Using the stresses as shown in part (b), we have, from equilibrium and assuming unit width,

For the distortion-energy criterion, it should be recognized that _x is now tensile, whereas in the text it is compressive. Therefore, Eq. (6.11) becomes

Thus When substituted into the equilibrium equation, one obtains Using the boundary conditions that _x = 0 (and therefore _y = Y 0) at x = 0, gives the value of C as Therefore, substituting into the expression for _y,

which is the same as Eq. (6.13) on p. 270. 6.63 Plot the force vs. reduction in height curve in open-die forging of a solid cylindrical, annealed copper specimen 2 in. high and 1 in. in diameter, up to a reduction of 70%, for the cases of (a) no friction between the flat dies and the specimen, (b) μ = 0.25, and (c) μ = 0.5. Ignore barreling and use average-pressure formulas. For annealed copper we have, from Table 2.3 on p. 37, K = 315 MPa = 46,000 psi and n = 0.54. The flow stress is where the absolute value of the strain is

From volume constancy, we have

Note that ro = 0.5 in and ho = 2 in. The forging force is given by Eqs. (6.18) and (6.19) on p. 272 as:

Some of the points on the curves are the following:

The curve is plotted as follows:

6.64 Use Fig. 6.9b to provide the answers to Problem 6.63. The force required for forging is the product of the average pressure and the instantaneous cross-sectional area. The average pressure is obtained from Fig. 6.9b on p. 272. Note that for μ = 0, pave/Yf = 1, and thus the answer is the same as that to Problem 6.63 given above. The following table can be developed where pave/Yf is obtained from Fig 6.9b, and h and r are calculated as in Problem 6.63. Note that Fig. 6.9b does not give detailed information for 2r/h < 10, which is where the data for this problem lies. However, the μ = 0.25 values (interpolated between the μ = 0.2 and μ = 0.3 curves) are noticeably above 1 by 2r/h = 5 or so, so we give the value 1.25, and all intermediate values are linearly interpolated from this reading. Similarly, for μ = 0.5, a value between μ = 0.3 and sticking suggests pave/Y is around 1.6 or so by 2r/h = 3. This is the basis for the numbers below.

Recall from Problem 6.63 that where the absolute value of the strain is

From this, the following forces are calculated (recall that F = paveA):

The results are plotted below. For comparison purposes, the results from Problem 6.63 are also included as dashed lines. As can be seen, the results are fairly close, even with the rough interpolation done in this solution.

6.68 Derive an expression for the average pressure in plane-strain compression under the condition of sticking friction. Sticking friction refers to the condition where a Tresca friction model is used with m = 1 [see Eq. (4.5) on p. 140]. Therefore, the following figure represents the applied stresses to an element of forging, which can be compared to Fig. 6.4b on p. 269. The approach in Section 6.2.2 starting on p. 269 is followed closely in this derivation.

From equilibrium in the x-direction, Solving for d_x,

Integrating,

where C is a constant. The boundary condition is that at x = a, _x = 0, so that

Therefore,

and

The die pressure is obtained by applying Eq. (2.36) on p. 64:

Note that this relationship is consistent with Fig. 6.10 on p. 273 for 0 < x < a. Since the relationship is linear, then we can note that

6.69 What is the magnitude of μ when, for plane-strain compression, the forging load with sliding friction is equal to the load with sticking friction? Use average-pressure formulas. The average pressure with sliding friction is obtained from Eq. (6.15) on p. 271, and for sticking friction it is obtained from the answer to Problem 6.68 using m = 1. Equating these two average pressures, we obtain

Therefore, μ = 0.5.

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