Effects of a bent structure on the linear viscoelastic response of diluted carbon nanotube suspensions

Abstract

Commonly isolated carbon nanotubes in suspension have been modelled as a perfectly straight structure. Nevertheless, single-wall carbon nanotubes (SWNTs) contain naturally side-wall defects and, in consequence, natural bent configurations. Hence, a semi-flexile filament model with a natural bent configuration was proposed to represent physically the SWNT structure. This continuous model was discretized as a non-freely jointed multi-bead–rod system with a natural bent configuration. Using a Brownian dynamics algorithm the dynamical mechanical contribution to the linear viscoelastic response of naturally bent SWNTs in dilute suspension was simulated. The dynamics of such system shows the apparition of new relaxation processes at intermediate frequencies characterized mainly by the activation of a mild elasticity. Storage modulus evolution at those intermediate frequencies strongly depends on the flexibility of the system, given by the rigidity constant of the bending potential and the number of constitutive rods.

Appendix A

Total bending force in a multi-bead–rod system writes:

$$ {\rm {\bf F}}_k^\phi =-\frac{\partial \phi }{\partial {\rm {\bf r}}_k }=\frac{K_{\rm b} }{a}\sum\limits_{i=2}^{n-1} {\frac{\partial \left({{\rm {\bf u}}_i \cdot {\rm {\bf u}}_{i-1}}\right)}{\partial {\rm {\bf r}}_k}} $$

(29)

Expanding Eq. 29 we have:

$$ {\rm {\bf F}}_k^\phi \!=\!\frac{K_{\rm b} }{a}\!\left(\begin{array}{l} \displaystyle \frac{\partial ( {{\rm {\bf u}}_2 \!\cdot {\rm {\bf u}}_1 } )}{\partial {\rm {\bf r}}_k }\!+\!...\!+\!\frac{\partial ({{\rm {\bf u}}_{k-1} \!\cdot {\rm {\bf u}}_{k-2} } )}{\partial {\rm {\bf r}}_k}\!+\!\frac{\partial ({{\rm {\bf u}}_k \cdot {\rm {\bf u}}_{k-1} })}{\partial {\rm {\bf r}}_k}\\[10pt]\displaystyle \!+\frac{\partial ({{\rm {\bf u}}_{k+1} \cdot {\rm {\bf u}}_k})}{\partial {\rm {\bf r}}_k }\!+\!...\!+\!\frac{\partial ( {{\rm {\bf u}}_{n-1} \cdot {\rm {\bf u}}_{n-2} } )}{\partial {\rm {\bf r}}_k } \end{array}\!\right) $$

(30)

$$ {\rm {\bf F}}_k^\phi \!=\!\frac{K_{\rm b} }{a}\!\left(\begin{array}{l} \displaystyle\frac{\partial {\rm {\bf u}}_2 }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_1 \!+\!\frac{\partial {\rm {\bf u}}_1 }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_2 ...\!+\!\frac{\partial {\rm {\bf u}}_{k-1} }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_{k-2} \\[10pt] \displaystyle \!+\frac{\partial {\rm {\bf u}}_{k-2} }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_{k-\!1} \!+\!\frac{\partial {\rm {\bf u}}_k }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_{k-1\!} \!+\!\frac{\partial {\rm {\bf u}}_{k-1} }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_k \!+\!\frac{\partial {\rm {\bf u}}_{k+1} }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_k \\[10pt] \displaystyle +\!\frac{\partial {\rm {\bf u}}_k }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_{k+1} \!+\!... \!+\!\frac{\partial {\rm {\bf u}}_{n-1} }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_{n-2} \!+\!\frac{\partial {\rm {\bf u}}_{n-2} }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_{n-1} \end{array}\!\right) $$

(31)

Equation 31 can be evaluated using the identity:

$$ \frac{\partial }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_i =\frac{1}{a}\left( {\delta _{k,i+1} -\delta _{k,i} } \right)\left( {{\rm {\bf I}}-{\rm {\bf u}}_i \otimes {\rm {\bf u}}_i } \right) $$

(32)

Observing identity (32) is clear that the derivative $\frac{\partial }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_i$ takes a non-zero value only when i is k − 1 or k. Hence, considering a bead k in the middle of an infinite chain, Eq. 31 becomes:

$$ \begin{array}{rll} {\rm {\bf F}}_k^\phi &=&\frac{K_{\rm b} }{a}\left( {\frac{\partial {\rm {\bf u}}_{k-1} }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_{k-2} } \right)+\frac{K_{\rm b} }{a}\left( {\frac{\partial {\rm {\bf u}}_k }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_{k-1} +\frac{\partial {\rm {\bf u}}_{k-1} }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_k } \right)\\ &&+\,\frac{K_{\rm b} }{a}\left( {\frac{\partial {\rm {\bf u}}_k }{\partial {\rm {\bf r}}_k }{\rm {\bf u}}_{k+1} } \right) \end{array}$$

(33)

Total bending force on bead k, as given in the previous equation, can be interpreted as the sum of three independent contributions. In order to explain the origin of those contributions consider a multi-bead–rod system containing n beads decomposed into n − 2 independent sub-sections of two consecutive rods as showed in Fig. 4. Taking each of those sub-sections as an independent system, the expressions for the bending forces associated to each sub-section are given by Eq. 29. For example, sub-section p (composed by beads p, p + 1 and p + 2) has the associated bending forces ${\rm {\bf F}}_{p,p}^\phi$, ${\rm {\bf F}}_{p+1,p}^\phi$ and ${\rm {\bf F}}_{p+2,p}^\phi$, where first sub-index refers to the bead and second sub-index refers to the sub-section. Mathematically, those bending forces write:

$$ {\rm {\bf F}}_{p,p}^\phi =\frac{K_{\rm b} }{a}\frac{\partial \left({{\rm {\bf u}}_{p+1} \cdot {\rm {\bf u}}_p } \right)}{\partial {\rm {\bf r}}_p }=\frac{K_{\rm b} }{a}\frac{\partial {\rm {\bf u}}_p }{\partial {\rm {\bf r}}_p }{\rm {\bf u}}_{p+1} $$

(34)

$$ \begin{array}{rll} {\rm {\bf F}}_{p+1,p}^\phi &=&\frac{K_{\rm b} }{a}\frac{\partial \left( {{\rm {\bf u}}_{p+1} \cdot {\rm {\bf u}}_p } \right)}{\partial {\rm {\bf r}}_{p+1} }=\frac{K_{\rm b} }{a}\frac{\partial {\rm {\bf u}}_{p+1} }{\partial {\rm {\bf r}}_{p+1} }{\rm {\bf u}}_p \\ &&+\frac{K_{\rm b} }{a}\frac{\partial {\rm {\bf u}}_p }{\partial {\rm {\bf r}}_{p+1} }{\rm {\bf u}}_{p+1} \end{array}$$

(35)

$$ {\rm {\bf F}}_{p+2,p}^\phi =\frac{K_{\rm b} }{a}\frac{\partial \left( {{\rm {\bf u}}_{p+1} \cdot {\rm {\bf u}}_p } \right)}{\partial {\rm {\bf r}}_{p+2} }=\frac{K_{\rm b} }{a}\frac{\partial {\rm {\bf u}}_{p+1} }{\partial {\rm {\bf r}}_{p+2} }{\rm {\bf u}}_p $$

(36)

Applying identity (32) to the three previous equations we have:

$$ {\rm {\bf F}}_{p,p}^\phi =-\frac{K_{\rm b} }{a^2}\left( {{\rm {\bf I}}-{\rm {\bf u}}_p \otimes {\rm {\bf u}}_p } \right){\rm {\bf u}}_{p+1} $$

(37)

$$ \begin{array}{rll} {\rm {\bf F}}_{p+1,p}^\phi &=&-\frac{K_{\rm b} }{a^2}\left( {{\rm {\bf I}}-{\rm {\bf u}}_{p+1} \otimes {\rm {\bf u}}_{p+1} } \right){\rm {\bf u}}_p \\ &&+\frac{K_{\rm b} }{a^2}\left({{\rm {\bf I}}-{\rm {\bf u}}_p \otimes {\rm {\bf u}}_p } \right){\rm {\bf u}}_{p+1} \end{array}$$

(38)

$$ {\rm {\bf F}}_{p+2,p}^\phi =\frac{K_{\rm b} }{a^2}\left({{\rm {\bf I}}-{\rm {\bf u}}_{p+1} \otimes {\rm {\bf u}}_{p+1} } \right){\rm {\bf u}}_p $$

(39)

From the previous equations it results clear that bending forces are in mechanical equilibrium. For this reason:

$$ {\rm {\bf F}}_{p+1,p}^\phi =-{\rm {\bf F}}_{p,p}^\phi -{\rm {\bf F}}_{p+2,p}^\phi $$

(40)

Now, using the previous results about the decomposition of a multi-bead–rod system is easy to explain the origin of different contributions in Eq. 33 as follows:

$$ {\rm {\bf F}}_{k}^\phi = {\rm {\bf F}}_{k,k-2}^\phi + {\rm {\bf F}}_{k,k-1}^\phi +{\rm {\bf F}}_{k,k}^\phi $$

(41)

In other words, total bending force on bead k can be interpreted as the sum of all the independent bending forces coming from sub-sections containing bead k.