Max steel plate curve without forming ?

[MikeJohns]

Does anyone have a rule of thumb that works when applied to the curve a steel plate can be pulled into fairly without resorting to forming or line heating ( Curving in one plane only not compound)

I have been told ( by a venerable old shipbuilder ) a radius of 700 times the plate thickness …. Seems reasonable to assume that a 10mm plate could be coaxed into a 7m radius, and 1mm into a 700mm radius.

I suppose the other way of approaching this is to assign a reasonable force to pull the plate in and calculate the resulting curve.

Any observations?

Cheers
Mike

[Frosty]

You mean by hand?

But say you could coax a 10mm plate in to 7 meter radius the centre of the plate would bend first due to the leverage advantage. The very ends of the plate would be straight, you would have a flat spot!!. I guess a huge hammer would cure that.

But to say push a steel plate into shape such as the bow --I would think your calcs are near.

[waikikin]

sometimes with steel, brute force & ignorance wins!

[MikeJohns]

I am intersted in the curve achievable with reasoable persuasion, that is pulling in with a come-along or screw setup to "sensible" amounts, past which you would need to line heat or roll form the plate to achieve the result.

I'll run this through the FEA package and get an idea from that.

Cheers

[Raggi_Thor]

No need for FEA :-)
FL^3 / 3EI should give you a hint.
See http://www.roymech.co.uk/Useful_Tabl...am_theory.html

Curvature should be smaller toward the ends, but that's normally not the case....

[M&M Ovenden]

Hi Mike,

There is no doubt steel can curl quite tight with enough persuasion. I just finished (or almost finished) plating my hull cold forming the 3/16 plate. The plates were pulled up against the hull using two chain lifts and getting it to curl was not that hard. My tightest radius on the hip is of about 18 in and that plate was one of the easiest one to get on the boat.
I decided to plate this manner as the radius is not constant, I think a constant radius would be a little more challenging to obtain.
The trick is not as much to get the plate to curl as to get it to curl right. It needs the longitudinals positioned properly to guide the plate threw the radius that is expected. Obviously the plat also has to cross a rib so it doesn't kink over the longitudinals. As Jack Frost mentions, the ends tend to be flatter but this is not a problem as long as it's what is expected. I drew the boat so the top and bottom of the “radius chine” plate faded into the “not as curved” top and bottom plate. The truly radiused plate starts a few inchs away from the edges and only after crossing a longitudinal. I didn't have chine bars but as an afterthoughts even though it would have been more welding it would of made easier to hold the seams in. Those are more the challenge then the curling of the plate.
So for somewhat of an answer to your question the radius is not the limiting factor to curve fairly the plate but more how you guide it.

Murielle


http://www.boatdesign.net/forums/att...2&d=1158239584

[westlawn5554X]

Yes ,weld attachment for chainblock point and tightening it work always if we dont have a roller.

[MikeJohns]

Thanks everyone

This is to re-plate a boat below the WL with 8mm plate ( around 1/3 inch ) this is a bit beyond the "Bend past yield" approach. Normally we would just pass the design on to the fabricator but this is a remote area with no rolling facility so I was hoping for an experienced builders rule of thumb for "Easy " plating radius without forming.

Note that When you bend past yield you are actually forming.

Raggi
The FEA package is my computer game :-)

Cheers
Mike

[Raggi_Thor]

Mike, :-)
Say your plate is 1m wide, 4m long and you need to bend it 1m at the bow while its fastened secutrly at the back end...:

I = 1000x8x8x8 / 12 = 42667 mm4
EI = 200.000N/mm x 40.000mm4 = 8 x 10^9 Nmm3 = 8000000000 mm4
L^3 = 4000^3 = 6.4 x 10^10 mm3 = 64000000000 mm3

FL^3 / 3EI = 1000mm
-> F = 1000mm * 3EI / L^3
F = 1000 * 3 * 8000000000 / 64000000000 = 375N , approx

[Frosty]

--As simple as that. I was going to say that.

[MikeJohns]

Originally Posted by Raggi_Thor

Say your plate is 1m wide, 4m long and you need to bend it 1m at the bow while its fastened secutrly at the back end...:

I = 1000x8x8x8 / 12 = 42667 mm4
EI = 200.000N/mm x 40.000mm4 = 8 x 10^9 Nmm3 = 8000000000 mm4
L^3 = 4000^3 = 6.4 x 10^10 mm3 = 64000000000 mm3

FL^3 / 3EI = 1000mm
-> F = 1000mm * 3EI / L^3
F = 1000 * 3 * 8000000000 / 64000000000 = 375N , approx

Raggi

admirable try and it's great to see people applying basics, my basics are so rusty these days.

A quick glance suggests that your numbers are out? Note that E of steel is a ratio of 200E9 ( your units are inconsistant in your EI calc...always calculate the units through as a basic check of your working :- ).

The model is more complex than simple cantilever beam theory , the fibre stresses carry on into the adjacent plate and so you cannot just assign fixity to one end. Also the plate is being pulled against a formwork.

A better model is to assign a force at either end and bend the plate over a pair of fixed supports equidistant from the centre. Then you can still apply the fundamental bending equation. Then we can put it into a spreadsheet and play with stresses and deflections wrt applied end forces. Want to have another go?

All the best
Mike

[Raggi_Thor]

OK, I should have used a spreadsheet :-)
I looked up the 200 or 210 x 10^9 N/mm2...
Do you think 400N or 40kg is realistic.
I'm firing up excel...

[Raggi_Thor]

First, isnt 210Gpa = 210.000N/mm2?

[MikeJohns]

That makes more sense. But units units units ! how about a few ^2 on some of those mm and m. You'll make life simpler for the quick casual reviewer.

At first glance above I thought you were trying to say thay you had a radius of 1 m with that force so I was looking for a few missed zero's... sorry.

I'll get back to this.... busy

[MikeJohns]

OK
I just worked this out. I still would have liked an experienced shipbuilders rule of thumb since these calcs always sound great in theory but when it comes to the job............

Anyway
An 8mm plate being formed over frames 1m apart can be curved to a radius of 3145mm (400 times its thickness) before the yield (250MPa) max fibre stress.

if the plate had 1m projecting past the last frame the force required would be around 2.7kN

If you applied 3kN the plate will start to yield and can be coaxed to a much tighter radius.

Sound about right Raggi?

Cheers