factor k development sheet.

[Cristian75]

Thanks cekkos,
I agree, but my problem is precisely the iter of use with sw, I mean, assuming that I have best unified the factor k... should I insert it while I create the sheet after inserting sp., radius, select factor k and insert it manually or does there exist any other method? ?
because even re_solidworks recommended me more or less the same thing, even unifying with only factor k and would be great.
thanks so much to all

Christ,

as you will certainly read the topic is very "hot"; If you're looking for the magic solution, don't exist.

do you have to do with different thickness / bending ratios? you will have different k factors.
use the first quarry you find to make the folds and/or every time is always different? in this case the factor is kaos, not k.

practically the only advice I give you is: try to uniform everything, and leave the tables alone. If you can get down to 3-4 sp/rint reports you will be appropriate. (by experimentation on real folds)

 

[re_solidworks]

Thanks to everyone, I know the subject is complex and customizable. . .
re_solidworks, I did a series of tests in production and I have a table with all 90° fold retreats and with various combinations of quarries.
You say there is a solution to find a unique k factor that can give me your results? How do I find the K factor with the information I already have? If you can tell what factor k you use?

the thing is easier to do than to say. fold a known size l, e.g. 100x100. drawings the same l in swx, with the correct radius and initially imposed the factor k to 0.35 (usually for human radius/thickness relationships you are around).
at this point you check the development that returns you swx, if it is too long you decrease the fatt. k, if it is too short increases the fatt. k.
When you find the correct value, you're fine. when I do this "research" I give the withdrawal from the bending machine (e.g. 0.3 mm per fold) and I carry out a piece with some folds (even 10) to eliminate even any decimals. I demand that the design be perfect, so I go to split the hair; In this way I know that if there are differences they should be sought elsewhere and are not a sum of design error + production.
the identified k factor will be valid for bending on a certain machine, with the combination of quarry and throat chosen and for a certain material. to every parameter that you change it must be reverified.

You remember that I asked you about the choice between inventor and sw... and you advised me sw... well that's exactly what I installed a week ago.
Thanks again

Well, I hope the data advice has been helpful to choose the right tool.

 

[Cristian75]

the thing is easier to do than to say. fold a known size l, e.g. 100x100. drawings the same l in swx, with the correct radius and initially imposed the factor k to 0.35 (usually for human radius/thickness relationships you are around).
at this point you check the development that returns you swx, if it is too long you decrease the fatt. k, if it is too short increases the fatt. k.
When you find the correct value, you're fine. when I do this "research" I give the withdrawal from the bending machine (e.g. 0.3 mm per fold) and I carry out a piece with some folds (even 10) to eliminate even any decimals. I demand that the design be perfect, so I go to split the hair; In this way I know that if there are differences they should be sought elsewhere and are not a sum of design error + production.
the identified k factor will be valid for bending on a certain machine, with the combination of quarry and throat chosen and for a certain material. to every parameter that you change it must be reverified.



Well, I hope the data advice has been helpful to choose the right tool.

Good evening, re_solidworks
I thank you so much for your precious advice, I followed your advice for the fk and I have to say that everything is working.
Only problem is to make it operational through a table (solidworks technicians recommended me t.gauge) but it doesn't work properly.
I better explain, the need to have a table is for the different internal rays that I find from the step files of the various customers, hoping not to have to modify them I created the table with every thickness, the various internal rays and for each combination the fk.
but unfortunately I do not know that fk goes to read, sure is not the one inserted in the table... result: depending on the sp. and the r.int. that I am facing do not select any table but simply the fk designated manually.
Do you think that's the only possible method? ?
Thanks again
Good evening
Christian

 

[re_solidworks]

Good evening, re_solidworks
I thank you so much for your precious advice, I followed your advice for the fk and I have to say that everything is working.
Only problem is to make it operational through a table (solidworks technicians recommended me t.gauge) but it doesn't work properly.
I better explain, the need to have a table is for the different internal rays that I find from the step files of the various customers, hoping not to have to modify them I created the table with every thickness, the various internal rays and for each combination the fk.
but unfortunately I do not know that fk goes to read, sure is not the one inserted in the table... result: depending on the sp. and the r.int. that I am facing do not select any table but simply the fk designated manually.
Do you think that's the only possible method? ?
Thanks again
Good evening
Christian

What version do you use? Is it 2013?

 

[Cristian75]

What version do you use? Is it 2013?

exact 2013 installed by 10 days

 

[andle11]

Good evening, re_solidworks
I thank you so much for your precious advice, I followed your advice for the fk and I have to say that everything is working.
Only problem is to make it operational through a table (solidworks technicians recommended me t.gauge) but it doesn't work properly.
I better explain, the need to have a table is for the different internal rays that I find from the step files of the various customers, hoping not to have to modify them I created the table with every thickness, the various internal rays and for each combination the fk.
but unfortunately I do not know that fk goes to read, sure is not the one inserted in the table... result: depending on the sp. and the r.int. that I am facing do not select any table but simply the fk designated manually.
Do you think that's the only possible method? ?
Thanks again
Good evening
Christian

time ago I got the ball to drive factor k with a table.
My needs are the same as yours, i.e. I get steps from customers and I have to develop sheet metal parts possibly by minimally intervening on the original model.
That said, I developed on the basis of the gauge table model of the installation of sw a mine table.
taken as primary factors the size of the matrix and the material to fold. as a practical example I created various tables:
a table "fe v12" for iron bent in 12 quarry (with in thicknesses and rays, relative retreats and k factor calculated by excel)
a table "fe v16" for iron bent in 16 quarry (with in thicknesses and rays, relative retreats and factor k calculated by excel)
a table "aisi v10" for bent in 10 quarry (with in thicknesses and rays, relative retreats and k factor calculated by excel)
... and away.. .

the values of the retreats took them from a well-made table that mike1967 gave to the community of this forum. I think back in this post is attached. otherwise look in the forum. x mike: thank you, if you go to my side I'll pay you a dinner!

I attach a table as an example. save it in the path of the gauge table files.
important! : If you miss a radius for a thickness, you must add it to every thickness of the table. you can do it simply by copying the column in excel.
View attachment FE V12.zip

 

[Cristian75]

Good morning andle11,
First of all, thank you 1000 for your answer, the fact that you have my own needs is rewarding, I thought I was pulling unnecessary mental saws. . .
Can it be that my withdrawal table doesn't work properly because I didn't insert the same rays for each thickness? ? ?
As soon as I can tie you up my retirement table, I had some trouble attaching it
Thanks again

 

[andle11]

Good morning andle11,
First of all, thank you 1000 for your answer, the fact that you have my own needs is rewarding, I thought I was pulling unnecessary mental saws. . .
Can it be that my withdrawal table doesn't work properly because I didn't insert the same rays for each thickness? ? ?
As soon as I can tie you up my retirement table, I had some trouble attaching it
Thanks again

to attach .xls files you have to zip, because it is not a allowed extension.
for the rest, if you read the online guide there is written how to do:http://help.solidworks.com/2013/italian/solidworks/sldworks/c_sheet_metal_gauge_tables.htm?format=pIf you attach your table I see if I understand where the problem is.
I see only now on the guide that it is enough that the rays are in increasing order. . .

 

[Cristian75]

Sorry, I didn't read the last line of the help.
I confirm, though, that if I am working on a bend angle not present in the table, the withdrawal is not handled and uses a default k factor?? I have to insert all angles like 30,8° and a similar variety that comes from step?
Thanks again, you're very kind.
table:

 

[andle11]

Sorry, I didn't read the last line of the help.
I confirm, though, that if I am working on a bend angle not present in the table, the withdrawal is not handled and uses a default k factor?? I have to insert all angles like 30,8° and a similar variety that comes from step?
Thanks again, you're very kind.
table:

I don't forget. in the sense that even working with decimal corners I did not encounter problems. I verified that development was equal when I imposed the k factor manually. but with open corners above 165° come out mess!!! I can't understand and overcome the problem. see below the image.
:4404:

boh... it was something I wanted to define later and then I would share it with my colleagues. . .
I think we're gonna move on by setting the K factor.

I did the tests with sw 2013 sp0 and excel 2013.

Your table seems to be fine at first glance. but also here with the .xlsx extension instead of the old .xls some problem I had.

Bye!

 

[andle11]

solved! !
was just a formatting problem in excel.
since I usually use the same k factor regardless of the fold angle, I put as corner values only the 0° and 180°. each fold angle always takes the desired k factor and calculated according to pickup, thickness and internal radius drawn.
withdrawal values are editable according to your needs and bending tests you perform.
in the file there is the formula: "k= (2*ritiro+4*r.int-r.int*pi.greco)/ (spessore*pi.greco)"
I attach part and 2 tables, one for the iron and another for the inox, both for plates of thicknesses workable in quarry from 12.
n.b.: 12 quarry is optimal for thickness 1.5 (width = 8 times thickness).

 

[Cristian75]

solved! !
was just a formatting problem in excel.
since I usually use the same k factor regardless of the fold angle, I put as corner values only the 0° and 180°. each fold angle always takes the desired k factor and calculated according to pickup, thickness and internal radius drawn.
withdrawal values are editable according to your needs and bending tests you perform.
in the file there is the formula: "k= (2*ritiro+4*r.int-r.int*pi.greco)/ (spessore*pi.greco)"
I attach part and 2 tables, one for the iron and another for the inox, both for plates of thicknesses workable in quarry from 12.
n.b.: 12 quarry is optimal for thickness 1.5 (width = 8 times thickness).

You're great! ! !
therefore, as we have the same way of work, if I have not understood badly enough to insert 0°/180° with the established fk and all the range of angle included is tracted with the same fk? ?
for my table, should I set for each sp the same rays, save it with xls format and everything should work? ?
Thank you very much.
Good lunch

 

[-Mc-]

this topic is particularly articulated and complicated as there are too many things (type of material and lotto, thickness, radius, quarry, knife...) to be identified in order to be able to work quickly and operationally by adopting every time (depending on the bend angle) a correct factor k;
it is years that I follow the topic both on the forum and through the various "official assistance" sw and no one has ever given a concrete and definitive answer;
for that reason, I am now opting for the "neutral" solution i.e. not to increase the sheet development, but keeping the fold size equal to the internal bending quota (inserting a note on the design where I go to indicate that the development is theoretical and who builds the piece, must adapt the development to the bending/machine/material method); I believe that in this field, only the experience is a mistress.
to you technicians I ask:
- what factor k and internal radius I have to set (depending on the thickness) to have a theoretical development equal to what would come out with the deduction of fold (i.e. you bring me the measurements of the piego interior of the finished piece?

Thank you.

mc
now I ask you

 

[Mike1967]

this topic is particularly articulated and complicated as there are too many things (type of material and lotto, thickness, radius, quarry, knife...) to be identified in order to be able to work quickly and operationally by adopting every time (depending on the bend angle) a correct factor k;
it is years that I follow the topic both on the forum and through the various "official assistance" sw and no one has ever given a concrete and definitive answer;
for that reason, I am now opting for the "neutral" solution i.e. not to increase the sheet development, but keeping the fold size equal to the internal bending quota (inserting a note on the design where I go to indicate that the development is theoretical and who builds the piece, must adapt the development to the bending/machine/material method); I believe that in this field, only the experience is a mistress.
to you technicians I ask:
- what factor k and internal radius I have to set (depending on the thickness) to have a theoretical development equal to what would come out with the deduction of fold (i.e. you bring me the measurements of the piego interior of the finished piece?

Thank you.

mc
now I ask you

your question needs an answer "particularly articulated and complicated as there are too many things to identify.
the question is simple in itself, but the answer you are looking for cannot be so simplified.
Many of us have come to produce "to measure" pieces by testing various factors, such as bend angle radius and so I can't give you any lyophilized recipe to explain the functioning of the sheet.
I wrote, personally, dozens of posts about it and if you "invest" some of your free time to learn something that
someone else made available by investing their time (free) you would benefit from it.
if you had read some posts you would have realized that what you ask is already published, I made available
excell tables with bending radius values, hollow width, materials and thicknesses etc... etc.
Good research.

 

[-Mc-]

I wrote, personally, dozens of posts about it and if you "invest" some of your free time to learn something that
someone else made available by investing their time (free) you would benefit from it.
if you had read some posts you would have realized that what you ask is already published, I made available
excell tables with bending radius values, hollow width, materials and thicknesses etc... etc.

good: that's why I took a August 2013 post because I felt the most suitable to write.
I am sorry to contradict you but now the posts you refer to and especially the various tables (some very doubtful) I have already downloaded and tried them.
to me it is interested to understand how the solidworks system reasons in order to get to eliminate the increase/reduction of sheet development; I am interested in zeroing everything as it used to do with autocad and with tables posted until now, especially on high thicknesses, I never got this result.
If you want I can post all the test parts I used.
I now look for your (which I don't remember) and then I will show you that, on my details (with many folds) it doesn't work.
I was not looking for a polemical answer, I also know that through auditions you get to the result, but knowing the forum I thought it was normal to post a question that no one had yet written.
you probably misinterpreted and gave the classic answer that, in a forum where technicians are looking for answers to work, unnecessarily occupies the web page...
I would ask a technical, not to a designer, an intervention, as I would like to understand, for my culture, as it reasons the system according to the elements we insert (which at the end are not many: thickness, radius, k).

 

[Mike1967]

I am sorry to contradict you but now the posts you refer to and especially the various tables (some very doubtful) I have already downloaded and tried them.
to me it is interested to understand how the solidworks system reasons in order to get to eliminate the increase/reduction of sheet development; I am interested in zeroing everything as it used to do with autocad and with tables posted until now, especially on high thicknesses, I never got this result.
If you want I can post all the test parts I used.
I now look for your (which I don't remember) and then I will show you that, on my details (with many folds) it doesn't work.
I was not looking for a polemical answer, I also know that through auditions you get to the result, but knowing the forum I thought it was normal to post a question that no one had yet written.
You probably misinterpreted.
I asked a technician, not a designer an intervention, as I would like to understand, for my culture, how he reasons the system according to the elements we insert (which in the end are not many: thickness, radius, k).

I'm sorry if I tell you that I'm not a tenacious... and for the truth not even a designer... .

I work with those tables and in 14 years of solidworks all I design, put on the table and step laser cutting
It works.
you have not read the discussion well, those parameters are those that I have extrapolated by making auditions (true clasp bent and measured with the caliber) for every thickness and type of material we use.
Pass me some files so that I compare it to developments that come out with my system.
I'll test you.
someone posted a technical scheme of how to calculate development mathematically. . Here, try, run
and then tell me if it actually worked. . .
Last thing, what base do you say that developments with my table don't match?
Do you get my table? (It is not ironic only that my colleagues and I know what a field or another means)
Maybe you didn't set the parameters as I considered them.
Maybe the used molds do not match those in my table. . .
if you think of providing the dxf file to the xyz firm and demand the pieces bent to the tenth... Well, forget it.
I give you an example of how the table works:
hypotheses, stainless steel sp 1.5 mm:
if the project with a bending radius of 0.001 mm (the fictitious value of which you have certainly read) imposed a factor k of 0.001.
if instead I want to draw it in the most likely way I will impose a 1.5 mm fold radius (usually the thickness goes well for the fold radius, at least it approaches a lot) with a fk di 0.273.
you will notice that the withdrawal value is 0 (zero) either with 0.001 radius or with 1.5 mm radius.if you want to delete stretches or retreats you simply set turn radius = to thickness and fk 0.273.
this way you will have developments at zero withdrawal/extension.I conclude here.
Hi.


ps: doubtful... .

ps2: the table is incomplete, it was being completed when I saved it in the home computer.

 

[Mike1967]

good: that's why I took a August 2013 post because I felt the most suitable to write.
I am sorry to contradict you but now the posts you refer to and especially the various tables (some very doubtful) I have already downloaded and tried them.
to me it is interested to understand how the solidworks system reasons in order to get to eliminate the increase/reduction of sheet development; I am interested in zeroing everything as it used to do with autocad and with tables posted until now, especially on high thicknesses, I never got this result.
If you want I can post all the test parts I used.
I now look for your (which I don't remember) and then I will show you that, on my details (with many folds) it doesn't work.
I was not looking for a polemical answer, I also know that through auditions you get to the result, but knowing the forum I thought it was normal to post a question that no one had yet written.
you probably misinterpreted and gave the classic answer that, in a forum where technicians are looking for answers to work, unnecessarily occupies the web page...
I would ask a technical, not to a designer, an intervention, as I would like to understand, for my culture, as it reasons the system according to the elements we insert (which at the end are not many: thickness, radius, k).

here you find the theory:
http://help.solidworks.com/2014/italian/solidworks/sldworks/c_bend_allowance_and_bend_deduction.htm

 

[-Mc-]

If instead I want to draw it in the most likely way I will impose a fold radius of 1.5 mm (usually the thickness goes well for the fold radius, at least it approaches a lot) with a fk of 0.273.
you will notice that the withdrawal value is 0 (zero) either with 0.001 radius or with 1.5 mm radius.
if you want to delete stretches or retreats you simply set turn radius = to thickness and fk 0.273.

hi mike;
thanks to the explanation and I perfectly agree with you;
the problem of your solution with k 0.273 with rpiega=thickness was the solution of the solidworks insurance that I am still using;
the problem is that it does not coincide with an increase in development equal to zero: on the last fold, in fact, increasing the thickness you will always have some tenth/millimeter extra: for now I am putting together with a cut the final development, but it does not seem to me a correct solution;
This is because I use often very high and I cannot afford to have too many tenths/mm more than theoretical development.
for this reason I asked for the help of a technician: I read these discussions and read them on the forum, but my question remains different.

do it
and then tell me if it actually worked. . .

You don't understand. read the discussion from the beginning.

I'll test you.
someone posted a technical scheme of how to calculate development mathematically. . Here, try, run

I tried all the tables. does not work on high thicknesses.

Maybe the used molds do not match those in my table. . .

You don't understand. read the discussion from the beginning:
I need a neutral k factor: means that development must be identical, equal, equal, superimposed to the theoretical one of sum of internal folds. machines, moulds etc... change by force of things having 5 production plants with completely different machinery...
I also conclude because I think you have not yet understood what I need and I always hope in the answer of a technician sw...
Thank you.
-Mc.

 

[Ste80vi]

Hi, mc.
I did not long ago bother to try to see - starting from an experimentally detected table containing the various shortening values - as varies the k factor at the varying angle (holding obviously all other parameters). had, the k factor does not remain constant, but tends to decrease from a substantial 0.5 to values close to 0.3 (approximately, I go to memory) depending on the decrease of the inner bending angle. so also the strategy to try to use a unique k factor that goes well for everything - that to me seems an absolute obscenity if I consider my bending experience - it does not work for nothing.
not knowing the reason why you want to have a development that is identical, equal, and superimposable to the theoretical one of the sum of the internal folds, you could try a folding table (also that standard of solidworks) using however not the system of bending deduction but rather that of the bending tolerance limit. According to this system, in fact, the development of the radius instead of being removed, is added. You can configure a folding table (in xls or btl format) with values all 0 for any internal radius of fold, any angle and any thickness. in this way to calculate the solidworks development will only submerge the naked and raw length of the sides, adding zero for each fold.
I hope I haven't misunderstood your intent and helped you to help.
If not, let me know,
Good work for everyone.

 

[-Mc-]

ciao ste80!

not knowing why you want to have a development that is identical, equal, and superimposable to the theoretical one of sum of the internal folds

because, having more suppliers (external and internal) it is better to provide theoretical development;
who builds the piece must adapt the development to the bending method, not the technical office; the technical office has the responsibility to design correct designs; providing development has become an additional task seen in recent years also given the simplicity in achieving it through 3d; the production has the task to adapt the sheet development to its bending method/machine/cave/coltelli/type of material.
For this reason I want to provide a neutral development, otherwise, as a technical office, powers incur in errors that would make documents questionable.
I as a technical office provide an excellent base (given the neutral development) that the lamierist will have to customize (which is very simple for him who knows all the production process);
a demonstration of this, mike gave us confirmation: for him quarries, knives, material, machinery, bending waste etc.. They are daily bread.
Therefore a practical person like him will take us much less time than humanity to make the optimal development to obtain the measures of the finite.
the assembly serves the almost perfect finish: the sheet development is a productive passage to reach the final goal.
I think that this concept is often neglected and that production, in many companies, takes advantage of its designers by downloading this task: the perfect development to design.. .

system of bending deduction but rather that of bending tolerance limit

the bending deduction works perfectly and would give the result to which I ambisk adopting a minimum bending radius (0,001), but this is not realistic and lacking to the design the real ingombs of the stools; In addition to this, with the ddp many sw functions you cannot use (see inclined folds);
idem for bending tolerance. . Unfortunately some functions (for what I have seen until today) are enabled only through use of the k factor.

I hope I haven't misunderstood your intent and helped you to help.

no, you have no misunderstood; But it bothers me who, reading in a hurry without understanding anything, throws answers at home, doing the expert in turn.
I have been running a U.T. for a long time and I follow the forum a lot; before answering I like to read I understand what other colleagues write; you interpreted perfectly.
soon and thank you.
-Mc.