Chapter 6: 1,2,10,62-64,68,69 6.1 How can you tell whether a certain part is forged or cast? Describe the features that you would investigate to arrive at a conclusion. Numerous tests can be used to identify cast vs. forged parts. Depending on the forging temperature, forged parts are generally tougher than cast parts, as can be verified when samples from various regions of the part are subjected to a tensile test. Hardness comparisons may also be made. Microstructures will also indicate forged vs. cast parts. Grain size will usually be smaller in forgings than in castings, and the grains will undergo deformation in specific directions (preferred orientation). Cast parts, on the other hand, will generally be more isotropic than forged parts. Surface characteristics and roughness are also likely to be different, depending on the specific casting processes used and the condition of the mold or die surfaces. 6.2 Why is the control of volume of the blank important in closed-die forging? If too large of a blank is placed into the dies in a closed-die forging operation, presses will (a) jam, (b) not complete their stroke, and (c) subject press structures to high loads. Numerous catastrophic failures in C-frame presses have been attributed to such excessive loads. If, on the other hand, the blank is too small, the desired shape will not be completely imparted onto the workpiece. 6.10 Why are end grains generally undesirable in forged products? Give examples of such products. As discussed in Section 6.2.5 starting on p. 283, end grains are generally undesirable because corrosion occurs preferentially along grain boundaries. Thus end grains present many grain boundaries at the surface for corrosion to take place. In addition, they may result in objectionable surface appearance, as well as reducing the fatigue life of the component because of surface roughness that results from corrosion. 6.62 In the free-body diagram in Fig. 6.4b, the incremental stress d_x on the element was shown pointing to the left. Yet it would appear that, because of the direction of frictional stresses, μp, the incremental stress should point to the right in order to balance the horizontal forces. Show that the same answer for the forging pressure is obtained regardless of the direction of this incremental stress. We will derive the pressure using the same approach as described in Section 6.2.2 starting on p. 269. The equivalent version of Fig. 6.4 on p. 269 is shown below.
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.
