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Sensitivity analysis-based full-scale bounds estimation for 2-D interval bi-modular problems

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Abstract

The full-scale bounds estimation for 2-D bi-modular problem with interval uncertain constitutive parameters is realized by means of sensitivity analysis. An efficient FE model is presented to solve the deterministic 2-D bi-modular problem by complementing a shear modulus identical with the coaxial condition required by the constitutive relationship, and the equations to calculate both the first- and second-order derivatives of displacements with respect to constitutive parameters are derived. When the interval scale of uncertain constitutive parameters is relatively small, two algorithms are developed for the bounds estimation by using the first-/second-order Taylor series approximation and interval arithmetic; when the interval scale is large, a rigorous bounds estimation can be achieved by using the first-order derivatives and a global searching technique. In addition, two second-order Taylor series approximation-based algorithms are proposed to reduce the computational expense in the process of optimization for bounds estimation. With the consideration of expansion order of Taylor series, interval scale of uncertainty, ratio of \(E^-/E^+\), etc., numerical examples are presented to illustrate the accuracy and efficiency of the proposed approach.

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Acknowledgements

The research leading to this paper is funded by NSF [11972109, 11572068] and NKBRSF [2015CB057804]

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  1. State Key Laboratory of Structural Analysis for Industrial Equipment, Department of Engineering Mechanics, Dalian University of Technology, Dalian, 116024, People’s Republic of China

    Chunjiang Ran & Haitian Yang

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Appendix A. Derivatives of \(\tilde{{\varvec{A}}}^b\) with respect to the bi-modular constitutive parameters.

Appendix A. Derivatives of \(\tilde{{\varvec{A}}}^b\) with respect to the bi-modular constitutive parameters.

Since \(\mu ^+\) and \(\mu ^-\) are related via Eq. (3), there are only three independent variables involved in the calculation of derivatives of \(\tilde{{\varvec{A}}}^b\); thus, two cases, i.e., (1) \(E^+\), \(E^-\) and \(\mu ^+\) are independent and (2) \(E^+\), \(E^-\) and \(\mu ^-\) are independent, which are considered in the derivation.

Assume \(E^+\), \(E^-\) and \(\mu ^+\) are independent variables,

$$\begin{aligned} \mu ^- = \mu ^+ \frac{E^-}{E^+} \end{aligned}$$
(A.1)

When \(\sigma _1^p \ge 0\) and \(\sigma _2^p \ge 0\),

$$\begin{aligned} \tilde{{\varvec{A}}}^b = \frac{1}{E^+} \begin{bmatrix} 1 &{} \quad -\mu ^+ &{} \quad 0 \\ -\mu ^+ &{} \quad 1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 2(1+\mu ^+) \\ \end{bmatrix} \end{aligned}$$
(A.2)

Then,

$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{ \partial E^+ }&= \frac{1}{(E^+)^2 } \begin{bmatrix} -1 &{} \quad \mu ^+ &{} \quad 0 \\ \mu ^+ &{} \quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2(1+\mu ^+) \\ \end{bmatrix} , \end{aligned}$$
(A.3a)
$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{ \partial \mu ^+ }&= \frac{1}{ E^+ } \begin{bmatrix} 0 &{} \quad -1 &{} \quad 0 \\ -1 &{} \quad \quad 0 &{} \quad 0 \\ 0 &{} \quad \quad 0 &{} \quad 2 \\ \end{bmatrix} , \quad \frac{\partial \tilde{{\varvec{A}}}^b}{ \partial E^- } = {\varvec{0}} . \end{aligned}$$
(A.3b)

Furthermore,

$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^+)^2 } = - \frac{2}{(E^+)^3} \begin{bmatrix} -1 &{} \quad \mu ^+ &{} \quad 0 \\ \mu ^+ &{} \quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2(1+\mu ^+) \\ \end{bmatrix} , \end{aligned}$$
(A.4a)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial \mu ^+ } = \frac{1}{(E^+)^2 } \begin{bmatrix} 0 &{} \quad 1 &{} \quad 0 \\ 1 &{} \quad \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2 \\ \end{bmatrix}, \end{aligned}$$
(A.4b)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial E^- } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^-)^2 } = {\varvec{0}} , \end{aligned}$$
(A.4c)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^- \partial \mu ^+ } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (\mu ^+)^2 } = {\varvec{0}} . \end{aligned}$$
(A.4d)

When \(\sigma _1^p \ge 0\) and \(\sigma _2^p < 0\),

$$\begin{aligned} \tilde{{\varvec{A}}}^b = \begin{bmatrix} \frac{1}{E^+} &{} \quad -\frac{\mu ^+}{E^+} &{} \quad 0 \\ -\frac{\mu ^+}{E^+} &{} \quad \frac{1}{E^-} &{} \quad 0 \\ 0&{} \quad 0&{} \quad \frac{ 2 \sigma _1^p }{ \left( \sigma _1^p - \sigma _2^p \right) E^+ } - \frac{ 2 \sigma _2^p }{ \left( \sigma _1^p - \sigma _2^p \right) E^-} + 2 \frac{ \mu ^+ }{ E^+ } \\ \end{bmatrix} \end{aligned}$$
(A.5)

Then,

$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{\partial E^+}&= \frac{1}{(E^+)^2 } \begin{bmatrix} -1 &{} \quad \mu ^+ &{} \quad 0 \\ \mu ^+ &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad - \frac{ 2 \sigma _1^p }{ \sigma _1^p - \sigma _2^p } - 2 \mu ^+ \\ \end{bmatrix} , \end{aligned}$$
(A.6a)
$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{\partial E^-}&= \frac{1}{(E^-)^2 } \begin{bmatrix} 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad \frac{ 2 \sigma _2^p }{ \sigma _1^p - \sigma _2^p } \\ \end{bmatrix} , \end{aligned}$$
(A.6b)
$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{\partial \mu ^+}&= \frac{1}{ E^+ } \begin{bmatrix} 0 &{} \quad -1 &{} \quad 0 \\ -1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 2 \\ \end{bmatrix} . \end{aligned}$$
(A.6c)

Furthermore,

$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^+)^2 } = - \frac{2}{(E^+)^3} \begin{bmatrix} -1 &{} \quad \mu ^+ &{} \quad 0 \\ \mu ^+ &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad - \frac{ 2 \sigma _1^p }{ \sigma _1^p - \sigma _2^p } - 2 \mu ^+ \\ \end{bmatrix}, \end{aligned}$$
(A.7a)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial \mu ^+ } = \frac{1}{(E^+)^2 } \begin{bmatrix} 0 &{} \quad 1 &{} \quad 0 \\ 1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2 \\ \end{bmatrix} , \end{aligned}$$
(A.7b)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^-)^2 } = -\frac{2}{(E^-)^3} \begin{bmatrix} 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad \frac{ 2 \sigma _2^p }{ \sigma _1^p - \sigma _2^p } \\ \end{bmatrix}, \end{aligned}$$
(A.7c)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial E^- } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^- \partial \mu ^+ } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (\mu ^+)^2 } = {\varvec{0}} . \end{aligned}$$
(A.7d)

When \(\sigma _1^p < 0\) and \(\sigma _2^p < 0\),

$$\begin{aligned} \tilde{{\varvec{A}}}^b = \begin{bmatrix} \frac{1}{E^-} &{} \quad -\frac{\mu ^+}{E^+} &{} \quad 0 \\ -\frac{\mu ^+}{E^+} &{} \quad \frac{1}{E^-} &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 2 \frac{ 1 }{ E^- } + 2 \frac{ \mu ^+ }{ E^+ } \\ \end{bmatrix} \end{aligned}$$
(A.8)

Then,

$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{ \partial E^+ }&= \frac{1}{(E^+)^2 } \begin{bmatrix} 0 &{} \quad \mu ^+ &{} \quad 0 \\ \mu ^+ &{} 0 &{} 0 \\ 0 &{} \quad 0 &{} \quad - 2 \mu ^+ \\ \end{bmatrix} , \end{aligned}$$
(A.9a)
$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{\partial E^-}&= \frac{1}{(E^-)^2 } \begin{bmatrix} -1 &{} \quad 0 &{} \quad \quad 0 \\ 0 &{} \quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2 \\ \end{bmatrix} , \end{aligned}$$
(A.9b)
$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{\partial \mu ^+}&= \frac{1}{ E^+ } \begin{bmatrix} 0 &{} \quad -1 &{} \quad 0 \\ -1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 2 \\ \end{bmatrix} . \end{aligned}$$
(A.9c)

Furthermore,

$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^+)^2 } = - \frac{2}{(E^+)^3} \begin{bmatrix} 0 &{} \quad \mu ^+ &{} 0 \\ \mu ^+ &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad - 2 \mu ^+ \\ \end{bmatrix}, \end{aligned}$$
(A.10a)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial \mu ^+ } = \frac{1}{(E^+)^2 } \begin{bmatrix} 0 &{} \quad 1 &{} \quad 0 \\ 1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2 \\ \end{bmatrix} , \end{aligned}$$
(A.10b)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^-)^2 } = -\frac{2}{(E^-)^3} \begin{bmatrix} -1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2 \\ \end{bmatrix}, \end{aligned}$$
(A.10c)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial E^- } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^- \partial \mu ^+ } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (\mu ^+)^2 } = {\varvec{0}} . \end{aligned}$$
(A.10d)

Assume \(E^+\), \(E^-\) and \(\mu ^-\) are independent variables,

$$\begin{aligned} \mu ^+ = \mu ^- \frac{E^+}{E^-} \end{aligned}$$
(A.11)

When \(\sigma _1^p \ge 0\) and \(\sigma _2^p \ge 0\),

$$\begin{aligned} \tilde{{\varvec{A}}}^b = \begin{bmatrix} \frac{1}{E^+} &{} \quad -\frac{\mu ^-}{E^-} &{} \quad 0 \\ -\frac{\mu ^-}{E^-} &{} \quad \quad \frac{1}{E^+} &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 2 \frac{ 1 }{ E^+ } + 2 \frac{ \mu ^- }{ E^- } \\ \end{bmatrix} \end{aligned}$$
(A.12)

Then,

$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{ \partial E^+ }&= \frac{1}{(E^+)^2 } \begin{bmatrix} -1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2 \\ \end{bmatrix} , \end{aligned}$$
(A.13a)
$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{ \partial E^- }&= \frac{1}{(E^-)^2 } \begin{bmatrix} 0 &{} \quad \mu ^- &{} \quad 0 \\ \mu ^- &{} 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2 \mu ^- \\ \end{bmatrix} , \end{aligned}$$
(A.13b)
$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{ \partial \mu ^- }&= \frac{1}{ E^- } \begin{bmatrix} 0 &{} \quad -1 &{} \quad 0 \\ -1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 2 \\ \end{bmatrix}. \end{aligned}$$
(A.13c)

Furthermore,

$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^+)^2 } = - \frac{2}{(E^+)^3} \begin{bmatrix} -1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2 \\ \end{bmatrix}, \end{aligned}$$
(A.14a)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^-)^2 } = -\frac{2}{(E^-)^3} \begin{bmatrix} 0 &{} \quad \mu ^- &{} \quad 0 \\ \mu ^- &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad \quad 0 &{} \quad -2 \mu ^- \\ \end{bmatrix} , \end{aligned}$$
(A.14b)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^- \partial \mu ^- } = \frac{1}{(E^-)^2 } \begin{bmatrix} 0 &{} \quad 1 &{} \quad 0 \\ 1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{}\quad -2 \\ \end{bmatrix}, \end{aligned}$$
(A.14c)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial E^- } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial \mu ^- } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (\mu ^-)^2 } = {\varvec{0}} . \end{aligned}$$
(A.14d)

When \(\sigma _1^p \ge 0\) and \(\sigma _2^p < 0\),

$$\begin{aligned} \tilde{{\varvec{A}}}^b = \begin{bmatrix} \frac{1}{E^+} &{} \quad -\frac{\mu ^-}{E^-} &{} \quad 0 \\ -\frac{\mu ^-}{E^-} &{} \quad \frac{1}{E^-} &{}\quad 0 \\ 0 &{} \quad 0 &{} \quad \frac{ 2 \sigma _1^p }{ \left( \sigma _1^p - \sigma _2^p \right) E^+ } - \frac{ 2 \sigma _2^p }{ \left( \sigma _1^p - \sigma _2^p \right) E^- } + 2 \frac{ \mu ^- }{ E^- } \\ \end{bmatrix} \end{aligned}$$
(A.15)

Then,

$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{\partial E^+}&= \frac{1}{(E^+)^2 } \begin{bmatrix} -1 &{} \quad 0 &{}\quad 0 \\ 0 &{} \quad 0 &{}\quad 0 \\ 0 &{} \quad 0 &{}\quad - \frac{ 2 \sigma _1^p }{ \sigma _1^p - \sigma _2^p } \\ \end{bmatrix} , \end{aligned}$$
(A.16a)
$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{\partial E^-}&= \frac{1}{(E^-)^2 } \begin{bmatrix} 0 &{} \quad \mu ^- &{} \quad 0 \\ \mu ^- &{}\quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{}\quad \frac{ 2 \sigma _2^p }{ \sigma _1^p - \sigma _2^p } - 2 \mu ^- \\ \end{bmatrix} , \end{aligned}$$
(A.16b)
$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{\partial \mu ^-}&= \frac{1}{ E^- } \begin{bmatrix} 0 &{} \quad -1 &{} \quad 0 \\ -1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 2 \\ \end{bmatrix}. \end{aligned}$$
(A.16c)

Furthermore,

$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^+)^2 } = - \frac{2}{(E^+)^3} \begin{bmatrix} -1 &{} \quad 0 &{} \quad \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad - \frac{ 2 \sigma _1^p }{ \sigma _1^p - \sigma _2^p } \\ \end{bmatrix}, \end{aligned}$$
(A.17a)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^-)^2 } = -\frac{2}{(E^-)^3} \begin{bmatrix} 0 &{} \quad \mu ^- &{} \quad 0 \\ \mu ^- &{} \quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad \frac{ 2 \sigma _2^p }{ \sigma _1^p - \sigma _2^p } - 2 \mu ^- \\ \end{bmatrix} , \end{aligned}$$
(A.17b)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^- \partial \mu ^- } = \frac{1}{(E^-)^2 } \begin{bmatrix} 0 &{} \quad 1 &{} \quad 0 \\ 1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2 \\ \end{bmatrix}, \end{aligned}$$
(A.17c)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial E^- } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial \mu ^- } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (\mu ^-)^2 } = {\varvec{0}} . \end{aligned}$$
(A.17d)

When \(\sigma _1^p < 0\) and \(\sigma _2^p < 0\),

$$\begin{aligned} \tilde{{\varvec{A}}}^b = \begin{bmatrix} \frac{1}{E^-} &{} \quad -\frac{\mu ^-}{E^-} &{} \quad 0 \\ -\frac{\mu ^-}{E^-} &{} \quad \frac{1}{E^-} &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 2 \frac{ 1 + \mu ^- }{ E^- } \\ \end{bmatrix} \end{aligned}$$
(A.18)

Then,

$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{\partial E^-}&= \frac{1}{(E^-)^2 } \begin{bmatrix} -1 &{} \mu ^- &{} 0 \\ \mu ^- &{} -1 &{} 0 \\ 0 &{} 0 &{} -2(1+\mu ^-) \\ \end{bmatrix} , \end{aligned}$$
(A.19a)
$$\begin{aligned} \frac{\partial \tilde{{\varvec{A}}}^b}{\partial \mu ^-}&= \frac{1}{ E^- } \begin{bmatrix} 0 &{} \quad -1 &{} \quad 0 \\ -1 &{} \quad \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 2 \\ \end{bmatrix} , \quad \frac{\partial \tilde{{\varvec{A}}}^b}{ \partial E^+ } = {\varvec{0}} . \end{aligned}$$
(A.19b)

Furthermore,

$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^-)^2 } = -\frac{2}{(E^-)^3} \begin{bmatrix} -1 &{} \quad \mu ^- &{} \quad 0 \\ \mu ^- &{} \quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{} -2(1+\mu ^-) \\ \end{bmatrix} , \end{aligned}$$
(A.20a)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^- \partial \mu ^- } = \frac{1}{(E^-)^2 } \begin{bmatrix} 0 &{} \quad 1 &{} \quad 0 \\ 1 &{} \quad \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -2 \\ \end{bmatrix}, \end{aligned}$$
(A.20b)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (E^+)^2 } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial E^- } = {\varvec{0}} , \end{aligned}$$
(A.20c)
$$\begin{aligned}&\frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial E^+ \partial \mu ^- } = {\varvec{0}} , \quad \frac{\partial ^2 \tilde{{\varvec{A}}}^b}{ \partial (\mu ^-)^2 } = {\varvec{0}} . \end{aligned}$$
(A.20d)

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Ran, C., Yang, H. Sensitivity analysis-based full-scale bounds estimation for 2-D interval bi-modular problems. Arch Appl Mech 91, 3011–3034 (2021). https://doi.org/10.1007/s00419-021-01945-x

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