Ch 13

Chapter 13

Rolling of Metals

QUALITATIVE PROBLEMS

13.16 Explain why the rolling process was invented and developed.

By the student. Machinery, structures, bridges, boilers, pressure vessels, etc. typically requiremetal plates or sheets. Consequently, there was urgent need for developing the rolling processwhich could economically deliver large amounts of the necessary plate. Note in Table I.2 onp. 3 that the word rolling first appears in the 1500s.

13.17 Flat rolling reduces the thickness of plates and sheets. It is possible, instead,to reduce their thickness simply by stretching the material? Would this be afeasible process? Explain.

By the student. Although stretching reduces the thickness of materials, there are severallimitations associated with it as compared to rolling. Stretching process is a batch processand it cannot be continuous as it is in rolling. The reduction in thickness is limited bynecking of the sheet, depending on its strain-hardening exponent, n (see Section 2.2.4 onp. 61). Furthermore, as the sheet is stretched, the surface finish becomes dull due to theorange-peel effect. Stretching the sheet requires some means of clamping the material at itsends which, in turn, will leave marks on the sheet.

13.18 Explain how the residual stress patterns shown in Fig. 13.9 become reversed whenthe roll radius or reduction-per-pass is changed.

As shown in Fig. 13.9a on p. 325, with small rolls and/or small reductions, the workpiece isdeformed, as expected, at its surfaces more than it is in the bulk. With large rolls and/orlarge reductions, the reverse is true. The large roll-strip contact area develops a situationsimilar to that shown in Fig. 13.9b, namely, that the material flows more along the insidewhile the surfaces are more constrained.

154

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Rolling of Metals 155

13.19 Explain whether it would be practical to apply the roller-leveling technique shownin Fig. 13.7a to thick plates.

It is doubtful that the roller-leveling process, shown in Fig. 13.7 on p. 324, can be applied toplates. In this process, the strip is flattened by repeatedly flexing it in opposite directions.To do the same with a plate would require much higher forces in order to develop stressesthat are of the same magnitude at the plate surface as they are in sheet. Also, unless it issufficiently ductile, the plate may develop cracks if bent to small radii.

13.20 Describe the factors that influence the magnitude of the roll force, F , in Fig. 13.2c.

By the student. As can be deduced by observing the equations on p. 319, the roll force, F , isinfluenced by the roll radius, strip width, draft (hence the roll-strip contact area), coefficientof friction, and the strength of the material at the rolling temperature. If the material isstrain-rate sensitive (i.e., high m value), the rolling speed would also influence the roll force;this is particularly important in hot rolling.

13.21 Explain how you would go about applying front and back tensions to sheet metalsduring rolling. How would you go about controlling these tensions?

Front tensions are applied and controlled by the take-up reel of a rolling mill (see Fig. 13.11on page 327). The greater the torque to this reel, the greater the front tension. Back tensionis applied by the pay-off reel of the rolling mill, whereby increasing the brake force on thepay-off reel increases the back tension.

13.22 What typically is done to make sure that the product in flat rolling is notcrowned?

To make sure that the product in flat rolling is not unreasonably crowned, a number ofstrategies can be followed, which basically compensate for roll bending. These include:

• The use of backing rolls.

• Using crowned rollers so that roll deflections are compensated by the geometry of theroller to produce a flat workpiece.

• Superimposing a deflection on the rolls by bending them; the elastic deformation of therollers is then compensated by the deflection from the bending moment.

• Using a front and/or back tension to reduce the rolling pressure, and hence the force onthe rolls.

13.23 Make a list of parts that can be made by (a) shape rolling and (b) thread rolling.

Parts that can be made by shape rolling include railroad rails, I-beams, and other structuralchannels. Note that there is a similarly-named process for sheet metals described in Section16.6 starting on p. 406 which uses sheet metal workpieces and can be used for gutters as wellas some structural channels. Thread rolling obviously produces bolts and screws, but alsocan produce threaded surfaces on anything that needs to be assembled through mechanicalfasteners.

13.24 Describe the methods by which roll flattening can be reduced. Which propertyor properties of the roll material can be increased to reduce roll flattening?




Flattening is elastic deformation of the roll and results in a larger contact length in the rollgap; therefore, the elastic modulus of the roll should be increased, for example, by makingit from materials with high modulus of elasticity, such as carbides (see Tables 2.1 on p. 56,2.2 on p. 58, and 22.1 on p. 602. Roll flattening also can be reduced by (a) decreasing thereduction per pass and (b) reducing friction at the roll-sheet interface.

13.25 It was stated that spreading in flat rolling increases with (a) a decreasing width-to-thickness ratio of the entering material, (b) decreasing friction, and (c) adecreasing ratio of the roll radius to the strip thickness. Explain why.

(a) If the width-to-thickness ratio is small, the material in the roll bite is less restrained bythe frictional force in the width direction and, as a result, spreading increases. (b) The lowerthe friction, the lower the resistance to relative motion between the rolls and the workpieceand, hence, the greater the spreading. (c) If the roll radius is large as compared to the stripthickness, there will be lower frictional resistance in the rolling direction than across it, andthus the material will flow more in the longitudinal direction, hence spreading will decrease.

13.26 Flat rolling can be carried out by front tension only, using idling rolls (Steckelrolling). Since the torque on the rolls is now zero, where, then, is the energycoming from to supply the work of deformation in rolling?

The energy for work of deformation in Steckel rolling (p. 322) is supplied by the front tensionrequired to pull the strip through the roll gap between the idling rolls. The product oftension and exiting strip velocity is power supplied in rolling. This power is provided by thecoil winder or draw bench.

13.27 Explain the consequence of applying too high a back tension in rolling.

If the back tension is too high, the rolls will begin to slip and no reduction in thickness willtake place. An analogy would be the slipping of the wheels of an automobile while pulling aheavy trailer.

13.28 Note in Fig. 13.3f that the driven rolls (powered rolls) are the third set fromthe work roll. Why isnt power supplied through the work roll itself? Is it evenpossible? Explain.

We note in Fig. 13.3d on p. 321 that the diameter of the rolls increases as we move away fromthe work (smallest) roll. The reason why power cannot be supplied through the work roll isthat the significant power required for this rolling operation will subject the work roll to ahigh torque. Since its diameter is small, the torsional stresses on the roll would be too high;the roll will either fracture or undergo permanent twist. With the setup shown in the figure,the power is applied to a larger-diameter roll, which can support a large torque.

13.29 Describe the importance of controlling roll speeds, roll gaps, temperature, andother process variables in a tandem rolling operation, as shown in Fig. 13.11.Explain how you would go about determining the optimum distance between thestands.

Referring to the tandem rolling operation shown in Fig. 13.11 on p. 326, we note that masscontinuity has to be maintained during rolling. Thus, if the roll speed is not synchronized with




the strip thickness in a particular stand, excessive tensions or slack may develop between thestands; some rolls may slip. Also, if the temperature is not controlled properly, strip thicknesswill change, thus affecting reduction per pass and, consequently, the roll forces involved. This,in turn, will also affect the actual roll gap and roll deflections. Complex control systems havebeen developed for monitoring and controlling such operations at high rolling speeds.

13.30 In Fig. 13.9a, if you remove the top compressive layer by, say, grinding, will thestrip remain flat? If not, which way will it curve and why?

We can model the residual stresses in the strip in Fig. 13.9a on p. 327 by three horizontal andparallel springs: compression spring (top), tension spring (middle), and compression spring(bottom). Note that the top layer is in compression, and when we remove the top spring, thebalance of internal moment and internal horizontal forces will be disturbed. The strip willthus distort, in a manner that it will hold water, i.e., like cupping your hand. The remainingresidual stresses in the strip will rearrange themselves to ensure balancing of the internalmoment and internal horizontal forces.

13.31 Name several products that can be made by each of the operations shown inFig. 13.1.

By the student. Examples of parts from cold rolled strip are car bodies and aluminum foilfor food packaging. Examples of plate are tractor and machinery frames and warship hulls.Rolled shapes include architectural beams and railroad rails.

13.32 List the possible consequences of rolling at (a) too high of a speed and (b) toolow of a speed.

There are advantages and disadvantages to each. Rolling at high speed is advantageous inthat production rate is increased, but it has disadvantages as well, including:

• The lubricant film thickness entrained will be larger, which can reduce friction and leadto a slick mill condition where the rolls slip against the workpiece. This can lead to adamaged surface finish on the workpiece.

• The thicker lubricant film associated with higher speeds can result in significant oil peel,or surface roughening.

• Because of the higher speed, chatter may occur, compromising the surface quality orprocess viability.

• There is a limit to speed associated with the motor and power source that drive therolls.

Rolling at low speed is advantageous because the surface roughness in the workpiece canmatch that of the rolls (which can be polished). However, rolling at too low a speed hasconsequences such as:

• Production rate will be low, and thus the cost per unit weight will be higher.

• Because a thick lubricant film cannot be developed and maintained, there is a danger oftransferring material from the workpiece to the roll (pickup), thus compromising surfacefinish.




• The workpiece may cool excessively before contacting the rolls. This is because a longbillet that is rolled slowly loses some of its heat to the environment and also throughconduction through the roller conveyor.

13.33 It is known that in thread rolling, as illustrated in Fig. 13.16, a workpiece mustmake roughly six revolutions to form the thread. Under what conditions (processparameters, thread geometry or workpiece properties) can deviation from thisrule take place?

By the student. If the force is increased, or speed is reduced, then fewer revolutions cantake place, but this will result in lower die life. It should be recognized that threads can beproduced with fewer revolutions, but at a cost in tolerances and form, and therefore threadstrength. With flat dies, more revolutions requires longer dies; fewer revolutions can causeexcessive heating and will require special cooling in order to achieve required tolerances.

13.34 If a rolling mill encounters chatter, what process parameters would you change,and in what order? Explain.

By the student. Chatter is briefly described on p. 323, and students should be encouragedto supplement the material with a literature review. As stated in the text, speed and lubri-cant type are the most important factors. Although not always practical to implement, italso has been suggested that chatter can be reduced by (a) increasing the distance betweenthe stands of the rolling mill, (b) increasing the strip width, (c) decreasing the reductionper pass (draft), (d) increasing the roll radius, (e) increasing the strip-roll friction, and (f)incorporating external dampers in the roll supports.

13.35 Can the forward slip ever become negative? Why or why not?

If a sufficient front tension is applied, then the forward slip can be negative. This is actuallyof interest, since there is an inverse correlation between workpiece surface roughness andforward slip. A negative forward slip would suggest that the roll asperities are smearing theworkpiece; this can lead to a smooth surface if the roll is highly polished.

QUANTITATIVE PROBLEMS

13.36 In Example 13.1, calculate the roll force and the power for the case in which theworkpiece material is 1100-O aluminum and the roll radius, R, is 8 in.

As discussed in Example 13.1 on p. 320, the roll-strip contact length, L, is given by

L =√R∆h =

√(8)(1.00− 0.8) = 1.265 in.

or L = 0.105 ft. Referring to Fig. 2.6 on p. 63 we find that for 1100-O aluminum the yieldstress is about 8,000 psi, and that at a true strain of 0.223, the true stress (flow stress) is




about 16,000 psi. Thus the average stress Yavg is 12,000 psi, and the roll force, F , is given byEq. (13.2) on p. 319 as

F = LwYavg = (1.265)(9)(12, 000) = 136, 600 lb

and the power is given by Eq. (13.4) on page 319 as:

P =2πFLN

33, 000 hp=

2π(136, 600)(0.105)(100)

33, 000= 273 hp

13.37 Calculate the individual drafts in each of the stands in the tandem-rolling oper-ation shown in Fig. 13.11.

The answers are:

• Stand 5: 2.25 - 1.45 = 0.80 mm, or 36%.

• Stand 4: 1.45 - 0.90 = 0.55 mm, or 38%.

• Stand 3: 0.90 - 0.56 = 0.34 mm, or 38%.

• Stand 2: 0.56 - 0.34 = 0.22 mm, or 39%.

• Stand 1: 0.34 - 0.26 = 0.08 mm, or 24%.

13.38 Estimate the roll force, F , and the torque for an AISI 1020 carbon-steel stripthat is 200 mm wide, 10 mm thick, and rolled to a thickness of 7 mm. The rollradius is 200 mm, and it rotates at 200 rpm.

The roll force is given by F = LwYavg, where L is the roll-strip contact length, w is the stripwidth, and Yavg is the average stress during the operation. As discussed in Example 13.1 onp. 320, L is given by

L =√R∆h =

√(0.2 m)(0.01 m− 0.007 m) = 0.0245 m

The true strain for this operation is

ε = ln(10/7) = 0.36

and the average flow stress, Yavg, is given by

Yavg =Kεn

n+ 1

For AISI 1020 carbon steel (from Table 2.3 on p. 61), K = 530 MPa and n = 0.26; therefore

Yavg = 323 MPa

and thus the roll force, F , is

F = LwYavg = (0.0245)(0.2)(323) = 1.58 MN

and the required torque, T , is

T = FL/2 = (1.58)(0.0245)/2 = 0.019 MN-m




13.39 A rolling operation takes place under the conditions shown in Fig. P13.39. Whatis the position, xn, of the neutral point? Note that there are front and backtensions that have not been specified. Additional data are as follows: Materialis 5052-O aluminum; hardened steel rolls; surface roughness of the rolls = 0.02µm; rolling temperature = 210◦C.

Note that more data is given than is needed to solve this problem. Assuming the material isincompressible, the velocity at the inlet is calculated as:

(2.0)(3 mm)w = Vi(5 mm)w

Therefore, Vi = 1.20 m/s. At the neutral point, the velocity is the roll velocity (or 1.5 m/s).Assuming incompressibility, we can compare the outlet and the neutral point:

(1.5)(h) = (2.0)(3) → h = 4.0 mm

Consider the sketch of the roll bite geometry given below.

R

x

Rcos

5/2=2.5 mm3/2=1.5 mm

θ can be calculated from:

75− (75) cos θ =4− 3

2

or θ = 6.62◦. Therefore,

xn = R sin θ = (75) sin 6.62◦ = 8.64 mm

13.40 Estimate the roll force and power for annealed low-carbon steel strip 200 mmwide and 10 mm thick, rolled to a thickness of 6 mm. The roll radius is 200 mm,and the roll rotates at 200 rpm; use µ = 0.1.

A low carbon steel such as AISI 1020 has K = 530 MPa and n = 0.26 (see Table 2.3 onp. 61). The strain is

ε = ln

(10

6

)= 0.511

Therefore,

Yavg =Kεn

n+ 1=

(530)(0.511)(0.26)

1.26= 353 MPa

The contact length is (see Example 13.1 on p. 320),

L =√R(ho − hf ) =

√0.2(0.01− 0.006) = 0.0283 m




The roll force is given by Eq. (13.3) on p. 319 as

F = LwYavg = (0.0283)(0.2)(353× 106) = 2.00 MN

The speed is N = 200 rpm, so from Eq. (13.3),

Power =2πFLN

60, 000=

2π(2× 106)(0.0283)(100)

60, 000= 5930 kW

13.41 A flat-rolling operation is being carried out where ho = 0.20 in., hf = 0.15 in.,wo = 10 in., R = 8 in., µ = 0.25, and the average flow stress of the materialis 40,000 psi. Estimate the roll force and the torque; include the effects of rollflattening.

The roll force can be estimated from Eq. (13.3) on p. 319, where the quantity L is obtainedfrom the relations in Example 13.1 on p. 320. Therefore,

L =√R∆h =

√(8)(0.20− 0.15) = 0.632 in.

and

have =0.20 + 0.15

2= 0.175 in.

In roll flattening (see Kalpakjian and Schmid, Manufacturing Processes for Engineering Ma-terials, 5th ed), the roll radius must be modified. The following is a reasonable approach,although other expressions can be found on the Internet:

F = LwY ′[1 +

µL

2have

]= (0.632)(10)(40, 000)

[1 +

(0.25)(0.632)

2(0.175)

]= 367, 000 lb

We check for roll flattening by using Eq. (6.48) on p. 299, where C = 1.6 × 10−7 in2/lb,assuming steel rolls, and

F ′ =F

w=

367, 000

10= 36, 700 lb/in.

Thus,

R′ = R

(1 +

CF ′

ho − hf

)= (8)

[1 +

(1.6× 10−7)(36, 700)

0.20− 0.15

]= 8.94 in.

Using this value in the force expression, we have L = 0.668 in. and F = 395, 000 lb. Thisforce predicts a flattened radius of R′ = 9.0 in. (Note that the expression is converging.)This radius predicts L = 0.671 and F = 397, 000 lb, which suggests a radius of R′ = 9.02 in.Therefore, the roll force is around 397,000 lb, with an effective roll radius of 9.0 in.




13.42 It can be shown that it is possible to determine µ in flat rolling without measuringtorque or forces. By inspecting the equations for rolling, describe an experimentalprocedure to do so. Note that you are allowed to measure any quantity otherthan torque or forces.

In this problem, we first measure the following quantities: vo, vf , vr, ho and hf . Fromthe available information and knowing R, we can calculate the magnitude of the angle ofacceptance, α. From the velocity distribution, as in Problem 13.39, we can now determine φnfrom which we obtain Hn, using a stress relationship or finite element model. To determinethe coefficient of friction, the friction value is modified until the correct value is obtained, asconfirmed by forces and stress distributions.

13.43 Assume that you are an instructor covering the topics described in this chapter,and you are giving a quiz on the numerical aspects to test the understanding ofthe students. Prepare two quantitative problems and supply the answers.

By the student. This is a challenging open-ended question and requires considerable focusand understanding on the part of the students, and has been found to be a very valuablehomework problem.

SYNTHESIS, DESIGN, AND PROJECTS

13.44 A simple sketch of a four-high mill stand is shown in Fig. 13.3c. Make a surveyof the technical literature and present a more detailed sketch for such a stand,showing the major components.

By the student. The results will vary widely depending on the age of the machine, the mate-rial, and the size of the plates rolled. For example, a fully automated aluminum rolling millwill have a complex system of sensors and controls, whereas a specialty jewelry manufacturermay have a manually powered (hand crank) four-high rolling mill for producing gold foil.

13.45 Obtain a piece of soft, round rubber eraser, such as that at the end of a pencil,and duplicate the process shown in Fig. 13.18b. Note how the central portion ofthe eraser will begin to disintegrate, producing a rough hole.

By the student. This is an interesting project, but is a little tricky to perform and mayneed several tries. Also, the hole needs to have the eroded material from the center removedperiodically, such as by brisk blowing, to make a well-defined hole.

13.46 If you repeat the experiment in Problem 13.45 with a harder eraser, such as thatused for erasing ink, you will note that the whole eraser will begin to crack andcrumble. Explain why.

By the student. The main reason for this behavior is that with an ordinary (tougher) eraser,the deterioration of the material starts at the center of the eraser and grows outward at a




slow rate. With a hard eraser (typically containing small abrasive particles such as fine sand),the crack growth is very fast, and fracture occurs before any noticeable cavity is formed.

13.47 Design a set of rolls to produce cross sections other than those shown in Fig. 13.12.

By the student. There are several possible designs, such as the following for producing railroadrails:

13.48 Design an experimental procedure for determining the neutral point in a flatrolling operation.

By the student. Problem 13.39 is useful; another is to place a reference mark on a work roll,such as a small dent. When the workpiece rolls, the distance between dent marks on the sheetwill be larger than the roll circumference. This allows measurement of the neutral point.

13.49 Using a rolling pin and any available dough (bread, cookie, etc), measuring 100mm by 100 mm by 8 mm, quantify the spreading in flat rolling for differentreductions in thickness.

By the student. Students should be encouraged to infer important variables from this systemand relate them to metal rolling. For example, the use of flour below and on top of the doughwill reduce friction and change results.

13.50 Derive an expression for the thickest workpiece that can be drawn between tworolls as a function of roll gap, roll radius, and coefficient of friction.

By the student. If a rectangular workpiece is brought in contact to two rolls that are separated,so that the contact angle is θ, then there is a normal force N , and an tangential force F .Taking x-components yields

N sin θ = F cos θ = µN cos θ

resulting in µ = tan θ.



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