Production Question and Answer Set 1

7/12/2011

Q. Wh Why y ev even en a bat batte tery ry operat operated ed pencil sharpener cannot be accepted as a machine tool?

Theory of Theory of Metal Metal Cutting

y

ns. n sp te o

av ng ng a ot er ma or eatu eature ress o

machine machine tools, tools, the sharpener sharpener is of low value. value.

By S K Mondal Compiled By: S K Mondal

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Compiled By: S K Mondal

IES‐2001 For cutting of brass with single point cutting tool on a lathe, tool tool should should have have (a) Negativ Negativee rake angle (b) Positiv Positivee rake angle (c) Zero Zero rake angle (d) (d) Zero Zero side relief angle Ans. (c)

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IES‐1995

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IES‐1993

GATE‐1995; 2008

Assertion (A): For a negative rake tool, the specific cutting cutting pressure pressure is smaller smaller than for a positive positive rake tool under under otherwise otherwise identical identical conditions conditions.. Reason (R): The shear strain undergone by the chip . (a) (a) Both Both A and and R are are indi indivi vidu dual ally ly true true and and R is the correc correctt explanati explanation on of A (b) Both A and R are individu individually ally true true but R is not the correc correctt explanati explanation on of A (c) A is true true but R is false (d) A is false false but R is true true Ans. (d)

Cutting power consumption in turning can be significantly reduced significantly reduced by (a) Increasing rake angle of the of the tool (b) Increasing the cutting angles of the of the tool (c) Widening the nose radius of the of the tool (d) Increasing the clearance angle Ans. (a) Compiled By: S K Mondal

Single point thread cutting tool should ideally have: a) Zero rake b) Positive rake c) Negative rake d) Normal rake Ans. (a)

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Page 1 of 79

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IES – IES – 2005

IES – IES – 2002

Assertion (A): Carbide tips are generally given negative negative rake rake angle. angle. Reason (R): Carbide tips are made from very hard materials. a ot an are n v ua y true an s t e correct correct explanati explanation on of A (b) Both A and R are individua individually lly true but but R is not the correct correct explanati explanation on of A (c) A is true true but R is false false (d) A is false but R is true true Ans. (b) Compiled By: S K Mondal Made Easy

Assertion (A): Negative rake is usually provided on carbide carbide tipped tools. Reason (R): Carbide tools are weaker aker in compression. a ot an are n v ua y true an s t e correc correctt explanatio explanation n of A (b) Both A and R are individu individually ally true true but R is not the correc correctt explanatio explanation n of A (c) A is true true but R is false (d) A is false false but R is true Ans. (c) Compiled By: S K Mondal Made Easy

IES 2011

IAS – IAS – 1994

Which one of the following statement is NOT correct correct with reference reference to the purposes and effects of rake angle of a cutting cutting tool? tool? (a) To guide the chip flow directio direction n (b) To reduce the friction between the tool flanks and the machined surface (c) To To add keenness keenness or sharpness sharpness to the cutting cutting edges. edges. (d) To To provide better thermal efficiency. efficiency. Ans. (b) Compiled By: S K Mondal

Consider the following characteristics 1. The cutting edge is normal to the cutting velocity cutting velocity.. 2. The cutting forces occur in two directions only. 3. The cutting edge is wider is wider than the depth of cut. of cut. The characteristics applicable to orthogonal cutting would include (a) (a) 1 and 2 (b) 1 and 3 (c) (c) 2 and 3 (d) 1, 1, 2 and 3 Ans. (d)

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IES‐2006

IES‐1995

Which of the of the following is a single point cutting tool? (a) Hacksaw Hacksaw blade b Mill Millin in cutter (c) Grinding Grinding wheel wheel (d) Parting Parting tool Ans. (d)

The angle between the face and the flank of the single single point point cuttin cutting g tool tool is known known as a) Rake angle angle Clearance angle b) Clearanceangle c) Lip angle angle Point angle angle d) Point Ans. (c)


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IES‐2006

Assertion (A): For drilling cast iron, the tool is prov provid ided ed with with a poin pointt angl angle e smal smaller ler than than that that required required for a ductile ductile material. material. Reason Reason (R): (R): Smalle Smallerr point point angle angle result resultss in lower lower ra e ang ang e. e. (a) Both Both A and R are are indivi individua dually lly true and R is the correc correctt explanati explanation on of A (b) Both A and R are individually true but R is not the correc correctt explanati explanation on of A (c) A is true but R is false (d) A is false but R is true Ans. (c) Compiled By: S K Mondal Made Easy


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IES‐1995

IES‐2009

Consid Consider er the follow following ing statem statement entss about about nose nose radius 1. It improves improves tool life 2. It reduces reduces the cutting cutting force force 3. It improves improves the surface surface finish. Select Select the correc correctt answer answer using the codes codes given given below: below: (a) 1 and 2 (b) 2 and 3 (c) 1 and 3 (d) 1, 2 and 3 Ans. (c)

Consider the following statements with respect to the effect effectss of a large large nose nose radius radius on the tool: tool: 1. It deteri deteriora oratessurfa tessurface ce finish. finish. 2. It increa increases ses the possib possibili ility ty of chatt chatter er.. 3. It improves tool life. Which of the above statements is/are correct? (a) 2 only (b) 3 only (c) (c) 2 and 3 only nly (d) 1, 2 and 3 Ans. (c) Compiled By: S K Mondal

IES‐2002

Consider Consider the following following statements statements:: The strength of a single point cutting tool depends upon 1. Rake Rake angle 2. Clearance Clearance angle 3. Lip angle Which of these statements are correct? (a) 1 and 3 (b) 2 and 3 (c) 1 and 2 (d) 1, 2 and 3 Ans. (d)

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IES‐2009

IES‐1994

Tool geometry of a single point cutting tool is specified by the following following elements: 1. Backra Backrakean keang gle 2. Siderak Siderakee angl anglee 3. Endcutting Endcutting edge edge angle angle 4. Side Side cuttin cutting g edge edge angle angle 5. Side Side relie relieff angle angle 6. End relief angle 7. Nose Nose radi radius us The corr correct ect seque sequenc ncee of these these tool tool eleme elements nts used used for for correctly correctly specifying the tool geometry is (a) (a) 1,2, 1,2,3, 3,6 6,5,4 ,5,4,7 ,7 (b) (b) 1,2,6 ,2,6,5 ,5,3 ,3,4 ,4,7 ,7 (c) 1,1,2,5,6,3,4,7 Compiled By: S K(Mondal d) 1, 2, Made 6, 3Easy , 5, 4,7 Ans. (b)

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The following following tool signature signature is specified specified for a single single point cutting cutting tool tool in American American system: system: 10, 12, 8, 6, 15, 20, 3 What does the angle 12 represent? (a) Side Side cuttin cutting g‐edge edge angle (b) Side rake rake angle angle (c) Back rake rake angle angle (d) Side clearanc clearancee angle angle Ans. (b) Compiled By: S K Mondal

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Page 3 of 79

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IES‐1993

GATE‐1995

In ASA System, if the tool nomenclature is 8 6 5 5 10 15 2 mm, then then the siderake angle angle will will be (a) 5° (b) 6° (c) 8° (d) 10° Ans. (b) ‐

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Plain milling of mild of mild steel plate produces (a) Irregular shaped discontinuous chips (b) Regular shaped discontinuous chip c on nuous nuous c ps w ps w ou u up e ge (d) Joined (d) Joined chips Ans. (b)

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IES 2007

GATE‐2002

During machining, excess metal is removed in the form of chip of chip as in the case of turning of turning on a lathe. Which lathe. Which of the of the following are correct? Continuous ribbon like chip is formed when formed when turning 1. At a hi her cuttin s eed eed 2. At a lower cutting speed 3. A brittle A brittle material 4. A ductile A ductile material Select the correct answer using the code given below: (a) 1 and 3 (b) 1 and 4 (c) 2 and 3 (d) 2 and 4 Compiled By: S K Mondal Made Easy Ans. (b)

A built A built up edge is formed while formed while machining ‐

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(a) Ductile materials at high speed (b) Ductile materials at low speed (c) Brittle materials at high speed (d) Brittle materials at low speed Ans. (b)


IES‐1997

Assertion (A) : For high speed turning of cast iron pistons, carbide carbide tool bitsare provide provided d with chipbreakers. chipbreakers. Reason (R): High speed turning may produce long, ribbon type continuous chips which must be broken into small lengths which otherwise would be difficult to handle and mayprove hazardo hazardous. us. a Bot an R ar are in iv ivi ua ua y t ru rue an an R is is t e correct explanation explanation of A (b) Both A and R are are individual individually ly true but but R is not the correct explanation explanation of A (c) A is is true true but R is false (d) A is false but R is true true

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GATE‐2001 During orthogonal cutting of mild steel with a 10° rake angle tool, the chip thickness ratio was obtained as 0.4. The shear angle (in de rees rees evaluat evaluated ed from from this data data is (a) 6.53 (b) 20.22 (c) 22.94 (d) 50.00 Ans. (c)

Ans. (d)


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IES‐1994

GATE 2011

The foll follow owing ing param paramet eters ers determ determine ine the model of continuous continuous chip formation: formation: 1. True rue feed feed 2. Cutt Cuttin in veloc velocit it 3. Chip Chip thickn thickness ess 4. Rake Rake angle angle of the the cuttin cutting g tool. tool. The parameters which govern the value of shear angle would would include (a) (a) 1,2 1,2 and and 3 (b) (b) 1,3 1,3 and and 4 (c) (c) 1,2 1,2 and and 4 (d) (d) 2,3 2,3 and and 4 Ans. (b)

A single – point cutting tool with 12° rake angle is used to machine a steel work – piece. The depth of cut, cut, i.e. i.e. uncu uncutt thic thickn knes esss is 0.81 0.81 mm. mm. The The chip chip thickness under orthogonal machining condition is . . (a) 22° (b) 26° (c) 56° (d) 76° Ans. (b) Compiled By: S K Mondal

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mpiled mpi ledBy: By: S K Mondal

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IES 2004

GATE 2009 ‐

Minimum shear strain in orthogonal turning with a cutting tool of zero rake angle is (a) 0.0 b 0. (c) 1.0 (d) 2.0

In a machining operation chip thickness ratio is 0.3 and the back rake angle of the tool is 10°. What is the value of the shear strain? a 0. 1 b 0.1 (c) 3.00 (d) 3.34

Ans. (d)

Ans. (d) Compiled By: S K Mondal

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GATE‐2007 In orthogonal turning of a low carbon steel bar of diamet diameter er 150 mm with with uncoat uncoated ed carbid carbide e tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. . . orthogonal rake angle is zero and the principal cuttin cutting g edge edge angle angle is 90°, 90°, the shear angle angle is degree degree is (a) 20.56 .56 (b) 26.56 (c) 30.56 .56 (d) 36.56 Ans. (b) Compiled By: S K Mondal

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GATE‐2008

In a single point turning tool, the side rake angle and orthogonal rake angle are equal. φ is the principal cutting edge angle and its range is 0 ≤ φ ≤ 90 . The chip flows in the orthogonal plane. o

o

(a) 00

(b) 450

(c) 600

(d) 900

Ans. (d)

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IES‐2006

IES‐2004

Consider the following statements with statements with respect to the relief angle relief angle of cutting of cutting tool: 1. This affects the direction of chip of chip flow 2. This reduces excessive friction between the tool an wor p ece 3. This affects tool life 4. This allows better access of coolant of coolant to the tool work piece interface Which of the of the statements given above are correct? (a) 1 and 2 (b) 2 and 3 (c) (c) 2 and 4 (d) By: S K3Mondal and 4 Made Easy Ans. (b) Compiled

Consider the following statements: 1. A large A large rake angle means lower strength of the of the cutting edge. 2. Cutt Cuttin ing g torque decreases with decreases with rake angle. Which of the of the statements given above is/are correct? (a) Only Only 11 (b) Only 2 (c) Both Both 1 and 2 (d) Ne Neither 1 nor 2 Ans. (c)

IES‐2004

IES‐2003


Match. List I with List II and select the correct answer using the codes given below the Lists: List I List II A. Plan approach angle 1. Tool face B. Rake Rakea angle 2. Tool flank C. Clear Clearanc ance e angle 3. Tool face and flank D. Wedge edge angle 4. Cutting edge 5. Tool nose A B C D A B C D (a) 1 4 2 5 (b) 4 1 3 2 (c) 4

1

2

3

(d)

1

4

3

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The The angl angle e of incl inclin inat atio ion n of the the rake rake face face with with resp respec ectt to the the tool tool base base meas measur ured ed in a plan plane e perpendicular to the base and parallel to the width of the tool tool is called called (b) Side rake rake angle angle (c) Side cutting cutting edge edge angle (d) End cuttin cutting g edge edge angle angle Ans. (b)

5

Ans. (c) Compiled By: S K Mondal

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IES‐2004


The rake angle of a cutting tool is 15°, shear angle angle 45° 45° and cuttin cutting g velo velocit city y 35 m/mi m/min. n. What is the velocity of chip along the tool face? (a) (a) 28.5 28.5 m/mi m/min n (b) (b) 27.3 27.3 m/mi m/min n (c) (c) 25.3 25.3 m/mi m/min n (d) (d) 23.5 23.5 m/mi m/min n Ans. (a)


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[IES‐2008]

Consider the following statements: In an orthogonal cutting the cutting ratio is found to be 0 75. The cutting speed is 60 m/min and depth of cut of cut 2 4 mm. Which of the of the following are correct? 1. Chip velocity will will be 45 m/min. 2. Chip velocity will will be 80 m/min. 3. Chip thickness will thickness will be 1∙8 mm. 4. Chip Chipthickness thickness will will be 3∙2 mm. Select the correct answer using the code given below: (a) (a) 1 and 3 (b) 1 and 4 (c) (c) 2 and 3 (d) 2 and 4 Ans. (b)

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∙

∙


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IES‐2001

If α is the rake rake angle angle of of the cuttin cutting g tool, tool, φ is the shear angle and V is the cutting velocity, then the velocity of chip sliding along the shear plane is given by (a) (c)

V cosα

cos(φ − α ) V cosα

sin(φ − α )

Ans. (a)

(b) (d)

V sin φ

cos (φ − α ) V sin α

sin(φ − α )


IES‐2003

An orthogonal cutting operation is being carried out under the following conditions: cutting speed = 2 m/s, depth of cut = 0.5 mm, chip chip thic thickn knes esss = 0.6 0.6 mm. mm. Then Then the the chip chip velocity is (a) (a) 2.0 2.0 m/s m/s (b) (b) 2.4 2.4 m/s m/s (c) (c) 1.0 1.0 m/s m/s (d) (d) 1.66 1.66 m/s m/s Ans. (d)

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IAS‐1997

IAS‐2003

Consider the following machining conditions: BUE will BUE will form in (a) Ductile Ductile mate materi rial al.. (b) (b) High High cutting speed. (c) Small Small rake angle. (d) (d) Small all uncut chip thickness.

Ans. (a)

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In orthogonal cutting, shear angle is the angle between (a) Shear Shear plane and the cutting velocity cutting velocity (b) Shear Shear plane and the rake plane (c) Shear Shear plane and the vertical the vertical direction ear p ane an t e rect on o e ongat on o crysta s n the chip

Ans. (a) Compiled By: S K Mondal

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IAS‐2002

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IAS‐2000

Ans. (d) Ans. (a) Compiled By: S K Mondal

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Page 7 of 79

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IAS‐1998

IAS‐1995

The cutting velocity in m/sec, for turning a work piece of diamet diameter er 100 mm at the spindl spindle e speed speed of 480 RPM is (a) 1.2 1.26 6 (b) 2.51 (c) 48 (d) 151

In an orthogonal cutting, the depth of cut is halved and the feed rate is double. If the chip thickness ratio is unaffected unaffected with the changed changed cutting cutting conditio conditions, ns, the actualchip thickness thickness will be a) Doubled b) ha halved (c) (c) Quad Quadru rup pled (d) Unch nchang anged.

Ans. (b) Ans. (b) Compiled By: S K Mondal

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Page 8 of 79

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ESE‐2003‐ Conventional

During turning a carbon steel rod of 160 mm diameter by a carbide tool of geometry; 0, 0, 10, 8, 15, 75, 0 (mm) at speed of 400 rpm, feed of 0.32 mm/rev and 4.0 mm depth of cut, the following observation were made. Tangential Tangential component of the cutting force, force, Pz = 1200 N , x Chip thickness thickness (after cut),α = 0.8mm. Forthe above above machining machining conditi condition on determine determine thevalues of (i) Friction Friction force, force, F and normal normal force, force, N acting at the chip tool tool interface. (ii) (ii) Yield Yield shears shears strength strength of the work mater materia iall under under this this machining condition. (iii) Cutting power power consumption in in kW. kW. Compiled by: S K Mondal Made Easy = 8 27 =

Fo r c e & Po w e r i n M e t a l Cu Cu t t i n g

2

By S K Mondal Compiled by: S K Mondal

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GATE – GATE – 1995 ‐Conventional

ESE ‐2000 (Conventional)

While turning a C ‐15 steel rod of 160 mm diameter diameter at 315 rpm, 2.5 mm depth of cut and feed of 0.16 mm/rev by a tool of geometry 0 0, 100, 80, 90,150, 750, 0(mm), 0(mm), the following following observationswere observationswere made. Tangential angential component component of the cutting cutting force force = 500 N Axial component of the cutting force = 200 N Chip Chip thickn thickness ess = 0.48 mm Draw schematically the Merchant’s circle diagram forthe cuttin cutting g force force in the presen presentt case. case. Ans. F = 2 9 1 , N = 457.6 57.677 N, Fn = 3 5 5.78 N, Fs = 40 8.31 N .49o Compiled by: S K Mondal Made Easy Friction Friction angle = 32

The following data from the orthogonal cutting test is available. Rake angle = 10 0, chip thickness ratio = 0.35, uncut chip thickness = 0.51, width of cut = 3 mm, yield stress stress of work work material material = 285 N/mm2, mean mean fricti friction on co‐effic efficie ient nt on tool tool forc force e = 0.6 , Determine (i) Cutt Cutting ing force force (Fc) (ii) (ii) Radial Radial force force (Ft) (iii) Normal Normal force force (N) on on tool tool and (iv) (iv) Shear Shear force force on the tool tool (F s ). Ans. F c = 1597 N; Ft = 678 N; Fs = 1265 N; F = 944.95 N, N = 1453.8 N Compiled by: S K Mondal Made Easy

ESE‐2005 Conventional

IAS‐2003 Main Examination

Mild Mild steel steel is being being machi machined ned at a cuttin cutting g speed of 200 m/min with a tool rake angle of 10. The width of cut and uncut thickness are 2 mm and 0.2 mm res ecti ective vell . If the aver avera a e value of co‐efficient of friction between the tool and the chip is 0.5 and the shear stress of the work material material is 400 N/mm2, Determine Determine (i) (i) shear shear angle angle and [Ans. [Ans. 36.7 36.7o (ii)Cut (ii)Cuttin ting g and and thrus thrustt compo componen nentt of the machine machine on force. force. [Ans. Fc = 420 N, F t = 125 N ]

During turning process with 7 ‐ ‐ 6 – 6 – 8 – 30 – 1 (mm) ASA tool the undeformed chip thickness of 2.0 mm and width of cut of 2.5 mm were used. The side rake angle of the tool was a chosen that the machining operation . tangential cutting force and thrust force were 1177 N and 560 N respectively. Calculate: [30 marks] (i) Theside rake rake angle angle [Ans [Ans.. 12 12o ] (ii) (ii) Co‐efficient of friction at the rake face [Ans. 0.82] 0.82] (iii) The dynamic dynamic shear shear strength strength of the work work materia materiall [Ans. 74.43 Mpa] Compiled by: S K Mondal Made Easy

Compiled by: S K Mondal

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Page 9 of 79

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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 of cut = mm Chi thickness = 0. mm Thrust force = 200 N Cutting force = 1200 N Assume Merchant's theory. The coefficient of friction 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

In an orthogonal machining operation: Uncut thickness = 0.5 mm Cutting speed = 20 m/min Rake angle = 15° Width of cut of cut = mm Chi thickness = 0. mm Thrust force = 200 N Cutting force = 1200 N Assume Merchant's theory. The percentage of total of total energy dissipated energy dissipated due to friction at the tool‐chip interface is (a) 30% (b) 42% (c) 58% (d) 70% Ans. (a)

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In an orthogonal machining operation: Uncut thickness = 0.5 mm Cutting speed = 20 m/min Rake angle = 15° Width of cut of cut = mm Chi thickness = 0. mm Thrust force = 200 N Cutting force = 1200 N Assume Merchant's theory. The values The values of shear of shear angle and shear strain, respectively, are (a) 30.3° 30.3° and 1.98 (b) 30.3° 30.3° and 4.23 (c) 40.2° and 2.97 (d) 40.2° and 1.65 Ans. (d) Compiled by: S K Mondal

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A cylinder is turned on a lathe with orthogonal orthogonal machining machining principle. principle. Spindle Spindle rotates rotates at 200 rpm. The axial feed rate is 0.25 mm per revolution. Depth of cut is 0.4 mm. The rake angle angle is 10°. In the analysis it is found t at t e s ear ear ang ang e s 27.7 7.75 The thickness of the of the produced chip is (a) 0.511 mm (b) 0.528 mm (c) 0.818 mm (d) 0.846 mm Ans. (a)

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GATE‐2008 Common Data Question (1)

A cylinder is turned on a lathe with orthogonal orthogonal machining machining principle. principle. Spindle Spindle rotates rotates at 200 rpm. The axial feed rate is 0.25 mm per revolution. Depth of cut is 0.4 mm. The rake angle angle is 10°. In the analysis analysis it is found t att es earang e s 27.75 In the above problem, the coefficient of friction of friction at the chip tool interface obtained using Earnest and Merchant theory is theory is (a) 0.18 (b) 0.36 (c) 0.71 (d) 0.908 Ans. (d) Compiled by: S K Mondal Made Easy

Orthogonal turning is performed on a cylindrical work piece piece with shear strength strength of 250 MPa. The following following conditions are used: cutting velocity is 180 m/min. feed is 0.20 mm/rev. depth of cut is 3 mm. chip thickness rat o = 0.5. e or ort og ogona ra e an ang e s 7 . pp y Merchant's Merchant's theory for analysis. The shear plane angle (in degree) and the shear force respectively are respectively are (a) 52: 320 N (b) 52: 400N (c) 28: 400N (d) 28:320N Ans. (d) Compiled by: S K Mondal Made Easy

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IES 2010

GATE‐2008 Common Data Question (2)

The relationship between the shear angle Φ, the friction angle β and cutting rake angle α is give given n as

Orthogonal turning is performed on a cylindrical work piece piece with shear strength strength of 250 MPa. The following following conditions are used: cutting velocity is 180 m/min. feed is 0.20 mm/rev. depth of cut is 3 mm. chip thickness ra o = 0.5. e or or ogona ra e an ang e s 7 . pp y Merchant's Merchant's theory for analysis. The cutting and frictional forces, respectively, are (a) 568N; 387N (b) 565N; 381N (c) 440N; 342N (d) 480N; 356N Ans. (b) Compiled by: S K Mondal

Ans. (b)

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IES‐2005

GATE‐1997

Which one of the following is the correct expression for the Merchant's machinability constant?

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

−

(b) 2φ γ α (c) 2φ γ α (d) φ + γ − α (Wher (Wheree φ = shear shear angle, angle,γ = fric frictio tion n angle angle andα = rake rake angle) angle) Ans. (a) −

−

+

−


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Ans. (c)

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IAS – IAS – 1999

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IES‐2003

In an orthogonal cutting process, rake angle of the tool tool is 20° 20° and and fric fricti tion on angl angle e is 25.5 25.5°. °. Usin Using g Merchant' Merchant'ss shear angle relationsh relationship, ip, the value of shear shear angle angle will will be . ° . ° (c) 47.75° (d) 50.5°

In orthogonal cutting test, the cutting force = 900 900 N, the the thru thrust st forc force e = 600 600 N and and chip chip o shear angle angle is 30 . Then Then the chip chip shear shear force force is is . . (c) 479.4 N (d) 69.6 N Ans. (c)

Ans. (b) Compiled by: S K Mondal

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IES‐2000

IES‐1996 Which of the following following forces are measured directly by strain strain gauges gauges or force force dynamome dynamometers ters during metal metal cutting ? 1. Force Force exer exerted ted by the the tool tool on the chip chip acting acting normall normallyy to the tool face. 2. Horizontal Horizontal cutting cutting force force exerted exerted by the tool tool on the work piece. 3. Friction Frictional al resistanc resistancee of the tool tool against against the chip chip flow acting along the tool face. 4. Verti ertica call forc forcee whic which h help helpss in hold holdin ing g the the tool tool in position. (a) 1 and 3 (b) 2 and 4 (c) 1 and 4 (d) by: S K2Mondal and 3Made Easy Ans. (b) Compiled

In an ortho orthogo gonal nal cuttin cutting g test, test, the cutting cutting forc forcee and thrust thrust force force were were observe observed d to be 1000N 1000N and 500 N resp respec ecti tive vely ly.. If the the rake rake angl anglee of tool tool is zero zero,, the the coefficie coefficient nt of friction friction in chip‐tool interface interface will be

(a)

1 2

( b) 2

( c)

1 2

( d)

2

Ans. (a) Compiled by: S K Mondal

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GATE‐2007

GATE‐2007

In orthogonal turning of medium of medium carbon steel. The specific machining energy is energy is 2.0 J/mm 2.0 J/mm 3. The cutting velocity, velocity, feed and depth of cut of cut are 120 m/min, 0.2 mm/rev and mm/rev and 2 mm respectively. The main cutting orce n s (a) 40 (b) 80 (c) 400 (d) 800

In orthogonal turning of low carbon steel pipe with principal cutting edge angle angle of 90°, the main cutting force is 1000 N and the feed force is 800 N. The shear angle angle is 25° and orthog orthogona onall rake rake angle angle is zero. zero. mp oy ng erc ant ant’s t eory eory,, t e rat rat o o r ct on force force to normal normal force acting acting on the cutting cutting tool is (a) 1.56 (b) 1.25 (c) 0.80 (d) 0.64

Ans. (d) Ans. (c) Compiled by: S K Mondal

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IES‐1997

IES‐1999

Consi Conside derr the foll follow owing ing forc forces es actin acting g on a finish turning tool: 1. Feed Feed force force 2. rus orce 3. Cutting Cutting force. force. The correct sequence of the decreasing order of the magnitudes of these forces is (a) 1, 2, 3 (b) 2, 3, 1 (c) 3, 1, 2 (d) 3, 2, 1 Ans. (c) Compiled by: S K Mondal

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The radial force in single ‐point tool during turning operation varies between (a) 0.2 to 0.4 times the main cutting force 0.4 o 0. mes e ma n cu ng orc orce (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‐1995

IES‐2002

The primary tool force force used in calcul calculatin ating g the total power consumption in machining is the (c) (c) Axia Axiall forc forcee

In a machini machining ng proces process, s, the percentag percentage e of heat carried away away by the chips is typically typically (a) 5% (b) 25 25%

(d) (d) Fric Fricti tion onal al forc forcee Ans. (d)

Ans. (b)


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IAS – IAS – 2003

IES‐1998

As the cutting speed increases (a) More More heat is transmitted to the work the work piece and less heat is transmitted to the tool (b) More More heat is carried away by away by the the chip and less heat is transm tte to t e too (c) More More heat is transmitted to both the chip and the tool (d) More More heat is transmitted to both the work the work piece and the tool

In metal cutting operation, the approximate ratio ratio of heat distribut distributed ed among among chip, chip, tool tool and work, work, in that order is (c) 20: 20: 60: 10 (d) 10: 10: 80 Ans. (a)


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Ans. (b)

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IAS – IAS – 1995

IES‐2001 Power ower consu onsump mpti tion on in meta metall cutt cuttin ing g is mainly due to (a) Tangential Tangential component component of the force force ong ong u na cco omponen o e or orce (c) Normal Normal component component of the force force (d) Friction at the metal ‐tool interface

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


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IES 2010 Consider Consider the following statements: statements: In an orthogonal, orthogonal, single‐point metal cutting, as the the side side‐cutting edge angle angle is increased, increased, 1. The tan ential force increases. increases. 2. The longitudinal force drops. 3. The radial force increases. increases. Which of these statements are correct? (a) (a) 1 and and 3 only only (b) (b) 1 and and 2 only only (c) (c) 2 and and 3 only only (d) (d) 1, 2 and and 3 Ans. (c) Compiled by: S K Mondal

IES‐1993 A 'Dynamometer' is a device used for the measurement measurement of (a) Chip thickness ratio (c) Wear of the the cutting cutting tool (d) Deflection Deflection of the cutting cutting tool tool Ans. (b)

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IAS – IAS – 2003

IES 2011

The heat eat generated in meta etal convenie conveniently ntly be determined determined by (a) Installing thermocouple thermocouple on the job (b) Installing thermocoupl thermocouplee on the tool (c) Calorimet Calorimetric ric set ‐up (d) Using radiation pyrometer pyrometer

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

cutting

can

Ans. (c)

Ans. (c)


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IES‐1998

IES‐2000

The gaug gauge e fact factor or of a resi resisti stive ve pick pick‐up of cutting force dynamometer is defined as the ratio ratio of (b) The proportional change in resistance to the applied strain (c) The resistanc resistancee to the applied applied strain strain (d) Change in resistance to the the applied strain Ans. (b) Compiled by: S K Mondal

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Assertion (A): In metal cutting, the normal laws of sliding friction friction are not applicable. applicable. Reas Reason on (R): (R): Very ery hig high temp temper erat atur ure e is . ‐ (a) Both A and R are individually true and R is the correct correct explanation explanation of A (b) Both A and R are individually true but R is not the correct correct explanatio explanation n of A (c) A is true but R is false (d) A is false but R is true Ans. (a) mpiled by: S K Mondal

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IES‐2004

GATE 1992 The effect effect of rake rake angl anglee on the mean fricti friction on angle angle in machining 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


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Assertion (A): The ratio of uncut chip thickness to actual chip thickness is always less than one and is termed termed as cutting cutting ratio ratio in orthogona orthogonall cutting cutting Reason (R): The frictional force is very high due to the occurren occurrence ce of stickin stickin fricti friction on rathe ratherr than than slidin slidin friction (a) Both A and R are individu individually ally true true and R is the correct correct explanation explanation of A (b) Both Both A and R are are individu individual ally ly true but R is not the correc correctt explanat explanation ion of A (c) A is true true but but R is false false Ans. (b) (d) A is false false but R is true Compiled by: S K Mondal Made Easy

GATE‐1993 The effect effect of rake rake angle on the the mean mean fricti friction on angle angle in machining machining can be explained explained by (a) Sliding (coulomb) model of friction (b) sticking and then then siding model of friction (c) Sticking friction (d) sliding and then sticking model of friction


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IES 2010 Flan Flank k wear wear occu occurs rs on the (a) Relief face face of the tool tool (b) Rake face

Tool Wear, Tool Life & Machinability

(d) Cutting edge Ans. (a)


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IES – IES – 2007

IES – IES – 2004

Flank wear Flank wear occurs mainly on mainly on which which of the of the following? (a) Nose Nose part and top face (b) Cutting edge only (c) (c) Nose Nose part, front relief face, and side relief face of the of the cutting tool (d) Face Face of the of the cutting tool at a shor shortt dist distan ance ce from the cutting edge

Consider Consider the following following statements statements:: Dur During the third stage of tool ool‐ wear, wear, rapid deteriorati deterioration on of tool edge takes takes place because because 1. Flank Flank wear wear is only only marg margina inall . 3. Temperat emperature ure of the tool tool increases increases graduall gradually y 4. Temperat emperature ure of the tool tool increa increases ses drastical drastically ly Which of the statements given above are correct? correct? (a) 1 an and 3 (b) 2 and 4 (c) 1 a nd nd 4 (d) 2 and 3 Ans. (b)

Ans. (c) Compiled by: S K Mondal

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IES – IES – 2002

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IAS – IAS – 2007

Crater wear Crater wear on tools always starts at some distance from the tool tip because at that point (a) Cutting Cutting fluid does not penetrate (b) Normal Normal stress on rake face is maximum (c) Temperatur Temperaturee is maximum (d) Tool strength is minimum Ans. (c)

Why does Why does crater wear crater wear start at some distance from the tool tip? (a) Tool strength is minimum at that region (b) Cutting fluid cannot penetrate that region (c) Tool temperature is maximum in that region (d) Stress Stress on rake face is maximum at that region Ans. (c)


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IES – IES – 2000

IES – IES – 1996

Crater wear Crater wear starts at some distance from the tool tip because (a) Cutting Cutting fluid cannot penetrate that region (b) Stress Stress on rake face is maximum at that region (c) Tool strength is minimum at that region (d) Tool temperature is maximum at that region Ans. (d)

Notch wear Notch wear at the outside edge of the of the depth of cut of cut is due to (a) Abrasive Abrasive action of the of the work work hardened chip material (b) Oxidation (c) Slip Slip‐stick action of the of the chip (d) Chipping. Chipping. Ans. (b)


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IES – IES – 1995

IES – IES – 1995

Match List I with List II and select the correct answer using the codes given below the lists: List I (Wear type) List II (Associated mechanism) A. Abrasive Abrasive wears 1. Galvanic action B. Adhesi Adhesive ve wears wears 2. Plou Plough ghin ing g acti action on C. Electrol Electrolytic ytic wear wear 3. Mole olecular ular transfer D. Diffus Diffusion ion wears wears 4. Plastic deformation [ Ans. (a)] 5. Metall allic bond Code: A Code: A B C D A B C D (a) 2 5 1 3 (b) 5 2 1 3 Compiled by: S K Mondal (c) 2 1 3 4 (dMade ) Easy 5 2 3 4

Crater wear Crater wear is predominant in (a) Carbon Carbon steel tools (b) Tungsten carbide tools (c) High High speed steel tools (d) Ceramic tools Ans. (a)


IES – IES – 1994 Assertion (A): Tool wear Tool wear is expressed in terms of flank wear flank wear rather than crater wear crater wear.. Reason (R): Measurement of flank of flank wear wear is simple and more accurate. a ot an are n v ua y true t rue an st e correct explanation of A A (b) Both A Both A and and R are R are individually true individually true but R is R is not the correct explanation of A A (c) (c) A is A is true but R is R is false (d) A is A is false but R is R is true Ans. (c) Compiled by: S K Mondal Made Easy

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IES – IES – 2008 What are the reasons for reduction of tool of tool life in a machining operation? 1. Tempera emperatur ture e rise of cutting of cutting edge 2. Chip Chippi ping ng of tool of tool edge due to mechanical impact 3. Gradual wears Gradual wears at tool point 4. Incr Increas ease e in feed of cut of cut at constant cutting force Select the correct answer using the code given below: (a) (a) 1, 2 and 3 (b) 2, 3 and 4 (c) (c) 1, 3 and 4 (d) 1, 2 and 4 Compiled by: S K Mondal Made Easy Ans. (a)

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IAS – IAS – 2002

IAS – IAS – 1999

Consider Consider the following following actions: actions: 1. Mecha echani nica call abr abrasio asion n 2. Diff Diffus usio ion n 3. Pla Plasticde sticdefform ormatio ation n 4. Oxida xidati tio on Which of the above are the causes of tool wear? (a) 2 and 3 (b) 1 and 2 (c) 1, 2 and and 4 (d) 1 and 3 Ans. (c)

The type of wear wear that occurs occurs due to the cuttin cutting g acti action on of the the part partic icle less in the the cutt cuttin ing g fluid fluid is referr referred ed to as (a) Attritionswear Attritionswear us on wear wear (c) Erosive Erosive wear wear (d) Corrosi Corrosive ve wear wear Ans. (a)


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IAS – IAS – 2003 Consider Consider the following following statements statements:: Chippi Chipping ng of a cuttin cutting g tool tool is due to 1. Tool material material being too brittle brittle 2. Hot Hot hardne hardness ss of the tool tool materi material. al. 3. High High positi positive ve rake rake angle angle of the tool. tool. Which of these statements are correct? (a) 1, 2 and 3 (b) 1 and 3 (c) 2 and 3 (d) 1 and 2 Ans. (b) Compiled by: S K Mondal

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IES – IES – 1992

IES‐1996

Chip equivalent equivalent is increased increased by (a) An increases in side ‐cutting edge edge angle of tool (b) An increase in nose radius and side cutting ed e an le of tool tool (c) Increasin Increasing g the plant plant area of cut (d) Increas Increasing ing the depth of cut. cut.

Tool life is generally specified generally specified by (a) Number Number of pieces of pieces machined (b) Volume Volume of metal of metal removed (c) Actual Actual cutting time (d) Any of Any of the the above Ans. (d)

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

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IES – IES – 2000

GATE‐2004

In a tool tool life life test test,, doubl doublin ing g the the cutt cuttin ing g spee speed d reduc reduces es the tool tool life to to 1/8th 1/8th of the origin original. al. The Taylor's aylor's tool life life index is

In a mach machin inin ing g oper operat atio ion, n, doub doubli ling ng the the 1 cutting cutting speed speed reduc reduces es the the tool tool life life to th of 8 the original value. The exponent n in Taylor's tool life life equation equation VTn = C, is ( a)

1 8

(b)

1

(c )

4

1 3

(d )

(a )

1

(b)

1

(c )

1

1 4

(d)

1

2

Ans. (b)


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IES – IES – 1999

IES – IES – 2008 In Taylor's tool life equation is VT is VT n = constant. What is the value the value of n of n for ceramic tools? (a) 0.15 0.15 to 0.25 (b) 0.4 to 0.55 (c) 0.6 to 0.75 (d) 0.8 to 0.9

In a single‐point turning operation of steel of steel with with a cemented carbide tool, Taylor's tool life exponent is 0.25. If the If the cutting speed is halved, the tool life will life will increase by (c) Eight times

(d)

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Si Sixteen times Ans. (c)

Ans. (d)


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IES – IES – 2006

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IES – IES – 1999

Which of the following values of index n is associated with carbide tools when Taylor's tool life equation, equation, V.Tn = constant constant is applied? applied? (a) (a) 0∙1 to 0∙15 (b) 0∙2 to 0∙4 c 0045 0045 to 0 ∙ 0∙ 5 to 0∙9

The The appr approx oxim imat atel ely y vari variat atio ion n of the the tool tool life life exponent exponent 'n' of cemented cemented carbide carbide tools is (a) (a) 0.0 0.03 to 0.08 0.08 (b) (b) 0.08 .08 to 0.20 0.20 (c) (c) 0.20 0.20 to 0.48 .48 (d) (d) 0.48 .48 to 0.70 .70

Ans. (c) Ans. (b)


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IAS – IAS – 1998

IES 2010

Match List ‐ I (Cutting tool material) with material) with List ‐ II (Typical value (Typical value of tool of tool life exponent 'n' in the Taylor's equation V equation V.T .Tn = C) and select the correct answer using the codes given below the lists: List – I List – II . 1. 0.1 B. Cast Castaalloy 2. 0. 0 .12 C. Ceramic 3. 0.25 D. Sinter Sintered ed carbide 4. 0.5 [ Ans. (d )] Codes: A Codes: A B C D A B C D (a) 1 2 3 4 (b) 2 1 3 4 (c) 2 1 4 3 (d) 1 2 4 3 Compiled by: S K Mondal

The above figure shows a typical relationship between tool life and cutt cuttin ing g spee speed d for for diff differ eren entt materials. materials. Match the graphs graphs for HSS, HSS, Carbide Carbide and Ceramic Ceramic tool mate materr a s an se ect ect t e corre correct ct answ answer er using using the the code code give given n belowthe lists: lists: Code Code:: HSS HSS Carb Carbid ide e Cera Cerami micc (a) 1 2 3 (b) 3 2 1 (c) 1 3 2 (d) 3 1 2 Compiled by: S K Mondal Made Easy

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GATE‐2010

GATE‐2003 A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1 mm. A similar batch of 10 tools of the same

For tool A, Taylor’s tool life exponent (n) is 0.45 and constant (K) is 90. Similarly for tool B, n = 0.3 and K = 60. The cutting speed (in m/min) above which tool A will have a higher tool life life than tool B is (a) (a) 26.7 26.7 (b) (b) 42.5 42.5 (c) (c) 80.7 80.7 (d) (d) 142. 142.9 9

while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut. How many comp compon onen ents ts can can be prod produc uced ed with with one one cutting tool at 60 rpm? (a) 29 (b) 31 [Ans. (a)] (c) 37 (d) 42


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For increasing the material removal rate in turning, without any constraints, any constraints, what what is the right sequence to adjust the cutting parameters? 1. Speed 2. Feed 3. Depth of cut of cut


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IES 2010

IES – IES – 1994, 2007

Select the correct answer using the code given below: (a) 1‐ 2‐ 3 (b) 2‐ 3‐ 1 (c) (c) 3‐ 2‐ 1 (d) 1‐ 3‐ 2 Ans. (c)

Ans. (a)

Tool life is affected affected mainly with (a) Feed (b) Depth Depth of cut cut (d) Cutting speed Ans. (d)

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IES – IES – 1997

IES – IES – 1992

Consider the following elements: 1. Nose radius 2. Cutting speed 3. Dept Depth h of c of cut 4. Feed The correct sequence of these of these elements in DECREASING order of their of their influence on tool life is (a) (a) 2, 4, 3, 1 (b) 4, 2, 3, 1 (c) (c) 2,4, 2,4, 1, 3 (d) 4, 2, I, 3

Tool life is generally better generally better when (a) Grain Grain size of the of the metal is large (b) Grain Grain size of the of the metal is small (c) Hard Hard constituents are present in the microstructure of the of the tool material (d) None of the of the above Ans. (a)


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IAS – IAS – 2003

IAS – IAS – 2002 Using the Taylor equation VT equation VT n = c, calculate the percentage increase in tool life when life when the cutting speed is reduced by 50% by 50% (n = 0∙5 and c = 400) (a) 300% (b) 400% c 100 50

The tool life curves for two tools A tools A and and B are shown in the figure and they follow they follow the tool life equation VT equation VT n = C. Consider the following statements: 1. 2. 3. 4.

Valueof ue of n n for both the tools is same. Value alue of C of C for both the tools is same. Value alueof of C C for tool A tool A will will be greater than that for the tool B. Value alueof of C C for tool B will be greater than that for the tool A. tool A.

Ans. (a)

Which of these of these statements is/are correct? (a) (a) 1 and 3 (b) 1 and 4 (c) (c) 2 only (d) 4 only

Ans. (a)


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IES 2010

IAS – IAS – 2002 Optimum Optimum cutting speed for minimum cost (V c min ) and and opti optim mum cut cutting ting spee speed d for maxi maximu mum m production production rate rate (V r max ) have have which which one of the the following relationships? cmin = r max c m in r max 2 (c) (c) V cmin < V r max (d) V = V max c min min r max max

With increasing cutting velocity, velocity, the total time for machini machining ng a compone component nt (a) Decreases (b Increases Increases (c) Remains unaffected (d) First decreases decreases and then increases increases

Ans. (c)

Ans. (d) Compiled by: S K Mondal

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IAS – IAS – 2000

IAS – IAS – 1997 In the Taylor's tool life equation, VT n = C, the value of n = 0.5. The tool has a life of 180 minutes at a cutting speed of 18 m/min. If the tool life is reduced to 45 minute minutes, s, then then the cuttin cutting g speed speed will will be

Consider the following statements: The tool life is increased by 1. Buil Builtt ‐up edge formation 2. Incre Increasi asing ng cutting velocity cutting velocity 3. Incre Increasi asing ng back rake angle up to certain value certain value Which of these of these statements are correct? (a) 1 and 3 (b) 1 and 2 (c) (c) 2 and 3 (d) 1, 1 , 2 and 3

(c) 36 m/min

(d)

72 m/min

Ans. (c)


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IAS – IAS – 1996

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IAS – IAS – 1995 In a single point turning operation with operation with a cemented carbide and steel combination having a Taylor exponent of 0.25, of 0.25, if the if the cutting speed is halved, then the tool life will life will become (a) Half Half (b) Two Two times (c) Eight times (d) Sixteen Sixteen times.

The tool life increases with increases with the (a) Increase Increase in side cutting edge angle (b) Decrease in side rake angle (c) Decrease Decrease in nose radius (d) Decrease in back rake angle Ans. (a)


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IAS – IAS – 1995

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IES – IES – 2006 conventional

Assertion (A): An increase in depth of cut shortens the tool life. life. Reason(R): Increases in depth of cut gives rise to relatively small increase in tool tool temperature. temperature. a ot an are n v ua ua y true an s t e correct correct explanati explanation on of A (b) Both A and R are individu individually ally true true but R is not the correct correct explanati explanation on of A (c) A is true true but R is false (d) A is false but R is true Ans. (a) Compiled by: S K Mondal Made Easy

An HSS tool is used for turning operation. The tool life is 1 hr. when turning is carried at 30 m/min. The tool life will be reduced to 2.0 min if the cutting speed is doubled. Find the suitable speed in RPM for turning 300 mm diameter so that tool life life is 30 min. [Ans. N = 36.66 rpm]


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IES 2009 Conventional

The following equation for tool life was obtained for HSS tool. A 60 min tool life was obtained using the following cutting condition VT 0.13f 0.6d0.3= C. v = 40 m/min, f = 0.25 mm, d = 2.0 mm. Calculate the effect on tool life if s eed, feed feed and de th of cut are are to ether ether incr increas eased ed b 25% and also if they are increased individually by 25%; where f = feed, d = depth of cut, v = speed.

Dete Determ rmin ine e the the opti optimu mum m cutt cuttin ing g spee speed d for for an operation on a Lathe machine using the following information: Tool change time: 3 min oo regr regr n s t me: me: 3 m n Machine Machine running running cost Re.0.50 Re.0.50 per min Depreciationof Depreciationof tool regrinds regrinds Rs. 5.0 The constants in the tool life equation are 60 and 0.2 [Ans. 26 m/min]

Ans. (2.3 min; 10.78 min; 21.42 min; 35.85 min)


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GATE‐2009 Linked Answer Questions (1)

In a certain machining operation with a cutting speed of 50 m/min, tool life of 45 minutes was observed. When the cutting speed was increased to 100 m/min, the tool life decreased to 10 min. Esti Estima mate te the the cutt cuttin ing g spee speed d for for maxi maximu mum m productivity productivity if tool change time is 2 minutes. [Ans. 195 m/min]


In a machining experiment, tool life was life was found to vary to vary with the cutting speed in the following manner: Cutting speed (m/min) Tool life (minutes) 60 81 90 36 The exponent (n) and constant (k) of the of the Taylor's tool life equation are (a) n = 0.5 and k = 540 (b) n= 1 and k=4860 (c) n = ‐1 and k = 0.74 (d) n‐0.5 and k=1.15 Ans. (a)

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GATE‐1999

GATE‐2009 Linked Answer Questions (2) In a machining experiment, tool life was life was found to vary to vary with the cutting speed in the following manner: Cutting speed (m/min) Tool life (minutes) 60 81 90 36 What is the percentage increase in tool life when life when the cutting speed is halved? (a) 50% (b) 200% (c) 300% (d) 400% Ans. (c)

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What is approximate percentage change is the life, t, of a tool with zero rake angle used in orthog orthogona onall cuttin cutting g wh when en its its clear clearan ance ce angle, α , is changed changed from 10o to 7o? (Hint: (Hint: Flank Flank wear wear rate rate is proport proportiona ionall to cot α (a) 30 % increase increase (b) 30%, decrease decrease (c) 70% increase increase (d) 70% decrease decrease Ans. (b)

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IAS – IAS – 2007

GATE‐2005


Contd…

A diagram related to machining economics with various cost components is given above. Match List I (Cost Element) with List II (Appropriate Curve) and select the correct correct answer answer using the the code given given below below the Lists: Lists: List I List II (Cost Elem Elemen ent) t) (App (Appro ropr pria iate te Curve) A. Machining cost 1. Curve‐l B. Tool ool cost 2. Curve‐2 C. Tool ool grinding cost 3. Curve‐3 D. Non‐productive cost 4. Curve‐4 5 . C urve‐5 Compiled by: S K Mondal Made Easy

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Contd………. From From previous previous slide

IES – IES – 1998 The The vari variab able le cost ost and and prod produc ucti tion on rate ate of a machining process against cutting speed are shown in the given given figure. figure. For For efficie efficient nt machin machining ing,, the range range of best best cutting cutting speed would would be between between .

Code:A (a) 3 (c) 3

B 2 1

C 4 4

D 5 2


(b) (d)

A 4 4

B 1 2

C 3 3

(b) (b) 1 and and 5 (c) 2 and and 4 (d) 3 and 5

D 2 5

Ans. (c)

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IES – IES – 1999 Cons Conside iderr the the foll follow owin ing g appr approa oach ches es norm normal ally ly applied applied for for the economic economic analysis analysis of machini machining: ng: 1. Maximu Maximum m produc productio tion n rate rate 2. Maximu Maximum m profit profit criter criterion ion 3. Minimum cost criterion The correct correct sequence sequence in ascending ascending order of optimum optimum cutting cutting speed speed obtained obtained by these these approach approaches es is (a) 1,1, 2, 3 (b) 1, 3, 2 (c) 3, 3, 2, 1 (d) 3, 1, 2 Ans. (c) Compiled by: S K Mondal

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IES 2011 The optimum cutting speed is one which should have: 1. High metal removal rate 2. High cutting tool tool life 3. Balance the metal removal rate and cutting tool life (a) 1, 2 and 3 (b) 1 and 2 only (c) 2 and 3 only Ans. (d) (d) 3 only Compiled by: S K Mondal

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IES – IES – 2000

IES – IES – 2004

The magnitude of the cutting speed for maximum profit profit rate rate must must be (a) (a) In betw betwee een n the the spee speeds ds for for mini minimu mum m cost cost and and maximum maximum producti production on rate rate g ert an t e spee spee ormax mum mum pro pro uct uct onrate onrate (c) Below Below the speed speed for for minimu minimum m cost cost (d) Equal Equal to the speed speed for for minimu minimum m cost cost Ans. (a) Compiled by: S K Mondal

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Consider the following following statements: 1. As the cuttin cutting g speed increa increases ses,, the cost of produc productio tion n initially initially reduces, reduces, then after an optimum optimum cutting cutting speed it increases 2. As the cuttin cutting g speed increase increasess the cost of product production ion a so incr increas eases es an a ter ter a crit critica ica va ue it re uces uces 3. Higher feed rate rate for the the same cuttin cutting g speed reduces reduces cost cost of production 4. Higher feed rate for the same cutting cutting speed increases increases the cost of productio production n Which of the statements given above is/are correct? (a) 1 and 3 (b) 2 and 3 Ans. (a) (c) 1 and 4 (d) 3 only Made Easy Compiled by: S K Mondal

IES – IES – 2002

IAS – IAS – 2007

In economics of machining, of machining, which which one of the of the following costs remains constant? (a) Machining Machining cost per piece (b) Tool changing cost per piece (c) Tool handling cost per piece (d) Tool cost per piece

Assertion (A): The optimum cutting speed for the minimum cost of machining may not maximize the profit. Reas Reason on (R): (R): The The profi profitt also also depen depends ds on rate rate of . (a) Both Both A and R are are individu individuall allyy true and R is the correc correctt explanatio explanation n of A (b) Both A and R are individua individually lly true true but R is not the correc correctt explanatio explanation n of A (c) A is true but R is false (d) A is false but R is true [ Ans. (a) ]

Ans. (c)


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IAS – IAS – 1997 In turning, the ratio of the of the optimum cutting speed for minimum cost and optimum cutting speed for maximum rate of production of production is always (a) Equal to 1 n t e range o 0. to 1 (c) (c) In the range of 0.1 of 0.1 to 0.6 (d) Greater than 1 Ans. (b) Compiled by: S K Mondal

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IES – IES – 1992

IES – IES – 2009

Ease of machining of machining is primarily judged judged by (a) Life Life of cutting of cutting tool between sharpening (b) Rigidity Rigidity of of work work ‐piece (c) Microstructure Microstructure of tool of tool material (d) Shape and dimensions of work work

Consider the following: 1. Tool ool life 2. Cutt Cutting ing forces 3. Surf Surfac acee finish Which of the of the above is/are the machinability criterion/criteria? (a) (a) 1, 2 and 3 (b) 1 and 3 only (c) (c) 2 and 3 only (d) 2 only

Ans. (a)


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IES – IES – 2007

IES – IES – 2003 Assertion (A): The machinability of steels improves by addi adding ng sulp sulphu hurr to obta obtain in so call called ed 'Fre 'Free e Machining Steels‘. Reason Reason (R): (R): Sulphu Sulphurr in steel steel forms forms mangan manganese ese sulphi sulphide de inclus inclusion ion which which helps helps to produc produce e thin thin ribbon like continuous continuous chip. chip. (a) Both Both A and R are are individu individuall allyy true and R is the correc correctt explanati explanation on of A (b) Both A and R are individua individually lly true true but R is not the correc correctt explanati explanation on of A (c) A is true true but R is false (d) (d) A is fals falsee but R is tru true [ Ans. (c) ]

Which of the following are the machinability criteria? 1. Tool ool life life 2. Cutting Cutting forces forces 3. Surf Surfac acee finis finish h Select Select the correc correctt answer answer using using the code code given given below below:: (a) 1, 2 and 3 (b) 1 and 2 only only (c) 1 and 3 only only (d) 2 and 3 only Ans. (a) Compiled by: S K Mondal

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IES – IES – 2009

Consider the following criteria in evaluating machinability: 1. Surf Surfac ace e finish 2. Type of chips of chips 3. Tool life 4. Po Power consumption In modern high speed CNC machining with machining with coated carbide tools, the correct sequence of these of these criteria in DECREASING order of their of their importance is (a) (a) 1, 2, 4, 3 (b) 2, 1, 4, 3 (c) (c) 1, 2, 3, 4 (d) 2, 1, 3, 4

Ans. (a)

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IES – IES – 1998

The elements which, added to steel, help in chip formation formation during during machining are (a) Sulphur, Sulphur, lead and phosphorous phosphorous (b) Sulphur, Sulphur, lead and cobalt (c) Aluminium, lead and copper (d) Aluminium, titanium and copper


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Ans. (c)


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11

7/14/2011

IES – IES – 1996 Which of the following machinability? 1. Smal Smallershe lershearangl arangle e 2. Highercutt Highercutting ing forces forces 3. Longer tool life 4. Bettersurfa Bettersurface ce finish. finish. (a) 1 an and 3 (b) 2 and 4 (c) 1 a nd nd 2 (d) 3 and 4 Ans. (d) Compiled by: S K Mondal

IES – IES – 1996 indicate

better

Smal Smalll amou amount ntss of whic which h one one of the the foll follow owin ing g element elements/p s/pair airss of elemen elements ts is added added to steel steel to increase increase its machinabilit machinability? y? (a) (a) Nick Nickel el (b) (b) Sulp Sulphu hurr and and phos phosph phor orus us c con anganese an an copper Ans. (b)

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IES – IES – 1995 In low carbon steels, presence of small of small quantities sulphur improves (a) (a) Welda eldabi bili lity ty (b) (b) Forma ormabi bili lity ty (c) (c) Mach Machin inab abil ilit ityy (d) (d) Hard Harden enab abil ilit ity y Ans. (c)


IES – IES – 1992 Machining of titanium of titanium is difficult due to (a) High High thermal conductivity of conductivity of titanium titanium (b) Chemical Chemical reaction between tool and work and work (c) Low Low tool‐chip contact area (d) None of the of the above Ans. (b)

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IAS – IAS – 1996 Assertion (A): The machinabilityof machinabilityof a material can be measured as an absolute quantity. Reason (R): Machinabili Machinability ty index indicates the case with which with which a material can be machined a ot an are n v ua y true t rue an st e correct explanation of A A (b) Both A Both A and and R are R are individually true individually true but R is R is not the correct explanation of A A (c) (c) A is A is true but R is R is false (d) A is A is false but R is R is true [ Ans. (d) ] Compiled by: S K Mondal

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Made Easy

GATE‐2009 Friction at the tool‐chip interface can be reduced by (a) decreasing the rake angle b increasin the de th of cut of cut (c) Decreasing the cutting speed (d) increasing the cutting speed Ans. (d)

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7/22/2011

IES 2011 Assertion (A): Lead, Zinc and Tin are always hot worked. Reason (R) : If they are worked in cold state they cannot cannot retain retain their mechanical mechanical properties properties.. (a) Both A and R are individually true and R is the correc correctt explanati explanation on of A (b) Both A and R are individually true but R is NOT the correc correctt explanat explanation ion of A (c) A is true but but R is false (d) A is false but R is true Ans. (b )


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GATE‐2003

GATE‐2002

Cold working working of steel steel is defined as working working (a) At its recrystallisation recrystallisation temperature temperature (b) Above its recrystallisation recrystallisation temperatur temperaturee (c) Below its recrystallisation recrystallisation temperature temperature (d) At two two thirds thirds of the melting melting temp temper eratu ature re of the metal Ans. (c)


Hotrolling Hotrolling of mild steel steel is carried carried out (a) At recrystallisation recrystallisation temperature temperature (b) Between 100°C 100°C to 150°C 150°C (c) Below recrystallisation recrystallisation temperature temperature (d) Above recrystallisation recrystallisation temperatu temperature re Ans. (d)

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IES – IES – 2006

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IES – IES – 2004

Which one of the following is the process to refine the grains of metal after it has been distorted by hammering hammering or cold working working?? (a) Annealing (b) Softening c e‐crysta z ng orma z ng Ans. (c)


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Consider Consider the following following statements statements:: In compari comparison son to hot working working,, in cold cold working, working, 1. Higher Higher force forcess are are requi require red d 2. No heatin heating g is requi require red d 3. Less Less duct ductil ilit ityy is requ requir ired ed 4. Better Better surface surface finish is obtained obtained Which of the statements given above are correct? correct? (a) (a) 1, 2 and and 3 (b) (b) 1, 2 and and 4 (c) 1 and 3 (d) 2, 3 and 4 Ans. (b)

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1

7/22/2011

IES – IES – 2009

IES – IES – 2008

Consider Consider the following following character characteristics istics:: 1. Porosity Porosity in the metal metal is largely largely eliminate eliminated. d. 2. Streng Strength th is decre decrease ased. d. 3. Close Close toleranc tolerances es cannot cannot be maintained maintained.. Which of the above characteristics characteristics of hot working is/are correct? (a) 1 o nl nly (b) 3 only (c) 2 and 3 (d) 1 and 3 Ans. (d)

Consider Consider the following following statements statements:: 1. Metal etal form formin ing g decr decrea ease sess harm harmfu full effe effect ctss of impurities and improves improves mechanical strength. 2. Metal Metal working working process process is a plasti plasticc deform deformati ation on process. 3. Very intrica intricate te shapes shapes can be produc produced ed by forging forging process process as compar compared ed to casting casting process. process. Which of the statements given above are correct? correct? (a) 1, 2 and 3 (b) 1 and 2 only (c) 2 and 3 only nly (d) 1 and 3 only Ans. (b) Compiled By: S K Mondal Made Easy


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IES – IES – 2008

IES – IES – 2004

Cold Cold forgin forging g result resultss in impro improved ved qualit quality y due to which of the following? 1. Better Better mechanica mechanicall properti properties es of the proces process. s. 2. Unbrok Unbroken en grain grain flow. flow. 3. Smooth Smoother er finishe finishes. s. 4. High High press pressur ure. e. Select Select the correc correctt answer answer using using the code code given given below below:: (a) (a) 1, 2 and and 3 (b) (b) 1, 2 and and 4 (c) (c) 2, 3 and and 4 (d) (d) 1, 3 and and 4 Ans. (a)

Assertion (A): Cold working of metals results in increase increase of strength strength and hardness hardness Reason (R): Cold working reduces the total number of disloca dislocatio tions ns per unit unit volume volume of the materi material al (a) ( ) Both Bot B th A and d R are are individu indiv i di iduall id ally lly true t and dR R is i the th th correc correctt explanatio explanation n of A (b) Both A and R are individua individually lly true true but R is not the correc correctt explanatio explanation n of A (c) A is true true but R is false (d) A is false but R is true Ans. (c) Compiled By: S K Mondal Made Easy


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IES – IES – 2003 Cold working working produces produces the following following effects: effects: 1. Stress Stresses es are are set up in the metal metal 2. Grain Grain struc structur turee gets gets disto distort rted ed 3. Strengt Strength h and hardness hardness of the metal metal are decrease decreased d 4. Surface Surface finish is reduced reduced Which of these statements are correct? correct? (a) 1a 1and 2 (b) 1, 2 and 3 (c) 3 and 4 (d) 1 and 4 Ans. (a) Compiled By: S K Mondal

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IES – IES – 2000 Assertion (A): To obtain large deformations by cold working intermediate annealing is not required. Reason (R): Cold working is performed below the recrystalli recrystallisation sation temperatur temperature e of the work material. material. (a) ( ) Both Bot B th A and d R are are individu indiv i di iduall id ally lly true t and dR R is i the th th correc correctt explanatio explanation n of A (b) Both A and R are individua individually lly true true but R is not the correc correctt explanatio explanation n of A (c) A is true true but R is false (d) A is false but R is true Ans. (d) Compiled By: S K Mondal Made Easy

Page 29 of 79

2

7/22/2011

IES – IES – 1997

IES – IES – 1996 Consider the following statements: When a metal or alloy is alloy is cold worked cold worked 1. It is worke is worked d below room temperature. 2. It is worke is worked d below recrystallisation recrystallisation temperature. temperature. 3. s ar ness ness an s reng ncrease. 4. Its hardness increases but strength does not increase. Of these Of these correct statements are (a) (a) 1 and 4 (b) 1 and 3 (c) (c) 2 and 3 (d) 2 and 4 Ans. (c)

In metals subjected to cold working, cold working, strain hardening effect is due to (a) Slip Slip mechanism (b) Twining mechanism (c) Dislocatio Dislocation n mechanism (d) Fracture Fracture mechanism Ans. (c)


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IES – IES – 2006

IES – IES – 1992

Assertion (A): In case of hot working of metals, the temperat temperature ure at which which the process process is finally finally stopped stopped should should not be above above the recrystal recrystallisat lisation ion temperat temperature. ure. Reas Reason on (R): (R): If the the proc proces esss is stop stoppe ped d abov above e the the recrystall recrystallisati isation on temperat temperature, ure, grain grain growth growth will take . (a) Both A and R are individ individuall uallyy true and R is the correct explanationof explanationof A (b) Both Both A and R are are indiv individu iduall allyy true but R is not the correct correct explanation explanation of A (c) A is true true but but R is false (d) A is false false but R is true true Ans. (d) Compiled By: S K Mondal

Specify the sequence sequence correctly correctly (a) Grain growth, recrystallisation, recrystallisation, stress stress relief relief (b) Stress relief, grain growth, recrystallisation recrystallisation (c) Stress relief, recrystallisation, recrystallisation, grain growth (d) Grain growth, stress relief, recrystallisatio recrystallisation n Ans. ( c)

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IAS – IAS – 1996

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IAS – IAS – 2004

For For mild mild steel, steel, the hot forgin forging g temper temperatu ature re range range is 0 0 (a) 400 C to 600 C (b) 7000C to 9000C (c) 1000 10000C to 1200 12000C 0 (d) 1300 Cto 15000C Ans. (c)


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Assertion (A): Hot working Hot working does not produce strain hardening. Reason (R): Hot working Hot working is done above the re‐ crystallization temperature. a ot an are n v ua y true t rue an st e correct explanation of A A (b) Both A Both A and and R are R are individually true individually true but R is R is not the correct explanation of A A (c) (c) A is A is true but R is R is false (d) A is A is false but R is R is true Ans. (a) Compiled By: S K Mondal Made Easy

Page 30 of 79

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7/22/2011

IAS‐2002 Assertion (A): There is good grain refinement in hot working. Reason (R): In hot working physical properties are generally improved. (a) are individu indiv tru ( ) Both B th A and d R are i di iduall id ally lly true t e and d R is i the th th correct correct explanati explanation on of A (b) Both A and R are individu individually ally true true but R is not the correct correct explanati explanation on of A (c) A is true true but R is false (d) A is false but R is true Ans. (b) Compiled By: S K Mondal Made Easy


IES Made Easy Made Easy

Page 31 of 79

4

7/24/2011

GATE‐2008 In a single pass rolling operation, a 20 mm thick plate with plate width of 100 mm, is reduced to 18 mm. The roller roller radius radius is 250 mm and rotation rotational al speed is 10 rpm. The The averag average e flow stress stress for the plate plate mate materi rial al is 00 MPa. MPa. The The ower ower re re uire uired d for the the rolling rolling operation operation in kW is closest closest to (a) 15.2 15.2 (b) 18.2 (c) 30.4 30.4 (d) 45.6 45.6 Ans. (a)

By S K Mondal

GATE‐2007

GATE‐2004

The thickness of a metallic sheet is reduced from an initial value of 16 mm to a final value of 10 mm in one single single pass pass rollin rolling g with with a pair pair of cylind cylindric rical al rollers each of diameter of 400 mm. The bite angle in de ree ree will will be (a) 5.936 (b) 7.936 (c) 8.936 8.936 (d) 9.936 Ans. (d)

In a rolling process, sheet of 25 mm thickness is rolled to 20 mm thickness. Roll is of diameter 600 mm and it rotates at 100 rpm. The roll strip contact length will be be (c) 78 78 mm Ans. (b)

(d)

120 mm

GATE‐1998

GATE‐2006

A strip with a cross‐section 150 mm x 4.5 mm is being rolled with 20% reduction of area using 450 mm diamet diameter er rolls. rolls. The angle subte subtende nded d by the deform deformati ation on zone zone at the roll roll centr centre e is (in radian radian)) . . (c) 0.03 0.03 (d) (d) 0.06 .06 Ans. (d)

A 4 mm thick sheet is rolled with 300 mm diameter rolls to reduce thickness without any charge in its width. The friction coefficient at the work‐roll interface is 0.1. The minimum possible thickness of the the sheetthatcan sheetthatcan be rodu roduce ced d in a sin sin le ass ass is (a) 1. 1.0 mm (b) 1.5 mm (c) 2. 2.5 mm (d) 3.7 mm Ans. (c)

Page 32 of 79

1

7/24/2011

IES – IES – 2003

IES – IES – 2002

Assertion (A): While rolling metal sheet in rolling mill, the edges are sometimes not straight and flat butare wavy wavy.. Reason (R): Non‐uniform mechanical properties of the flat material rolled out result i n waviness of the edges. (a) Both A and R are are individu individuall allyy true true and R is the correct correct explanati explanation on of A (b) Both A and R are individuall individuallyy true but R is not the correct correct explanati explanation on of A (c) A is true true but R is false (d) A is false but R is true [ Ans. (c)]

In rolling a strip between two rolls, the position of the neutral neutral point point in the arc of conta contact ct does not depend depend on (a) Amount Amount of reduc reductio tion n (b) Diame Diamete terr of the rolls rolls c oe c ent o r ct on ater a o t e ro s Ans. (d)

IES – IES – 2001

IES – IES – 2001

Which of the following assumptions are correct for cold rolling? 1. The material material is plastic. 2. The arc of contact is circular with a radius greater than the radiu radiuss of the roll. roll. Coeffici icient ent of frict friction ion is const constant ant over over the arc arc of 3. Coeff contact and acts in one direction throughout the arc of contact. Select Select the correc correctt answer answer using the codes codes given given below below:: Codes: (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1, 2 and 3 [ Ans. (a)]

A strip is to be rolled from a thickness of 30 mm to 15 mm using a two‐high high mill mill havi having ng roll rollss of diamete diameterr 300 mm. The coeffic coefficien ientt of fricti friction on for unaided bite should should nearly nearly be . . (c) 0.25 (d) 0.07 Ans. (a)

IES – IES – 2000

IES – IES – 1999

In the rolling process, roll separating force can be decreased by (a) Reducing Reducing the the roll diameter diameter (b) Increasing the roll diameter (c) Providing back‐up rolls (d) Increasing Increasing the friction friction between between the rolls and the metal Ans. (a)

Assertion (A): In a two high rolling mill there is a limit to the possible reduction in thickness in one pass. Reason (R): The reduction possible in the second . (a) Both Both A and A and R are R are individually true individually true and R is R is the correct explanation of A A (b) Both A Both A and and R are R are individually true individually true but R is R is not the correct explanation of A A (c) (c) A is A is true but R is R is false (d) A is A is false but R is R is true [ Ans. (b)]

Page 33 of 79

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7/24/2011

IES – IES – 1993

IES – IES – 1993

In order to get uniform thickness of the plate by rolling rolling process, process, one provides provides (a) Camber Camber on the the rolls rolls (b) Offset Offset on the rolls rolls (c) Hardening Hardening of the rolls rolls (d) Antifrictio Antifriction n bearings bearings

The blank diameter diameter used in in thread thread rolling rolling will be be (a) Equal to to minor minor diameterof diameterof the thread thread (b) Equal to pitch pitch diameter diameter of the thread thread (c) A little large than the minor diameter of the thread thread (d) A little little large largerr than the the pitch pitch diameter diameter of the threa thread d Ans. (d)

Ans. (a)

IES – IES – 1992

IAS – IAS – 2004

Thread rolling is restricted to (a) Ferrous Ferrous materials (b) Ductile Ductile materials (c) Hard Hard materials (d) None of the of the above

Assertion (A): Rolling requires high friction which increases increases forces forces and power consumpti consumption. on. Reason (R): To prevent damage to the surface of the rolled rolled products, lubricants lubricants should should be used. (a) ( ) Both Bot B th A and d R are are individu indiv i di iduall id ally lly true t and dR R is i the th th correc correctt explanatio explanation n of A (b) Both A and R are individua individually lly true true but R is not the correc correctt explanatio explanation n of A (c) A is true true but R is false (d) A is false but R is true [ Ans. (c )]

Ans. (b)

IAS – IAS – 2001

IAS – IAS – 2000

Consider Consider the following following character characteristics istics of rolling rolling process: 1. Shows Shows work work hardening hardening effect effect 2. Surface Surface finish is not good good 3. Heavy Heavy reduc reductio tion n in areas areas can be obtai obtained ned Which of these characteristics characteristics are associated with hot rolling? (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1, 2 and 3 Ans. (c)

Rolling very Rolling very thin thin strips of mild of mild steel requires (a) Large Large diameter rolls (b) Small Small diameter rolls (c) High High speed rolling (d) Rolling Rolling without without a lubricant Ans. (b)

Page 34 of 79

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7/24/2011

IAS – IAS – 1998

IAS – IAS – 2007

Match List ‐ I (products) with List ‐ II (processes) and select the correct answer using the codes given below below the lists: lists: List – I List ‐II . . . ang es an an c anne s 1. e ng B. Carburetors 2. Forging C. Roof trusses 3. Casting D. Gear wh wheels 4. Rolling [ Ans. [ Ans. (d)] Codes:A B C D A B C D (a) 1 2 3 4 (b) 4 3 2 1 (c) 1 2 4 3 (d) 4 3 1 2

IAS – IAS – 2003

Match List I with List II and select the correct correct answer using the codegiven below below the Lists: Lists: List I List II (Typ (Type e of Roll Rollin ing g Mill Mill)) (Cha (Chara ract cter eris isti tic) c) A. Two Two high non‐reversing mills 1. Middle roll rotates by friction . ree g m s 2. y sma wo wor ng ro , power for rolling is reduced reduced C. Four our high igh mills ills 3. Rollsof equa equall siz size are rotated only in one direction D. Clus te ter mills 4. Diame te ter of working roll is very small [ Ans. (d)] Code:A B C D A B C D (a) 3 4 2 1 (b) 2 1 3 4 (c) 2 4 3 1 (d) 3 1 2 4

IAS – IAS – 2007

In one setting of rolls in a 3 ‐high rolling mill, one gets (a) One reductio reduction n in thickn thickness ess (b) Two Two reductions in thickness thickness (c) Three Three reductio reductions ns in thickness thickness (d) Two Two or three three reductio reductions ns in thickness thickness depending depending upon upon the settin setting g Ans. (b)

Consider Consider the following following statements statements:: Roll forces in rolling can be reduced by 1. Reduc Reducing ing fricti friction on 2. Using Using large diameter diameter rolls rolls to increase increase the contact contact area. 3. Taking aking smalle smallerr reduc reductio tions ns per pass pass to reduce reduce the contact area. Which of the statements given above are correct? correct? (a) (a) 1 and 2 only nly (b) 2 and and 3 onl only (c) 1 and 3 only (d) 1, 2 and 3 [ Ans. (c)]

GATE 2011 The maximum possible draft in cold rolling of sheet increases with the (a) increase increase in coefficie coefficient nt of friction (b) decrease in coefficient of friction (c) decrease decrease in roll radius (d) increase in roll velocity Ans. (a)

Page 35 of 79

4

Analysis of Rolling

Fig. Geometry of Rolling Process Total reduction or “draft” taken in rolling.

h = he - h1 = 2 (R - R cos cos a) a) = D (1 - cos cos a) a) Usually, the reduction in blooming mills is about 100 mm and in slabbing mills, about 50 to 60 mm. The projected length if the arc of contact is,

l = R.sin a BC 2 - CE 2

or l =

Now BC = l=

R. h and CE = R (1 - cos a) (1 - cos a) = 0.5 h

R. h - 0.5 h

2

P= Usually, 0.5 h l

R h

2

is < R

h

1/2

Assumption Assumption in Rolling

`

1. Rolls are straight, rigid cylinders. 2. Strip is wide compared with its thickness, so that no widening of strip occurs (plane strain conditions). 3. The material is rigid perfectly plastic (constant yield strength). 4. The co-efficient of friction is constant over the tool- work interface.

Fig. Stress Equilibrium of an Element in Rolling Considering the thickness of the element perpendicular to the plane of paper to be unity, Page 36 of 79 We get equilibrium equation in x-direction as,

- σ l h + ( σ σ l +dσ +dσ ) +2 τ R dθ dθ cos θ = 0 l ) (h + dh) - 2P R dθ sin θ +2 τ l

For sliding friction, τ = μ p . Simplifying and neglecting second order terms, we get l

d ( σx h ) dθ p − σx

= 2pR(θ ± μ ) 2

=

3

σ0

d ⎡h ( p − σ '0 )⎤ ⎦ dθ ⎣

= σ '0 = 2pR ( θ ± μ )

d ⎡ ' ⎛ p ⎢σ 0 h ⎜ ' dθ ⎣⎢ ⎝ σ0

⎞⎤ − 1 ⎟ ⎥ = 2 pR ( θ ± μ ) ⎠ ⎦⎥ ⎞ d ' d ⎛ p⎞ ⎛ p σ '0 h ⎜ ' ⎟ + ⎜ ' − 1 ⎟ ( σ 0 h ) = 2 pR ( θ ± μ ) dθ ⎝ σ 0 ⎠ ⎝ σ 0 ⎠ dθ Due to cold rolling, σ '0 increases as h decreases, thus σ '0 h nearly a constant and its derivative zero.

d p / σ '0 dθ p / σ 0 h = hf

2R (θ ± μ ) h

=

+ 2R (1 − cos θ ) = h f + Rθ2

d ( p / σ '0 )

(p / σ ) ' 0

=

2R ( θ ± μ ) dθ h f + Rθ2

Inte Integr grat atin ing g both both side side 2Rθdθ ∓ Rθ2 f +

∫h

ln ( p / σ '0 ) =

∫h

2Rμ dθ Rθ2 f +

= I ∓ II I=

2Rθdθ 2 f + Rθ

∫h

=

h/R

∫

=

h f R

2Rθdθ h

=

2θdθ

⎛h⎞

∫ h / R = ln ⎜⎝ R ⎟⎠

+ θ2

d ⎛ h f ⎞ ⎜ ⎟ = 2θ dθ ⎝ R ⎠ II =

∫h

=

∫h

= 2μ ∴

2Rμ dθ 2 f + Rθ 2μ dθ 2 f / R + θ

R . tan −1 hf

R .θ h f

⎛ h ⎞ ∓ 2μ R . tan −1 ⎟ hf ⎝R⎠

ln ( p / σ '0 ) = ln ⎜

R .θ + lnC ln C hf

h ∴ p = C σ '0 ⎛⎜ ⎞⎟ e ⎝R⎠

∓ μH

where H

=2

R . tan −1 hf

R .θ. hf

Now No w atentr atentry, y, θ = α Hence H = H0 with θ replaced by At exit θ = 0 , H = H1

nabove equation ∝ inab

=0

There for p = σ '0 In the ent entry zone p = C. σ '0

⎛ h o ⎞ e − μH ⎜R⎟ ⎝ ⎠

o

Page 37 of 79

C=

R μHo .e ho

p = σ '0

h μ H −H . e ( 0 ) h0

In the exit zone p = σ '0

⎛ h ⎞ μH ⎜ ⎟ .e ⎝ h f ⎠

hn μ H −H . e ( 0 n) h0 ho h f

or

or Hn

= eμ ( H

0

hn

−2Hn )

⎛ h ⎞⎤ 1⎡ 1 = ⎢ H 0 − ln ⎜ 0 ⎟ ⎥ 2 ⎣⎢ μ ⎝ hf ⎠⎦⎥ R . tan −1 hf

from H = 2

∴ θn =

hn . eμ Hn hl

=

hf . tan R

R .θ. hf

⎛ h f Hn ⎞ . ⎜⎜ ⎟⎟ R 2 ⎝ ⎠

= h f + 2R (1 − cos θn )

Maximum Draft. It has already been proved that if the strip is to enter the rolls unaided then, the following relation has to be satisfied between the angle of bite and co-efficient of friction between the roll and material surfaces.

μ > tan a Now, from Fig. 13.12, the projected length of are of contact,

l = R. h, and l

tan a =

=

R h

h R 05 h 2 h, it can be written that R -

Since R > > 0.5

tan a = Since

h R

tan a

The maximum draft is given by

h R or,

h

max

=

2

R

Q.1. In rolling rolling process, 25 mm thick plate is rolled to 20 mm in a four high mill. Determine the the coefficient of friction if this is the maximum reduction possible. Roll diameter is 500 mm. Find neutral Section, Back word and forward slip sad maximum pressure, σo = 100 N / m m2 for hot rolls of middle steel at about 1100oC. Solution: (i). Δh = μ2R

or μ =

Δh R

=

( 25 − 20 ) 250

= 0.142

Δh = 2R (1 − cos α ) or 5 = 500 (1 − cos α )

and

α = 8.110 = 0.1429

Page 38 of 79

(ii)

⎛ R ⎞ .α ⎟ ⎜⎜ ⎟ h ⎝ f ⎠ ⎛ 250 ⎞ 250 . tan −1 ⎜ 0.1429 ⎟ = 3.306 = 2. × ⎜ 20 ⎟ 20 ⎝ ⎠ ⎛ h ⎞⎤ 1⎡ 1 Hn = ⎢ H0 − log e ⎜ 0 ⎟ ⎥ 2⎣ μ ⎝ hf ⎠⎦

H0

R . tan −1 hf

=2

1⎡ 1 ⎛ 25 ⎞ ⎤ = 0.8678 . l og e ⎜ = ⎢3.306 − ⎟⎥ 2⎣ 0.142 ⎝ 20 ⎠ ⎦ ⎛ h f Hn ⎞ hf . tan ⎜ θn = ⎜ R . 2 ⎟⎟ R ⎝ ⎠ ⎧ 20 ⎛ 0.8678 ⎞ ⎫⎪ 250 = × tan ⎪⎨ ×⎜ ⎟⎬ 20 ⎩⎪ 250 ⎝ 2 ⎠ ⎭⎪ 0.0349 9 rad = 0.034 hn = h f + 2R (1 − cos θn )

= h f + Rθn 2 2

(iii)

= 20 + 250 × (0.0349 ) = 20.3 mm V − V0 V h 20.3 =1− 0 =1− n =1− = 18.8% Backward slip = r Vr

Forward slip =

Vf − Vr Vr

Vr

=

Vf Vr

−1 =

h0

hn hf

25

−1 = 1 −

Vo

Vr

20.3 20

= 1.5%

Vf

N

(iv) pmax

= pn = σ′0 =

2 3

hn μ Hn .e h f .100 ×

20.3 0.142 ×0.8678 .e 20

= 132.4 N / mm2

Q2. Sheet steel is reduced from 4.05 mm to 3.55 mm with 500 mm diameter rolls having a co2 efficient of fiction of 0.04. The mean flow stress in tension is 210 N/mm . Neglect work hardening and roll flattening. (a) Calculate the roll pressure at the entrance to the rolls, the n eutral plane, and the roll exit. (b) If the co-efficient of friction is 0.40, determine the roll pressure at the neutral point. 2 (c) If 35 N/mm front tensions are applied in the problem find the roll pressure at the neutral point. Solution: Given ho = 4.05 mm h f = 3.55 mm R = 250 mm, μ = 0.04,

Page 39 of 79

σ0 = 210 N / mm2

(a) The roll pressure at entry and exit, 2 p = σ′0 = 42.5N / mm2 σ0 = 242.5N 3 Now

H0

=2

⎛ R ⎞ R . tan −1 ⎜ ⎜ hf α ⎟⎟ hf ⎝ ⎠

H0

=2

⎛ 250 250 . tan −1 ⎜ ⎜ 3.55 3.55 ⎝

⎞ × 0.0447 ⎟⎟ ⎠

= 6.02 ⎛ h ⎞⎤ 1⎡ 1 H n = ⎢ H 0 − lo g e ⎜ o ⎟ ⎥ 2⎣ μ ⎝ hf ⎠⎦ 1⎡ 1 4.05 ⎞ ⎤ = ⎢6.02 − × log e ⎛⎜ ⎟ ⎥ = 1.363 2⎣ 0.04 ⎝ 3.55 ⎠ ⎦ pn

= σ′0 .

h n μ Hn .e h f

⎛ h f Hn ⎞ hf . tan ⎜ ⎜ R . 2 ⎟⎟ = R ⎝ ⎠

⎛ 3.55 3.55 . tan ⎜ ⎜ 250 250 ⎝

Now

θn =

And

Δh = 2R (1- cosα) (4.05-3.55) = 2 × 250 × (1- cos α)

hn

= h + 2R (1 − cos θ n ) f

or

⎞ × 0.6815 ⎟⎟ = 0.009672 rad. = 0.554 0 ⎠

∝ = 2.56o = 0.0447 rad.

= 3.55 +2 × 250 (1- cos 0.554 ) o

= 3.5734 mm h pn = σ ′0 . n .eμ Hn h f

= 242.5 ×

3.5734 0.04×1.363 e 3.55

= 257.78 N / mm2

( b ) H0 = 6.02 ( earlier ) μ = 0.4 1⎡ 1 ⎛ 4.05 ⎞ ⎤ = 2.845 lo g e ⎜ = ⎢6.02 − ⎟⎥ 2⎣ 0.4 ⎝ 3.55 ⎠ ⎦ ⎛ 3.55 ⎞ 3.55 tan ⎜ 1.4225 ⎟ = 0.02rad θn = × ⎜ 250 ⎟ 250 ⎝ ⎠ 2 hn = h f + Rθn

then Hn

2

= 3.55 + 250 × (0.02 ) = 3.65 mm pn

= σ′0 .

hn h f

. eμ Hn

3.65 0.04×2.845 ×e 3.55 h (c ) pn = ( σ′0 − σ f ) . n . eμ Hn hf

= 242.5 ×

= ( 242.5 − 35 )

= 777.9N / mm 2

3.5734 × e0.04×1.363 3.55

= 220.57 N / mm 2

Q 3. A wide-strip is rolled to a final thickness of 6.35 mm will a reduction of 30 percent. The roll radius is 50 cm and the co-efficient of friction is 0.2. Determine the neutral plane. Page 40 of 79 Solution:

hf = 6.35mm, R = 50cm = 500mm, μ = 0.2 100 = 9.07mm ho = hf × 70 Δh = h0 − hf = 9.07 − 6.35 = 2.72mm

Δh = 2R (1 − cos α) 2.72 = 2 × 500 × (1 − cos α ) α = 4.230 = 0.0738 rad. ⎛ R ⎞ R Now H0 = 2. . tan −1 ⎜ ⎜ hf .α ⎟⎟ hf ⎝ ⎠ ⎛ 500 ⎞ 500 = 2× × tan −1 ⎜⎜ × 0.0738 ⎟⎟ 6.35 ⎝ 6.35 ⎠ = 10.29. ⎛ h ⎞⎤ 1 ⎡ 1⎡ 1 1 9.07 ⎞ ⎤ now Hn = ⎢H0 − log e ⎜ 0 ⎟ ⎥ = ⎢10.29 − × log e ⎛⎜ ⎟⎥ = 4.26 2⎣ 0.2 μ ⎝ 6.35 ⎠⎦ ⎝ h f ⎠ ⎦ 2 ⎣ ⎛ h H ⎞ h θn = f . tan ⎜⎜ f . n ⎟⎟ R ⎝ R 2 ⎠ ⎛ 6.35 ⎞ 6.35 = × tan ⎜⎜ × 2.13 ⎟⎟ = 0.0273 rad = 1.550 500 ⎝ 500 ⎠ or

Q.4. A metal strip is to be rolled from an initial wrought thickness of 3.5 3.5 mm to a final rolled from an initial wrought thickness of 2.5 mm in a single pass rolling mill having rolls of 250 mm diameter. The strip is 450 mm wide. The average co-efficient of friction in the roll gap is 0.08. Taking plain strain flow stress of 140 MPa, for the metal and assuming neglecting spreading, estimate the roll separating force. [GATE-1997] p l . b p Solution Hint: We know p= = m m Use.

pm

=

0 1 ⎡ n ⎢ pdh + pdh + Δh ⎢ hl hn

⎣

h

h

∫

∫

∫

h0

hb

⎤

p.dh ⎥

⎥⎦

Torque and Power The power is spent principally in four ways 1) The energy needed to deform the metal. 2) The energy needed to overcome the frictional force. 3) The power lost in the pinions and power-transmission system. system. 4) Electrical losses in the various motors and generators. Remarks: Losses Losses in the windup reel and uncoiler must also be considered.

Page 41 of 79

The total rolling load is distributed over the arc of contact in the typical friction-hill pressure distribution. However the total rolling load can be assumed to be concentrated at a point along the act of contact at a distance a from the line of centres of t he rolls. The ratio of the moment arm a to the projected length of the act of contact Lp can be given as

=

a a = LP R h

Where λ is 0.5 for hot-rolling and 0.45 for cold-rolling. The torque MT is equal to the total rolling load P multiplied by the effective moment arm a. Since there are two work rolls, the torque is given by MT = 2Pa During one revolution of the top roll the resultant rolling load P moves along the circumference of a circle equal to 2πa. Since there are two work rolls, the work done W is equal to

Work = 2(2 a)P Since power is defined as the rate of doing work, i.e., 1 W = 1 J s -1, the power (in watts) needed to operated a pair of rolls revolving at N Hz (s -1) in deforming metal as it flows through the roll gap is given by W = 4 aPN Where P is in Newton’s and a is in metre.

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7/24/2011

GATE‐2007 In open‐die forging, a disc of diameter 200 mm and height 60 mm is compressed without any barreling effect. The final diameter of the disc is 400 mm. The The true true strain strain is . . (c) 1.386 (d) 0.602

Forging

Ans. (c)

By S K Mondal

GATE‐1994

GATE‐1998

Match 4 correct pairs between List I and List II for the questions List I gives a number of processes and List List II gives gives a numberof numberof produc products ts List I List II a nvestment ca c ast ng 1. ur ne ne r ot otors (b) Die ccaasting 2. Turbine blades (c) (c) Centr entrif ifug ugal al cast castin ing g 3. Conne onnect ctin ing g rods ods (d) Drop forging 4. Galvanized iron pipe (e) Extrusion 5. Cast iron pi p ipes (f) Shell moulding 6. Carburettor body Ans. (a) ‐ 2, (b) ‐ 6, (c) ‐ 5, (d) – 3

List I List II (A) Alumin Aluminium ium brak brakee shoe shoe (1) Deep Deep drawin drawing g (B) (B) Plas Plasti ticc wate waterr bott bottle le (2) (2) Blow Blow moul mouldi ding ng (C) (C) Stai Stainl nles esss stee steell cups cups (3) (3) Sand Sand cast castin ing g (D) Soft drink can (alumini (aluminium) um) (4) Centr Centrifu ifugal gal castin casting g (5) Impac Impactt extrus extrusion ion (6) (6) Upse Upsett forg forgin ing g Ans. (A) ‐3, (B) ‐2, (C) ‐1, (D) – 5

IES‐2008

IES – IES – 2006

Which one of the following is correct? Malleab Malleabili ility ty is the property property by whi which ch a metal metal or alloy can be plastically plastically deformed deformed by by applying applying (a) Tensile stress (b) Be Bending stress (c) Shearstr rstress (d) Compressivestr estress Ans. (d)

Assertion (A): Forging dies are provided with taper or draft draft angles angles on vertic vertical al surfac surfaces. es. Reason Reason (R): (R): It facili facilitat tates es compl complete ete filling filling of die cavity cavity and favourab favourable le grain flow. flow. a ot an are n v ua y true an s t e correc correctt explanatio explanation n of A (b) Both A and R are individu individually ally true true but R is not the correc correctt explanatio explanation n of A (c) A is true true but R is false (d) A is false false but R is true Ans. (c)

Page 43 of 79

1

7/24/2011

IES – IES – 2005

IES – IES – 1996

Consider Consider the following following statements statements:: Forging reduces the grain size of the metal, which 1. result resultss in a decreas decrease e in streng strength th and toughne toughness. ss. Forged ed compo componen nents ts can be provide provided d with with thin thin 2. Forg sections, sectio witho reduc stren gth. ti ns, without ith utt reducing d ing i the th strength. t th Which of the statements given above is/are correct? (a) Only 1 (b) On Only 2 (c) Both Both 1 and 2 (d) (d) Neithe itherr 1 nor nor 2 Ans. (b)

Which one of the following is an advantage of forging? (a) Good surface finish (b) Low tooling tooling cost cost (c) Close Close toleranc tolerancee (d) Improved Improved physical physical property Ans. (d)

IES – IES – 1993

IES – IES – 1997 Assertion (A): In drop forging besides the provision for flash, provision is also to be made in the forging die for additional additional space space called gutter gutter.. Reason (R): The gutter helps to restrict the outward flow of metal thereby helping to fill thin ribs and basesin basesin the the uppe upperr die. die. (a) (a) Both Both A and and R are are indi indivi vidu dual ally ly true true and and R is the correc correctt explanatio explanation n of A (b) Both A and R are individu individually ally true true but R is not the correc correctt explanatio explanation n of A (c) A is true true but R is false (d) A is false but R is true [ Ans. (c)]

Which one of the following manufacturing processes processes requires requires the provision provision of ‘gutters’ ‘gutters’?? (a) Closed die forging (b) Centrifug Centrifugal al casting casting (c) Investm Investment ent casting casting (d) Impact Impact extrusion extrusion Ans. (a)

IES – IES – 2004

IES – IES – 2003

Mat Match List List I (Dif (Diffe fere rent nt syst system ems) s) with with List List II (Assoc (Associat iated ed termin terminolo ology) gy) and select select the correc correctt answerusing answerusing the codes codes given given below below the Lists: Lists: List I List II A. Riveted oints 1. Ni in B. We Welded joints 2. Angular movement C. Leaf springs 3. Fullering D. Knuckle joints 4. Fusion A B C D A B C D (a) 3 2 1 4 (b) 1 2 3 4 (c) 1 4 3 2 (d) 3 4 1 2 Ans. (d)

A forging method for reducing the dia meter of a bar and in the proces processs making making it longeris longeris termed termed as (a) (a) Full Fuller erin ing g (b) (b) Punc Punchi hing ng (c) (c) Upse Upsett tting ing (d) (d) Extr Extrud udin ing g Ans. (a)

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7/24/2011

IES – IES – 2002

IES – IES – 1999

Consider the following steps involved in hammer forging forging a connecting connecting rod rod from bar stock: stock: 1. Blocking 2. Trimming 3. Finis inishi hing ng 4. Fulle ullerring ing 5. Edg Edging ing Which of the following is the correct correct sequence of operations? (a) 1, 4, 3, 2 and 5 (b) 4, 5, 1, 3 and and 2 (c) 5, 4, 3, 2 and 1 (d) 5, 1, 4, 2 and 3 [ Ans. (b)]

Conside Considerr the follo followin wing g operat operation ionss invo involve lved d in forging a hexagonal bolt from a round bar stock, whose diameter is equal to the bolt diameter: 1. Flat Flatte teni ning ng 2. Upset psetti ting ng 3. wag ng 4. am er ng The correc correctt sequence sequence of these these operatio operations ns is (a) (a) 1, 2, 3, 4 (b) (b) 2, 3, 4, 1 (c) 2, 1, 3, 4 (d) 3, 2, 1, 4 Ans. (a)

IES – IES – 2003

IES – IES – 2005

Consider Consider the following following steps steps in forging forging a connectin connecting g rod from from the bar stock: stock: 1. Blocking 2. Trimming 3. Finishing 4. Edging Select the correct sequence of these operations using the codes given below: below: Codes: (a) 1‐2‐3‐4 (b) 2‐3‐4‐1 (c) (c) 3‐4‐1‐2 (d) 4‐1‐3‐2 Ans. (d)

The process of removing the burrs or flash from a forged forged compo componen nentt in drop drop forgin forging g is called: called: (a) (a) Swag Swagin ing g (b) (b) Perf Perfor orat atin ing g (c) Trimm Trimming ing (d) Fettl Fettling ing Ans. (c)

IES 2011

IES – IES – 2008

Which of the following processes processes belong to forging operation operation ? 1. Fullering 2. Swaging 3. Welding (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (b) 1, 2 and 3 only [ Ans. (a)]

The balls of the ball bearin bearings gs are manufact manufacture ured d from steel rods. The operations operations involve involved d are: 1. Grou Ground nd 2. Hot Hot forge forged d on hammer hammerss 3. Heat Heat trea treate ted d 4. Polis Polished hed What is the correct sequence of the above operations operations from start? start? (a) (a) 3‐2‐4‐1 (b) 3‐2‐1‐4 (c) (c) 2‐3‐1‐4 (d) 2‐3‐4‐1 Ans. (None) Correct sequence is 2 – 1 – 3 ‐ 4

Page 45 of 79

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IES – IES – 2001 In the forging operation, fullering is done to (a) Draw Draw out the material (b) Bend the material (c) Upset Upset the material (d) Extruding Extruding the material Ans. (a)

IES – IES – 1998

IES 2011 Consider the following statements statements : 1. Any metal will require some time to undergo complete plastic deformation particularly if deforming metal has to fill cavitie cavitiess and corners corners of small radii. radii. 2. For For larg larger er work work piec piecee of meta metals ls that that can can reta retain in toughness at forging temperature it is preferable to use forge press rather rather than forge hammer hammer.. (a) 1 and and 2 are correc correctt and 2 is the the reason reason for for 1 (b) 1 and 2 are are correc correctt and 1 is the reason reason for for 2 (c) 1 and 2 are correct correct but unrelated (d) 1 only correct [ Ans. (b)]

The The bend bendin ing g forc force e requ requir ired ed for for V ‐bending bending,, U‐ bending bending and Edge Edge‐bend bendin ing g will will be in the the rati ratio o of (a) (a) 1 : 2 : 0.5 0.5 (b) (b) 2: 1 : 0.5 0.5 (c) 1: 1: 2 : 1 (d) 1: 1 : 1 Ans. (a)

IES – IES – 2005

IES – IES – 2008

Match List I (Type of Forging) with List II (Operation) and select select the corre correct ct answe answerr using using the code code give given n below below the Lists: List I List II A. Drop Forging Forging 1. Metal is gripped in the dies and pres pressu sure re isapp ie on t e eate eate en B. Press Press Forg Forging ing 2. Squeez Squeezing ing actio action n C. Upset Upset Forging Forging 3. Metal Metal is placed placed betw between een roller rollerss and pushed D. Roll Roll Forg Forging ing 4. Repea Repeate ted d hammerblow hammerblowss [ Ans. (c)] A B C D A B C D (a) 4 1 2 3 (b) 3 2 1 4 (c) 4 2 1 3 (d) 3 1 2 4

Match List‐I with List‐II and select the correct answer using the code given given below below thelists: List‐I ( Fo Forg in in g Te Tec h hn n iq iq ue ue) Li st st‐II (Process) A. Smith Forging 1. Material is only upset to get the desired shape B. Drop Drop Forg Forgin ing g 2. Carr Carrie ied d out out manu manual allyope lyopen n dies dies C. Press Press Forg Forging ing 3. Done Done inclosed inclosed impre impressi ssiondies ondies by hammer hammerss in blows D. Machin Machinee Forg Forging ing 4. Donein Donein closed closed impre impressi ssiondies ondies bycontinu bycontinuous ous squeezing force Code: A B C D (a) 2 3 4 1 (b) 4 3 2 1 (c) 2 1 4 3 (d) 4 1 2 3

Ans. (a)

Page 46 of 79

4

7/24/2011

IES – IES – 1998

IES – IES – 1994

Which one of the following processes is most common commonly ly used for the forgin forging g of bolt bolt heads heads of hexagonal hexagonal shape? (a) Closed Closed die drop forging forging (b) die for (b) Open O di upsett forging f ging i (c) Close die press press forging forging (d) Open Open die progress progressive ive forging forging Ans. (c )

In drop drop forgin forging, g, forging forging is done done by droppi dropping ng (a) The work work piece piece at high velocity velocity (b) The hammer hammer at high velocit velocityy. (c) The die with with hammerat hammerat high high veloc velocity ity (d) (d) a weig weight ht on hamm hammer er to prod produc ucee the the requ requis isit itee impact. Ans. (c)

IES – IES – 2009

IAS – IAS – 2001

Match List‐I with List‐II and select the correct answer using the codegiven below below the Lists: Lists: List‐I List‐II (Article) (Processing Method) A. Disposable coffee cups 1. Rotomoulding Rotomoulding B. Lar Lar e wate waterr tank tankss 2. Ex and andab ablebea lebead d moul mouldi din n C. Plastic s he heets 3. Th The rm rmoforming D. Cushion pads 4. Bl ow ow moulding 5. Calen Calendar daring ing Code: (a) A B C D (b) A B C D 3 5 1 2 4 5 1 2 (c) A B C D (d) A B C D 4 3 3 1 3 1 5 2 Ans. (d)

Match List I (Forging operations) with List II (Descriptions) and selectthe correct correct answer answer using the codesgiven belowthe Lists: Lists: List I List II A. Flattening 1. Thickness is reduced continuously at different sections along length B. Draw Drawin ing g 2. Meta Metall isdisp isdispla lace ced d away way from from cent centre re,, reducin reducin thickness thickness in middle middle and increasi increasin n length C. Fuller llerin ing g 3. Rod is pulle ulled d thr throug ough adie adie D. Wire Wire drawi drawing ng 4. Pressu Pressurea rea workp workpiec iecee betwee between n two two flatdies Codes: A B C D A B C D (a) 3 2 1 4 (b) 4 1 2 3 (c) 3 1 2 4 (d) 4 2 1 3 Ans. (b)

IAS – IAS – 2000

IAS – IAS – 1998

Drop forging forging is used to produce produce (a) Small components (b) Large Large compone components nts (c) Identical Components Components in large large numbers (d) Medium Medium‐size components Ans. (a)

The forging defect due to hindrance to smooth flow of meta metall in the the comp compon onen entt call called ed 'Lap' 'Lap' occu occurs rs because (a) The corner corner radius radius provided provided is too too large large e corne ornerr ra us prov prov e s toosma (c) Draft Draft is not provided provided (d) The shrinkag shrinkagee allowanc allowancee is inadequate inadequate Ans. (b)

Page 47 of 79

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7/24/2011

IAS‐1996

IAS – IAS – 2002

Compound Compound die performs performs (a) Two or mor moree opera operatio tions ns at one statio station n in onestroke onestroke (b) Two Two or more operations at different stations in one stroke (c) Only Only oneoperati oneoperations ons and that that too too at onework stat station ion (d) Two Two operations at two different work work stations in the same stroke stroke Ans. (a)

Consider the following statements related to forging: 1. Flash lash is excess material added to stock which stock which flows around parting line. 2. as e ps n ng o t n r s an osses n upper die. 3. Amou Amount nt of flash of flash depends upon forging force. Which of the of the above statements are correct? (a) (a) 1, 2 and 3 (b) 1 and 2 (c) (c) 1 and 3 (d) 2 and 3 [Ans. (b)]

IES – IES – 2006 ‐ Conventional

IES 2011

A certain disc of lead of radius 150 mm and thickness 50 mm is reduced to a thickness of 25 mm by open die forging. forging. If the co‐efficient of friction between between the job and die is 0.25, determine the maximum forging force. The aver averaa e shea shearr ield ield stre stress ss of lead lead can be taken taken as N/mm2. [10 – Marks]

Assertion (A) : Hot tears occur during forging because of inclusions in the blank material material Reason (R) : Bonding between the inclusions and the the pare parent nt mate materi rial al is thro throug ugh h phys physic ical al and and chemical bonding. (a) Both A and R are individually true and R is the correct correct explanation explanation of A (b) Both A and R are individually true but R is NOT the correct correct explanation explanation of A (c) A is true but R is false (d) A is false but R is true Ans. (c)

IES – IES – 2007 Conventional A cylinder of height 60 mm and diameter 100 mm is forged at room temperature between two flat dies. Find the die load at the end of compression to a height 30 mm, using slab method of analysis. The yield strength of the work work mater material ial is iven as 120 N mm2 and the coefficie coefficient nt of friction friction is 0.05. Assume that volume volume is constant after deformation. There is no sticking. Also [20‐Marks] find mean mean die pressure pressure..

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8/5/2011

Extrusion y

1 Extrusion 2 Die backer 3 Die 4 Billet 5 Dummy bl Dummy bloc ock k 6 Pressing stem 7 Container liner 8 Container body


Made Easy


Direct Extrusion y

Made Easy

Indirect Extrusion

A soli solid d ram ram driv drives es the the enti entire re bill billet et to and and thro throug ugh h a stat statio iona nary ry die die and and must must prov provid idee addi additi tion onal al powe powerr to overcome the frictional resistance between the surface of the moving moving billetand the confining confining chamber. chamber.

y

y


The extrusion process is like sq ueezing toothpaste out of a tube. tube.

Made Easy

A hollow ram drives the die back through a stationary, confined billet.

Since no relative motion, friction between the billet and the chamber chamber is eliminate eliminated. d. Compiled By: S K Mondal

Made Easy

Impact Extrusion

The basic principles of forward and forward and backward backward cold extrusion using open and closed dies. Compiled By: S K Mondal

Made Easy


Made Easy

Page 49 of 79

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8/5/2011

GATE‐2006

JWM 2010 Assertion (A) : Extrusion speed depends on work material. Reason (R) : High extrusion speed causes cracks in the material. material. correct correct explanati explanation on of A (b) Both A and R are individuall individuallyy true but R is not the correct correct explanati explanation on of A (c) A is true true but R is false (d) A is false but R is true Ans. (a) Compiled By: S K Mondal

In a wire drawing drawing operation, operation, diameter diameter of a steel wire is reduced from 10 mm to 8 mm. The mean flow stress of the material is 400 MPa. The ideal force requ requir ired ed for for draw drawin ing g (ign (ignor orin ing g fric fricti tion on and and redund redundantwork antwork is (a) 4.48 kN (b) 8.97 kN (c) 20.11 kN (d) 31.41 kN Ans. (b)

Made Easy


Made Easy

GATE‐2001

GATE‐2003

For rigid perfectly ‐plastic work material, negligible interf interface ace fricti friction on and no redund redundant ant work, work, the theoretical theoretically ly maximum maximum possible possible reduction reduction in the wire drawing operation is . . (c) 1.00 (d) 2.72 Ans. (b)

A brass billet is to be extruded from its initial diameter of 100 mm to a final diameter of 50 mm. The The worki orking ng temp temper erat atur ure e of 700° 700°C C and and the the extrusion constant is 250 MPa. The force required forextrusio forextrusion n is (a) (a) 5.44 5.44 MN (b) (b) 2.72 2.72 MN (c) (c) 1.36 1.36 MN (d) (d) 0.36 .36 MN Ans. (b)


Made Easy


GATE‐1996

GATE‐1994

A wire of 0.1 mm diameter is drawn from a rod of 15 mm diameter. Dies giving reductions of 20%, 40% and 80% are available. For minimum error in the final size, the number of stages and reduction at each each sta sta e res res ecti ective vell woul would d be (a) 3 stages stages and and 80% reduction for all three stages (b) 4 stages stages and 80% reduction reduction for first three three stages followed followed by a finishing stage stage of 20% reduction reduction (c) 5 stages and reduction of of 80%, 80%.40%, 40%, 20% in a sequen sequence ce (d) none of of the ab above Ans. (b) Compiled By: S K Mondal

Made Easy

The proces processs of hot extrus extrusion ion is used used to produc produce e (a) Curtain Curtain rods made of alumini aluminium um (b) Steel Steel pipes/ordomesticwater pipes/ordomesticwater supply supply (c) Stainless steel tubes used in furniture (d) Large Large she pipes used in city water water mains Ans. (a)

Made Easy


Made Easy

Page 50 of 79

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8/5/2011

IES – IES – 2007

IES – IES – 2007

Which one of the following is the correct statement? (a) Extrusion Extrusion is used for the manufactur manufacturee of seamless seamless tubes. b) Extrusion is used for reducin reducin the diameter diameter of round bars and tubes by rotating dies which open and close rapidly rapidly on the work? work? (c) Extrusion is used to improve improve fatigue fatigue resistance resistance of the metal metal by setting setting up compres compressive sive stresses stresses on its surface surface (d) Extrusi Extrusion on compr comprise isess press pressing ing the metal metal inside inside a chamber to force it out by high pressure through an orifice which is shaped to provide the desired from of the finished pa part. Compiled By: S K Mondal Made Easy Ans. (d)

Assertion (A): Greater force on the plunger is required in case of direct direct extrusion extrusion than indirect indirect one. Reason (R): In case of direct extrusion, the direction of the force applied on the plunger and the direction of the move movemen mentt of the extrud extruded ed metalare thesame. (a) Both A and R are individuall individuallyy true and R is the correct explanation explanation of A (b) Both Both A and R are indivi individua duall llyy true but R is not the correct correct explanation explanation of A (c) A is true true but but R is false (d) A is false false but R is true Ans. (b) Compiled By: S K Mondal Made Easy

IES – IES – 1993

IES – IES – 1994

Assertion (A): Direct extrusion requires larger force force than indirect indirect extrusion. extrusion. Reason (R): In indirect extrusion extrusion of cold steel, zinc phosphate phosphate coating coating is used. a ot an are n v ua ua y true an s t e correct correct explanati explanation on of A (b) Both A and R are individu individually ally true true but R is not the correct correct explanati explanation on of A (c) A is true true but R is false (d) A is false but R is true Ans. (b) Compiled By: S K Mondal Made Easy

Meta Metall extr extrus usio ion n proc proces esss is gene genera rall lly y used used for for producing (a) Uniform Uniform solid sections sections (b) Uniform hollow sections sections (c) Uniform solid and hollow hollow sections (d) Varying solid and hollow sections. sections. Ans. (c)

IES – IES – 2009

IES – IES – 1999

Which one of the following statements is correct? (a) In extrusion extrusion process, process, thicker walls walls can be obtained obtained by increasing the forming pressure (b) Extrusion is an ideal process for obtaining obtaining rods from meta meta av ng poor poor ens ens ty (c) As compare compared d to roll formi forming, ng, extruding extruding speed is is high (d) Impact extrusion is quite similar to Hooker's Hooker's process process including including the flow of metal being being in the same direction direction Ans. ( c) Compiled By: S K Mondal


Made Easy

Which one of the following is the correct temperatur temperature e range range for hot extrusion extrusion of aluminium? aluminium? (a) 300‐340 340°C (b) (b) 350‐400°C (c) 430‐480° 480°C C (d) (d) 550‐650°C Ans. (c)

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8/5/2011

IES – IES – 2000

IES – IES – 2009

Consider the following following statements: In forward forward extrusion extrusion process process 1. The ram and the extrude extruded d produ product ct travel travel in the same direction. 2. The ram and the extruded product travel in the opposite direction. 3. The speed of travel of the extruded product is same as that of the ram. ram. 4. The speed of travel of the extruded product is greater than that that of the the ram. ram. Which of these Statements are correct? (a) 1 and 3 (b) 2 and 3 (c) 1 and 4 (d) By: S K2Mondal and Made 4 Easy Ans. (c) Compiled

What is the major problem in hot extrusion? (a) Design of punch (b) Design of die (c) (c) Wearand Wearand tear tear of die die (d) (d) Wear ear of punc punch h Ans. (b)


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IES – IES – 2008

IES – IES – 2003

Which one of the following methods is used for the manufactur manufacture e of collapsibl collapsible e tooth‐paste tubes? (a) (a) Impa Impactex ctextr trus usio ion n (b) (b) Dir Direct ect extr extrus usio ion n (c) Deep d ra rawing (d) Pi Piercing

The extrusion process (s) (s) used for the production of toothpaste toothpaste tube is/are is/are 1. Tubeextrusio ubeextrusion n 2. Forwar Forward d extrusio extrusion n 3. Impactextru Impactextrusio sion n Select Select the correc correctt answer answer using the codes codes given given below: below: Codes: (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 3 only Ans. (d)

Ans. (a)


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IES – IES – 2001

Made Easy

IES – IES – 2006

Which of the following following statements are the salient features of hydrostatic hydrostatic extrusion? 1. It is suitable suitable for soft and ductile ductile material. material. 2. It is suitable suitable for high high‐‐strength strength super‐ super‐alloys. 3.The billet is inserted into the extrusion chamber and pressure is applied applied by a ram toextrude toextrude the billet billet throug through h the die. die. 4. The billet billet is inserted inserted into the extrusion chamber where where it is surrou surrounde nded d by a suitab suitable le liquid liquid.. The billet billet is extrude extruded d through through the die by applying applying pressure pressure to the liquid. liquid. Select the correct correct answer answer using the codes given given below: below: Codes: (a) 1 and 3 (b) 1 and 4 (c) 2 and 3 ( d) By: S K2Mondal and Made 4 Easy Ans. (d) Compiled

What does hydrostatic pressure in extrusion process improve? (a) Ductility (b) Compressive s tr trength (c) Brittleness (d) Tensile s tr trength Ans. (a)


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8/5/2011

IES 2010

IES – IES – 2009

Assertion (A): Pickling and washing and washing of rolled of rolled rods is carried out before wire before wire drawing. Reason (R): They lubricate They lubricate the surface to reduce friction while friction while drawing wires. drawing wires. (a) Both A and R are individually true and R is the correct correct explanati explanation on of A (b) Both A and R are individually true but R is NOT the correct correct explanati explanation on of A (c) A is true but R is false (d) A is false but R is true Ans. (c)

Which one of the following stress is involved involved in the wire drawing process? (a) (a) Comp Comprressi essivve (b) (b) Tensi ensile le (c) Shear (d) Hydrostatic s tr tress


Ans. (b)

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IES – IES – 2000

IES – IES – 1999

Match List I (Components of a table fan) with List II (Manufact (Manufacturing uring processes) processes) and select the correct correct answerusing answerusing the codes codes given given below below the Lists: Lists: List I List II A. Base with stand 1. Stam in and pressing B. Blade 2. Wire drawing C. Arma Armatu turre coil oil wir wire 3. Turnin rning g D. Armature shaft 4. Casting [ Ans. (d)] Codes:A B C D A B C D (a) 4 3 2 1 (b) 2 1 4 3 (c) 2 3 4 By: S K1Mondal Made (dEasy ) 4 1 2 3 Compiled

Match Match List List‐I with with List‐II and sele select ct the the corr correc ectt answerusing answerusing the codes codes given given below below the Lists: Lists: List‐I List‐II A. Drawing 1. Soap solution . . C. Wire dr drawing 3. Pilots D. Sheet Sheet metal metal opera operatio tions ns using using 4. Crate Craterr progressive d ie ies 5. Ironing Ans. (d) Code:A B C D A B C D (a) 2 5 1 4 (b) 4 1 5 3 (c) 5 2 3 4 (d) 5 2 1 3 Compiled By: S K Mondal

IES – IES – 1996 Match List List I with List II and select the correct correct answer answer List I (Metal/forming (Metal/forming process) process) List II (Associated (Associated force) force)

A. Wire drawing B. Extrusion C. Blanking D. Bending Codes:A B C (a) 4 2 1 (c) 2 3 1 Ans. (c)

1. 2. 3. 4. D 3 4


(b) (d)

Shear force Tensile fo force Co C ompressive fo f orce Spring back force A B C D 2 1 3 4 4 3 2 1

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IES – IES – 1996 In wire drawing process, the bright shining surface on the wire wire is obtain obtained ed if if one one (a) does not use a lubricant lubricant (b) uses solid powdery powdery lubricant. (c) uses thick thick paste lubricant (d) uses thin thin film lubricant lubricant Ans. (d)

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IES – IES – 1994

IES – IES – 1993

Match List I with List II and select the correct answer using the codes given below the Lists:

Matc Match h List List I with with List List II and and sele select ct the the corr correc ectt answerusing answerusing the codes codes given given below below the lists: lists: List I (Mechanical (Mechanical property) property) List II (Related (Related to) A. Malleability 1. Wire drawing . . C. Resilience 3. Cold rolling D. Isotropy 4. In Indentation 5. Dire Direct ctio ion n [Ans [Ans.. (b)] (b)] Codes:A B C D A B C D (a) 4 2 1 3 (b) 3 4 2 5 (c) 5 4 2 3 (d) 3 2 1 5

List I (Metal farming process) List II (A similar (A similar process)

A. Blanking . o n ng C. Extrusion D. Cup Cup drawing Codes odes:A :A (a) 2 (c) 3

B 3 2

1. 2. 3. 4. 5.

C D 4 1 (b) Compiled By: S K Mondal Made 1 5 (dEasy )

Wire drawing erc ng Em E mbossing Rolling Bending [Ans. (d)] A B C D 2 3 1 4 2 3 1 5


IES – IES – 2007 Which metal forming process manufactur manufacture e of long steel wire? (a) (a) Deepdr Deepdraawing wing (b) (b) Forgi orging ng (c) Drawing (d) Extrusion

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IES – IES – 2005 is

used

for

Which of the following types of stresses is/are involved involved in the wire‐drawing drawing operation? operation? (a) Tensile ensile only (b) Compres Compressiv sivee only (c) A combinat combination ion of tensile tensile and compre compressiv ssivee stresses stresses (d) A combinat combination ion of tensile, tensile, compressiv compressivee and shear stresses

Ans. (c)

Ans. (a) Compiled By: S K Mondal

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IES – IES – 2000

IES – IES – 1998

Which one of the following lubricants is most suitable suitable for drawing drawing mild mild steel wires? wires? (a) Sodium stearate (b) Water (c) Lime Lime‐ water (d) Kerosene Ans. (c)


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Assertion (A): The first draw in deep drawing operation can have up to 60% reduction, the second draw up to 40% reductionand, reductionand, the third third drawof about 30% only. only. Reason (R): Due to strain hardening, the subsequent dra draws in a dee dee dra drawin win o erat eratio ion n hav have redu reducced percentages. (a) Both A and R are individuall individuallyy true and R is the correct explanation explanation of A (b) Both Both A and R are indivi individua duall llyy true but R is not the correct correct explanation explanation of A (c) A is true true but but R is false (d) A is false but R is true Ans. (a) Compiled By: S K Mondal

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8/5/2011

IES – IES – 1993

IES – IES – 2002

A moving mandrel is used in (a) Wire drawing drawing (b) Tube drawing drawing (c) Metal cutting (d) Forging

Matc Match h List List I with with List List II and and sele select ct the the corr correc ectt answer: List List I (Par (Parts ts))

A. Seamless tubes 1. Roll forming . ccurate a n sm smoot tu tu es 2. ot p ee een ng C. Su Surfa rfaces havi having ng high higheer 3. Forg orging ing hardne hardness ss and fatigu fatiguee streng strength4. th4. Cold Cold formin forming g Codes: A B C A B C (a) 1 4 2 (b) 2 3 1 (c) 1 3 2 (d) 2 4 1 Compiled By: S K Mondal Made Easy Ans. (a)

Ans. (b)


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IAS – IAS – 2004

IAS – IAS – 1995

Assertion (A): Indirect extrusion operation can be performed either by moving ram or by moving the container. Reason (R): Advantage in indirect extrusion is less . (a) Both A and R are are individu individuall allyy true true and R is the correct correct explanati explanation on of A (b) Both A and R are individuall individuallyy true but R is not the correct correct explanati explanation on of A (c) A is true true but R is false (d) A is false but R is true Ans. (d) Compiled By: S K Mondal

List List II (Man (Manuf ufac actu turi ring ng proc proces esse ses) s)

The follow following ing operat operation ionss are perfo performe rmed d whi while le prepar preparing ing the billet billetss forextrusio forextrusion n proces process: s: 1. Alkali Alkaline ne cleani cleaning ng 2. Phosph Phosphat atee coati coating ng 3. Pick Pickli ling ng 4. Lubricati Lubricating ng with reactivesoap. reactivesoap. The correc correctt sequence sequence of these operatio operations ns is (a) (a) 3, 1, 4, 2 (b) (b) 1, 3, 2, 4 (c) (c) 1, 3. 4, 2 (d) (d) 3, 1, 2, 4 Ans. (d)

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IAS – IAS – 2001

Made Easy

IAS – IAS – 1997

Matc Match h List List I (Pro (Produ duct cts) s) with with List List II (Sui (Suita tabl ble e processes) and select the correct answer using the codes codes given given below below the Lists: Lists: List I List II . onnect ng ro ro s 1. e ng ng B. Pressure v es essels 2. Extrusion C. Mac Machine tool beds 3. Forging D. Co Collapsible tubes 4. Casting Ans. (a) Codes:A B C D A B C D (a) 3 1 4 2 (b) 4 1 3 2 Compiled (c) 3 2 4 By: S K1Mondal Made (dEasy ) 4 2 3 1

Extrusion Extrusion force force DOES NOT depend upon the (a) Extrusion Extrusion ratio ratio (b) Type Type of extrusion extrusion process (c) Materia Materiall of of the die (d) Working Working temperature temperature Ans. (c)


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IAS – IAS – 2000

IAS – IAS – 2002

Assertion (A): Brittle materials such as grey cast iron cannot be extruded extruded by hydrostati hydrostaticc extrusion. extrusion. Reaso Reason( n(R) R):: In hydr hydros osta tati ticc extr extrus usion ion,, bill billet et is uniformlycompresse uniformlycompressed d from all sides sides by the liquid. liquid. a ot an are n v ua ua y true an s t e correct correct explanati explanation on of A (b) Both A and R are individu individually ally true true but R is not the correct correct explanati explanation on of A (c) A is true true but R is false (d) A is false but R is true Ans. (d) Compiled By: S K Mondal

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Assertion (A): In wire‐drawin drawing g proce process, ss, the rod cross‐sectio section n is reduc reduced ed gradu graduall ally y by drawin drawing g it several several times in successiv successively ely reduced reduced diameter diameter dies. Reason (R): Since each drawing reduces ductility of the wire wire,, so afte afterr fina finall draw drawin ing g the the wir wire is normalized. (a) Both Both A and R are are individu individuall allyy true and R is the correc correctt explanatio explanation n of A (b) Both A and R are individua individually lly true true but R is not the correc correctt explanatio explanation n of A (c) A is true true but R is false (d) A is false but R is true Ans. (b) Compiled By: S K Mondal

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IES 2011 Match List –I with –I with List –II and select the correct answer using the code given below the lists :

List –I

List –II

A. Connecting rods

1. Welding Welding

B. Pressure vessels Pressure vessels

2. Extrusion

C. Machine tool beds

3. Forming

D. Collapsible tubes

4. Casting

Codes A (a) 2 (c) 2

B 1 4

C D 4 3 1 By: S K 3Mondal Compiled

(b) (dE)asy Made

A 3 3

[Ans. (b)] B 1 4

C 4 1

D 2 2

Page 56 of 79

10

Analysis of Extrusion

For Tube

1 + B ⎡ ⎛ h1 ⎞ ⎢1 − ⎜ ⎟ σd = σ 0 B ⎢ ⎝ h0 ⎠

B

⎣

But

B=

⎤ ⎥ ⎥⎦

μ1 + μ2 tan α − tan β

In the case of moving mandrel B=

μ1 − μ2 tan α − tan β

Maximum reduction possible

1 + B ⎡ ⎛ h1 ⎢1 − ⎜ B ⎢ ⎝ h0

⎤ ⎥ =1 ⎥⎦ ⎣ if μ1 = μ 2 = 0.05, α = 150 , β = 0 ⎛h ⎞ then ⎜ 1 ⎟ = 0.4275 ≈ 43% ⎝ h0 ⎠max ⎞ ⎟ ⎠

B

max

σ0 (1 + B ) ⎡ ⎛ r0 ⎞ ⎢1 − ⎜ ⎟ Extrusion σxo = B ⎢⎣ ⎝ rf ⎠ Extrusion pressure (pt) = σxo + p Extension load = pt × πr02 ⎛ A ⎞ Real condition P = k.A o ln ⎜ o ⎟ ⎝ A f ⎠ where k = extrusion extrusion constant constant

2B

⎤ ⎥ ⎥⎦

f

Reduction in Area (RA) =

r02

− r12 r02

2

2

⎛ D1 ⎞79 ⎛r ⎞ = 1 − ⎜ 1 ⎟ =Page 1 − ⎜57 of ⎟ ⎝ r1 ⎠ ⎝ D0 ⎠

Analysis of wire/Rod Drawing

⎛ ( σx + dσx ) π ( r + dr ) − σx πr2 + τ x cos α ⎜ 2πr 2

⎝

or

dx ⎞ dx ⎞ + Px . sin α ⎛⎜ 2π r ⎟ ⎟=0 cos α ⎠ cos α ⎠ ⎝

dx + Px .2rdx ta tan α = 0 σx 2rdr + dσx r + 2rτ xdx 2

dx = cot α, and and devid devide e both both by r2 dr dr dσx 2 2τ + ( σx + Px ) + x cot α = 0 or dr r r Vertical component of Px ≈ Px and that of τx can be neglected due to small half die angles. There fore only or

two principal stress

σx and Px

Tresca’s condition σx + Px = σ0

τx = μPx = μ ( σ 0 − σ x ) dσ x 2σ0 2μ + + ( σ0 − σx ) cot α = 0 Therefore

dr r Taking μ cot α = B dσx dr Or

r

2

+ ⎡⎣Bσx − (1 + B ) σ 0 ⎤⎦ r dσx

Bσx

2

− (1 + B) σ0

= dr r

Integration both side

loge ⎡⎣ Bσx

− (1 + B) σ0 ⎤⎦ B

or

log e ⎡⎣Bσx

or

Bσ x

at

= 2loge rC 2B

rC) where,C = integretion constant − (1 + B) σ0 ⎤⎦ = loge ( rC

− (1 + B ) σ0 = ( rC ) r = r0 , σx = σ b

2B

2B

∴ Bσb − (1 + B ) σ0 = ( r0C ) or

( Bσ − (1 + B ) σ ) C= x

1 2B

0

r0

2B

or

or

⎛r⎞ ∴ Bσx − (1 + B ) σ0 = ⎜ ⎟ ⎡⎣Bσb − (1 + B ) σ0 ⎤⎦ ⎝ r0 ⎠ 2B 2B σ0 (1 + B) ⎡ ⎛ r ⎞ ⎤ ⎛ r ⎞ ⎢1 − ⎜ ⎟ ⎥ + ⎜ ⎟ σb σx = B ⎢⎣ ⎝ r0 ⎠ ⎥⎦ ⎝ r0 ⎠ Page 58 of 79

Drawing stress,

σ (1 + B ) ⎡ ⎛ r ⎞ σd = 0 ⎢1 − ⎜ ⎟ B ⎢⎣ ⎝ r0 ⎠

2B

⎤ ⎛ r1 ⎞ 2 B ⎥ + ⎜ ⎟ .σb ⎥⎦ ⎝ r0 ⎠

σd > σ0 (in ideal case), therefore, maximum reduction can be found out,

Now

Die Pressure σ (1 + B ) ⎡ ⎛ h1 ⎞ ⎢1 − ⎜ ⎟ σd = 0 B ⎢⎣ ⎝ h0 ⎠ ⎛ A P = A1σ ln ⎜ 0 ⎝ A 1

B

⎤ ⎛ h ⎞B ⎥ + ⎜ 1 ⎟ .σb ⎥⎦ ⎝ h 0 ⎠

⎞ ⎟ ⎠

Maximum Reduction or Draft per pass σd =1 σ0 For zero back stress, the condition will be

(1 + B ) ⎡

1 − (1 − RA )

⎣

B

B

⎤ =1 ⎦

In wire and rod drawing, co-efficient of friction of the order 0.1 are usually obtained (by the use of proper lubrication) Now B = μ cot α

μ = 0.1 and α = 6 B = 0.1 × 9.51 .515 = 0.95 .9515 D

From hence, we will get the limited maximum reduction RA=50.5% Example: Calculate the drawing load required to obtain 30% reduction in area on a 12 mm diameter copper wire. The following data is given σ0 =240 N/mm2, 2 α =120, =0.10 Calculate the power of the electric motor if the drawing speed is 2.3 m/s. Take efficient of motor is 98%. Solution: RA = 0.30 B = μ cot α = 0.1 × cot σ = 0.95

σ (1 + B ) ⎡ ⎛ r1 ⎞ ⎢1 − ⎜ ⎟ σd = 0 B ⎢⎣ ⎝ r0 ⎠

2B

⎤ ⎥ ⎥⎦

2

⎛r ⎞ RA = 1 − ⎜ 1 ⎟ ⎝ r0 ⎠ 2 ⎛ r1 ⎞ ∴ ⎜ ⎟ = 0.7 or r1 = 0.7 × 6 = 5.02 mm mm ⎝ r0 ⎠ 1.95 ⎡ 0.95 1 − ( 0.7 ) ⎤ = 141.60 N / mm2 ⎦ 0.95 ⎣ 2 Drawing load = 141.60 × π × r1 =11.21 kN

σd = σ0 ×

Power =

11.2 11.21 1 × 2.3 2.3

η

=

25.7 25.78 8 kW = 26.3 26.31kW 1kW 0.98

Page 59 of 79

8/1/2011

Piercing (Punching) and Blanking

Sheet Metal Operation


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Made Easy

Punching Press


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Made Easy

Blanking

Punching Compiled by: S K Mondal

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Example Estimate the blanking force to cut a blank 25 mm wide and 30 mm long from a 1.5 mm thick metal strip, if the ultimate shear stress of the material is 450 N/mm 2. Also determine the work done if the percentage percentage penetration penetration is 25 percent of material thickness.

[Ans. 74.25 kN and 27.84 J] Compiled by: S K Mondal

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GATE‐2010 Statement Linked 1

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GATE‐2010 Statement Linked 2

Statement Statement for Linked Linked Answer Answer Questions: Questions: In a shear cutting operation, a sheet of 5 mm thickness is cut along a length of 200 mm. The cutting blade blade is 400 mm long and zero‐shear (S = 0) is provided on the edge. edge. The ultimate shear stren th of the sheet is 100 MPa and penetrati penetration on to thickness thickness ratio ratio is 0.2. Neglec Neglectt friction. friction.

Statement Statement for Linked Linked Answer Answer Questions: Questions: In a shear cutting operation, a sheet of 5mm thickness is cut along a length of 200 mm. The cutting blade blade is 400 mm long and zero‐shear (S = 0) is provided on the edge. edge. The ultimate shear strength of the sheet is 100 MPa and penetrati penetration on to thickness thickness ratio ratio is 0.2. Negle Neglect ct friction. friction. 400

400

S

S

Assuming force vs displacement curve to be rectangular, rectangular, the work work done (in (in J) is is (a) 10 100 (b) Compiled 200by: S K(Mondal c) 2Made 50Easy (d) 300 [Ans. (a)]

Bolster plate


A shear of 20 mm (S = 20 mm) is now provided on the blade. blade. Assumi Assuming ng force force vs displ displace acemen mentt curve curve to be trapezoid trapezoidal, al, the maximum maximum force force (in kN) exerted exerted is (a) 5 (b) Compiled 10 by: S K(Mondal c) 2Made 0 Easy (d) 40 [A [Ans. (b)]

Stripper

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Knockout y

Dowel pin

Knockout Knockout is a mechanism mechanism,, usually usually connect connected ed to and operated by the press ram, for freeing a work piece from a die. die.


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GATE‐2007

GATE 2011

The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is ‘t’ and diameter of the blanked part is ‘d’. For the same work material, if the diameter of the blanked blanked art is increa increased sed to to 1. d and thickness thickness is reduced reduced to 0.4 t, the new blankin blanking g force force in kN is (a) 3.0 (b) 4.5 (c) 5.0 (d) 8.0 Ans. (a)

The shear strength of a sheet metal is 300 MPa. The blanking force required to produce a blank of 100 mm diameter diameter from a 1.5 mm thick thick sheet is is close close to (a) 45 kN (b) 70 kN (c) 141 kN (d) 3500 kN Ans. (c)


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GATE‐2004

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GATE‐2003

10 mm diameter holes are to be punched in a steel sheet sheet of 3 mm thicknes thickness. s. Shear Shear strength strength of the material is 400 N / mm 2 and penetration is 40%. Shear provided on the punch is 2 mm. The blanking force force durin durin the o eratio eration n will will be (a) 22 22.6 kN (b) 37.7 kN (c) 61 61.6 kN (d) 94.3 kN

A metal disc of 20 mm diameter is to be punched from a sheet of 2 mm thickness. The punch and the die cleara clearance nce is 3%. The requir required ed punch punch diamet diameter er is (a) (a) 19.8 19.88 8 mm (b) (b) 19.9 19.94 4 mm c 20 2 0.0 mm 20.12 mm Ans. (a)

Ans. (a)


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GATE‐2002

GATE‐2001

In a blanking operation, the clearance is provided on (a) (a) The The die die (b) Both Both the the die and the punch punch equall equally y (c) (c) The punch punch (d) Brittl Brittlee the punch punch nor nor thedie Ans. (c)

The The cutt cuttin ing g forc force e in punc punchi hing ng and and blan blanki king ng operat operation ionss mainlydepend mainlydependss on (a) The modulus modulus of elasti elasticity city of metal metal (b) The shear shear streng strength th of metal metal (c) The bulk bulk modulus modulus of metal (d) The yield strength strength of metal metal Ans. (b)


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GATE‐1996

Made Easy

IES – IES – 1994

A 50 mm diameter d iameter disc is to be punched out from a carbon steel sheet 1.0 mm thick. The diameter of the punch punch should should be (a) 49.92 49.9255 mm (b) 50.00 50.00 mm ((c) (c) 50.0 50.075 75 mm ((d) d) d) noneof noneoff the th the abov ab bovee

In sheet sheet meta metall blan blanki king ng,, shear shear is prov provide ided d on punc punche hess and and dies dies so that that (a) Press Press load load is reduced reduced (b) Good Good cut edge edge is obtain obtained. ed. (c) Warping Warping of sheet sheet is minimiz minimized ed (d) Cut blanks blanks are straigh straight. t.

Ans. (d ) Ans. (a)


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IES – IES – 2002

Made Easy

IAS – IAS – 1995

Cons Conside iderr the the foll follow owin ing g stat statem emen ents ts rela relate ted d to piercing piercing and blanking: 1. Shear Shear on the the punch punch reduces reduces the maxim maximum um cutting cutting force 2. ear ear ncr ncrease easess t e capa capacc tyo t e pres presss nee nee e 3. Shear Shear increase increasess the the life of the the punch 4. The total total energy energy needed needed to make the cut remai remains ns unaltered unaltered due due to provisio provision n of shear shear Which of these statements are correct? correct? (a) 1 and 2 (b) 1 and 4 (c) 2 and 3 (d) by: S K3Mondal and 4Made Easy [Ans. (b)] Compiled

In blanki blanking ng operation operation the clearance clearance provided provided is is (a) 50% on punc punch h and 50% 50% on on die (b) (b) On die die (c) (c) On punc punch h (d) On die or punch punch depend depending ing upon upon design designer’ er’ss choic choicee

Ans. (c)


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IES – IES – 2006

IES – IES – 2004

In which which one of the following following is a flywheel general generally ly employed? (a) Lathe (b) Electric mo motor (c) (c) Pun Punch chin ing g mach machin inee (d) (d) Gear Gearbo boxx

Which one of the following statements is correct? If the the size size of a flywhe flywheel el in a punc punchi hing ng mach machin inee is increased (a) Then Then the fluctuati fluctuation on of speed speed and fluctuat fluctuation ion of energy w ot ecrease (b) Then the fluctuation fluctuation of speed speed will decrease decrease and the fluctuation fluctuation of energy will increase increase (c) Then the fluctuation fluctuation of speed will increase increase and the fluctuation fluctuation of energy will decrease decrease (d) Then the fluctuati fluctuation on of speed speed and f luctua luctuatio tion n of ener energybot gyboth h will will incr increa ease se [Ans [Ans.. (a)] (a)]

Ans. (c)


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IES – IES – 1999

Made Easy

IES – IES – 1997

A hole is to be punched in a 15 mm thick plate having ultimate shear strength of 3N ‐mm‐2. If the allowable crushing stress in the punch is 6 N ‐mm‐2, the diameter diameter of the smallest smallest hole whi which ch can be unched unched is e ual to (a) 15 mm (b) 30 mm (c) 60 60 mm (d) 120 mm

For 50% penetration of work material, a punch with singleshear singleshear equal equal to thickn thickness ess will will (a) Reduce Reduce the punch load to half the value (b) Increase Increase the punch load by half half the the value (c) Mainta Maintain in the same same punch punch load load (d) Reduc Reducee the punch punch load load to quart quarter er load load Ans. (a)

Ans. (b)


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IAS – IAS – 2000

Made Easy

IAS – IAS – 1994

A blank of 30 mm diameter is to be produced out of 10 mm thic thick k sheet sheet on a simple simple die. If 6% clear clearanc ance e is recommended, then the nominal diameters of pie and punch punch are respectiv respectively ely . . (b) 30.6 mm and 30 mm mm (c) 30 mm mm and 29.4 29.4 mm (d) 30 mm mm and 28.8 28.8 mm mm

In a blanking operation to produce steel washer, washer, the maximum punch load used in 2 x 10 5 N. The plate thickness is 4 mm and percentage penetration is 25. Thework done done during during this sheari shearing ng operat operation ion is (c) 600 J

(d)

8 00 J

Ans. (a)


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Made Easy

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IAS – IAS – 2002

IAS – IAS – 2007

In deciding the clearance between punch and die in press work in shearing, shearing, the following following rule is helpfu helpful: l: (a) Punch size controls hole size die size controls blank size unc s ze c on ontro s ot o e s ze a n an s ze (c) Die size contro controls ls both hole hole size and blank blank size (d) Die size controls hole size, punch size controls blank size

For punching operation the clearance is provided on which which one of the the follow following ing?? (a) The punch punch (b) (b) The The die die (c) 50% on the the punch punch and 50% 50% on on the the die (d) 1/3rd 1/3rd on the the punch punch and 2/3rd 2/3rd on the the die Ans. (b)


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IAS – IAS – 1995

IES – IES – 2002

Assertion (A): A flywheel fly wheel is attached to a punching pres presss so as to redu reducce its its spee speed d fluct fluctua uati tion ons. s. Reason Reason(R) (R):: The flywheel flywheel stores stores energy energy whe when n its speed increase. correct correct explanati explanation on of A (b) Both A and R are individ individually ually true true but R is not the correct correct explanati explanation on of A (c) A is true true but R is false false (d) A is false but R is true [Ans. (a)] Compiled by: S K Mondal

Made Easy

Which one is not a method method of reduci reducing ng cuttin cutting g forcesto forcesto preven preventt the overl overloadi oading ng of press? press? (a) Provid Providing ing shear shear on die (b) Providing Providing shear on punch (c) Increasi Increasing ng die clearanc clearancee (d) Stepping Stepping punches punches Ans. (c)

Made Easy


IAS – IAS – 2003

Made Easy

IES – IES – 2000

Match List I (Press ‐part) with List II (Function) and select the correct correct answer answer using thecodes given given belowthe lists: List‐I List ‐II (Press‐part) (Function) (A) (A) Punch Punch plat platee 1. Assi Assist stin ing g with withdr draw awal al of the the punc punch h (B) (B) Strip trippe perr 2. Advanc vancin ing g thewor hework k‐piece piece through through correct correct s ance (C) Stopper 3. Ej ec ection o f the w or ork‐piecefrom piecefrom diecavity diecavity (D) (D) Knoc Knocko kout ut 4. Hold Holdin ing g the the smal smalll pun punch ch in the the pro prope perr position Codes: A B C D A B C D (a) 4 3 2 1 (b) 2 1 4 3 (c) 4 1 2 3 (d) 2 3 4 1

Best position of crank for blanking operation in a mechanical mechanical press is (a) Top dead centre centre (b) 20 degrees degrees below top dead centre centre (c) 20 degrees degrees before before bottom bottom dead centre centre (d) Bottom Bottom dead centr centree Ans. (b)


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Made Easy

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IES – IES – 1999

IAS – IAS – 2003

Assertion (A): In sheet metal blanking operation, clea cleara ranc nce e must must be give given n to the the die. die. Reas Reason on (R): (R): The The blan blank k shou should ld be of requ requir ired ed dimensions. a ot an are n v ua y true an s t e correct correct explanati explanation on of A (b) Both A and R are individua individually lly true but but R is not the correct correct explanati explanation on of A (c) A is true true but R is false false (d) A is false but R is true true Ans. (d) Compiled by: S K Mondal Made Easy

The 'spring 'spring back' back' effect effect in press working working is (a) Elastic Elastic recovery recovery of the sheet metal after removal removal of the load (b) Regaining Regaining the original original shape shape of the sheet metal metal (c) Release Release of stored stored energy energy in the the sheet sheet metal metal (d) Partial Partial recov recovery ery of the sheet sheet metal metal Ans. (a)


Made Easy


Made Easy

Drawing Compiled by: S K Mondal

Made Easy

IES – IES – 1997 A cup of 10 cm height and 5 cm diameter is to be made from a sheet metal of 2 mm thickness. The number number of deductions deductions necessary necessary will be (a) One wo (c) Three Three (d) Four Four Ans. (c) Compiled by: S K Mondal

Made Easy


Made Easy

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GATE‐2008

Surface scratches y

In the deep drawing of cups, blanks show a tendency to wrinkle up around the periphery (flange). The most likely cause and remedy of the phenomenon are, respectively,

Die or punch not having a smooth surface, insufficient lubrication


blank holder holder pressure (B) High High blank blank holder holder pressure pressure and high friction; friction; Reduce blank blank holder holder pressur pressuree and apply lubrican lubricantt (C) High temperature temperature causing increase in circumferential circumferential length: Apply coolant to blank blank (D) Buckling due to circumferential compression; decrease blank blank holder holder pressur pressuree [Ans. [Ans. (a)] (a)]

Made Easy


GATE‐1999

GATE‐2003

Identify Identify the stress stress ‐ state in the FLANCE portion portion of a PARTIALL PARTIALLYDRAWN YDRAWN CYLINDRICAL CUP when deep ‐ drawing drawing without without a blank holder (a) Tensile ensile in all three three directio directions ns o st stress n t e ang ange at at a , ecaus causee t ere ere s no no blank‐holder (c) Tensile ensile stress stress in one direction direction and compre compressiv ssivee in theone other other direct direction ion (d) Compressi Compressive ve in two directions directions and tensile tensile in the thir third d dir directi ection on [Ans [Ans.. (b)] (b)] Compiled by: S K Mondal

A shell of 100 mm diameter and 100 mm height with the corner radius of 0.4 mm is to be produced by cup drawin drawing. g. The requir required ed blank blank diamete diameterr is (a) 11 118 mm (b) 161 mm c 224 mm 312 mm Ans. (c)

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GATE‐2006

Made Easy

IES – IES – 2008

Match Match the items items in columns columns I and II. Column I Column I P. Wrinkling 1. Yield pointelo telongation Q. Orange p ee eel 2. Anisotropy . tretc erst erstrra ns 3. argegra egra n s ze S. Earing 4. Insufficient bl blank ho holding force [Ans. (d)] 5. Finegr Finegrai ain n siz size 6. Exce Excessi ssive ve blank blank holdi holding ng forc forcee (a) (a) P – 6, Q – 3, R – 1, S – 2 (b) (b) P – 4, Q – 5, R – 6, S – 1 (c) (c) P – 2, Q – 5, R – 3, S – 4 (d) (d) P – 4, Q – 3, R – 1, S – 2 Compiled by: S K Mondal

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A cylindrical vessel with flat bottom can be deep drawn by (a) Shallow Shallow drawing drawing (b) Single Single action action deep drawing drawing (c) Doubleactio Doubleaction n deep deep draw drawing ing (d) Triple Triple action action deep drawing drawing Ans. (c)

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Made Easy

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IES – IES – 1999

IES – IES – 1993

Conside Considerr the follow following ing statem statement ents: s: Earrin Earring g in a draw drawn n cup cup can can be be due due to non non‐uniform 1. Spee Speed d of the the pres presss 2. Clearanc Clearancee between between tools tools 3. Material Material properti properties es 4. Blank Blank holdin holding g Which of these statements are correct? correct? (a) 1, 2 and and 3 (b) (b) 2, 3 and and 4 (c) (c) 1, 3 and and 4 (d) (d) 1, 2 and and 4 Ans. (b) Compiled by: S K Mondal

Tandem andem drawin drawing g of wires wires and tubes tubes is necess necessary ary because (a) It is not not possib possible le to to reduc reducee at one stage stage (b) Annealing is needed needed between between stages (c) Acc Accurac uracyy in dimensions dimensions is not possible possible otherwise otherwise (d) Surface Surface finish improve improvess after after every every drawing drawing stage stage Ans. (a)

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IES – IES – 1994

IAS – IAS – 2007

For For obtain obtaining ing a cup of diamete diameterr 25 mm and and height height 15 mm by drawing, the size of the round blank should he approximat approximately ely (a) 42 mm (b) 44 mm c 4 mm 4 mm Ans. (c)


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In draw drawin ing g oper operat atio ion, n, prop proper er lubr lubric icat atio ion n essential essential for whichof the following following reasons? reasons? 1. To impro improvedie vedie life life 2. To reduc reducee drawin drawing g force forcess 3. To reduce reduce tempera temperature ture 4. To improvesurface improvesurface finish Select Select the correc correctt answer answer using the code given given below: below: (a) (a) 1 and 2 only nly (b) 1, 3 and 4 onl only (c) (c) 3 and and 4 only nly (d) 1, 2, 3 and and 4 Ans. (d)

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is

Made Easy

IAS – IAS – 1997

IAS – IAS – 1994

Which one of the following factor promotes the tendency tendency for wrinking wrinking in the process process of drawing? drawing? (a) Increase Increase in the ratio of thickness thickness to blank diameter diameter of work material ecr ecrease ease n t e rat o t c ness ness to an amet ameter er o work material material (c) Decrease in the holding force force on the blank (d) Use of solid solid lubricant lubricantss

Consider Consider the following following factors factors 1. Clear Clearanc ancee between between the punch punch and and the die is too small. 2. The finish finish at thecorners thecorners of the punch punch is poor poor. 3. The finish finish at thecorners thecorners of thedie is poor poor. 4. The punch punch and die alignm alignment ent is not prope properr. The factors responsible for the vertical lines parallel to the axis notice noticed d on the outside outside of a drawn drawn cylindri cylindrical cal cup would include. (a) (a) 2, 3 and and 4 (b) (b) 1 and and 2 (c) 2 and 4 (d) by: S K1Mondal , 3 andMade 4 Easy [Ans. (d)] Compiled


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Spinning


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Made Easy

GATE‐1992

tc

The thickness of the of the blank needed to produce, by power spinning a missile cone of thickness of thickness 1.5 mm and half cone half cone angle 30°, angle 30°, is (a) 3.0 mm (b) 2.5 mm c 2.0 mm 1.5 mm

= tb sinα

Ans. (a)


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IES – IES – 1994

Made Easy

IES – IES – 2006

The The mode mode of defo deform rmat atio ion n of the the meta metall duri during ng spinning is (a) Bending Bending (b) Stretching Stretching (c) Rolling and stretching stretching (d) Bending Bending and stretch stretching. ing.

Which one of the following is a continuous bending process in which opposing rolls are used to produce long sections of formed shapes from coil or strip stock? (c) Roll bending

(d)

Spinning

Ans. ( b)

Ans. (d)


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Made Easy

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Underwater Explosions

High Energy Energy Rate Rate Forming(HERF)


Made Easy


Made Easy


Made Easy


Made Easy

IES 2011 High energy rate forming process used for forming components from thin metal sheets or deform thin tubes is: (a) Petro ‐forming (b) Magnetic Magnetic pulse forming forming (c) Explosive forming (d) electro ‐hydraulic forming Ans. (b) Compiled by: S K Mondal

JWM 2010 Assertion (A) : In magnetic pulse‐forming method, magnetic field field produ produced ced by eddy eddy curren currents ts is used to create create force force between between coiland workpiec workpiece. e. Reason (R) : It is necessary for the workpiece material to have magnetic magnetic properties properties.. a o an are n v ua y rue an s e correc explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true Ans. (c) The workpiece has to be electrically conductive but need not be magnetic.

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IES 2010

IES – IES – 2007

Assertion (A) : In the high energy rate forming method, the explosive forming has proved to be an excellent method of utilizing energy at high rate and utilizes utilizes both the high explosive explosivess and low explosive explosives. s. Reason Reason R): The as ressur ressure e and and rate rate of deto detonat nation ion can be controlle controlled d for both types of explosiv explosives. es. (a) Both A and R are individually true and R is the correct explanati explanation on of A (b) Both A and R are individually true but R is NOT the correctexplanat correctexplanation ion of A (c) A is true but R is false Ans. (c) (d) A is false butCompiled R isby:true S K Mondal Made Easy

Which one of the following metal forming proces processes ses is nota hig high h energy energy rate rate formin forming g proces process? s? (a) Electro Electro‐mechanical forming (b) Roll‐forming (c) Explosive Explosive forming forming (d) Electro Electro‐hydraulic forming Ans. (b)


IES – IES – 2009

Made Easy

IES – IES – 2005

Which one of the following is a high energy rate forming forming process? process? (a) Roll forming forming (b) Electro ‐hydraulic forming (c) Rotary forging forging (d) Forward Forward extrusion

Magnet Magnetic ic formin forming g is an exampl example e of: (a) Cold forming (b) Hot fo f orming (c) (c) High High ener energy gy rate rate form formin ing g (d) (d) Roll Roll form formin ing g Ans. (c)

Ans. (b)


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Stretch Forming


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GATE‐2000

Contd......

A 1.5 mm thick sheet is subject to unequal biaxial stretching and the true strains in the directions of stretching stretching are 0.05 and 0.09. The final thickness thickness of the sheet sheet in mm mm is . . (c) 1 362 (d) 289 Ans. (b)


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Made Easy

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Ironing


Contd....

Made Easy


Bending Force

Example

2

F

=

Klσ ut t

y

w

Where Where l =Bend =Bend length =widthof the stock,mm stock,mm σ

2

ut

= Ultimatetensil Ultimatetensilestrength,MPa(N/mm estrength,MPa(N/mm )

t=

Made Easy

Calculate the bending force for a 45 o bend in aluminium blank. Blank thickness, 1.6 mm, bend length = 1200 mm, Die opening = 8t, UTS = 455 MPa, Die opening factor = 1.33

an t c nes ness,m s,mm m

w = width width of die-ope die-opening,mm ning,mm K =die-openingfactor, =die-openingfactor, (canbe usedfollowintable) usedfollowintable) Condition

V-Bending

U-Bending

y

W < 16t

1.33

2.67

0.67

W > = 16t

1.20

2.40

0.6

For U or channel bending force required is double than V than V –– bending For edge bending it will it will be about one‐half that half that for V for V ‐ bending Compiled by: S K Mondal

Ans. (145.24 kN)

Edge-Bending

Made Easy


Made Easy


Made Easy

GATE‐2005 A 2 mm thick metal sheet is to be bent at an angle of one radian with a bend radius of 100 mm. If the stretch stretch factor factor is 0.5, 0.5, the the bend allowanc allowance e is (a) 99 99 mm (b) 100 mm c 101 mm 102 mm 2mm

1 radian

Ans. (c)


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GATE‐2007

GATE‐2004

Match the correct combination for following metal working processes. Processes Associated state o f stress P. Blanking 1. Tension . . R. Coining 3. Shear [Ans. (d)] S. DeepDra DeepDrawi wing ng 4. Tensi ensionand onand Comp Compre ress ssio ion n 5. Tensi ensionand onand Shea Shearr Codes:P Q R S P Q R S (a) 2 1 3 4 (b) 3 4 1 5 (c) 5 4 3 1 (d) 3 1 2 4 Compiled by: S K Mondal

Match the following following Product Process P. Moulde ulded d lugg luggag agee 1. Inje Inject ctio ion n mould ouldin ing g Q. Packaging Packaging container containerss for liquid 2. Hot rolling rolling R. Long Long struct structur ural al shapes shapes 3. Impactextru Impactextrusio sion n S. Coll ollaps apsible ible tub tubes 4. Transfe nsferr mould ouldin ing g 5. Blo Blow moul mouldi ding ng 6. Coining [Ans. (b)] (a) (a) P‐1 Q‐4 R ‐6 S ‐3 (b) P‐4 Q‐5 R ‐2 S ‐3 (c) (c) P‐1 Q‐5 R ‐3 S ‐2 (d) P‐5 Q ‐1 R ‐2 S ‐2

Made Easy


Made Easy

IAS – IAS – 1999

IAS – IAS – 1997

Match List I (Process) with List II (Production of parts) and select the correct correct answer answer using the codes given below the lists: [Ans. (d)] List‐I List‐II A. Rolling 1. Discrete parts B. Forging 2. Rod a nd nd Wire C. Extr Extrus usio ion n 3. Wide Wide vari variet etyy of of shap shapes es with with thi thin n walls D. Dra Drawin wing 4. Fla Flat plate ates and and she sheets ets 5. Soli Solid d and and holl hollo ow part partss Codes:A B C D A B C D (a) 2 5 3 4 (b) 1 2 5 4 Compiled (c) 4 1 3 by: S K Mondal 2 (Made d) Easy 4 1 5 2

Match List ‐I (metal (metal forming forming process) with List ‐II (Associated feature) and select the correct answer using using the codes codes given given below below the Lists: Lists: List‐l List‐ II [ Ans. (c)] . an an ng 1. ear ang e B. Fl Flow forming 2. Coiled stock C. Roll forming 3. Mandrel D. Embossing 4. Closed matching dies Codes:A B C D A B C D (a) 1 3 4 2 (b) 3 1 4 2 Compiled by: S K Mondal Made Easy (c) 1 3 2 4 (d) 3 1 2 4

IES 2010 Consider the following statements: statements: The The mate materi rial al prop proper erti ties es wh whic ich h prin princi cipa pall lly y determin determine e how well a metal metal may be drawn drawn are 1. Ratio of yield stress stress to ultimate ultimate stress. stress. 2.Ra 2.Rate te of incr increa ease se of yiel yield d stre stress ss rela relati tivve to progressive progressive amounts of cold work. work. 3. Rate of work hardening. [Ans. (d) (d)] Which of the above statements is/are correct? (a) 1 and 2 only (b) 2 and 3 only (c) 1 only (d) 1, 2 and 3 Compiled by: S K Mondal

Made Easy


Made Easy

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Atomization using a gas stream

Powder Metallur

By S K Mondal

Compacting

IES – IES – 2007 Conventional y

Metal powders powders are compacted compacted by many methods, methods, but sintering is required to achieve which property? What s ot so so‐stat c press ng ng [ 2 Marks]

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Oil‐impregnated Porous Bronze Bearings

IES – IES – 2011 Conventional y y

What is isostatic isostatic pressing pressing of metal of metal powders ? What are its advantage ? [ 2 Marks]

IES 2010

IES 2010

Consider Consider the following following parts: 1. Grinding wheel 2. Brake lining ‐ . Which of these parts are made by powder metallurgy metallurgy technique? technique? (a) 1, 2 and 3 (b) 2 only (c) 2 and 3 only only (d) 1 and 2 only only Ans. (c)

Metallic Metallic powders powders can be produced produced by (a) Atomization (b) Pulverization Pulverization ‐ (d) All of of the above above Ans. (d)

IES – IES – 2002

IES – IES – 2001

The rate of production of a powder metallurgy part depends depends on (a) Flow Flow rate rate of powder powder (b) Green Green strength strength of compa compact ct (c) Apparent Apparent density density of compact compact (d) Compressibility Compressibility of powder Ans. (c)

Match List‐I (Components) with List‐II (Manufactu (Manufacturing ring Processes) Processes) and select the correct correct answerusing answerusing the codes codes given given below below the lists: lists: List I List II . ar o y meta 1. ac n ng ns. B. Clutch lining 2. Casting ing C. Gears 3. Sheet metal pressing D. Eng Engin inee block lock 4. Powder der metal etalllurgy urgy Codes:A B C D A B C D (a) 3 4 2 1 (b) 4 3 1 2 (c) 4 3 2 1 (d) 3 4 1 2

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IES – IES – 1998

GATE 2011 The operati operation on in which which oil is perme permeat ated ed into into the pores pores of a powder metallur metallurgy gy product product is known as (a) mixing (b) sintering (c) impregnation (d) Infiltration Ans. (c)

In powder metallurgy, the operation carried out to impro improve ve the bearin bearing g proper property ty of a bush bush is called called (a) infiltra infiltratio tion n (b) (b) impre impregna gnatio tion n (c) (c) plat platin ing g (d) (d) heattre heattreat atme ment nt Ans. (b)

IES – IES – 1997

IES – IES – 1999

Which of the following components can be manufactur manufactured ed by powder powder metallurgy metallurgy methods? methods? 1. Carbide tool tips 2. Bearings 3. Filters 4. Brake linings Select Select the correc correctt answer answer using using the codes codes given given below below:: (a) 1, 3 and and 4 (b) 2 and 3 (c) (c) 1, 2 and and 4 (d) (d) 1, 2, 3 and and 4 Ans. (d)

The correct correct sequen sequence ce of the given given proces processes ses in manufactur manufacturing ing by powder powder metallurgy metallurgy is (a) Blending, Blending, compacti compacting, ng, sinter sintering ing and sizing sizing (b) Blending, compacting, compacting, sizing and sintering sintering (c) Compacting, Compacting, sizing, blending and sintering (d) Compacting, Compacting, blending, sizing and and sintering Ans. (a)

IES – IES – 2001

IES – IES – 1999

Carbide‐tipped cutting tools are manufactured by powder ‐ metal metal tech techno nolo logy gy proc proces esss and and have have a composition composition of (a) Zirconium Zirconium‐Tungsten (35% ‐65%) ung ungstenca stencarr e ‐ o a t 90 ‐ 10 (c) Aluminium Aluminium oxide oxide‐ Silica (70% ‐ 30%) (d) Nickel Nickel‐Chromium ‐ Tungsten (30% ‐ 15% ‐ 55%) Ans. (b)

Assertion (A): In atomization process of manufacture of metal powder powder,, the molten metal is forced forced through a small orifice and broken up by a stream of compressed air. Reaso eason n (R): (R): The meta metall llic ic owder der obta obtain ine ed b atomizati atomization on process process is quiteresistant to oxidatio oxidation. n. (a) Both A and R are individuall individuallyy true and R is the correct explanation explanation of A (b) Both Both A and R are indivi individua duall llyy true but R is not the correct correct explanation explanation of A (c) A is true true but but R is false Ans. Ans. (c) (d) A is false false but R is true

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IES – IES – 2007

IES – IES – 2006

What are the advantages of powder metallurgy? 1. Extre Extreme me purity purity produ product ct 2. Low Low labourcost labourcost 3. Low Low equi equipm pmen entt cost cost.. Select Select the correc correctt answer answer using using the the code code given given below below (a) 1, 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only Ans. (b)

Which of the following are the limitations of powder powder metallurgy? metallurgy? 1. High High tooli tooling ng and equipm equipmentcost entcosts. s. 2. Wastag astagee of mater material ial.. 3. It cannot cannot be autom automate ated. d. 4. Expensive Expensive metallic metallic powders powders.. Select Select the correc correctt answer answer using the codes codes given given below: below: (a) On Only 1 and 2 (b) Only 3 and 4 (c) Onl Only 1 and 4 (d) Only 1, 2 and 4 Ans. (c)

IES – IES – 2004

IES – IES – 2009

Consider Consider the following following factors: factors: 1. Size and shape that can be produce produced d economic economically ally 2. Porosityof Porosityof the parts produce produced d 3. Availab vailable le press press capaci capacity ty 4. Highdens Highdensit ity y Which of the above are limitations of powder metallurgy? (a) 1, 3 and and 4 (b) (b) 2 and 3 (c) 1, 2 and and 3 (d) 1 and 2 Ans. (a)

Which of the following cutting tool bits are made by powder powder metallurgy metallurgy process? process? (a) Carbon Carbon steel steel tool tool bits bits (b) Stell Stellitetool itetool bits bits (c) (c) Cer Ceram amictoo ictooll bits bits (d) (d) HSS HSS tool tool bits bits Ans. (c)

IAS – IAS – 2003

IAS – IAS – 2003

Which of the following are produced by powder metallurgy metallurgy process? process? Cemented d carbide carbide dies 1. Cemente 2. Porous bearings . Parts with intricate intricate shapes shapes 4. Parts Select Select the correc correctt answer answer using the codes codes given given below below:: Codes: (a) (a) 1, 2 and and 3 (b) (b) 1, 2 and and 4 (c) (c) 2, 3 and and 4 (d) (d) 1, 3 and and 4 Ans. (a)

In parts produced produced by powder powder metallurgy process, process, pre‐sinter sintering ing is done to (a) Increase Increase the toughness toughness of the the compone component nt (b) Increase Increase the density density of the componen componentt (c) Facilita Facilitate te bonding bonding of non‐metallic particles (d) Facilitate Facilitate machining of the part Ans. (d)

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IAS – IAS – 2000

IAS – IAS – 1997

Consider Consider the following following processes: processes: 1. Mechanic Mechanical al pulveriza pulverization tion 2. Atomi Atomizat zation ion 3. Chemical Chemical reduction reduction 4. Sinte Sinterin ring g Which of these processes processes are used for powder preparation preparation in powder metallurgy? (a) (a) 2, 3 and and 4 (b) (b) 1, 2 and and 3 (c) (c) 1, 3 and and 4 (d) (d) 1, 2 and and 4 Ans. (b)

Assertion (A): Close dimensional tolerances tolerances are NOT possible possible with isostatic isostatic pressing pressing of metal powder powder in powder metallurgy technique. Reason Reason (R): In the process process of isostatic isostatic pressing, pressing, the press pressur ure e is equal equal in all direct direction ionss whi which ch permit permitss . (a) Both A and R are individuall individuallyy true and R is the correct correct explanation explanation of A (b) Both Both A and R are indivi individua duall llyy true but R is not the correct correct explanation explanation of A (c) A is true true but but R is false (d) A is false false but R is true Ans. (d)

IAS – IAS – 1998

IAS – IAS – 1996

Thro Throwa wawa way y tung tungst sten en manufactured by (a) Forging (c) (c) Powder der meta metall llur urgy gy Ans. (c)

carb carbid ide e (b) (d) (d)

tip tip

tool toolss

are are

Brazing Extr Extrus usio ion n

Which one of the following processes is performed in powder metallurgy to promote self ‐lubricating properties properties in sintered sintered parts? parts? (a) Infiltrat Infiltration ion (b) Impregna Impregnation tion c at ng rap t zat on Ans. (b)

IAS – IAS – 2007

IAS – IAS – 2004

Assertion (A): Mechanical disintegration of a molten metal stream into into fine particles by means of a jet of compre compressed ssed air is known known as atomiz atomizati ation. on. Reason Reason (R): (R): In atomiz atomizati ation on proce process ss inert inert‐gas gas or water cannot be used as a substitute for compressed air. (a) Both A and R are are individu individuall allyy true true and R is the correct correct explanati explanation on of A (b) Both A and R are individu individually ally true true but R is not the correct correct explanati explanation on of A (c) A is true true but R is false (d) A is false but R is true Ans. (c)

The follow following ing are the consti constitue tuent nt steps steps in the process process of powder metallurgy: metallurgy: 1. Powd Powder er condi conditio tionin ning g 2. Sint Sinter erin ing g 3. Produ Producti ction on of metall metallic ic powd powder er 4. Pressing Pressing or compact compacting ing into the desired desired shape Indentify the correct order in which they have to be performed and select the correct answer using the codes given below: (a) (a) 1‐2‐3‐4 (b) 3‐1‐4‐2 (c) (c) 2‐4‐1‐3 (d) 4‐3‐2‐1 Ans. (b)

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8/14/2011

IAS – IAS – 2003

IAS – IAS – 2007

Assertion (A): Atomization method for production of metal powders consists of mechanical disintegration of molten molten stream stream into fine particles particles.. Reason (R): Atomization method is an excellent means of makin owder owderss from from hi hi h tem eratur erature e metals metals.. (a) Both A and R are individ individuall uallyy true and R is the correct explanationof explanationof A (b) Both Both A and R are are indiv individu iduall allyy true but R is not the correct correct explanation explanation of A (c) A is true true but but R is false Ans. Ans. (c) (d) A is false false but R is true true

Consider the following basic steps involved in the production production of porous bearings: bearings: 1. Sint Sinter erin ing g 2. Mixi Mixing ng 3. Repr Repress essing ing 4. Impre Impregna gnatio tion n 5. Cold‐die‐compaction Which one of the following is the correct correct sequence of the above above steps? steps? Ans. (b)

Conventional Questions

Conventional Questions

1. Explain why metal powders are blended. Describe what hap happensdu pensduri ring ng sint sinter erin ing g. [IES [IES‐2010, 2 Marks]

1. Disc Discus usss the the terms erms fine finene ness ss and and part partic icle le size size distribution in powder powder metallurgy. metallurgy. [IES ‐2010, 2 Marks] Ans. Fineness: Is the diameter of spherical shaped particle and mean mean diameter diameter of non‐s herica hericall sha ed articl article. e.

Particle Particle size distribution distribution:: Geometri Geometricc standard standard deviat deviation ion (a measur measuree for the bredt bredth h or width of a distribut distribution), ion), is the ratio of particle particle size diameters diameters taken taken at 84.1 and 50% of the cumulative cumulative undersize undersized d weight plot, respectively respectively and mean mass diameter define the particle particle size distribut distribution. ion.

Conventional Questions Enumerate the steps involved in “powder metallurgy” process. Discuss these steps. Name the materials used in “powder “powder metallurgy” metallurgy”. What are the limitations limitations of powder m et etallurgy? [IES ‐2005, 10 Marks]

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Production Question and Answer Set 1

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