War in space

Forum: Spacesettlers
Thread: War in space

# 7297 byhappygallimore@... on Jan. 13, 2006, 5:15 a.m.
Member since 2021-10-03

Without going into too much math, could someone explain why the circumferential stress would be 2x the longitudinal tensile stress?

At the moment I was just thinking about the cylinder length. At the L/2 you would have maximum moment force loading the two ends. The ends have tremendous tensile loading, but so would the cylinder at mid-length. The image in my mind was that you might need some cables to give structural support midway. However, this is one of those things that I am certain it has been considered by many before from the 70s forward.

Thus my question if someone knows why, in laymen's terms, why the radial loading is greater than the tensile longitudinal loading?

From: Ian Woollard
To: spacesettlers@yahoogroups.com
Sent: Thursday, January 12, 2006 12:52:31 PM
Subject: [spacesettlers] Re: War in space

On 1/12/06, Dan Wylie-Sears wrote:
> > Actually there's also longitudinal stress, 50% of the
> circumferental
> > stress, which is caused by air pressure on the end caps.
>
> How does it come out so simple?

I don't know of any deep reason for it, but I know how to prove it.

It's easy to work out how much metal you need to stop a pressure
vessel exploding. You cut it with imaginary planes and work out how
much force (=pressure*gasArea) there is trying to blow the two halves
apart and then you make sure you have enough metal to stop it doing
that (tensile strength*metalArea/safetyFactor) in that plane.

In this case, the amount of metal needed to stop a cylinder exploding
lengthwise is half that needed to stop it bursting circumferentially.

> It seems as though it has to be
> something more complicated, because the circumferential stress
> depends on the g's whereas the longitudinal stress depends only on
> the pressure.

To a first order approximation the rotation can be neglected. The
atmospheric pressure is 5-10 tonnes, but the weight of the floor is
only about 10-20% of that.

> > Well, you've reduced the load on the structure by more than the
> size
> > of the hole.
>
> I still don't get it. How do those quantities even have the same
> dimensions?

You're removed the hole, which means the strength has gone down, but
you've also removed the load where the hole used to be. Also the air
pressure around the hole is lower too, so you've reduced the load
surrounding the hole as well.

-Ian Woollard

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