Employing Cold Spray to Alter the Residual Stress Distribution of Workpieces: A Case Study on Fusion-Welded AA2219 Joints

Abstract

Eight-millimeter-thick high-strength AA2219 aluminum alloy plates were successfully welded by variable polarity tungsten inert gas welding, and the effect of cold spraying (CS) process on the residual stress, microstructure and mechanical properties of the welded joints were investigated. The CS experiment was performed under different driving gas temperatures and pressures and with different powder types. The microstructure of the as-welded joints and as-coated joints was analyzed by optical microscopy and scanning electron microscopy, while the residual stress was measured by the blind-hole-drilling method. The results show that the CS process could significantly modify and re-distribute the residual stress of the welded joints. The maximum residual tensile stress of the welded joint has been significantly reduced by 111.9% and distributed more uniformly compared to the as-welded state. With the increase in gas temperature and/or pressure in CS, the residual tensile stress of joints was reduced, and even some compressive stress appeared. Among the powders of Cu, Al, and Ni, Cu particles have the best-improving effect. In addition, double-sided spraying treatment can reduce the residual tensile stress of both sides more effectively. Besides, the grain size of the joint surface after CS was refined.

Appendix

The residual stress is calculated by the following formulas:

$$\sigma_{x} = \frac{{\sigma_{1} + \sigma_{2} }}{2} + \frac{{\sigma_{1} - \sigma_{2} }}{2}\cos 2\gamma ,\quad \sigma_{y} = \frac{{\sigma_{1} + \sigma_{2} }}{2} - \frac{{\sigma_{1} - \sigma_{2} }}{2}\cos 2\gamma$$

(1)

$$\sigma_{{1}} { = }\frac{{E\left( {\varepsilon_{{0^{ \circ } }} + \varepsilon_{{90^{ \circ } }} } \right)}}{4A} + \frac{E}{4B}\sqrt {\left( {\varepsilon_{{0^{ \circ } }} + \varepsilon_{{90^{ \circ } }} } \right)^{2} + \left( {2\varepsilon_{{45^{ \circ } }} - \varepsilon_{{0^{ \circ } }} - \varepsilon_{{90^{ \circ } }} } \right)^{2} }$$

(2)

$$\sigma_{2} { = }\frac{{E\left( {\varepsilon_{{0^{ \circ } }} + \varepsilon_{{90^{ \circ } }} } \right)}}{4A} - \frac{E}{4B}\sqrt {\left( {\varepsilon_{{0^{ \circ } }} + \varepsilon_{{90^{ \circ } }} } \right)^{2} + \left( {2\varepsilon_{{45^{ \circ } }} - \varepsilon_{{0^{ \circ } }} - \varepsilon_{{90^{ \circ } }} } \right)^{2} }$$

(3)

$$\gamma = \frac{1}{2}\arctan \frac{{\varepsilon_{{0^{ \circ } }} - 2\varepsilon_{{45^{ \circ } }} + \varepsilon_{{90^{ \circ } }} }}{{\varepsilon_{{0^{ \circ } }} - \varepsilon_{{90^{ \circ } }} }}$$

(4)

$$A = - \frac{1 + \mu }{2}\frac{{r^{2} }}{{l_{1} l_{2} }},{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} B = - \frac{{8r^{2} }}{{l_{1} l_{2} }}\left[ {1 - \frac{1 + \mu }{4}\frac{{\left( {l_{1}^{2} + l_{1} l_{2} + l_{2}^{2} } \right)r^{2} }}{{l_{1} l_{2} }}} \right]$$

(5)

where $\sigma_{x}$ and $\sigma_{y}$ denote the transverse and longitudinal stresses, respectively, $\sigma_{1}$ and $\sigma_{2}$ are the first and second principal stresses, $\varepsilon$ represents the strain with the subscripts 0°, 45° and 90° standing for the angles of rosette strain-gauge, E is the elastic modulus, $\gamma$ is the angle between transverse direction and $\sigma_{1}$, finally, A and B are the release factors calculated by Eq. 5, where $\mu$ is the Poisson’s ratio, $l_{1}$ and $l_{2}$ are the distances from both ends of rosette strain-gauges to hole centers, and $r$ is the hole radius.