cable tension

[mosk]

horizontal cable between 2 points
known:
pretension of cable
cable weight/unit length
span

add a static known mass to center of span

how do I calculate tension of cable?

 

[desertfox]

Hi mosk

Try this link:-

http://www.slideruleera.net/miscellaneous.html

then go for Uss Wire rope book.

desertfox

 

[rb1957]

sounds like a student post to me ...

 

[MiketheEngineer]

This has been discussed a few times in these forums. Search on "catenary"

 

[zekeman]

Can you assume the cable weightless?
If so it is easy to do it with energy conservation.

If not, then post your thinking.

 

[GregLocock]

Here's a thought experiment, it may be wrong.

Imagine a stndard catenary, tension T, mass per unit length m.

Now cut the centre L1 length of the cable out such that m*L1=M where M is the mass of the single mass you want to hang at the centre.

Then cut and shut the catenary curve around the missing part.

It looks ok on a sketch, it sort of has the right shape.

page 45 in desertfox's ref has a calculation, albeit with a not very reassuring preamble.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[Denial]

Greg.

Thinking about your cable before it is cut, the horizontal component of the cable's tension is everywhere the same (this being a fundamental property of the catenary). The vertical component at each of the two places you cut it will be equal to half of the weight of the mass that is to be removed (ie m*L1/2). Bringing the two remnant pieces together and attaching the lumped mass will not change these two force components, so therefore the shape and force-state of the remnant cable pieces will be unchanged by the operations you have performed. Thus you will have everything in a state of equilibrium.

But will it be the correct state of equilibrium? Only if the total unstressed-length of cable matches. Even if we simplify the situation by assuming that the cable is inextensible, I have my doubts. If your thought experiment is correct, then the horizontal component of tension in Mosk's cable is unchanged when Mosk hangs his mass on its mid point.

A bit of serious maths would answer this one way or the other (for the inextensible cable case). I might give it a go some time, but not today. In the meantime, in the spirit of the current horse racing season, my money's on "no".

 

[mosk]

Not a student post.
I am having a discussion with another engineer who is designing a fall protection system for my plant.
I found formulas in books and online, but not with the parameters I neeeded worked out. I figured I needed to rearrange the equations to find my answers, but thought I'd check to see if someone here knew where it might be done already.
With all my other duties, I'm having a hard time concentrating on the math.
Thanks everyone for the help and I will keep plugging at it.

 

[rb1957]

ok, sorry about that ...
will the cable react like a shock cord ?

can you negelct the weight of the cable, so all you end up with is the ccable, bent like a "V", and the component of cable tension is reacting the load ?

 

[MiketheEngineer]

OK

Here is the equation:

H = wL^2/8 d

H = Tension at mid span
W = loading
L = Length
D = initial sag

Watch units

If point load - just use trig and statics

 

[zekeman]

I don't think so. The curve should be in the form

y=F(x)

For this problem I got

(1) To*cos(@o)*y=To*sin(@o)*x+w*x^2/2 which is parabolic
(obtained from moment equilibrium for rope from 0,0 to x,y about
the point x,y.

To and @o are constants obtained from

boundary at x=L/2, y=D

(2) To*cos(@o)*D=To*sin(@o)*L/2+w*L^2/8

and
(3) To*sin(@o)=Q/2, force equilibrium on point load


where

To is tension at x=0+delta, a small distance from x=0

@o= angle of rope at x=0+ delta

Q= point load at x=0

L= span

D= center sag

Assumption no flexure, no shear in rope

 

[zekeman]

Oh, I forgot to answer the OP question.

After all the above. you get T from

T*cos(@) =To*cos(@o)

which is statement of horizontal equilibrium of rope

To evaluate:

To from previous post and @ from

tan(@)=[w*x+To*sin(@o)]/[To*cos(@o)]

 

[chicopee]

The way I would look at this problem would in two steps. First determine the cable to be massless and figure the tension based on the center load. Secondly figure the cable as a catenary without a center load and determine the tension solely on the cable weight. Then add the two tensions for an approximate value.