Rolling Fig Q a Analysis

7/24/2011

GATE 2008 ‐

By S K Mondal

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)

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 diameter rolls to reduce thickness without any charge in its width width.. The fricti friction on coeffi coefficie cient nt at the work 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)

‐

‐

‐

‐

1

Titles you can't find anywhere else

Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.





7/24/2011

IES – IES – 2003 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)] ‐

IES – IES – 2002 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 en o r c on a er a o e ro s Ans. (d)

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

IES – IES – 2000 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 ‐

IES – IES – 2001 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 – 1999 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





7/24/2011

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

IES – IES – 1993 The blank diameter diameter used in in thread thread rolling rolling will be be (a) Equal Equal to minordiameter minordiameter of the the thread thread (b) Equal 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 Thread rolling is restricted to (a) Ferrous Ferrous materials (b) Ductile Ductile materials (c) Hard Hard materials (d) None of the of the above Ans. (b)

IAS – IAS – 2004 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. (b)]

IAS – IAS – 2001 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 are associated with hot rolling?

IAS – IAS – 2000 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)





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 ‐

‐

‐

Match List I with List II and select the 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 high non reversing reversing mills 1. Middle roll rotates 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 equal ual siz sizeare eare rotated rotated only in one direction direction D. Clus te ter mills 4. Diameter of working roll is very very small 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 – 2003

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 Which of the statemen statements ts given given above above are correc correct? t? (a) 1 and 2 only (b) 2 and 3 only (c) 1 an 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





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 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.





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

= σ '0

d ⎡h ( p − σ '0 ) ⎤ = 2pR ( θ ± μ ) ⎦ dθ ⎣

⎡ ' ⎛p ⎞⎤ ⎢ σ 0 h ⎜ ' − 1 ⎟ ⎥ = 2 pR ( θ ± μ ) dθ ⎣⎢ ⎝ σ 0 ⎠ ⎦⎥ ⎞ d ' d ⎛ p ⎞ ⎛ p σ '0 h ⎜ ' ⎟ + ⎜ ' − 1 ⎟ ( σ 0 h ) = 2 pR ( θ ± μ ) dθ ⎝ σ 0 ⎠ ⎝ σ 0 ⎠ dθ d

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 ln ( p / σ '0 ) =

2Rθdθ ∓ Rθ2 f +

∫h

∫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

= = 2μ ∴

2Rμ dθ 2 f + Rθ

∫h

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 ⎜

h ∴ p = C σ '0 ⎛⎜ ⎞⎟ e ⎝R⎠

∓ μH

R .θ + lnC ln C hf







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 or

ho h f

or Hn

= eμ ( H

0

hn

hn . eμ Hn hl

−2Hn )

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

from H = 2

∴ θn =

=

R . tan −1 hf

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 Since R > > 0.5 h, it can be written that R -

tan a = Since

tan a

h R





(ii)

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

H0

=2

R . tan −1 hf

1⎡ 1 ⎛ 25 ⎞ ⎤ = 0.8678 . l og e ⎜ = ⎢3.306 − ⎟⎥ 2⎣ 0.142 ⎝ 20 ⎠ ⎦ ⎛ h f Hn ⎞ hf θn = . tan ⎜ ⎜ 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 N

20.3 20

Vf

= 1.5%





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

⎛ R ⎞ ⎜ hf α ⎟⎟ hf ⎝ ⎠ ⎛ 250 ⎞ 250 H0 = 2 . tan −1 ⎜ 0.0447 ⎟ × ⎜ 3.55 ⎟ 3.55 ⎝ ⎠ = 6.02 ⎛ h ⎞⎤ 1⎡ 1 H n = ⎢ H 0 − lo g e ⎜ o ⎟ ⎥ 2⎣ μ ⎝ hf ⎠⎦ H0

Now

=2

R

. tan −1 ⎜

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

∝ = 2.56o = 0.0447 rad.

= 3.55 +2 × 250 (1- cos 0.554 )

= 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 / mm 2

( b ) H0 = 6.02 ( earlier ) μ = 0.4 then Hn

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

1⎡ 1 ⎛ 4.05 ⎞ ⎤ 2.845 lo g e ⎜ = ⎢6.02 − ⎟⎥ = 2⎣ 0.4 ⎝ 3.55 ⎠ ⎦ ⎛ 3 55 ⎞ 3 55

o





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⎣ h 2 0 . 2 6 . 3 5 μ ⎝ ⎠⎦ ⎣ ⎝ f ⎠ ⎦ ⎛ 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] Solution Hint: We know p= p = l. bm pm Use.

pm

=

h ⎡h ⎤ h p d h p d h p . d h + + ⎢∫ ⎥ ∫ ∫h Δh ⎣⎢ h h ⎦⎥

1

n

0

0

b

l

n

Torque and Power The power is spent principally in four ways





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.

Rolling Fig Q a Analysis

Recommend Documents