The document discusses three exercises involving calculations for cables. Exercise 1 involves determining the tension and minimum tension for a symmetric cable supported at both ends. Exercise 2 involves calculating the tension at supports and maximum tension for a cable with a uniform load subjected to a deflection in the middle. Exercise 3 asks to calculate the total length of a cable following a given equation between two points. Solutions are provided for each exercise involving integration, vector expressions, and the hyperbolic cosine function.

1.
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Verification of the value of s and
Exercise 1: The equation of a cable subjected to
its own weight (catenary) and supported at
. sinh
200. sinh
two points located at the same height (thus, the
200
cable is symmetric), is
0,02.
.
.
Tension in a point where the tangent forms
,
.
60° with the horizontal is 4 kN. Determine:
120
200
127,3
. 254,6
2
10
2
1271,57
As it is seen, both methods lead to the
same result.
Consider now a distance of 240 m
Exercise 2: A cable is subjected to a uniform
between the supports. Calculate the
continuous load (parabolic cable) of 2 kN/m
maximum tension at the cable and its
b)
Minimum tension (T0) and weight per
unit of length (p).
a)
and its deflection at the middle‐span is 0,1 m.
length.
a)
Solution:
a)
and D knowing the distance between them
cosh /
b)
200 cosh 0,02
4. cos 60° →
200
Solution:
2 ;
2
→
cosh
Ty
C
/
Ty
D
f
T0
120
200 cosh
.
120
≅ 237
200
2 kN/m
2370
a)
Tension
≅ 1271,57
0,1
127,33
.
2
2m
T0
We work with half of the cable, so x is:
2
0,1 ;
/
Tension and length
240
2
Vectorial expression of the tension at C
and D.
200 .
cos
b)
is 2 m.
Minimum tension
then
Determine the tension at the supports C
,
2
2
2
.
2

2.
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2
.2
2
.
2
10,19
b)
Vectorial expressions.
,
,
Exercise 3: A cable following the equation
y=(2/3).x1,5 is suspended between two points
located at x=0 and x=1. Calculate its total
Solution:
→
1
length.
4√2
3
2
→
3
1
/
1
,
3
√
2