Rolling Fig Q A Analysis 5i4171

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GATE‐2008

Rolling By  S K Mondal

In a single rolling operation, a 20 mm thick plate with plate width of 100 mm, is reduced to 18 mm. The roller radius is 250 mm and rotational speed is 10 rpm. The average flow stress for the plate material is 300 MPa. The power required for the rolling operation in kW is closest to (a) 15.2 (b) 18.2 (c) 30.4 (d) 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 rolling with a pair of cylindrical rollers each of diameter of 400 mm. The bite angle in degree will be (a) 5.936 (b) 7.936 (c) 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 length will be (a) 5 mm (b) 39 mm (c) 78 mm (d) 120 mm Ans. (b)

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 diameter rolls. The angle subtended by the deformation zone at the roll centre is (in radian) (a) 0.01 0 01 (b) 0.02 0 02 (c) 0.03 (d) 0.06 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 sheet that can be produced in a single is (a) 1.0 mm (b) 1.5 mm (c) 2.5 mm (d) 3.7 mm Ans. (c)

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IES – 2003 Assertion (A): While rolling metal sheet in rolling mill, the edges are sometimes not straight and flat but are wavy. Reason (R): Non‐uniform mechanical properties of the flat material rolled out result in waviness of the edges. (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. (c)]

IES – 2002 In rolling a strip between two the neutral point in the arc depend on (a) Amount of reduction (b) (c) (d) ( ) Coefficient C ffi i t off friction f i ti

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

Diameter of the rolls Material M t i l off the th rolls ll

Ans. (d)

IES – 2001 Which of the following assumptions are correct for cold rolling? 1. The material is plastic. 2. The arc of is circular with a radius greater than the radius of the roll. 3. Coefficient of friction is constant over the arc of and acts in one direction throughout the arc of . 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)]

rolls, the position of of does not

IES – 2001 A strip is to be rolled from a thickness of 30 mm to 15 mm using a two‐high mill having rolls of diameter 300 mm. The coefficient of friction for unaided bite should nearly be (a) 0.35 0 35 (b) 0.5 05 (c) 0.25 (d) 0.07 Ans. (a)

IES – 1999 Assertion (A): In a two high rolling mill there is a  limit to the possible reduction in thickness in one  . Reason (R): The reduction possible in the second   is less than that in the first .  is less than that in the first  (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 – 1993 In order to get uniform thickness of the plate by rolling process, one provides (a) Camber on the rolls (b) Offset on the rolls (c) Hardening of the rolls (d) Antifriction bearings

IES – 1993 The 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)

Ans. (a)

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

IAS – 2004 Assertion (A): Rolling requires high friction which increases forces and power consumption. Reason (R): To prevent damage to the surface of the rolled products, lubricants should be used. ( ) Both (a) B th A and d R are individually i di id ll true t and d R is i the th 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)]

IAS – 2001 Consider the following characteristics of rolling process: 1. Shows work hardening effect 2. Surface finish is not good 3. Heavy reduction in areas can be obtained Which of these 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)

IAS – 2000 Rolling 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 – 1998 Match List ‐ I (products) with List ‐ II (processes) and select the correct answer using the codes given below the lists: List – I List ‐II A. 1. Welding A M.S. M S angles l and d channels h l W ldi B. Carburetors 2. Forging C. Roof trusses 3. Casting D. 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 – 2003 In one setting of rolls in a 3‐high rolling mill, one gets (a) One reduction in thickness (b) Two reductions in thickness (c) Three reductions in thickness (d) Two or three reductions in thickness depending upon the setting Ans. (b)

IAS – 2007 Match List I with List II and select the correct answer using the 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 friction B. Three high 2. By roll, B Th hi h mills ill B small ll working ki ll power for rolling is reduced C. Four high mills 3. Rolls of equal size are rotated only in one direction D. 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 – 2007 Consider the following statements: Roll forces in rolling can be reduced by 1. Reducing friction 2. Using large diameter rolls to increase the area. 3. Taking smaller reductions per to reduce the 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 2011 The maximum possible draft in cold rolling of sheet increases 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)

4

Analysis of Rolling

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

Δ h = he - h1 = 2 (R - R cos a) = D (1 - cos 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 is,

l = R.sin a

BC 2 - CE 2

or l =

Now BC = R.Δh and CE = R (1 - cos a) (1 - cos a) = 0.5 Δh ∴

l=

R.Δh - ( 0.5 Δh )

2

P=σ 2

Usually, ( 0.5Δh ) is < R Δ h

∴ l ≅

( R Δh )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.

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,

- σl h + (σl +dσl ) (h + dh) - 2P R dθ sin θ +2 τ l R dθ cos θ = 0 For sliding friction, τ = μp . Simplifying and neglecting second order , we get

d (σx h ) dθ p − σx =

l

= 2pR(θ ± μ) 2 3

σ 0 = σ '0

d ⎡ h p − σ '0 ⎤ = 2pR ( θ ± μ ) ⎦ dθ ⎣ ⎞⎤ d ⎡ ' ⎛ p ⎢σ 0 h ⎜ ' − 1 ⎟ ⎥ = 2pR ( θ ± μ ) dθ ⎣⎢ ⎝ σ0 ⎠ ⎦⎥

(

)

⎞ d ' d ⎛ p⎞ ⎛ p σ 0 h = 2pR ( θ ± μ ) ⎜ ' ⎟ + ⎜ ' −1⎟ 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.

(

σ '0 h

)

d p / σ '0 2R = (θ ± μ ) dθ p / σ 0 h h = h f + 2R (1 − cos θ ) = h f + Rθ2

(

d p / σ '0

(p / σ ) ' 0

)=

2R ( θ ± μ ) dθ h f + Rθ2

Integrating both side

(

)

ln p / σ '0 = ∫

2Rθdθ ∓ h f + Rθ2

∫h

2Rμ dθ 2 f + Rθ

= I ∓ II I=

2Rθdθ = 2 f + Rθ

∫h

h/R =

∫

2Rθdθ = h

2θdθ

∫ h/R

⎛h⎞ = ln ⎜ ⎟ ⎝R⎠

hf + θ2 R

d ⎛ hf dθ ⎜⎝ R

⎞ ⎟ = 2θ ⎠ 2Rμ II = ∫ dθ h f + Rθ2 =

∫h

R R .tan −1 .θ hf hf

= 2μ

∴

f

2μ dθ / R + θ2

⎛h⎞ ln p / σ '0 = ln ⎜ ⎟ ∓ 2μ ⎝R⎠

(

)

R .tan −1 hf

R .θ + ln C hf

⎛h⎞ ∴ p = C σ '0 ⎜ ⎟ e∓ μH ⎝R⎠ where H = 2

R .tan −1 hf

R .θ. hf

Now at entry ,θ = α Hence H = H0 with θ replaced by ∝ in above equation At exit θ = 0 ,H = H1 = 0 There for p = σ '0

⎛h ⎞ In the entry zone p = C.σ '0 ⎜ o ⎟ e− μHo ⎝R⎠

C=

R μHo .e ho

p = σ '0

h μ H −H . e ( 0 ) h0

In the exit zone ⎛ h p = σ '0 ⎜ ⎝ hf hn μ H −H . e ( 0 n) h0 or

⎞ μH ⎟ .e ⎠ h = n . e μ Hn hl

ho μ H − 2H = e ( 0 n) hf

or Hn =

⎛ h ⎞⎤ 1⎡ 1 ⎢H0 − ln ⎜ 0 ⎟ ⎥ 2 ⎣⎢ μ ⎝ h f ⎠ ⎦⎥

from H = 2

R .tan −1 hf

R .θ. hf

⎛ h f Hn ⎞ hf .tan ⎜ . ⎜ R 2 ⎟⎟ R ⎝ ⎠ hn = h f + 2R (1 − cos θn )

∴ 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 ,

l = R.Δh, and l

tan a =

=

R Δh R − 05 Δh

Δh 2 Since R > > 0.5 Δh, it can be written that R-

tan a =

Δh R

Since μ ≥ tan a ∴ The maximum draft is given by

μ ≥ or,

( Δh )max

Δh R

= μ2R

Q.1. In rolling process, 25 mm thick plate is rolled to 20 mm in a four high mill. Determine 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 / mm 2 for hot rolls of middle steel at about 1100oC. Solution: Δh = μ2 R (i).

( 25 − 20 ) = 0.142 Δh = R 250 and Δh = 2R (1 − cos α ) or μ =

or 5 = 500 (1 − cos α ) α = 8.110 = 0.1429

(ii)

⎛ R ⎞ .α ⎟ ⎜⎜ ⎟ ⎝ hf ⎠ ⎛ 250 ⎞ 250 . tan −1 ⎜ = 2. × 0.1429 ⎟ = 3.306 ⎜ ⎟ 20 ⎝ 20 ⎠

H0 = 2

Hn = = θn = =

R .tan −1 hf

⎛ h0 ⎞ ⎤ 1⎡ 1 ⎢H0 − log e ⎜ ⎟ ⎥ 2⎣ μ ⎝ hf ⎠⎦ 1⎡ 1 ⎛ 25 ⎞ ⎤ 3.306 − .log e ⎜ ⎟ ⎥ = 0.8678 ⎢ 2⎣ 0.142 ⎝ 20 ⎠ ⎦ ⎛ h f Hn ⎞ hf .tan ⎜ . ⎜ R 2 ⎟⎟ R ⎝ ⎠ ⎧ 250 ⎪ 20 ⎛ 0.8678 ⎞ ⎫⎪ × tan ⎨ ×⎜ ⎟⎬ 20 ⎩⎪ 250 ⎝ 2 ⎠ ⎭⎪

= 0.0349 rad hn = h f + 2R (1 − cos θn ) = h f + Rθn 2 = 20 + 250 × ( 0.0349 ) = 20.3mm 2

(iii)

Backward slip = Forward slip =

Vr − V0 V h 20.3 =1− 0 =1− n =1− = 18.8% Vr Vr h0 25

Vf − Vr Vf h 20.3 = −1 = n −1 = 1 − = 1.5% Vr Vr hf 20

Vo Vr

Vf

N

h n μ Hn .e hf 2 20.3 0.142 ×0.8678 .100 × .e = = 132.4 N / mm2 20 3

(iv) pmax = pn = σ′0

Q2. Sheet steel is reduced from 4.05 mm to 3.55 mm with 500 mm diameter rolls having a coefficient 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 hf = 3.55 mm R = 250 mm, μ = 0.04, σ0 = 210 N / mm2

(a) The roll pressure at entry and exit, 2 p = σ′0 = σ0 = 242.5N / mm2 3 ⎛ R ⎞ R H0 = 2 .tan −1 ⎜ α Now ⎜ h ⎟⎟ hf f ⎝ ⎠ ⎛ 250 ⎞ 250 H0 = 2 .tan −1 ⎜ × 0.0447 ⎟ ⎜ ⎟ 3.55 ⎝ 3.55 ⎠ = 6.02

⎛ ho ⎞⎤ 1⎡ 1 ⎢H0 − log e ⎜ ⎟ ⎥ 2⎣ μ ⎝ hf ⎠⎦

Hn = =

1⎡ 1 ⎛ 4.05 ⎞ ⎤ 6.02 − × log e ⎜ ⎟ ⎥ = 1.363 ⎢ 2⎣ 0.04 ⎝ 3.55 ⎠ ⎦

pn = σ′0 .

h n μ Hn .e hf

⎛ h f Hn ⎞ ⎛ 3.55 ⎞ hf 3.55 .tan ⎜ . .tan ⎜ × 0.6815 ⎟ = 0.009672 rad. = 0.5540 ⎟⎟ = ⎜ ⎟ ⎜ R 250 ⎝ 250 ⎠ ⎝ R 2 ⎠

Now

θn =

And

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

h n = h f + 2R (1 − cos θn )

= 3.55 +2 × 250 (1- cos 0.554o)

= 3.5734 mm h pn = σ′0 . n .eμ Hn hf

= 242.5 ×

3.5734 0.04×1.363 = 257.78 N / mm2 e 3.55

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

1⎡ 1 ⎛ 4.05 ⎞ ⎤ 6.02 − log e ⎜ ⎢ ⎟ ⎥ = 2.845 2⎣ 0.4 ⎝ 3.55 ⎠ ⎦ ⎛ 3.55 ⎞ 3.55 × 1.4225 ⎟ = 0.02rad tan ⎜ ⎜ ⎟ 250 ⎝ 250 ⎠

h n = h f + Rθn2 = 3.55 + 250 × ( 0.02 ) = 3.65 mm 2

pn = σ′0 .

hn . eμ Hn hf

3.65 0.04×2.845 ×e = 777.9N / mm2 3.55 h (c) pn = ( σ′0 − σf ) . n . eμ Hn hf = 242.5 ×

= ( 242.5 − 35 )

3.5734 0.04×1.363 ×e = 220.57 N / mm2 3.55

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 = 0.2 100 ho = hf × = 9.07mm 70 Δh = h0 − h f = 9.07 − 6.35 = 2.72mm

Δh = 2R (1 − cos α )

2.72 = 2 × 500 × (1 − cos α ) or α = 4.230 = 0.0738 rad. Now

⎛ R ⎞ R .tan −1 ⎜ .α ⎟ ⎜ h ⎟ hf f ⎝ ⎠

H0 = 2. = 2×

⎛ 500 ⎞ 500 × tan −1 ⎜ × 0.0738 ⎟ ⎜ ⎟ 6.35 ⎝ 6.35 ⎠

= 10.29.

now Hn =

⎛ h0 ⎞ ⎤ 1 ⎡ 1⎡ 1 1 ⎛ 9.07 ⎞ ⎤ × log e ⎜ ⎢H0 − log e ⎜ ⎟ ⎥ = ⎢10.29 − ⎟ ⎥ = 4.26 2⎣ 0.2 μ ⎝ 6.35 ⎠ ⎦ ⎝ hf ⎠⎦ 2 ⎣

θn = =

⎛ h f Hn ⎞ hf .tan ⎜ . ⎜ R 2 ⎟⎟ R ⎝ ⎠ ⎛ 6.35 ⎞ 6.35 × tan ⎜ × 2.13 ⎟ = 0.0273 rad = 1.550 ⎜ ⎟ 500 ⎝ 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 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 h0 ⎤ h0 1 ⎡ n ⎢ ∫ pdh + ∫ pdh + ∫h p.dh ⎥ b Δh ⎢⎣ hl ⎥⎦ hn

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 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 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 Lp can be given as

λ=

a = LP

a R Δh

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(2π 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 = 4π aPN Where P is in Newton’s and a is in metre.