Use of Bending Force Chart
General
Reading the bending force chart
Four Relationships
Calculation of Bending Tonnage
Complement of the bending force chart
Table of Contents
21 General A bending force chart is one of the necessary instruments in bending operations. It can easily be found as a nameplate on the press brake or in the manufacturer's catalogue and instruction manual. This chapter explains the use of the force chart for mastery purposes. The operator should understand the chart completely in order to enhance job efficiency and technical ability. It is recommended that this chapter be reread if there are any questions concerning its contents.
22 Reading the bending force chart Look at Table 21 which is the air bending force chart. The first or leftmost column shows the sheet thickness (t) of the work. The three rows at the top indicate the Vwidth of the die (V), minimum flange length (b), and the inside bending radius (ir) from the top. The figures shown are the required tonnage per meter of work. The following items can be read from this table:
221 Explanation of symbols Symbol (t) in the first column indicates the sheet thickness of the work, ranging from 0.5 mm to 30 mm. Symbol V in the top row indicates the width of the die Vgroove, or Vwidth. The figures in this row are standard commercial sizes. However, the standard size of the Vwidth vary from manufacturer to manufacturer.
t : Sheet thickness (mm) (tensile strength F: Bending force per meter (ton/m) i: inside bending radius (mm) b: Minimum flange length (mm) V: Vdie width (MM)
The following can be read from this chart if the Sheet thickness of the material and the inside radius of the bend are determined.
Symbol b in the middle row indicates the minimum flange length. Symbol ir stands for the inside bending radius of the product. Here let us examine the bending force chart to see if ir= V/6. When V is 10 mm, ir is 1.6 mm, and when V is 32 mm, ir is 5 mm. Therefore, ir is approximately 1/6 the Vwidth. F indicates the "required tonnage" used to bend one meter of the work.
222 Minimum flange length As previously mentioned, if the Vwidth is known from the sheet thickness, the minimum flange length can be found from the bending force chart. Let us examine why the word "minimum" is used for the flange length. Until bending is completed, the work must be securely supported at the shoulder of the Vdie groove. If not, as bending progresses, the work slides off the shoulder, and the bending line changes irregularly. Good bending accuracy cannot be obtained in such a bending operation and the operation itself becomes dangerous. The "minimum flange length" is necessary in order to bend the work with sufficient accuracy and safety. Fig. 21 shows the minimum flange length calculation. According to the right isosceles triangle ABC, the length of side b is V2times V/2. That is, the minimum flange length can be calculated by b=(2)½( V /2).
If a 6 mm Vwidth die is used to bend work, b is 4 mm according to the chart. Similarly, when the Vwidth is 25 mm, then b becomes 17.5 mm. If the flange size in the work drawing is larger than b, then it can be assumed the Vwidth is correct. However, if it is less than b, the Vwidth is inadequate for the bending and a smaller Vwidth must be used.
2.2.3 Required tonnage The statement, "tensile strength: 4550 kg/mm², is shown in Table 21. This 4550 kg/mm² refers to mild steel. As already stated, the value of F (bending force) is the required tonnage per meter of work. It is also the required tonnage in bottoming. Generally, the bending force charts indicate the required tonnage needed to bend mild steel by bottoming. The following steps should be taken to find the necessary F. First, determine the Vwidth according to work thickness t. Next, follow the horizontal row of the necessary t to the right, and read the figure in the position where it meets the vertical row of the determined V. For example, if a Vwidth of 12 mm is used to bend work that is 2 mm in thickness, you will find the required tonnage F is equal to 22 tons. Likewise, when t is 3.2 mm and V is 25 mm, then F should read 27 tons. When t is 6 mm and V is 50 mm, F should be 48 tons. These F values are the required tonnage per meter of work. Go to Top
23 Four Relationships The four relationships refer to those between the required tonnage and Vwidth, sheet thickness, bending length, and tensile strength. In many cases, the bending force chart is not utilized effectively due to the operator not understanding it. Effective use of the chart is being able to make the most of these four relationships, or those between F and V, F and t, F and and F and b.
231 Relationship between F and V Table 22 shows the required tonnage used per meter when 1 mm and 2 mm mild steel sheets are bent in different Vwidths. As shown in the chart, if the Vwidth is doubled, the required tonnage will be reduced to 1/2. In short, the required tonnage F is inversely proportional to the Vwidth V. This inverse proportion is expressed as F=k11V.
Table 22
t=1.0 mm
t=2.0 mm
V(mm)
6
12
25
F(ton)
11
22
232 Relationship between F and t The relationship between the required tonnage and the sheet thickness is apt to be misunderstood. A conversation, "The sheet is double in thickness, so the tonnage also becomes double," is often heard in the production site. This statement is incorrect. Table 23 shows changes in the required tonnage when the Vwidth was kept constant and the sheet thickness was varied. When t doubles, F becomes approximately fourfold if the Vwidth is the same. That is, bending force F increases in proportion to the square of sheet thickness. This is not a simple proportional relationship. For this reason, die designers and expert operators caution by saying "Bending force is affected by the square of sheet thickness." Therefore, the above conversation on the job site must be corrected. Operators should know that if sheet thickness doubles, the tonnage will become fourfold. The relationship between F and t is expressed by F=k.t².
Table 23
t=2.0 mm"
t(mm)
1
2
1.2
2.3
23
233 Relationship between F and L LP is the bending length of work, which is proportional to F. The tonnage shown in the bending force chart are values based on the unit bending length of one meter. To obtain the tonnage for a certain length of work, convert the length into meters and multiply by the related tonnage. The relationship between F and t is given by F=kt.
234 Relationship between F and bb F and tensile strength bb are proportional to each other, just as F and e above. Thus, if only b is known, the required tonnage to bend materials other than SS41 can be determined by using this relationship.
Table 24 shows the tensile strength of commonly used sheet materials. Since a variety of materials are available, the bb of the confirmed when calculating F.
The relationship between F and bb is expressed by F=kb. Go to Top
Table 24 Tensile Strength Of Various Materials
Material
Tensile strength, kg/mm²
Soft
Hard
Lead
2.54

Tin
45
Aluminium(99.0%)
9.3
171
Hightension aluminium alloy Type 4
48
Duralumin
26
Zinc
15
Copper
2228
3040
Brass (70:30)
33
53
Brass (60:40)
38
49
Phosphor bronze
4050
5075
Bronze
Nickel silver
3545
5570
Cold rolled iron sheet
3238
Steel, 0.1%C
32
40
Steel, 0.2%C
50
Steel, 0.3%C
45
60
Steel, 0.4%C
56
72
Steel, 0.6%C
90
Steel, 0.8%C
110
Steel, 1.0%C
100
130
Silicon steel sheet
55
65
Stainless steel sheet
6570
Nickel
4450
5763
24 Calculation of Bending Tonnage The bending tonnage serves as the basis for judgement in selecting a new machine, as well as determining whether an alreadyinstalled press brake can achieve a bending operation from the standpoint of capacity. The required bending tonnage can be calculated by the four relationships, based on the F value in the bending force chart. Examples of bending tonnage calculations and their solutions are given below. The reader should try to answer them in order to familiarize himself with the four relationships.
[Example 1] How many tons are required to bend 4 meters of a 1.5mm thick SUS304 stainless steel sheet (tensile strength: 60 kg/mm²)? [Solution] The sought Vwidth is calculated as V=6XI.5 . .10 (6xl.5=9, but V=9 mm is not a standard size so V=10 mm is substituted). In this case, t=1.5 mm is not in the bending force chart, so the bending force (F=17 tons) should be read for V=to mm and t = 1. 6 mm (the closest t). The calculating formula, which takes into consideration the sheet thickness, tensile strength, and bending length, will be given as follows: F.17X(1.5/1.6)2X60/45 x 4= 80 The sought solution is 80 tons. In this calculation, raise the fractions to units. [Example 2] How many tons are necessary to bend a 15 mm thick and 3,100 mm long sheet of rolled SS41 steel? The flange length is 120 mm. [Solution] The sought Vwidth is V=t2xl5=180. Although a standard Vwidth of 160 mm or 200 mm is acceptable, a Vwidth of 200 mm is not suitable from the viewpoint of length b (flange length). This is because the minimum flange length for a Vwidth Of 200 mm is 14o mm, which is greater than the required 120 mm. Therefore, V is set at 160 mm. Also, t=15 mm is not shown in the chart, so 107 tons is obtained when the F for t=16 mm and V=160 mm is read. Thus, we will have the following formula: F= 107 x (15/16)2 x 3. 1=292 Consequently, the sought answer is 292 tons. [Example 3] Calculate the required tonnage when bending SPCC of t3.2 mm and =2,400 mm with a V width of 18 mm. 6b is 32 kg/mm². [Solution] The sought Vwidth is V=8x3.2=25, and V=25 mm is adequate. However, when reducing the V width for any reason (e.g., when the flange length is short or the inside radius of the product is small), you may lower by one rank (a blank space where the tonnage is not stated). We don't recommend lowering more than one rank because satisfactory precision cannot be obtained with a Vwidth of 16 mm. Because F=34 tons can be read as t=3.2 mm and V=20 mm, the required tonnage will be given as follows: F= 34 x 20/18 x 32/45 x 2.4=65 Thus, the required tonnage is 65 tons.
[Example 1] How many tons are required to bend 4 meters of a 1.5mm thick SUS304 stainless steel sheet (tensile strength: 60 kg/mm²)?
[Solution] The sought Vwidth is calculated as V=6XI.5 . .10 (6xl.5=9, but V=9 mm is not a standard size so V=10 mm is substituted). In this case, t=1.5 mm is not in the bending force chart, so the bending force (F=17 tons) should be read for V=to mm and t = 1. 6 mm (the closest t). The calculating formula, which takes into consideration the sheet thickness, tensile strength, and bending length, will be given as follows: F.17X(1.5/1.6)2X60/45 x 4= 80 The sought solution is 80 tons. In this calculation, raise the fractions to units.
[Example 2] How many tons are necessary to bend a 15 mm thick and 3,100 mm long sheet of rolled SS41 steel? The flange length is 120 mm.
[Solution] The sought Vwidth is V=t2xl5=180. Although a standard Vwidth of 160 mm or 200 mm is acceptable, a Vwidth of 200 mm is not suitable from the viewpoint of length b (flange length). This is because the minimum flange length for a Vwidth Of 200 mm is 14o mm, which is greater than the required 120 mm. Therefore, V is set at 160 mm. Also, t=15 mm is not shown in the chart, so 107 tons is obtained when the F for t=16 mm and V=160 mm is read. Thus, we will have the following formula: F= 107 x (15/16)2 x 3. 1=292 Consequently, the sought answer is 292 tons.
[Example 3] Calculate the required tonnage when bending SPCC of t3.2 mm and =2,400 mm with a V width of 18 mm. 6b is 32 kg/mm².
[Solution] The sought Vwidth is V=8x3.2=25, and V=25 mm is adequate. However, when reducing the V width for any reason (e.g., when the flange length is short or the inside radius of the product is small), you may lower by one rank (a blank space where the tonnage is not stated). We don't recommend lowering more than one rank because satisfactory precision cannot be obtained with a Vwidth of 16 mm. Because F=34 tons can be read as t=3.2 mm and V=20 mm, the required tonnage will be given as follows: F= 34 x 20/18 x 32/45 x 2.4=65 Thus, the required tonnage is 65 tons.
[Exercises] 1. Determine the tonnage required to bend SPCC of t=1.6 mm and =3,000 mm. 2. How many tons are required to bend an anticorrosive aluminium alloy sheet with t=3.0 mm, =2,500 mm, and bb=22 kg/mm² 3. Determine the tonnage required to bend SS4] of t=20 mm and e=600 mm. 4. Calculate the tonnage required to bend soft copper (6b=25 kg/mm²) of t=8 mm and =2,500 mm. 5. Find the tonnage of F when bending SUS304 (t=8 mm and l=3,000 mm) at V=63 mm. (See page 25 for the solutions.) Go to Top
2.5 Complement of the bending force chart So far, we have described in the previous subsections the necessary fundamentals of how to read the bending force chart, the four relationships, and the calculation of bending tonnage. The information already given will enable you to properly use the chart in your job site. The inquires we have received mainly concern the usage of the chart. These inquires seem to suggest the chart is not always effectively utilized. If so, that is indeed an unfortunate situation. The above fundamentals may appear complicated, but each of them can be broken down into simpler factors. They are all made up of correlative simple factors. Just start with the first step, and you will find that it is easy to master the usage of the bending force chart. Now, let us consider a certain calculating formula that is mentioned in almost all books regarding sheet bending. It is the formula,
F=(c x b x l x t²)/(V x 1000)
At a glance, you will find that this formula involves all the factors of the four relationships. The denominator V is inversely proportional to F, and the numerator b, e and t² is proportional to F. The major problem with the above formula is knowing how to determine the value of coefficient C. Usually, the value of C is somewhere between 1.00 and 2.00, varying with V/t. The smaller V/t, the larger value C becomes. When V is 8t, C is said to be 1.33. However it is the value of just one datum. Another datum (by the same researcher) shows a difference of 15 percent in the value C. Such being the case, it causes US to question this calculating formula, and the true value for C varying with V/t. This means the calculated tonnage may turn out to be extremely inaccurate. Our experience reveals the calculation using the tonnage in the chart (as in the exercises) is more practical than the use of calculating formula. This account for the fact that we have calculated the bending tonnage according to the bending force chart , not using the calculating formula.
[Solutions] 1. F=17x32/45x3=37 2. F =24x22/45x2.5=30 3. F=125x(20/19)²x0.6=84 4. F=52x(8/7)²x25/45x2.5=95 5. F=(52(8/7)²x60/45x3=272
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