Ch 13

Chapter 13

Rolling of Metals QUALITA QUALIT ATIVE PROBLEMS

13.16  Explain why the rolling process was invented and developed. By the student. Machinery, structures, bridges, boilers, pressure vessels, etc. typically require metal plates or sheets. Consequently, there was urgent need for developing the rolling process which which could economically economically deliver deliver large amounts amounts of the necessary necessary plate. Note in Table I.2 on p. 3 that the word rolling first appears in the 1500s. 13.17   Flat Flat rolling rolling reduces reduces the thick thicknes nesss of plates plates and sheets. sheets. It is possible, possible, instea instead, d, to reduce reduce their thickne thickness ss simply simply by stretch stretching ing the materi material? al? Would ould this this be a feasible feasible process? Explain. Explain. By the student. Although Although stretching stretching reduces reduces the thicknes thicknesss of materials, materials, there are several limitations limitations associated with it as compared compared to rolling. rolling. Stretchin Stretching g process is a batch process and and it cann cannot ot be cont contin inuo uous us as it is in rolli rolling ng.. The The redu reduct ction ion in thic thickn knes esss is limite limited d by necking of the sheet, depending on its strain-hardening exponent, n   (see Section 2.2.4 on p. 61). 61). Further urthermor more, e, as the sheet sheet is stretc stretched hed,, the surface surface finish finish becomes becomes dull due to the orange-peel orange-peel effect. Stretching Stretching the sheet requires some means of clamping clamping the material material at its ends which, in turn, will leave marks on the sheet. 13.18 Explain Explain how the residual residual stress patterns shown shown in Fig. 13.9 become reversed reversed when the 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 is deform deformed, ed, as expecte expected, d, at its surfac surfaces es more than it is in the bulk. bulk. With With large large rolls and/or and/or large reductions, reductions, the reverse reverse is true. The large roll-strip roll-strip contact area develops develops a situation situation similar to that shown in Fig. 13.9b, namely, that the material flows more along the inside while the surfaces are more constrained.

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13.19  Explain whether it would be practical to apply the roller-leveling technique shown in 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 to plates. 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 stresses that are of the same magnitude at the plate surface as they are in sheet. Also, unless it is sufficiently 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 , is influenced by the roll radius, strip width, draft (hence the roll-strip contact area), coefficient of friction, and the strength of the material at the rolling temperature. If the material is strain-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 metals during 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. 13.11 on page 327). The greater the torque to this reel, the greater the front tension. Back tension is applied by the pay-off reel of the rolling mill, whereby increasing the brake force on the pay-off reel increases the back tension. 13.22   What typically is done to make sure that the product in flat rolling is not crowned? To make sure that the product in flat rolling is not unreasonably crowned, a number of  strategies 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 the roller to produce a flat workpiece.

•  Superimposing a deflection on the rolls by bending them; the elastic deformation of the rollers 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 on the 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 structural channels. Note that there is a similarly-named process for sheet metals described in Section 16.6 starting on p. 406 which uses sheet metal workpieces and can be used for gutters as well as some structural channels. Thread rolling obviously produces bolts and screws, but also can produce threaded surfaces on anything that needs to be assembled through mechanical fasteners. 13.24  Describe the methods by which roll flattening can be reduced. Which property or properties of the roll material can be increased to reduce roll flattening?

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Flattening is elastic deformation of the roll and results in a larger contact length in the roll gap; therefore, the elastic modulus of the roll should be increased, for example, by making it 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 the reduction 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 widthto-thickness ratio of the entering material, (b) decreasing friction, and (c) a decreasing 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 by the frictional force in the width direction and, as a result, spreading increases. (b) The lower the friction, the lower the resistance to relative motion between the rolls and the workpiece and, hence, the greater the spreading. (c) If the roll radius is large as compared to the strip thickness, there will be lower frictional resistance in the rolling direction than across it, and thus 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 (Steckel rolling). Since the torque on the rolls is now zero, where, then, is the energy coming 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 tension required to pull the strip through the roll gap between the idling rolls. The product of  tension and exiting strip velocity is power supplied in rolling. This power is provided by the coil 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 will take place. An analogy would be the slipping of the wheels of an automobile while pulling a heavy trailer. 13.28   Note in Fig. 13.3f that the driven rolls (powered rolls) are the third set from the work roll. Why isnt power supplied through the work roll itself? Is it even possible? Explain. We note in Fig. 13.3d on p. 321 that the diameter of the rolls increases as we move away from the work (smallest) roll. The reason why power cannot be supplied through the work roll is that the significant power required for this rolling operation will subject the work roll to a high 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, and other process variables in a tandem rolling operation, as shown in Fig. 13.11. Explain how you would go about determining the optimum distance between the stands. Referring to the tandem rolling operation shown in Fig. 13.11 on p. 326, we note that mass continuity has to be maintained during rolling. Thus, if the roll speed is not synchronized with

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the strip thickness in a particular stand, excessive tensions or slack may develop between the stands; some rolls may slip. Also, if the temperature is not controlled properly, strip thickness will 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 have been 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 the strip 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 and parallel 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, the balance of internal moment and internal horizontal forces will be disturbed. The strip will thus distort, in a manner that it will hold water, i.e., like cupping your hand. The remaining residual stresses in the strip will rearrange themselves to ensure balancing of the internal moment and internal horizontal forces. 13.31   Name several products that can be made by each of the operations shown in Fig. 13.1. By the student. Examples of parts from cold rolled strip are car bodies and aluminum foil for 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) too low of a speed. There are advantages and disadvantages to each. Rolling at high speed is advantageous in that production rate is increased, but it has disadvantages as well, including:

•  The lubricant film thickness entrained will be larger, which can reduce friction and lead

to a slick mill condition where the rolls slip against the workpiece. This can lead to a damaged 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 or process viability.

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

Rolling at low speed is advantageous because the surface roughness in the workpiece can match that of the rolls (which can be polished). However, rolling at too low a speed has consequences 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 of  transferring material from the workpiece to the roll (pickup), thus compromising surface finish.

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•  The workpiece may cool excessively before contacting the rolls.

This is because a long billet that is rolled slowly loses some of its heat to the environment and also through conduction through the roller conveyor.

13.33  It is known that in thread rolling, as illustrated in Fig. 13.16, a workpiece must make roughly six revolutions to form the thread. Under what conditions (process parameters, thread geometry or workpiece properties) can deviation from this rule take place? By the student. If the force is increased, or speed is reduced, then fewer revolutions can take place, but this will result in lower die life. It should be recognized that threads can be produced with fewer revolutions, but at a cost in tolerances and form, and therefore thread strength. With flat dies, more revolutions requires longer dies; fewer revolutions can cause excessive 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 encouraged to supplement the material with a literature review. As stated in the text, speed and lubricant type are the most important factors. Although not always practical to implement, it also has been suggested that chatter can be reduced by (a) increasing the distance between the stands of the rolling mill, (b) increasing the strip width, (c) decreasing the reduction per 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 actually of interest, since there is an inverse correlation between workpiece surface roughness and forward slip. A negative forward slip would suggest that the roll asperities are smearing the workpiece; 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 the workpiece 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 yield stress is about 8,000 psi, and that at a true strain of 0.223, the true stress (flow stress) is

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about 16,000 psi. Thus the average stress Y avg  is 12,000 psi, and the roll force, F , is given by Eq. (13.2) on p. 319 as F  = LwY avg = (1.265)(9)(12, 000) = 136, 600 lb and the power is given by Eq. (13.4) on page 319 as: P  =

2πFLN  2π(136, 600)(0.105)(100) = = 273 hp 33, 000 hp 33, 000

13.37   Calculate the individual drafts in each of the stands in the tandem-rolling operation 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 strip that is 200 mm wide, 10 mm thick, and rolled to a thickness of 7 mm. The roll radius is 200 mm, and it rotates at 200 rpm. The roll force is given by F  = LwY avg , where L  is the roll-strip contact length, w  is the strip width, and Y avg  is the average stress during the operation. As discussed in Example 13.1 on p. 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

 = ln(10/7) = 0.36 and the average flow stress, Y avg , is given by Y avg =

K n n+1

For AISI 1020 carbon steel (from Table 2.3 on p. 61), K  = 530 MPa and n  = 0.26; therefore Y avg  = 323 MPa and thus the roll force, F , is F  = LwY avg = (0.0245)(0.2)(323) = 1.58 MN and the required torque,  T , is T  = F L/2 = (1.58)(0.0245)/2 = 0.019 MN-m

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

Note that more data is given than is needed to solve this problem. Assuming the material is incompressible, the velocity at the inlet is calculated as: (2.0)(3 mm)w = V i (5 mm)w Therefore, V i = 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

Rcos   x 

5/2=2.5 mm 3/2=1.5 mm

θ can be calculated from: 75 − (75)cosθ =

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 mm wide 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 on p. 61). The strain is 10  = ln = 0.511 6

 

Therefore, Y avg

K n   (530)(0.511)( 0.26) = = = 353 MPa n+1 1.26

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

    L = R(h − h ) = 0.2(0.01 − 0.006) = 0.0283 m o

f 

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The roll force is given by Eq. (13.3) on p. 319 as F  = LwY avg  = (0.0283)(0.2)(353 × 106 ) = 2.00 MN The speed is N   = 200 rpm, so from Eq. (13.3), Power =

2πFLN  2π(2 × 106 )(0.0283)(100) = = 5930 kW 60, 000 60, 000

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 material is 40,000 psi. Estimate the roll force and the torque; include the effects of roll flattening. The roll force can be estimated from Eq. (13.3) on p. 319, where the quantity L is obtained from the relations in Example 13.1 on p. 320. Therefore,

√ 

L =

 (8)(0.20 − 0.15) = 0.632 in.

R∆h =

and

0.20 + 0.15 = 0.175 in. 2 In roll flattening (see Kalpakjian and Schmid, Manufacturing Processes for Engineering Materials, 5th ed), the roll radius must be modified. The following is a reasonable approach, although other expressions can be found on the Internet: have =

  ¯

µL F  = LwY  1 + 2have 

 

(0.25)(0.632) = (0.632)(10)(40, 000) 1 + 2(0.175) = 367, 000 lb



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

F  =

7

−

in2 /lb,

F  367, 000 = = 36, 700 lb/in. 10 w

Thus, 



R

=

 CF   R 1+  h (1.6− ×h 10 o

= (8) 1 + = 8.94 in.

f 

7 )(36, 700)

−

0.20 − 0.15



Using this value in the force expression, we have L = 0.668 in. and F   = 395, 000 lb. This force 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. 



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13.42  It can be shown that it is possible to determine  µ  in flat rolling without measuring torque or forces. By inspecting the equations for rolling, describe an experimental procedure to do so. Note that you are allowed to measure any quantity other than torque or forces. In this problem, we first measure the following quantities: vo , vf , vr , ho and hf . From the available information and knowing R, we can calculate the magnitude of the angle of  acceptance, α. From the velocity distribution, as in Problem 13.39, we can now determine φ n from which we obtain H n , using a stress relationship or finite element model. To determine the coefficient of friction, the friction value is modified until the correct value is obtained, as confirmed 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 of  the students. Prepare two quantitative problems and supply the answers. By the student. This is a challenging open-ended question and requires considerable focus and understanding on the part of the students, and has been found to be a very valuable homework problem.

SYNTHESIS, DESIGN, AND PROJECTS

13.44  A simple sketch of a four-high mill stand is shown in Fig. 13.3c. Make a survey of 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 material, and the size of the plates rolled. For example, a fully automated aluminum rolling mill will have a complex system of sensors and controls, whereas a specialty jewelry manufacturer may 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 of  the 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 may need several tries. Also, the hole needs to have the eroded material from the center removed periodically, 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 that used for erasing ink, you will note that the whole eraser will begin to crack and crumble. 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

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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 railroad rails:

13.48   Design an experimental procedure for determining the neutral p oint in a flat rolling 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 sheet will 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 100 mm by 100 mm by 8 mm, quantify the spreading in flat rolling for different reductions in thickness. By the student. Students should be encouraged to infer important variables from this system and relate them to metal rolling. For example, the use of flour below and on top of the dough will reduce friction and change results. 13.50   Derive an expression for the thickest workpiece that can be drawn between two rolls 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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