Several methods of analysis have been applied for determining the punch pressure in direct extrusion.
These are as follows:
(i) Slab method.
(ii) Slip line method.
1. Slab Method:
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For small die angles, the slab method already discussed in case of wire drawing may be applied to extrusion as well. The metal flow in both the wire drawing and extrusion processes is similar. The difference is that in wire drawing the wire is pulled through the die while in extrusion it is pushed.
However, the die angles used in extrusion are generally very large as compared to those used in wire drawing. Therefore, the radial component of frictional stress which was neglected in wire drawing is now taken into account in the following analysis.
Consider the Fig. (10.17). A slab in the conical portion of die is acted upon by pressures pex + dpex and pex on its flat faces perpendicular to die-axis, pressure px and frictional stress τx on the die-billet interface. The equation of equilibrium of forces in axial direction is simplified to following by neglecting terms with more than one differential and taking τx = µ px.
where C is the constant of integration. C can be evaluated from the stress condition at the exit end of the die, i.e.
where σ01 and σ02 are the values of yield strength of material before and after the extrusion. For determining extrusion stress we have to take into account the shear deformation at entry and exit from the die.
Effect of Shear Deformation at Entry and Exit:
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The material suffers a shear deformation at the entry to the die and also at the exit from the conical portion of die. The increase in stress σs due to shear deformation at entry and exit is determined in a way similar to that done in case of wire drawing, however, in the following we simplify the calculation by simply adding the increase in stress at both at entry and exit to the pressure.
Effect of Friction on Cylinder Wall:
The effect of friction on cylinder wall is taken into consideration in a way similar to the effect of land portion of die in wire drawing. Consider the stresses acting on a slab in the material in the cylinder as shown in (Fig. 10.18). The equilibrium of forces results in the following equation.
where C is the constant of integration. It may be evaluated with help of boundary condition at the entry to the conical portion of die, i.e.,
Example 1:
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A 50 mm diameter bar is extruded to 25 mm diameter bar. Determine the value of pe/σ0 if the co-efficient of friction on cylinder and die interface is 0.2 and semi die angle is 30°. If the length of billet is 50 mm, determine the value of relative punch pressure (pt/σ0).
Solution:
2. Slip Line Solutions:
Slip line method has been used extensively by Hill, Johnson and his co-workers and other researchers for determining punch pressures for a large variety of extrusion problems. These solutions have the following drawbacks.
(a) They are applicable to plane strain deformation condition only. This is a severe restriction to applicability of these solutions because in the industrial extrusions, billets as well as containers are cylindrical in shape. Besides, the extrude may be having circular or non- circular shape.
(b) The material is assumed to be rigid perfectly plastic. The real materials have tendency to strain harden as well as their flow stress is also affected by strain rate. In case of hot extrusions, the re-crystallization is also present. All these factors cannot be taken into considerations in slip line solutions.
(c) The boundary conditions, particularly the friction conditions are never known exactly. Therefore, the slip line solutions are also approximate and some of them are by intuition.
We cannot discuss all the slip line solutions here. It is better to discuss the results of these solutions, which show that the relative extrusion pressure pe/σ0 in all types of extrusion processes may be expressed as given in Eqn. (10.23).
where r is the relative reduction in area, a and b are constants and σ0 is the yield strength of material in compression.
Based on slip line solutions for extrusion through 90 degree dies, Johnson has given the following expressions for extrusion pressures.
Example 2:
Recalculate the relative extrusion pressure for the data of Example 1 with solutions of slip line method.
Solution:
According to Eqn. (10.26) the relative extrusion pressure = 0.8 + 1.5 ln 4 = 2.88
Relative punch pressure (Eqn. 10.21) = 5.184
The value (2.88) is less than that given by slab method with µ = 0.2 (see Example 1), it agrees approximately when µ = 0.15 in which case slab method gives pe/σ0 = 2.918
Empirical Formulations for Extrusion Pressure:
Hydrostatic Extrusion of Aluminum Alloys:
For hydrostatic extrusion of aluminum and its alloys Hakon and Mikkelesen have determined the values of a and b for the Eqn. (10.23) experimentally. The values for some materials are given in Table 10.1.
Hot Extrusion of Al-Alloys:
Sheppart and Raybould have worked on extrusion of super pure aluminum, on Al-Zn alloy and Al-Zn-Mg alloy in the temperature range of 50°C to 500°C. According to them the above mentioned formulae by Johnson, i.e. Eqn. (10.25, 10.26) give values of extrusion pressure lower than the ones observed experimentally. The authors have also given an upper bound solution for extrusion. According to them the following formulae fit the experimental values better.
Hot Extrusion of Steel:
Vincent Depierre has worked on hot extrusion of steel with glass as lubricant. According to him the extrusion pressure is given by-
The S and k are interface frictional shear stresses on die and liner respectively. Both these factors depend on the temperature and speed of extrusion, and the lubrication conditions. S also depends on die angle. The value of S/σ0 varies from 0.2 to 0.6 for glass coated billet with bare dies, while for glass coated billet and zirconia coated dies it varies from 0.15 to 0.28. Authors carried out experiments in temperature range of about 870°C (1600° F) to 1204°C (2200°F). The value of k similarly varies with speed and temperature of billet and liner temperature.
Forces in Cold Extrusion of Steel:
In order to estimate the press load in cold extrusion steel for different types of components, one of the options is to carry out experiments on soft metal like lead and relate this load to that for extrusion of steel by a factor which is the ratio of yield strength of the two metals.
The method would be highly approximate if the frictional conditions and strain hardening characteristics of the two metals are not identical. By taking strain hardening rule σ̅ = σ0 ԑ̅n, Ronald et al. determined the values of n for lead (Pb 99.9, Ag 0.01 and Cu 0.07) and for 15 steels which are generally cold extruded such as SAE 1010, 1012, 1.18. 1040 etc. and found that the value of n = 0.2 is approximately true for the steels and as well as for the lead. However, other authors have given different value of n for same steels. According to Ronald,
The above formulation should be verified by actually carrying out experiment on steels included in the formulation. Generally the extrusions are carried out in dies with 120 included angle. The multiplying factor for other angles may be determined by α-factor in Equation (10.30).
The maximum pressure for backward extrusion of steel for extrusion ratios 1.65 to 4.25 for billets with length/diameter ratio of 0.6 is given by the following relation.
Example 3:
Compare the influence die angle in cold extrusion of steel for die angle of 30, 60, 90, 120, 150, 180 degrees.
Solution:
Referring to Eqn. 10.30, the influence of die angle is given by the factor α0.375 where ‘α’ is semi die angle. In steel extrusion generally, die angle of 120° is used. Therefore, we compare other die angles in terms of 120°. In the following we evaluate multiplying factor = (die angle/120)0.375
Upper Bound Solutions for Plane Strain Extrusion:
Plane Strain Extrusion through Square Dies:
The method based on the concept of slip planes is described below. Figure 10.19(a) illustrates the process. The punch pushes the material with velocity V1. Metal flow is symmetrical about the centre line, therefore let us consider the flow in upper half only.
The material on encountering slip plane AB suffers a shear deformation and starts traveling parallel to AC. The material in the top corner ACS is stationary and forms the dead zone. The material in triangle ABC shears along AC. On reaching the line BC the material again suffers shear deformation along BC and starts traveling parallel to the axis of bar.
Similar deformations occur in the material below the center-line of the bar. The corresponding slip planes are BM, BN and MN. For determination of extrusion pressure we need to determine the total rate of work (power) being dissipated along all the slip planes. This power is supplied by ram of the press and is equal to extrusion load × V1. Thus by equating the two we can determine the extrusion load.
Here b is the width of container and the extrude. The factor 2 is due to two regions i.e. one above the center line and other below the center line.
For different values of α and β the above expression would give different values of extrusion pressure. We have to determine such values of α and β for which the extrusion pressure is the minimum. The optimum values have been calculated numerically for different values of reductions. These are given in Table 10.4 below.
Plain Strain Extrusion with Inclined Dies:
If the die consists of inclined planes with semi die angle more than 6 as calculated above, then the metal flow after crossing first slip plane would be parallel to die surface. The shear deformations and velocity diagram is shown in (Fig. 10.20).
In such a case the optimum values of α and β are to be determined with value of θ equal to semi die angle. Also the shear stress along the die face will be determined by the frictional condition. Therefore, we may take it equal to mK where m is the friction factor and K is the yield strength of material in shear. The value m varies from 0 to 1. The values 0 and 1 correspond to frictionless case and sticking friction respectively.
The formulation given by Eqn. (10.39) has to be slightly modified for application to this case. The second term on the R.H.S. is multiplied by m. The Eqn. (10.39) becomes
For different values of m and reductions the optimum values of α and β for which the extrusion pressure is minimum are required to be determined.