Fragor

6. 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?

Answer. (c).

4. 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.

Solution: (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

4. Determine the blanking force required in Problem 2, if the shear strength of the steel = 325 MPa and the tensile strength is 450 MPa.

Solution: F = StL t = 2.0 mm from Problem 2 L = πD = 75π = 235.65 mm F = 325(2.0)(235.65) = 153200 N

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

Solution: 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 mm3 /min = 12500 mm3 /s Pc = RMRU Pc = (12500 mm3/s) (2.38 J/mm3 ) = 29750 J/s = 29750 W = 29.75 kW Pg = Pc / E Pg = 29.75/0.90 = 33.06 kW

3. A circular sheet-metal slug produced in a hole punching operation will have the same diameter as (a) the die opening or (b) the punch?

Answer. (a)

6. The following are measures of feasibility for several proposed cup- drawing operations; which of the operations are likely to be feasible (three best answers): (a) DR = 1.7, (b) DR = 2.7, (c) r = 0.35, (d) r = 0.65, and (e) t/D = 2%?

Answer. (a), (c), and (e).

1. Most sheet metalworking operations are performed as which one of the following: (a) cold working, (b) hot working, or (c) warm working?

Answer. (a).

2. In a sheet-metal-cutting operation used to produce a flat part with a hole in the center, the part itself is called a blank, and the scrap piece that was cut out to make the hole is called a slug: (a) true or (b) false?

Answer. (a).

5. Which of the following are variations of sheet-metal-bending operations (two best answers): (a) coining, (b) flanging, (c) hemming, (d) ironing, (e) notching, (f) shear spinning, (g) trimming, and(h) tube bending?

Answer. (b) and (c).

4. Which of the following descriptions applies to a V-bending operation as compared to an edge bending operation (two best answers): (a) costly tooling, (b) inexpensive tooling, (c) limited to 90° bends or less, (d) used for high production, (e) used for low production, and (f) uses a pressure pad to hold down the sheet metal?

Answer. (b) and (e).

5. 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?

Answer. (b).

7. 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?

Answer. (b).

3. 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?

Answer. (c).

7. The holding force in drawing is most likely to be (a) greater than, (b) equal to, or (c) less than the maximum drawing force?

Answer. (c).

4. Which of the following are characteristics of the orthogonal cutting model (three best answers): (a) a circular cutting edge is used, (b) a multiple-cutting- edge 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?

Answer. (d), (f), and (g).

1. 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?

Answer. (e) and (f).

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

Answer. (e).

7. Explain the difference between roughing and finishing operations in machining.

Answer. 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.

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

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

6. What are some of the possible defects in drawn sheet-metal parts?

Answer. Drawing defects include (1) wrinkling, (2) tearing, (3) earing, and (4) surface scratches

4. Define drawing in the context of sheet metalworking.

Answer. Drawing is a sheet metalworking operation used to produce cup- shaped or box-shaped, or other complex-curved, hollow parts. Drawing is accomplished by placing a piece of sheet metal over a die cavity and then using a punch to push the metal into the cavity.

2. Describe each of the two types of sheet-metal-bending operations: V- bending and edge bending.

Answer. In V-bending, the sheet metal is bent between a V-shaped punch and die. Included angles ranging from very obtuse to very acute can be made with V-dies. Edge bending involves cantilever loading of the sheet metal. A pressure pad is used to apply a force Fh to hold the base of the part against the die, while the punch forces the part to yield and bend over the edge of the die.

2. What distinguishes machining from other manufacturing processes?

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

5. What are some of the simple measures used to assess the feasibility of a proposed cup-drawing operation?

Answer. Measures of drawing feasibility include (1) drawing ratio DR = D/Dp; (2) reduction r = (D - Dp)/D; and (3) thickness-to-diameter ratio, t/D; where t = stock thickness, D = blank diameter, and Dp = punch diameter.

3. For what is the bend allowance intended to compensate?

Answer. The bend allowance is intended to compensate for stretching of the sheet metal that occurs in a bending operation when the bend radius is small relative to the stock thickness. In principle the bend allowance equals the length of the bent metal along its neutral axis.

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

Answer. The four forces that act upon the chip are (1) friction force, (2) normal force to friction, (3) shear force, and (4) normal force to shear.

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

Answer. 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, and (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.

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

Answer. 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; and (4) it can create good surface finishes.

1. What are the three basic categories of material removal processes?

Answer. The three basic categories of material removal processes are (1) conventional machining, (2) abrasive processes, and (3) nontraditional processes.

1. Identify the three basic types of sheet metalworking operations.

Answer. The three basic types of sheet metalworking operations are (1) cutting, (2) bending, and (3) drawing.

4. Name the three most common machining processes.

Answer. The three common machining processes are (1) turning, (2) drilling, and (3) milling.

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

Answer. 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.

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

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

10.A cup-drawing operation is performed in which the inside diameter = 80 mm and the height = 50 mm. The stock thickness = 3.0 mm, and the starting blank diameter = 150 mm. Punch and die radii = 4 mm. Tensile strength = 400 MPa and yield strength = 180 MPa for this sheet metal. Determine (a) drawing ratio, (b) reduction, (c) drawing force, and (d) blankholder force.

Solution: (a) DR = 150/80 = 1.875 (b) r = (Db - Dp)/Db = (150 - 80)/150 = 70/150 = 0.46 (c) F = πDpt(TS)(Db/Dp - 0.7) = π(80)(3)(400)(150/80 - 0.7) = 354418 N. (d) Fh = 0.015Yπ(Db 2 - (Dp + 2.2t + 2Rd) 2 ) Fh = 0.015(180)π(1502 - (80 + 2.2 x 3 + 2 x 4)2 ) = 0.015(180)π(1502 - 94.62 ) Fh = 114942 N

9. A cup is to be drawn in a deep drawing operation. The height of the cup is 75 mm and its inside diameter = 100 mm. The sheet-metal thickness = 2 mm. If the blank diameter = 225 mm, determine (a) drawing ratio, (b) reduction, and (c) thickness-to-diameter ratio. (d) Does the operation seem feasible?

Solution: (a) DR = Db/Dp = 225/100 = 2.25 (b) r = (Db - Dp)/Db = (225 - 100)/225 = 0.555 = 55.5% (c) t/Db = 2/225 = 0.0089 = 0.89% (d) Feasibility? No, DR is too large (greater than 2.0), r is too large (greater than 50%), and t/D is too small (less than 1%).

9. Solve Problem 8 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.

Solution: (a) From Table 19.2, U = 0.7 N-m/mm3 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 ) mm3 /s. Pc = U RMR = 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

8. 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.

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

5. The foreman in the pressworking section comes to you with the problem of a blanking operation that is producing parts with excessive burrs. (a) What are the possible reasons for the burrs, and (b) what can be done to correct the condition?

Solution: (a) Reasons for excessive burrs: (1) clearance between punch and die is too large for the material and stock thickness; and (2) punch and die cutting edges are worn (rounded) which has the same effect as excessive clearance. (b) To correct the problem: (1) Check the punch and die cutting edges to see if they are worn. If they are, regrind the faces to sharpen the cutting edges. (2) If the die is not worn, measure the punch and die clearance to see if it equals the recommended value. If not, die maker must rebuild the punch and die.

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.

Solution: (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 103 mm/m) (0.3) (2.6) = 1404 mm3/s

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.

Solution: (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

3. 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.

Solution: (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

5. 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.

Solution: (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

11.A deep drawing operation is to be performed on a sheet-metal blank that is 5/16 cm thick. The height (inside dimension) of the cup = 9.5 cm and the diameter (inside dimension) = 12.5 cm. Assuming the punch radius = 0, compute the starting diameter of the blank to complete the operation with no material left in the flange. Is the operation feasible (ignoring the fact that the punch radius is too small)?

Solution: Cup area = wall area + base area = πDph + πDp 2 /4 = (π x12.5x9.5)+ π(12.5)2 x0.25 = π 157.8125 cm 2 Blank area = πDb 2 /4 = 0.25πDb 2 Blank area = cup area: 0.25πDb 2 = 157.8125π Db 2 = 157.8125/0.25 = Db = 25.1246 cm

3. A compound die will be used to blank and punch a large washer out of 6061ST aluminum alloy sheet stock 3.50 mm thick. The outside diameter of the washer is 50.0 mm and the inside diameter is 15.0 mm. Determine (a) the punch and die sizes for the blanking operation, and (b) the punch and die sizes for the punching operation.

Solution: From Table 18.1, Ac = 0.060. Thus, c = 0.060(3.50) = 0.210 mm (a) Blanking punch diameter = Db - 2c = 50 - 2(0.21) = 49.58 mm Blanking die diameter = Db = 50.00 mm (b) Punching punch diameter = Dh = 15.00 mm Punching die diameter = Dh + 2c = 30 + 2(0.210) = 15.42 mm

1. A power shears is used to cut soft cold-rolled steel that is 4.75 mm thick. At what clearance should the shears be set to yield an optimum cut?

Solution: From Table 18.1, Ac = 0.060. c = Act = 0.060(4.75) = 0.285 mm

2. A blanking operation is to be performed on 2.0 mm thick cold-rolled steel (half hard). The part is circular with diameter = 75.0 mm. Determine the appropriate punch and die sizes for this operation.

Solution: From Table 18.1, Ac = 0.075. Thus, c = 0.075(2.0) = 0.15 mm. Punch diameter = Db - 2c = 75.0 - 2(0.15) = 74.70 mm Die diameter = Db = 75.0 mm

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 RMR = vfd = (200 m/min)(103 mm/m)(0.25 mm)(7.5 mm) = 375,000 mm3 /min = 6250 mm3 /s Pc = RMRU Pc = (6250 mm3 /s) (2.8 J/mm3 ) = 17500 J/s = 17500 W = 17.5 kW Pg = Pc/E Pg = 17.5/0.90 = 19.44 kW

10. Orthogonal cutting is performed on a metal whose mass specific heat = 1.0 J/g-C, density = 2.9 g/cm3 , and 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.

Solution: 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 N- m/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

What is springback?

Springback is when the bent sheet bends back a bit after the punch is removed. This occurs because the sheet is a little elastic which means there will be residual stresses in the sheet that presses it towards its previous form. There is two ways to compensate sprinback there bottoming and overbending. Bottoming is when you squeeze the bottom of the bent sheet. Overbending is when you bend more than the angle you want, so that after the springback you will get the desired angle.