Question

5. A spool of Cu wire has a starting diam. Of 2.5 mm. It is drawn through a die with an opening of 2.1 mm. Coefficient of friction at the work-die interface is 0.08. The pure Cu has a strength coefficient of 300 MPa & a strain hardening exponent of 0.5. The operation is done at room temperature. Determine: (A) area reduction, (B) draw stress & C) draw force required for the operation. Assume die angle to be 100.

Answer 1

(a) Area Reduction(r) r = A_0 - A_f/A_0 = 0.25(pi)(2.5)^2 - 0.25(pi)(2.1)62/0.25 pi(2.5)^2 = 4.91 - 3.46/4.91 r = 0.294 b) Draw stress (sigma) E = ln(A_0/A_E) = ln(4.91/3.46) = 0.349 sigma^derivative_0 = 300 (E)^0.5 = 300(0.349)^0. = 177.22 Mpa Given strength coefficient rightarrow 30 Mpa strain hardening exponent rightarrow 0.5 phi = 0.88 + 0.12(D/L_c) D = 2.5 + 2.1/2 = 2.3 L_c = 2.5 - 2.1/2_(sin 18) = 0.647 (Assuming entrance angle as 18 degree) phi = 0.88 + 0.12(2.3/0.647) = 1.31 sigma = sigma^derivative_0(1 + mu/tan alpha)phi ln(A_0/Af) = 177.22(1 + 0.08/tan 18)(1.31)ln(4.91/3.46) = 100.97 Mpa. C) Draw Force (F) F = af sigma = 3.46(100.97) = 349.35 N

Answer 2

·             A) r=()=()=0,293

B) =ln= =0,348
Yf===271Mpa
D===2,3
Lc===0,647mm
=0,88+0,12()=1,3
d=Yf(1+()ln = d=271(1+()1,3*ln=153 MPa
C) f=Af* d=3,46*153=528,64N

Answer 3

·             A) r=()=()=0,293

B) =ln= =0,348
Yf===271Mpa
D===2,3
Lc===0,647mm
=0,88+0,12()=1,3
d=Yf(1+()ln = d=271(1+()1,3*ln=153 MPa
C) f=Af* d=3,46*153=528,64N