8. Appendix#

8.1. Auxiliary problems table#

We can find in the following table the sequence of \(342\) auxiliary problems calculation management, for membrane, bending and bond-sliding cases, for the different values of damage variables, used in the automated DHRC parameters identification procedure.

EPI value

PRE_EPSI value

Calculations number

Membrane index

Bending index

RVE

Damage

Sliding

Membrane aux. field

-1	1	1	9	0	8	0	8			1X and 2X	Compression X	Yes	nr	\({\chi }_{c}^{\mathrm{xx}}\)	\({\xi }_{c}^{\mathrm{xx}}\)
1	-1	10	18	9	17	9	17			1X and 2X	X voltage	Yes	nr	\({\chi }_{t}^{\mathrm{xx}}\)	\({\xi }_{t}^{\mathrm{xx}}\)
-1	1	19	27	18	26	18	26			1X and 2X	Y compression	Yes	nr	\({\chi }_{c}^{\mathrm{yy}}\)	\({\xi }_{c}^{\mathrm{yy}}\)
1	-1	28	36	27	35	27	35			1X and 2X	Y voltage	Yes	nr	\({\chi }_{t}^{\mathrm{yy}}\)	\({\xi }_{t}^{\mathrm{yy}}\)
0.5	-0.5	37	45	36	44	36	44			1X and 2X	Shear	Yes	nr	\({\chi }^{\mathrm{xy}}\)	\({\xi }^{\mathrm{xy}}\)
-1	1	46	54	45	53	45	53			1Y and 2Y	Compression X	Yes	nr	\({\chi }_{c}^{\mathrm{xx}}\)	\({\xi }_{c}^{\mathrm{xx}}\)
1	-1	55	63	54	62	54	62			1Y and 2Y	X voltage	Yes	nr	\({\chi }_{t}^{\mathrm{xx}}\)	\({\xi }_{t}^{\mathrm{xx}}\)
-1	1	64	72	63	71	63	71			1Y and 2Y	Y compression	Yes	nr	\({\chi }_{c}^{\mathrm{yy}}\)	\({\xi }_{c}^{\mathrm{yy}}\)
1	-1	73	81	72	80	72	80			1Y and 2Y	Y voltage	Yes	nr	\({\chi }_{t}^{\mathrm{yy}}\)	\({\xi }_{t}^{\mathrm{yy}}\)
0.5	-0.5	82	90	81	89	81	89			1Y and 2Y	Shear	Yes	nr	\({\chi }^{\mathrm{xy}}\)	\({\xi }^{\mathrm{xy}}\)
-1	1	91	99	90	98	90	98			1T and 2T	Compression X	Yes	nr	\({\chi }_{c}^{\mathrm{xx}}\)	\({\xi }_{c}^{\mathrm{xx}}\)
1	-1	100	108	99	107	99	107			1T and 2T	X voltage	Yes	nr	\({\chi }_{t}^{\mathrm{xx}}\)	\({\xi }_{t}^{\mathrm{xx}}\)
-1	1	109	117	108	116	108	116			1T and 2T	Y compression	Yes	nr	\({\chi }_{c}^{\mathrm{yy}}\)	\({\xi }_{c}^{\mathrm{yy}}\)
1	-1	118	126	117	125	117	125			1T and 2T	Y voltage	Yes	nr	\({\chi }_{t}^{\mathrm{yy}}\)	\({\xi }_{t}^{\mathrm{yy}}\)
0.5	-0.5	127	135	126	134	126	134			1T and 2T	Shear	Yes	nr	\({\chi }^{\mathrm{xy}}\)	\({\xi }^{\mathrm{xy}}\)
0	0	136	144	135	143			0	8	1X and 2X		nr	1X
0	0	145	153	144	152			9	17	1X and 2X		nr	2X
0	0	154	162	153	161			18	26	1X and 2X		nr	1Y
0	0	163	171	162	170			27	35	1X and 2X		nr	2Y
0	0	172	180	171	179			36	44	1Y and 2Y		nr	1X
0	0	181	189	180	188			45	53	1Y and 2Y		nr	2X
0	0	190	198	189	197			54	62	1Y and 2Y		nr	1Y
0	0	199	207	198	206			63	71	1Y and 2Y		nr	2Y

8.2. Convexity of the strain energy density function with discontinuity in damage function#

As presented in section 3.1.2, the macroscopic free energy density \(W(E,K,{D}^{\zeta },{E}^{\eta \zeta })\) is continuous when its variables are taken in the whole strains and sliding strains space. However, the discontinuity in microscopic damage functions \(\xi (d)\) — due to the assumed distinction between tensile and compressive states, see Hyp 14 — induces a piece-wise definition of the elasticity tensors \({A}^{\mathrm{mm}}(D)\) and \({A}^{\mathrm{ff}}(D)\), while \({A}^{\mathrm{mf}}(D)\) components don’t include this dissymmetry. We recall that, according to Remark 16, \(B(D)\) and \(C(D)\) tensors don’t involve any tension-compression dissymmetry. In domains far enough from this discontinuity, it can be easily shown the convexity of the strain energy density function from its quadratic expression. Conversely, either side of this discontinuity, we have to proceed as proposed for instance by (Curnier, He, & Zysset, 1995).

As being assumed at Hyp 19, the discontinuity is defined by the hyperplane \(E+{Q}^{m\zeta }(D)\mathrm{.}{E}^{\eta \zeta }=0\) and \(K+{Q}^{f\zeta }(D)\mathrm{.}{E}^{\eta \zeta }=0\), for \(D\) being fixed, with \({Q}^{\zeta }(D)={A}^{(-1)}(D):{B}^{\zeta }(D)\) defined by (3.2-10).

We take advantage that the stress resultants (3.2-11 and -12) obtained by derivation of the strain energy density function are continuous by construction with the discontinuity of \({A}^{\mathrm{mm}}(D)\) and \({A}^{\mathrm{ff}}(D)\) defined by the sign of \(x\). Nevertheless, as we need to express the updated values of \({A}^{\mathrm{mm}}(D)\) and \({A}^{\mathrm{ff}}(D)\) components to determine \({Q}^{\zeta }(D)\) at each step increment, we decided to input the previous value \({D}^{\text{-}}\), in an explicit algorithm way, see section 4.

8.3. Proof of the zero-valued tensor \(B\) if microscopic damage field is homogeneous in the RVE#

Let us recall that the tensor \(B\) components are defined by: \({B}_{\mathrm{\alpha \beta \gamma }}^{m\zeta }(D)={\langle \langle {a}_{\mathrm{\alpha \beta kl}}(d):{\epsilon }_{\mathrm{kl}}({\chi }^{{\eta }_{\gamma }^{\zeta }})\rangle \rangle }_{\Omega }\) and \({B}_{\mathrm{\alpha \beta \gamma }}^{f\zeta }(D)={-\langle \langle {x}_{3}\mathrm{.}{a}_{\mathrm{\alpha \beta kl}}(d):{\epsilon }_{\mathrm{kl}}({\chi }^{{\eta }_{\gamma }^{\zeta }})\rangle \rangle }_{\Omega }\), see (3.2-7 and -8). Let us assume that the components of the concrete damaged elastic tensor \({a}_{\mathrm{pqrs}}^{c}({d}^{\zeta })\) are symmetric with respect to the plane \({x}_{1}=0\), i.e. \({\Gamma }_{s}\) see; moreover, according to section 2.1.4, this tensor fulfils the symmetries with respect to the reference frame. Let us consider the auxiliary problem (3-1.5) concerning the corrector \({\chi }^{{\eta }_{1}^{\zeta }}\) associated with sliding in the \({x}_{1}\) direction. Necessarily, we have the following properties: \({\chi }_{1}^{{\eta }_{1}^{\zeta }}\) is symmetric with respect to the plane \({x}_{1}=0\), whereas \({\chi }_{2}^{{\eta }_{1}^{\zeta }}\) and \({\chi }_{3}^{{\eta }_{1}^{\zeta }}\) are antisymmetric. Therefore, \({a}_{\mathrm{iikl}}(d):{\epsilon }_{\mathrm{kl}}({\chi }^{{\eta }_{1}^{\zeta }})\) and \({a}_{\mathrm{23kl}}(d):{\epsilon }_{\mathrm{kl}}({\chi }^{{\eta }_{1}^{\zeta }})\) are antisymmetric with respect to the plane \({x}_{1}=0\), whereas \({a}_{\mathrm{1jkl}}(d):{\epsilon }_{\mathrm{kl}}({\chi }^{{\eta }_{1}^{\zeta }})\) are symmetric, for \(j\ne 1\) and \({a}_{\mathrm{1jkl}}(d):{\epsilon }_{\mathrm{kl}}({\chi }^{{\eta }_{1}^{\zeta }})\) are antisymmetric with respect to the plane \({x}_{2}=0\).

Performing the integration on the whole RVE, we deduce that: \({B}_{\mathrm{\alpha \beta 1}}^{m\zeta }(D)\) and \({B}_{\mathrm{\alpha \beta 1}}^{f\zeta }(D)\) vanish. The same reasoning holds for the other direction in the mid-plane of the RVE.

Since the tensor \(B\) manages the coupling between RC plate kinematics and steel bond sliding, we conclude that bond sliding can occur only if symmetry loss appears in the RVE, due to damage.