Example: 1 A 300-mm-wide strip 25 mm thick is fed through a rolling mill with two powered rolls each of radius = 250 mm. The work thickness is to be reduced to 22 mm in one pass at a roll speed of 50 rev/min. The work material has a flow curve defined by K = 275 MPa and n = 0.15, and the coefficient of friction between the rolls and the work is assumed to be 0.12. Determine if the friction is sufficient to permit the rolling operation to be accomplished. If so, calculate the roll force, torque, and horsepower. The draft attempted in this rolling operation is d = 25 22 = 3 mm From Eq. (1), the maximum possible draft for the given coefficient of friction is 0.12 250 3.6mm Since the maximum allowable draft exceeds the attempted reduction, the rolling operation is feasible. To compute rolling force, we need the contact length L and the average flow stress Y f The contact length is given by Eq, (2): Y f is determined from the true strain: 250 25 22 27.4 mm 25 22 275 0.128. 1.15 Rolling force is determined from Eq. (3): 0.128 175.7 MPa F = 175.7(300)(27.4) = 1,444,786 N Torque required to drive each roll is given by Eq. (4): T = 0.5(1,444,786) (27.4) (10-3 ) = 19,786 N-m and the power is obtained from Eq. (5):
P = 2π (50) (1,444,786) (27.4) (10-3 ) = 12,432,086 N m/min = 207,201 N- m/s (W) For comparison, let us convert this to horsepower (we note that one horsepower = 745.7 W): 207,201 745.7 278hp It can be seen from this example that large forces and power are required in rolling. Inspection of Eqs. (3) and (5) indicates that force and/or power to roll a strip of a given width and work material can be reduced by any of the following: (1) using hot rolling rather than cold rolling to reduce strength and strain hardening (K and n) of the work material; (2) reducing the draft in each pass; (3) using a smaller roll radius R to reduce force; and (4) using a lower rolling speed N to reduce power.
Example: 2 A cylindrical workpiece is subjected to a cold upset forging operation. The starting piece is 75 mm in height and 50 mm in diameter. It is reduced in the operation to a height of 36mm. The work material has a flow curve defined by K = 350 MPa and n = 0.17. Assume a coefficient of friction of 0.1. Determine the force as the process begins, at intermediate heights of 62 mm. 49 mm and at the final height of 36 mm. Work piece volume V = 75π (50 2/ 4) = 147,262 mm 2. At the moment contact is made by the upper die, h = 75 mm and the force F = 0. At the start of yielding, h is slightly less than 75 mm, and we assume that strain = 0.002, at which the flow stress is yf n = K = 350(0.002) 0.17 = 121.7 MPa The diameter is still approximately D = 50 mm and area A = π (50 2 /4) = 1963.5 mm 2. For these conditions, the adjustment factor K f is computed as. The forging force is 0.4(0.1)(50) K f = 1+ = 1.027 75 F = 1.027 (121.7) (1963.5) = 245,410 MPa 75 = In = I n(1.21) = 0.1904 62 Y = = MPa f.17 350(0.1904) 264.0 Assuming constant volume, and neglecting barre line, 2 A 147, 264/ 62 2375.2mm andd 55.0mm = = = 0.4(0.1)(55) K f = 1+ = 1.035 62 F = 1.035(264)9237.2) = 649,303N Similarly at h = 49 mm, F = 955,642 N. and at h = 36 mm, F = 1,467, 422 N. The load- stroke curve in Figure 19.5 was developed from the values in this example.
Example 3 Wire is drawn through a draw die with entrance angle = 15 o. Starting diameter is 2.5 mm and final diameter = 2.0 mm. The coefficient of friction at the work-die interface = 0.07. The metal has a strength coefficient K = 205 MPa and a strain hardening exponent n = 0.20. Determine the draw stress and draw force in this operation. The values of D and L, for Eq. (6) can be determined using Eqs. 19.34. D = 2.25 mm and L, = 1.0 mm. Thus, 0.88 0.12 2.25 1.0 1.15 The areas before and after drawing are computed as A o = 4.91 mm 2 and A f = 3.14 mm 2. The resulting true strain = In (4.91/3.14)= 0.446. And the average flow stress in the operation is computed: Draw stress is given by Eq. (7): 205 0.446. 1.20 145.4MPa 145.4 1. 1.15 0.446 94.1 MPa Finally, the draw force is this stress multiplied by the cross-sectional area of the exiting wire: F = 94.1(3.14) = 295.5 N. Maximum Reduction per Pass A question that may occur to the reader is: Why is more than one step required to achieve the desired reduction in wire drawing? Why not take the entire reduction in a single pass through one die, as in extrusion? The answer can be explained as follows. From the preceding equations, it is clear that as the reduction increases, draw stress increases. If the reduction is large enough, draw stress will exceed the yield strength of the exiting metal. When that happens, the drawn wire will simply elongate instead of new material being squeezed through the die opening. For wire drawing to be successful, maximum draw stress must be less than the yield strength of the exiting metal. It is a straightforward matter to determine this maximum draw stress and the resulting maximum possible reduction that can be made in one pass, under certain assumptions.
Let us assume a perfectly plastic metal (n = 0), no friction, and no redundant work. In this ideal case, the maximum possible draw stress is equal to the yield strength of the work material. Expressing this using the equation for draw stress under conditions of ideal deformation, Eq. (8), and setting Y f = Y (because n = 0), 1.0 1 1 This means that In (A o /A f ) = In(l/(l-r)) = 1. Hence, A o /A f = 1/(1-r) must equal the natural logarithm base e. That is, the maximum possible strain is 1.0: The maximum possible area ratio is (9) 2.7183 (10) and the maximum possible reduction is 0.632 (11) The value given by Eq. (11) is often used as the theoretical maximum reduction possible in a single draw, even though it ignores the effects of friction and redundant work, which would reduce the' maximum possible value, and strain hardening, which would increase the maximum possible reduction because the exiting wire would be stronger than the starting metal. In practice, draw reductions per pass are quite below the theoretical limit. Reductions of 0.50 for single-draft bar drawing and 0.30 for multiple-draft wire drawing seem to be the upper limits in industrial operations.
Example 4 A round disk of 150mm diameter is to be blanked from a strip of 3.2mm, half-hard cold-rolled steel whose shear strength = 310 MPa. Determine (a) the appropriate punch and die diameters, and (b) blanking force. (a) The clearance allowance for half-hard cold-rolled steel is a = 0.075. Accordingly, C = 0.075 (3.2 mm) = 0.24 mm The blank is to have a diameter = 150 mm, and die size determines blank size. Therefore, Die opening diameter = 150.00 mm Punch diameter = 150-2(0.24) = 149.52 mm (b) To determine the blanking force, we assume that the entire perimeter of the part is blanked at one time. The length of the cut edge is And the force is 150 471.2 mm F = 310(471.2)(3.2) = 467,469 N (This is about 53 tons.)