Tillverkningsteknik - A2 Metal cutting

Cutting Temperature Orthogonal cutting is performed on a metal whose mass specific heat = 1.0 J/g-C density = 2.9 g/cm3 thermal diffusivity = 0.8 cm2/s. The cutting speed is 4.5 m/s, uncut chip thickness is 0.25 mm, and width of cut is 2.2 mm. The cutting force is measured at 1170 N. Using Cook's equation, determine the cutting temperature if the ambient temperature = 22°C.

ρC = (2.9 g/cm3 )(1.0 J/g-°C) = 2.90 J/cm3 -°C = (2.90x10-3 ) J/mm3 -°C K = 0.8 cm2 /s = 80 mm2 /s U = Fcv/RMR = 1170 N x 4.5 m/s/(4500 mm/s x 0.25 mm x 2.2 mm) = 2.127 Nm/mm3 T = 0.4U/(ρC) x (vto/K) 0.333 T = 22 + (0.4 x 2.127 N-m/mm3 /(2.90x10-3 ) J/mm3 -C) [4500 mm/s x 0.25 mm/80 mm2 /s]0.333 T = 22 + (0.2934 x 103 C)(14.06).333 = 22 + 293.4(2.41) = 22° + 707° = 729°C

Low carbon steel having a tensile strength of 300 MPa and shear strength of 220 MPa is cut in a turning operation with a cutting speed of 3.0 m/s. The feed is 0.20 mm/rev and the depth of cut is 3.0 mm. The rake angle of the tool is 5° in the direction of chip flow. The resulting chip ratio is 0.45. Using the orthogonal model as an approximation of turning, determine (a) the shear plane angle, (b) shear force, (c) cutting force and feed force.

(a) φ = tan-1 (0.45 cos 5/ (1 - 0.45 sin 5)) = tan-1 (0.4666) = 25.0° (b) As = tow/sin φ = (0.2) (3.0)/sin 25.0 = 1.42 mm2. Fs = AsS = 1.42(220) = 312 N (c) β = 2(45) + α - 2(φ) = 90 + 5 - 2(25.0) = 45.0° Fc = Fscos (β - α)/cos (φ + β - α) Fc = 312 cos (45 - 5)/cos (25.0 + 45.0 - 5) = 566 N Ft = Fssin(β - α)/cos(φ + β - α) Ft = 312 sin(45 - 5)/cos(25.0 + 45.0 - 5) = 474 N

The cutting force and thrust force in an orthogonal cutting operation are 1470 N and 1589 N, respectively. The rake angle = 5°, the width of the cut = 5.0 mm, the chip thickness before the cut = 0.6, and the chip thickness ratio = 0.38. Determine (a) the shear strength of the work material and (b) the coefficient of friction in the operation.

(a) φ = tan^-1(0.38 cos 5/ (1 - 0.38 sin 5)) = tan^-1(0.3916) = 21.38° Fs = 1470 cos 21.38 - 1589 sin 21.38 = 789.3 N As = (0.6) (5.0)/sin 21.38 = 3.0/.3646 = 8.23 mm2 S = 789.3/8.23 = 95.9 N/mm2 = 95.9 MPa (b) φ = 45 + α/2 - β/2; rearranging, β = 2(45) + α - 2φ β = 2(45) + α - 2(φ) = 90 + 5 - 2(21.38) = 52.24° μ = tan 52.24 = 1.291

Which of the following are characteristics of the orthogonal cutting model (three best answers): (a) a circular cutting edge is used, (b) a multiple-cuttingedge tool is used, (c) a single-point tool is used, (d) only two dimensions play an active role in the analysis, (e) the cutting edge is parallel to the direction of cutting speed, (f) the cutting edge is perpendicular to the direction of cutting speed, and (g) the two elements of tool geometry are rake and relief angle?

(d), (f), and (g)

Name the three most common machining processes.

turning, drilling, milling

PROBLEMS - Chip formation and forces in machining 1. In an orthogonal cutting operation, the tool has a rake angle = 15°. The chip thickness before the cut = 0.30 mm and the cut yields a deformed chip thickness = 0.65 mm. Calculate (a) the shear plane angle and (b) the shear strain for the operation.

(a) r = to/tc = 0.30/0.65 = 0.4615 φ = tan^-1 (0.4615 cos 15/(1 - 0.4615 sin 15)) = tan^-1(0.5062) = 26.85° (b) Shear strain γ = cot 26.85 + tan(26.85 - 15) = 1.975 + 0.210 = 2.185

An orthogonal cutting operation is performed using a rake angle of 15°, chip thickness before the cut = 0.03cm and width of cut = 0.25 cm. The chip thickness ratio is measured after the cut to be 0.55. Determine (a) the chip thickness after the cut, (b) shear angle, (c) friction angle, (d) coefficient of friction, and (e) shear strain.

(a) r = to/tc, tc = to/r = 0.03/0.55 = 0.0545 cm (b) φ = tan^-1(0.55 cos 15/(1 - 0.55 sin 15)) = tan^-1(0.6194) = 31.8° (c) β = 2(45) + α - 2(φ) = 90 + 15 - 2(31.8) = 41.5° (d) μ = tan 41.5 = 0.88 (e) γ = cot 31.8 + tan(31.8 - 15) = 1.615 + 0.301 = 1.92

What are the three basic categories of material removal processes?

(1) conventional machining (2) abrasive processes (3) nontraditional processes.

Identify the four forces that act upon the chip in the orthogonal metal cutting model but cannot be measured directly in an operation.

(1) friction force, (2) normal force to friction (3) shear force (4) normal force to shear.

A turning operation is carried out on aluminum (100 BHN). Cutting speed = 5.6 m/s, feed = 0.25 mm/rev, and depth of cut = 2.0 mm. The lathe has a mechanical efficiency = 0.85. Based on the specific energy values in Table 19.2, determine (a) the cutting power and (b) gross power in the turning operation, in Watts.

(a) From Table 19.2, U = 0.7 Nm/mm^3 for aluminum. RMR = vfd = 5.6(103 ) (0.25) (2.0) = 2.8(103) mm3 /s. Pc = URMR = 0.7(2.8) (103) = 1.96(103) N-m/s = 1960 W (b) Pg = Pc / E Pg = 1960/0.85 = 2306 W

A lathe is used to perform which one of the following manufacturing operations: (a) broaching, (b) drilling, (c) lapping, (d) milling, or (e) turning?

(e)

Which of the following manufacturing processes are classified as material removal processes (two correct answers): (a) casting, (b) drawing, (c) extrusion, (d) forging, (e) grinding, (f) machining, (g) molding, (h) pressworking, and (i) spinning?

(e) and (f).

Explain the difference between roughing and finishing operations in machining.

A roughing operation is used to remove large amounts of material rapidly and to produce part geometry close to the desired shape. A finishing operation follows roughing and is used to achieve the final geometry and surface finish.

In using the orthogonal cutting model to approximate a turning operation, the chip thickness before the cut to corresponds to which one of the following cutting conditions in turning: (a) depth of cut d, (b) feed f, or (c) speed v?

B

The chip thickness ratio is which one of the following: (a) tc/to, (b) to/tc, (c) f/d, or (d) to/w, where tc = chip thickness after the cut, to = chip thickness before the cut, f = feed, d = depth, and w = width of cut?

B

A roughing operation generally involves which one of the following combinations of cutting conditions: (a) high v, f, and d; (b) high v, low f and d; (c) low v, high f and d; or (d) low v, f, and d, where v = cutting speed, f = feed, and d = depth?

C

Which one of the four types of chip would be expected in a turning operation conducted at low cutting speed on a brittle work material: (a) continuous, (b) continuous with built-up edge, (c) discontinuous, or (d) serrated?

C

What are the parameters of a machining operation that are included within the scope of cutting conditions?

Cutting conditions include speed, feed, depth of cut, and whether or not a cutting fluid is used.

Solve Problem 25 but with the following changes: cutting speed = 1.3 m/s, feed = 0.75 mm/rev, and depth = 4.0 mm. Note that although the power used in this operation is only about 10% greater than in the previous problem, the metal removal rate is about 40% greater.

From Table 19.2, U = 0.7 N-m/mm^3 From figure 19.14. For f = 0.75 mm/rev Correction factor = 0.80. RMR = vfd = 1.3(103) (0.75) (4.0) = 3.9(103) mm^3/s. Pc = URMR = 0.8(0.7) (3.9) (103) = 2.184(103) N-m/s = 2184 W (b) Gross power Pg = Pc/E Pg = 2184/0.85 = 2569 W

Power and Energy in Machining 6. In a turning operation on stainless steel with hardness = 200 HB, the cutting speed = 200 m/min, feed = 0.25 mm/rev, and depth of cut = 7.5 mm. How much power will the lathe draw in performing this operation if its mechanical efficiency = 90%. Use Table 19.2 to obtain the appropriate specific energy value. Solution: From Table 19.2, U = 2.8 N-m/mm3 = 2.8 J/mm3

From Table 19.2, U = 2.8 N-m/mm^3 = 2.8 J/mm^3 RMR = vfd = (200 m/min)(10^3 mm/m)(0.25 mm)(7.5 mm) = 375,000 mm^3/min = 6250 mm^3/s Pc = RMRU Pc = (6250 mm3/s) (2.8 J/mm^3) = 17500 J/s = 17500 W = 17.5 kW Pg = Pc/E Pg = 17.5/0.90 = 9.44 kW

What distinguishes machining from other manufacturing processes?

In machining, material is removed from the workpart so that the remaining material is the desired part geometry.

Name and briefly describe the four types of chips that occur in metal cutting.

The four types are; (1) discontinuous, in which the chip is formed into separated segments; (2) continuous, in which the chip does not segment and is formed from a ductile metal; (3) continuous with built-up edge, which is the same as (2) except that friction at the tool-chip interface causes adhesion of a small portion of work material to the tool rake face, (4) serrated, which are semi-continuous in the sense that they possess a saw-tooth appearance that is produced by a cyclical chip formation of alternating high shear strain followed by low shear strain.

Identify some of the reasons why machining is commercially and technologically important.

The reasons include the following: (1) it is applicable to most materials; (2) it can produce a variety of geometries to a part (3) it can achieve closer tolerances than most other processes (4) it can create good surface finishes.

What are the two basic categories of cutting tools in machining? Give two examples of machining operations that use each of the tooling types.

The two categories are (1) single-point tools, used in operations such as turning and boring; and (2) multiple-edge cutting tools, used in operations such as milling and drilling.

Identify the two forces that can be measured in the orthogonal metal cutting model.

The two forces that can be measured in the orthogonal metal cutting model are (1) cutting force and (2) thrust force.

In Problem 6, compute the lathe power requirements if feed = 0.50 mm/rev.

Using Figure 19.14, for f = 0.50 mm, correction factor = 0.85. From Table 19.2, U = 2.8 J/mm3 . With the correction factor, U = 2.8(0.85) = 2.38 J/mm3 RMR = vfd = (200 m/min)(103 mm/m)(0.50 mm)(7.5 mm) = 750000 mm^3/min = 12500 mm^3/s Pc = RMRU Pc = (12500 mm^3/s) (2.38 J/mm3) = 29750 J/s = 29750 W = 29.75 kW Pg = Pc / E Pg = 29.75/0.90 = 33.06 kW

2. In a turning operation, spindle speed is set to provide a cutting speed of 1.8 m/s. The feed and depth of cut of cut are 0.30 mm and 2.6 mm, respectively. The tool rake angle is 8°. After the cut, the deformed chip thickness is measured to be 0.49 mm. Determine (a) shear plane angle, (b) shear strain, and (c) material removal rate. Use the orthogonal cutting model as an approximation of the turning process.

a) r = to/tc = 0.30/0.49 = 0.612 φ = tan^-1(0.612 cos 8/(1 - 0.612 sin 8)) = tan^-1(0.6628) = 33.6° (b) γ = cot 33.6 + tan (33.6 - 8) = 1.509 + 0.478 = 1.987 (c) RMR = vfd RMR = (1.8 m/s x 10^3 mm/m) (0.3) (2.6) = 1404 mm^3 /s