Residual stresses are often observed when metals are quenched from elevated temperatures. Rapid quenching will, in the absence of phase transformations, produce a residual compressive stress on the surface. An imbalance of the internal forces associated with residual stress within a part (tensile and compressive) causes distortion.
A very simple explanation of the formation of residual stress during quenching can be demonstrated by considering two concentric cylinders (see Fig. 1) of equal cross-sectional area (AS=AC). During quenching, the outer shell will shorten faster than the center as a result of a temperature gradient and thermal contraction. This will place the surface in tension and the center in compression while the center is still hot.
A simple calculation can be made to determine the stress. Consider that the cylinders are the same length to maintain continuity. The strain in the surface cylinder will be a combination of thermal contraction and strain from the residual stress. The strain in the central cylinder will only be a result of the residual stress since we will assume that the temperature has not changed. The strains for each cylinder may be calculated from the following equations:
where (alpha) is the coefficient of thermal expansion,Eis the modulus, (sigma) is the stress and (epsilon) is the strain. Subscripts indicate the quantity expressed is either for the surface (S) or the center (C). The force balance can be converted to a stress balance since the areas of the two cylinders are assumed to be equal.
Now consider steel quenched rapidly from below the A1 (727°C) to room temperature. In this example, a phase transformation will not occur since there is no austenite prior to quenching. If we assume for simplicity the outer cylinder is cooled to room temperature while the center cylinder is maintained at a temperature just below the A1, the temperature differential between surface and center will be approximately 700°C. The elastic modulus of steel, just below the A1, is approximately half the room temperature value of 207 GPa (30x10^6 psi). Using a coefficient of thermal expansion for steel of 11.4x10^-6 /°C, a residual stress of 550 MPa (80,000 psi) will develop. This stress will be tensile in the surface cylinder and compressive at the center. When the center cools and contracts, the equations are modified to show that the elastic modulus is the same for both surface and center to give the following
Assuming that strain in the surface and center cylinders remain equal and no yielding occurs, the new stress calculated is simply added to the stress from the first set of equations. A stress reversal is expected since the surface cylinder will resist the contraction of the center, but now the center has a higher elastic modulus. This will leave a residual compressive stress of 275 MPa (40,000 psi) in the surface cylinder and will be balanced by a tensile residual stress in the center of the same magnitude. Fig. 1 shows a schematic of how the stress changes with time during quenching. The temperature of the stress reversal will be dependent upon yielding of the surface or center during quenching.
The addition of a phase transformation can greatly complicate the analysis of the final residual stress. Transformation of austenite will in general produce a positive volume change. For a 0.45 weight percent carbon steel, the austenite to martensite transformation produces a volume expansion of 4.4% whereas a 3.6% volume expansion is observed for the austenite to upper bainite transformation. To develop the highest possible compressive residual stress on the surface, the center must transform before the thermal stress reversal and the surface must transform after the stress reversal. This sequence of transformation is often observed in carburized components where the martensite start temperature of the high carbon case is lower than that of the low carbon base metal.