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# Re: Cable Tension Force

• To: seaint(--nospam--at)seaint.org
• Subject: Re: Cable Tension Force
• From: "mat sanchez" <matsanchezksa(--nospam--at)hotmail.com>
• Date: Mon, 09 Dec 2002 14:31:11 +0300

I believe it is feasible to eliminate the sag completely specially if there are intermediate supports wherein it could run over it. However, eliminating the sag means enducing a greater tension to the cable. The lesser the sag, the larger tension the cables will carry and vice versa. You should investigate the additional Stress and strain it would cause, getting rid of the sag.

>From: C Chan
>To: seaint(--nospam--at)seaint.org
>Subject: Re: Cable Tension Force
>Date: Sun, 8 Dec 2002 23:39:37 -0800 (PST)
>
>
>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 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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