Effect of Strain Rate on the Formation of Strain-Induced Martensite in AISI 304L Stainless Steel

Abstract

The effect of strain rate on the stress–strain response and the austenite to strain-induced α′-martensite transformation of austenitic steel 304L was studied. Compression tests were carried out at room temperature in the strain rate range of 10⁻³ to 10³ s⁻¹ and the evolution of martensite was quantified using a ferritoscope. Higher strain rates resulted in lower strain-induced α′-martensite. Strain incremental tests were carried out at 1 s⁻¹ to simulate isothermal tests and to delineate effect of adiabatic heating on the α′-martensite transformation. Strain rate change tests were also carried out to determine the effect of prior strain rate history on the strain-induced α′-martensite content. These experiments showed that apart from the effect of adiabatic heating at high strain rates on the α′-martensite content, there is an additional effect of strain rate. While the adiabatic heating effect could be based on the increase in stacking fault energy (reduction of stacking fault width) with temperature, the additional effect due to strain rate was explained based on the expected reduction in stacking fault width with increasing strain rate.

Appendix A

The temperature rise in the sample due to plastic deformation can be estimated by considering the heat dissipation to the environment by the specimen as a one dimension heat transfer problem as shown in Figure A1. The specimen is touching the platens at either end of its length, and it is taken that the ends are always at ambient temperature of 25 °C. There is a constant heat source Q present in the specimen due to the work being done on it that is getting converted to heat. In actual experiments the specimen length decreases with strain, however for sake of simplifying the problem the length of sample is kept fixed. Thus, as the conduction path is slightly longer than actual, the temperature solved for may be a small overestimation.

Fig. A1

Schematic of 1-D specimen with the ends held at ambient temperature (25 °C)

The heat transfer equation in 1-dimension is given by

$$ \frac{\partial T}{\partial t} = \frac{K}{{\rho \cdot C_{\text{p}} }} \cdot \frac{{\partial^{2} T}}{{\partial x^{2} }} + \frac{Q}{{\rho \cdot C_{\text{p}} }}, $$

(A1)

where ρ is mass density (kg/m³), C_p is specific heat capacity (J/kg K), K is thermal conductivity (W/K m), and Q is heat generation rate per unit volume (W/m³). In the present case Q is derived from the incremental work done per unit volume dw on the specimen and is given by

$$ {\text{d}}w = \sigma \cdot {\text{d}}\varepsilon $$

(A2)

Of the plastic work increment (Eq. [A2]) some will be consumed in increasing the defect density (internal energy) and the remaining will generate heat (factor β taken here as 0.9). Thus the rate of heat generation is

$$ Q = \beta \frac{{{\text{d}}w}}{{{\text{d}}t}} = \beta {\kern 1pt} \sigma \frac{{{\text{d}}\varepsilon }}{{{\text{d}}t}} = \beta \,{\kern 1pt} \sigma \,\dot{\varepsilon } $$

(A3)

For the sake of estimating an average number for Q, it can be assumed as a first approximation that σ varies about linearly with strain, and thus an average value of σ over the strain range should suffice to get an average value of Q. From Figure 1 of text, σ varies from ~300 to 1200 MPa at a strain of 0.5 for 10⁻³ to 1 s⁻¹ and varies from ~600 to 1200 MPa for 10³ s⁻¹. Hence an average value of 750 MPa for strain rate up to 1 s⁻¹ and 900 MPa for strain rate of 10³ s⁻¹ was taken. Thus the values of Q using Eq. [A3] for strain rates of 10⁻³, 10⁻¹, 1 and 10³ s⁻¹ are 6.75×10⁵, 6.75×10⁷, 6.75×10⁸, and 8 × 10¹¹ W/m³, respectively.

Equation [A1] needs to be solved numerically using a finite difference method for which time steps need to be given. Solving for different strain rates would require different time steps. To overcome this, the variable of time t was converted to strain ε, as the strain increments can be kept same for all strain rates used. The variable t is replaced by ε using

$$ \frac{\partial T}{\partial t} = \frac{\partial T}{\partial \varepsilon }\frac{\partial \varepsilon }{\partial t} = \frac{\partial T}{\partial \varepsilon }\dot{\varepsilon }. $$

(A4)

Substituting Eq. [A4] in Eq. [A1], the 1D heat transfer equations can be re-written in terms of ε as

$$ \frac{\partial T}{\partial \varepsilon } = \left( {\frac{K}{{\dot{\varepsilon }\,\rho \,C_{P} }}} \right)\frac{{\partial^{2} T}}{{\partial x^{2} }} + \frac{Q}{{\dot{\varepsilon }\,\rho \,C_{P} }}. $$

(A5)

The initial condition is T(x, ε = 0) = 25 °C, and the boundary conditions are T(0, ε) = 25 °C and T(L, ε) = 25 °C. For SS 304 ρ is 7900 kg/m³, C_p is 500 J/(kg K), K is 16 W/(m K). Figure A2 shows the results of the numerical solution of partial differential equation (Eq. [A5]).

Fig. A2

Temperature rise at the center of the specimen as a function of strain (a), and temperature rise at a strain of 0.5 as a function of distance in the specimen from one end at 0 mm to another end at 7 mm (b)

There was no temperature rise at 10⁻³ s⁻¹ making it a completely isothermal test. The temperature rise at 10³ s⁻¹ was the highest of about 100 °C and was near constant throughout its length making it a completely adiabatic test. At 10⁻¹ s⁻¹ there was a small temperature rise of about 20 °C, while that at 1 s⁻¹ there was a temperature rise of 80 °C. The strain rates 10⁻¹ and 1 s⁻¹ are in the transition zone between isothermal and adiabatic conditions, with 10⁻¹ s⁻¹ being a near isothermal test, while 1 s⁻¹ being a near adiabatic test.