59920082 Force Power in Metal Cutting

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Force & Power in Metal Cutting

By S K MondalCompiled by: S K Mondal Made Easy

ESE‐2003‐ ConventionalDuring turning a carbon steel rod of 160 mm diameter by acarbide tool of geometry; 0, 0, 10, 8, 15, 75, 0 (mm) at speed of400 rpm, feed of 0.32 mm/rev and 4.0 mm depth of cut, thefollowing observation weremade.

Tangential component of the cutting force, Pz = 1200 NAxial component of the cutting force P = 800 NAxial component of the cutting force, Px = 800 NChip thickness (after cut),

For the abovemachining condition determine the values of(i) Friction force, F and normal force, N acting at the chip toolinterface.(ii) Yield shears strength of the work material under thismachining condition.(iii) Cutting power consumption in kW.

Ans F = 800 N N = 1200 N 256 7 Mpa 4 021 KW

=2 0.8mm.α

Compiled by: S K Mondal Made Easy

GATE – 1995 ‐ConventionalWhile turning a C‐15 steel rod of 160 mm diameter at315 rpm, 2.5 mm depth of cut and feed of 0.16mm/rev by a tool of geometry 00, 100, 80, 90,150, 750,0(mm), the following observationsweremade.Tangential component of the cutting force = 500 N

Axial component of the cutting force = 200 NChip thickness = 0.48 mm

Draw schematically the Merchant’s circle diagramfor the cutting force in the present case.

Ans. F = 284 N, N = 457.67 N, Fn = 348.78 N, Fs = 410.31 NFriction angle = 32oCompiled by: S K Mondal Made Easy

ESE ‐2000 (Conventional)The following data from the orthogonal cutting testis available. Rake angle = 100, chip thickness ratio =0.35, uncut chip thickness = 0.51, width of cut = 3mm, yield stress of work material = 285 N/mm2,mean friction co‐efficient on tool force = 0.65,5,Determine

(i) Cutting force (Fc)(ii) Radial force (Ft)(iii) Normal force (N) on tool and(iv) Shear force on the tool (Fs ).

Ans. Fc = 1597 N; Ft = 678 N; Fs = 1265 N; F = 944.95 N, N =1453.8 N


ESE‐2005 ConventionalMild steel is being machined at a cuttingspeed of 200 m/min with a tool rake angle of10. The width of cut and uncut thickness are 2mm and 0.2 mm respectively. If the averagep y gvalue of co‐efficient of friction between thetool and the chip is 0.5 and the shear stress ofthework material is 400 N/mm2,

Determine (i) shear angle and [Ans. 36.7o

(ii)Cutting and thrust component of themachine on force. [Ans. Fc = 420 N, Ft = 125 N ]


IAS‐2003 Main ExaminationDuring turning process with 7 ‐ ‐ 6 – 6 – 8 – 30 – 1 (mm)ASA tool the undeformed chip thickness of 2.0 mm andwidth of cut of 2.5 mm were used. The side rake angle ofthe tool was a chosen that the machining operationcould be approximated to be orthogonal cutting Thecould be approximated to be orthogonal cutting. Thetangential cutting force and thrust force were 1177 N and560 N respectively. Calculate:[30 marks](i) The side rake angle [Ans. 12o ](ii) Co‐efficient of friction at the rake face [Ans. 0.82](iii) The dynamic shear strength of the work material

[Ans. 74.43 Mpa] Compiled by: S K Mondal Made Easy

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GATE‐2006 Common Data Questions(1)In an orthogonal machining operation:Uncut thickness = 0.5 mm Cutting speed = 20 m/min Rake angle = 15°Width of cut = 5 mm Chip thickness = 0.7 mmWidth of cut 5 mm Chip thickness 0.7 mmThrust force = 200 N Cutting force = 1200 NAssume Merchant's theory.The coefficient of friction at the tool‐chip interface is (a) 0.23 (b) 0.46 (c) 0.85 (d) 0.95

Ans. (b)Compiled by: S K Mondal Made Easy

GATE‐2006 Common Data Questions(2)In an orthogonal machining operation:Uncut thickness = 0.5 mm Cutting speed = 20 m/min Rake angle = 15°Width of cut = 5 mm Chip thickness = 0.7 mmWidth of cut 5 mm Chip thickness 0.7 mmThrust force = 200 N Cutting force = 1200 NAssume Merchant's theory.The percentage of total energy dissipated due to friction at the tool‐chip interface is

(a) 30% (b) 42% (c) 58% (d) 70% Ans. (a)


GATE‐2006 Common Data Questions(3)In an orthogonal machining operation:Uncut thickness = 0.5 mm Cutting speed = 20 m/min Rake angle = 15°Width of cut = 5 mm Chip thickness = 0.7 mmWidth of cut 5 mm Chip thickness 0.7 mmThrust force = 200 N Cutting force = 1200 NAssume Merchant's theory.The values of shear angle and shear strain, respectively, are

(a) 30.3° and 1.98 (b) 30.3° and 4.23 (c) 40.2° and 2.97 (d) 40.2° and 1.65 Ans. (d)


GATE‐2003 Common Data Questions(1)A cylinder is turned on a lathe with orthogonalmachining principle. Spindle rotates at 200 rpm. Theaxial feed rate is 0.25 mm per revolution. Depth of cut is0.4 mm. The rake angle is 10°. In the analysis it is foundth t th h l i °that the shear angle is 27.75°

The thickness of the produced chip is(a) 0.511 mm (b) 0.528 mm (c) 0.818 mm (d) 0.846 mmAns. (a)


GATE‐2003 Common Data Questions(2)A cylinder is turned on a lathe with orthogonalmachining principle. Spindle rotates at 200 rpm. Theaxial feed rate is 0.25 mm per revolution. Depth of cut is0.4 mm. The rake angle is 10°. In the analysis it is foundth t th h l i °that the shear angle is 27.75°In the above problem, the coefficient of friction at the chip tool interface obtained using Earnest and Merchant theory is (a) 0.18 (b) 0.36 (c) 0.71 (d) 0.908

Ans. (d)Compiled by: S K Mondal Made Easy

GATE‐2008 Common Data Question (1)Orthogonal turning is performed on a cylindrical workpiece with shear strength of 250 MPa. The followingconditions are used: cutting velocity is 180 m/min. feedis 0.20 mm/rev. depth of cut is 3 mm. chip thicknessti Th th l k l i o A lratio = 0.5. The orthogonal rake angle is 7o. Apply

Merchant's theory for analysis.The shear plane angle (in degree) and the shear force respectively are (a) 52: 320 N (b) 52: 400N (c) 28: 400N (d) 28:320N Ans. (d)


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GATE‐2008 Common Data Question (2)Orthogonal turning is performed on a cylindrical workpiece with shear strength of 250 MPa. The followingconditions are used: cutting velocity is 180 m/min. feedis 0.20 mm/rev. depth of cut is 3 mm. chip thicknessti Th th l k l i o A lratio = 0.5. The orthogonal rake angle is 7o. Apply

Merchant's theory for analysis.The cutting and frictional forces, respectively, are (a) 568N; 387N (b) 565N; 381N (c) 440N; 342N (d) 480N; 356NAns. (b)


IES 2010The relationship between the shear angle Φ,the friction angle β and cutting rake angle αis given as

Ans. (b) Compiled by: S K Mondal Made Easy

IES‐2005Which one of the following is the correctexpression for the Merchant's machinabilityconstant?(a) 2φ γ α+(a)(b)(c)(d)(Where = shear angle, = friction angleand = rake angle) Ans. (a)

2φ γ α+ −2φ γ α− +

2φ γ α− −

φ γ α+ −

φ γα


GATE‐1997In a typical metal cutting operation, using a cutting tool of positive rake angle = 10°, it was observed that the shear angle was 20°. The friction angle is g(a) 45° (b) 30°(c) 60° (d) 40°

Ans. (c)


IAS – 1999In an orthogonal cutting process, rake angle of thetool is 20° and friction angle is 25.5°. UsingMerchant's shear angle relationship, the value ofshear angle will be(a) 39 5° (b) 42 25°(a) 39.5 (b) 42.25(c) 47.75° (d) 50.5°

Ans. (b)


IES‐2003In orthogonal cutting test, the cutting force =900 N, the thrust force = 600 N and chipshear angle is 30o. Then the chip shear force is(a) 1079 4 N (b) 969 6 N(a) 1079.4 N (b) 969.6 N(c) 479.4 N (d) 69.6 N

Ans. (c)


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IES‐2000In an orthogonal cutting test, the cutting force and

thrust force were observed to be 1000N and 500 N

respectively. If the rake angle of tool is zero, the

coefficient of friction in chip‐tool interface will be

( ) ( ) ( ) ( )1 1a b 2 c d 2 2 2

Ans. (a)Compiled by: S K Mondal Made Easy

IES‐1996Which of the following forces are measured directly bystrain gauges or force dynamometers during metalcutting ?1. Force exerted by the tool on the chip acting normally tothe tool face.2. Horizontal cutting force exerted by the tool on the workpiece.3. Frictional resistance of the tool against the chip flowacting along the tool face.4. Vertical force which helps in holding the tool inposition.(a) 1 and 3 (b) 2 and 4(c) 1 and 4 (d) 2 and 3 Ans. (b)Compiled by: S K Mondal Made Easy

GATE‐2007In orthogonal turning of medium carbon steel. The specific machining energy is 2.0 J/mm3. The cutting velocity, feed and depth of cut are 120 m/min, 0.2 mm/rev and 2 mm respectively. The main cutting f i N iforce in N is(a) 40 (b) 80 (c) 400 (d) 800

Ans. (d)


GATE‐2007In orthogonal turning of low carbon steel pipe withprincipal cutting edge angle of 90°, themain cuttingforce is 1000 N and the feed force is 800 N. The shearangle is 25° and orthogonal rake angle is zero.E l i M h t’ th th ti f f i tiEmploying Merchant’s theory, the ratio of frictionforce to normal force acting on the cutting tool is(a) 1.56 (b) 1.25(c) 0.80 (d) 0.64

Ans. (c)


IES‐1997Consider the following forces acting on afinish turning tool:1. Feed force2 Thrust force2. Thrust force3. Cutting force.The correct sequence of the decreasing order ofthe magnitudes of these forces is(a) 1, 2, 3 (b) 2, 3, 1(c) 3, 1, 2 (d) 3, 2, 1 Ans. (c)


IES‐1999The radial force in single‐point tool duringturning operation varies between(a) 0.2 to 0.4 times the main cutting force(b) 0 4 to 0 6 times the main cutting force(b) 0.4 to 0.6 times the main cutting force(c) 0.6 to 0.8 times the main cutting force(d) 0.5 to 0.6 times the main cutting force

Ans. (a)


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IES‐1995The primary tool force used in calculatingthe total power consumption in machining isthe(a) Radial force (b) Tangential force(a) Radial force (b) Tangential force(c) Axial force (d) Frictional force

Ans. (b)


IES‐2002In a machining process, the percentage ofheat carried away by the chips is typically(a) 5% (b) 25%(c) 0% (d) %(c) 50% (d) 75%

Ans. (d)


IES‐1998In metal cutting operation, the approximateratio of heat distributed among chip, tooland work, in that order is(a) 80: 10: 10 (b) 33: 33: 33(a) 80: 10: 10 (b) 33: 33: 33(c) 20: 60: 10 (d) 10: 10: 80

Ans. (a)


IAS – 2003As the cutting speed increases(a) More heat is transmitted to the work piece and less heat is transmitted to the tool(b) More heat is carried away by the chip and less heat is t itt d t th t ltransmitted to the tool(c) More heat is transmitted to both the chip and the tool(d) More heat is transmitted to both the work piece and the tool


IES‐2001Power consumption in metal cutting ismainly due to(a) Tangential component of the force(b) Longitudinal component of the force(b) Longitudinal component of the force(c) Normal component of the force(d) Friction at the metal‐tool interface

Ans. (a)


IAS – 1995Thrust force will increase with the increase in(a) Side cutting edge angle(b) Tool nose radius (c) Rake angle(d) End cutting edge angle

Ans. (a)


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IES 2010Consider the following statements:In an orthogonal, single‐point metal cutting,as the side‐cutting edge angle is increased,

1. The tangential force increases.g2. The longitudinal force drops.3. The radial force increases.Which of these statements are correct?(a) 1 and 3 only (b) 1 and 2 only(c) 2 and 3 only (d) 1, 2 and 3 Ans. (c)


IES‐1993A 'Dynamometer' is a device used for themeasurement of(a) Chip thickness ratio(b) Forces during metal cutting(b) Forces during metal cutting(c) Wear of the cutting tool(d) Deflection of the cutting tool

Ans. (b)


IES 2011The instrument or device used to measure the cutting forces in machining is :(a) Tachometer(b) Comparator(c) Dynamometer(d) Lactometer

Ans. (c)


IAS – 2003The heat generated in metal cutting canconveniently be determined by(a) Installing thermocouple on the job(b) Installing thermocouple on the tool(c) Calorimetric set‐up(d) Using radiation pyrometer

Ans. (c)


IES‐1998The gauge factor of a resistive pick‐up ofcutting force dynamometer is defined as theratio of(a) Applied strain to the resistance of the wire(a) Applied strain to the resistance of the wire(b) The proportional change in resistance to theapplied strain(c) The resistance to the applied strain(d) Change in resistance to the applied strain


IES‐2000Assertion (A): In metal cutting, the normallaws of sliding friction are not applicable.Reason (R): Very high temperature isproduced at the tool‐chip interfaceproduced at the tool‐chip interface.(a) Both A and R are individually true and R isthe correct explanation of A(b) Both A and R are individually true but R isnot the correct explanation of A(c) A is true but R is false(d) A is false but R is true Ans. (b)Compiled by: S K Mondal Made Easy

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GATE 1992The effect of rake angle on the mean friction angle inmachining can be explained by(A) sliding (Coulomb) model of friction(B) sticking and then sliding model of friction(C) sticking friction(D) Sliding and then sticking model of friction

Ans. (b)


IES‐2004Assertion (A): The ratio of uncut chip thickness toactual chip thickness is always less than one and istermed as cutting ratio in orthogonal cuttingReason (R): The frictional force is very high due to theoccurrence of sticking friction rather than slidingg gfriction(a) Both A and R are individually true and R is the correctexplanation of A(b) Both A and R are individually true but R is not thecorrect explanation of A(c) A is true but R is false(d) A is false but R is true Ans. (b)


GATE‐1993The effect of rake angle on themean friction angle inmachining can be explained by(a) Sliding (coulomb) model of friction(b) sticking and then siding model of frictiong g(c) Sticking friction(d) sliding and then sticking model of friction

Ans. (d)

Compiled by: S K Mondal Made Easy Compiled by: S K Mondal Made Easy

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59920082 Force Power in Metal Cutting