1. Catenary Tutorial Part-1
Engineering 25
Catenary Tutorial Part-1
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer

2. 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
Where

3. UNloaded Cable → Catenary (2)
Next, relate horizontal distance, x, to cable-length s
But by Force Balance Triangle
Also From last slide recall
Thus

4. UNloaded Cable → Catenary (3)
Factoring Out c
Finally the Integral Eqn
Integrate Both Sides using Dummy Variables of Integration:
σ: 0→x η: 0→s

5. UNloaded Cable → Catenary (4)
Using σ: 0→x η: 0→s
Now the R.H.S. AntiDerivative is the argSINH
Noting that

6. UNloaded Cable → Catenary (5)
Thus the Solution to the Integral Eqn
Then
Solving for s in terms of x

7. UNloaded Cable → Catenary (6)
Finally, Eliminate s in favor of x & y. From the Diagram
From the Force Triangle
And From Before
So the Differential Eqn

8. UNloaded Cable → Catenary (7)
Recall the Previous Integration That Relates x and s
Using s(x) above in the last ODE
Integrating with Dummy Variables:
Ω: c→y σ: 0→x

9. UNloaded Cable → Catenary (8)
Noting that cosh(0) = 1
Solving for y yields the Catenary Equation in x&y:
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)

10. Catenary Tension, T(y) With Hyperbolic-Trig ID: cosh2 – sinh2 = 1
Thus:
Recall From the Differential Geometry

11. y = c Catenary Cabling Contraption
Shape is defined by the Catenary Equation
Note that the ORIGIN for y is the Distance “c” below the HORIZONTAL Tangent Point
y = c

12. 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)