10.21. Molybdenum rod that alligatored during rolling under high conditions. The “teeth”
formed by edge cracking are not common in alligatored bars.
diameter. Furthermore, ingots often have porosity and second phases concentrated near the centerline. Hot rolling under favorableconditions tends to close such porosity.
Residual tension at the surface is particularly undesirable because it leads to increased susceptibility to fatigue and stress-corrosion cracking.
Despite the general usefulness of slip-line field theory in metal forming, it does not adequately explain the observed patterns of residual stress. For deformation under high conditions, the theory predicts that during deformation, the level of hydrostatic stress is highest at the centerline and most compressive at the surface. The theory allows for no plastic deformation once the material leaves the deformation zone, so unloading must be perfectly elastic. The normal stress must go to zero and the horizontal force must go to zero. It is difficult to rationalize how there can be complete reversal of stresses on unloading.
If the deformation is rapid enough to be nearly adiabatic, the surface will be hotter than the interior. The surface and interior must undergo the same contraction on cooling so the surface will be left under residual tension.
Very low reductions lead to an opposite pattern of residual stress with the surface being left in residual compression. Figure10.22 illustrates this. Note, however, that as soon as the reduction reaches 0.8% the normal pattern of residual tension on the surface is observed.
10.7 COMPARISON OF PLANE-STRAIN AND AXISYMMETRIC
10.7. COMPARISON OF PLANE-STRAIN AND AXISYMMETRIC DEFORMATION 179
10.22. Residual stresses in lightly drawn steel rods. From W. M. Baldwin,op. cit.after H. B ¨uhler and H. Buchholtz,Archiv. f ¨ur des Eisenh ¨uttenwesen, 7 (1933–4), pp. 427–30.
Table 10.1. Comparison of plane-strain and axisymmetric drawing
Plane strain Axisymmetry Comment Ratio of mean cross-
sectional area to tool–
workpiece contact area
/2 /4 Equal for samerandα.
Frictional contribution to σd/2kaccording to slab analysis with shear stress
m/sin 2α m/sin 2α Equal for samerandα, but with same, axisymmetry constant has twice the effect.
Frictional contribution to σd/2kaccording to slab analysis with friction coefficient
≈µcotα(1−ε/2) ≈µcotα(1−ε/2) Equal for samerandα, but with same, axisymmetry constant has twice the effect.
Redundant strain contribution toσd/2k according to upper- bound slab analysis
≈/4 ≈/6 For samerandα, less
redundant strain in plane strain but for sameless for axisymmetry.
Redundant strain according to slip-line field theory
≈0.23(−1) no solution for axismmetry Redundant strain from
Burke and Backofen experiments†
≈0.21(−1) ≈0.12(−1) For same, nearly twice as much for plane strain but for sameαandr, roughly equal.
Inhomogeneity factor [1+(tanα)/εh]n−1 Approximately equal for axisymmetry and plane strain at the sameα.
†Source: W. A. Backofen,op. cit.
if comparisons are made at equal ratios of mean cross-sectional area to contact area, the values ofφare slightly higher for axisymmetry.
NOTE OF INTEREST
W. A. Backofen was born December 8, 1928, in Rockville, Connecticut. He studied at MIT from 1943 to 1950, which he received his doctorate and was appointed as an assistant professor. He did much innovative research in his 25 years at MIT. He introduced the basic concepts of superplasticity and of texture strengthening and made clear the importance of deformation-zone geometry. He retired in 1975 to a farm in New Hampshire where he raised Christmas trees, apples, and blueberries. Backofen died in 2007.
REFERENCES
B. Avitzur,Metal Forming: Processes and Analysis, McGraw-Hill, 1968.
W. A. Backofen,Deformation Processing, Addison-Wesley, 1972.
E. M. Meilnik,Metalworking Science and Engineering, McGraw-Hill, 1991.
PROBLEMS
10.1. The XYZ Special Alloy Fabrication Company has an order to roll a 4×4×15 in billet of a nickel-base superalloy to1/2in thickness. They tried rolling it with a 5%
reduction per pass, but it split on the fourth pass. At a meeting the process engi- neer suggested using a lower reduction per pass, the consultant suggested apply- ing forward and backward tension, and the shop foreman was in favor of heavier reductions per pass. With whom, if anyone, do you agree? Defend your position.
10.2. The Mannesmann process for making tubes from cylindrical billets is illustrated in Figure 10.23. It involves passing the billet between two nonparallel rolls adjusted for a very small reduction onto a mandrel positioned in the middle.
Explain why the axial force on the mandrel is low and why the mandrel, which is long and elastically flexible, follows the center of the billet.
10.23. The Mannesman process for making tubes.
10.3. a) Show that as r andα approach zero equations10.11 and10.12 reduce to φ=1+/4 for plane strain andφ=1+/6 for axisymmetry.
b) What percent error does this simplification introduce atr=0.5 andα=30◦?
PROBLEMS 181
10.4. Consider wire drawing with a reduction ofr=0.25 andα=6◦.
a) Calculate the ratio of average cross-sectional area to the contact area between wire and die.
b) What is the value of?
c) For strip drawing withα= 6◦, what reduction would give the same value of?
d) For the reduction in (c), calculate the ratio of of average cross-sectional area to the contact area and compare with your answer to (a).
e) Explain why, for the same, the frictional drag is greater for axisymmetric drawing.
10.5. Some authorities defineas the ratio of the length of an arc through the middle of the deformation zone, centered at the apex of the cone or wedge formed by extrapolating the die, to the contact length.
a) Show that for wire drawing with this definition=α(1+√
1−r)2/r. b) Calculate the ratio offrom this definition to the value offrom equation
10.4for (i)α=10◦,r=0.25; (ii)α=10◦,r=0.50; (iii)α=45◦,r=0.50;
(iv)α=90◦,r=0.50.
10.6. Backofen developed the following equation for the mechanical efficiency, η, during wire drawing:η=[1+C(−1)+µtanα]−1 for >1, whereC is an empirical constant equal to about 0.12 andµis the friction coefficient. Note that the termC(−1) representswr/wiand the termµ/sinαrepresentswf/wi, soη=(1+wr/wi+wf/wi)−1=(wa/wi)−1.
a) Evaluate this expression forµ=0.05 andα=0.3 for 2≤α ≤20◦. Note that for small angles <1 so the redundant strain termC(−1) is zero.
b) Plotηvs.αand find the optimum die angle,α∗.
c) Derive an expression forα∗as a function ofµandr. Doesα∗ increase or decrease withµ? withr?
10.7. A slab of annealed copper with dimensions 2.5 cm×1.5 cm×8 cm is to be compressed between rough platens until the 2.5 cm dimension is reduced to 2.2 cm. With a sketch, describe, as fully as possible the resulting inhomogeneity of hardness.
10.8. A high-strength steel bar must be cold reduced from a diameter of 2 cm to 1.2 cm by drawing. A number of schedules have been proposed. Which schedule would you choose to avoid drawing failure and minimize the likelihood of centerline bursts? Assumeη=0.5 and give your reasoning.
a) A single pass through a die of semi-angle 8◦.
b) Two passes (2.0 cm to 1.6 cm and 1.6 to 1.2 cm) through a die of semi-angle 8◦.
c) Three passes (2.0 cm to 1.72 cm, 1.72 cm to 1.44 cm, and 1.44 cm to 1.2 cm) through a die of semi-angle 8◦.
d) Four passes (2.0 cm to 1.8 cm, 1.8 cm to 1.6 cm, 1.6 cm to 1.4 cm, and 1.4 cm to 1.2 cm) through a die of semi-angle 8◦.
(e,f,g,h) Same reductions as in schedules a, b, c, and d except through a die of semi-angle 16◦.
10.9. Determinefrom the data of Hundy and Singer (Figure10.6) for reductions of 2.3%, 6.5%, 13.9%, and 26%. Make a plot of I.F. versus.
11 Formability
An important concern in forming is whether a desired process can be accomplished without failure of the work material. Forming limits vary with material for any given process and deformation-zone shape. As indicated in Chapter10, central bursts may occur at a given level ofin some materials and not in others. Failure strains for a given process depend on the material.