Can I Flame Straighten 2'' Plate?

[Jim Coke]

Greetings Mac,

 

Sorry you did not get the results that you wanted..  I would use much caution using the air syphon.   At that temp you might get quite a steam flash..  That's a lot of iron to cool...

 

Forge on and make beautiful things

Jim

[yahoo2]

The industry standard for heat straightening 50mm plate is oxy/acet with a size #8 tip or size #4 rosebud, if you cant do it with that your heating technique is faulty. Cooling the plate quickly with water will just make it brittle it makes no difference to the final shape, compressed air is fine. Once the temp is below 315^o C you can dump as much water on it as you like.

 

You are supposed to hold the tip still and just inch it along not wave it around all over the place. Go back and read my last post and try and get your head around the idea that you are trying to upset the metal vertically with the cold metal in front and around the sides of the expanding hot spot clamping and squeezing it and forcing it up.

 

the only time a large area is heated is if the piece has torsion stresses that need to be relieved before starting the job.

[macbruce]

I worked in a big fab shop back in 1980 and learned a thing or two about this.......When I watched the old timers they never put the water spray directly on the heated area, rather they would spray around the hot spot and gradually work the cool towards the hot spot and that's how I'll do it when my new toy arrives......Harris Rosebud 2290hp-4 Cutting Welding Torch 2393-3F / 43-2 Heating sku 193   

I was able to get this on eb for less than the Victor tip alone would cost ($180+tax).....Always wanted a Harris rig anyhow, plus the tips are WAY cheaper than Victor.......... B)

The main reason for this query is that I've never tackled big thick plate before, it's a little daunting but doable I think..... :unsure:

[forgemaster]

I would second the advice of Yahoo2, its called line heating for a reason, its not called just get the whole plate hot heating. I have pulled 3" plate back into line using just a cutting torch with oxy/lpg and a small plastic bottle with a hole at the bottom (to trickle the water onto the line of heat as I am going) filled up from a bucket of water as I went.

Phil


This may help to explain the principles behind line heating, hope it is able to be read.

Phil

[macbruce]

Got it! I'll have a crack at it in the AM............ B)

[macbruce]

Sucess! I ran the monster slowly across the plate on the high side for 20 min and the worse gap in the rt side foreground has shrunk from 3/8'' to 1/8'' and it's not even cool yet. Might run another pass tomorrow if needed............. B)

[Frosty]

Uh Bruce: The torch is aimed the wrong way in the first picture!

 

Well done, I love it when a plan works.

 

Frosty The Lucky.

[macbruce]

That torch is so KA it just need to be close...... :P ....It operates at 25-30 psi fuel and 80-90 psi oxyen....... I reckon if I let go of it, it would whip around the shop like firey Dervish so I keep a firm grip on it............Tried a second pass with it yesterday but I ran out of ox.......go figure...... :mellow:

[ianinsa]

That torch is so KA it just need to be close...... :P ....It operates at 25-30 psi fuel and 80-90 psi oxyen....... I reckon if I let go of it, it would whip around the shop like firey Dervish so I keep a firm grip on it............Tried a second pass with it yesterday but I ran out of ox.......go figure...... :mellow:

You should not have used the bottle from the porta-pac? Grin.

[rivet]

Put your torch on the high side. At the same time use a water siphon on the opposite side.When hot quinch quickly . Large rags,cold water.The hot side becomes the

short side .  by Mr. John Adolph

 

 

 

 .

[jkw]

hello all.

 

this has been interesting as i have a 3/16 plate 24x36 inches that i cut from a larger piece that was once part of a coil.  Anyway, how they taught me in engineering school was as follows.

 

say you have a cube of steel for example.  and you stick it in a vice like below

 

------||    ||-----  sorry  i cant draw a top and bottom of the cube of steel but pretend they are there and NOT held by the vice.  prevend also that ----|   and |------- are the vice.

 

Now, say you heat JUST the cube and can keep the vice the same temp as originally of the cube and vice before heating.  what happens is the coefficent of linear expansion and expansion take place in the UP and DOWN directions ONLY.  Remember the sides are in a vice.  Pretend it is a 3 dimensional vice too so  ALL 4 sides are held the same and cant expand but the top and bottom do expand as it is heated.  Google linear expansion and you will get the formula.....it depends on the DIFFERENCE in temp beteeen the part that is not heated and the part heated.

 

Now for what happens..........as the cube starts to cool, ALL sides contract.  the tops and bottoms AND ALL the sides contract since there is nothing holding it any  longer.  so what should happen in theory is the TOP and BOTTOM should go back to where they were originally, BUT the sides will contract by the amount given in the equation for linear expansion and it will fall on the floor since nothing is holding it any longer.

 

So, what does this say?  In the theory, it should not matter if you cool it off sooner.....that really has nothing to do with it.  However, what MAY happen is it PREVENTS the further spread of heat to areas the torch didnt touch and acts as our hypothetical vice we used above. 

Thats just what i was thinking as i read this.......not real sure thats what is taking place as water is applied to the hot side.

 

Just my 2 cents

       |    |

[JNewman]

In this theoretical situation your top and bottom would not return to the original position as the cube would now be bigger in the height.  The metal displaced from the sides has to go somewhere.  

[jkw]

well in theory the top and bottom should return to the original, or close to original, position. Why?  Because ALL side are now free to shrink as they are not held by a vice or anything and are a heated  temp, call it 'B'.  We simply apply the formula for expansion in the reverse.  The delta temp variable is now ( new temp - the steady state temp[room temp]) and the expansion (on ALL faces) is now negative.

[JNewman]

You can't compress a solid.  Where does the metal from the sides go? Assuming immovable vice the material draws out/extrudes to the top and bottom.

 

[jkw]

well it has expanded and then contracted.  nothing out of the ordinary here.  here is an explanation :  https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Physical_Chemical/ThermalExpansion.htm

 

if you take a cube and heat it and it is not held on ANY side, it will expand as given in the explanation above.  ALL sides expand. then as it cools down, it contracts on all sides. By the same amout given in the eqation in the link.

 

just because it is considered a sold, does not mean it cannot expand or contract.  Here is a good example....at times when i try and fit a new bearing race into a hub it is very tight.  so i will just stick the race into the freezer for awhile.  then it will fall right into the hub with little effort.  why?  because the race has 'contracted' as given by the formula in the link above.  as it heats to room temp again. it will expand and be snug in the hub.

[ThomasPowers]

So when they use shaped charges to compress the plutonium in an atomic bomb it's not compressing? "The convergent shock wave of an implosion can compress solid uranium or plutonium by a factor of 2 to 3. " http://nuclearweaponarchive.org/Library/Implsion.html

[jkw]

so i wonder if user MACBRUCE had great success with heating with a large rosebud?  it sounds like he did. i am just wondering if the plate was fairly straight the whole length/width of it.  what i am trying to say is if it came out wavey in any way after it totally cooled.  or was in fairly smooth and of the same height along the whole plane of steel.

 

why i am interested is i have the plates i mentioned above, BUT i dont have my oxy tanks with me.  i am a snow bird and in FL now.  anyway, i was wondering if using a MAP torch would work or i would just be wasting time and fuel trying it.  or if someone knows of a mechanical way of straightening that would be cool...i am all ears.  i just dont have any more plate with me so i dont want to botch what i have trying all types of experiments. right now the plates have a symetric curve/bulge to them so if i could get rid of the curve that would be great.

[JNewman]

I fully understand that material expands and contracts will temperature change. Any working day I am not forging I am anticipying the changes in size of constrained and unconstrained metal. I still have to disagree about the change in length. I will try to explain better.

I hope we can all agree that if you squeeze a cube of steel on 4 sides beyond its elastic limit it will get longer.  The volume remains the same.  This is the basic concept of constant volume forging.

Heating the cube with it constrained on 4 sides is squeezing the cube, the force is applied by the temperature increase rather than by moving the constraints like you would with a press or hammer. This will cause the cube to lengthen on the unconstrained sides exactly as if you squeezed it with a press.

To explain it another way if you had a 1x1x1 cube at room temperature.  This is 1 cubic inch. Then we Did your constrained heat. Let's say that the sides end up 0. 01"smaller at room temperature so that it falls out of your vise which I fully agree will happen at the return to room temperature. If the cube does not get longer the volume at room temperature will only be 0.9801 cubic inches.  What I am saying is that the average length will be 1.02" long

 

As to your 3/16 plate I don't think you will be very successful flattening the amount of curve you will have from a coil.  That is a lot of curvature. I have straightened an I beam somewhat with heat but plate is something of an art. It is easy turn the plate into a potato chip. 3/16 plate is not very expensive, I suspect that you will spend much more money on gas than you would on a new piece of plate even if you were able to flatten it

[jkw]

thanks for the reply about the 3/16 plate thats what i was afraid of.

 

 

let try and use the formula to dive deeper into the cube
 

 

The linear coefficient of thermal expansion ( a) describes the relative change in length of a material per degree temperature change. As shown in the following equation, a is the ratio of change in length ( Dl) to the total starting length (l_i) and change in temperature ( DT).

for mild steel 

12.0 x 10 -6  the units are very important...hence:  a (m/m/oK)

 

so alpha(a) is in units of (meter* degree Kelvin)/meter.  the conversion from Kelvin to Farenheight is

℉ =(K - 273.15)* 1.8000+ 32.00    or rearanging:  (F -32)/1.8 +273.15  =K    so lets start at 70 F  => 294.26K  and 1000 F  is => 810 K

 

so rearanging the formula above:

 

a(l*deltaT)= delta L  for our cube that is 1 meter.

 

12 x10^-6 *1* (516.5) = .00619 meter or 6.19 mm  now remember that is th total length from top, to bottom of the cube held by teh 4 sided hypothetical vice that doesnt heat up or move in anyway.

 

Or cube started at 70 F and is now 1000 F.  Now lets cool it back down:

 

our delta T is the same ( 810-294.26 = 516.5 K)  the original length of the top to bottom of the cube is now 1.00619 meter.  remember, it expanded.  the other sides of the cube are still 1 meter since they didnt expand.

 

so, for the top to bottom: 12x10^-6 * 1.00619 * 516.5  =Delta L  or .006236   NOTE:  this is slightly differnt from the expansion number since we started at a length that was expanded.  So the amount that it shrank from top to bottom is 1.00619 - 006236= .999953 meter.

 

now for the sides:

 

12x10^-6 * 1* 516.5 = .00619 meter  that it will contract.  OR  1 -.00619 = .9981 meter  from one side to the opposite.

 

So what can we determine on a theoretical basis?  That the cube is now no longer a cube.  the sides measure LESS than the distance from the top to bottom.  So yes, there is a difference in volume...instead of 1 x1x1 = 1meter cube, we have a .9981 x .9981 x .999953= .99615 meter cube.

 

 

 

[JNewman]

4 hours ago, jkw said:

our delta T is the same ( 810-294.26 = 516.5 K)  the original length of the top to bottom of the cube is now 1.00619 meter.  remember, it expanded.  the other sides of the cube are still 1 meter since they didnt expand..k

Except they did expand, however being constrained by the immovable vice the metal is forced to go up and down. exactly like if you were to squeeze it with a press.  

So your formula is missing data you need to add the volume displaced from the sides to the top to bottom dimension

 

 

So what can we determine on a theoretical basis?  That the cube is now no longer a cube.  the sides measure LESS than the distance from the top to bottom.  So yes, there is a difference in volume...instead of 1 x1x1 = 1meter cube, we have a .9981 x .9981 x .999953= .99615 meter cube.

So where did the extra metal go?  It has to go somewhere. Can we agree steel cannot be compressed?