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RollingRolling

By  S K Mondal

GATE‐2008In a single pass rolling operation, a 20 mm thickplate with plate width of 100 mm, is reduced to 18mm. The roller radius is 250 mm and rotationalspeed is 10 rpm. The average flow stress for the platematerial is 300 MPa. The power required for thematerial is 300 MPa. The power required for therolling operation in kW is closest to(a) 15.2(b) 18.2(c) 30.4(d) 45.6

Ans. (a)

GATE‐2007The thickness of a metallic sheet is reduced from aninitial value of 16 mm to a final value of 10 mm inone single pass rolling with a pair of cylindricalrollers each of diameter of 400 mm. The bite anglein degreewill bein degreewill be(a) 5.936(b) 7.936(c) 8.936(d) 9.936

Ans. (d)

GATE‐2004In a rolling process, sheet of 25 mm thickness isrolled to 20 mm thickness. Roll is of diameter 600mm and it rotates at 100 rpm. The roll strip contactlengthwill be(a) 5 mm (b) 39 mm(a) 5 mm (b) 39 mm(c) 78 mm (d) 120 mm

Ans. (b)

GATE‐1998A strip with a cross‐section 150 mm x 4.5 mm isbeing rolled with 20% reduction of area using 450mm diameter rolls. The angle subtended by thedeformation zone at the roll centre is (in radian)(a) 0 01 (b) 0 02(a) 0.01 (b) 0.02(c) 0.03 (d) 0.06

Ans. (d)

GATE‐2006A 4 mm thick sheet is rolled with 300 mm diameterrolls to reduce thickness without any charge in itswidth. The friction coefficient at the work‐rollinterface is 0.1. The minimum possible thickness ofthe sheet that can be produced in a single pass isthe sheet that can be produced in a single pass is(a) 1.0 mm (b) 1.5 mm(c) 2.5 mm (d) 3.7 mm

Ans. (c)

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IES – 2003Assertion (A): While rolling metal sheet in rollingmill, the edges are sometimes not straight and flatbut arewavy.Reason (R): Non‐uniform mechanical properties ofthe flat material rolled out result in waviness of theedges.(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 – 2002In rolling a strip between two rolls, the position ofthe neutral point in the arc of contact does notdepend on(a) Amount of reduction (b) Diameter of the rolls( ) C ffi i t f f i ti (d) M t i l f th ll(c) Coefficient of friction (d) Material of the rolls

Ans. (d)

IES – 2001Which of the following assumptions are correct forcold rolling?

1. The material is plastic.2. The arc of contact is circular with a radius greater than

the radius of the roll.3. Coefficient of friction is constant over the arc of

contact and acts in one direction throughout the arc ofcontact.

Select the correct answer using the codes given below:Codes:(a) 1 and 2 (b) 1 and 3(c) 2 and 3 (d) 1, 2 and 3 [Ans. (a)]

IES – 2001A strip is to be rolled from a thickness of 30 mm to15 mm using a two‐high mill having rolls ofdiameter 300 mm. The coefficient of friction forunaided bite should nearly be(a) 0 35 (b) 0 5(a) 0.35 (b) 0.5(c) 0.25 (d) 0.07

Ans. (a)

IES – 2000In the rolling process, roll separating force can bedecreased by(a) Reducing the roll diameter(b) Increasing the roll diameter(c) Providing back‐up rolls(d) Increasing the friction between the rolls and themetal

Ans. (a)

IES – 1999Assertion (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 pass is less than that in the first passpass is less than that in the first pass.(a) Both A and R are individually true and R is the correct 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. (b)]

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IES – 1993In order to get uniform thickness of the plate byrolling process, one provides(a) Camber on the rolls(b) Offset on the rolls(c) Hardening of the rolls(d) Antifriction bearings

Ans. (a)

IES – 1993The blank diameter used in thread rolling will be(a) Equal to minor diameter of the thread(b) Equal to pitch diameter of the thread(c) A little large than the minor diameter of the thread(d) A little larger than the pitch diameter of the thread

Ans. (d)

IES – 1992Thread rolling is restricted to(a) Ferrous materials(b) Ductile materials(c) Hard materials(d) None of the above

Ans. (b)

IAS – 2004Assertion (A): Rolling requires high friction whichincreases forces and power consumption.Reason (R): To prevent damage to the surface of therolled products, lubricants should be used.( ) 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. (b)]

IAS – 2001Consider the following characteristics of rollingprocess:1. Shows work hardening effect2. Surface finish is not good3. Heavy reduction in areas can be obtainedWhich of these characteristics are associated with hotrolling?(a) 1 and 2 (b) 1 and 3(c) 2 and 3 (d) 1, 2 and 3

Ans. (c)

IAS – 2000Rolling very thin strips of mild steel requires(a) Large diameter rolls(b) Small diameter rolls(c) High speed rolling(d) Rolling without a lubricant

Ans. (b)

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IAS – 1998Match List ‐ I (products) with List ‐ II (processes)and select the correct answer using the codes givenbelow the lists:

List – I List ‐IIA M S l d h l W ldiA. M.S. angles and channels 1. WeldingB. Carburetors 2. ForgingC. Roof trusses 3. CastingD. Gear wheels 4. Rolling [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 – 2007Match List I with List II and select the correct answer usingthe code given below the Lists:

List I List II(Type of Rolling Mill) (Characteristic)A. Two high non‐reversing mills 1. Middle roll rotates by frictionB Th hi h ill B ll ki llB. Three high mills 2. By small working roll, power

for rolling is reducedC. Four high mills 3. Rolls of equal size are

rotated only in one directionD. Cluster mills 4. Diameter 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 – 2003In one setting of rolls in a 3‐high rolling mill, onegets(a) One reduction in thickness(b) Two reductions in thickness(c) Three reductions in thickness(d) Two or three reductions in thickness dependingupon the setting

Ans. (b)

IAS – 2007Consider the following statements:Roll forces in rolling can be reduced by1. Reducing friction2. Using large diameter rolls to increase the contactarea.3. Taking smaller reductions per pass to reduce thecontact area.Which of the statements given above are correct?(a) 1 and 2 only (b) 2 and 3 only(c) 1 and 3 only (d) 1, 2 and 3 [Ans. (c)]

GATE 2011The maximum possible draft in cold rolling of sheetincreases with the

(a) increase in coefficient of friction(b) decrease in coefficient of friction(c) decrease in roll radius(d) increase in roll velocity

Ans. (a)

Analysis of Rolling

Fig. Geometry of Rolling Process

Total reduction or “draft” taken in rolling. e 1 h = h - h = 2 (R - R cos a) = D (1 - cos a)Δ

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,

= R.sin al

2 2or l = BC - CE Now BC = R. h and CE = R (1 - cos a) (1 - cos a) = 0.5 hΔ Δ

( )2R. h∴ Δ Δl = - 0.5 h P = σ

( )2Usually, 0.5 h is < R hΔ Δ

( )1/2∴ ≅ Δ l R h

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, We get equilibrium equation in x-direction as,

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

For sliding friction, lτ = μp . Simplifying and neglecting second order terms, we get

( )

( ) ( )

( )

( ) ( )

x

'x 0 0

'0

'0 '

0

' '0 0' '

0 0

d σ h2pR(θ μ)

dθ2p σ σ σ3

d h p σ 2pR θ μdθd pσ h 1 2pR θ μdθ σ

d p p dσ h 1 σ h 2pR θ μdθ dθσ σ

= ±

− = =

⎡ ⎤− = ±⎣ ⎦

⎡ ⎤⎛ ⎞− = ±⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦⎛ ⎞ ⎛ ⎞

+ − = ±⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Due to cold rolling, '0σ increases as h decreases, thus '

0σ h nearly a constant and its derivative zero.

( )

( )( )( ) ( )

( )

= ±

= + − = +

= ±+

=+ +

=

⎛ ⎞= = = = ⎜ ⎟+ ⎝ ⎠

= +

⎛ ⎞ =⎜ ⎟⎝ ⎠

∫ ∫

∫ ∫ ∫

'0

02

f f

'0

2'f0

'0 2 2

f f

2f

2f

f

p / σd 2R θ μdθ p / σ hh h 2R 1 cosθ h Rθ

d p / σ 2R θ μ dθh Rθp / σ

Integrating both side2Rθdθ 2Rμln p / σ dθ

h Rθ h RθI II

2Rθdθ 2Rθdθ 2θdθ hI lnh Rθ h h / R R

hh / R θR

hd 2θdθ R

∓

∓

−

=+

=+

=

∫

∫

2f

2f

1

f f

2RμII dθh Rθ

2μ dθh / R θR R2μ .tan .θh h

( ) −

−

⎛ ⎞∴ = +⎜ ⎟⎝ ⎠

⎛ ⎞∴ = ⎜ ⎟⎝ ⎠

=

== ∝

= = =

=

' 10

f f

' μH0

1

f f

0

1'0

h R Rln p / σ ln 2μ .tan .θ lnCR h h

hp Cσ eR

R Rwhere H 2 .tan .θ.h h

Now atentry ,θ αHence H H with θ replaced by inabove equation

At exit θ 0,H H 0There for p σ

∓

∓

In the entry zone p = oμH' o0

hC.σ eR

−⎛ ⎞⎜ ⎟⎝ ⎠

( )

o

0

μH

o

μ H H'0

0

' μH0

f

RC .eh

hp σ . eh

In the exit zonehp σ .eh

−

=

=

⎛ ⎞= ⎜ ⎟

⎝ ⎠

( )

( )

( )

−

−

−

=

=

⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

=

⎛ ⎞∴ = ⎜ ⎟⎜ ⎟

⎝ ⎠= + −

0 n n

0 n

μ H H μHn n

0 l

μ H 2Ho

f

0n 0

f

1

f f

f f nn

n f n

h h. e . eh h

hor eh

h1 1or H H ln2 μ h

R Rfrom H 2 .tan .θ.h h

h h Hθ .tan .R R 2

h h 2R 1 cosθ

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,

R. h, andR htan h R hR

2

Δ

ΔΔ − Δ

l =

l a = = 05 -

Since h,it canΔ R > > 0.5 be written that

hRΔtan a =

Since ≥∴

tan aThe maximum draft is given by

μ

hRΔ

≥ μ

( ) 2maxor, hΔ = Rμ

Q.1. In rolling process, 25 mm thick plate is rolled to 20 mm in a four high mill. Determine the co-efficient 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, oσ = 100 2N / mm for hot rolls of middle steel at about 1100oC. Solution: (i). 2h RΔ = μ

( )

( )( )

−Δμ = = =

Δ = − α

= − α

α = =0

25 20hor 0.142R 250

and h 2R 1 cosor 5 500 1 cos

8.11 0.1429

(ii) − ⎛ ⎞= α⎜ ⎟⎜ ⎟

⎝ ⎠

10

f f

R RH 2 .tan .h h

− ⎛ ⎞= × =⎜ ⎟⎜ ⎟

⎝ ⎠⎡ ⎤⎛ ⎞

= −⎢ ⎥⎜ ⎟μ ⎝ ⎠⎣ ⎦⎡ ⎤⎛ ⎞= − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎛ ⎞θ = ⎜ ⎟⎜ ⎟

⎝ ⎠⎧ ⎫⎪ ⎪⎛ ⎞= × ×⎨ ⎬⎜ ⎟

⎝ ⎠⎪ ⎪⎩ ⎭=

= + −

1

0n 0 e

f

e

f f nn

n f

250 2502. . tan 0.1429 3.30620 20

h1 1H H log2 h

1 1 253.306 .log 0.86782 0.142 20

h h H.tan .R R 2

250 20 0.8678tan20 250 2

0.0349 radh h 2R 1 cos( )

( )

θ

= + θ

= + × =

n

2f n

2

h R

20 250 0.0349 20.3mm

(iii) Backward slip = r 0 0 n

r r 0

V V V h 20.31 1 1 18.8%V V h 25−

= − = − = − =

Forward slip = −= − = − = − =f r f n

r r f

V V V h 20.31 1 1 1.5%V V h 20

Vo Vr Vf

N

(iv) μ′= = σ nHnmax n 0

f

hp p .eh

0.142 0.8678 22 20.3.100 .e 132.4N / mm203

×= × =

Q2. Sheet steel is reduced from 4.05 mm to 3.55 mm with 500 mm diameter rolls having a co-efficient of fiction of 0.04. The mean flow stress in tension is 210 N/mm2. Neglect work hardening and roll flattening. (a) Calculate the roll pressure at the entrance to the rolls, the neutral plane, and the roll exit. (b) If the co-efficient of friction is 0.40, determine the roll pressure at the neutral point. (c) If 35 N/mm2 front tensions are applied in the problem find the roll pressure at the neutral point. Solution: Given ho = 4.05 mm fh = 3.55 mm

R = 250 mm, 200.04, 210N / mmμ = σ =

(a) The roll pressure at entry and exit,

p = 0′σ = 20

2 242.5N / mm3σ =

Now − ⎛ ⎞= α⎜ ⎟⎜ ⎟

⎝ ⎠

10

f f

R RH 2 .tanh h

−

μ

⎛ ⎞= ×⎜ ⎟⎜ ⎟

⎝ ⎠=

⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟μ ⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞= − × =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

′= σ n

10

on 0 e

f

e

Hnn 0

f

250 250H 2 .tan 0.04473.55 3.55

6.02h1 1H H log

2 h

1 1 4.056.02 log 1.3632 0.04 3.55

hp . . eh

Now ⎛ ⎞ ⎛ ⎞

θ = = × = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

f f 0nn

h h H 3.55 3.55.tan . .tan 0.6815 0.009672 rad.R R 2 250 25

0.5540

And Δh = 2R (1- cosα) (4.05-3.55) = 2 × 250 × (1- cos α) or ∝ = 2.56o = 0.0447 rad.

f nθ= + −n h h 2R (1 cos ) = 3.55 +2 × 250 (1- cos 0.554o) = 3.5734 mm

μ

×

′= σ

= × =

nHnn 0

f

0.04 1.363 2

hp . .eh

3.5734242.5 e 257.78N / mm3.55

( ) ( )

( )n

0

n e

n

2n f n

2

Hnn 0

f

0.04 2.845 2

b H 6.02 earlier0.4

1 1 4.05then H 6.02 log 2.8452 0.4 3.55

3.55 3.55tan 1.4225 0.02rad250 250

h h R

3.55 250 0.02 3.65mmhp . . eh

3.65242.5 e 777.9N / mm3.55

μ

×

=

μ =

⎡ ⎤⎛ ⎞= − =⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

⎛ ⎞θ = × =⎜ ⎟⎜ ⎟

⎝ ⎠= + θ

= + × =

′= σ

= × × =

( )

( )

μ

×

′= σ − σ

= − × =

nHnn 0 f

f

0.04 1.363 2

h(c) p . . eh

3.5734242.5 35 e 220.57N / mm3.55

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. Solution:

hf = 6.35mm, R = 50cm = 500mm, μ = 0.2

ho = hf × 100 9.07mm70

=

( )

( )

Δ = − = − =

Δ = − α

= × × − α

α = =

0 f

0

h h h 9.07 6.35 2.72mmh 2R 1 cos

2.72 2 500 1 cosor 4.23 0.0738rad.

Now − ⎛ ⎞= α⎜ ⎟⎜ ⎟

⎝ ⎠

10

f f

R RH 2. .tan .h h

1500 5002 tan 0.0738

6.35 6.3510.29.

− ⎛ ⎞= × × ×⎜ ⎟⎜ ⎟

⎝ ⎠=

⎡ ⎤⎛ ⎞ ⎡ ⎤⎛ ⎞= − = − × =⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥μ ⎝ ⎠⎣ ⎦⎝ ⎠⎣ ⎦0

n 0 e ef

h1 1 1 1 9.07now H H log 10.29 log 4.262 h 2 0.2 6.35

⎛ ⎞θ = ⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞

= × × = =⎜ ⎟⎜ ⎟⎝ ⎠

f f nn

0

h h H.tan .R R 2

6.35 6.35tan 2.13 0.0273rad 1.55500 500

Q.4. A metal strip is to be rolled from an initial wrought thickness of 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= = m mp l.b p

Use. ⎡ ⎤

= + +⎢ ⎥Δ ⎢ ⎥⎣ ⎦

∫ ∫ ∫0n

0

bl n

hhh

m hh h

1p pdh pdh p.dhh

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. 4) Electrical losses in the various motors and generators.

Remarks: Losses in the windup reel and uncoiler must also be considered.

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 the rolls. The ratio of the moment arm a to the projected length of the act of contact Lp can be given as

P

a aL R hΔ

= = λ

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 TM = 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)Pπ 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 aPNπ Where P is in Newton’s and a is in metre.