taut flat cable parallel to roof to support a roof? 12

[IJR]

Pals

Will a set of cables, parallel to a simple roof, work just like purlins to support a flexible roof material?

For a cable to work I think I need a profile with a sag, so that no further sag can occur under load

But the architect says no sag.

Anyway around this. Seems impossible to me.

respects
ijr

 

[MikeHalloran]

Another alternative is to take a field trip with the architect, get the car stuck in a ditch, and pull it out with a rope stretched across the road to a tree that just happens to be there. Demonstrate how easy it is to move the car just a little by pushing laterally on a straight rope, and how quickly the lateral force has to go up to keep the car moving.

You can explain the principles once again while you wait for the tow truck.



Mike Halloran
Pembroke Pines, FL, USA

 

[dhengr]

IJR:
It’s not a matter of your (or our) respecting or disrespecting the arch. or of knowing he calls the shots. If he has enough money to throw at this roof, you can do almost anything. You can make it carry the roof load and make it safe, you must. But you are his consulting engineer, not just his calculating machine. And, given the fact that the arch. doesn’t seem to understand the physical laws by which we must design and live, you must go on record with him about what he is asking for, and what you are doing to try to accommodate his wishes, and what the results will be in the way of forces, deflections, cable and fabric sag, etc. Since there is no such thing as laterally loaded cables and fabrics which do not have any sag or deflection. This should be in the form of a letter and report and calcs. which explains what the final results will be, with proof that he received the letter and understands the report. Then you may still be sorry because the argument with him continues; but at least you won’t be sorry because you ended up in court facing him, because he doesn’t understand you (and he by playing arch. and wishing hard enough) can’t defy gravity and you can’t do the impossible, just because HE says so. The whole thing may be just a matter of degree. Does he mean you can’t have a cable sag equal to 1/3 of the cable length; and do you think he means you can’t have a cable sag of 6" in 30'?

One of your biggest problems may be designing the cable restraint structural system in this rectilineal structure. My second paragraph, above, still stands, you have just rotated my system by 90° and now claim a sloped flat roof. You still have the potential of ponding (growing rain filled swimming pools) on the tension fabric skin.

 

[Bobber1]

If you don't get the frequency of this roof in the right range, the roof could experience wind induced vibrations.

And what about deflections of the cable supports?

 

[IJR]

Pals

From the posts here I have fully understood that all friends here agree that taut horizontal cables can in no way support gravity loads.

I am going to use good purlins and let cables stand there basically idle. so I wont have a load carrying or a frequency problem

respects
ijr

 

[rb1957]

IMHO that's not what we're saying ... we're saying that cables will deflect under transverse load ... we're objecting to the architect saying the cables can carry the load without deflection; i suspect he means "without significant deflection" which is of course a whole different thing.

 

[dik]

Bobber1... if the cable is stretched tight enough... only the dogs will complain! <G>

Dik

 

[slickdeals]

@Dik (23 Apr 10 12:08)
Just wanted to clarify that for a 100 lb on 1 in 100 slope the tension would be 5000 lb and not 10000 lb?
T = PL/4D or P = W/2 tan [&theta;]

We are Virginia Tech
Go HOKIES

 

[paddingtongreen]

@slickdeals, I'm not sure I agree, I could be mixing up the symbols though and there is not enough info to be sure.

Let the horizontal component be H. If the total load is W, the span is L, and the sag is D, then:

H=(W/2)(L/2)/D, and the Tension = the square root of the sum of the squares of H and (W/2). The tangent of the instantaneous slope at the support is (W/2)/H.

Once the tension is determined, the intermediate sags can be calculated.

As a check on a program result, I would put that on a spreadsheet to check tensions for various sags.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[slickdeals]

I did not clarify my symbols. My mistake.
P = W/2 tan ?
P = Tension in Cable
W = Point load in the middle
[&theta;] = slope

T = PL/4D
T = Tension in the cable
P = Point load in the middle
L = Total length of cable
D = Sag at location of point load

We are Virginia Tech
Go HOKIES

 

[paddingtongreen]

@slickdeals, gotcha. I was thinking uniform load, that's why I said "instantaneous slope".

Michael.
Timing has a lot to do with the outcome of a rain dance.