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Calculating tension on chain fence with someone sitting on it.

  1. Jan 5, 2008 #1

    I'm hoping someone can show me a simple formula to calculate the tension or force on a chain fence with someone sitting on it.

    I imagine that the variables are;
    - the distance between the fence posts
    - the arc of the chain
    - the persons weight

  2. jcsd
  3. Jan 5, 2008 #2


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    I don't think I am quite understanding your question. The way you have it worded now, it's a basic statics problem, but I don't think that is what you are getting at.

    Can you be a bit more specific as to what you are wanting to see as a result?
  4. Jan 5, 2008 #3
    Thanks for your quick reply.

    I want to be able to calculate the tension or chain strength needed in kgs to support a persons weight sitting on a fence given the length and arc/radius of the chain.
    Each section of the chain linked fence is chain suspended between 2 posts, I would like to calculate the stress/tension put upon the chain should someone or something be put on or hung from it, this would then able correct chain size/strength.

    PS I'm not an engineer or an engineering student so this problem may seem easy for most people on this forum.
  5. Jan 5, 2008 #4


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    Generally speaking, the tension in a rope, cord, chain is found using a static equilibrium approach (Fred eluded to this).

    Can you clarify what you mean by arc exactly?

    If you have the weight of the person sitting on the chain and the distance from him to the ends AND the distance from the horizontal that he "sinks" while sitting on it you can find the answer. Or, if you know the angle the chain makes with the horizontal (this what you will find with the distances mentioned above if you don't know or set the angle) plus the weight you can determine the tension as well.

    The basic equation is found using Newton's Second Law, Fnet = Ma.

    So you would have this initially:

    [tex] T_1 + T_2 + Mg = Ma [/tex]

    But since it is in static equilibrium you have:

    [tex] T_1 + T_2 + Mg = 0 [/tex]

    And inserting a standard cartesian coordinate system you have:

    Note: this assumes the guy is in the middle of the chain and the angle he makes with the horizontal is the same on both sides. If not, the you'll just need to use two different angles (theta1 and theta2).

    x components: [tex]-T_1 \cdot cos(\theta) + T_2 \cdot cos(\theta) + 0 = 0[/tex]

    y components: [tex]T_1 \cdot sin(\theta) + T_2 \cdot sin(\theta) - Mg = 0[/tex]

    where [tex]Mg[/tex]= the person's weight

    As you can see, if the angle ([tex] \theta [/tex] is the same), for the x components [tex] T_1 = T_2 [/tex].

    Just substitute that fact into the second (y component) equation and solve using the other knows.

    If you don't know the angle, you can find it using basic trigonometry.

    Last edited: Jan 5, 2008
  6. Jan 5, 2008 #5


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    Ah. Stewart beat me to it. That's exactly the approach I was thinking about. A basic statics approach. The tension will give you the required load the chain needs handle.

    Don't worry about having an arc shape in stead of a "V" shape when someone sits on the chain. Also, if this is in a public place, or for any other reason, make sure to at least double the chain to give you an acceptable factor of safety.
  7. Jan 5, 2008 #6
    I'm not sure I am following this.

    If the angle the chain sinks to 15 degrees (even for both sides) and the persons weight is 80kgs then this is the formula as I understand it from above.
    T1 . sin(0) + T2 . sin(0) - Mg = Ma
    15 . 0 + 15 . 0 - 80

    I am sure I am doing something wrong here as I expect the greater the angle the less the tension and that only at a 90degree angle would the tension be the same as the persons weight.

    I hope someone can clarify where I am going wrong.

  8. Jan 5, 2008 #7


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    Take another look at what I wrote, you have to use the y component equation to find [tex]T_1 [/tex] (and [tex]T_2[/tex] since they are equal per the x component equation).

    Using the y component equation (remembering that [tex]T_1[/tex] and [tex]T_2[/tex] are equal) you end up with:

    [tex]T_1 \cdot sin(\theta) + T_1 \cdot sin(\theta) - Mg = 0 [/tex]

    which then gives:

    [tex]T_1 \cdot sin(\theta) + T_1 \cdot sin(\theta) = Mg [/tex]

    then solving for [tex] T_1 [/tex] gives:

    [tex]T_1 = \frac{Mg}{2 \cdot sin(\theta) } [/tex]

  9. Jan 5, 2008 #8
    Thanks very much Stewart.

    It makes perfect sense now its morning.

    thanks for all your help.
  10. Jan 5, 2008 #9
    BTW, munga - perhaps you've already figured this out, but just in case ...

    When you talk about the arc that a chain takes (for an ideally flexible chain it takes the form of a catenary curve), that comes from considering the weight of the chain itself. The statics approach that the other posters above give neglects the weight of the chain, so in effect it forms two straight lines from its ends to the person's butt in the middle. Once you make that approximation, then the angle just comes from the geometry and is independent of the person's weight; you just draw a triangle using the known lengths of the chain on either side of the person as well as the horizontal distance between the ends of the chain.

    If you really wanted to do a more accurate calculation, e.g. if the weight of the chain were heavy enough compared to the person's weight that you would see noticeable sag, then the calculation becomes more difficult. The only way I know how to do it is to use the calculus of variations, which usually shows up only in more advanced classes in Mechanics.
  11. Jan 5, 2008 #10


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    As long as the mass of the chain is relatively small as compared to the body, then you can assume it to be negligible, hence a massless chain. The other condition that must be true is that it be unstretchable. Both of those seem to be the case here, or at least that was my interpretation of the problem.

  12. Jan 6, 2008 #11


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    We can think of a thousand ways to Sunday to complicate the problem. Yes, it is a simplified approach. However, take a look at the return on investment for the time spent solving a much more complicated problem versus the extra degree of accuracy you would get. It wouldn't be worth the time.

    That's one of the reasons for putting in a factor of safety.
  13. Jan 6, 2008 #12


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    Yes the safety factor would take care of everything else. No need to analyze a guy sitting on a chain fence to death.
  14. Jan 7, 2008 #13
    I must not have been clear, but I didn't meant to suggest at all that it would be preferable to consider the weight of the chain or otherwise to calculate its shape. I mentioned the catenary solution only because the OP asked about the arc of the chain, so I thought I'd make mention of what it would take to do that. I also meant to suggest the reason why in all likelihood it is completely unnecessary to do this.
  15. Jan 7, 2008 #14


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    Actually, the chain wouldn't even form a catenary since it has a point load in the middle. It would only form a catenary if it was of uniform density and under its own self weight.

    Last edited: Jan 7, 2008
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