There are questions often arising in the working process. We are going into some details suggesting solutions to them.
Question: Bending radius is mostly formed to be equal or more than the thicknesses of materials. Yet, the radius might be smaller compared to the thickness. The question is what is the reason for the inner bending radius to be equal or larger than the density of materials. How do the k factors and neutral axles determine this process?
Answer: Most sheets have got the least bending radius designated by the material producers. These metals contain aluminum fusions, that bear listed adviseable minimum inner bending radius depending on the composition and specifications of materials.
Inner bending radii considered as acute bending might bring about fractures throughout outer radii. These strained fractures appear in the form of particles in the metal being ruptured.
Materials get extended along outer bending, contracted towards inner side. The zone which dpes not extends or compress appears to be the neutral axles. Its span does not change, yet its location does not stay the same. It starts moving towards the bending inner side.
Its shifting extent is dependent on the sheet types and formation methods: air formation, bottomong or coining. You can see a factor in Fig.1 and applying it to the sheet thickness, you will identify the position of the shifted neutral axles. The average, usually applied ratio for k-factors is 0.446. This should be multiplied by the sheet thickness to detect the neutral axles shift span tawards inner radii in the bend process. In case the sheet is of 0.062 inches, we get 0.02752 outcome (0.446×0.062) by multiplying by k-factor. We can predict that the nuetral axles moves from 0.031 inches to 0.027652 inches or 0.003348 inches near the inner radii (0,03-0,027652=0,003348).
This seems sufficient to extend bending. That is why the flat parts appear less compared to the entire outer sizes of already shaped pieces. (Fig.2)
If the inner radii is sharp, the compression over the sheet increases upon outer bending. Because of it the likelihood of cracks and breakdowns of parts is great within field. Having equivalent or larger than the sheet density bending radius, the outer side of bending is exposed to relatively less compression. In addition, in case of excessively little and acute bending radii, excessively great tonnage for formation, the punch tip radii bend the centeral part of the bend. It reinforces angle changes occuring due to variations in material features, like thicknesses, grain directions as well as rigidity.
Bending radii closest to the sheet density provides the most proper formation, as well as bending angle consistency, complete sizes. This is the reason to consider it the most stable bending.
All these provide good reasoning to keep the inner bending radii greater than an acute bending and to make use of K-factor ratio, which corresponds to industrial norms and practice.
Question: After reading one of your articles on tonnage matter, I had a question related to this formula.
{[575 × (Sheetl thickness)2]/ Die breadth}/12 = Tonnage per inch.
I would ask you to clarify how 575 emerges and how the demanded angle is taken into account in this formula. Does the tonnage shift in case of bending 85 degrees compared to 125 degrees bending?
Answer: The origins of the formula are not identified. Yet, it goes back to 1950s and has come from building industries. The figure 575 relates to yielding force of soft cold-roll steels. Nowadays there exists a variaty of materials, and the ton formula results are multiplied by the material factors to gain exact outcome.
The mentioned formula is mainly suitable for air-formation of cold-roll steels. In case of bottom forming, tonnage appears 3-5 times higher than that of air formation, in case of coin forming, it appears 10 times more.
Cold-roll steels are our base lines, bearing 60 KSI force. It is adviseable to review the extensile force of different material types to make corresponding adjustments for your task. Just devide the material extensile force by 60 KSI base line (or 60.000 PSI). For instance, in case the extensile force of the material is 120 KSI, you will gain 2.0 factor by deviding120 with 60. In case of mild aluminum of 30 KSI extensile force, the factor will be 0.5. (30/60)
The work tonnage is calculated at some point upon stress strained curving, at which the proportionate range of materials is located. It is followed by a slight growth in loading when it comes to the material yeilding point. This ratio is used to calculate the formation tonnage.
Sure, the tonnage to make 85 degree bending shifts when making a bending of 125 degrees. As soon as you make bending 20 degrees aditional, the better part of the formation tonnage will apear in use.
Actually, nearly 80 % of the entire molding tonnage is reached through the initial 20 degrees of bending angles. Thus, even a small angle, bending might produce enormous pressure upon tools and machine.
Question: We make bending of fusion sheet materials with 0.0393-0.0787 inch thickness bearing radii equivalent to sheet thicknesses. Punches are of 88 degrees, dies are of 60 degrees. It has never been possible to gain a bending of 90 degrees with no compensation. What will be the best combination of angles? Should the die hole be 8 times as great as the sheet thicknesses to be considered ideal?
Answer: The operation you have described involves many dangers, particularly when used accuracy-ground tools. Nevertheless, selsction of a secure tool will provide a solution to the trouble.
In case punching angle appears the greater one, it locates tonnage loading on the upper radius of die, i.e both upper ridges upon each part of the hole. (Fig. 3)
If so, you produce a sufficient number of side impacts that, if not already done, will explode the punch, die or both of them.
In case of applying conventional plane style tools, an explosion may happen, the tool might fall onto your feet. In case of accuracy tooling, be careful, as they bear Rockwell rigidity rate of 60 and even more. If explosed, shrapnels get thrown quite a long distance.
The first thing to do is to make sure that the punch fits the die or it is smaller compared to it. Besides, a gap is needed amongst punches and dies. Just use a die hole, which is 8 times as big as the sheet thickness. Yet, your sheet is of 0.0393 inches and inner radii does not differ. This produces a 1:1 ratio amongst inner radii of bending and the material thicknesses, which is called ideal bending. This is the most steady ratio of radii to materials and their thicknesses.
Note this very formula to achieve an ideal die hole: Outer radii of materials in inch x 3,29435 equals an ideal die hole.
Add the inner radii to sheet thicknesses to identify the outer radii. Then decide on the die hole which appears the closest to your values. Based on that figure estimate the formation tonnage, inner radii as well as bending deductions. Watch out: a wider than required die will cause the loading rise, in some cases severely.
I assume your operation deals with cold-roll steels of 60 KSI extensile force, that shape their radii at nearly 16 % of a die hole.
For 0.0393 inch thick materials possessing 0.0393 inch inner bend radius, the outer bending radius will be 0.0786 inch. One more time it appears to be the inner inner radius + sheet thicknesses. Insert the outer radii in the formula to gain the proper die hole with 0.2589 inch. Here air form is recommended with more than 0.275 inch die hole possessing die angles of 90, 88, 85 degree. In air forming it would lead to producing inner radii of 0.04 inch in a work piece. (0.04 is 16% of 0,275, i.e 0.075×0.16=0.04) Use this very figure to estimate the bending deductions. Suppose you make an air form, then the tonnage will be 0.178 US tons for each inch or 2.133 US tons for each foot.
As to your punch, 1 mm-nose radii is adviseable as well as an angle two degrees smaller compared to the die angle chosen by you. This is going to reduce the form tonnage allowing to gain a bending of 90 degrees easily.
For materials of 0.0787 ich thickness, the same tips for tooling angles and punch nose radii are applicable. In this case die breadth of 0.472 inch si preferable, as it would provide about 0.285 US tons for each inch or 3.416 US tons for each foot in case of an air form. You will gain inner radii of 0.075 inch (one more time 16% of the die hole), that is later used to estimate bending deductions.