Longitudinal bending, or bending with the materialgrain, increases the required minimum inside radius ofthe bend.
Q: In a previous Bending Basics article, you mention a “material’s tensile reduction percentage” used for calculating the minimum inside bend radius of various metals. You described a rough rule to find a steel’s minimum bend radius: Divide 50 by the material’s reduction percentage, as specified by the material supplier, subtract by 1, and then multiply by the plate thickness.
To state your example, “If the steel has a tensile reduction value of 10 percent, divide 50 by that value: 50/10 = 5. Next, subtract 1 from that answer: 5 – 1 = 4. Now, multiply that answer by the plate thickness. If the material is 0.5 inch thick: 4 × 0.5 = 2. So in this case, the minimum inside bend radius is 2 times the material thickness.”
If I am looking at properties for A36 steel, I see elongation properties between 18 and 21 percent. Is this the property to which you refer? Using the median of 20 percent and material thickness of 0.5 in., would this equate to [(50/20) – 1] × 0.5 = 0.75. Then I’d multiply this answer by the material thickness: 0.75 × 0.5 = 0.375 in. However, a bend allowance chart I have from our press brake manufacturer gives different information. For 0.5-in.-thick plate, it shows an inside bend radius of either 0.781 in. for a 5-in. die width, or 0.625 in. for a 4-in. die width.I know you mentioned that the calculation you gave is a rough guide, but the discrepancy between the formula and the brake manufacturer’s bend allowance chart seems significant; it’s off by a factor of 2. Is the 20 percent number correct, or did I apply the equation incorrectly?
A:You have the formula correct at 0.375 in. However, I believe you misinterpreted the meaning and how it relates to your bend allowance chart. The calculated 0.375-in. radius is the minimum producible inside radius for this material and not the recommended inside radius that you see on the bend allowance chart.
It’s true that the harder and thicker the plate is, the greater the minimum bend radius. The minimum inside bend radius is even larger when bending with the grain. In steel between 0.5 and 0.8 in. thick, grade 350 and 400 may have a minimum bend radius of 2.5 times the material thickness when transverse bending, while longitudinal bending may require a minimum bend radius that’s 3.75 times the material thickness (see Figure 1).
Also note that all these are just examples; actual minimum radius values vary by grade. For steel, aluminum, and stainless, you will find a variety of minimum bend radii-to-thickness ratios, and you will need to research these values in data provided by your material supplier.
But again, all this refers to the minimum inside bend radius—what’s physically possible if you have press brakes and tooling that can handle the high tonnage. The recommended inside bend radius is what’s optimal in a typical application.
The radii from the bend allowance chart (see Figure 2) is based on 15 percent of the die opening and represent basically a one-to-one relationship between the material thickness and the inside radius when using the recommended die opening of 5 in. By using the 5-in. die and thereby producing that one-to-one inside radius-to-material-thickness ratio, you will achieve the best results in your product and the most stable bend angles.
Assuming you have the tonnage and could force the material into a die opening only 2.5 in. wide—just five times the material thickness—you would achieve the minimum inside radius that’s possible in an air form. However, for many reasons I wouldn’t recommend you do this. Nonetheless, some folks do this if they have high-tonnage machines.
This portion of a bend allowance chart gives the recommended inside radius, not the minimum inside radius that canbe produced. Also note that this is not a bend deduction chart. The bend allowance is used to figure the bend deductionsyou need for the flat layout. Editor’s Note: This chart is for illustrative purposes only and should not replace bendallowance charts provided by the press brake manufacturer.
The 5-in. die is 10 times the material thickness, and, as a general rule at these thicknesses, 10 to 12 times the material thickness is correct and common. I, along with your press brake manufacturer, would not recommend using a smaller die opening, if for no other reason than to avoid damage to your press brake and tooling. If your die opening is less than 8 times the material thickness for sheet metal, or less than 10 to 12 times the material thickness for plate, you will see a dramatic increase in the amount of pressure required to bend the material.
Also, lifting the workpiece during forming can increase the tonnage required dramatically. Figure 3 shows 10-gauge (0.135-in.) material; that’s technically sheet metal and a long way from 0.5-in. plate, but the principle is the same. The material will need to be lifted during forming over a 1.125-in. die opening.
To calculate the additional tonnage, you need the weight of steel per square foot for the given thickness, which you can get from your supplier or on gauge charts for common materials. The weight of 10-ga. carbon steel per square foot is 5.625 pounds, so for 10-ft.-wide material that extends out in front of the press by roughly 10 ft., the total weight of the sheet in front of the press is:
Material length (ft.) × Material width (ft.) × Weight of steel per square foot (lbs.) =
Total weight (lbs.)
10 × 10 × 5.625 = 562.50 lbs.
Next, multiply the total weight by 60 (one-half of the sheet width in inches) and divide that by half the die opening width. Finally, divide the bending force by 2,000 lbs. (1 U.S. ton), and you get the total additional tonnage required for lifting the part.
Total weight (lbs.) × Half of sheet width (in.) /
Half of die opening width (in.) =
Bending force (lbs.)
(562.5 × 60) / 0.5625 = 60,000 lbs.
of bending force
60,000 / 2,000 = 30 U.S. tons
Lifting the workpiece during forming can increase the tonnage required.
Next, use this formula to calculate forming tonnage for mild steel (all variables are in inches unless otherwise noted):
[(575 × Material thickness2)/Die opening] × Bend length in feet = Forming tonnage
[575 × 0.1352/1.125] × 10 = 93.15 tons to bend
Finally, add the tonnage required to lift the part:
93.15 + 30 = 123.15 tons
to bend and lift this part
You can reduce the tonnage by using a larger die opening, but this will require you to take into account a larger inside bend radius and, therefore, a larger bend deduction and bend allowance for the flat-blank calculations. For instance, a die opening of 1.25 in. will require 88.84 tons to form and 27.00 tons to lift, for a total of 115.84 tons. A die opening of 1.50 in. will require 69.86 tons to form and 22.50 tons to lift, for a total of 92.36 tons.
And this is just for 10-ga. material. Going back to the original question, you can see what would happen if you tried to form 0.5-in.-thick material in a die opening that’s only five times the material thickness.
Factors which influence the minimum bending radius include the cable size, the cable construction, the conductor type and the sheathing and insulation types used. The bending radius is normally expressed as a factor of the overall dimension of the cable for example, 6D or 6x the outer diameter of the cable.What should my bend radius be? ›
We recommend a minimum bend radius of 1t for all sheet metal parts. Thus the smallest radius of any bend in a sheet should be at least equal to the thickness of the sheet. For example, if the thickness of the sheet is 1 mm, the minimum bend radius should be 1 mm.What is the minimum wire bend radius? ›
The conductor shall not be bent to a radius less than 8 times the overall diameter for nonshielded conductors or 12 times the overall diameter for shielded or lead-covered conductors during or after installation.What is the minimum bending radius for other bending machines and devices? ›
Generally speaking, the minimum bending radius should not be less than 2-2.5 times the outer diameter of the pipe, and the shortest straight distance should not be less than 1.5-2 times the outer diameter of the pipe, except in special cases.How do you find the minimum bend radius for sheet metal? ›
You described a rough rule to find a steel's minimum bend radius: Divide 50 by the material's reduction percentage, as specified by the material supplier, subtract by 1, and then multiply by the plate thickness.What is the minimum bend diameter? ›
The minimum bend radius refers to the lowest radius at which a cable can be bent. So a smaller bend radius means a cable is more flexible. The bend diameter is equal to twice the bend radius.What is the minimum bending length? ›
For bends, the minimum distance between the inside edge of the bend and the outside of the hem should be 5 times material thickness plus bend radius plus hem radius.What is the bend radius formula? ›
Input everything into the bend allowance formula: BA = angle × (π/180) × (radius + K-factor × thickness) .What are the minimum and maximum bend radius of copper cable? ›
The inside radius of bends should be no tighter than 4X the outer diameter (OD) of the cable. For trueCABLE F/UTP solid copper shielded cable the inside radius of the cable should not be tighter than 7X the OD of the cable.What do you mean by minimum bend radius in bending operation? ›
The minimum bend radius is the radius below which an object such as a cable should not be bent.
The normal recommendation for fiber optic cable bend radius is the minimum bend radius under tension during pulling is 20 times the diameter of the cable. When not under tension (after installation), the minimum recommended long term bend radius is 10 times the cable diameter.Why is bending radius important? ›
Why is Bend Radius Important? It is important to make sure your cable support has the proper bend radius because it aligns with the ANSI/TIA 568.0-D for Generic Telecommunications Cabling standard.What is the minimum bending radius for cold bending? ›
In cold-rolled mild steel, the minimum bend radius is 63 percent of the material thickness. For example, a piece of 12-gauge (0.104-in., 2.64-mm) material turns sharp at 0.065 in. (1.66 mm), making any punch radius less than that value, by definition, a sharp bend, such as 0.062 in. (1/16 in., 1.58 mm), 0.032 in.What is the minimum bend radius for 1/4 OD? ›
|Inside Diameter (inches)||Bend Diameter (inches)*||Bend Radius (inches)|
The radius of bends in the conductor shall not be less than the lead diameter or lead thickness. The minimum distance from the part body or seal to the start of the bend in a part lead shall be a minimum of 2 lead diameters for round leads, and 0.5 mm (0.020 in.)What is the bend allowance formula? ›
Bend Allowance = Angle(Π/180)(Bend Radius+KFactor(Thickness)) Take this equation and plug in your own values for Angle, Bend Radius, and K-Factor.What is minimum diameter of bending former? ›
|Nominal size of bar, d, mm||Minimum radius for scheduling, r||Minimum diameter of bending former, M|
There are generally four (4) different methods of tube/pipe bending that can also be classified according to complexity. You'll encounter the following: compression bending, rotary draw bending, roll bending, and mandrel tube bending.How is bend measured? ›
Use the correct formula to calculate the circumference of the bend. For example, the formula used when making a 90˚ bend is: Circumference = (pi*2*r) / (360/90).What is K factor in bending? ›
The K factor is defined as the ratio between the material thickness (T) and the neutral fibre axis (t), i.e. the part of the material that bends without being compressed nor elongated. Bend allowance is a fundamental parameter to calculate sheet elongation.
What are the recommended minimum bend radii when installing indoor copper premise cabling? The recommended minimum bend radii for unshielded horizontal cables (6 pair or smaller) is 4 times the cable diameter.What is the minimum bend radius of copper cable after installation? ›
General rules and recommendations for selecting the radius
If this is achieved by a loop, the cable must be provided with a bend radius of at least 10 times the diameter of the cable. The larger the radius, the less stress is exerted on the cables, which ensures a longer service life.
The bending angle of the sheet metal is the key factor affecting the bending allowance. For example, if bending the sheet metal with large radius, more bending allowance will be deducted; if bending acute angle, less bending allowance shall be deducted.How do you determine the minimum bend radius with regard to pipe diameter thickness and material? ›
The MBR depends on diameter and wall thickness, but is generally 3 to 5 times the pipe diameter. This means that an 18 in. pipe designed with a 3D radius will have a radius to centerline of 54 in. The angle of the bend must be such that the riser follows the platform batter in the plane of the riser and pipeline.Why do optical fibers have a minimum bend radius? ›
There are two reasons for maintaining minimum bend radius protection: enhancing the fiber's long-term reliability; and reducing signal attenuation. Bends with less than the specified minimum radius will exhibit a higher probability of long-term failure as the amount of stress put on the fiber grows.What is the maximum bend radius for fiber optic cable? ›
After completion of the pull, the cable should not have any bend radius smaller than 10 times the cable diameter. When a fiber cable is bent excessively, the optical signal within the cable may refract and escape through the fiber cladding. Bending can also permanently damage the fiber by causing micro-cracks.What is the minimum bending radius of types use and SE cable? ›
Bends in Types USE and SE cable shall be so made that the cable will not be damaged. The radius of the curve of the inner edge of any bend, during or after installation, shall not be less than five times the diameter of the cable.Why do you need to calculate bend allowance? ›
The bend allowance defines the material you will need to add to the actual leg lengths of the part in order to get the flat pattern cut to a correct size.Why bend deduction is required? ›
Considering the Bend Deduction and Bend Allowances is a critical first step in designing sheet metal parts as it affects nearly every following step in the fabrication process. More so, it will allow you to achieve the correct size and dimensions needed in the flat pattern.What is the effect of bend radius on the loss factor? ›
It has been found that the loss increases as the bending radius and number of turns increase. The result also showed that elliptical shaped bending configuration produced more loss in contrast to that of sinusoidal shaped at bending angles of 180° and 360°.
The bend radius ratio b is defined as the ratio of the bend centerline radius, r, to the outer diameter of the pipe, D (see Fig. ... Context 2. ... the azimuth u is the angular coordinate of the damage in degrees (see Fig. 1(b)).What is a 1 D Bend? ›
1D would be a centerline bend radius that is one times the diameter of the pipe. This means a 1D bend on a 6-inch pipe would be a 6-inch centerline radius. Short radius bends are specified to be 1D bends. 1.5D bends and up are considered long radius bends.How much force does it take to bend 1/4 steel? ›
EXAMPLES: Question: How many tons is required for a brake press machine to bend 10 feet of 1/4 inch thick mild steel using a standard 2 inch wide bottom die? Answer: 15.3 tons/foot x 10 feet = 153 tons. Therefore a 175 tons would be a good minimum size brake press machine if you are making more than just a few parts.What is a 2 bend radius? ›
The bending radius is expressed in relation to the pipe's outer diameter. If the radius is equal to the pipes diameter then the radius is 1D. The radius is 2 times greater than the outer diameter then the pipes bends is 2D.What three factors determine the minimum bend radius that can be made in a piece of aluminum alloy? ›
The composition of the metal, its temper, and its thickness determine the minimum radius that may be successfully generated in the part through bending without adversely affecting the material.How do you find the minimum bend radius of tubing? ›
Standard draw bend radius is 2 x D
What that means is that if you have a tube OD (outside diameter) of 20 mm then the bend radius to choose, if you can, is 40 mm.
Springback in sheet metal bending depends upon different variable parameters like sheet metal material, sheet metal thickness, punch angle, punch height, punch radius, punch load, die opening, die lip radius, sheet metal material grain direction, pre bend strip condition etc.What determines bending strength? ›
This finding strongly suggests that bending strength is determined by the yield strain. The yield stress in tension, which might be expected to predict the bending strength, underestimates the true bending strength by approximately 40 %.How is bend factor calculated? ›
To calculate bend allowance: Obtain the properties of the bend (bend radius, angle, and method used). Obtain the characteristics of your material (thickness and K-factor for this specific bend). Input everything into the bend allowance formula: BA = angle × (π/180) × (radius + K-factor × thickness) .What is the minimum bending radius for steel pipe in cold bending? ›
In cold-rolled mild steel, the minimum bend radius is 63 percent of the material thickness. For example, a piece of 12-gauge (0.104-in., 2.64-mm) material turns sharp at 0.065 in. (1.66 mm), making any punch radius less than that value, by definition, a sharp bend, such as 0.062 in. (1/16 in., 1.58 mm), 0.032 in.