Engineering 25
Catenary
Tutorial Part-1
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
ENGR/MTH/PHYS25: Computational Methods
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt
UNloaded Cable → Catenary

Consider a cable uniformly
loaded by the cable itself,
e.g., a cable hanging under
its own weight.
•
With loading on the cable from
lowest point C to a point D given
by W = ws, the Force Triangle on
segment CD reveals the internal
tension force magnitude, T

T  T w s  w T
2
0
2 2
– Where
2

w  s w c  s
2
2
2
2
c  T0 w
ENGR/MTH/PHYS25: Computational Methods
2
2
0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt
UNloaded Cable → Catenary (2)

Next, relate horizontal
distance, x, to cable-length s
dx  ds cos 

T0
But by Force
cos  
Balance Triangle
T

Also From last slide recall
T  w c2  s2
and T 0 wc
 Thus
T0
wc
c
dx  ds cos   ds
 ds

ds
T
w c2  s2
c2  s2
ENGR/MTH/PHYS25: Computational Methods
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt
UNloaded Cable → Catenary (3)

Factoring Out c
c
c
dx 
ds 
ds
c2  s2
c c2 c2  s2 c2

Finally the Integral Eqn
dx 

1
1 s2 c2
Integrate Both Sides using
Dummy Variables of Integration:
•
σ: 0→x
ENGR/MTH/PHYS25: Computational Methods
4
ds
η: 0→s
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt
UNloaded Cable → Catenary (4)

Using σ: 0→x
 x


 x

0

0
 s
d  
 0
η: 0→s
1
d
1 2 c2
Now the R.H.S. AntiDerivative is the argSINH
d   
 x
 0
 s

 0
 s

  
d  c  arg sinh  
2
2
 c  0
1  c

1
Noting that
arg sinh 0  sinh 1 0  0
ENGR/MTH/PHYS25: Computational Methods
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt
UNloaded Cable → Catenary (5)

Thus the Solution to the Integral Eqn
 
 s
 x
 0

  
1  s 
 x  0  c  arg sinh    c  sinh    0
 c  0
c

x
s
1  s 
 Then c  sinh    x  sinh   
c
c c
1

Solving for s in terms of x
 x
s  c sinh  
c
ENGR/MTH/PHYS25: Computational Methods
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt
UNloaded Cable → Catenary (6)

Finally, Eliminate s in favor
of x & y. From the Diagram
dy  dx tan 

From the
Force Triangle

And From Before
W  ws and T0  wc

So the Differential Eqn
W
tan  
T0
W
ws
s
dy  dx tan   dx 
dx  dx
T0
wc
c
ENGR/MTH/PHYS25: Computational Methods
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt
UNloaded Cable → Catenary (7)

 x
Recall the Previous Integration
s  c sinh  
That Relates x and s
c

Using s(x) above in the last ODE
1
1
 x
 x
dy  dx tan   sdx  c sinh  dx  sinh  dx
c
c
c
c

Integrating with Dummy Variables:
•

 y
 c
Ω: c→y
σ: 0→x
d  
 y
 c

  sinh 
 0
c
ENGR/MTH/PHYS25: Computational Methods
8
 x



d  c cosh 

c

 x


  0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt
UNloaded Cable → Catenary (8)

Noting that cosh(0) = 1
 x

  
 y
c

 y  c  c cosh  
 c  0

 x
 c cosh    c
c
Solving for y yields the
Catenary Equation in x&y:
y  c cosh x c 
•
Where
– c = T0/w
– T0 = the 100% laterally directed force at the ymin point
– w = the lineal unit weight of the cable (lb/ft or N/m)
ENGR/MTH/PHYS25: Computational Methods
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt
c  T0 w
Catenary Tension, T(y)

y  c cosh x c 
With Hyperbolic-Trig ID: cosh2 – sinh2 = 1
y  s  c cosh x c   c sinh x c 
2
2
2
2
2

2

 y  s  c cosh x c   sinh x c   c
2
2
2
2
2

Thus: y  s  c

Recall From the Differential Geometry
2
2
or c  s  y
2
2
2
2
T c, s   w c  s  w y  wy  T  y 
2
2
2
T  y   wy
ENGR/MTH/PHYS25: Computational Methods
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt
2
 Catenary Cabling
Contraption
 Shape is defined by
the Catenary
Equation
y  c cosh x c 
y=c
ENGR/MTH/PHYS25: Computational Methods
11
• Note that the
ORIGIN for y is the
Distance “c” below
the HORIZONTAL
Tangent Point
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt
The Problem
 An 8m length of chain
has a lineal unit mass of
3.72 kg/m. The chain is
attached to the Beam at
pt-A, and passes over a
small, low friction
pulley at pt-B.
 Determine the value(s)
of distance a for which
the chain is in equilibrium
(does not move)
ENGR/MTH/PHYS25: Computational Methods
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_Catenary_Tutorial_Part-1.ppt