The aim of this work is the development of an analytical method for the calculation of metal plasticity during isothermal multistage forging.
According to the data from Reference Mykhalevych (1998) the more intensive plasticity recovery occurs at the beginning of the pause. It was shown in Mykhalevych (1998) that further increases in pause duration leads to minor increases in plasticity recovery (Fig. 2).
Curves of recovery plasticity can be described by the next equation:
$$\Delta \psi = K(1 - e^{ - t \cdot n} ) ,$$
(1)
where К—coefficient which shows the maximum recovery of plasticity; t—pause duration, sec; n—coefficient which shows velocity of plasticity recovery.
Results of Eq. (1) match with those of the more complex equation (Mykhalevych 1998) when the pauses are small:
$$\Delta \psi = \frac{{\ln (1 + e^{{s\psi_{1} }} (e^{{s\Delta_{p} }} - 1))}}{s} - \Delta_{p} ,$$
(2)
where s—material coefficient; Δ
p
—relative duration of pause; ψ
1—used plasticity before pause.
Equation (2) has a major fault with long pauses. This equation was obtained based on the hypothesis of full reversibility in the process of metal damage accumulation in the hot forging. The consequence of this hypothesis is that, during a long pause, the plasticity has a full recovery independent of the level of used plasticity.
In cold forging there is critical value of metal damage. After this damage level is reached, the metal cannot recover its initial plasticity after heat treatment (Matviychuk and Aliev 2009). These assumptions can be expressed for hot forging schemes. Suppose that maximum value of plasticity recovery (Δψ) with low used plasticity ψ
i
on i-th stage of deformation is equal ψ
i
and with a high value of used plasticity (close to 1) the recovery plasticity approaches 0. In this case the coefficient K of Eq. (1) takes the next form:
$$K = m \cdot \psi_{i} , and \left\{ {\begin{array}{*{20}c} {\psi_{i} = 0 \to m = 1} \\ {\psi_{i} = 1 \to m = 0} \\ \end{array} } \right.$$
Graphically it can be expressed by the next way (Fig. 3).
In this case Eq. (1) can be expressed as:
$$\Delta \psi = \psi_{i} \cdot m(1 - e^{ - t \cdot n} ) .$$
(3)
In general, parameters m and n are determined by material properties and deformation temperature. From the previous analysis it can be found that parameter m can be expressed as follows:
$$m = \left( {1 - \psi_{i} } \right)^{{f\left( {\psi_{i} ,\,T} \right)}} ,$$
(4)
where function f(ψ, T), characterizes the material, and can be found experimentally. To determine the function f(ψ, T) it is necessary to conduct a set of experiments. Specimens should be deformed under tensile conditions with certain prescribed temperature and velocity. Experiments should include deformation of the specimen with a value of ψ
i
in the range of 0.1–0.9. This allows the determination of parameter Δψ and parameter m that simply define f(ψ, T) for Eq. (4). Equation (3) can be transformed to the following expression:
$$\Delta \psi = \psi_{i} \cdot \left( {1 - \psi_{i} } \right)^{{f\left( {\psi_{i} ,\,T} \right)}} \left( {1 - e^{ - t \cdot n} } \right)$$
(5)
Experimental determination of recovery plasticity value for calculation coefficients of Eq. (5) must be found by the next equation:
$$\Delta \psi = \psi_{1} + \psi_{2*} - 1 ,$$
(6)
where \(\psi_{1} = {{\varepsilon_{u1} } \mathord{\left/ {\vphantom {{\varepsilon_{u1} } {\varepsilon_{p} }}} \right. \kern-0pt} {\varepsilon_{p} }}\)—used plasticity before the pause; \(\varepsilon_{p} = \ln \left( {{{l_{p} } \mathord{\left/ {\vphantom {{l_{p} } {l_{0} }}} \right. \kern-0pt} {l_{0} }}} \right)\)—logarithmic strain fracture during monotonic tensile test; \(\varepsilon_{u1} = \ln \left( {{{l_{l} } \mathord{\left/ {\vphantom {{l_{l} } {l_{1} l_{0} \cdot l_{0} }}} \right. \kern-0pt} {l_{1} l_{0} \cdot l_{0} }}} \right)\)—logarithmic strain before the pause in the tensile test; \(\varepsilon_{p2} = \ln \left( {{{l_{p2} } \mathord{\left/ {\vphantom {{l_{p2} } {_{0} }}} \right. \kern-0pt} {_{0} }}} \right)\)—logarithmic strain of the fracture in the tensile test after the pause; \(\psi_{2*} = \frac{{\varepsilon_{p2} - \varepsilon_{u1} }}{{\varepsilon_{p} }}\)—residual plasticity after the pause; l
0; l
1—initial and pre-pause tensile test-derived lengths of the billet respectively, mm; \(l_{p} ;\,l_{p2}\)—maximum length of the billet in monotonic and multistage tensile test respectively, mm.
If recovery of plasticity does not occur i.e. Δψ = 0 then ψ
1 + ψ
2* = 1 or
$$\frac{{\varepsilon_{p2} - \varepsilon_{u1} }}{{\varepsilon_{p} }} + \frac{{\varepsilon_{u1} }}{{\varepsilon_{p} }} = 1 .$$
(7)
In this case, the length of the fracture billet during the tensile test after the pause must be equal to the length of fracture billet during the monotonic tensile test without a pause \(\varepsilon_{p2} = \varepsilon_{p}\).
Then Eq. (7) takes the next form:
$$\frac{{\varepsilon_{p} }}{{\varepsilon_{p} }} - \frac{{\varepsilon_{u1} }}{{\varepsilon_{p} }} + \frac{{\varepsilon_{u1} }}{{\varepsilon_{p} }} = 1$$
If all previously described equations are substituted in Eq. (6) and some mathematic simplification is done, we obtain:
$$\Delta \psi = \frac{{\ln \left( {\frac{{l_{p2} }}{{l_{0} }}} \right)}}{{\ln \frac{{l_{p} }}{{l_{o} }}}} - 1$$
A practical application of upsetting a cylindrical billet is shown in (Fig. 4). In the upsetting process point T is the point at which the fracture will most likely occur. Parameter ψ
1 must be determined for this point. This parameter ψ
1 depends on relative height deformation (ratio of the upsetting stroke to the initial billet height \(\varepsilon_{h} = \Delta h/\Delta h_{0}\)). At this point the values of logarithmic strain ɛ
i
and used plasticity \(\psi_{ 1} = \varepsilon_{ 1} /\varepsilon_{p}\) can be determined, where ɛp—logarithmic strain of the fracture.
After the first stage of upsetting and pause the recovery of plasticity can be calculated by the next equation:
$$\Delta \psi = \psi_{1} \cdot \left( {1 - \psi_{1} } \right)^{{f\left( {\psi_{1} ,\,T} \right)\,}} \left( {1 - e^{ - t \cdot n} } \right)$$
Used plasticity after the first pause can be found as a difference between recovery plasticity at the end of pause and used plasticity before the pause:
$$\psi_{1}^{1} = \psi_{1} - \Delta \psi_{1}.$$
During the pause after the second stage of upsetting, plasticity will be recovered which was used during the second stage of upsetting ψ
2 and some metal damage also accumulated after the first stage of upsetting. Assuming that used plasticity is additive (allowing the possibility of a simple summation) the value of recovery plasticity and used plasticity after two stages of upsetting can be found as follows:
$$\Delta \psi_{2} = \left( {\psi_{1}^{1} + \psi_{2} } \right) \cdot (1 - \psi_{1}^{1} - \psi_{2} )^{{f(\psi_{1}^{1} + \psi_{2} ,\,\,T)}} (1 - e^{ - t \cdot n} ); \psi_{2}^{1} = \psi_{1}^{1} + \psi_{2} - \Delta \psi_{2}$$
For the k stages of upsetting
$$\psi_{k}^{1} = \psi_{k - 1}^{1} + \psi_{k} - \Delta \psi_{k} ,$$
(8)
and \(\psi_{k - 1}^{1} + \psi_{k} \le 1\), in another case recovery of plasticity is not calculated.