Even with sheets of plastic or soft metal used to create a low shear-stress interface, the edges of the work piece cut through the film at low strains. Thus for axially symmetric compression neither the assumption of a constant coefficient of friction nor constant shear-stress interface is correct.
7.16 RING FRICTION TEST
A simple test for friction in compression involves compressing a ring as shown in Figure7.24. If there were no friction the inner diameter would increase by the same percentage as the outer diameter. With high friction, there is a no-slip location between the inner and outer diameters so the inner diameter must decrease during compression.
Dio Do
Di
Di
original ring
high friction
low friction
7.24.Ring compression test: original specimen (top); after compression with high friction (middle);
and after compression with low friction (bottom).
0.1 0.2 0.3 0.4
0.3
0.2
0.1
-0.1
-0.2
-0.3
Fractional change of inner diameter
Axial compression, ∆h/ho µ = 0.1 sticking friction
frictionless 7.25.Inner diameter change calculated for
a specimen and with an outer radius twice the inner radius and with a height 1.33 times as high as its inner radius. Data from J. B.
Hawkyard and W. Johnson,Int. J. Mech. Sci, 17 (1967), pp. 163–182.
Figure7.25shows the change of inner diameter as a function of the friction coefficient for a ring with an outer diameter twice the inner diameter.
REFERENCES
W. Johnson and P. B. Mellor,Plasticity for Mechanical Engineers, Van Nostrand, 1973.
G. T. van Rooyen and W. A. Backofen,J. Mech. Phys. Solids,7 (1959), pp. 163–68.
PROBLEMS
7.1. A coil of steel, 252 mm wide and 3 mm thick, is drawn though a pair of dies of semi-angle 8◦to a final thickness of 2.4 mm in a single pass. The outlet speed is 3.5 m/s. The average yield strength is 700 MPa and the friction coefficient is 0.06. Calculate the power in kW consumed.
7.2. An efficiency of 65% was found in a rod-drawing experiment with a reduction of 0.2 and a semi-die angle of 6◦.
a) Using Sachs’ analysis, find the coefficient of friction.
b) Using the value of η found in (a), what value of efficiency should be predicted from the Sachs’ analysis fora=6 andr=0.4?
c) The actual value ofηfound for the conditions in (b) was 0.80. Explain.
7.3. Estimate the force required to coin a U.S. 25c/piece. Assume that the mean flow stress is 30,000 psi, the diameter is 0.95 in, and the thickness after forming is 0.060 in.
7.4. Figure7.26shows a billet before and after hot forging from an initial size of 2.5 mm×2.5 mm×25 mm to 5 mm×1.25 mm×10 mm. This is accomplished by using a flat-face drop hammer. Sticking friction can be assumed. For the rate of deformation and the temperature, a flow stress of 18 MPa can be assumed.
a) Find theforcenecessary.
b) Find theworkrequired. (Remember that work=ʃFdLand thatFchanges withL.)
2.5 mm
1.25 mm
5 mm 2.5 mm
25 mm
25 mm
7.26.Compression in Problem 7.4.
PROBLEMS 107
c) From what height would the hammer have to be dropped?
d) Compute the efficiency,η.
7.5. Two steel plates are brazed as shown in Figure7.27. The steel has a tensile yield strength of 70 MPa and the filler material has a tensile yield strength of 7 MPa. Assume that the bonds between the filler and the steel do not break.
Determine the force necessary to cause yielding of the joint.
steel
steel
brazed
joint
2 mm
50 mm
3 mm 7.27. Brazed joint for Problem 7.5.
7.6. Figure 7.28 shows a thin lead ring being used as a gasket. To insure an acceptable seal the gasket must be compressed to a thickness of 0.25 mm.
Assume that the flow stress of lead is 15 MPa and strain hardening is negligible.
Find the required force.
25 cm
2 mm 0.5 mm
F F
lead gasket
7.28. Lead gasket for Problem 7.6.
7.7. The derivation of equation 7.10 assumed a constant coefficient of friction.
Derive an equivalent expression assuming sticking friction.
7.8. Magnetic permalloy tape is produced by roll flattening of drawn wire, as shown in Figure7.29. The final cross section is 0.2 mm by 0.025 mm. It is physically possible to achieve this cross section with different rolling schedules. However, it has been found that the best magnetic properties result with a maximum amount of lateral spreading. For production the rolling direction must be parallel to the wire axis. Describe how you would vary each of the parameters below to achieve the maximum spreading:
a) roll diameter, b) reduction per pass, c) the friction,
d) back and front tension.
rolling direction
0.2 mm
0.025 mm
7.29. Permalloy tape (Problem 7.8).
7.9. A metal with a flow stress of 35 MPa is to be drawn from a diameter of 25 mm to 20 mm through a die of 15◦ semi-angle. Calculate the necessary drawing stress if
a) the conditions are frictionless, b) there is sticking friction, and c) µ=0.20.
7.10. Consider the rolling of a sheet 15 cm wide from a thickness of 1.8 mm to 1.2 mm in a single pass by steel rolls 20 cm in diameter. Assume a friction coefficient of 0.10 and a flow stress of 125 MPa.
a) Calculate the roll pressure if roll flattening is neglected.
b) Calculate the roll pressure taking into account roll flattening.
c) Estimate the minimum thickness that could be achieved.
7.11. Use equation7.14to predict how the ratio ofwf/wi depends onµ,αandεh. (Realize that equation7.14neglects redundant work. Expand the exponential term after simplifying and assume that εh is small enough so higher order terms can be neglected.) Describe in words howwidepends onµ,αandεh.
PROBLEMS 109
7.12. In the force balance in the slab analysis for frictional effects in plane-strain compression,Pwas assumed to be a principal stress, even though with finite friction it can’t be. Examine this assumption by assuming a constant shear stress interface withτ =mk, and derive an expression for the angle,θ, between the principal axis, 1, and thex-axis. Express your answer in terms ofm,x,h, L, and 2k. (Not all of these need be in the final expression.) See Figure7.30.
σxh (σx+dσx)h
dx Pdx
τdx
h
x
L/2
1 2
x y
θ θ
7.30. Sketch for Problem 7.12.
8 Upper-Bound Analysis
Calculation of exact forces to cause plastic deformation in metal forming processes is often difficult. Exact solutions must be bothstatically andkinematicallyadmissible.
That means they must be geometrically self-consistent as well as satisfying the required stress equilibrium everywhere in the deforming body. Frequently it is simpler to use limit theorems that allow one to make analyses that result in calculated forces that are known to be either correct or higher or lower than the exact solution.
Lower bounds are based on satisfying stress equilibrium, while ignoring geometric self-consistency. They give forces that are known to be either too low or correct. As such they can assure that a structure is “safe.” Conditions in whichη=0 are lower bounds.
Upper-bound analyses, on the other hand, predict stress or forces that are known to be too large. These are usually more important in metal forming. Upper bounds are based on satisfying yield criteria and geometric self-consistency. No attention is paid to satisfying equilibrium.