Review Questions
22.1 Discuss the differences between rotational parts and prismatic parts in machining.
Answer. Rotational parts are cylindrical or disk-shaped and are machined on a turning machine;
prismatic parts are block-shaped or flat and are generally produced on a milling machine, shaper, or planer.
22.2 Distinguish between generating and forming when machining workpart geometries.
Answer. Generating refers to the creation of work geometry due to the feed trajectory of the cutting tool; examples include straight turning, taper turning, and profile milling. Forming involves the creation of work geometry due to the shape of the cutting tool; common examples include form turning and drilling.
22.3 Give two examples of machining operations in which generating and forming are combined to create workpart geometry.
Answer. Two examples are thread cutting on a lathe and slot milling; both are described in Article 25.6.1.
22.4 Describe the turning process.
Answer. Turning is a machining process in which a single point tool removes material from the surface of a rotating cylindrical workpiece, the tool being fed in a direction parallel to the axis of work rotation.
22.5 What is the difference between threading and tapping?
Answer. A threading operation is performed on a turning machine and produces an external thread, while tapping is normally performed on a drilling machine and produces an internal thread.
22.6 How does a boring operation differ from a turning operation?
Answer. Boring produces an internal cylindrical shape from an existing hole, while turning produces an external cylindrical shape.
22.7 What is meant by the designation 12 x 36 inch lathe?
Answer. A 12 x 36 lathe has a 12 inch swing (maximum work diameter that can be
accommodated) and a 36 inch distance between centers (indicating the maximum work length that can be held between centers).
22.8 Name the various ways in which a workpart can be held in a lathe.
Answer. Methods of holding the work in a lathe include: (1) between centers, (2) chuck, (3) collet, and (4) face plate. See Article 25.1.4.
22.9 What is the difference between a live center and a dead center, when these terms are used in the context of workholding in a lathe?
Answer. A center holds the work during rotation at the tailstock end of the lathe. A live center is mounted in bearings and rotates with the work, while a dead center does not rotate - the work rotates about it.
22.10 How does a turret lathe differ from an engine lathe?
Answer. A turret lathe has a toolholding turret in place of a tailstock; the tools in the turret can be brought to work to perform multiple cutting operations on the work without the need to change tools as in operating a conventional engine lathe.
22.11 What is a blind hole?
Answer. A blind hole does not exit the work; by comparison, a through hole exits the opposite side of the workpart.
22.12 What is the distinguishing feature of a radial drill press?
Answer. A radial drill has a long radial arm along which the drill head can be positioned to allow the drilling of large workparts.
22.13 What is the difference between peripheral milling and face milling?
Answer. In peripheral milling, cutting is accomplished by the peripheral teeth of the milling cutter and the tool axis is parallel to the work surface; in face milling, cutting is accomplished by the flat face of the cutter whose axis is perpendicular to the work surface.
22.14 Describe profile milling.
Answer. Profile milling generally involves the milling of the outside periphery of a flat part.
22.15 What is pocket milling?
Answer. Pocket milling uses an end milling cutter to machine a shallow cavity (pocket) into a flat workpart.
22.16 Describe the difference between up milling and down milling?
Answer. In up milling, the cutter speed direction is opposite the feed direction; in down milling, the direction of cutter rotation is the same as the feed direction.
22.17 How does a universal milling machine differ from a conventional knee-and-column machine?
Answer. The universal milling machine has a worktable that can be rotated about a vertical axis to present the part at any specified angle to the cutter spindle.
22.18 What is a machining center?
Answer. A machining center is a CNC machine tool capable of performing multiple types of cutting operations involving rotating spindles (e.g., milling, drilling); the machine is typically equipped with automatic tool-changing, pallet shuttles to speed workpart changing, and automatic workpart positioning.
22.19 What is the difference between a machining center and a turning center?
Answer. A machining center is generally confined to rotating spindle operations (e.g., milling, drilling); while a turning center performs turning type operations, generally with single point tools.
22.20 What can a mill-turn center do that a conventional turning center cannot do?
Answer. The mill-turn center has the capacity to position a rotational workpart at a specified angular location, permitting milling or drilling to be performed at a location on the periphery of the part.
22.21 How do shaping and planing differ?
Answer. In shaping, the work is stationary during the cut, and the speed motion is performed by the cutting tool; while in planing, the cutting tool is stationary, and the workpart is moved past the tool in the speed motion.
22.22 What is the difference between internal broaching and external broaching?
Answer. Internal broaching is accomplished on the inside surface (hole) of a workpart; while external broaching is performed on one of the outside surfaces of the part.
22.23 Identify the three basic forms of sawing operation?
Answer. The three forms of sawing are: (1) hacksawing, (2) bandsawing, and (3) circular sawing.
Multiple Choice Questions
There are a total of 20 correct answers in the following multiple choice questions (some questions have multiple answers that are correct). To attain a perfect score on the quiz, all correct answers must be given, since each correct answer is worth 1 point. For each question, each omitted answer or wrong answer reduces the score by 1 point, and each additional answer beyond the number of answers required reduces the score by 1 point. Percentage score on the quiz is based on the total number of correct answers.
22.1 Which of the following are examples of generating the workpart geometry in machining, as compared to forming the geometry (more than one)? (a) broaching, (b) contour turning, (c) drilling, and (d) profile milling.
Answer. (b) and (d).
22.2 In a turning operation, the change in diameter of the workpart is equal to which one of the following? (a) 1 x depth of cut, (b) 2 x depth of cut, (c) 1 x feed, or (d) 2 x feed.
Answer. (b)
22.3 A lathe can be used to perform which of the following machining operations (more than one)? (a) boring, (b) broaching, (c) drilling, (d) milling, (e) planing, or (f) turning.
Answer. (a), (c), and (f)
22.4 A facing operation is normally performed on which of the following machine tools (one best answer)? (a) drill press, (b) lathe, (c) milling machine, (d) planer, or (e) shaper.
Answer. (b)
22.5 Knurling is performed on a lathe, but it is a metal forming operation rather than a metal removal operation: (a) true or (b) false?
Answer. (a)
22.6 Which of the following cutting tools can be used on a turret lathe (more than one)? (a) broach, (b) cut-off tool, (c) drill bit, (d) single point turning tool, or (e) threading tool.
Answer. (b), (c), (d), and (e).
22.7 Which of the following turning machines permits very long bar stock to be used (one best
answer)? (a) chucking machine, (b) engine lathe, (c) screw machine, (d) speed lathe, or (e) turret lathe.
Answer. (c)
22.8 Reaming is used for which of the following functions (more than one)? (a) accurately locate a hole position, (b) enlarge a drilled hole, (c) improve surface finish on a hole, (d) improve tolerance on hole diameter, and (e) provide an internal thread.
Answer. (b), (c), and (d).
22.9 End milling is most similar to which one of the following? (a) face milling, (b) peripheral milling, (c) plain milling, or (d) slab milling.
Answer. (a)
22.10 The basic milling machine is which one of the following: (a) bed type, (b) knee-and-column, (c) profiling mill, (d) ram mill, and (e) universal milling machine.
Answer. (b)
22.11 A broaching operation is best described by which one of the following: (a) a rotating tool moves past a stationary workpart, (b) a tool with multiple teeth moves linearly past a stationary workpart, (c) a workpart is fed past a rotating cutting tool, or (d) a workpart moves linearly past a stationary single point tool.
Answer. (b)
22.12 A planing operation is best described by which one of the following: (a) a single point tool moves linearly past a stationary workpart, (b) a tool with multiple teeth moves linearly past a stationary workpart, (c) a workpart is fed linearly past a rotating cutting tool, or (d) a workpart moves linearly past a single-point tool.
Answer. (d) Problems
Turning and Related Operations
22.1 A cylindrical workpart 125 mm in diameter and 900 mm long is to be turned in an engine lathe.
Cutting conditions are: v = 2.5 m/s, f = 0.3 mm/rev, and d = 2.0 mm. Determine: (a) cutting time, and (b) metal removal rate.
Solution: (a) N = (2.5 m/s)/.125π = 6.366 rev/s.
fr = 6.366(.3) = 1.91 mm/s
Tm = 900/1.91 = 471.2 s = 7.85 min.
(b) MRR = vfd = (2.5 m/s)(103)(.3 mm)(2.0 mm) = 1500 mm3/s
22.2 In a production turning operation, the foreman has decreed that the single pass must be completed on the cylindrical workpiece in 5.0 min. The piece is 400 mm long and 150 mm in diameter. Using a feed = 0.30 mm/rev and a depth of cut = 4.0 mm, what cutting speed must be used to meet this machining time requirement?
Solution: Starting with Eq. (22.4): Tm = L/fr
Substitute Eq. (22.3) (fr = Nf) into the denominator to obtain Tm = L/Nf
Then substituting for N from Eq. (22.1) we get Tm = πDoL/vf (this equation is later used in Chapter 24). Rearranging to determine cutting speed: v = πDoL/fTm
Tm = π(0.4)(0.15)/(0.30)(10-3)(5.0) = 0.1257(103) m/min = 125.7 m/min
22.3 A tapered surface is to be turned on an automatic lathe. The workpiece is 750 mm long with minimum and maximum diameters of 100 mm and 200 mm at opposite ends. The automatic controls on the lathe permit the surface speed to be maintained at a constant value of 200 m/min by adjusting the rotational speed as a function of workpiece diameter. Feed = 0.25 mm/rev and depth of cut = 3.0 mm. The rough geometry of the piece has already been formed, and this operation will be the final cut. Determine (a) the time required to turn the taper and (b) the rotational speeds at the beginning and end of the cut.
Solution: (a) MRR = vfd = (200 m/min)(103 mm/m)(0.25 mm)(3.0 mm) = 150,000 mm3/min Area of frustrum of cone A = π(R1 + R2){h2 + (R1 – R2)2}0.5
Given R1 = 100 mm, R2 = 50 mm, and h = 750 mm,
A = π(100 + 50){7502 + (100 – 50)2}0.5 = 150π(565,000)0.5 = 354,214 mm2
Given depth of cut d = 3.0 mm, volume cut V = Ad = (354,214 mm2)(3.0 mm) = 1,062,641 mm3 Tm = V/MRR = (1,062,641 mm3)/(150,000 mm3/min) = 7.084 min
(b) At beginning of cut (D1 = 100 mm), N = v/πD = 200,000/100π = 636.6 rev/min At end of cut (D2 = 200 mm), N = 200,000/200π = 318.3 rev/min
22.4 In the taper turning job of previous Problem 22.3, suppose that the automatic lathe with surface speed control is not available and a conventional lathe must be used. Determine the rotational speed that would be required to complete the job in exactly the same time as your answer to part (a) of that problem.
Solution: At a constant rotational speed and feed, feed rate fr is constant and Eqs. (22.3) and (22.4) can be used. Combining, Tm = L/Nf and then rearranging to obtain rotational speed N = L/fTm
Given L = 750 mm, f = 0.25 mm/rev, and Tm = 7.084 min from Problem 22.3, N = 750/(0.25)(7.084) = 423.5 rev/min
22.5 A workbar with 5.0 in diameter and 48 in length is chucked in an engine lathe and supported at the opposite end using a live center. A 40.0 in portion of the length is to be turned to a diameter of 4.75 in one pass at a speed = 400 ft/min and a feed = 0.012 in/rev. Determine: (a) the required depth of cut, (b) cutting time, and (c) metal removal rate.
Solution: (a) depth d = (5.00 - 4.75)/2 = 0.125 in.
(b) N = 400 x 12/5π = 305.5 rev/min.
fr = 305.5(0.012) = 3.67 in/min Tm = 40/3.67 = 10.91 min.
(c) MRR = (400 x 12)(0.125)(0.012) = 7.2 in3/min.
22.6 A 4.00 in diameter workbar that is 25 in long is to be turned down to 3.50 in diameter in two passes on an engine lathe using the following cutting conditions: v = 300 ft/min, f = 0.015 in/rev, and d = 0.125 in. The bar will be held in a chuck and supported on the opposite end in a live center. With this workholding setup, one end must be turned to diameter; then the bar must be reversed to turn the other end. Using an overhead crane available at the lathe, the time required to load and unload the bar is 5.0 minutes, and the time to reverse the bar is 3.0 minutes. For each turning cut an allowance must be added to the cut length for approach and overtravel. The total
allowance (approach plus overtravel) = 0.50 in. Determine the total cycle time to complete this turning operation.
Solution: First end: cut 15 in. of 25 in. length.
N = 300 x 12/4π = 286.4 rev/min., fr = 286.4(0.015) = 4.297 in/min.
Tm = (15 + 0.5)/4.297 = 3.61 min.; this reduces diameter to 3.75 in.
N = 300 x 12/3.75π = 305.5 rev/min., fr = 305.5(0.015) = 4.583 in/min.
Tm = 15.5/4.583 = 3.38 min’ this reduces diameter to 3.50 in.
Reverse bar which takes 3.0 min. cut remaining 10 in. of 25 in. length.
N = 300 x 12/4π = 286.4 rev/min., fr = 286.4(0.015) = 4.297 in/min.
Tm = (10 + 0.5)/4.297 = 2.44 min.; this reduces diameter to 3.75 in.
N = 300 x 12/3.75π = 305.5 rev/min., fr = 305.5(0.015) = 4.583 in/min.
Tm = 10.5/4.583 = 2.29 min’ this reduces diameter to 3.50 in.
Loading and unloading bar takes 5.0 min.
Total cycle time = 5.0 + 3.61 + 3.38 + 3.0 + 2.44 + 2.29 = 19.72 min.
22.7 The end of a large tubular workpart is to be faced on a NC vertical boring mill. The part has an outside diameter = 45.0 in and inside diameter = 25 in. If the facing operation is performed at a rotational speed = 30 rev/min, feed = 0.020 in/rev, and depth = 0.150 in, determine: (a) the cutting time to complete the facing operation, (b) the cutting speeds and metal removal rates at the beginning and end of the cut.
Solution: (a) Distance traveled L = (Do - Di)/2 = (45 - 25)/2 = 10 in.
fr = (30 rev/min)(0.020 in/rev) = 0.6 in/rev Tm = 10/0.6 = 16.67 min
(b) at Do = 45 in., v = (30 rev/min)(45π/12) = 353.5 ft/min.
MRR = (353.5 x 12)(0.020)(0.15) = 12.73 in3/min at Di = 25 in., v = (30 rev/min)(25π/12) = 196.4 ft/min.
MRR = (196.4 x 12)(0.020)(0.15) = 7.07 in3/min
22.8 Solve previous Problem 22.7 except that the machine tool controls operate at a constant cutting speed by continuously adjusting rotational speed for the position of the tool relative to the axis of rotation. The rotational speed at the beginning of the cut = 30 rev/min, and is continuously increased thereafter to maintain a constant cutting speed.
Solution: (a) Total metal removed = 0.25πd(Do2
- Di2) = 0.25π(0.15)(452 - 252)
= 164.96 in3
MRR is constant throughout cutting if v is constant.
v = (30 rev/min)(45π/12) = 353.5 ft/min
MRR = (353.5 x 12)(0.020)(0.15) = 12.73 in3/min
Tm = (total metal removed)/MRR = 164.96/12.73 = 12.96 min.
Drilling
22.9 A drilling operation is to be performed with a 25.4 mm diameter twist drill in a steel workpart. The hole is a blind-hole at a depth = 50 mm, and the point angle = 118°. Cutting conditions are: speed = 25 m/min, feed = 0.25 mm/rev. Determine: (a) the cutting time to complete the drilling operation, and (b) metal removal rate during the operation, after the drill bit reaches full diameter.
Solution: (a) A = 0.5(25.4) tan(90 – 118/2) = 12.7 tan 31 = 7.65 mm
N = 25(103)/25.4π = 313.3 rev/min fr = 313.3(0.25) = 78.3 mm/min Tm = (50 + 7.63)/78.3 = 0.736 min
(b) MRR = 0.25π(25.4)2(78.3) = 39,675.2 mm3/min
22.10 A NC drill press is to perform a series of through-hole drilling operations on a 1.75 in thick aluminum plate that is a component in a heat exchanger. Each hole is 3/4 in diameter. There are 100 holes in all, arranged in a 10 by 10 matrix pattern, and the distance between adjacent hole centers (along the square) = 1.5 in. The cutting speed = 300 ft/min, the penetration feed (z-direction) = 0.015 in/rev, and the feed rate between holes (x-y plane) = 15.0 in/min. Assume that x-y moves are made at a distance of 0.5 in above the work surface, and that this distance must be included in the penetration feed rate for each hole. Also, the rate at which the drill is retracted from each hole is twice the penetration feed rate. The drill has a point angle = 100 degrees. Determine the time required from the beginning of the first hole to the completion of the last hole, assuming the most efficient drilling sequence will be used to accomplish the job.
Solution: Time to drill each hole:
N = 300 x 12/0.75π = 1527.7 rev/min.
fr = 1527.7(0.015) = 22.916 in/min.
Distance per hole = 0.5 + A + 1.75
A = 0.5(0.75) tan(90 - .5 x 100) = 0.315 in.
Tm = (0.5 + 0.315 + 1.75)/22.916 = 0.112 min.
Time to retract drill from hole = 0.112/2 = 0.056 min.
All moves between holes are at a distance = 1.5 in. using a back and forth path between holes.
Time to move between holes = 1.5/15 = 0.1 min. With 100 holes, the number of moves between holes = 99.
Total cycle time to drill 100 holes = 100(0.112 + 0.056) + 99(0.1) = 26.7 min.
22.11 A gundrilling operation is used to drill a 7/16-in diameter hole to a certain depth. It takes 4.5 minutes to perform the drilling operation using high pressure fluid delivery of coolant to the drill point. The cutting conditions are: N = 3000 rev/min at a feed = 0.002 in/rev. In order to improve the surface finish in the hole, it has been decided to increase the speed by 20% and decrease the feed by 25%. How long will it take to perform the operation at the new cutting conditions?
Solution: fr = 3000 rev/min(0.001 in/rev) = 3.0 in/min.
Hole depth d = 4.5 min(3.0 in/min.) = 13.5 in.
New speed v = 3000(1 + 0.20) = 3600 rev/min.
New feed f = 0.001(1- 0.25) = 0.00075 in/min.
New feed rate fr = 3600(0.00075) = 2.7 in/min.
New drilling time Tm = 13.5 in/2.7 in/min = 5.0 min.
Milling
22.12 A peripheral milling operation is performed on the top surface of a rectangular workpart which is 300 mm long by 100 mm wide. The milling cutter, which is 75 mm in diameter and has four teeth, overhangs the width of the part on both sides. Cutting conditions are: v = 80 m/min, f = 0.2 mm/tooth, and d = 7.0 mm. Determine: (a) the time to make one pass across the surface, and (b) the material removal rate during the cut.
Solution: N = 80,000 mm/75π = 339.5 rev/min.
fr = 339.5(4)(0.2) = 271.6 mm/min.
A = (d(D-d)).5 = (7(75-7)).5 = 21.82 mm Tm = (300 + 21.82)/271.6 = 1.185 min.
MRR = 100(7)(271.6) = 190,120 mm3/min.
22.13 A face milling operation is used to machine 5 mm from the top surface of a rectangular piece of aluminum 400 mm long by 100 mm wide. The cutter has four teeth (cemented carbide inserts) and is 150 mm in diameter. Cutting conditions are: v = 3 m/s, f = 0.27 mm/tooth, and d = 5.0 mm.
Determine: (a) time to make one pass across the surface, and (b) metal removal rate during cutting.
Solution: (a) N = (3000 mm/s)/150π = 6.366 rev/s fr = 6.366(4)(.27) = 6.875 mm/s
A = D/2 = 150/2 = 75 mm
Tm = (400 + 75)/6.875 = 80 s = 1.333 min.
(b) MRR = 100(5)(6.875) = 3437.5 mm3/s
22.14 A slab milling operation is performed to finish the top surface of a steel rectangular workpiece 10.0 in long by 3.0 in wide. The helical milling cutter, which has a 2.5 in diameter and eight teeth, is set up to overhang the width of the part on both sides. Cutting conditions are: v = 100 ft/min, f = 0.009 in/tooth, and d = 0.250 in. Determine: (a) the time to make one pass across the surface, and (b) the metal removal rate during the cut.
Solution: N = 100 x 12/2.5π = 152.8 rev/min.
fr = 152.8(8)(0.009) = 11.0 in/min.
A = (d(D-d)).5 = (.25(2.5-.25)).5 = 0.75 in Tm = (10.0 + 0.75)/11 = 0.98 min.
MRR = 3.0(.25)(11.0) = 8.25 in3/min.
22.15 A face milling operation is performed to finish the top surface of a steel rectangular workpiece 12.0 in long by 2.0 in wide. The milling cutter has four teeth (cemented carbide inserts) and a 3.0 in diameter. Cutting conditions are: v = 500 ft/min, f = 0.010 in/tooth, and d = 0.150 in. Determine:
(a) the time to make one pass across the surface, and (b) the metal removal rate during the cut.
Solution: N = 500 x 12/3π = 636.5 rev/min.
fr = 636.5(4)(0.010) = 25.46 in/min.
A = O = 3/2 = 1.5 in.
Tm = (12.0 + 2 x 1.5)/25.46 = 0.59 min.
MRR = 2.0(.150(25.46) = 7.64 in3/min.
22.16 Solve previous Problem 22.15 except that the workpiece is 5.0 in wide and the cutter is offset to one side so that the swath cut by the cutter = 1.0 in wide.
Solution: N = 500 x 12/3π = 636.5 rev/min.
fr = 636.5(4)(0.010) = 25.46 in/min.
A = O = (1(3-1)).5 = 1.414 in
Tm = (12.0 + 2 x 1.414)/25.46 = 0.58 min.
MRR = 1.0(.150(25.46) = 3.82 in3/min.
Other Operations
22.17 An open side planer is to be used to plane the top surface of a rectangular workpart, 25.0 in by 40.0 in. Cutting conditions are: v = 25 ft/min, f = 0.020 in/pass, and d = 0.200 in. The length of the
stroke across the work must be set up so that 10 in are allowed at both the beginning and end of the stroke for approach and overtravel. The return stroke, including an allowance for acceleration and deceleration, takes 75% of the time for the forward stroke. How long will it take to complete the job, assuming that the part is oriented in such a way as to minimize the time?
Solution: Orient work so that its length (L = 40 in.) is in direction of stroke. This will minimize the number of passes required which will minimize time in this case. Time per forward stroke = (10 + 40 + 10)/(25 x 12) = 0.2 min.
Time per reverse stroke = 0.75(.2) = 0.15 min.
Total time per pass = 0.20 + 0.15 = 0.35 min.
Number of passes = 25.0/0.020 = 1250 passes Total time = 1250(0.35) = 437.5 min.
Check: orient work so that its width (w = 25 in.) is in direction of stroke.
Time per forward stroke = (10 + 25 + 10)/(25 x 12) = 0.15 min.
Time per reverse stroke = 0.75(.15) = 0.1125 min.
Total time per pass = 0.15 + 0.1125 = 0.2625 min.
Number of passes = 40.0/0.020 = 2000 passes Total time = 2000(0.2625) = 525.0 min.