Báo cáo bài tập lớn kỹ thuật chế tạo gia công cơ khí chương 21theory of metal machining

orthogonal cutting operation is performed using a tool with a rake angle = 20° at the following cutting conditions: cutting speed = 100 ft/min, chip thickness before the cut = 0.015 in,

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Chương 21: THEORY OF METAL MACHINING

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

the friction angle remains the same, determine (a) the shear plane angle, (b) the chip thickness, and (c) the shear strain for the operation.

Equation, Eq (21.16): φ = 45 + α/2 - β/2; rearranging, β = 2(45) + α - 2φ β = 2(45) + α - 2(φ) = 90 + 15 – 2(26.85) = 51.3° Now, with α = 0 and β remaining the same

21.3  In an orthogonal cutting operation, the 0.250 in wide tool has a rake angle of 5° The lathe is set so the chip thickness before the cut is 0.010 in After the cut, the deformed chip thickness is measured to be 0.027 in Calculate (a) the shear plane angle and (b) the shear strain for the operation.

rake angle is 8 After the cut, thedeformed chip thickness is measured to be 0.49

mm Determine (a) shear plane angle, (b) shearstrain, and (c) material removal rate Use the orthogonal cutting model as an approximation of theturning process.

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.

1470 cos 21.38 – 1589 sin 21.38 = 789.3 N As = (0.6)(5.0)/sin 21.38 = 3.0/.3646 =

(b) φ = 45 + α/2 - β/2; rearranging, β = 2(45) + α - 2φ β = 2(45) + α - 2(φ) = 90 + 5

width of cut = 0.200 in, chip thickness before the cut = 0.015, and chip thickness ratio = 0.4 Determine (a) the shear strength of the work material and (b) the

chip thickness before the cut = 0.012 in and width of cut = 0.100 in 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,

before the cut = 0.2 mm and width of cut = 4.0 mm The chip ratio = 0.4.

Determine (a) the chip thickness after the cut, (b) shear angle, (c) friction angle, (d) coefficient of friction, and (e) shear strain.

orthogonal cutting operation is performed using a tool with a rake angle = 20° at the following cutting conditions: cutting speed = 100 ft/min, chip thickness before the cut = 0.015 in, and width of cut = 0.150 in The resulting chip thickness ratio = 0.50 Determine (a) the shear plane angle; (b) shear force.

Solution: (a) φ = tan -1 (0.5 cos 20/(1 - 0.5 sin 20)) = tan -1 (0.5668) = 29.5°

 (b) As = (0.015)(0.15)/sin 29.5 = 0.00456 in 2 Fs = As S = 0.00456(50,000) =

228 lb

to -5° and the resulting chip thickness ratio = 0.35

18.7°

(a) As = (0.015)(0.15)/sin 18.7 = 0.00702 in 2 Fs = As S = 0.00702(50,000) =

351 lb

21.12. A carbon steel bar 7.64 inch in diameter has a tensile strength of 65,000 lb/in2 and a shear strength of 45,000 lb/in2 The diameter is reduced using a turning operation at a cutting speed of 400 ft/min The feed is 0.011 in/rev and the depth of cut is 0.120 in The rake angle on the tool in the direction of chip flow is 13° The cutting conditions result in a chip ratio of 0.52 Using the orthogonal model as an approximation of turning, determine

(a) the shear plane angle, (b) shear force, (c) cutting force and feed force, and (d) coefficient of friction between the tool and chip.

an approximation of the turning process.

Solution :

φ = tan-1(0.4 cos 10/(1 - 0.4 sin 10)) = tan-1(0.4233) = 22.9

   As = (0.010)(0.10)/sin 22.9 = 0.00257 in2

Solution : Begin with the definition of the chip ratio, Eq (21.2):

r =t o/t c = sin φ /cos ( φ - α )Rearranging,r cos ( φ - α ) = sin φ

Using the trigonometric identity cos( φ - α ) = cos φ cos α +

sin φ sin α r (cos φ cos α + sin φ sin α ) = sin φ

 Dividing both sides by sin φ , we obtain r cos α /tan φ +r sin α = 1

r cos α /tan φ = 1 -r sin α

Rearranging, tan φ =r cos α /(1 -r sin α )

21.16:   Show how Eq (21.4) is derived from Figure 21.6.

Solution : In the figure,

γ = AC / BD = ( AD + DC )/ BD = AD/ BD + DC / BD

 AD/ BD = cot φ and DC / BD= tan ( φ - α )Thus,

γ = cot φ + tan ( φ - α ).

21.18:

  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 21.3 to obtain theappropriate specific energy value.

Solution : From Table 21.3,

U = 2.8 N-m/mm3 = 2.8 J/mm3

 RMR =vfd = (200 m/min)(10 mm/m)(0.25 mm)(7.5 mm) = 375,000 mm3/min = 6250mm3/s

 P c = (6250 mm3/s)(2.8 J/mm3) = 17,500 J/s = 17,500 W = 17.5 kWAccounting for mechanical efficiency,

cutting speed = 900 ft/min, feed = 0.020 in/rev, and depth of cut = 0.250 in What

horsepower is required of the drive motor, if the lathe has a mechanical efficiency

= 87%? Use Table 21.2 to obtain the appropriate unit horsepower value.

Solution:

From Table 21.3, HPu = 0.25 hp/(in 3 /min) for aluminum Since feed is greater than 0.010 in/rev in the table, a correction factor must be applied from Figure 21.14 For f = 0.020 in/rev = t o , correction factor = 0.9 HPc = HPu x R MR, HP g = HPc /E R MR = vfd = 900 x 12(.020)(0.250) = 54 in 3 /min HPc = 0.9(0.25)(54) = 12.2 hp HP g = 12.2/0.87 = 14.0 hp.

21.22  :A turning operation is to be performed on a 20 hp lathe that has an 87% efficiency rating Theroughing cut is made on alloy steel whose hardness is in the range 325 to 335 HB The cuttingspeed is 375 ft/min The feed is 0.030 in/rev, and the depth of cut is 0.150 in Based on these values,can the job be performed on the

20 hp lathe? Use Table 21.3 to obtain the appropriate unithorsepower value.

Solution :

  From Table 21.3,

 HP u = 1.3 hp/(in3/min)Since the uncut chip thickness (0.030 in) is different from the tabular value of 0.010, a correctionfactor must be applied From Figure 21.14, the correction factor is 0.7 Therefore, the corrected

 E = 87%, available horsepower = 0.87(20) = 17.4 hp  

Since required horsepower exceeds available horsepower, the job cannot be

accomplished on the 20hp lathe, at least not at the specified cutting speed of 375 ft/min.

21.23: Suppose the cutting speed in Problems 21.7 and 21.8 is cutting speed = 200 ft/min From youranswers to those problems, find (a) the horsepower consumed in the operation, (b) metal removalrate in in3/min, (c) unit horsepower (hp-min/(in3), and (d) the specific energy (in-lb/in3).

Solution : (a) From Problem 21.8,

 F 

= 155 lb.

(d)U = 155(200)/2.88 = 10,764 ft-lb/in3 = 129,167 in-lb/in3

21.24 :For Problem 21.12, the lathe has a mechanical efficiency = 0.83 Determine (a) the horsepowerconsumed by the turning operation; (b) horsepower that must be generated by the lathe; (c) unithorsepower and specific energy for the work

material in this operation.

 = 1.0 hp/(in3/min) for stainless steel Since feed is lower than0.010 in/rev in

the table, a correction factor must be applied from Figure 21.14 For 

  Solve Problem 21.27 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 virtually the same as inthe previous problem, the metal removal rate is about 40% greater.

on horsepower, compute your best estimate of the cutting force for this turning operation.Use the orthogonal cutting model as an approximation of the turning  process.

horsepower for this material under these conditions, and (c) unit horsepower as it would belisted in Table 21.3 for a t oof 0.010 in Use the orthogonal cutting model

as an approximation of theturning process.

Solution :

  (a) Must find

 F c and   v to determine HP

 F c = 101 cos(31 - 6)/cos(32.5 + 31.0 – 6) = 170 lb.

 HP c = F cv/33,000 = 170(700)/33,000 = 3.61 hp.

(b)

 R MR = 700 x 12(0.0075)(0.075) = 11.3 in3/min

 HP u = HP c / R MR= 3.61/11.3 = 0.319 hp/(in3 /min)

(c) Correction factor = 0.85 from Fig 21.14 to account for the fact that

 f

= 0.015 in/rev instead of0.010 in/rev Taking this correction factor into account,

 HP u = 0.375/0.85 = 0.441 hp/(in3 /min)

as itwould appear in Table 21.3 for a feed (

t o) = 0.010 in/rev.

21.31:

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

Solution :

 ρ C = (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 = F cv/ R MR = 1175 N x 4.5 m/s/(4500 mm/s x 0.25 mm x 2.2 mm) = 2.135 m/mm3

N-T = 0.4U /( ρ C ) x (vt o/ K )0.333

T = 22 + (0.4 x 2.135 N-m/mm/(2.90x10-3) J/mm3-C)[4500 mm/s x 0.25 mm/80 mm2/s]0.333

T =

(70.4 250,000 /110 70 0.008 / 0. x ) x 14

= ( ) ( )0.333

Solution : From Table 21.3,

U for alloy steel (310 BHN) = 320,000 in-lb/in3.Since

 f = 0.005 in/rev, correction factor = 1.25.Therefore

U = 320,000(1.25) = 400,000 in-lb/in3.

v = 500 ft/min x 12 in/ft/60 sec/min = 100 in/sec.

T =T a + (0.4U /  ρ C )(vt o/ K )0.333 = 88 + (0.4 x 400,000/210)(100 x

0.005/0.16)0.333 =88 + (762)(3.125)0.333 = 88 + 1113 = 1201F

  21.35:

  An orthogonal machining operation removes metal at 1.8 in3/min The cutting force in the process =300 lb The work material has a thermal diffusivity = 0.18 in2/sec and a volumetricspecific heat =124 in-lb/in3 F If the feed  f =t o = 0.010 in and width of cut = 0.100 in, use the Cook formula tocompute the cutting

temperature in the operation given that ambient temperature = 70 ° F.

v =160 m/min,T = 592 ° C Determine an equation for temperature as a function of cutting speed that isin the form of the Trigger equation, Eq (21.23).

22, Turning and Related Operations

22.2. In a production turning operation, the foreman has decreed that a single pass must becompleted on the cylindrical workpiece in 5.0 min The piece is 400 mm long and 150 mm indiameter Using a feed = 0.30 mm/rev and a depth of cut = 4.0 mm, what cutting speed must beused to meet this machining time requirement?

22.3. A facing operation is performed on an engine lathe The diameter of the cylindrical part is 6

in and the length is 15 in The spindle rotates at a speed of 180 rev/min.Depth of cut= 0.110in,and feed=0.008in/ rev Assume the cutting tool moves from the outer diameter of the workpiece

to exactly the centerata constant velocity Determine (a) the velocity of the tool as it moves fromthe outer diameter towards the center and (b) the cutting time

Answer  

a diameter of 4.25in one pass at as peed of 450ft/min.Themetal removal rate should be 6.75

22.11. A two-spindle drill simultaneously drills a 1/2 in hole and a 3/4 in hole through a

workpiece that is 1.0 in thick Both drills are twist drills with point angles of 118o

 .Cutting speedfor the material is 230 ft/min The rotational speed of each spindle can be set individually Thefeed rate for both holes must be set to the same value because the two spindles lower at the samerate The feed rate is set so the total metal removal rate does not exceed 1.50 in3/ min Determine(a) the maximum feed rate (in/ min) that can be used, (b) the individual feeds (in/ rev) that resultfor each hole, and (c) the time required to drill the holes

Answer

Determine the time required from the beginning of the first hole to the completion of the lasthole, assuming the most efficient drilling sequence will be used to accomplish the job.

A 0.5 0.75 tan 90 50 0.315 in

T 0.5 0.315 1.75 / 22.916 0.112 minTime to retract drill from hole 0.112 / 2 0.056 min

p

All moves between holes are at a distance = 1.5 in using back and forth path between rows ofholes Time to move between holes =1.15/1.15=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

22.13. A gun-drilling operation is used to drill a 9/64in diameter hole to a certain depth It takes4.5 minutes to perform the drilling operation using high pressure fluid delivery of coolant to thedrill point The current spindle speed = 4000 rev/min, and feed = 0.0017 in/rev In order toimprove the surface finish in the hole, it has been decided to increase the speed by 20% anddecrease the feed by 25% How long will it take to perform the operation at the new cuttingconditions?

Answer

r 

r  m

22.15. A face milling operation is used to machine 6.0mm from the top surface of a rectangular piece of aluminum 300 mm long by 125 mm wide in a single pass Thecutter follows a path that

is centered over the workpiece It has four teeth and is 150 mm in diameter Cutting speed = 2.8m/s, and chip load = 0.27 mm/tooth Determine (a) the actual machining time to make the passacross the surface and (b) the maximum metal removal rate during cutting

an overtravel distance is provided at the end of the pass equal to the cutter radius plus 0.5 in,what is the duration of the feed motion

0.006in/tooth,anddepthofcut=0.150 in Determine (a) the actual cutting time to make one passacross the surface and (b) the maximum metal removal rate during the cut (c) If an additionalapproach distance of 0.5 in is provided atthe beginning of the pass (before cutting begins), and anovertravel distance is provided at the end of the pass equal to the cutter radius plus 0.5 in, what isthe duration of the feed motion.

Machining and Turning Centers

22.21. A three-axis CNC machining center is tended by a worker who loads and unloads parts between machining cycles The machining cycle takes 5.75 min, and the worker takes 2.80 minusing a hoist to unload the part just completed and load and fixture the next part onto the

machine worktable A proposal has been made to install a twoposition pallet shuttle at the

machine so that the worker and the machine tool can perform their respective tasks

simultaneously rather than sequentially The pallet shuttle would transfer the parts between themachine work table and the load/ unload station in 15 sec Determine (a) the current cycle timefor the operation and (b) the cycle time if the proposal is implemented What is the percentageincrease in hourly production rate that would result from using the pallet shuttle?

Answer

c c

 p  p

a T 5.75 2.8 8.55 min

 b T Max 5.75, 2.8 0.25 6 min

c The current hourly production rate R 60 / 8 55 7.02 pc / hr  The production rate u nder the proposal R 60 / 0.6 10 pc / hr  This is an increase of 10 7.02 42.5%

Other Operations

and has a tensile strength of 270 MPa and a Brinell hardness of 165 HB The starting dimensions

of the part are 750 mm x 450 mm x 50 mm The cutting speed is 0.125 m/sec and the feed is 0.40mm/pass The shaper ram is hydraulically driven and has a return stroke time that is 50% of thecutting stroke time An extra 150 mm must be added before and after the part for accelerationand deceleration to take place Assuming the ram moves parallel to the long dimension of the part, how long will it take to machine?

Answer

  Time per forward stroke =

150 750 100

8 s125

Time per reverse stroke =

0.4 6 2.4 s

Total time per pass = 8+2.4 = 10.4s = 0.17 min  Number of passes = 450/0.5 = 900 passes

  Total time

m

T 900 0.17 153 min

23 CUTTING-TOOL TECHNOLOGY

PROBLEMS

Tool Life and the Taylor Equation

23.1Flank wear data were collected in a series of turning tests using a coated carbide tool on hardenedalloy steel at a feed of 0.30 mm/rev and a depth of 4.0 mm At a speed of 125 m/min, flank wear = 0.12

mm at 1 min, 0.27 mm at 5 min, 0.45 mm at 11 min, 0.58 mm at 15 min, 0.73 at 20 min, and 0.97 mm at Problems 581 E1C23 11/09/2009 16:40:55 Page 582 25 min At a speed of 165 m/min, flank wear= 0.22

mm at 1 min, 0.47 mm at 5 min, 0.70 mm at 9 min, 0.80 mm at 11 min, and 0.99 mm at 13 min The lastvalue in each case is when final tool failure occurred (a) On a single piece of linear graph paper, plot flank wear as a function of time for both speeds Using 0.75 mm of flank wear as the criterion of tool failure, determine the tool lives for the two cutting speeds (b) On a piece of natural log-log paper, plotyour results determined in the previous part From the plot, determine the values of n and C in the TaylorTool Life Equation (c) As a comparison, calculate the values of n and C in the Taylor equation solvingsimultaneous equations Are the resulting n and C values the same? 

determine the values of n and C in the Taylor Tool Life Equation (c) As a comparison, calculate the values

of n and C in the Taylor equation solving simultaneous equations Are the resulting n and C values thesame? 

C=291.15m/min

23.5 Tool life tests on a lathe have resulted in the following data: (1) at a cutting speed of 375 ft/ min,the tool life was 5.5 min; (2) at a cutting speed of 275 ft/min, the tool life was 53 min (a) Determine the parameters n and C in the Taylor tool life equation (b) Based on the n and C values, what is the likely tool material used in this operation? (c) Using your equation, compute the tool life that corresponds to acutting speed of 300 ft/min (d) Compute the cutting speed that corresponds to a tool life T = 10 min.Solution :

Using the graph at 350 ft/min the tool last about 6.2min , at 450 ft/min in lasts 19 min

b) the points are graphed in excel and the line connecting the two points is extended to the axis

C is read from the Y-intercept and is approximately 680ft/min th slope , n can be determined bytaking the ln of the x and y coordinates of any 2 points and determining ∆Y/∆Y It is positivebecause the taylor tool life equation is deried assuming the slope is negative Using the points(1.68) and (19.350) the slope is about 0.226

c) Depending on the values of tool life read from the flank wear graph , the values of n and C willvary two equations 1) 350(19)n=C and 450(6.2)n

1 and 2

350(19)n=450(6.2)n

n=0.2241)C=6772)C=677

n=0.1971) C=5972)C= 597C=597

23 7Turning tests have resulted in 1-min tool life at a cutting speed = 4.0 m/s and a 20-min tool life at aspeed =2.0 m/s (a) Find the n and C values in the Taylor tool life equation (b) Project how long the toolwould last at a speed of 1.0 m/s

60(T)0.283=288

T=254 min23.8.  A 15.0-in 2.0-in-workpart is machined in a face milling operation using a 2.5-in diameter fly cutterwith a single carbide insert The machine is set for a feed of 0.010 in/tooth and a depth of 0.20 in If acutting speed of 400 ft/min is used, the tool lasts for three pieces If a cutting speed of 200 ft/min is used,the tool lasts for 12 parts Determine the Taylor tool life equation

Solution : a) two equation : 100(9)n=75(35)n

n = 0.212C=100(9)0.212

=159.3C= 75(35)0.212

= 159.4Use C=159

b) 110T0.212

=159T=5.7 min

c) ν(20)0.212=159

ν= 84m/min

23.9. In a production turning operation, the workpart is 125 mm in diameter and 300 mm long A feed of0.225 mm/rev is used in the operation If cutting speed = 3.0 m/s, the tool must be changed every fiveworkparts; but if cutting speed= 2.0 m/s, the tool can be used to produce 25 pieces between toolchanges Determine the Taylor tool life equation for this job

Use C=448 the tool life equation isν(T)0.149 =448

b)comparing these values of n and C with those in table 22.2 , the likely tool material is high speedsteel

c) at ν= 300ft/min , T=(C/ν)1/n=( 448/300)1/0.149=14.75min

d) for T=10 min , ν=C/Tn=318ft/min

23.10. For the tool life plot of Figure 23.5, show that the middle data point (v = 130 m/min, T = 12 min) isconsistent with the Taylor equation determined in Example Problem 23.1

Solution :

N1=ν/πD=500(12)/6.0π=318.3 rev/min

The tool life equation is vT0.267=769.8

23.11. In the tool wear plots of Figure 23.4, complete failure of the cutting tool is indicated by the end ofeach wear curve Using complete failure as the criterion of tool life instead of 0.50 mm flank wear, theresulting data are: (1) v = 160 m/min, T = 5.75 min; (2) v = 130 m/min, T = 14.25 min; and (3) v =100m/min, T = 47 min Determine the parameters n and C in the Taylor tool life equation for this data.Solution:

(1)Tm=π(102mm)(400mm)/(3*103 mm/s)(0.26mm)=164.33s = 2.739min

T=5(2.739)=13.69 min

1) v= 3m/s = 180m/min2) v = 2m/s = 160m/min

180(13.69)n=C

120(90.38)n=C

180(13.69)n=120(90.38)n

n=0.215C=316 m/min

23.12. The Taylor equation for a certain set of test conditions is vT.25 = 1000, where the U.S customaryunits are used: ft/min for v and min for T Convert this equation to the equivalent Taylor equation in theInternational System of units (metric), where v is in m/sec and T is in seconds Validate the metric