Measurement of Residual Stresses in Nuclear-grade Zircaloy-4(R) Tubes—Effect of Heat Treatment

Abstract

Nuclear-grade Zircaloy-4(R) tubes are produced by a unique manufacturing process known as pilgering, which leaves the material in a work-hardened state containing a pattern of residual stresses. Moreover, such tubes exhibit elastic anisotropy as a result of the pilgering process. Therefore, standard equations originally proposed by Sachs (Z Met Kd, 19: 352–357, 1927; Sachs, Espey, Iron Age, 148: 63–71, 1941). for isotropic materials do not apply in this situation. Voyiadjis et al. (Exp Mech, 25: 145–147, 1985) proposed a set of equations for treating elastically anisotropic materials, but we have determined that there are discrepancies in their equations. In this paper, we present the derivation for a set of new equations for treating elastically anisotropic materials, and the application of these equations to residual stress measurements in Zr-4(R) tubes. To this end, through thickness distribution of residual stress components in as-received and heat treated (500°C) Zr-4(R) tubes was measured employing the Sachs’ boring-out technique in conjunction with electrochemical machining as the means of material removal, and our new equations. For both as-received and the heat treated materials, the axial and tangential residual stresses were significantly higher than the radial and shear residual stresses. The largest residual stress was the tangential stress component in the as-received material, showing a tensile value at the outer surface and a compressive value at the inner surface. At high values of von Mises equivalent stress, the principal directions of residual stress coincided with the principal axes of the tube for the as-received material, as well as for the material heat treated at 500°C.

Appendices

Appendix A

Voyiadjis et al. [3] presented a set of equations for calculation of residual stresses in materials exhibiting cylindrical anisotropy. Their equations were based on utilization of experimental data obtained from Sachs’ boring-out technique. We have determined that for the case of material removal from the outer surface, their equations contain some discrepancies (both in terms and sign), which could have a significant effect on the results. Below, we present the derivation of equations for residual stresses in cylindrically anisotropic materials based on Sachs’ boring-out technique, leading to equations (1.1) through (1.4).

Material Removal From the Outside

A cylinder subjected to material removal from either the inner and outer surfaces can be thought of as a cylinder subjected to internal or external pressures, P _i and P ₀, respectively. In such a case, the radial and tangential stresses in the cylinder are given by [3, 9]

$$ \sigma _{r} = {\left[ {\frac{{P_{i} C^{{k + 1}}_{0} - P_{0} }} {{{\left( {1 - C^{{2k}}_{0} } \right)}}}} \right]}{\left( {\frac{r} {b}} \right)}^{{k - 1}} - {\left[ {\frac{{P_{i} - P_{0} C^{{k - 1}}_{0} }} {{{\left( {1 - C^{{2k}}_{0} } \right)}}}} \right]}C^{{k + 1}}_{0} {\left( {\frac{b} {r}} \right)}^{{k + 1}} $$

(10.1)

$$ \sigma _{\theta } = {\left[ {\frac{{P_{i} C^{{k + 1}}_{0} - P_{0} }} {{{\left( {1 - C^{{2k}}_{0} } \right)}}}} \right]}k{\left( {\frac{r} {b}} \right)}^{{k - 1}} + {\left[ {\frac{{P_{i} - P_{0} C^{{k - 1}}_{0} }} {{{\left( {1 - C^{{2k}}_{0} } \right)}}}} \right]}kC^{{k + 1}}_{0} {\left( {\frac{b} {r}} \right)}^{{k + 1}} $$

(10.2)

where $C_{0} = {\left( {\frac{a} {b}} \right)}$ and $k = {\sqrt {\frac{{E_{\theta } }} {{E_{r} }}} }$

Radial Stress

Knowing that the radial pressure at the inner surface vanishes (P _i  = 0) for the case of material removal from the outer surface, and using $U = a \mathord{\left/ {\vphantom {a r}} \right. \kern-\nulldelimiterspace} r$ and noting that the outer radius is reducing from b to r, the radial stress at the outer surface,σ _r , can be expressed as;

$$ \sigma _{r} = {\left( { - \frac{{P_{0} }} {{1 - U^{{2k}} }}} \right)}{\left( {\frac{r} {b}} \right)}^{{k - 1}} - {\left( { - \frac{{P_{0} U^{{k - 1}} }} {{1 - U^{{2k}} }}} \right)}U^{{k + 1}} {\left( {\frac{b} {r}} \right)}^{{k + 1}} $$

(11)

Since b = r for material removal from the outer surface, we obtain

$$ \sigma _{r} {\left( U \right)} = - P_{0} $$

(12)

The tangential stress at the inner surface (r = a) can be expressed as a function of the radial stress at the outer surface (b = r), by replacing (r / b) in equation (10.2) with (a / r) and noting that P _i  = 0 and $U = a \mathord{\left/ {\vphantom {a r}} \right. \kern-\nulldelimiterspace} r$, one can write

$$ \sigma _{\theta } = {\left( { - \frac{{P_{0} U^{{k - 1}} k}} {{1 - U^{{2k}} }}} \right)} + {\left( { - \frac{{P_{0} U^{{k - 1}} k}} {{1 - U^{{2k}} }}} \right)}U^{{k + 1}} \frac{1} {{U^{{k + 1}} }} $$

(13)

The tangential stress at the inner surface due to a radial stress at the outer surface can be obtained as;

$$ \sigma _{\theta } = - \frac{{2P_{0} U^{{k - 1}} k}} {{1 - U^{{2k}} }} $$

(14)

Before the material is removed, σ _r (U) = −P ₀ at the outer surface according to equation (12), after the material is removed, the internal pressure required to cancel the tangential surface strain is equal to radial stress at the original surface (σ _r (U) = P ₀). The tangential stress at the inner surface as a function of radial stress at the outer surface is therefore given by:

$$ \sigma _{\theta } = - \frac{{2kU^{{k - 1}} }} {{1 - U^{{2k}} }}\sigma _{r} {\left( U \right)} $$

(15)

The tangential stress can also be expressed in terms of the strains measured at the inner surface as

$$ \sigma _{\theta } = E^{\prime }_{\theta } {\left( {\varepsilon _{\theta } + \nu _{{z\theta }} \varepsilon _{z} } \right)} = E^{\prime }_{\theta } \theta _{i} $$

(16)

Combining equations (15) and (16) yields the radial stress,

$$ \sigma _{r} {\left( U \right)} = - E^{\prime }_{\theta } {\left( {\frac{{1 - U^{{2k}} }} {{2kU^{{k - 1}} }}} \right)}\theta _{i} $$

(17)

The requirement of equilibrium in the radial direction is governed by the tangential stress equation as a function of the radial position through the tube wall as follows:

Tangential Stress

$$ \sigma _{\theta } = \frac{\partial } {{\partial r}}r\sigma _{r} $$

(18)

To obtain the tangential stress as a function of U, equation (18) first needs to be written in terms of U by noting that $U = a \mathord{\left/ {\vphantom {a r}} \right. \kern-\nulldelimiterspace} r$;

$$ \partial U = - \frac{a} {{r^{2} }}\partial r $$

(19.1)

Thus, from equation (18),

$$ \partial r = - \frac{{r^{2} }} {a}\partial U $$

(19.2)

$$ \sigma _{\theta } = - \frac{\partial } {{\partial U}}U\sigma _{r} $$

(19.3)

Inserting equation (17) into equation (19.3) gives the following expressions which need to be differentiated with respect to U,

$$ \sigma _{\theta } {\left( U \right)} = - \frac{\partial } {{\partial U}}U{\left\{ { - E^{\prime }_{\theta } {\left( {\frac{{1 - U^{{2k}} }} {{2kU^{{k - 1}} }}} \right)}\theta _{i} } \right\}} $$

(20.1)

$$ \sigma _{\theta } {\left( U \right)} = - E^{\prime }_{\theta } {\left[ {\frac{{{\text{d}}\theta _{i} }} {{{\text{d}}U}}{\left( {\frac{{U^{{2k}} - 1}} {{2kU^{{k - 2}} }}} \right)} + \theta _{i} \frac{{\text{d}}} {{{\text{d}}U}}{\left( {\frac{{U^{{2k}} - 1}} {{2kU^{{k - 2}} }}} \right)}} \right]} $$

(20.2)

The term ${\left( {\frac{{U^{{2k}} - 1}} {{2kU^{{k - 2}} }}} \right)}$ can be differentiated with respect to U as follows:

$$ \frac{{\text{d}}} {{{\text{d}}U}}{\left( {\frac{{U^{{2k}} - 1}} {{2kU^{{k - 2}} }}} \right)} = \frac{{{\left( {2kU^{{2k - 1}} } \right)}{\left( {2kU^{{k - 2}} } \right)} - {\left[ {{\left( {k - 2} \right)}{\left( {2kU^{{k - 3}} } \right)}{\left( {U^{{2k}} - 1} \right)}} \right]}}} {{4k^{2} U^{{2k - 4}} }} $$

(21.1)

$$ \frac{{\text{d}}} {{{\text{d}}U}}{\left( {\frac{{U^{{2k}} - 1}} {{2kU^{{k - 2}} }}} \right)} = \frac{{{\left( {k + 2} \right)}U^{{2k}} + k - 2}} {{2kU^{{k - 1}} }} $$

(21.2)

Inserting the derived term given in equation (21.2) into equation (20.2), the tangential stress in its final form is,

$$ \sigma _{\theta } {\left( U \right)} = E^{\prime }_{\theta } U^{2} {\left\{ {{\left( {\frac{{1 - U^{{2k}} }} {{2kU^{k} }}} \right)}{\left( {\frac{{{\text{d}}\theta _{i} }} {{{\text{d}}U}}} \right)} + \theta _{i} {\left[ {\frac{{2 - k - {\left( {k + 2} \right)}U^{{2k}} }} {{2kU^{{k + 1}} }}} \right]}} \right\}} $$

(22)

Axial Stress

The axial stress can simply be written as:

$$ \sigma _{z} = \frac{F} {{\pi r^{2} {\left( {1 - U^{2} } \right)}}} $$

(23)

Considering the basic elasticity equations and the distribution of axial stress, equation (23) becomes,

$$ \sigma _{z} = \frac{{{\int\limits_b^r {2\pi \sigma _{z} {\left( \rho \right)}\rho {\text{d}}\rho } }}} {{\pi r^{2} {\left( {1 - U^{2} } \right)}}} = E^{\prime }_{z} \phi _{i} $$

(23.1)

$${\int\limits_b^r {2\pi \sigma _{z} {\left( \rho \right)}\rho {\text{d}}\rho = E^{'}_{z} \phi _{i} \pi r^{2} {\left( {1 - U^{2} } \right)}} }$$

(23.2)

Knowing $U = \frac{a} {r}$, and differentiating equation (23.2) with respect to r the axial stress can be obtained in terms of the radial position as follows:

$$ - 2\pi \sigma _{z} {\left( r \right)}r = E^{\prime}_{z} \pi \frac{{\text{d}}} {{{\text{d}}r}}{\left[ {\phi _{i} r^{2} {\left( {1 - \frac{{a^{2} }} {{r^{2} }}} \right)}} \right]} $$

(24.1)

Note that equation (23.2) shows an integral from b to r, which explains the reason for the negative sign (−) in equation (24.1), which may be rewritten as,

$$ - 2\sigma _{z} {\left( r \right)}r = E^{\prime }_{z} \frac{{\text{d}}} {{{\text{d}}r}}{\left[ {{\left( {\phi _{i} r^{2} - \phi _{i} a^{2} } \right)}} \right]} $$

(24.2)

$$ \sigma _{z} {\left( r \right)} = - E^{\prime }_{z} {\left[ {\frac{{{\text{d}}\phi _{i} }} {{{\text{d}}r}}{\left( {\frac{{r^{2} - a^{2} }} {{2r}}} \right)} + \phi _{i} } \right]} $$

(24.3)

Noting that $U = \frac{a} {r}$, then ${\text{d}}U = - \frac{a} {{r^{2} }}{\text{d}}r$, and ${\text{d}}r = - \frac{{r^{2} }} {a}{\text{d}}U$, which after substitution in above equation yields,

$$ \sigma _{z} {\left( U \right)} = - E^{\prime }_{z} {\left[ { - \frac{{{\text{d}}\phi _{i} }} {{{\text{d}}U}}{\left( {\frac{a} {{r^{2} }}} \right)}{\left( {\frac{{r^{2} - a^{2} }} {{2r}}} \right)} + \phi _{i} } \right]} $$

(25)

or

$$ = - E^{\prime }_{z} {\left[ { - \frac{{{\text{d}}\phi _{i} }} {{{\text{d}}U}}{\left( {\frac{U} {2} - \frac{{U^{3} }} {2}} \right)} + \phi _{i} } \right]} $$

Yielding the equation for axial stress in its final form as,

$$ \sigma _{z} {\left( U \right)} = - E^{\prime }_{z} {\left[ { - {\left( {\frac{{1 - U^{2} }} {2}} \right)}U{\left( {\frac{{{\text{d}}\phi _{i} }} {{{\text{d}}U}}} \right)} + \phi _{i} } \right]} $$

(26)

Shear Stress

The shear stress due to a torque moment can be expressed as:

$$ \tau _{{z\theta }} = \frac{{2M_{z} a}} {{\pi r^{4} {\left( {1 - U^{4} } \right)}}} $$

(27)

Using equation (27), shear strain is given as follows.

$$ \gamma _{{z\theta }} {\left( U \right)} = \frac{{2M_{z} a}} {{\pi r^{4} {\left( {1 - U^{4} } \right)}G_{{z\theta }} }} $$

(28)

The moment can be expressed in terms of shear stress as a function of radial position.

$$ M_{z} = {\int\limits_b^r {2\pi \tau _{{z\theta }} {\left( \rho \right)}\rho ^{2} {\text{d}}\rho } } $$

(29)

Inserting equation (29) into equation (28) gives the relationship between the shear strain, the shear stress, and the shear modulus as follows:

$$ \gamma _{{z\theta }} {\left( U \right)} = \frac{{4\pi a{\int\limits_b^r {\tau _{{z\theta }} {\left( \rho \right)}\rho ^{2} {\text{d}}\rho } }}} {{\pi r^{4} {\left( {1 - U^{4} } \right)}G_{{z\theta }} }} $$

(30.1)

$$ 4\pi a{\int\limits_b^r {\tau _{{z\theta }} {\left( \rho \right)}\rho ^{2} {\text{d}}\rho = \pi G_{{z\theta }} \gamma _{{z\theta }} r^{4} {\left( {1 - U^{4} } \right)}} } $$

(30.2)

Knowing that $U = \frac{a} {r}$, and differentiating equation (30.2) with respect to r, the shear stress can now be evaluated as follows. Note that equation (30.2) shows an integral from b to r, which explains the reason for the negative sign (−) in following equation.

$$ - 4a\tau _{{z\theta }} {\left( r \right)}r^{2} = G_{{z\theta }} \frac{{\text{d}}} {{{\text{d}}r}}{\left[ {\gamma _{{z\theta }} r^{4} {\left( {1 - \frac{{a^{4} }} {{r^{4} }}} \right)}} \right]} $$

(31)

$$ \tau _{{z\theta }} {\left( r \right)} = - G_{{z\theta }} {\left[ {\frac{{{\text{d}}\gamma _{{z\theta }} }} {{{\text{d}}r}}{\left( {\frac{{r^{4} - a^{4} }} {{4ar^{2} }}} \right)} + \frac{r} {a}\gamma _{{z\theta }} } \right]} $$

(32)

Noting that $U = \frac{a} {r}$, then ${\text{d}}U = - \frac{a} {{r^{2} }}{\text{d}}r$, and ${\text{d}}r = - \frac{{r^{2} }} {a}{\text{d}}U$, which after substitution in above equation yields,

$$ \tau _{{z\theta }} {\left( U \right)} = - G_{{z\theta }} {\left[ { - \frac{{{\text{d}}\gamma _{{z\theta }} }} {{{\text{d}}U}}{\left( {\frac{{r^{4} - a^{4} }} {{4r^{4} }}} \right)} + \frac{r} {a}\gamma _{{z\theta }} } \right]} $$

(33)

yielding the shear stress equation in its final form as,

$$ \tau _{{z\theta }} {\left( U \right)} = - G_{{z\theta }} {\left[ { - {\left( {\frac{{1 - U^{4} }} {4}} \right)}{\left( {\frac{{{\text{d}}\gamma _{{z\theta }} }} {{{\text{d}}U}}} \right)} + \frac{1} {U}\gamma _{{z\theta }} } \right]} $$

(34)

Appendix B

Strain parameters ϕ, θ, and the shear strain $\gamma _{{z\theta }}$ and their derivatives with respect to U are used in equations (1.1)–(1.4) to evaluate the residual stresses. To calculate the derivatives of strain parameters with respect to U, the variation of the strain parameters as a function of U was first determined by fitting a fourth order polynomial curve through the data points. After the equation of the curve was determined, the derivatives of these curves were used in equations (1.1)–(1.4) to calculate the residual stresses. Table 2 shows the polynomial coefficients and R ² coefficients for the fitted curves.

Table 2 Polynomial coefficients and R ² coefficients of the fourth order polynomial curves for as-received and annealed specimens