Re: We started the 100-foot long 10-foot wide deck high up in the California redwoods

John Robertson wrote, on Fri, 17 Oct 2014 23:46:50 -0700:

Better check the specs of the wire rope, the owner may be going by the > rated maximum capacity (12,200 lb for 3/8" wire rope) and only derating > it 50%. Note the manufacturer recommends only a 20% load factor

I asked the owner about the weight of the deck, where this came back:

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Hello Danny:

Thanks for your help last week. I couldn't have done it without you.h

You seemed skeptical about my estimate of the first bridge section. Given a 2x10 weighs 3.37 lbs per foot. Given a 2x6 weighs 2.00 lbs per foot.

3x16x3.37 = 162 lbs for the three long boards 2x10x3.37 = 68 lbs for the two end boards

2x6 2.00 lbs/ft

10x2x32 = 640 lbs for the decking

870 lbs per 16 foot section.

So if there are 5 sections (80 feet), or 6 sections (96 feet) we have

4,350 lbs or 5,220 lbs for the deck.

The cables can support 28,000 lbs.

That gives us 22,780 lbs for the house and occupants, assuming the only support is the cable.

If we allow that some of the weight is supported by the posts and the trees, we have even more leeway.

If the house is 24 feet long and ten feet wide, it will weigh about 8,000 pounds.

Add 4,000 pounds for furnishings and appliances. That gives us 10,780 pounds of leeway for occupants.

But I plan to have most of the house weight supported by the redwoods, not by the cable.

Reply to
Danny D.
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It again all hinges on the diameter of the cable (wire rope) they are using and how it is secured. If it is 3/8" then I wouldn't go near it assuming he is going by the maximum load.

Based on his reasoning (28,000 load spread over two cables) he needs at least 7/8" (12,900 lb safe load each), but 1" (16,700 lb) would be better.

Attaching the house to the tree will save a lot of weight, however trees grow larger in diameter so the support must allow for that somehow.

Also his securing of the wire rope must be flawless, and the fact that he has already stressed the rope in a few places with the clamps has weakened the original wire rope significantly.

However he doesn't appear to be willing to get an inspector, so I'll assume I'll read about this upcoming disaster in the newspaper in the near future (next couple of years).

Just because you are clever doesn't mean you are right. Some of us can be quite smug...(ducking)

John :-#(#

Reply to
John Robertson

---------------------------------------------------------- You can buy them books, BUT if they eat the covers.

Lew

Reply to
Lew Hodgett

Now days Lew you buy them books and they eat the Teacher!

Reply to
Leon

You just don't get it. 28,000 is how much the cables will support if the weight were hanging straight down towards the center of the earth. If you pull the cables parallel to the center of the earth, with some sag even, then pull down on the middle of the cable with the weight of this deck, the pull on the cables is going to be 10 times or more the force (weight of the deck) you are pulling down with.

That comes before the working load is decreased because of turns in the cables, or weakening because of being clamped with a cable clamp.

If I was out there, I would turn you in to the authorities to get a stop put to the project.

Someone is going to get hurt, or die. Mark my words. It will fail. It has to. Physics.

Reply to
Morgans

The problem as others have been saying generally and that Morgans points out in another posting is this doesn't account for the geometry.

Consider for simplicity the condition if the cables were mounted on telephone poles on level ground 100-ft apart. Say you tension the cable so there's 10-ft sag in the middle. Using the tabulated weight for 3/8" cable, I estimated that it takes only about 30 lbf to achieve that sag so we'll ignore that for the time being.

With 10 ft drop at midpoint of 100 ft run, and considering that the applied load will essentially straighten the cable, if the load were at the center rather than distributed the angle between the horizontal and the cable is invtan(10/50) --> angle ~11 degrees.

Now to support that load, the vertical component of the tension has to balance the weight of approximately 5000 lb. That vertical component is Tv=T sin(angle) or the cable tension T is Ty/sin(angle). Substituting numbers and noting that for small angles sin(theta)~theta, the tension to support that 5000 lb is

T=W/sin(angle) = (W/2)/sin(11) --> 2500/0.2 --> 12,500 lb

So, you've taken up roughly half the total strength of the cable simply by the decking. That's only a safety factor of ~2X and minimum generally accepted is 3X while for overhead rigging and personal safety of support 5X is considered prudent.

As said, this isn't going to approach that kind of margin and is extremely risky going forward without more serious engineering than has happened to date.

There are excellent design guides in the various handbooks that have been linked to extensively before in the previous threads; no point in reproducing them yet again if they're not going to be heeded.

But, your friend really needs some input from an engineer who understands statics well enough to make some reasonable calculations for the actual geometry and design.

Reply to
dpb

On 10/23/2014 7:36 AM, dpb wrote: ...

...

ERRATUM: I forgot to divide the cable limit by the two above -- the actual limit per cable is (optimistically) as used by your friend 14000, not 28000.

So, the decking alone is roughly 90% of the rated breaking strength and adding a 200 lb person is 100/0.2-->500 lb.

You'll gain a little by considering the decking as a distributed instead of point load, but that'll be only a marginal improvement and you'll likely lose some (and potentially a lot) for the non-uniform geometry on the downhill side as that side may be almost perfectly horizontal so the amplification factor of 1/sin(theta) goes way up as Morgans feared (hence his 10X estimate).

THIS IS VERY BAD...

Reply to
dpb

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On the last point -- the angle at which the total tension amplification factor reaches 10X the applied load is invsin(10) --> 5.7 degrees as opposed to the 11 degrees. That's not a lot of difference.

Reply to
dpb

I see a complete failure to account for wind loads. Wind loads on this tree house will be larger than the dead loads. Ask the owner to calculate the wind loads as well.

?-)

Reply to
josephkk

Danny D. wrote, on Tue, 21 Oct 2014 20:10:10 +0000:

We finished rigging up the second 16 foot section, which missed the next set of redwood trees by about a foot or two.

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Unfortunately, those two redwoods straddling the end of the 32-foot suspended section are just a tad under ten feet apart.

So, we're gonna have to engineer a slight bevel inward, to squeeze in between those two trees, and then it's on to the next three or four 16-foot long 10-feet wide sections, all of which is suspended by ropes and temporary cables, at the moment, as we build it as we walk out to the edge...

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Reply to
Danny D.

josephkk wrote, on Thu, 23 Oct 2014 18:23:37 -0700:

This is a good point so I will mention it to him.

We worked on the second floating 16-foot section today, by the way.

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So now we're suspended 32 feet straight out.

Only 60 or so feet to go!

Reply to
Danny D.

dpb wrote, on Thu, 23 Oct 2014 07:36:12 -0500:

Wow. Those were wonderful calculations. I forwarded it all to my friend, and will reply back with his response.

Meanwhile, we worked on the second section today, and we ended up stopping about 2 feet away from redwoods which we need to squeeze through.

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Reply to
Danny D.

CRNG wrote, on Thu, 23 Oct 2014 12:40:15 -0500:

I think the owner is taking heed, it's just that he's a third party to this conversation (he doesn't know Usenet). :)

BTW, here's a view from below today, when I dropped my glasses and had to climb down the steep hill to retrieve them.

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Reply to
Danny D.

dpb wrote, on Thu, 23 Oct 2014 13:35:20 -0500:

I'm sorry I haven't responded in a while. I hurt my back and was laid up but hopefully I'm better now ...

The whole thing is supposed to hang from the cables, but we did anchor one end because we needed a way for people to get "on" the decking.

Here's where we left it today...

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Reply to
Danny D.

CRNG wrote, on Thu, 23 Oct 2014 08:05:45 -0500:

I'm sorry I haven't been able to respond lately.

We had to readjust all the cables today, with a set of 5 winches, as we had to re-balance everything once the second 16-foot section was planked.

Unfortunately, I ruined my clothing, as I hadn't expected the oil to still be soaking wet ... even though it was drying outside for a day ...

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Reply to
Danny D.

dpb wrote, on Fri, 24 Oct 2014 08:22:06 -0500:

I'm forwarding all this to the owner.

The only thing I can say is that it "seems" sturdy when we're both on it, and that's almost 500 pounds of people alone ...

It's being supported, at the moment, by 6 separate winched vertical cables (the winches are for level adjustments).

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Reply to
Danny D.

VinnyB wrote, on Wed, 22 Oct 2014 05:58:58 -0500:

You forgot the screw manufacturers!

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Notice we gave up on the lower screw (the one with the longer thread).

It was just too hard to drive into the wood.

Even with this nice pile driver thing from Harbor Freight!

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The screw on top, with the shorter thread, goes in without pounding!

Reply to
Danny D.

"Danny D." wrote in news:m2mk88$c0c$ snipped-for-privacy@dont-email.me:

See The chances of your bridge deck fluttering in a blow are fairly high.

Reply to
Ian Malcolm

Ian Malcolm wrote, on Tue, 28 Oct 2014 00:21:08 +0000:

We actually joked, a few times, about the Tacoma Narrows bridge, and, yes, I think just about everyone has seen that video of the car on the bridge and the person getting out and making it just in time.

This bridge isn't nearly as long. It's only about 100 feet long, by 10 feet wide, supported on one end on the ground and on the far end about 40 or 50 feet up in a tree on a (very) steep slope.

Here's what the first two 16-foot-long sections looked like today, when we ran out of oiled wood:

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I'm currently learning how to wash good clothes to get the oil out! :(

Reply to
Danny D.

Where are the supporting cables in the picture--on top or just hidden by the view?

What I'd like to know is can you measure the height differential between the upper and lower mounting locations and the approximate distance from the straight line between them to the low point and where that point turns out to be between the two end points. In a _very_ crude sketch..

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Reply to
dpb

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