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# Re: Cable Tension Force

• To: seaint(--nospam--at)seaint.org
• Subject: Re: Cable Tension Force
• From: desi j kiss <djkiss55(--nospam--at)juno.com>
• Date: Mon, 9 Dec 2002 06:42:44 -0800

```See the 3rd Edition of Structural Steel Designer's Handbook by
Roger Brockenbrough and Frederick Merritt.

Desi J. Kiss, PE
http://djkeng.tripod.com/

On Sun, 8 Dec 2002 23:39:37 -0800 (PST) C Chan <seaint_list(--nospam--at)yahoo.com>
writes:
>
> I wonder if it is feasible to eliminate sag completely.
> Cables carriying uniform load along its length (e.g. self weight)
> have shape of a catenery.
> For a symmetrical cable, max. tension = w(S+c), where S is sag, w is
> weight of cable per unit length, c is solved from the equation of
> catenery: S+c = c cosh(a/c). Where a= half span.
> Instead, if the cable is assumed parabolic, solution will be
> simpler.
>
>
>  Oshin Tosounian <oshin(--nospam--at)kcematrix.com> wrote:Does anyone have the
> formula to calculate the tension force in a cable to
> keep it from sagging under its own weight handy?
> I have a single span condition and a multiple span condition where
> the cable
> is free to slide over the intermediate supports. The supports are at
> varying
> heights.
>
>
> Oshin Tosounian, S.E.
> Los Angeles, CA
>
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