A study of the residual stress induced by shot peening for an isotropic material based on Prager’s yield criterion for combined stresses

Abstract

Prager developed a yield criterion for a generalized isotropic material that is more capable of predicting material behavior that occurs from combined loading. A less studied topic is the effect of combined loading on the residual stress state. The purpose of the present work is to develop a theoretical model for a residual stress based on Prager’s relation and establish how the predictions are influenced. Prager’s yield criterion which includes the third deviatoric invariant, J3, provides a way to estimate the influence J3 has on the residual stress. The analyzed residual stress originates from shot peening because much experimental data exists to compare and validate the theoretical results.

Appendix: Equivalent elastic–plastic stresses

Starting from Eq. 8 and the Von Mises yield function

$$\begin{aligned} de_{ij}^p=\bar{G}\partial {f}\frac{\partial {f}}{\partial {\sigma _{ij}}}=\bar{G}s_{ij}^pdJ_2 \end{aligned}$$

(35)

We follow the same procedure outlined in textbook [36] to obtain an expression of $\bar{G}$ in terms of $s_{ij}^p$. Begin by taking the product of Eq. 35 with itself

$$\begin{aligned} de_{ij}^pde_{ij}^p=\bar{G}^22J_2dJ_2^2 \end{aligned}$$

(36)

By taking the root of both sides we have

$$\begin{aligned} d\epsilon _i^p=\frac{2}{3}\bar{G}\sigma ^{p}_idJ_2= \frac{4}{9}\bar{G}\sigma ^{p2}_id\sigma _i^p \end{aligned}$$

(37)

Solving for $\bar{G}$ yields

$$\begin{aligned} \bar{G}=\frac{9}{4}\frac{1}{H_p\sigma _i^{p2}}=\frac{3}{4H_pJ_2} \end{aligned}$$

(38)

Upon substituting this expression into Eq. 35 gives

$$\begin{aligned} de_{ij}^p=s_{ij}^p\bar{G}dJ_2=s_{ij}^p\frac{3a(2n+1)}{4b} \left( \frac{3}{b^2}\right) ^nJ_2^{n-1}dJ_2 \end{aligned}$$

(39)

where an $H_p=\frac{b}{a(2n+1)}\left( \frac{b^2}{3J_2}\right) ^n$ has been used. Find the first principal plastic strain of Eq. 39 and write the equation in terms of the stress components with $J_2=\frac{1}{2}s_{ij}^ps_{ij}^p=3s_{11}^{p2}$ and $dJ_2=s_{ij}^pds_{ij}^p=6s_{11}^pds_{11}^p$

$$\begin{aligned} de_{11}^p&= \frac{3a(2n+1)}{4b}\left( \frac{3}{b^2}\right) ^n(3s_{11}^{p2})^{n-1}6s_{11}^{p2}ds_{11}^p=\nonumber \\&\frac{6a(2n+1)}{4b}\left( \frac{3}{b^2}\right) ^n(3s_{11}^{p})^{2n}ds_{11}^p \end{aligned}$$

(40)

Integrating this result gives $e_{11}^p$

$$\begin{aligned} e_{11}^p=\frac{6a}{4b}\left( \frac{3}{b}\right) ^{2n}(s_{11}^{p})^{2n+1} \end{aligned}$$

(41)

By making use of Eq. 14, we have the following

$$\begin{aligned} \frac{1}{3}(1+\nu )\epsilon _i^p=\frac{1}{3}(1+\nu ) \left[ a\left( \frac{\sigma _i^p}{b}\right) ^{2n+1}\right] =\frac{6a}{4b} \left( \frac{3}{b}\right) ^{2n}(s_{11}^{p})^{2n+1} \end{aligned}$$

(42)

Now, by solving for $s_{11}^p$ in terms of $\sigma _i^p$, the desired result is obtained

$$\begin{aligned} s_{11}^p=\left[ \frac{4b}{6a}\left( \frac{b}{3} \right) ^{2n}\frac{1}{3}(1+\nu )a \left( \frac{\sigma _i^p}{b}\right) ^{2n+1} \right] ^{\frac{1}{2n+1}}=\frac{1}{3}\sigma _i^p \end{aligned}$$

(43)