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Material Handbook 9th Editition, Volume 14
Forming and Forging, pg 331
Wire, Rod, and Tube Drawing
Aproach Angle	Δ
deformation zone	Δ	( α rad/γ)*(1+(1-γ)^.5 ) ^2
die angle	α	The half angle in radians
	5	(Deg/180)*3.14156	typical 5-15 degree half angles
D0	2.25	in
D1	2.20	in
α	0.0872222	B12/180*3.14
reduction	γ	1 - (A1/Ao)
A0	3.9740625	3.14 *$B$20*$B$20/4
A1	3.7994	3.14 *$B$21*$B$21/4
γ	0.0439506	1-($B$18/$B$17)
Δ	7.7627778	( α rad/γ)*(1+(1-γ)^.5 ) ^2
Minumum draw stress	Δmin	=4.9(μ / (LN(1/(1-γ))))^.5
coefficient of friction	μ
μ	0.11	guess
Δmin	7.6656377
draw stress	σd
avg flow stress	σa	average work per volume
Yield Strength	σy	psi
actual ys	200,000	lb/in*in
drawing stress ratio	Σ	[(3.2/Δ )+0.9](α+ μ)	Another Book 5.13
	σd	σy * (3.2/ Δ ) + 0.9)  (α+ μ)
psi Draw Stress	572,776	$B$38 * (3.2/($B$24 + 0.9)) * ($B$16+ $B$33)
Pounds Force		Stress * Area	572,776 * 3.799
Force	2,176,205	=$B$41 * $B$19
Draw Bench is called 100,000 pound?

Can the Actual tested Yield be used in these calculations?

Where do find where my math went wrong?

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------

Original Message:
Sent: 12/14/2023 6:29:00 PM
From: David Kirchner
Subject: Draw Bench math

Material Handbook 9th Editition, Volume 14
Forming and Forging, pg 331
Wire, Rod, and Tube Drawing
Aproach Angle	Δ
deformation zone	Δ	( α rad/γ)*(1+(1-γ)^.5 ) ^2
die angle	α	The half angle in radians
	5	(Deg/180)*3.14156	typical 5-15 degree half angles
D0	2.25	in
D1	2.20	in
α	0.0872222	B12/180*3.14
reduction	γ	1 - (A1/Ao)
A0	3.9740625	3.14 *$B$20*$B$20/4
A1	3.7994	3.14 *$B$21*$B$21/4
γ	0.0439506	1-($B$18/$B$17)
Δ	7.7627778	( α rad/γ)*(1+(1-γ)^.5 ) ^2
Minumum draw stress	Δmin	=4.9(μ / (LN(1/(1-γ))))^.5
coefficient of friction	μ
μ	0.11	guess
Δmin	7.6656377
draw stress	σd
avg flow stress	σa	average work per volume
Yield Strength	σy	psi
actual ys	200,000	lb/in*in
drawing stress ratio	Σ	[(3.2/Δ )+0.9](α+ μ)	Another Book 5.13
	σd	σy * (3.2/ Δ ) + 0.9)  (α+ μ)
psi Draw Stress	572,776	$B$38 * (3.2/($B$24 + 0.9)) * ($B$16+ $B$33)
Pounds Force		Stress * Area	572,776 * 3.799
Force	2,176,205	=$B$41 * $B$19
Draw Bench is called 100,000 pound?

Can the Actual tested Yield be used in these calculations?

Where do find where my math went wrong?

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------

The context I was writing yesterday looked good to me, but it see now it did not flow well to this formatting.

I am looking to find out how to apply mechanical engineering math to my draw bench. While some engineers are trying to get humans to Mars, I am trying to help by cold drawing to the next level.  This data was recorded from testing done till the carriage grip broke. Now I am slowing down and checking the math.  Input was a highly cold worked stainless at 2.25" diameter and it was drawn to 2.20" diameter. This die was bought as FF RA15 6" 5° 2.200. The 5° may be the full angle, not the half angle.

Testing results:

2.250" diameter 212 UTS, 193Yld, 17EL, 60RA Lot G4818

2.220" diameter estimate 200Yld

2.150" diameter 230UTS 210Yld 14EL 54EL 

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------

Original Message:
Sent: 12-15-2023 07:34
From: David Coulston
Subject: Draw Bench math

David,

Are you really drawing a 2.25 inch diameter bar down to 2.20 inches?! That is very large material, but a very light reduction (<5% RA).

Dave Coulston

Original Message:
Sent: 12/14/2023 6:29:00 PM
From: David Kirchner
Subject: Draw Bench math

Material Handbook 9th Editition, Volume 14
Forming and Forging, pg 331
Wire, Rod, and Tube Drawing
Aproach Angle	Δ
deformation zone	Δ	( α rad/γ)*(1+(1-γ)^.5 ) ^2
die angle	α	The half angle in radians
	5	(Deg/180)*3.14156	typical 5-15 degree half angles
D0	2.25	in
D1	2.20	in
α	0.0872222	B12/180*3.14
reduction	γ	1 - (A1/Ao)
A0	3.9740625	3.14 *$B$20*$B$20/4
A1	3.7994	3.14 *$B$21*$B$21/4
γ	0.0439506	1-($B$18/$B$17)
Δ	7.7627778	( α rad/γ)*(1+(1-γ)^.5 ) ^2
Minumum draw stress	Δmin	=4.9(μ / (LN(1/(1-γ))))^.5
coefficient of friction	μ
μ	0.11	guess
Δmin	7.6656377
draw stress	σd
avg flow stress	σa	average work per volume
Yield Strength	σy	psi
actual ys	200,000	lb/in*in
drawing stress ratio	Σ	[(3.2/Δ )+0.9](α+ μ)	Another Book 5.13
	σd	σy * (3.2/ Δ ) + 0.9)  (α+ μ)
psi Draw Stress	572,776	$B$38 * (3.2/($B$24 + 0.9)) * ($B$16+ $B$33)
Pounds Force		Stress * Area	572,776 * 3.799
Force	2,176,205	=$B$41 * $B$19
Draw Bench is called 100,000 pound?

Can the Actual tested Yield be used in these calculations?

Where do find where my math went wrong?

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------

The context I was writing yesterday looked good to me, but it see now it did not flow well to this formatting.

I am looking to find out how to apply mechanical engineering math to my draw bench. While some engineers are trying to get humans to Mars, I am trying to help by cold drawing to the next level.  This data was recorded from testing done till the carriage grip broke. Now I am slowing down and checking the math.  Input was a highly cold worked stainless at 2.25" diameter and it was drawn to 2.20" diameter. This die was bought as FF RA15 6" 5° 2.200. The 5° may be the full angle, not the half angle.

Testing results:

2.250" diameter 212 UTS, 193Yld, 17EL, 60RA Lot G4818

2.220" diameter estimate 200Yld

2.150" diameter 230UTS 210Yld 14EL 54EL 

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------

Original Message:
Sent: 12-15-2023 07:34
From: David Coulston
Subject: Draw Bench math

David,

Are you really drawing a 2.25 inch diameter bar down to 2.20 inches?! That is very large material, but a very light reduction (<5% RA).

Dave Coulston

Original Message:
Sent: 12/14/2023 6:29:00 PM
From: David Kirchner
Subject: Draw Bench math

Material Handbook 9th Editition, Volume 14
Forming and Forging, pg 331
Wire, Rod, and Tube Drawing
Aproach Angle	Δ
deformation zone	Δ	( α rad/γ)*(1+(1-γ)^.5 ) ^2
die angle	α	The half angle in radians
	5	(Deg/180)*3.14156	typical 5-15 degree half angles
D0	2.25	in
D1	2.20	in
α	0.0872222	B12/180*3.14
reduction	γ	1 - (A1/Ao)
A0	3.9740625	3.14 *$B$20*$B$20/4
A1	3.7994	3.14 *$B$21*$B$21/4
γ	0.0439506	1-($B$18/$B$17)
Δ	7.7627778	( α rad/γ)*(1+(1-γ)^.5 ) ^2
Minumum draw stress	Δmin	=4.9(μ / (LN(1/(1-γ))))^.5
coefficient of friction	μ
μ	0.11	guess
Δmin	7.6656377
draw stress	σd
avg flow stress	σa	average work per volume
Yield Strength	σy	psi
actual ys	200,000	lb/in*in
drawing stress ratio	Σ	[(3.2/Δ )+0.9](α+ μ)	Another Book 5.13
	σd	σy * (3.2/ Δ ) + 0.9)  (α+ μ)
psi Draw Stress	572,776	$B$38 * (3.2/($B$24 + 0.9)) * ($B$16+ $B$33)
Pounds Force		Stress * Area	572,776 * 3.799
Force	2,176,205	=$B$41 * $B$19
Draw Bench is called 100,000 pound?

Can the Actual tested Yield be used in these calculations?

Where do find where my math went wrong?

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------

Hi,

For those seeking additional context, these equations are found in the following article in the current ASM Handbook series:

Wire, Rod, and Tube Drawing, Metalworking: Bulk Forming, Vol 14A, ASM Handbook, Edited By S.L. Semiatin, ASM International, 2005, p 448–458, https://doi.org/10.31399/asm.hb.v14a.a0004008

The relevant excerpt is shown in the image.

------------------------------
Scott Henry
Senior Content Engineer
ASM International
------------------------------

Original Message:
Sent: 12-14-2023 18:28
From: David Kirchner
Subject: Draw Bench math

Material Handbook 9th Editition, Volume 14
Forming and Forging, pg 331
Wire, Rod, and Tube Drawing
Aproach Angle	Δ
deformation zone	Δ	( α rad/γ)*(1+(1-γ)^.5 ) ^2
die angle	α	The half angle in radians
	5	(Deg/180)*3.14156	typical 5-15 degree half angles
D0	2.25	in
D1	2.20	in
α	0.0872222	B12/180*3.14
reduction	γ	1 - (A1/Ao)
A0	3.9740625	3.14 *$B$20*$B$20/4
A1	3.7994	3.14 *$B$21*$B$21/4
γ	0.0439506	1-($B$18/$B$17)
Δ	7.7627778	( α rad/γ)*(1+(1-γ)^.5 ) ^2
Minumum draw stress	Δmin	=4.9(μ / (LN(1/(1-γ))))^.5
coefficient of friction	μ
μ	0.11	guess
Δmin	7.6656377
draw stress	σd
avg flow stress	σa	average work per volume
Yield Strength	σy	psi
actual ys	200,000	lb/in*in
drawing stress ratio	Σ	[(3.2/Δ )+0.9](α+ μ)	Another Book 5.13
	σd	σy * (3.2/ Δ ) + 0.9)  (α+ μ)
psi Draw Stress	572,776	$B$38 * (3.2/($B$24 + 0.9)) * ($B$16+ $B$33)
Pounds Force		Stress * Area	572,776 * 3.799
Force	2,176,205	=$B$41 * $B$19
Draw Bench is called 100,000 pound?

Can the Actual tested Yield be used in these calculations?

Where do find where my math went wrong?

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------

David,

I am a little bit confused by this calculation:

σy * (3.2/ Δ ) + 0.9)  (α+ μ)

I am probably not grabbing the right numbers... but when I plug values in, I get an answer that is about 1/10 of the 572,776 value you ended up with.

Another Dave

------------------------------
David Coulston
Niles MI
------------------------------

Original Message:
Sent: 12-14-2023 18:28
From: David Kirchner
Subject: Draw Bench math

Material Handbook 9th Editition, Volume 14
Forming and Forging, pg 331
Wire, Rod, and Tube Drawing
Aproach Angle	Δ
deformation zone	Δ	( α rad/γ)*(1+(1-γ)^.5 ) ^2
die angle	α	The half angle in radians
	5	(Deg/180)*3.14156	typical 5-15 degree half angles
D0	2.25	in
D1	2.20	in
α	0.0872222	B12/180*3.14
reduction	γ	1 - (A1/Ao)
A0	3.9740625	3.14 *$B$20*$B$20/4
A1	3.7994	3.14 *$B$21*$B$21/4
γ	0.0439506	1-($B$18/$B$17)
Δ	7.7627778	( α rad/γ)*(1+(1-γ)^.5 ) ^2
Minumum draw stress	Δmin	=4.9(μ / (LN(1/(1-γ))))^.5
coefficient of friction	μ
μ	0.11	guess
Δmin	7.6656377
draw stress	σd
avg flow stress	σa	average work per volume
Yield Strength	σy	psi
actual ys	200,000	lb/in*in
drawing stress ratio	Σ	[(3.2/Δ )+0.9](α+ μ)	Another Book 5.13
	σd	σy * (3.2/ Δ ) + 0.9)  (α+ μ)
psi Draw Stress	572,776	$B$38 * (3.2/($B$24 + 0.9)) * ($B$16+ $B$33)
Pounds Force		Stress * Area	572,776 * 3.799
Force	2,176,205	=$B$41 * $B$19
Draw Bench is called 100,000 pound?

Can the Actual tested Yield be used in these calculations?

Where do find where my math went wrong?

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------

In Excel I typed drawing stress formula as:

σy * (3.2/( Δ  + 0.9))  (α+ μ)  = 572,776

σy * ((3.2/ Δ ) + 0.9)  (α+ μ) = 51,760

Draw Stress	51,760	$B$31 * (3.2/$B$21) + 0.9) * ($B$13+ $B$25)
		200,000 * 1.312 * .1972
Pounds Force		Draw Stress * Area
		51,760 * 3.799
Draw Force	196,657 lb	=$B$40 * $B$17

Thanks, you were right about the error.

I have the updated 14A now, thanks Scott Henry.

Can I use the actual Yield as the Average flow stress?

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------

Original Message:
Sent: 12-15-2023 13:36
From: David Coulston
Subject: Draw Bench math

David,

I am a little bit confused by this calculation:

σy * (3.2/ Δ ) + 0.9)  (α+ μ)

I am probably not grabbing the right numbers... but when I plug values in, I get an answer that is about 1/10 of the 572,776 value you ended up with.

Another Dave

------------------------------
David Coulston
Niles MI

Original Message:
Sent: 12-14-2023 18:28
From: David Kirchner
Subject: Draw Bench math

Material Handbook 9th Editition, Volume 14
Forming and Forging, pg 331
Wire, Rod, and Tube Drawing
Aproach Angle	Δ
deformation zone	Δ	( α rad/γ)*(1+(1-γ)^.5 ) ^2
die angle	α	The half angle in radians
	5	(Deg/180)*3.14156	typical 5-15 degree half angles
D0	2.25	in
D1	2.20	in
α	0.0872222	B12/180*3.14
reduction	γ	1 - (A1/Ao)
A0	3.9740625	3.14 *$B$20*$B$20/4
A1	3.7994	3.14 *$B$21*$B$21/4
γ	0.0439506	1-($B$18/$B$17)
Δ	7.7627778	( α rad/γ)*(1+(1-γ)^.5 ) ^2
Minumum draw stress	Δmin	=4.9(μ / (LN(1/(1-γ))))^.5
coefficient of friction	μ
μ	0.11	guess
Δmin	7.6656377
draw stress	σd
avg flow stress	σa	average work per volume
Yield Strength	σy	psi
actual ys	200,000	lb/in*in
drawing stress ratio	Σ	[(3.2/Δ )+0.9](α+ μ)	Another Book 5.13
	σd	σy * (3.2/ Δ ) + 0.9)  (α+ μ)
psi Draw Stress	572,776	$B$38 * (3.2/($B$24 + 0.9)) * ($B$16+ $B$33)
Pounds Force		Stress * Area	572,776 * 3.799
Force	2,176,205	=$B$41 * $B$19
Draw Bench is called 100,000 pound?

Can the Actual tested Yield be used in these calculations?

Where do find where my math went wrong?

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------

Original Message:
Sent: 12/15/2023 2:55:00 PM
From: David Kirchner
Subject: RE: Draw Bench math

In Excel I typed drawing stress formula as:

σy * (3.2/( Δ  + 0.9))  (α+ μ)  = 572,776

σy * ((3.2/ Δ ) + 0.9)  (α+ μ) = 51,760

Draw Stress	51,760	$B$31 * (3.2/$B$21) + 0.9) * ($B$13+ $B$25)
		200,000 * 1.312 * .1972
Pounds Force		Draw Stress * Area
		51,760 * 3.799
Draw Force	196,657 lb	=$B$40 * $B$17

Thanks, you were right about the error.

I have the updated 14A now, thanks Scott Henry.

Can I use the actual Yield as the Average flow stress?

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------

Original Message:
Sent: 12-15-2023 13:36
From: David Coulston
Subject: Draw Bench math

David,

I am a little bit confused by this calculation:

σy * (3.2/ Δ ) + 0.9)  (α+ μ)

I am probably not grabbing the right numbers... but when I plug values in, I get an answer that is about 1/10 of the 572,776 value you ended up with.

Another Dave

------------------------------
David Coulston
Niles MI

Original Message:
Sent: 12-14-2023 18:28
From: David Kirchner
Subject: Draw Bench math

Material Handbook 9th Editition, Volume 14
Forming and Forging, pg 331
Wire, Rod, and Tube Drawing
Aproach Angle	Δ
deformation zone	Δ	( α rad/γ)*(1+(1-γ)^.5 ) ^2
die angle	α	The half angle in radians
	5	(Deg/180)*3.14156	typical 5-15 degree half angles
D0	2.25	in
D1	2.20	in
α	0.0872222	B12/180*3.14
reduction	γ	1 - (A1/Ao)
A0	3.9740625	3.14 *$B$20*$B$20/4
A1	3.7994	3.14 *$B$21*$B$21/4
γ	0.0439506	1-($B$18/$B$17)
Δ	7.7627778	( α rad/γ)*(1+(1-γ)^.5 ) ^2
Minumum draw stress	Δmin	=4.9(μ / (LN(1/(1-γ))))^.5
coefficient of friction	μ
μ	0.11	guess
Δmin	7.6656377
draw stress	σd
avg flow stress	σa	average work per volume
Yield Strength	σy	psi
actual ys	200,000	lb/in*in
drawing stress ratio	Σ	[(3.2/Δ )+0.9](α+ μ)	Another Book 5.13
	σd	σy * (3.2/ Δ ) + 0.9)  (α+ μ)
psi Draw Stress	572,776	$B$38 * (3.2/($B$24 + 0.9)) * ($B$16+ $B$33)
Pounds Force		Stress * Area	572,776 * 3.799
Force	2,176,205	=$B$41 * $B$19
Draw Bench is called 100,000 pound?

Can the Actual tested Yield be used in these calculations?

Where do find where my math went wrong?

------------------------------
David Kirchner
COO
High Performance Alloys, Inc
Tipton IN
(765) 945-8230
------------------------------