Tension Force in Wire

Minus the X length...

http://www.globalspec.com/reference/36205/203279/appendix-5-1-conductor-sag-and-tension-calculations

SolarEagle, thank you for your reply! Can I simply use the second last expression D=WL^2/8T and solve for T? Since I have exactly "half" the situation of a parabolic wire with even supports (between A and C in your link), the boundary condition at sag bottom should be the same as in my case? Right? And T in D=WL^2/8T is at point A or C in your example, i.e. at point A in my case?

If the mass is constant per length of cable and not per span, the curve will be better approximated as catenary, instead of parabolic.

Here is a derivation:

http://home.earthlink.net/~w6rmk/math/catenary.htm

I don't know if that derivation is correct, but it looks like a set of brackets was left out in this term:

Fh = w / (8*h) * (S^2 - 4*h^2)

In order for the answer to be in terms of Force, the equation would have to be:

Fh =[ w / (8*h)] * (S^2 - 4*h^2).

Force = [(Force/Length)/Length] * (Length squared)

Force = Force

I don't know if that derivation is correct, but it looks like a set of brackets was left out in this term:

Fh = w / (8*h) * (S^2 - 4*h^2)

I don't think the brackets are necessary. The two expressions are equivalent. The '*' between (8*h) and (S^2-4*h^2) is on the same level as '/'. Both terms would have to be bracketed together to both be in the denominator.

In mathematical operations, multiplication is done before division.

The mnemonic for remebering the order of operations is 'Bless My Dear Aunt Sally' == brackets, multiplicaton, division, addition, subtraction.

Thus, without brackets the expression is equivalent to

w /[(8*h) * (S^2 - 4*h^2]

How many points is the question worth?

This problem is not from any exam, I just had a discussion with a friend regarding installing a zipline :-)

A slack would do a great job to ease the tension of a wire rope, promise

Resolve for mg as vertical component of "T" vector and then determine necessary horizontal vector component to resolve.

(Assumes frictionless contact at point "A" AND perfect alignment of vector "T")

Or I could be way off track!!!!! LOOOONG time since I did statics.

It's just like the analogy of the picture below.

Weight = 2Ty ; Ty = T cosθ ==> T = mg/2cosθ

Taking the limit of the equation as θ ->89.99⁰, you probably get a large number for Tension

The angle the line makes at A (with mg) is what is needed. Maximum tension will be at A. From there just realize horizontal component at A will be equal/opposite of B (which only has a horizontal component). The vertical component at A will be mg.

The components are related as the sin and cos of the angle at A. The magnitude will be the square root of the sum of the squares of the components.

"The truth is not a compromise", I realized that it could be solved this way. Nice though to have someone else confirm it. The answer becomes very close to the one given by the expression T=WL^2/8D which had been suggested in one of the links posted in this thread earlier.

Edit....erased already answered question.

Just stretch it as tight as you can and put turn buckles on both ends for tension adjustment.

Next question?

Tension force T at point A may be calculated by the following formula:

T=(mg/L)*h

Tension force at any point along the wire:

T=(mg/L)*y (y is vertical distance measured from point B)

Your solution for tension at A would suggest tension at A is always less than mg (since L is always longer than h).

Since B does not carry any vertical load, tension at A will always be at least mg.

Care to revise anything?

Vertical component of T at point A must be equal to mg.

Anything to revise?

So, since the vertical component is mg, then tension at A will never be less than mg....

...yet, since L is always greater than h, your equation would have tension at A be less than mg, since h/L<1.

...also tension at B is not zero, and your second equation would give that result, since B is at zero height.

You are right "truth is not a compromise". I apologize for this mistake.

The right solution should be as follow :

Let's define a=H*L/mg (distance between y=0 line and the lowest point (B) of wire)

Following equation containing hyperbolic Cosine to be solved to find the only unknown H:

(H*L/mg)+h-((H*L/mg)*COSH(w*mg/(H*L)))=0 (In order to solve this equation "Goal Seek" in "What If Analysis" of Excel may be used).

Q may be calculated after H is calculated:

TAN(Q)=SINH(w*H*l/mg) (Q is the angle between horizontal and the direction of T at point A)

T maybe calculated after Q is calculated:

T=H/COS(Q)