# Worked Examples ```6.7 Cables: Catenaries
6.7 Cables: Catenaries Example 1, page 1 of 4
1. An electric power line of length 140 m and mass per
unit length of 3 kg/m is to be suspended between two
towers 120 m apart and of the same height. Determine
the sag and maximum tension in the power line.
120 m
A
B
6.7 Cables: Catenaries Example 1, page 2 of 4
1 Many problems involving catenary cables can be
solved using the following formulas:
s = c sinh (x/c)
(1)
y2
(2)
s2 = c2
W = ws
(3)
y = c cosh (x/c)
(4)
To = wc
(5)
T = wy
(6)
where
y
T
s
(x, y) To
w = weight of cable per unit length of cable,
and
c
W = weight of length of cable from low
point to a point a distance s along
the cable.
Other quantities are defined in the figure to the
right.
x
tension
in cable
horizontal
component
of tension
6.7 Cables: Catenaries Example 1, page 3 of 4
sB
A
y
(140 m)/2
70 m
B
2 The sag, h, can be found from Eq. 2, provided that we
can determine the distance, c:
yB2
Sag
xB
(120 m)/2
60 m
h
sB2 = c2
or,
(h + c)2
yB
(70 m)2 = c2
(7)
The distance, c, can be determined from Eq. 1:
c
sB = c sinh (xB/c)
x
120 m
(Eq. 2 evaluated at point B)
(Eq. 1 evaluated at point B)
or,
70 m = c sinh (60 m/c)
(8)
This equation must be solved numerically for c. An initial
estimate for c, when the solver on a calculator is to be
used, could be
c = sB = 70 m
The solution to Eq. 8 is
c = 61.45 m
Another possible solution is c = 61.45 m, but this has no
physical meaning.
6.7 Cables: Catenaries Example 1, page 4 of 4
3 Substituting c = 61.45 m into Eq. 7 gives
(h + c)2
(70 m) = c2
(h + 61.45 m)2
4
(Eq. 7 repeated)
The maximum tension, Tmax, occurs where the cable
has its steepest slope, point B ( or point A).
(70 m)2 = (61.45 m)2
Tension
A
y
B
Solving gives the sag
h
h = 154.6 m
B
T
max
31.70 m
and
h = 31.70 m
yB
Ans.
c
61.45 m
The negative root has no physical meaning.
x
5 The maximum tension can be calculated from Eq. 6:
Tmax = wyB
(Eq. 6 evaluated at point B)
w is given.
= (3 kg/m)(9.81 m/s2) (31.70 m + 61.45 m)
= 2740 N
= 2.74 kN
Ans.
6.7 Cables: Catenaries Example 2, page 1 of 4
2. A cable is supported at two points 400 ft apart and at
the same elevation. If the sag is 40 ft and the weight per
unit length of the cable is 4 lb/ft, determine the length of
the cable and the tension at the low point, C.
400 ft
40 ft
A
B
C
6.7 Cables: Catenaries Example 2, page 2 of 4
1
Many problems involving catenary cables can be
solved using the following formulas:
s = c sinh (x/c)
(1)
y2
(2)
s2 = c2
W = ws
(3)
y = c cosh (x/c)
(4)
To = wc
(5)
y
T = wy
T
(6)
where
s
(x, y) To
w = weight of cable per unit length of cable,
and
c
W = weight of length of cable from low
point to a point a distance s along
the cable.
Other quantities are defined in the figure to the
right.
x
tension
in cable
horizontal
component
of tension
6.7 Cables: Catenaries Example 2, page 3 of 4
2 The length of cable, sB, from the low point to point B can be
found from Eq. 1, provided that we can determine the
distance c:
sB = c sinh (xB/c)
(Eq. 1 evaluated at point B)
= c sinh (200 ft/c)
y
sB
yB = c cosh (xB/c)
(Eq. 4 evaluated at point B)
or,
c + 40 ft = c cosh (200 ft/c)
(7)
(8)
This equation must be solved numerically. An initial
estimate for c, when the solver on a calculator is to be
used, could be
400 ft
A
3 The distance c can be determined from Eq. 4:
B
xB
c = sag = 40 ft.
(400 ft)/2
Sag
200 ft
A
y
40 ft
B
c
xB
x
200 ft
c
yB
x
6.7 Cables: Catenaries Example 2, page 4 of 4
4
The solution to Eq. 8 is
c = 506.53 ft
Using this value of c in Eq. 7 gives
sB = c sinh (200 ft/c)
(Eq. 7 repeated)
= (506.53 ft) sinh (200 ft/506.53 ft)
= 205.237 ft
By symmetry, the total length of cable is twice sB:
sTotal = 2
205.237 ft
= 410 ft
Ans.
Because the tension at the low point of the cable is
horizontal, it can be found from Eq. 5:
To = wc
w is given.
(Eq. 5 repeated)
= (4 lb/ft)(506.53 ft)
= 2,025 lb
Ans.
6.7 Cables: Catenaries Example 3, page 1 of 4
3. A 20-m chain is suspended between two points at
the same elevation and with a sag of 6 m as shown.
If the total mass of the chain 45 kg, determine the
distance between the supports. Also determine the
maximum tension.
A
B
6m
6.7 Cables: Catenaries Example 3, page 2 of 4
1 Many problems involving catenary cables can be
solved using the following formulas:
s = c sinh (x/c)
(1)
y2
(2)
s2 = c2
W = ws
(3)
y = c cosh (x/c)
(4)
To = wc
(5)
T = wy
(6)
where
y
T
s
(x, y) To
w = weight of cable per unit length of cable,
and
c
W = weight of length of cable from low
point to a point a distance s along
the cable.
Other quantities are defined in the figure to the
right.
x
tension
in cable
horizontal
component
of tension
6.7 Cables: Catenaries Example 3, page 3 of 4
B
y
A
sB
y
A
sB
(20 m)/2
10 m
B
10 m
Sag
6m
xB
c
yB
x
2 The distance between the supports is 2xB, and xB can be
found from Eq. 1, provided that we can determine the
distance c;
sB = c sinh (xB/c)
c
x
3 The distance, c, can be determined from Eq. 2:
(Eq. 1 evaluated at point B)
10 m
yB2
sB2 = c2
or
This equation can be solved explicitly for xB by
rearranging it as
(6 m + c)2
(10 m)2 = c2
or
sinh (xB/c) = 10 m/c
36 + 12 c + c2
which implies
100 = c2
The c2 terms cancel and the resulting linear equation
has the solution
xB/c = sinh-1(10 m/c)
so
c = 5.333 m
xB = c sinh-1(10 m/c)
(Eq. 2 evaluated at point B)
(7)
6.7 Cables: Catenaries Example 3, page 4 of 4
4 Substituting this value of c into Eq. 7 gives
xB = c sinh-1(10 m/c)
(Eq. 7 repeated)
= (5.333 m) sinh-1(10 m/5.333 m)
= 7.393 m
Thus the distance between supports, 2xB, can be found:
2xB = 2(7.393 m)
= 14.786 m
Ans.
The maximum tension occurs where the slope of the cable is a
maximum, at end B (or at A). Eq. 6 gives
Tmax = wyB
T
A
y
B
(Eq. 6 evaluated at point B)
Sag
=
Total weight of the cable
y
Total length of the cable B
=
(45 kg)(9.81 m/s2)
(6 m + 5.333 m)
20 m
= 250 N
Tmax
Ans.
6m
yB
c
5.333 m
x
6.7 Cables: Catenaries Example 4, page 1 of 6
4. A certain cable will break if the maximum tension
exceeds 500 N. If the cable is 50-m long and has a
mass of 50 kg, determine the greatest span possible.
Also determine the sag.
A
B
Supports A and B are at the same elevation.
6.7 Cables: Catenaries Example 4, page 2 of 6
1 Many problems involving catenary cables can be
solved using the following formulas:
s = c sinh (x/c)
(1)
y2
(2)
s2 = c2
W = ws
(3)
y = c cosh (x/c)
(4)
To = wc
(5)
T = wy
(6)
y
where
T
s
(x, y) To
w = weight of cable per unit length of cable,
and
c
W = weight of length of cable from low
point to a point a distance s along
the cable.
Other quantities are defined in the figure to the
right.
x
tension
in cable
horizontal
component
of tension
6.7 Cables: Catenaries Example 4, page 3 of 6
2 The maximum tension has been specified (500 N), so
a good place to start our solution is to see how we can
use the fact that Tmax = 500 N. Eq. 6 relates the
tension, T, to the y coordinate of a point on the curve:
T = wy
(Eq. 6 repeated)
Since Tmax occurs where the curve has its steepest
slope, Eq. 6 gives Tmax at point B:
Tmax = wyB
(Eq. 6 evaluated at point B)
T
A
3 Thus because we know the maximum tension, we can
compute yB:
Tmax
y
yB = Tmax
w
B
=
yB
=
x
Tmax
Total weight of the cable
Total length of the cable
500 N
(50 kg)(9.81 m/s2)
Given
= 50.97 m
50 m
(7)
6.7 Cables: Catenaries Example 4, page 4 of 6
4 The distance between supports is 2xB, so we need to use the
value of yB to determine xB. This can be done by using Eq. 4,
provided that we can determine c:
yB = c cosh (xB/c)
(Eq. 4 evaluated at point B)
We can solve this equation explicitly for xB by rewriting it as
A
2 xB
(By symmetry)
y
B
cosh (xB/c) = yB/c
so
xB/c = cosh-1(yB/c)
c
or,
xB = c cosh-1(yB/c)
xB
yB
(8)
x
6.7 Cables: Catenaries Example 4, page 5 of 6
5 The distance, c, can be calculated from Eq. 2:
yB2
sB2 = c2
(Eq. 2 evaluated at point B)
or,
sB
A
(50.97 m)2
(50 m)/2
25 m
y
B
(25 m)2 = c2
The solution is
c = 44.42 m
The negative root has no physical meaning.
c
yB
50.97 m
Substituting c = 44.42 m and yB = 50.97 m in Eq. 8
gives
xB = c cosh-1(yB/c)
(Eq. 8 repeated)
-1
= (44.42 m) cosh (50.97 m/44.42 m)
= 23.836 m
So the distance between supports, 2xB, is known:
2xB = 2(23.836 m)
= 47.7 m
Ans.
x
6.7 Cables: Catenaries Example 4, page 6 of 6
6 Since c and yB are known, the sag h can be computed:
h = yB
c
= (50.97 m)
(44.42 m)
= 6.55 m
A
Ans.
y
B
Sag
c
44.42 m
h
yB
50.97 m
x
6.7 Cables: Catenaries Example 5, page 1 of 4
5. The cable is attached to a fixed support at A and a
moveable support at B. If the cable is 80-ft long, weighs
0.3 lb/ft, and spans 50 ft, determine the force F holding
the moveable support in place. Also determine the sag.
A
B
F (force)
6.7 Cables: Catenaries Example 5, page 2 of 4
1
Many problems involving catenary cables can be
solved using the following formulas:
s = c sinh (x/c)
(1)
y2
(2)
s2 = c2
W = ws
(3)
y = c cosh (x/c)
(4)
To = wc
(5)
T = wy
(6)
y
T
where
s
(x, y) To
w = weight of cable per unit length of cable,
and
c
W = weight of length of cable from low
point to a point a distance s along
the cable.
Other quantities are defined in the figure to the
right.
x
tension
in cable
horizontal
component
of tension
6.7 Cables: Catenaries Example 5, page 3 of 4
2
The force F acting on the moveable support at B equals the
horizontal component, T , of tension in the cable, F = T .
Eq. 5 can be used to calculate T , provided that we can
determine the distance, c:
T = wc
y
(Eq. 5 repeated)
Given
B
A
= (0.3 lb/ft)c
(7)
sB
The distance, c, can be calculated from Eq. 1:
sB = c sinh (xB/c)
(80 ft)/2
(Eq. 1 evaluated at point B)
c
or,
x
40 ft = c sinh (25 ft/c)
This equation must be solved numerically. An initial
estimate for c could be c = xB = 25 m, when finding c with
the use of the solver on a calculator. The solution to Eq. 8 is
c = 14.229 ft
The negative root has no physical meaning.
xB
(8)
50 ft
(50 ft)/2
25 ft
40 ft
6.7 Cables: Catenaries Example 5, page 4 of 4
3 Using c = 14.229 ft in Eq. 7 gives
T = (0.3 lb/ft)c
(Eq. 7 repeated)
= (0.3 lb/ft)(14.229 ft)
= 4.27 lb
Since F = T , we have
F = 4.27 lb
Ans.
y
The sag, h, can be calculated from Eq. 4 and the known
value of c:
xB
A
h = yB
25 ft
B
c
Sag
By Eq. 4
h
yB
= c cosh (xB/c)
c
= (14.229 ft) cosh (25 ft/14.229 ft)
= 28.2 ft
Ans.
14.229 ft
c
14.229 ft
x
6.7 Cables: Catenaries Example 6, page 1 of 4
6. The cable is attached to a fixed support at
A and a moveable support B. If the cable is
40-m long and has a mass of 0.4 kg/m,
determine the span and sag.
A
B
50 N
6.7 Cables: Catenaries Example 6, page 2 of 4
1 Many problems involving catenary cables can be
solved using the following formulas:
s = c sinh (x/c)
(1)
y2
(2)
s2 = c2
W = ws
(3)
y = c cosh (x/c)
(4)
To = wc
(5)
T = wy
(6)
where
y
T
s
(x, y) To
w = weight of cable per unit length of cable,
and
c
W = weight of length of cable from low
point to a point a distance s along
the cable.
Other quantities are defined in the figure to the
right.
x
tension
in cable
horizontal
component
of tension
6.7 Cables: Catenaries Example 6, page 3 of 4
2 The span is 2xB. The coordinate, xB, can be found from
Eq. 1, provided that the distance c can be found:
sB = c sinh (xB/c)
(Eq. 1 evaluated at point B)
This equation can be rewritten to give xB explicitly:
sinh (xB/c) = sB/c
so
A
xB/c = sinh-1 (sB/c)
2 xB
(by symmetry)
y
B
sB
or
xB = c sinh-1 (sB/c)
= c sinh-1 (20 m/c)
c
xB
(7)
x
(40 m)/2
20 m
6.7 Cables: Catenaries Example 6, page 4 of 4
3 Because the 50-N force acting on the moveable
support equals the horizontal component, T , of the
tension in the cable, Eq. 5 with T = 50 N can be used
to solve for c:
T = wc
(Eq. 5 repeated)
or
50 N = (0.4 kg/m)(9.81 m/s2) c
Solving gives
c = 12.742 m
Using this value of c in Eq. 7 gives
xB = c sinh-1 (20 m/c)
(Eq. 7 repeated)
= (12.472 m) sinh-1 (20 m/12.472 m)
= 15.708 m
So the span is
Span = 2xB
= 2(15.708 m)
= 31.4 m
Ans.
6.7 Cables: Catenaries Example 7, page 1 of 4
7. A cable goes over a frictionless pulley at B and
supports a block of mass m. The other end of the cable
is pulled by a horizontal force P. If the cable has a mass
per length of 0.3 kg/m, determine values of P and m that
will maintain the cable in the position shown.
B
5m
A
m
P (horizontal)
10 m
6.7 Cables: Catenaries Example 7, page 2 of 4
1 Many problems involving catenary cables can be
solved using the following formulas:
s = c sinh (x/c)
(1)
y2
(2)
s2 = c2
W = ws
(3)
y = c cosh (x/c)
(4)
To = wc
(5)
T = wy
(6)
y
T
where
s
(x, y) To
w = weight of cable per unit length of cable,
and
c
W = weight of length of cable from low
point to a point a distance s along
the cable.
Other quantities are defined in the figure to the
right.
x
tension
in cable
horizontal
component
of tension
6.7 Cables: Catenaries Example 7, page 3 of 4
y
2 The force P equals T , the horizontal component of
the cable tension given by Eq. 5:
xB = 10 m
T = wc
B
(Eq. 5 repeated)
so, with T = P
P = wc
5m
(7)
Here,
A
w = (0.3 kg/m)(9.81 m/s )
= 2.943 N/m
yB
P
2
(8)
c
The value of c in Eq. 7 can be found from Eq. 4:
yB = c cosh (xB/c)
x
(Eq. 4 repeated)
or,
5 m + c = c cosh (10 m/c)
Solving numerically gives
c = 10.743 m
3
Using this value of c in Eq. 7 gives
P = wc
(Eq. 7 repeated)
by Eq. 8
= (2.943 N/m)(10.743 m)
= 31.617 N
Ans.
6.7 Cables: Catenaries Example 7, page 4 of 4
Pulley B
4 The cable tension at B must equal the weight, mg:
TB = mg
TB
Thus the mass is
m = TB/g
mg
by Eq. 6
TB
y
= wyB/g
mg
B
by Eq. 8
5m
= (2.943 N/m)(5 m + 10.743 m) /9.81 N/m2
= 4.72 kg
Ans.
P
A
yB
c
10.743 m
x
6.7 Cables: Catenaries Example 8, page 1 of 5
8. A chain makes angles of 30° and 60° at its
supports as shown. Determine the location of the
low point C of the chain relative to A. Also
determine the tension at support A, if the cable has
a mass per length of 0.6 kg/m.
Supports A and B
are at different
elevation.
30°
A
C
20 m
60°
B
6.7 Cables: Catenaries Example 8, page 2 of 5
1 Many problems involving catenary cables can be
solved using the following formulas:
s = c sinh (x/c)
(1)
y2
(2)
s2 = c2
W = ws
(3)
y = c cosh (x/c)
(4)
To = wc
(5)
T = wy
(6)
where
y
T
s
(x, y) To
w = weight of cable per unit length of cable,
and
c
W = weight of length of cable from low
point to a point a distance s along
the cable.
Other quantities are defined in the figure to the
right.
x
tension
in cable
horizontal
component
of tension
6.7 Cables: Catenaries Example 8, page 3 of 5
y
2 The geometric data are shown in the figure. To
determine the location of the low point C relative to A,
we need to determine the coordinates xA and yA. We
can get an equation for xA by using the fact that the
slope is known at A:
dy
tan 30° =
dx A
60°
B (xB, yB)
30°
A (xA, yA)
C
by Eq. 4
c
d
= dx c cosh (x/c)
A
x
= sinh (xA/c)
Solving for xA gives
xA = c sinh-1( tan 30°)
A
(7)
Similarly at point B, we have
xB = c sinh-1(tan 60°)
(8)
dx
30°
20 m
dy (Slope is negative)
6.7 Cables: Catenaries Example 8, page 4 of 5
3
The coordinates xA and xB are related to the 20-m span
through the equation
xB
y
60°
B (xB, yB)
xA = 20 m
30°
A (xA, yA)
or,
by Eq. 8
c sinh-1(tan 60°)
by Eq. 7
C
c sinh-1( tan 30°) = 20
c
Since this equation is linear in c, it is easily solved to
give c = 10.717 m. Eq. 7 then gives
xA = c sinh-1( tan 30°)
(Eq. 7 repeated)
20 m
= (10.717 m) sinh-1( tan 30°)
= 5.887 m
Ans.
The y coordinate of point A can now be calculated from
Eq. 4:
yA = c cosh (xA/c) (Eq. 4 evaluated at point A)
= (10.717 m) cosh ( 5.887 m/10.717 m)
= 12.375 m
x
(9)
6.7 Cables: Catenaries Example 8, page 5 of 5
4 The vertical distance between support A and the low
point C is given by
y
B
5.887 m
d = yA c
d
by Eq. 9
= 12.375 m
A
10.717 m
C
= 1.658 m
Ans.
yA
c = 10.717 m
The tension at A is given by Eq. 6:
TA = wyA
(w given)
(Eq. 6 evaluated at point A)
by Eq. 9
= (0.6 kg/m)(9.81 m/s2) (12.375 m)
= 72.8 N
Ans.
x
6.7 Cables: Catenaries Example 9, page 1 of 4
9. A wire weighing 0.2 lb/ft is attached to a moveable
support at A and makes an angle of 55° at a fixed
support at B. Supports A and B are at different
elevations. Determine the location of the low point C
of the wire relative to support B. Also, determine the
tension in the wire at C.
A
8 lb
B
C
55°
6.7 Cables: Catenaries Example 9, page 2 of 4
1 Many problems involving catenary cables can be
solved using the following formulas:
s = c sinh (x/c)
(1)
y2
(2)
s2 = c2
W = ws
(3)
y = c cosh (x/c)
(4)
To = wc
(5)
T = wy
(6)
where
y
T
s
(x, y) To
w = weight of cable per unit length of cable,
and
c
W = weight of length of cable from low
point to a point a distance s along
the cable.
Other quantities are defined in the figure to the
right.
x
tension
in cable
horizontal
component
of tension
6.7 Cables: Catenaries Example 9, page 3 of 4
y
2 To determine the location of the low point, C, relative to
the support at B, we need to determine the coordinates xB
and yB. We can get an equation for xB by using the fact
that the slope is known at B:
dy
tan 55° =
dy
55°
dx B
B
dx
by Eq. 4
d
= dx c cosh (x/c)
B
A
55°
B (xB, yB)
C
c
x
= sinh (xB/c)
Thus
xB = c sinh-1( tan 55°)
(7)
The value of c occurring in Eq. 7 can be found by
observing that the 8-lb force acting at support A equals T ,
the horizontal component of tension at A, so Eq. 5 gives
A
To
8 lb
B
T = wc
8 lb
(Eq. 5 repeated)
w
0.2 lb/ft
C
Solving gives
c = 40 ft
0.2 lb/ft
(8)
6.7 Cables: Catenaries Example 9, page 4 of 4
3
y
Using this result, c = 40 ft, in Eq. 7 gives
xB = c sinh-1(tan 55°)
A
(Eq. 7 repeated)
xB
46.169 ft
-1
= (40 ft) sinh (tan 55°)
= 46.169 ft
B
Ans.
d
The vertical distance between B and C is
yB
C
d = yB
c
c = 40 ft
by Eq. 4
= c cosh (xB/c)
x
c
= (40 ft) cosh (46.169 ft/40 ft)
= 29.7 ft
40 ft
Ans.
Free body diagram of portion AC of cable
Since point C is the low point of the cable, the tension
there is horizontal and so must equal the horizontal
component of tension at A which is known to be 8 lb.
That is,
TC = 8 lb
A
8 lb
Weight of portion
AC of cable
Ans.
Vertical force
acting on support
C
TC
6.7 Cables: Catenaries Example 10, page 1 of 6
10. Determine the location of the low point C relative to
the support at A. Also determine the tension at C, if the
mass of the cable per unit length is 1 kg/m.
A
B
2m
1m
C
10 m
6.7 Cables: Catenaries Example 10, page 2 of 6
1 Many problems involving catenary cables can be
solved using the following formulas:
s = c sinh (x/c)
(1)
y2
(2)
s2 = c2
W = ws
(3)
y = c cosh (x/c)
(4)
To = wc
(5)
T = wy
(6)
where
y
T
s
(x, y) To
w = weight of cable per unit length of cable,
and
c
W = weight of length of cable from low
point to a point a distance s along
the cable.
Other quantities are defined in the figure to the
right.
x
tension
in cable
horizontal
component
of tension
6.7 Cables: Catenaries Example 10, page 3 of 6
2 The geometric data are shown in the figure.
To locate the low point, C, relative to support A, we
must determine the coordinate xA. We know that xA is
related to yA through Eq. 4:
A(xA, yA)
2m+1m
y
B(xB, yB)
3m
1m
C
yA = c cosh (xA/c) (Eq. 4 evaluated at point A)
And this can be solved for xA to give
-1
xA = c cosh (yA/c)
c
(7)
The minus sign must be inserted here to make xA
negative because we will consider "cosh-1 " to be the
principal value of the inverse cosh function, and the
principal value is always positive.
x
10 m
cosh-1 v
(principal value)
cosh u
1
Branch of cosh function
corresponding to the
principal value of the inverse
function
u
v
1
Note that cosh-1 v is not defined for v < 1
6.7 Cables: Catenaries Example 10, page 4 of 6
3
An equation analogous to Eq. 7 can be written for point B:
xB = c cosh-1(yB/c)
(8)
A
Here no minus sign has to be inserted, because xB is positive.
The figure shows that
xB
y
B
3m
By Eq. 8
yA
By Eq. 7
-1
c cosh (yB/c)
1m
C
xA = 10 m
c
-1
c cosh (yA/c) = 10 m
yB
or
c cosh-1(
1m+c
c
x
)
+ c cosh-1( 3 mc + c
10 m
)
10 m = 0
(9)
This equation must be solved numerically for c. It can be difficult to
solve with the solver on a calculator because the cosh-1 (u) function is
undefined for u 1, and, as a result, the solver may fail if, during the
solution procedure, an attempt is made to evaluate cosh-1 (u) for u 1.
6.7 Cables: Catenaries Example 10, page 5 of 6
4
One approach to finding the solution of Eq. 9 is to define
the left-hand side of the equation as a function,
f(z)
z cosh-1 (
1+z
z
) + z cosh-1 ( 3 +z z )
10
And then use the calculator to plot f(z) versus z. The point
z = zr where the f(z) curve crosses the z axis is the
approximate root of Eq. 9, that is, where f(zr) = 0.
Another approach to finding the solution of Eq. 9 is simply
to experiment with initial estimates for c while using the
solver on a calculator. For example, since the span was
given as 10 m, we might try c = 10 m, 50 m, 100 m as
successive initial estimates.
The solution to Eq. 9 is
A
c = 7.044 m
y
B
3m
Eq. 7 then gives
C
xA = c cosh (yA/c)
= c cosh ( 3 + c
c
(Eq. 7 repeated)
)
= (7.044 m) cosh ( 3 + 7.044 m )
7.044 m
= 15.5 m
Ans.
yA
c = 7.044 m
x
6.7 Cables: Catenaries Example 10, page 6 of 6
A
5 At the low point of the cable, the tension is horizontal, so
it can be calculated from Eq. 5:
y
B
To = wc
(Eq. 5 repeated)
C
To
Given
= [(1 kg/m)(9.81 m/s2)](7.044 m)
= 69.1 N
c
7.044 m
Ans.
x
```