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Re: reducing rod deflection with tension

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Ken,

As both Roger and Thomas pointed out, you may be more governed by the
desired deflection in the beam.  In otherwords, in pointing the tension in
the cable that you want to achieve the desired amount of sag in the rod,
you may end up putting too much compression in the beam causing it to
camber too much.  So you should be careful about just looking at the
tension in the cable to minimize the sag.  As a result, Thomas' suggestion
of use temperature in a computer program is a good one.

After saying that, if you want to just deal with the sag in the cable
without considering the impact on the beam, then it becomes basically a
matter of statics.  I have dealt with a similar but inverted problem in
the past...I had a rod for which I wanted tensioned to a certain level.
The problem was how was the contractor supposed to know when he had
reached that desired tension level.  The solution was to tell the
contractor how much sag to have in the rod, which then would produce about
the desired tension.

As I said, it becomes an issue of statics.  You would especially have some
insight if you have dealt with equivalent loading methods in prestressed
concrete.

Image your rod.  It is going to deflection in approimately a
parabolic/circular/arc shape due to its own self weight.  If the ends are
free to move (as in your case), then the moment at the midspan will
approximately be that of a simply supported beam...w*L^2/8.  Now, since
it is not really a simply supported beam but rather has a tensile force in
it, we know that there is a tension force at each end (call it P) of the
rod that acts at a tangent to the deflected shape at the end.  This P
force will be predominately horizontal with a small vertical component.
Since, the slope/angle at the ends is rather small, however, we can
approximate the horizontal component of the force as the overall force
itself.  So, we can approximate the horizontal force as being P.  Now, at
the midspan, that horizontal force is acting through a moment arm of the
deflection at midspan...call it e (as in eccentricity in prestressing, but
you can use delta if it gives you warm fuzzies).  So, you have a moment
due to the horizontal force at midspan of P*e.  This moment must be equal
to the moment at midspan due to self-weight.  Thus, you get P*e=w*L^2/8.
You know w, L and are defining e, so solve for P.

So, by this method, with w=2.67 lb/ft (what I quickly calculated for the
self-weight), L=22 ft (per your message) and desired sag of 1/2", you
would need about 3900 lbs of tension in the rod (unless I goofed
somewhere...always a possibility).  So, now you would have consider what
effect that would have on your top chord.

HTH,

Scott
Ypsilanti, MI


On Thu, 28 Aug 2003, Roger Turk wrote:

> Ken,
>
> It is probably better to specify the deflection of the beam than to specify
> the tension in the rod.
>
> For example, the deflection of the beam under applied loading without the
> Queen post is calculated for queen post locations/center of span.  Calculate
> the load that the queen posts must apply to bring this deflection back to
> what you want.  From geometry, you can calculate what force is in the rod,
> but the contractor does not have to worry about that.  You just have to make
> sure that the rod is big enough to take that force.
>
> If you want to specify the force in the rod to prevent the rod from sagging
> more than 1/2", you might end up with the beam looking like an arch.
>
> HTH
>
> A. Roger Turk, P.E.(Structural)
> Tucson, Arizona
>
> Ken Peoples wrote:
>
> . > I have a queen post truss with 1" rods for the bottom chord.  The spacing
> . > between the panel points is 22 feet.  According to our calculations, the
> . > 1" rods would sag about 10" under their own weight if they are not in
> . > tension and free to move at their ends.  I would like to find out how
> . > much tension will need to be applied to reduce the self-weight deflection
> . > to a reasonable amount - say 1/2".
>
> . > I see from Roark's formulas - Table 12 Case 3 that if the ends can't
> . > move, then the force on the ends would be about 1100 pounds and the
> . > deflection of the rod would be about 1 1/2".
>
> . > While this is helpful, this formula does not get me the force required to
> . > bring the rod up that remaining 1" to get the deflection down to 1/2".  I
> . > understand that it is not possible to get the deflection to zero and
> . > expect that as one approaches zero the force qets quite high.  What I
> . > don't want to do is to specify that the erector have to take more sag out
> . > of the rod than is reasonable - thus inducing huge tension forces.
>
> . > Thanks in advance for your help.
>
> . > Best regards,
>
> . > Ken
>
> . > Kenneth S. Peoples, P. E.
> . > Lehigh Valley Technical Associates, Inc.
> . > 1584 Weaversville Road
> . > Northampton, PA 18067-9039
> . > Phone: (610) 262-6345
> . > Fax: (610) 262-8188
> . > e-mail: kpeoples(--nospam--at)lvta.net
>
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