[email protected] ENGR-25_Catenary_Tutorial_Part-1.ppt 1 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods Bruce Mayer, PE Registered Electrical

[email protected] • ENGR-25_Catenary_Tutorial_Part-1.ppt1

Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 25

Catenary Catenary Tutorial Tutorial Part-1Part-1



UNloaded Cable → CatenaryUNloaded Cable → Catenary Consider a cable uniformly

loaded by the cable itself, e.g., a cable hanging under its own weight.

222220

22220 scwswTwswTT

• 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

wTc 0– Where



UNloaded Cable → Catenary (2)UNloaded Cable → Catenary (2) Next, relate horizontal

distance, x, to cable-length s

cosdsdx But by Force

Balance Triangle

Thus

T

T0cos

Also From last slide recall

wcTscwT 022 and

dssc

c

scw

wcds

T

Tdsdsdx

2222

0cos



UNloaded Cable → Catenary (3)UNloaded Cable → Catenary (3) Factoring Out c

Integrate Both Sides using Dummy Variables of Integration: • σ: 0→x η: 0→s

dscsccc

cds

sc

cdx

222222

Finally the Integral Eqn

dscs

dx221

1



UNloaded Cable → Catenary (4)UNloaded Cable → Catenary (4) Using σ: 0→x η: 0→s

Now the R.H.S. AntiDerivative is the argSINH

Noting that

sxd

cd

0 220 1

1

s

sxx

ccd

cd

0

0 2200sinharg

1

1

00sinh0sinharg 1



UNloaded Cable → Catenary (5)UNloaded Cable → Catenary (5) Thus the Solution to the Integral Eqn

Then

Solving for s in terms of x

0sinhsinharg0 1

0

0

c

sc

ccx

sx

c

x

c

sx

c

sc

11 sinhsinh

c

xcs sinh



UNloaded Cable → Catenary (6)UNloaded Cable → Catenary (6) Finally, Eliminate s in favor

of x & y. From the Diagram

So the Differential Eqn

From the Force Triangle

tandxdy

0

tanT

W

And From Before

wcTwsW 0and

dxc

sdx

wc

wsdx

T

Wdxdy

0

tan



UNloaded Cable → Catenary (7)UNloaded Cable → Catenary (7) Recall the Previous Integration

That Relates x and s

Integrating with Dummy Variables: • Ω: c→y σ: 0→x

x

xyc

y

c ccd

cd

00

coshsinh

c

xcs sinh

Using s(x) above in the last ODE

dxc

xdx

c

xcc

sdxc

dxdy

sinhsinh

11tan



UNloaded Cable → Catenary (8)UNloaded Cable → Catenary (8) Noting that cosh(0) = 1

• 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)

Solving for y yields theCatenary Equation in x&y:

cxcy cosh

cc

xc

cccy

xyc

coshcosh

0



Catenary Tension, Catenary Tension, TT((yy))

With Hyperbolic-Trig ID: cosh2 – sinh2 = 1

Recall From the Differential Geometry

Thus:

cxcy cosh

222222

222222

sinhcosh

sinhcosh

ccxcxcsy

cxccxcsy

222222 or ysccsy

yTwyywscwscT 222,

wTc 0

wyyT



Catenary Cabling Contraption

Shape is defined by the Catenary Equation

cxcy cosh

• Note that the ORIGIN for y is the Distance “c” below the HORIZONTAL Tangent Point

y = c



The ProblemThe 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)

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[email protected] ENGR-25_Catenary_Tutorial_Part-1.ppt 1 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods Bruce Mayer, PE Registered Electrical