Ironworker/Research Development - Revision history

Marcin: /* Overview */

2013-04-05T19:47:37Z

Overview

← Older revision		Revision as of 19:47, 5 April 2013
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	=Overview=		=Overview=
	[[Brianna Kufa]] ~~is currently~~ (as of 3/17) developing an ironworker which imitates a 50T Piranha, but with increased punching and shearing capacities. See [http://osebri.wordpress.com/ her blog] for the latest updates on progress.		[[Brianna Kufa]] was (as of 3/17/12) developing an ironworker which imitates a 50T Piranha, but with increased punching and shearing capacities. See [http://osebri.wordpress.com/ her blog] for the latest updates on progress.

	[[Media:Strippercalcs1.pdf\|Here]] are some calculations made my Mischa John for the punch stripper element. The analysis proved the current design on that element was insufficient. We are now working on another one which will provide enough resistance.		[[Media:Strippercalcs1.pdf\|Here]] are some calculations made my Mischa John for the punch stripper element. The analysis proved the current design on that element was insufficient. We are now working on another one which will provide enough resistance.

Bkufa: /* Overview */

2012-03-17T19:24:15Z

Overview

← Older revision		Revision as of 19:24, 17 March 2012
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	=Overview=		=Overview=
	[[Brianna Kufa]] is currently developing an ironworker which imitates a 50T Piranha, but with increased punching and shearing capacities. See [http://osebri.wordpress.com/ her blog] for the latest updates on progress.		[[Brianna Kufa]] is currently (as of 3/17) developing an ironworker which imitates a 50T Piranha, but with increased punching and shearing capacities. See [http://osebri.wordpress.com/ her blog] for the latest updates on progress.

			[[Media:Strippercalcs1.pdf\|Here]] are some calculations made my Mischa John for the punch stripper element. The analysis proved the current design on that element was insufficient. We are now working on another one which will provide enough resistance.

Bkufa at 06:10, 20 January 2012

2012-01-20T06:10:25Z

← Older revision		Revision as of 06:10, 20 January 2012
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	=Overview=		=Overview=
	~~We are~~ currently ~~in the design stages of the shear prototype. There's quite~~ a ~~few things we need to know before the design can continue~~. ~~Please answer any of the following questions if you can:~~		[[Brianna Kufa]] is currently developing an ironworker which imitates a 50T Piranha, but with increased punching and shearing capacities. See [http://osebri.wordpress.com/ her blog] for the latest updates on progress.
	*What is the necessary force to shear 1" thick x 12" wide metal?

	~~according to:~~ [http://~~web.mit~~.~~edu/2~~.~~810~~/~~www/lecture2010/9.pdf an MIT presentation~~] :
	~~F = 0.7 T L (UTS)~~

	~~T = sheet thickness~~
	~~L = total length sheered~~
	~~UTS = Ultimate Tensile Strength of material (so I think it depends~~ on ~~the kind of metal)~~

	~~''Note:'' This formula does not consider the effect of ''blade shear angle''; increasing the shear angle trades off force against stroke~~.

ChuckH: /* Overview */

2012-01-14T21:55:12Z

Overview

← Older revision		Revision as of 21:55, 14 January 2012
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	[http://www.metalprocorp.com/viewProduct.jsp?id=24 Metalpro's 45ton ironworker] looks about as simple as you can get. It's got about 25tons at the flat shear. So, if you replaced the hydraulic cylinder with a linkage down to a lever underneath the table, and had a longer cylinder under the table, you could get 5-10X mechanical advantage and increase the tonnage at the shear until it could cut through 1x12" plate. [https://www.youtube.com/watch?v=vr1D4qPrYd4 Here is a good overview] of how it functions.		[http://www.metalprocorp.com/viewProduct.jsp?id=24 Metalpro's 45ton ironworker] looks about as simple as you can get. It's got about 25tons at the flat shear. So, if you replaced the hydraulic cylinder with a linkage down to a lever underneath the table, and had a longer cylinder under the table, you could get 5-10X mechanical advantage and increase the tonnage at the shear until it could cut through 1x12" plate. [https://www.youtube.com/watch?v=vr1D4qPrYd4 Here is a good overview] of how it functions.

	[http://www.thefabricator.com/article/shearing/cutting-structural-steel-to-length--the-sheer-facts-of-shear-tooling-and-processes ~~An article about shear design~~].		Articles about shear design [http://www.custompartnet.com/wu/sheet-metal-shearing here] and [http://www.thefabricator.com/article/shearing/cutting-structural-steel-to-length--the-sheer-facts-of-shear-tooling-and-processes here].

	=Concepts=		=Concepts=

ChuckH: /* Concepts */

2012-01-14T21:52:41Z

Concepts

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	*Shear for 8-12" of 1" steel		*Shear for 8-12" of 1" steel
	*Angle cutter for up to 6x6 angle		*Angle cutter for up to 6x6 angle
	*DIY shear blade - made from hardening of mild steel, at $40 in materials		*DIY shear blade - made from case hardening of mild steel, at $40 in materials
			** what is trade-off of DIY material against the [http://cartech.ides.com/datasheet.aspx?i=101&c=TechArt&E=114 S-7] class of steel usually used for commercial shear blades?
	*May involve [[Induction Furnace]] for hardening		*May involve [[Induction Furnace]] for hardening

ChuckH: /* Overview */

2012-01-14T21:45:15Z

Overview

← Older revision		Revision as of 21:45, 14 January 2012
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	The Scotsman ironworker provided as an example was validated on A36 steel, which has a UTS of 400MPa.<ref>[http://en.wikipedia.org/wiki/Ultimate_tensile_strength tensile strength on wikipedia]</ref> That works out to about 58kpsi.<ref>[http://www.calculatenow.biz/conversions/pressure.php?bar=&mbar=&hPa=&Pa=400000000&psi=&cmHg=&cmHg=&cmWC=&inWC=&atm=&submit=Convert calculator at calculate now]</ref> So the amount of force necessary to shear 12 square inches (1 by 12) of a material with 58kpsi UTS should be 487,200 pounds or 245 short tons.		The Scotsman ironworker provided as an example was validated on A36 steel, which has a UTS of 400MPa.<ref>[http://en.wikipedia.org/wiki/Ultimate_tensile_strength tensile strength on wikipedia]</ref> That works out to about 58kpsi.<ref>[http://www.calculatenow.biz/conversions/pressure.php?bar=&mbar=&hPa=&Pa=400000000&psi=&cmHg=&cmHg=&cmWC=&inWC=&atm=&submit=Convert calculator at calculate now]</ref> So the amount of force necessary to shear 12 square inches (1 by 12) of a material with 58kpsi UTS should be 487,200 pounds or 245 short tons.

	I believe the grabCAD competition specified a 5" hydraulic cylinder. Assuming it's operating at 2500psi that should give it around 25 short tons (49k pounds). So, to get a 120T ironworker you'd need at least 5X mechanical advantage. And to get 250 short tons at the shear I guess you'd need 10X mechanical advantage.		I believe the grabCAD competition specified a 5" hydraulic cylinder. Assuming it's operating at 2500psi that should give it around 25 short tons (49k pounds). So, to get a 120T ironworker you'd need at least 5X mechanical advantage. And to get 250 short tons at the shear I guess you'd need 10X mechanical advantage. (10X advantage means the cylinder will travel 10" while cutting the 1" workpiece -- it might be worthwhile to increase the cylinder diameter.)

	Also, you'd want the motion to be such that the hydraulic cylinder is extending, because that direction will be stronger than pulling.		Also, you'd want the motion to be such that the hydraulic cylinder is extending, because that direction will be stronger than pulling.

ChuckH: /* Overview */

2012-01-14T21:41:36Z

Overview

← Older revision		Revision as of 21:41, 14 January 2012
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	L = total length sheered		L = total length sheered
	UTS = Ultimate Tensile Strength of material (so I think it depends on the kind of metal)		UTS = Ultimate Tensile Strength of material (so I think it depends on the kind of metal)

			''Note:'' This formula does not consider the effect of ''blade shear angle''; increasing the shear angle trades off force against stroke.


	The Scotsman ironworker provided as an example was validated on A36 steel, which has a UTS of 400MPa.<ref>[http://en.wikipedia.org/wiki/Ultimate_tensile_strength tensile strength on wikipedia]</ref> That works out to about 58kpsi.<ref>[http://www.calculatenow.biz/conversions/pressure.php?bar=&mbar=&hPa=&Pa=400000000&psi=&cmHg=&cmHg=&cmWC=&inWC=&atm=&submit=Convert calculator at calculate now]</ref> So the amount of force necessary to shear 12 square inches (1 by 12) of a material with 58kpsi UTS should be 487,200 pounds or 245 short tons.		The Scotsman ironworker provided as an example was validated on A36 steel, which has a UTS of 400MPa.<ref>[http://en.wikipedia.org/wiki/Ultimate_tensile_strength tensile strength on wikipedia]</ref> That works out to about 58kpsi.<ref>[http://www.calculatenow.biz/conversions/pressure.php?bar=&mbar=&hPa=&Pa=400000000&psi=&cmHg=&cmHg=&cmWC=&inWC=&atm=&submit=Convert calculator at calculate now]</ref> So the amount of force necessary to shear 12 square inches (1 by 12) of a material with 58kpsi UTS should be 487,200 pounds or 245 short tons.

ChuckH: /* Overview */

2012-01-14T21:37:19Z

Overview

← Older revision		Revision as of 21:37, 14 January 2012
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	[http://www.metalprocorp.com/viewProduct.jsp?id=24 Metalpro's 45ton ironworker] looks about as simple as you can get. It's got about 25tons at the flat shear. So, if you replaced the hydraulic cylinder with a linkage down to a lever underneath the table, and had a longer cylinder under the table, you could get 5-10X mechanical advantage and increase the tonnage at the shear until it could cut through 1x12" plate. [https://www.youtube.com/watch?v=vr1D4qPrYd4 Here is a good overview] of how it functions.		[http://www.metalprocorp.com/viewProduct.jsp?id=24 Metalpro's 45ton ironworker] looks about as simple as you can get. It's got about 25tons at the flat shear. So, if you replaced the hydraulic cylinder with a linkage down to a lever underneath the table, and had a longer cylinder under the table, you could get 5-10X mechanical advantage and increase the tonnage at the shear until it could cut through 1x12" plate. [https://www.youtube.com/watch?v=vr1D4qPrYd4 Here is a good overview] of how it functions.

			[http://www.thefabricator.com/article/shearing/cutting-structural-steel-to-length--the-sheer-facts-of-shear-tooling-and-processes An article about shear design].

	=Concepts=		=Concepts=

Matt Maier at 22:14, 28 December 2011

2011-12-28T22:14:58Z

← Older revision		Revision as of 22:14, 28 December 2011
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	*What is the necessary force to shear 1" thick x 12" wide metal?		*What is the necessary force to shear 1" thick x 12" wide metal?

	according to: [web.mit.edu/2.810/~~ts_temp~~/~~sheetmetal~~.~~ppt~~] :		according to: [http://web.mit.edu/2.810/www/lecture2010/9.pdf an MIT presentation] :
	F = 0.7 T L (UTS)		F = 0.7 T L (UTS)

Matt Maier at 22:10, 28 December 2011

2011-12-28T22:10:07Z

← Older revision		Revision as of 22:10, 28 December 2011
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	Also, you'd want the motion to be such that the hydraulic cylinder is extending, because that direction will be stronger than pulling.		Also, you'd want the motion to be such that the hydraulic cylinder is extending, because that direction will be stronger than pulling.

			[http://www.metalprocorp.com/viewProduct.jsp?id=24 Metalpro's 45ton ironworker] looks about as simple as you can get. It's got about 25tons at the flat shear. So, if you replaced the hydraulic cylinder with a linkage down to a lever underneath the table, and had a longer cylinder under the table, you could get 5-10X mechanical advantage and increase the tonnage at the shear until it could cut through 1x12" plate. [https://www.youtube.com/watch?v=vr1D4qPrYd4 Here is a good overview] of how it functions.
	=Concepts=		=Concepts=