### ENGR36_Tutorial_Catenary_Cables

```Engineering 25
Tutorial:
Catenary
Cables
Bruce Mayer, PE
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Engineering-36: Engineering Mechanics - Statics
1
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
Partial Catenary
 The cable
has a mass of
0.5 kg/m and
is 25 m long.
 Determine
the vertical
and
horizontal
components
of force it
exerts on the
top of the
tower.
Engineering-36: Engineering Mechanics - Statics
2
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
Eqns Used
dy
 sinh
dx
T0
dy
dx
µx
 sinh
µx
p
T0
x x p
µx Q
T0 
µx P
 sinh
S 
 sinh

µ 
T0
T0
 µx p
T p  T 0 cosh 
 T0




Engineering-36: Engineering Mechanics - Statics
3
tan  p  sinh
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx




µx p
T0
Engineering-36: Engineering Mechanics - Statics
4
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
Engineering-36: Engineering Mechanics - Statics
5
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
Engineering-36: Engineering Mechanics - Statics
6
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
Engineering-36: Engineering Mechanics - Statics
7
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
MATLAB Code
% Bruce Mayer, PE
% ENGR36 * 22Jul2
%
ENGR36_Tutorial_Partial_Catenary_H
13e_P7_119_1207.m
%
K1 = tand(30)
K2 = asinh(K1)
S0 = @(z)
(z/K2)*(sinh(K2*(z+15)/z)-K1) - 25
xB = fzero(S0, 10)
xA = xB+15
u = 0.5*9.81
TO = u*xB/K2
TA = TO*cosh(u*xA/TO)
QA = atand(sinh(u*xA/TO))
WA = TO*tand(QA)
Engineering-36: Engineering Mechanics - Statics
8
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
MATLAB Results
xB =
8.2804
xA =
23.2804
u =
4.9050
TO =
73.9396
TA =
181.0961
QA =
65.9026
WA =
165.3141
Engineering-36: Engineering Mechanics - Statics
9
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
Chain Lift Problem
 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 A
and B that 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 h
must he lift the chain? Hint: The
slopes at A and B are zero.
Engineering-36: Engineering Mechanics - Statics
10
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
Eqns Used
µx Q
T0 
µx P
 sinh
S 
 sinh

µ 
T0
T0

T0 
x
 cosh

y 

1


 
T0

 µx p
T p  T 0 cosh 
 T0
tan  p  sinh
Engineering-36: Engineering Mechanics - Statics
11




µx p
T0
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx




Engineering-36: Engineering Mechanics - Statics
12
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
Engineering-36: Engineering Mechanics - Statics
13
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
Engineering-36: Engineering Mechanics - Statics
14
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
MATLAB Code
% Bruce Mayer, PE
% ENGR36 * 22Jul2
%
ENGR36_Tutorial_Chain_Lift_Catenar
y_H13e_P7_124_1207.m
%
Zf1 = @(q) (q/3)*sinh(75/q) - 26
TO = fzero(Zf1, 150)
h = (TO/3)*(cosh(75/TO) - 1)
Q = atand(78/TO)
Th = 3*h + TO
Tup = Th*sind(Q)
Engineering-36: Engineering Mechanics - Statics
15
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
MATLAB Results
Zf1 =
@(q)(q/3)*sinh(75/q)-26
TO =
154.0033
h =
6.2088
Q =
26.8614
Th =
172.6297
Tup =
78
Engineering-36: Engineering Mechanics - Statics
16
Bruce Mayer, PE
ENGR36_H13_Tutorial_Catenary_Cables.pptx
```