Hot extrusion of nanocrystalline yttria-stabilized tetragonal zirconia polycrystals


Hot indirect extrusion behaviors of fully dense 3 mol% Y2O3-stabilized ZrO2 polycrystals with a grain size of 80 nm were investigated by using a conical graphite die. During early extrusion at 1843 K and a constant compression stress of 60–80 MPa, the relative moving billet velocity (v0) shows a sharp maximum, followed by a continuous decrease until it reaches a steady-state v0 value. Visual observations of the material flow by using graphite foil markers indicate that the deformation zone at the steady-state v0 is well represented by a spherical velocity field. The decrease in v0 corresponds to the transition from heterogeneous to homogeneous deformation. The stress exponent (n) of 2.07 was obtained from the log–log plots of the steady-state v0 versus the external applied stress that is compensated by threshold and sliding friction stresses. This n value suggests plastic deformation through grain-boundary sliding in indirect extrusion.

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FIG. 1
FIG. 2
FIG. 3
FIG. 4
FIG. 5
FIG. 6


A :

Dimensionless constant …

A 1 :

Coefficient \({A_1} = A{{Gb} \over k}{\rm{\cdot}}{{{d^p}} \over {{G^n}}}{D^*}\) from Eq. (E3) [mp K/(Pan s)]

B :

Coefficient \(B = {{{\sigma _{{\rm{th}}}}d} \over {\exp \left( { - {Q_{{\rm{th}}}}/RT} \right)}}\) from Eq. (3a) [Pa m]

b :

Burgers vector [m]

C :

Coefficient \(C = {{{{\dot \varepsilon }_{\rm{e}}}} \over {\sigma _{\rm{e}}^n}}\) from Eq. (A5) and \(C = {{{A_1}} \over {T{d^p}}}\exp \left( { - {Q \over {RT}}} \right)\) from Eq. (E5) [1/(Pan s)]

D :

Diameter [m]


Pre-exponential factor for diffusion, frequency factor [m2/s]

D :

Grain size [m]

G :

Shear modulus [Pa]

K :

Dimensionless constant from Eq. (D3)

k :

Boltzmann constant k = 1.3806 × 10−23 [J/K]

L 0 :

Length, instantaneous length of pre-extruded cylindrical portion of billet [Fig. 1(a)] [m]

L f :

Length, instantaneous length of extruded rod [Fig. 1(a)] [m]

n :

Stress exponent …

p :

Grain-size exponent …

Q :

Activation energy for plastic deformation [J/mol]

Q th :

Activation energy for threshold stress [J/mol]

r, θ, φ:

Axes of spherical coordinate system …

r :

Radius, instantaneous radius [m]

R :

Molar gas constant R = 8.3145 [J/(K mol)]


Surface of velocity discontinuities [m2]


Conical surface [m2]


Cylindrical surface [m2]

T :

Absolute temperature [K]

U :

Displacement, relative billet displacement [Fig. 1(a)] [m]

U ch :

Cross-head displacement [m]

U die :

Die displacement obtained by subtracting thermal expansion of die and rams from \({U_{{\rm{ch}}}}\) [Fig. 1(a)] [m]

\(\dot U\) :

Velocity [m/s]

V :

Volume [m3]

v :

Velocity, relative moving velocity of billet [m/s]

w :

Work per unit volume [J/m]

\(\dot w\) :

Power (work per unit time) per unit volume [J/(m3 s)]

W :

Work [J]

\(\dot W\) :

Power (work per unit time) [J/s]


Strain …

\(\dot \varepsilon \) :

Strain-rate [1/s]

\({\dot \varepsilon _{\rm{e}}}\) :

Equivalent strain rate [1/s]


Angle [Degree]

\({\mu _{{\rm{sf}}}}\) :

Coefficient of sliding friction [Pa]


Relative density …


Initial relative density …

\(\dot \rho \) :

Densification rate, \(\dot \rho = {{{\rm{d}}\rho } \over {{\rm{d}}t}}\) [1/s]


Stress [Pa]

\({\sigma _{\rm{e}}}\) :

Equivalent stress [Pa]


Efficient stress [Pa]


External stress required to overcome an interfacial sliding friction stress by Eqs. (D5a) and (D5b) [Pa]


External stress required to overcome an inhibiting factor (Y segregation) of grain-boundary sliding by Eq. (3a) [Pa]


Interfacial sliding friction stress [Pa]

\(\emptyset \) :

Die angle [Degree]


Pre-extruded region of billet …


Extruded region of billet …


Equivalent …


Efficient …


Sliding friction …


Threshold …

ij :

Tensor component …


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The author would like to acknowledge the experimental support of Mr Ryuta Nakano and Ms Kaori Nakase, National Defense Academy.

Author information



Corresponding author

Correspondence to Tatsuo Kumagai.



A. Internal power of deformation

When materials flow toward the apex in the deformation zone between SI and SII (spherical velocity field10) in Fig. A, the relative moving velocity of the billets is expressed as follows:

$${\dot U_{\rm{r}}} = v = - {v_{\rm{f}}}\cos \,\theta {\left( {{{{r_{\rm{f}}}} \over r}} \right)^2},\>\>{\dot U_\theta } = {\dot U_\varphi } = 0\quad .$$

Schematic of spherical velocity field proposed by Avitzur.10

From Eq. (A1), the strain-rates as a function of velocity components are

$${\dot \varepsilon _{rr}} = {{\partial {{\dot U}_r}} \over {\partial r}} = 2{v_{\rm{f}}}r_{\rm{f}}^2{{\cos \theta } \over {{r^3}}}\quad ,$$
$${\dot \varepsilon _{\theta \theta }} = {{{{\dot U}_r}} \over r} + {{\partial {{\dot U}_\theta }} \over {r\partial \theta }} = - {v_{\rm{f}}}r_{\rm{f}}^2{{\cos \theta } \over {{r^3}}}\quad ,$$
$${\dot \varepsilon _{\varphi \varphi }} = {{{{\dot U}_r}} \over r} + {{\partial {{\dot U}_\varphi }} \over {r\partial \theta }} = - {v_{\rm{f}}}r_{\rm{f}}^2{{\cos \theta } \over {{r^3}}}\quad ,$$
$${\dot \varepsilon _{rr}} = - 2{\dot \varepsilon _{\theta \theta }} = - 2{\dot \varepsilon _{\varphi \varphi }}\quad ,$$
$${\dot \varepsilon _{r\theta }} = {1 \over 2}{\rm{\cdot}}{1 \over r}{\rm{\cdot}}{{\partial {{\dot U}_r}} \over {\partial \theta }} = {1 \over 2}{v_{\rm{f}}}r_{\rm{f}}^2{{\sin \,\theta } \over {{r^3}}}\quad ,$$
$${\dot \varepsilon _{\theta \varphi }} = {\dot \varepsilon _{\varphi r}} = 0\quad .$$

The equivalent strain-rate is defined as follows:

$$\eqalign{ & \dot \varepsilon _{\rm{e}}^2 = \>{2 \over 9}\left[ {{{\left( {{{\dot \varepsilon }_{rr}} - {{\dot \varepsilon }_{\theta \theta }}} \right)}^2} + {{\left( {{{\dot \varepsilon }_{\theta \theta }} - {{\dot \varepsilon }_{\varphi \varphi }}} \right)}^2} + {{\left( {{{\dot \varepsilon }_{\varphi \varphi }} - {{\dot \varepsilon }_{rr}}} \right)}^2}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\, + {4 \over 3}\left( {\dot \varepsilon _{r\theta }^2 + \dot \varepsilon _{\theta \varphi }^2 + \dot \varepsilon _{\varphi r}^2} \right) \cr} $$
$$ = \dot \varepsilon _{rr}^2 + {4 \over 3}\dot \varepsilon _{r\theta }^2\quad .$$

Because of volume constancy (\({v_0}r_0^2 = {v_{\rm{f}}}r_{\rm{f}}^2\)), the equivalent strain rate (Eq. A3) with Eqs. (A2d) and (A2e) can be rewritten as follows:

$${\dot \varepsilon _{\rm{e}}} = 2{v_{\rm{f}}}r_{\rm{f}}^2{1 \over {{r^3}}}\sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } = 2{v_0}r_0^2{1 \over {{r^3}}}\sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \quad .$$

The constitutive equation for steady-state plastic deformation at high temperature is represented as follows3:

$${\dot \varepsilon _{\rm{e}}} = C\sigma _{\rm{e}}^n\quad .$$

Because \({{{{\dot \varepsilon }_{ij}}} \over {{\sigma _{ij}}}} = {3 \over 2}{{{{\dot \varepsilon }_{\rm{e}}}} \over {{\sigma _{\rm{e}}}}}\), the stress component is rewritten by using Eq. (A5) as follows:

$${\sigma _{ij}} = {2 \over 3}{\left( {{{{{\dot \varepsilon }_{\rm{e}}}} \over C}} \right)^{{1 \over n}}}{{{{\dot \varepsilon }_{ij}}} \over {{{\dot \varepsilon }_{\rm{e}}}}}\quad .$$

Thus, by using Eq. (A6), the internal power per unit volume within the deformation zone can be represented as follows:

$${\dot w_1} = {\sigma _{ij}}{\dot \varepsilon _{ij}} = {1 \over {\root n \of C }}{\left( {{{\dot \varepsilon }_{\rm{e}}}} \right)^{{{n + 1} \over n}}}\quad .$$

After integration of Eq. (A7),

$${\dot W_1} = \int_V {{\sigma _{ij}}{{\dot \varepsilon }_{ij}}{\rm{d}}V} = {1 \over {\root n \of C }}\int_V {{{\left( {{{\dot \varepsilon }_{\rm{e}}}} \right)}^{{{n + 1} \over n}}}\,{\rm{d}}V} \quad ,$$

where \({\rm{d}}V = 2\pi r\sin \,\theta r\,{\rm{d}}\theta \,{\rm{d}}r\). By using Eq. (A4), Eq. (A8a) is rewritten as follows:

$${\dot W_1} = {1 \over {\root n \of C }}\int_0^\theta {{{\int_{{r_f}}^{{r_0}} {\left[ {{{2{v_0}r_0^2} \over {{r^3}}}\sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } } \right]} }^{{{n + 1} \over n}}}2\pi r\sin \,\theta r\,{\rm{d}}\theta \,{\rm{d}}r} $$
$$ = 2\pi {{{{\left( {2{v_0}r_0^2} \right)}^{{{n + 1} \over n}}}} \over {\root n \of C }}\int_0^\phi {\left[ {{{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{{{n + 1} \over {2n}}}}\sin \theta \int_{{r_{\rm{f}}}}^{{r_0}} {{{\left( {{1 \over {{r^3}}}} \right)}^{{{n + 1} \over n}}}{r^2}\,{\rm{d}}r} } \right]} \,{\rm{d}}\theta $$

Because \(\int_{{r_{\rm{f}}}}^{{r_0}} {{{\left( {{1 \over {{r^3}}}} \right)}^{{{n + 1} \over n}}}{r^2}\,{\rm{d}}r = {n \over 3}{{\left( {{1 \over {{r_0}}}} \right)}^{{3 \over n}}}\left[ {{{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)}^{{3 \over n}}} - 1} \right]} \), and r0 sin ϕ = D0/2, Eq. (A8b) is rewritten as follows:

$$\eqalign{ & {{\dot W}_1} = \>{{n\pi } \over 3}{v_0}D_0^2{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}\left[ {{{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)}^{{3 \over n}}} - 1} \right] \cr & \times \,\,{\sin ^{{1 \over n} - 2}}\phi \int_0^\phi {{{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{{{n + 1} \over {2n}}}}\sin \theta \>{\rm{d}}\theta } \quad . \cr} $$

An approximation of \(1 - {{11} \over {12}}{\sin ^2}\,\theta \simeq 1 - {\sin ^2}\,\theta \) leads to

$$\eqalign{ & {{\dot W}_1} = {{n\pi } \over 3}{v_0}\,D_0^2{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}\left[ {{{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)}^{{3 \over n}}} - 1} \right] \cr & \,\,\,\,\,\,\,\,\,\,\, \times {\sin ^{{1 \over n} - 2}}\phi \left[ {{{1 - {{\cos }^{2 + {1 \over n}}}\phi } \over {\left( {2 + 1/n} \right){{\sin }^{2 - {1 \over n}}}\phi }}} \right]\quad . \cr} $$

B. Power loss because of velocity discontinuities

The stress component along SI is represented by Eq. (A6) as follows:

$${\sigma _{r\theta }} = {2 \over 3}{\left( {{{{{\dot \varepsilon }_{\rm{e}}}} \over C}} \right)^{{1 \over n}}}{{{{\dot \varepsilon }_{r\theta }}} \over {{{\dot \varepsilon }_{\rm{e}}}}}\quad .$$

The strain-rate component along SI is represented by Eq. (A2e) as follows:

$${\dot \varepsilon _{r\theta }}\left( {{S_{\rm{I}}}} \right) = {1 \over 2}{v_0}r_0^2{{\sin \theta } \over {r_0^3}} = {{{v_0}} \over {{D_0}}}\sin \phi \sin \theta \quad .$$

The equivalent strain-rate component along SI is represented by Eq. (A4) as follows:

$$\eqalign{ & {{\dot \varepsilon }_{\rm{e}}}\left( {{S_{\rm{I}}}} \right) = 2{v_0}{r_0}{1 \over {r_0^3}}\sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4{{{v_0}} \over {{D_0}}}\sin \phi \sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \quad . \cr} $$

Modification of Eq. (B3) leads to

$${v_0} = {{{D_0}} \over {4\sin \phi }}{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)^{ - {1 \over 2}}}{\dot \varepsilon _{\rm{e}}}\quad .$$

Because \({\rm{d}}S = 2\pi {r^2}\sin \theta \,{\rm{d}}\theta \), power loss of velocity discontinuity along SI is obtained through Eqs. (B1), (B2), (B3), and (B4) as follows:

$${\dot W_2} = \int_{{S_{\rm{I}}}} {\left| {\Delta {v_{{S_{\rm{I}}}}}} \right|{\sigma _{r\theta }}{\rm{d}}S} = \int_{{S_{\rm{I}}}} {\left| {{v_0}\sin \theta } \right|{\sigma _{r\theta }}{\rm{d}}S} $$
$$\eqalign{ & = - \int_0^\phi {\left[ {{{{D_0}} \over {4\,\sin \phi }}{{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{ - {1 \over 2}}}} \right.} \cr & \,\,\,\,\,\,\, \times \,\,\sin \theta {2 \over 3}{\left( {4{{{v_0}} \over {C{D_0}}}\sin \phi \sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } } \right)^{{1 \over n}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \left( {{{{v_0}} \over {{D_0}}}\sin \phi \sin \theta } \right)2\pi r_0^2\sin \theta \,{\rm{d}}\theta \cr} $$
$$ = - {{\pi {v_0}D_0^2} \over {12}}{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}{\sin ^{{1 \over n} - 2}}\phi \int_0^\phi {{{\sin }^3}\,\theta {{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{{{1 - n} \over {2n}}}}{\rm{d}}\theta } \quad .$$

An approximation by \(1 - {{11} \over {12}}{\sin ^2}\,\theta \simeq 1 - {\sin ^2}\,\theta \) leads to

$${\dot W_2} = - {{n\pi {v_0}D_0^2} \over {12}}{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}{\sin ^{{1 \over n} - 2}}\,\phi \left( {{{{{\cos }^{{{2n + 1} \over n}}}\phi } \over {2n + 1}} - {{\cos }^{{1 \over n}}}\,\phi {{2n} \over {2n + 1}}} \right)\quad .$$

Power loss of velocity discontinuity along SII (\({\dot W_3}\)) is obtained in the same manner as \({\dot W_2}\). The strain-rate component along SII is represented by Eq. (A2e) as follows:

$${\dot \varepsilon _{r\theta }}\left( {{S_{{\rm{II}}}}} \right) = {1 \over 2}{v_{\rm{f}}}r_{\rm{f}}^2{{\sin \theta } \over {r_{\rm{f}}^3}} = {{{v_{\rm{f}}}} \over {{D_{\rm{f}}}}}\sin \phi \sin \theta \quad .$$

The equivalent strain-rate component along SII is represented by Eq. (A4) as follows:

$$\eqalign{ & {{\dot \varepsilon }_{\rm{e}}}\left( {{S_{{\rm{II}}}}} \right) = 2{v_{\rm{f}}}{r_{\rm{f}}}{1 \over {r_{\rm{f}}^3}}\sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4{{{v_{\rm{f}}}} \over {{D_{\rm{f}}}}}\sin \phi \sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \quad . \cr} $$

Modification of Eq. (B7) leads to

$${v_{\rm{f}}} = {{{D_{\rm{f}}}} \over {4\sin \phi }}{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)^{ - {1 \over 2}}}{\dot \varepsilon _{\rm{e}}}\quad .$$

Thus, power loss because of velocity discontinuity along SII is

$${\dot W_3} = \int_{{S_{{\rm{II}}}}} {\left| {\Delta {v_3}} \right|{\sigma _{r\theta }}{\rm{d}}S} = \int_{{S_{{\rm{II}}}}} {\left| {{v_{\rm{f}}}\sin \theta } \right|{\sigma _{r\theta }}{\rm{d}}S} $$
$$ = {{\pi {v_{\rm{f}}}D_{\rm{f}}^2} \over {{\rm{12}}}}{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}{\sin ^{{1 \over n} - 2}}\phi \int_0^\phi {{{\sin }^3}\,\theta {{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{{{1 - n} \over {2n}}}}{\rm{d}}\theta } $$

Because \({v_{\rm{f}}}r_{\rm{f}}^2 = {v_0}r_0^2\) and Df = D0(rf/r0), \({{{v_{\rm{f}}}} \over {{D_{\rm{f}}}}} = {{{v_0}} \over {{D_0}}}{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)^3}\), Eq. (B9a) can be rewritten as follows:

$${\dot W_3} = {{\pi {v_0}D_0^2} \over {12}}{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}{\sin ^{{1 \over n} - 2}}\phi {\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)^{{3 \over n}}}\int_0^\phi {{{\sin }^3}\,\theta {{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{{{1 - n} \over {2n}}}}{\rm{d}}\theta } \quad .$$

Approximation by \(1 - {{11} \over {12}}{\sin ^2}\,\theta \simeq 1 - {\sin ^2}\,\theta \) leads to

$${\dot W_3} = {{n\pi {v_0}D_0^2} \over {12}}{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}{\sin ^{{1 \over n} - 2}}\phi {\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)^{{3 \over n}}}\left( {{{{{\cos }^{{{2n + 1} \over n}}}\phi } \over {2n + 1}} - {{\cos }^{{1 \over n}}}\phi + {{2n} \over {2n + 1}}} \right)\quad .$$

C. Power loss because of sliding friction

Power loss because of sliding friction along the conical portion of SIII is represented as follows:

$${\dot W_4} = \int_{{S_{{\rm{III}}}}} {{\tau _{{\rm{sf}}}}\left| {\Delta {v_{{\rm{III}}}}} \right|{\rm{d}}S} = \int_{{r_{\rm{f}}}}^{{r_0}} {{\tau _{{\rm{sf}}}}{v_0}\cos \,\emptyset {{\left( {{{{r_0}} \over r}} \right)}^2}2\pi r\sin \,\emptyset \,{\rm{d}}r} $$
$$ = \left[ {{\tau _{{\rm{sf}}}}2\cot \,\emptyset \,{\rm{ln}}\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)} \right]\pi {\left( {{{{D_0}} \over 2}} \right)^2}{v_0}\quad .$$

For indirect extrusion, there is no need to consider power loss because of sliding friction along the cylindrical portion of SIV (\({\dot W_5}\)). However, for direct extrusion, \({\dot W_5}\) is an important factor that affects extrusion behavior. \({\dot W_5}\) is represented as follows:

$$\eqalign{ & {{\dot W}_5} = \int_{{S_{{\rm{IV}}}}} {{\tau _{{\rm{sf}}}}\left| {\Delta {v_{{\rm{IV}}}}} \right|{\rm{d}}S} = \int_0^L {{\tau _{{\rm{sf}}}}{v_0}\pi {D_0}\,{\rm{d}}L} \cr & \,\,\,\,\,\, = {\tau _{{\rm{sf}}}}{v_0}\pi {D_0}L = \left[ {{\tau _{{\rm{sf}}}}{{4L} \over {{D_0}}}} \right]\pi {\left( {{{{D_0}} \over 2}} \right)^2}{v_0}\quad . \cr} $$

D. Applied power

When the applied power equals to the upper bound energies,10 the applied stress (σ0) is represented as follows:

$${\sigma _0}\pi {\left( {{{{D_0}} \over 2}} \right)^2}{v_0} = {\dot W_1} + {\dot W_2} + {\dot W_3} + {\dot W_4} + {\dot W_5}\quad .$$

Equation (D1) is rewritten through Eq. (A8d), (B5b), (B9c), (C1), and (C2) as follows:

$$\left[ {{\sigma _0} - {\tau _{{\rm{sf}}}}2\,\cot \emptyset \ln \left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right) - {\tau _{{\rm{sf}}}}{{4L} \over {{D_0}}}} \right]\pi {\left( {{{{D_0}} \over 2}} \right)^2}{v_0}$$
$$\eqalign{ & = \pi n{v_0}D_0^2{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}\left[ {{{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)}^{{3 \over n}}} - 1} \right]{\sin ^{{1 \over n} - 2}}\phi \left\{ {{1 \over 3}\left[ {{{1 - {{\cos }^{2 + {1 \over n}}}\phi } \over {\left( {2 + 1/n} \right){{\sin }^{2 - {1 \over n}}}\phi }}} \right]} \right. \cr & \left. {\,\, + {1 \over {12}}\left( {{{{{\cos }^{{{2n + 1} \over n}}}\phi } \over {2n + 1}} - {{\cos }^{{1 \over n}}}\phi + {{2n} \over {2n + 1}}} \right)} \right\}\quad . \cr} $$

Equation (D2) is rewritten as follows:

$${{{v_0}} \over {{D_0}}} = {C \over {4{{\left( {4nK} \right)}^n}}}\sigma _{{\rm{eff}}}^n\quad ,$$


$$\eqalign{ & K = \left[ {{{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)}^{{3 \over n}}} - 1} \right]{\sin ^{{1 \over n} - 2}}\phi \left\{ {{1 \over 3}\left[ {{{1 - {{\cos }^{2 + {1 \over n}}}\phi } \over {\left( {2 + 1{\rm{/}}n} \right){{\sin }^{2 - {1 \over n}}}\phi }}} \right]} \right. \cr & \,\,\,\,\,\,\,\,\left. {\, + {1 \over {12}}\left( {{{{{\cos }^{{{2n + 1} \over n}}}\phi } \over {2n + 1}} - {{\cos }^{{1 \over n}}}\phi + {{2n} \over {2n + 1}}} \right)} \right\} \cr} $$


$${\sigma _{{\rm{eff}}}} = {\sigma _0} - {\sigma _{{\rm{sf}}}}\quad .$$

The external stresses required to overcome the interfacial sliding friction stress (σsf) in Eq. (D4) are given as follows:

For direct extrusion,

$${\sigma _{{\rm{sf}}}} = {\tau _{{\rm{sf}}}}2\,\cot \,\emptyset \,\ln \left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right) - {\tau _{{\rm{sf}}}}{{4L} \over {{D_0}}}\quad .$$

For indirect extrusion,

$${\sigma _{{\rm{sf}}}} = {\tau _{{\rm{sf}}}}2\cot \,\emptyset \,{\rm{ln}}\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)\quad .$$

E. Uniaxial creep deformation

The constitutive equation for steady-state uniaxial creep deformation obtained experimentally4 is

$${\dot \varepsilon _{rr}} = A{{Gb} \over {kT}}{\left( {{b \over d}} \right)^p}\left[ {{D^*}\exp \left( { - {Q \over {RT}}} \right)} \right]{\left( {{{{\sigma _{rr}}} \over G}} \right)^n}\quad .$$

When the stress exponent (n) and grain size exponent (p) are constant during uniaxial creep deformation, Eq. (E1) can be rewritten as follows:

$${\dot \varepsilon _{rr}} = {{{A_1}} \over {T{d^p}}}\left[ {\exp \left( { - {Q \over {RT}}} \right)} \right]\sigma _{rr}^{\,\,\,n}\quad ,$$


$${A_1} = A{{Gb} \over k}{\rm{\cdot}}{{{b^p}} \over {{G^n}}}{D^*}\quad .$$

In uniaxial creep deformation, \({\dot \varepsilon _{rr}}\) and \({\sigma _{rr}}\) equal \({\dot \varepsilon _{\rm{e}}}\) and \({\sigma _{\rm{e}}}\), respectively, and Eq. (E2) can be rewritten as follows:

$${\dot \varepsilon _{\rm{e}}} = {{{A_1}} \over {T{d^p}}}\left[ {\exp \left( { - {Q \over {RT}}} \right)} \right]\sigma _{\rm{e}}^{\,\,\,n}\quad .$$

A comparison of Eqs. (A5) and (E4) yields

$$C = {{{A_1}} \over {T{d^p}}}\exp \left( { - {Q \over {RT}}} \right)\quad .$$

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Kumagai, T. Hot extrusion of nanocrystalline yttria-stabilized tetragonal zirconia polycrystals. Journal of Materials Research 31, 3290–3302 (2016).

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