Latest revision as of 00:03, 30 January 2012

Overview

Ironworker Hole puncher calculations.

Calculations

Force Required to Shear the metal

According to an Ironworker Blade manufacturer (he prefers not to be mentioned by name) with a 6 degree rake in the blade, you can shear 1"x10" steel with 100 tons of force.

No data has been found regarding how much force to shear 1"x12" as is the desired spec for the machine, so we will assume that there is a direct relaitonship between length of cut and force to shear it. Following this logic, 100T/10"=xT/12" Solving for x, we get Force to shear 1"x12"=120T.

We will include a safety factor to account for any errors we have made. A force of 130T or more should suffice.

The required moment to act on blade

Force Diagram for design option 1

Force Diagram for design 2

To find this, we must make a few basic assumptions:

The fulcrum supports are 8" wide
The blade is 12" wide and placed as close as possible to the fulcrum supports.
The force will be applied at the center of the blade and distributed evenly.
The angle blade, and the flat blade are mounted on the same mount.

Blade moment

Since the force on the blade(B) is known, as is its distance from the fulcrum, we can calculate the necessary moment, M. M=Force(B) x Distance from the pivot(d) (10")

For the capability of cutting 1"x12", B=130 tons.

M=130 tons x 10" = 1300 inch tons

Throw Requirements

Throw is the vertical movement required for the blade. In this case, It's 2.75".

Calculating the arm length and cylinder size

The arm length will control how many times magnification we get from the cylinder force.

If building an ironworker whose max cutting ability is under 1" x 12", you could scale the cylinder size down according to your desires.

Since the such high force is required to shear the desired metal, we will use the 6" bore cylinder with a 12" stroke available at surplus center.

Arm length must satisfy throw and force requirements. Force generates a minimum arm length requirement, whereas throw generates a maximum arm length.

Throw requires that the cylinder rod is long enough, or spaced properly, so that the blade has enough vertical movement. This is effected adversely by arm length. IE, the longer the arm is, the longer rod you will need.
Force requires that the lever arm is long enough to magnify the force from the cylinder to enough to shear the metal. The longer the arm is, the less force you will need.

Minimum Arm Length

To calculate the arm length using this cylinder, we must set the cylinder force (C) multiplied by the lever arm length (A) equal to the required moment (M), so A X C = M. Solving for A, we get A=M/C

In this particular design, the cylinder will be contracting while shearing, which means the cylinder tonnage we are using must be on the contracting side. See Cylinder Tonnage Calculations for how to find the contracting strength. The calculated contracting tonnage (C) for this specific cylinder is 31.8 tons.

So, A= 1300inT/31.8T =
A =40.86" Minimum.

Maximum Arm Length

This is calculated by the ratios of throw to rod length and arm length. The distance (D) any point on the arm travels is inversley related to its distance (L) from the fulcrum. Or: D1/L1=D2/L2. 1 corresponds to the cylinder, and 2 corresponds to the blade. L1=12", as it is the rod length on the cylinder. L2=2.75", as it is the required throw for the blade. D2 is 10", the distance from the blade to the fulcrum. D1 is unknown.

Solving for D1, we get D1= D2L1/L2= 10"*12"/2.75"= 43.63"=A max

Calculation of the size of the pin

Vertical Support Thicknesses

Arm Thickness

Since the arm is absorbing twice the force from the pin as the Vertical Supports, It needs to be twice as thick, or 3" thick.

Arm Dimensions

Making shear and bending moment diagrams will help us determine how thick the arm should be at various locations.

@@ Line 4: / Line 4: @@
 =Calculations=
 ==Force Required to Shear the metal==
-Using the shear tonnage calculator at [http://multicyl.biz/calculations.html Multicyl], the force required to shear 80KSI Tensile strength A36 Steel is:
+According to an Ironworker Blade manufacturer (he prefers not to be mentioned by name) with a 6 degree rake in the blade, you can '''shear 1"x10" steel''' with '''100 tons''' of force.
-*'''1"x12"'''=480 Tons
-*'''1"x10"'''=400 Tons
-On [http://www.cutsmart.com/pages/articles/die_cutting_tonnage_needed.html Cut Smart]'s website, the formula to calculate the required pounds of force to shear any material is: "Straight from the from Machinery's Handbook 25th ED. P1924 “P = periphery x thickness x tensile strength(PSI) where P is cutting force in pounds.”
+No data has been found regarding how much force to shear 1"x12" as is the desired spec for the machine, so we will assume that there is a direct relaitonship between length of cut and force to shear it. Following this logic, '''100T/10"=xT/12"''' Solving for x, we get Force to shear ''' 1"x12"=120T.'''
-So, using the periphery(the length of the cut) as 12", the thickness as 1", and the tensile strength as 80,000 PSI (common upper end value for A36 steel), we get,
+We will include a safety factor to account for any errors we have made. A force of '''130T''' or more should suffice.
-P=12"x1"x80,000PSI= 960,000 pounds = 480 tons
-*Given that the two websites match, as well as a few others I could find, this calculation is valid. The question is, how do we generate such high tonnage? Do we really want to make the machine that powerful?
-*Likewise, we can calculate the required tonnage to cut the '''6"x6"x1/2"''' angle:
-P=12"x1/2"x80,000PSI= 480,000 pounds= 240 tons
-*Alternately, If 6x6 is too hard to cut, we could cut '''4"x4"x1/2"'''
-P=8"x1/2"x80,000PSI= 320,000 pounds= 160 tons
-*'''Shear Angle''' - is the angle from the horizontal with which a cutting blade contacts the workpiece. The above calculations assume a shear angle of zero degrees. With a shear angle of just 5 degrees from the horizontal, the ironworker needs significantly less force to shear the aforementioned steel.
-*'''Angle Calculation Note''' - The force calculation for angles should yield significantly lower values due to the support piece for the angle (the angle is cut with the two edges at the top, the cross-section area being sheared has no support from the rest of the angle, hence requires much less force than calculated above).
 ==The required moment to act on blade==
@@ Line 32: / Line 18: @@
 *The blade is 12" wide and placed as close as possible to the fulcrum supports.
 *The force will be applied at the center of the blade and distributed evenly.
-*The required moment will be whichever is highest for a particular design, the blade force, or the angle shear force.
+*The angle blade, and the flat blade are mounted on the same mount.
-*In design 1, the angle shear is between the blade and the cylinder
-*In design 2, the angle shear is between the cylinder and the fulcrum supports.
-*As the distance of the blade to the fulcrum doesn't change, the blade moment is the same for design 1 and design 2.
 ===Blade moment===
@@ Line 41: / Line 24: @@
 M=Force(B) x Distance from the pivot(d) (10")
-'''For the capability of cutting 1"x12",''' B=480 tons.
+'''For the capability of cutting 1"x12",''' B=130 tons.
-*M=480 tons x 10" = 4800 inch tons
+*M=130 tons x 10" = 1300 inch tons
-'''For the capability of cutting 1"x10", '''B=400 tons.
-*M=400 tons x 10" = 4000 inch tons
-===Angle Shear moment===
+===Throw Requirements===
-B=240 tons<br/>
+[[Image:throwironworker.png|thumb|400px]]
-'''Design 1, 6x6 angle:''' d=21.5"
+'''Throw''' is the vertical movement required for the blade. In this case, It's 2.75".
-*M=240 tons x 21.5" = 5160 inch tons<br/>
-'''Design 1, 4x4 angle:'''
-*M=160 tons x 21.5" = 3440 inch tons.<br/>
-'''Design 2:''' d= 9"
-*M=240 tons x 9" = 2160 inch tons
-===Required Moments===
-'''Design 1, 6x6:''' As the angle moment is the highest, this value is the requirement, or M=5160 inch tons. This means this machine would have the capacity for 1"x12" flat cuts.<br/>
-'''Design 1, 4x4:''' The blade moment is highest, so for the '''12" cut, M=4800''' inch tons. For the '''10" cut, M=4000''' inch tons.<br/>
-'''Design 2:''' The Blade moment is highest, so for the 12" cut, M=4800 inch tons. For the 10" cut, M=4000 inch tons.
 ==Calculating the arm length and cylinder size==
-There are two possible designs for this; the blade being between the cylinder and the fulcrum supports(Design 1), or the blade being on the other side, along with the punch (Design 2)
+The arm length will control how many times magnification we get from the cylinder force.
-*If building an ironworker whose max cutting ability is under 1" x 10", you could scale the cylinder size down according to your desires.
+*If building an ironworker whose max cutting ability is under 1" x 12", you could scale the cylinder size down according to your desires.
-Since the such high force is required to shear the desired metal, we will use the [https://www.surpluscenter.com/item.asp?item=9-4907-08&catname=hydraulic largest available cylinder] with an 8" stroke available at surplus center.
+Since the such high force is required to shear the desired metal, we will use the [https://www.surpluscenter.com/item.asp?item=9-1144-12&catname=hydraulic 6" bore cylinder] with a 12" stroke available at surplus center.
-To calculate the arm length using this cylinder, we must set the cylinder force (C) multiplied by the lever arm (A) equal to the required moment (M), so A X C = M. Solving for A, we get A=M/C
+Arm length must satisfy '''throw''' and '''force''' requirements. '''Force''' generates a minimum arm length requirement, whereas '''throw''' generates a maximum arm length.
+* '''Throw''' requires that the cylinder rod is long enough, or spaced properly, so that the blade has enough vertical movement. This is effected adversely by arm length. IE, the longer the arm is, the longer rod you will need.
+* '''Force''' requires that the lever arm is long enough to magnify the force from the cylinder to enough to shear the metal. The longer the arm is, the less force you will need.
+===Minimum Arm Length===
+To calculate the arm length using this cylinder, we must set the''' cylinder force (C)''' multiplied by the '''lever arm length (A)''' equal to the '''required moment (M),''' so A X C = M. Solving for A, we get '''A=M/C'''
-===Design 1===
+In this particular design, the cylinder will be contracting while shearing, which means the cylinder tonnage we are using must be on the contracting side. See [[Cylinder Tonnage Calculations]] for how to find the contracting strength. The calculated contracting tonnage (C) for this specific cylinder is 31.8 tons.
-In this particular design, the cylinder will be contracting while shearing, which means the cylinder tonnage we are using must be on the contracting side. See [[Cylinder Tonnage Calculations]] for how to find the contracting strength. The calculated contracting tonnage (C) for this specific cylinder is 56.52 tons.
-'''6x6''' M=5160 inch tons, so A=5160 inch tons / 56.52 tons = 91.30 inches.<br/>
+So, A= 1300inT/31.8T =<br/>
-'''4x4, 12"''' M=4800 inch tons, so A=4800 inch tons / 56.52 tons = 84.93"<br/>
+'''A =40.86" Minimum.'''
-'''4x4, 10"''' M=4000 inch tons, so A=4000 inch tons / 56.52 tons = 70.77""<br/>
-===Design 2===
+===Maximum Arm Length===
-In this design, the cylinder is expanding while shearing the 1"x12", so we use the expanding cylinder tonnage for the 8" bore cylinder, or 75.36 tons.
+This is calculated by the ratios of throw to rod length and arm length. The distance (D) any point on the arm travels is inversley related to its distance (L) from the fulcrum.  Or: '''D1/L1=D2/L2.''' 1 corresponds to the cylinder, and 2 corresponds to the blade. L1=12", as it is the rod length on the cylinder. L2=2.75", as it is the required throw for the blade. D2 is 10", the distance from the blade to the fulcrum. D1 is unknown.
-A=M/C
+Solving for D1, we get D1= D2L1/L2= 10"*12"/2.75"=  '''43.63"=A max'''
-'''For the capability of cutting 1"x12", M=4800 Inch tons.'''
-*A=4800 inch tons/ 75.36 tons = 63.69 inches
-'''For the capability of cutting 1"x10", M= 4000 inch tons'''
-*A=4000 inch tons / 75.36 tons = 53.08 inches
-'''We should also double check to make sure the punch still has enough power'''
-We must find the required punch moment, and insure it is less than the required blade moment.
-*M= Punch force (P) x distance (d). P= 150 tons, d=23"
-M= 150 tons x 23" = 3450 inch tons.
-This is less than the blade moment, so we are OK.
-==Design analysis==
-As we can see, both designs are going to require really long overall arms. Here are the calculated lengths:<br/>
-'''Design 1''' will require A+6", including the end amounts (this depends on how close the cylinder attaches to the end, and how far the punch is away from the end). <br/>
-*'''6x6'''= 91.3"+6" = 97.3 inches
-*'''4x4, 12"'''= 84.9"+6" = 90.9" including the end amounts.
-*'''4x4, 10"'''= 70.7"+6" = 76.7", including the end amounts.
-'''Design 2''' will require A+23", including the end amounts.
-*'''For 12":''' 63.36+23"= 96.36", including the end amounts.
-*'''For 10":''' 53.08+23"= 66.08", including the end amounts.
 ==Calculation of the size of the pin==
-[[image:armforcediagram.jpg|thumb|400px|right|the forces acting on the arm and the pin]]
+[[Image:pindiameter.jpg|600px]]
-To find the size of the pin, we must first find how much force is acting on it when it is most stressed. In this case, it is when we are cutting thru the 1"x10" metal.
-Since the arm is not accelerating upwards, we can do a statics calculation, and assume the sum of all the forces is zero. So: '''-31.5T+400T+P=0'''
-Solving for P, we get '''P=368.5T.'''
-Now, we must use this to find the necessary cross sectional area of the pin to prevent it from warping.
-*At each side, where the pin goes thru the pin holders, a force of P/2 is applied to it via the fulcrum supports. This means that there is a '''shear force of P/2, or 184.25T''' acting on the pin.
-*The allowable shear for A36 steel is 7.2TSI.
-*To calculate the radius of the pin, we must set the product of the cross sectional area (A) and the allowawble shear (S) equal to the shear force (P/2). or, '''AxS=P/2'''
-**We can replace A with PixR^2, since the cross sectional area is a circle. So '''PixR^2xS=P/2'''. Solving for R, we get '''R=Root[(P/2)/Pi/S]'''
-**Plugging it all in, we get R=Root[184.25T/3.14/7.2TSI]=2.86"
-**To find the diameter of the pin, we simply multiply Rx2. so '''D=5.71'''
-Now that's a big pin!
+==Vertical Support Thicknesses==
+[[Image:supportthickness.jpg|600px]]
+==Arm Thickness==
+Since the arm is absorbing twice the force from the pin as the Vertical Supports, It needs to be twice as thick, or 3" thick.
+==Arm Dimensions==
+Making shear and bending moment diagrams will help us determine how thick the arm should be at various locations.
+[[Image:bendingdiagram.jpg|600px]]
+===Height at pin===
+[[Image:armdimensionatpin.jpg|600px]]
+===Cross sectional areas due to shear===
+[[Image:xcshear.jpg|600px]]
 ==See Also==
 {{Ironworker}}
+[[Category:Ironworker]]

Hole Puncher Calculations: Difference between revisions

Latest revision as of 00:03, 30 January 2012

Contents

Overview

Calculations

Force Required to Shear the metal