Reply To: Doubts in Manufacturing & Industrial Engineering :(

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  1. @Dinesh: Ok I wants to say that I am not too good in this Forming section but still I will try to solve this; Lets see how much can I help you !!!
    Say, Original c/s Area = O.A.
    Final c/s Area = F.A.
    Area After First Die = A1
    Original Length = O.L.
    Final Length = F.L.
    Now O.A. = (pi/4)*(10^2) = 78.5398 mm2
    A1 = (1-0.35)*(O.A.) = 0.65*(O.A.) = 51.0508 mm2
    Similarly for the area after 7th die
    = 0.65*0.65*0.65*0.65*0.65*0.65*0.65*(O.A.) … .. (For 7 dies)
    = (0.65)^7 * (O.A.)
    = 3.8502 mm2 = F.A.
    Now, as True strain is instantaneous % of change in length of specimen & it is given by,
    True strain (TS) = Natural log of (O.A./F.A.)
    = ln (78.5398/3.8502)
    = 3.0155 ———————————–ANS
    Final length can be calculated by using constant volume principle as,
    Initial Volume = Final Volume
    (O.A.)*(O.L.) = (F.A.)*(F.L.)
    78.5398 * 100 = 3.8502 * F.L.
    F.L. = 2039.8888
    = 2040 mm ————————————-ANS
    Now, For force we have to calculate a Draw Stress after first die,
    using uniform energy deformation method,
    Draw Stress (D.S.) = Y * ln(O.A./A1) …… … (Y = Yield Strength)
    (Note this formula can be used for only Ideal Conditions i.e. neglecting friction and redundant work)
    Hence, D.S. = 200 * ln (78.5398/51.0508)
    = 86.1568 N/mm2
    Now, Drawing Force through first die is,
    Drawing Force (F) = (D.S.)*(A1)
    = 86.1568 * 51.0508
    = 4398.3735 N
    = 4.4 kN ———————————ANS

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