2) becomes
(3)
at and use the result From Eq. (1)
Rearranging Eq. (1), we have
(4)
Substituting Eq. (4) into (3) yields
Performing the integration
(5)
at From Eq. (5) thus, C
3
=-
F
H
3
0=F
H
3cosh 0 +C
3
,x=0.y=0
y=F
H
3cosh 3
F
H
x+C
3
dy
dx =sinh
¢
3
F
H
x
≤
s=F
H
3sinh
¢
3
F
H
x
≤
x=F
H
3
b
sinh
–1
B
1
F
H
10+02
R
+C
2
r
C
2
=0
C
1
=0.x=0s=0
dy
dx =tan u=3s
F
H
0=1
F
H
10+C
1
2C
1
=0s=0.
dy
dx =0
dy
dx =1
F
H
Lw
0
ds =1
F
H
13s+C
1
2
x=F
H
3
b
sinh
–1
B
1
F
H
13s+C
1
2
R
+C
2
r
31+11>F
H
2
211w
0
ds2
2
4
2
The man picks up the 52-ft chain and holds it just high
enough so it is completely off the ground. The chain has
points of attachment Aand Bthat are 50 ft apart. If the chain
has a weight of 3 lb/ft, and the man weighs 150 lb,determine
the force he exerts on the ground. Also,how high hmust he
lift the chain? Hint:The slopes at Aand Bare zero.
AB
h
25 ft 25 ft
*7–116. Continued
Then, Eq. (5) becomes
(6)
at From Eq. (4)
By trial and error
at From Eq. (6)
Ans.
From Eq. (3)
The vertical force that each chain exerts on the man is
26 ft
tan 3 26
154.003 0.5065
26.86°
154.003
3cosh 3
154.003 25 1 6.21 ft
25 ft.
154.003 lb
26 3sinh 325
25 ft.26 ft
3cosh 31
7–117.
The cable has a mass of 0.5 and is 25 m long.
Determine the vertical and horizontal components of force
it exerts on the top of the tower.
kg
>
m,
SOLUTION
x=
ds
2
1
2
30
B
A
7–118.
From Example 7–13:
Solving,
Ans.
Then,
Ans.T
max
=F
H
cos u
max
=200
cos 84.3° =2.01 kip
u
max
=tan
–1
B
79.9 (25)
200
R
=84.3°
dy
dx
`
max
=tan u
max
=w
0
s
F
H
Total weight =w
0
l=79.9 (50) =4.00 kip
w
0
=79.9 lb>ft
50
2=200
w
0
sinh aw
0
200 a15
2bb
s=F
H
w
0
sinh aw
0
x
F
H
b
A 50-ft cable is suspended between two points a distance of
15 ft apart and at the same elevation. If the minimum
tension in the cable is 200 lb, determine the total weight of
the cable and the maximum tension developed in the cable.
742
7–119.
S
h
ow t
h
at t
h
e
d
ef
l
ect
i
on curve of t
h
e ca
bl
e
di
scusse
d
i
n
Example 7–13 reduces to Eq. 4 in Example 7–12 when the
hyperbolic cosine function is expanded in terms of a series
and only the first two terms are retained. (The answer
indicates that the catenary may be replaced by a parabola in
the analysis of problems in which the sag is small. In this
case, the cable weight is assumed to be uniformly
distributed along the horizontal.)
SOLUTION
Substituting into
Using Eq. (3) in Example 7–12,
We get QE
D
y=4h
L2x2
F
H= w0L2
8h
= w0x2
2F
H
=F
H
w0
B
1+w2
0x2
2F2
H
+Á–1
R
y=F
H
w0
B
cosh
¢
w0
F
H
x
≤
–1
R
cosh x=1+x2
21 +Á
*7–120.
A telephone line (cable) stretches between two points which
are 150 ft apart and at the same elevation.The line sags 5 ft
and the cable has a weight of 0.3 lb ft. Determine the
length of the cable and the maximum tension in the cable.
SOLUTION
From Example 7–13,
Ans.
Ans.L=2s=150 ft
s=169.0
0.3 sinhc0.3
169.0 1752d =75.22
T
max =F
H
cos umax
=169
cos 7.606° =170 lb
umax =tan–1c
sinh
¢
7510.32
169
≤
d=7.606°
dy
dx
`
max
=tan umax = sinh
¢
w
F
H
x
≤
`
x=75 ft
F
H=169.0 lb
5=F
H
w
B
cosh
¢
75w
F
H
≤
–1
R
w=0.3 lb>fty=5 ft,At x=75 ft,
y=F
H
w
B
cosh
¢
w
F
H
x
≤
–1
R
s=F
H
w sinh
¢
w
F
H
x
≤
w=0.3 lb>ft
>
7–121.
A cable has a weight of 2 lb ft. If it can span 100 ft and has
a sag of 12 ft, determine the length of the cable.The ends of
the cable are supported from the same elevation.
From Eq. (3) of Example 7–13:
Ans.l=104 ft
l
2=212.2
2 sinh a21502
212.2 b
s=F
H
w0
sinh
¢
w0
F
H
x
≤
F
H=212.2 lb
24 =F
H
B
cosh
¢
100
F
H
≤
–1
R
12 =F
H
2
B
cosh
¢
211002
2 F
H
≤
–1
R
w0
2F
H
>
7–122.
SOLUTION
A cable has a weight of 3 lb ft and is supported at points
that are 500 ft apart and at the same elevation. If it has a
length of 600 ft, determine the sag.
7–123.
Acable has a weight of 5 lb/ft.If it can span 300 ft and has a
sag of 15 ft, determine the length of the cable. The ends of the
cable are supported at the same elevation.
747
*7–124.
The cable is suspended between the supports
and . If the cable can sustain a maximum tension of
and the maximum sag is , determine the
maximum distance between the supports
SOLUTION
Applying the boundary equation m at we have
The maximum tension occurs at either points Aor Bwhere the cable makes the
greatest angle with the horizontal. From Eq. (1),
By referring to the geometry shown in Fig. b, we have
hus,
(3)
Solving Eqs. (2) and (3) yields
Ans.
F
H=1205.7 N
L=16.8 m
1500 =F
H cosh
¢
49.05L
F
H
≤
T
max =FH
cos umax
cos umax =1
A1+ sinh2
¢
49.05L
F
H
≤
=1
cosh
¢
49.05L
F
H
≤
tan umax =sinh
¢
49.05L
F
H
≤
3=F
H
98.1
B
cosh
¢
49.05L
F
H
≤
–1
R
x=L
2,y=3
y=F
H
98.1
B
cosh
¢
98.1x
F
H
≤
–1
R
y=F
H
w0
B
cosh
¢
w0
F
H
x
≤
–1
R
L
3 m1.5 kN
B
A10 kg m
AB
L
3 m