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    Spring steel loose laminate prod questions

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    JoaoLS
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    Spring steel loose laminate prod questions

    Post by JoaoLS on Mon Apr 20, 2015 4:36 am

    Hi guys!

    I've been making experiments with loose laminates for a while, mainly with old saw blades, and the results seem promising. With <1mm thick backsaw blades (300mmx3mm) i managed to get a neat little 22# @ 185mm (7-1/4') prod for a pistol bow. The low weight is sort of balanced by the relatively long power stroke of 5-1/4', and it is a sweet shooting toy. Link to that here:http://thearbalistguild.forumotion.com/t1365p15-small-pistol-crossbow#13133

    So now i'm thinking on doing a more serious version of the loose laminate prod, for a medium sized bow i'm planning. I've got access to 2mm and 3mm thick spring steel strip (C67 S), the thing is I'm not sure what kind of draw weight to expect from a single strip, around 650mm ntn (25').  


    I'm aiming for around 100#, so my question is, anyone got any idea how much power one can squeeze out a 3mm or 2mm thick strip of spring steel (50mm wide for the 2mm and 30mm wide for the 3mm)?


    Any advice, on that or on the loose laminate general topic, is most welcome!


    Joao
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    hullutiedemies
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    Re: Spring steel loose laminate prod questions

    Post by hullutiedemies on Mon Apr 20, 2015 8:16 am

    1.4 Joules per cubic cm without problems. 2J/cm3 with problems.

    A single 30cm piece of 50x2mm strip will store over 40 Joules. (if evenly bent )

    If you assemble 30cm bow same way as your pistol, 100 pounds might be around 14 cm power stroke, About .8 kN (180#) at 19cm is about maximum you should get away with it. -Give or take. Exact numbers depend from exact limb&string geometry.
    Longer bow makes lighter and longer draw.
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    JoaoLS
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    Re: Spring steel loose laminate prod questions

    Post by JoaoLS on Mon Apr 20, 2015 4:19 pm

    hullutiedemies thanks for the input!



    By "same way as your pistol" you mean 6 strips of 2mm, staggered, right?  



    I was hoping to go for a longer bow, maybe 60cm, to fit a medium sized stock. What do you reckon it'll take to get to around 100#, and what kind of draw length could i expect to safely draw?


    Would you recommend some kind of width tapering on the longest lamination, from the ends of the second to the nocks? I'm guessing the answer will have to do with the distance from the 2nd's end to the nock, but i'm not really sure.  scratch


    If i'm thinking correctly, if i know the draw weight of one 60cm piece, of either the 2x50mm or 3x30mm, i could calculate the number of laminations i'd need to get to any desired weight, right? Tho...how do i factor in the decreasing length of the laminations? Is it linear like 60cm+30cm lams would be (1x#)+(.5x#)? 


    Sorry for the battering of questions, but you seem to know your physics and i'm an avid newbie  study

    Andy.
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    Re: Spring steel loose laminate prod questions

    Post by Andy. on Tue Apr 21, 2015 12:38 am

    Hey JoaoLS, will watch your build with interest!

    Where are you accessing  2x 50mm and 3x 30mm spring steel strip from??

    Sound like very convenient stock sizes for prod builds.
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    JoaoLS
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    Re: Spring steel loose laminate prod questions

    Post by JoaoLS on Tue Apr 21, 2015 2:53 am

    Hey Andy! 

    I found it in a store called Açometais Lda. in Porto, Portugal. Initially though I tried to contact BSS Steel Strip http://www.steelstrip.co.uk/BSS/ , but got no answer, so i rummaged a bit around manufacturers in Portugal and came up with this one http://www.acometais.pt/index.php?option=com_phocagallery&view=category&id=1&Itemid=68

    Their site is a bit crappy, but they know their steel and sell spring steel by the meter, although i don't think they ship small quantities anywhere.  Sad You can always mail and ask, they seem to be very nice people and i got a reply in a couple of hours.

    Btw, the two strips, 3x30x2000mm and 2x50x2500mm cost me about 28€, roughly 28$ nowadays.
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    JoaoLS
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    Re: Spring steel loose laminate prod questions

    Post by JoaoLS on Sun May 03, 2015 3:50 pm

    So...


    I'm in no way an expert on physics, so if anyone notices anything wrong in the following post, please please correct me, I don't want to pass on wrong information.


    I have found a couple of interesting formulae regarding leaf springs.

    As it turns out, knowing your material's Young modulus and dimensions, you can calculate the spring tip deflection (and thus draw length) for a given force applied to the spring, like this:

    s = (4 x F x L3)/(E x n x b x t3) x K


    Where:




    S - tip deflection - mm
    F - force applied to the tip of the spring - Newtons
    L - working length of spring - mm
    E - Young's Modulus - MPa (megapascal)
    n - number of leaves in the spring
    b - width of the leaves - mm
    t - thickness of the leaves - mm
    K - form coefficient




    We'll deal with form coefficient in a minute.




    Also, i found this formula applies to a spring that has a fixed point ( the thickest point) and the force is applied at the tip, so for calculating a prod bend, you must assume there are 2 springs (one for each limb) that just happened to be connected at the middle. 




    This translates into L being your (ntn)/ 2. 


    For example, if you want to know how much tip deflection you get from applying 100kg of pull to a 70 cm ntn prod, composed of 4 evenly spaced leaves of 3x30mm steel, you get:




    100kg = 980.6Newtons




    Divide by two to calculate force on each tip:




    980N / 2 = 490N




    Determine your L:




    70cm / 2 = 35cm = 350mm




    Look up Y. modulus of spring material in question:




    E spring steel = 210000 MPa




    Now apply values to the formula:




    s = (4 x 490 x 3503) / (210000 x 4 x 30 x 33x K






    s = (84,035,000,000) / (680,400,000) x K




    s = 123.5 x K




    Now K has a simple formula that translates how the number of leaves affects bending:




    K = 3 / 2 + ((n' + 1) / n)




    Where;
    n' - is the number of additional full length leaves (in this case, 0)
    n - is the total number of leaves (in this case, 4)
     
    So,

    K = 3 / 2 + ((0 + 1) / 4)

    K = 1.33


    Now we can know how much tip deflection we have:


    s = 123.5 x 1.33


    s = 164.25mm




    Alright, now we got tip deflection for our spring, how do we calculate draw length? Easiest part.

    You either got fond of maths again all of a sudden and you go into a trigonometric rampage, or you make a model. I used a CAD program.




    Draw a 700mm line.

    Offset that line by 50mm and 164.25mm. 

    "Bend" the first line into an arc so the tips touch the second (50mm) line. 

    Make sure arc length hasn't stretched and remains 700mm. 

    Now connect both tips with another line. That's your string at 50mm (around 2") brace.   

    Bend the arc a bit more so tips touch 164.25mm line. Check again for arc length and tinker with it until it's exactly 700mm

    Now draw 2 circles that have a diameter equal to your string length, centered on each tip. 


    Measure from the furthest point where the 2 circles meet to the center point of your bent 700mm line.




    There's your draw length.




    In this case, we have around 345mm of draw length from a 50mm brace height. This means 295mm (11.6") power stroke at around 220#.




    There is now the question of the material limitations. Can this steel take this kind of bend at these dimensions? My calculations say yes, but i'm tired of typing, tomorrow i'll post how i found this out.




    Any input or correction is welcome!




    study  Phew...
    By the way, i have my steel now, and things are progressing well, so the build will be coming soon, lest we grow tired of all the numbers ;P
    Joao

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    Re: Spring steel loose laminate prod questions

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