Beam calculations .. well sort of;!...

Anyone know how to sort this please?..

We have a steel mast structure like a small electrically pylon.

It comprises of Four lattice parts that are bolted together. Its 4 sections are 7.5 metres long but taper from around 1.4 metres across faces at the bottom to 300 mm at the top.

The weights are Top is 127 KG, next down is 200 next 290 and the bottom is 400 respectively.

By what method would you find the mid point where it would be balanced thats to say if it were lying flat on the ground where would the point be that if you picked it up it would be balanced as it were?..

Any help appreciated...

cheers...

Reply to
tony sayer
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?? - are you just after the centre of gravity ?? - secondary school methods should work??

Reply to
Tim Ward

If you wish to find the balance point of a piece of any long item with a moderately smooth surface, think of a bit of 2x1, you can rest it on both your index fingers with your hands wide apart, then slowly move your hands together. When your hands meet, you have found the balance point to within a finger-thickness!

Obviously you can adapt the technique to circumstances if your hands do not stretch to 7.5 metres apart!

Or are you saying that you need to find this out without laying it down horizontally?

Reply to
polygonum

Total mass 1017kg, so half of that is 508.5kg.

So, you need to be at about 508.5-400=108.5kg up the second section, it doesn't taper that much so it will be roughly 1/3 of the way up, as in

108.5/290. I make that about 2.8m so it should balance at ~10.3m from the base.
Reply to
Brian Morrison

the centre of gravity of any assembly is the sum of the respective centers of gravity times their distance from the final centre of gravity.

use algebra to define a simple linear equation and solve it.

.e.e if cG is x and you have 4 chunks at whose weight is w[1-4]and whose CG is at X[1-4] from one end of the structure, then the equation is that X1*W1 +X2*W2 +X3*W3 +X4*W4

-------------------------- = x where x is the overall CG position., (W1+W2+W3+W)

For first level approximations, assume the CG of each chunk is in the middle of it

Reply to
The Natural Philosopher

AFAICR from A-Level Applied Maths (I was always more interested in Pure), the formula you need is: "The moment of the sum = The sum of the moments"

If we take h=0 as being the bottom of the base, and H as being the height or length along of the CoG, and h(i) is ditto for the i-th component:

H*(400+290+200+127) = 400*h(0) + 290*h(1) + 200*h(2) + 127*h(3)

If we assume as a rough first guess approximation that h(i) = (2*i+1)/2 - that is half-way along its length - we get: H = 11.6m

This will be an overestimate, but h>

Reply to
Java Jive

h(i) = (2*i+1)/2 * 7.5

Reply to
Java Jive

Did you miss the bit where the all up weight was over a tonne? ;-)

Reply to
John Rumm

But you could adapt the technique by resting in on two steel supports - and pushing them closer.

Reply to
polygonum

tony sayer considered Thu, 21 Mar 2013 11:14:13

+0000 the perfect time to write:

Ever wonder why crane operators normally use two slings around any asymmetrical load that is not fitted with lifting points?

Now you know.

Reply to
Phil W Lee

No, but I do wonder how cranes are erected. It seems clear to me that a crane must be at least a bit taller than the thing being built. Therefore it is impossible for any crane to be higher than a man is tall.

Reply to
Frederick Williams

Quite often they are built using mobile cranes. However you can lift a thing above your head. And you can lift an object with a crane so that its CoG is almost at the top of the jib - which means that its top may be well above the jib.

Andy

Reply to
Andy Champ

They fly. Hence babies can be higher than men. :-)

You can build a square section tower that is smaller than an existing such tower and entirely within the existing. Then rack up the inner tower.

Reply to
polygonum

Simple, the top section just below the boom/control deck is fatter than the tower with an open side. This climbs up the tower with the jib being used to raise and insert new sections as it goes up. See

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Mike

Reply to
Muddymike

There are a number of ways of dealing with this. One that I don't think is used very often is to insert components at the base, jacking the crane up from underneath.

More commonly, the upper parts of the crane are lifted by a mobile crane, which consists of a set of telescoping beams.

-patrick.

Reply to
Patrick Gosling

There's multiple solutions, but you can jack one up from the bottom, quite a lot just get built using a larger telescopic crane.

Reply to
Duncan Wood

This is the one that got hit by a helicopter, and needed a pretty big mobile crane to recover the damaged sections.

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Chris

Reply to
Chris J Dixon

In article , tony sayer scribeth thus

Thanks to all who replied. This is at preset a Vertical still standing structure it isn't Horizontal as yet!.

What we need to know is at what point up the structure could you pick it up and lift it off its mountings which would need a vertical lift of say

100 mm to clear the mounting bolts and when lifted it would be still be hanging vertical and not topple over.

So it seems that from the consensus of opinion as long as its greater than 11.6, say 12 meters up the structure then that will remain in balance.

Of course each section will have a slightly displaced off centre of gravity owing to the tapering of the parts and the whole as sections bolted together.

Cheers..

Reply to
tony sayer

Wouldn't it be safest just pick it up from the top, or can the top of the structure not be trusted to carry the weight of all of it? In which case why not unbolt and lift one section from the top at a time?

Yes, but, unless you drop the point of suspension down through the middle of it before you attach it at 12m, then the act of pulling it up is likely to cause forces that will tend to make it tilt over, and this might lock it on its anchor bolts, or have other side-effects which those of us not knowing the situation can only guess at, but at any rate are more likely to be undesirable/dangerous than otherwise.

I would have expected that it would have been constructed in such a way as to keep the CoG over its anchorage point throughout its height, but you can see the situation, I can't.

Reply to
Java Jive

Something else you might want to bear in mind is that when you take it off the mounting you will lose some of the bracing effect that keeps the tower in its square cross-section and stops it folding up. You may want to add internal cross-braces if not already present.

Reply to
Al Grant

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