Power required and die pressure in rod drawing

A round rod of annealed brass 70-30 is being drawn from a diameter of 6 mm to 3 mm at a speed of 0.6 m/s. Assume that the frictional and redundant work together constitute 35% of the ideal work of deformation.(a) Calculate the power required in this operation, and (b) calculate the die pressure at the exit of the die.

Question

Power required and die pressure in rod drawing

A round rod of annealed brass 70-30 is being drawn from a diameter of 6 mm to 3 mm at a speed of 0.6 m/s. Assume that the frictional and redundant work together constitute 35% of the ideal work of deformation.(a) Calculate the power required in this operation, and (b) calculate the die pressure at the exit of the die.

Accepted Answer

The true strain that the material undergoes in this operation is

[latex]\epsilon _{1}=\ln \left\lgroup\frac{6^2}{3^2} ight group=0.6931 [/latex]

From Table 2.3,

Typical Values for K and n in Eq. (2.11)([latex]\sigma =K\epsilon ^n[/latex] )at Room Temperature
Material	K (MPa)	n
Aluminum, 1100-O	180	0.20
2024-T4	690	0.16
5052-O	210	0.13
6061-O	205	0.20
6061-T6	410	0.05
7075-O	400	0.17
Brass, 7030, annealed	895	0.49
85–15, cold rolled	580	0.34
Bronze (phosphor), annealed	720	0.46
Cobalt-base alloy, heat treated	2070	0.50
Copper, annealed	315	0.54
Molybdenum, annealed	725	0.13
Steel, low carbon, annealed	530	0.26
1045 hot rolled	965	0.14
1112 annealed	760	0.19
1112 cold rolled	760	0.08
4135 annealed	1015	0.17
4135 cold rolled	1100	0.14
4340 annealed	640	0.15
17–4 P-H, annealed	1200	0.05
52100, annealed	1450	0.07
304 stainless, annealed	1275	0.45
410 stainless, annealed	960	0.10

we find that for this material and condition, K = 895 MPa and [latex]n =0.49.[/latex] Hence,

[latex]\overline{Y}=\frac{K\epsilon ^{n}_{1} }{n+1} =\frac{(895 )(0. 6931)^{0.49}}{1.49} =502 \ MPa [/latex]

From Eq. (6.64), the drawing force is

[latex]F=\overline{Y}A_{f}\ln \left\lgroup \frac{A_{0}}{A_{f}} ight group, [/latex]

where

[latex]F=(502)\left[\frac{\pi }{4} \left\lgroup\frac{3}{1000} ight group^2 imes 0 .6931 ight] [/latex]

Hence,

F = [latex]0.002459[/latex] MN

and

Power= [latex] (F)(V_{f}) =(0.002459)(0.6)[/latex]

[latex]= 0.001475[/latex] MN m/s

[latex]=0.001475 [/latex] MW = [latex]1475 [/latex] W

and the actual power will be 35% higher, or

Actual power = [latex](1.35)(1475)=1.992[/latex]kW

The die pressure can be obtained by noting from Eq. (6.70) that

[latex]p =Y_{f} \ - \ \sigma[/latex] ,

where [latex] Y_{f} [/latex] represents the flow stress of the material at the exit of the die. Thus,

[latex]Y_{f}= K \epsilon^{n}_{1} =(895)(0.6931)^{0.49}=748 [/latex] MPa

In this equation, [latex] \sigma [/latex] is the drawing stress, [latex] \sigma_{d} [/latex]. Hence, using the actual force, we have

[latex]\sigma _{d}=\frac{F}{A_{f}} =\frac{(1 .35)(0. 002459) imes 1000^2 imes 4}{\pi imes 3^2} =470[/latex] MPa

Therefore, the die pressure at the exit is

[latex] p=748 – 470= 278[/latex] MPa.

Question 6.7: Power required and die pressure in rod drawing A round rod o...

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