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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 (li) 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.

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.

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.

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.

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 | |