Forging Fig Q a Analysis

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Forging

By  S K Mondal

GATE‐2007In open‐die forging, a disc of diameter 200 mm andheight 60 mm is compressed without any barrelingeffect. The final diameter of the disc is 400 mm. Thetrue strain is(a) 1 986 (b) 1 686(a) 1.986 (b) 1.686(c) 1.386 (d) 0.602

Ans. (c)

GATE‐1994Match 4 correct pairs between List I and List II forthe questions List I gives a number of processes andList II gives a number of products

List I List II( ) I t t ti T bi t(a) Investment casting 1. Turbine rotors(b) Die casting 2. Turbine blades(c) Centrifugal casting 3. Connecting rods(d) Drop forging 4. Galvanized iron pipe(e) Extrusion 5. Cast iron pipes(f) Shell moulding 6. Carburettor body

Ans. (a) ‐ 2, (b) ‐ 6, (c) ‐ 5, (d) – 3

GATE‐1998List I List II

(A) Aluminium brake shoe (1) Deep drawing(B) Plastic water bottle (2) Blow moulding(C) Stainless steel cups (3) Sand casting(D) Soft drink can (aluminium)

(4) Centrifugal casting(5) Impact extrusion(6) Upset forging

Ans. (A) ‐3, (B) ‐2, (C) ‐1, (D) – 5

IES‐2008Which one of the following is correct?Malleability is the property by which a metal oralloy can be plastically deformed by applying(a) Tensile stress (b) Bending stress(c) Shear stress (d) Compressive stress

Ans. (d)

IES – 2006Assertion (A): Forging dies are provided with taperor draft angles on vertical surfaces.Reason (R): It facilitates complete filling of diecavity and favourable grain flow.( ) B th A d R i di id ll t d R i th(a) Both A and R are individually true and R is thecorrect explanation of A(b) Both A and R are individually true but R is not thecorrect explanation of A(c) A is true but R is false(d) A is false but R is true

Ans. (c)

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IES – 2005Consider the following statements:Forging reduces the grain size of the metal, whichresults in a decrease in strength and toughness.Forged components can be provided with thin

ti ith t d i th t thsections, without reducing the strength.Which of the statements given above is/are correct?(a) Only 1 (b) Only 2(c) Both 1 and 2 (d) Neither 1 nor 2

Ans. (b)

IES – 1996Which one of the following is an advantage offorging?(a) Good surface finish(b) Low tooling cost(c) Close tolerance(d) Improved physical property

Ans. (d)

IES – 1993Which one of the following manufacturingprocesses requires the provision of ‘gutters’?(a) Closed die forging(b) Centrifugal casting(c) Investment casting(d) Impact extrusion

Ans. (a)

IES – 1997Assertion (A): In drop forging besides the provisionfor flash, provision is also to be made in the forgingdie for additional space called gutter.Reason (R): The gutter helps to restrict the outwardflow of metal thereby helping to fill thin ribs andbases in the upper die.(a) Both A and R are individually true and R is thecorrect explanation of A(b) Both A and R are individually true but R is not thecorrect explanation of A(c) A is true but R is false(d) A is false but R is true [Ans. (c)]

IES – 2004Match List I (Different systems) with List II(Associated terminology) and select the correctanswer using the codes given below the Lists:

List I List IIA. Riveted Joints 1. NippingJ pp gB. Welded joints 2. Angular movementC. Leaf springs 3. FulleringD. 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)

IES – 2003A forging method for reducing the diameter of a barand in the process making it longer is termed as(a) Fullering (b) Punching(c) Upsetting (d) Extruding

Ans. (a)

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IES – 2002Consider the following steps involved in hammerforging a connecting rod from bar stock:1. Blocking 2. Trimming3. Finishing 4. Fullering 5. EdgingWhich of the following is the correct sequence ofoperations?(a) 1, 4, 3, 2 and 5(b) 4, 5, 1, 3 and 2(c) 5, 4, 3, 2 and 1(d) 5, 1, 4, 2 and 3 [Ans. (b)]

IES – 1999Consider the following operations involved inforging a hexagonal bolt from a round bar stock,whose diameter is equal to the bolt diameter:1. Flattening 2. Upsetting

S i C b i3. Swaging 4. CamberingThe correct sequence of these operations is(a) 1, 2, 3, 4 (b) 2, 3, 4, 1(c) 2, 1, 3, 4 (d) 3, 2, 1, 4

Ans. (a)

IES – 2003Consider the following steps in forging a connectingrod from the bar stock:1. Blocking 2. Trimming3. Finishing 4. EdgingSelect the correct sequence of these operations using thecodes given below:Codes:(a) 1‐2‐3‐4 (b) 2‐3‐4‐1(c) 3‐4‐1‐2 (d) 4‐1‐3‐2

Ans. (d)

IES – 2005The process of removing the burrs or flash from aforged component in drop forging is called:(a) Swaging (b) Perforating(c) Trimming (d) Fettling

Ans. (c)

IES 2011Which of the following processes belong to forgingoperation ?

1. Fullering2. Swaging3. Welding

(a) 1 and 2 only(b) 2 and 3 only(c) 1 and 3 only(b) 1, 2 and 3 only [Ans. (a)]

IES – 2008The balls of the ball bearings are manufacturedfrom steel rods. The operations involved are:1. Ground2. Hot forged on hammers3. Heat treated4. PolishedWhat is the correct sequence of the aboveoperations from start?(a) 3‐2‐4‐1 (b) 3‐2‐1‐4(c) 2‐3‐1‐4 (d) 2‐3‐4‐1

Ans. (None) Correct sequence is 2 – 1 – 3 ‐ 4

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VideoVideo

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

Ans. (a)

IES 2011Consider the following statements :1. Any metal will require some time to undergo completeplastic deformation particularly if deforming metal hasto fill cavities and corners of small radii.2. For larger work piece of metals that can retaintoughness at forging temperature it is preferable to useforge press rather than forge hammer.(a) 1 and 2 are correct and 2 is the reason for 1(b) 1 and 2 are correct and 1 is the reason for 2(c) 1 and 2 are correct but unrelated(d) 1 only correct [Ans. (b)]

IES – 1998The bending force required for V‐bending, U‐bending and Edge‐bending will be in the ratio of(a) 1 : 2 : 0.5 (b) 2: 1 : 0.5(c) 1: 2 : 1 (d) 1: 1 : 1

Ans. (a)

IES – 2005Match List I (Type of Forging) with List II (Operation)and select the correct answer using the code givenbelow the Lists:

List I List IIA. Drop Forging 1. Metal is gripped in the dies and

i li d h h d dpressure is applied on the heated endB. Press Forging 2. Squeezing actionC. Upset Forging 3. Metal is placed between rollers and

pushedD. Roll Forging 4. Repeated hammer blows [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

IES – 2008Match List‐I with List‐II and select the correct answer using the codegiven below the lists:List‐I (Forging Technique) List‐II (Process)A. Smith Forging 1. Material is only upset to get the desired shapeB. Drop Forging 2. Carried out manually open diesC. Press Forging 3. Done in closed impression dies by hammers in

blowsD. Machine Forging 4. Done in closed impression dies by continuous

squeezing forceCode: 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)

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IES – 1998Which one of the following processes is mostcommonly used for the forging of bolt heads ofhexagonal shape?(a) Closed die drop forging(b) O di t f i(b) Open die upset forging(c) Close die press forging(d) Open die progressive forging

Ans. (b)

IES – 1994In drop forging, forging is done by dropping(a) The work piece at high velocity(b) The hammer at high velocity.(c) The die with hammer at high velocity(d) a weight on hammer to produce the requisiteimpact.

Ans. (c)

IES – 2009Match List‐I with List‐II and select the correct answer usingthe code given below the Lists:

List‐I List‐II(Article) (Processing Method)

A. Disposable coffee cups 1. RotomouldingB. Large water tanks 2. Expandable bead mouldingg p gC. Plastic sheets 3. ThermoformingD. Cushion pads 4. Blow moulding

5. CalendaringCode:(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 2Ans. (d)

IAS – 2001Match List I (Forging operations) with List II (Descriptions) andselect the correct answer using the codes given below the Lists:

List I List IIA. Flattening 1. Thickness is reduced continuously at

different sections along lengthB. Drawing 2. Metal is displaced away from centre,

reducing thickness in middle and increasingreducing thickness in middle and increasinglength

C. Fullering 3. Rod is pulled through a dieD. Wire drawing 4. Pressure a workpiece between two flat diesCodes: 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 – 2000Drop forging is used to produce(a) Small components(b) Large components(c) Identical Components in large numbers(d) Medium‐size components

Ans. (a)

IAS – 1998The forging defect due to hindrance to smooth flowof metal in the component called 'Lap' occursbecause(a) The corner radius provided is too large(b) Th di id d i t ll(b) The corner radius provided is too small(c) Draft is not provided(d) The shrinkage allowance is inadequate

Ans. (b)

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IAS‐1996Compound die performs(a) Two or more operations at one station in one stroke(b) Two or more operations at different stations in onestroke(c) Only one operations and that too at one work station(d) Two operations at two different work stations in thesame stroke

Ans. (a)

IAS – 2002Consider the following statements related to forging:1. Flash is excess material added to stock which flows around parting line.

Fl h h l  i  filli   f thi   ib   d b  i    2. Flash helps in filling of thin ribs and bosses in upper die.3. Amount of flash depends upon forging force.Which of the above statements are correct?(a) 1, 2 and 3 (b) 1 and 2(c) 1 and 3 (d) 2 and 3 [Ans. (b)]

IES 2011Assertion (A) : Hot tears occur during forgingbecause of inclusions in the blankmaterialReason (R) : Bonding between the inclusions andthe parent material is through physical andchemical bonding.(a) Both A and R are individually true and R is thecorrect explanation of A(b) Both A and R are individually true but R is NOT thecorrect explanation of A(c) A is true but R is false(d) A is false but R is true

Ans. (c)

IES – 2006 ‐ ConventionalA certain disc of lead of radius 150 mm and thickness 50mm is reduced to a thickness of 25 mm by open dieforging. If the co‐efficient of friction between the job anddie is 0.25, determine the maximum forging force. Theaverage shear yield stress of lead can be taken as 4average shear yield stress of lead can be taken as 4N/mm2. [10 – Marks]

IES – 2007 ConventionalA cylinder of height 60 mm and diameter 100 mm isforged at room temperature between two flat dies. Findthe die load at the end of compression to a height 30mm, using slab method of analysis. The yield strength ofthe work material is given as 120 N/mm2 and thethe work material is given as 120 N/mm and thecoefficient of friction is 0.05. Assume that volume isconstant after deformation. There is no sticking. Alsofind mean die pressure. [20‐Marks]

Analysis of Forging

Fig. Open die forging

( )σ + σ − σ − τ =

σ τ− =

x x x x

x x

d Bh Bh 2 dx B 0d 2 0dx h

For Sliding Friction τ = μ

σ μ− =

x

nx

pd 2 por 0 ..........eq Adx h

Assuming xσ and p as principal stresses, and taking Tresca’s yield theory; we have for plan strain, σ − σ = σ =1 3 0 2k

x 0pσ + = σ ←p being compressive σ

= −

μ= −

x

n

d dpdx dxfrom eq A

dp 2 dxp h

( )

( )

μ= − +

= σ =

=σ

μσ = − +

μ= σ +

e

x

0

e 0

e 0

integrating2log p x Ch

Now at x L, 0, stress free suface ,so we havep

2or log L Ch2or C log Lh

( )

( )

( )

( )

μ −

μ −

μ −

⎛ ⎞ μ= −⎜ ⎟σ⎝ ⎠

= σ

= =σ

⎡ ⎤σ = σ − = σ −⎣ ⎦

e0

2 L x /h0

2 L x /h

0

2 L x /hx 0 0

p 2log L xh

or p ep por e

2k

p 1 e

Distribution of p and iσ

( )2 L

h

2 Lh

max 0

max

x 0max

p ep L1 2

2k h

1 eμ

μ

= σ

⎛ ⎞ + μ⎜ ⎟⎝ ⎠

⎡ ⎤σ = σ −⎢ ⎥

⎣ ⎦

As h decreases, p will increase for the same value of μp is directly proportional to μ. Now total forging load

= ∫L

t 0p 2B pdx

Average pressure ( ) = = ∫L

ta

0

p 1p pdx2BL L

μ⎡ ⎤σ= − +⎢ ⎥μ ⎣ ⎦

2 L0 h

ahp 1 e

L 2

As if 2 Lhμ is small then,

⎡ ⎤μ μ= = + + −⎢ ⎥σ μ ⎣ ⎦

μ⎡ ⎤= +⎢ ⎥⎣ ⎦

2 2a a

20

p p h 2 L 4 L1 12k 2 L h 2h

L1h

Sticking Friction When there is a condition of sticking friction, the work piece material does not slide along the die face and actually becomes a part of the die face and there is sub surface flow of metal. The frictional shear stress at the die interface cannot be more than the yield shear stress of the material.

So, for sticking friction στ = = 0

x k (Tresca condition)2

σσ− =

σ− − =

0x

0

dor 0dx h

dp 0dx h

σ= − +0or p x c

h

Now usually the sliding friction exists near the edges of the work piece (x = L), where the pressure is low, but at some distance nearer to the centre line, sticking fiction may exist. Let sticking occurs at x = xs, where.

τs = 0p2σ

μ =

μ −σ= μσ

⎛ ⎞= − ⎜ ⎟μ μ⎝ ⎠= =

σ= − +

σ= +

s2 (L x )/h00

S e

S S

0S S

0S S

e2

h 1or x L log2 2

at x x ,p p

now p .x Ch

C p .xh

In the sticking region ( )0S Sp p x x

hσ

= + −

s2 (L x )/hS 0p e μ −= σ

Hot Forging STICKING FRICTION IS HIGH and sticking regime extends over the whole interface, for this situation.

στ = =

σ + = −σ

σ = −

− − =∴

= − +

0x

x o

x

k2

now por d dp

hdp 2kdx 0int regratingdp x c2k h

Now at x = L, p = 2k = 0σ (Since xσ = 0)

c = 1 + Lh

( )∴ = + −

⎛ ⎞ = + =⎜ ⎟⎝ ⎠max

p L1 L x2k h

p Lor 1 (at centrex 0)2k h

Average pressure will be ⎛ ⎞= +⎜ ⎟⎝ ⎠

aLp 2k 12h

Note: The value of co-efficient of friction at sticking condition;

μ = 0.5 Tresca yield condition = 0.577 for Von-mises yield condition.

Axi-symmetric forging σ τ

− =

τ = μ = −μσ = −σ

σ μσ+ =

r

Z Z

r Z

d 2 0dr h

p (where p )d 2 0dr h

Now using Von-Mises’s yield condition,

( ) ( ) ( )

( ) ( ) ( )

( )

2 2 220 1 2 2 3 3 1

r z2 2 22

0 r z z r

r22

0 r z

0 r z r

r

e

r Z 0

e 0

2Taking , and as principal stress,we get

2and

2 2or por d dpdp 2 rp h

2 ror log p Ch

At outer surface r R, 0, p2C log

θ

θ θ

θ

σ = σ − σ + σ − σ + σ − σ

σ σ σ

σ = σ − σ + σ − σ + σ − σ

σ = σ

σ = σ − σ

σ = σ − σ = σ +

σ = −

μ= −

μ= − +

= σ = = −σ = σ

μ∴ = σ +

( )

( )

( )

2 2

e0

2 R rh

0R

0a 2

2 Rh

ma 0

R0 h

x

rh

p 2log R rh

or p e

2p rdr

h 2 Re

Average pressure will be, pR

maximum pressur

2 R

e

1

p

h

e

μ

μ−

μ

⎡ ⎤σ ⎛ ⎞ μ= − −⎢ ⎥⎜ ⎟μ

⎛ ⎞ μ= −⎜ ⎟σ⎝ ⎠

= σ

π=

π

= σ

⎝ ⎠ ⎣ ⎦

∫

Sticking Friction When sticking friction occurs over a portion of the disc, the problem can be analyzed on the same lines as for plane strain forging. Thus, sticking radius will be given as,

⎛ ⎞= − ⎜ ⎟μ μ⎝ ⎠

⎛ ⎞= − ⎜ ⎟μ μ⎝ ⎠

S e

e

h 1R R log2 2

h 1R log2 3

Pressure at the sticking radius, μ −= σ s

S

2 (R R )/h0p e

In the sticking region,

( )0S Sp p R R

hσ

= + −

Pressure at the centre R = 0

0C S Sp p .R

hσ

= +

For Hot Forging τ = k

( )

σ τ∴ − =

σ− =

σ = −

= − +

= σ = = σ

σ= σ +

σ=σ + −

−= +

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

r

r

r

r 0

00

00

max

a

d 2 0dr h

d 2k 0dr h

d dp2kor p .r ch

r R, 0and p

or c .Rh

p R rh

p R r12k h

Rp 2k 1h

Rp 2k 12h

Example-1: A piece of lead 25 mm × 25 mm × 150 mm having a yield stress of 7N/mm2 is to be pressed between flat dies a size of approximately 6.25mm × 100 mm × 150 mm, μ = 0.25. Determine pressure distribution and the total forging load. Solution: Final Dimension h = 6.25mm, 2R = 100mm

Sticking radius, ⎛ ⎞= − ⎜ ⎟μ μ⎝ ⎠

S eh 1R R log2 2

⎛ ⎞= − =⎜ ⎟×⎝ ⎠e

6.25 150 log 41.34mm0.5 0.25 2

It is clear that on the major portion of piece, Sticking conditions exist, in the sliding portion

μ −

μ −

= σ ∴ < <

= σ

2 (L x)/h0

2 (R r)/h0

p e 41.34 x 50e

At r= R i.e. outer surface p =σ0 At r = Rs,

p = 0σ× ×

= σ10.5 8.66

6.250e 2

In the sticking region, ( )0S Sp p R R

hσ

= + −

( )002 41.34 R

6.25σ

= σ + −

At the centre, R = 0 p = 2 0σ + 0.16 0σ × 41.34 = 8.61 0σ

∴Forging load per unit width = 12×100×7.61 0σ

=380.5 0σ Total forging load = 380.5 0σ × 150 = 380.5×7× 150 = 399.5 kN Example-2: A strip of lead with initial dimension 24 mm × 14 mm × 150 mm is forged between two flat dies to a final size 6 mm × 96 mm × 150 mm. If the co-efficient of friction μ = 0.25. Determine the maximum forging force. The average yield stress of lead in tension 7 N/mm2. Solution; Final dimensions: h = 6mm, 2R = 96mm

Sticking radius, ⎛ ⎞= − ⎜ ⎟μ μ⎝ ⎠

S eh 1R R log2 2

⎛ ⎞= − ⎜ ⎟× ×⎝ ⎠=

e6 148 log

2 0.25 2 0.539.68mm

It is clear that on the major portion of piece, sticking conditions exist. In the sliding portion

2 (L x)/h

02 (R r)/h

0

p ee

μ −

μ −

= σ

= σ

At r = k i.e. outer surface p = 0σ

( )2 0.5 48 39.68 /6S 0p e × −= σ

=2 0σ

( )

( )

σ= + −

σ= σ + × −

== σ + σ = σ

0S S

00

max 0 0 0

p p R RR

2 39.68 R6

at R 0p 2 6.61 8.61

Forging load per unit width = 12× 96 × 7.61 0σ

=365.28 0σ Total forging load = 365.28 0σ × 150 = 383.5 kN