Comparison of forming and machining energies

You are given two pieces of annealed 304 stainless-steel rods, each with a diameter of 10 mm and a length of 125 mm, and you are asked to reduce their diameters to 0.5 mm (a) for one piece by pulling it in tension and (b) for the other by machining it on a lathe (see Fig. 8.8) in one pass. Calculate the respective amounts of work involved, and explain the reasons for the difference in the energies dissipated.

Question

Comparison of forming and machining energies

You are given two pieces of annealed 304 stainless-steel rods, each with a diameter of 10 mm and a length of 125 mm, and you are asked to reduce their diameters to 0.5 mm (a) for one piece by pulling it in tension and (b) for the other by machining it on a lathe (see Fig. 8.8) in one pass. Calculate the respective amounts of work involved, and explain the reasons for the difference in the energies dissipated.

Accepted Answer

(a) The work done in pulling the rod is (see Section 2.12)

[latex]W_{tension} = (u)[/latex](volume),

where

[latex]u = \int_{0}^{\epsilon _{1}}{\sigma d\epsilon } .[/latex]

The true strain is found from the expression

[latex]\epsilon _{1}=\ln \left\lgroup\frac{10}{9.5} ight group^2 =0.1025.[/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	0.2
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 obtain the following values for K and n for this material:

K = 1275 MPa and n = 0.45.

Thus

[latex]u=\frac{K\epsilon ^{n+1}_{1}}{n+1} =\frac{\left(1275 imes 10^6 ight) \left(0.1025 ight)^{1.45} }{1.45} [/latex]

[latex]u=32.3 imes 10^{6} \ m.N m^3 [/latex]

and thus

[latex]W_{tension}=\left(32.3 imes 10^{6} ight) \pi \left\lgroup\frac{5^2}{1000^2} ight group\left\lgroup\frac{125}{1000} ight group =317[/latex] N.m.

(b) From Table 8.3,

Approximate Specific-Energy Requirements in Machining Operations
Material	Specific energy* (W-s/mm[latex]^{3}[/latex])
Aluminum alloys	0.4–1.1
Cast irons	1.6–5.5
Copper alloys	1.4–3.3
High-temperature alloys	3.3–8.5
Magnesium alloys	0.4–0.6
Nickel alloys	4.9–6.8
Refractory alloys	3.8–9.6
Stainless steels	3.0–5.2
Steels	2.7–9.3
Titanium alloys	3.0–4.1
*At drive motor, corrected for 80% efficiency; multiply the energy by 1.25 for dull tools.

let’s estimate an average value for the specific energy in machining stainless steels as 4.1 W-S/[latex]mm^3[/latex]. The volume of material machined is

Volume[latex]=\frac{\pi }{4}\left[10^2-9.5^2 ight](125) =957.20 \ mm^3 [/latex]

Hence the work done in machining is

[latex]W_{mach}= (4.1)(957.2) = 3924[/latex] N.m

Note that the work done in machining is about 12 times higher than that for tension. The reasons for the difference between the two energies are that tension involves very little strain and that there is no friction. Machining, however, involves friction, and the material removed (even though relatively small in volume)has undergone much higher strains than the bulk material undergoing tension.
Assuming, from Tables 8.1

Data on Orthogonal Cutting of 4130 Steel
[latex]\alpha [/latex]	[latex]\phi [/latex]	[latex]\gamma [/latex]	[latex]\mu [/latex]	[latex]\beta [/latex]	[latex]F_{c}[/latex] (N)	[latex]F_{t}[/latex] (N)	[latex]u_{t}[/latex]	[latex]u_{s}[/latex]	[latex]u_{f}[/latex]	[latex]u_{f}/u_{t}[/latex]%
[latex]\alpha [/latex]	[latex]\phi [/latex]	[latex]\gamma [/latex]	[latex]\mu [/latex]	[latex]\beta [/latex]	[latex]F_{c}[/latex] (N)	[latex]F_{t}[/latex] (N)	[latex](m-N/m^{3} imes 10^{6})[/latex]	[latex](m-N/m^{3} imes 10^{6})[/latex]	[latex](m-N/m^{3} imes 10^{6})[/latex]	[latex]u_{f}/u_{t}[/latex]%
25°	20.9°	2.55	1.46	56	1691	997	2207	1442	765	35
35	31.6	1.56	1.53	57	1130	454	1476	772	704	48
40	35.7	1.32	1.54	57	1032	316	1345	648	697	52
45	41.9	1.06	1.83	62	1032	303	1345	517	828	62
[latex]t_{o}[/latex] = 0.0635 mm; [latex]w[/latex]= 12.065 mm; V = 27.423 m/min; tool: high-speed steel. Source: After E.G. Thomsen.

and 8.2,

Data on Orthogonal Cutting of 9445 Steel
[latex]\alpha [/latex]	V	[latex]\phi [/latex]	[latex]\gamma [/latex]	[latex]\mu [/latex]	[latex]\beta [/latex]	[latex]F_{c}[/latex] (N)	[latex]F_{t}[/latex] (N)	[latex]u_{t}[/latex]	[latex]u_{s}[/latex]	[latex]u_{f}[/latex]	[latex]u_{f}/u_{t}[/latex]%
[latex]\alpha [/latex]	V	[latex]\phi [/latex]	[latex]\gamma [/latex]	[latex]\mu [/latex]	[latex]\beta [/latex]	[latex]F_{c}[/latex] (N)	[latex]F_{t}[/latex] (N)	[latex](m-N/m^3 imes 10^6)[/latex]	[latex](m-N/m^3 imes 10^6)[/latex]	[latex](m-N/m^3 imes 10^6)[/latex]	[latex]u_{f}/u_{t}[/latex]%
+10	197	17	3.4	1.05	46	1646	1215	2759	2014	745	27
	400	19	3.1	1.11	48	1602	1259	2690	1834	856	32
	642	21.5	2.7	0.95	44	1464	966	2455	1717	738	30
	1186	25	2.4	0.81	39	1348	748	2262	1552	710	31
-10	400	16.5	3.9	0.64	33	1851	1713	3104	2359	745	24
	637	19	3.5	0.58	30	1709	1451	2862	2152	710	25
	1160	22	3.1	0.51	27	1584	1170	2655	1993	662	25
[latex]t_{o}[/latex] =0.9398 mm; [latex]w[/latex] = 6.35 mm; tool: cemented carbide. Source: After M.E. Merchant.

an average shear strain of 3, which is equivalent to an effective strain of 1.7 [see Eq. (2.58)],

[latex]\overline{\epsilon } =\frac{\sqrt{2} }{3} \left[\left(\epsilon _{1}-0 ight)^{2} +(0+\epsilon _{1})^{2}+(\epsilon _{1}-\epsilon _{1})^{2} ight]^{{1}/{2}} =\frac{2}{\sqrt{3} } \epsilon _{1}=\frac{1}{\sqrt{3} }\gamma . [/latex] (2.58)

the material removed in machining is subjected to a strain of [latex]1.7/0.1025 = 17 [/latex] times higher than that in tension. These differences explain why machining consumes much more energy than reducing the diameter of this rod by stretching. It can be shown, however, that as the diameter of the rod decreases, the difference between the two energies becomes smaller, assuming that the same depth of material is to be removed. This result can be explained by noting the changes in the relative volumes involved in machining vs. tension as the diameter of the rod decreases.

Question 8.2: Comparison of forming and machining energies You are given t...

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