Rigging puzzle

[messabout]

Help;
I have been pondering a math/physics problem possibly related to rigging or maybe anchoring.

Stretch a line or wire between two rigid points lying on the same horizontal plane. Attachment points lie at a distance, (D). Apply some tension to the line, call it (T). Now hang a concentrated weight in the center of the taut line, weight will be (W). The line will sag in the middle by an amount Y. The Y dimension will be dependant on T, D, W, and the elastic modulous of the line (e)

for example; Let
D=200 inches
W=10 pounds
T=n...... (variable, let it be 10 lbs for the example)
e= 5 x 10^6
Resolve dimension Y

Can anyone provide a workable equation that describes Y when T is the only variable ? The appropriate equation will allow me to play with different values for the above fixed elements. :confused: Grandma knew, intuitively, about how far her clothesline would sag when she hung out her wash. In this case I'd like a bit more scientific approach.

Ignore the weight of the line to avoid catenary complication. Line weight is insignifigant with respect to W.

Thanks a bunch!

[Phil Stevo]

I hate being analytical but I am willing to have a go:
You need to add the cross sectional area of the wire A to the mess. And also ad a term TL for the tensionin the loaded (distorted) wire.

The wire lengthens: the new half length from anchor to strut is
SQRT((D/2)sq+(y)sq) (Form 1)
So the strain of the wire is (SQRT((D/2)sq+(y)sq) - (D/2)sq)/(D/2)sq)
which also = the change in stress/e = (TL-T)A/e (form 2)

The new load in the wire is TL = W*y/(SQRT((D/2)sq+(y)sq) (form 3 derived by vectors)

so put TL from formula 3 into formula 2 you get
(SQRT((D/2)sq+(y)sq) - (D/2)sq)/(D/2)sq) =
((W*y/(SQRT((D/2)sq+(y)sq))-To)A/e

Solve for T
T = e/A * (SQRT((D/2)sq+(y)sq) - (D/2)sq)/(D/2)sq) - yW/(SQRT((D/2)sq+(y)sq)

Sorry its too long ago for me to solve for y. But you can work backwards.
It looks better with square root signs and supertex squares. I hope I got all the brackets right.

Phil S

[formsys]

See
http://en.wikipedia.org/wiki/Tensile_structure#Simple_mathematics_of_cables

[water addict]

Are you a student looking for free answers to your homework?

[messabout]

Phil and Formsys;
Thanks a bunch for your information. One of the best things about this forum is that there are a lot of very bright people involved. Even better, many of them share their knowledge generously.

Phil; Several hours after my query it struck me that the area of the wire or line was an integral part of the solution.

Water Addict;
Good question. The answer is yes and no. I have not seen the inside of a school for several decades, so I am not that kind of student. I am a student in the sense that I continue to try to learn things. Call it mental gymnastics if you like. I believe that mental excercise is as desireable as physical excercise. There are some studies out there suggesting that brisk mental activity can delay the onset of senility. I need that.

To set the record straight I will actually apply my newfound information to a real world project. TA.

[water addict]

ok, just checking for fun. frequently there are similarly worded questions on here, and often the reason is because some student has a semester design project due Friday, and hasn't done anything yet...

[messabout]

WA;
I've noticed that too. Aside from some students impending deadline. There are numerous fellows, apparently young ones, who ask questions but do not want to hear the answers. They argue that such and such is the way they want to do something. Their reasons are often discovered to be "because they like it". You know,.... things like a bat wing sail on a coal barge.

Nonetheless, I think we'd ought to be helpful within reason.

An astute geezer is said to have said "Youth is wasted on the young".

[Tim B]

The obvious solution is a Finite-Element approach, with the wire either side of the load being described as a tension-only element.

Shouldn't be that hard to derive. I'll post a solution when I've done it.

Tim B.

[messabout]

Tim;
Thanks, I'll look for your further post. Phil's solution is based on the familiar Pythagorean method and while useful, it does not quite do what I am after.