Splitting and curling performance of metal foam-filled circular tubes

Abstract

In this paper, analytical and experimental methods are used to study the deformation mechanisms of metal foam-filled circular tubes in axial splitting and curling modes. Metal foam-filled tubes with prefabricated cracks are placed on a die, and axial compressive experiments are conducted. In the experiments, the force–displacement curves of metal foam-filled tubes are measured, and the deformation modes of metal foam-filled tubes are observed. An analytical model is established for metal foam-filled circular tubes in axial splitting and curling modes. In the analytical model, energy is dissipated by splitting, curling, and plastic bending of the tubes, as well as by friction between the tubes and die and foam compression. The analytical, numerical and experimental results are compared. The analytical model captures the experimental results reasonably well, and the numerical results are in good agreement with experimental ones. It is found that, in the stage of steady deformation, the axial compressive force increases with the increase in the number of prefabricated cracks, the ratio of radius to the wall-thickness and the semi-angle of conical die. The axial compressive force decreases with the increase in the ratio of the prefabricated-crack length to the wall-thickness. Foam-filled tubes have a higher specific energy absorption than sandwich tubes.

Appendices

Appendix 1: Derivation of solving for M _p

This appendix is the derivation of solving for the bending moment M_p of a circular tube to its PNA [31].

The centre angle of each strip is $\frac{2\pi }{n}$ for the foam-filled tube. Take one strip $FGG^{^{\prime}} F^{^{\prime}}$ and take the centre O of the foam-filled tube as the origin to establish the coordinate system, as shown in Fig. 

Fig. 14

Sketch of the section for the circular tube between prefabricated cracks. a The case of y_C > y_E > y_F, b the case of y_F > y_C > y_E, c the case of y_C > y_F > y_E, and d the case of y_E > y_C > y_F

14a, where the y-axis is the angular bisector of the centre angle, $r_{1}^{^{\prime\prime}}$ is the inner radius of the metal tube, $r_{1}^{^{\prime}}$ is the outer radius of the metal tube, and $r_{1} = \frac{{r_{1}^{^{\prime}} + r_{1}^{^{\prime\prime}} }}{2}$_. The line $EE^{^{\prime}}$ is tangential to the arc $GG^{^{\prime}}$ at point D. Points $B$ and $B^{^{\prime}}$ are points on the arc $EE^{^{\prime}}$, and the angle BOA is $\frac{2\pi }{n}$. The bending moment of each part of section $FGG^{^{\prime}} F^{^{\prime}}$ to straight line $BB^{^{\prime}}$ is given as follows. For the specimens of different geometric sizes, the relative positions of lines $CC^{^{\prime}}$, $EE^{^{\prime}}$ and $FF^{^{\prime}}$ are different, and y_C, y_E, and y_F are used to represent the y-direction coordinates of lines $CC^{^{\prime}}$, $EE^{^{\prime}}$ and $FF^{^{\prime}}$, respectively.

When $y_{C} > y_{E} > y_{F}$, as shown in Fig. 14a, the total bending moment of all the strips of the tube is

$$M_{p} = n\left( {M_{1} + M_{2} + M_{3} + M_{4} } \right),$$

(A1)

where

$$M_{1} = 2\sigma_{t} \int_{0}^{{\omega_{1} }} {\left( {r_{1}^{^{\prime}} \cos \omega - r_{1} \cos \frac{\pi }{2n}} \right)} \left( {r_{1}^{^{\prime}} \sin \omega } \right)^{2} {\text{d}} \omega ,$$

$$M_{2} = 2\sigma_{t} \int_{{\omega_{1} }}^{{\omega_{2} }} {\left( {r_{1} \cos \frac{\pi }{2n} - r_{1}^{^{\prime}} \cos \omega } \right)} \left( {r_{1}^{^{\prime}} \sin \omega } \right)^{2} {\text{d}} \omega ,$$

$$M_{3} = 2\sigma_{t} \int_{{\omega_{2} }}^{{\frac{\pi }{n}}} {\left( {r_{1} \cos \frac{\pi }{2n} - r_{1}^{^{\prime}} \cos \omega } \right)} \left( {r_{1}^{^{\prime}} \sin \omega - r_{1}^{^{\prime\prime}} \,\sqrt {1 - \left( {\frac{{r_{1} \cos \omega }}{{r_{1}^{^{\prime\prime}} }}} \right)^{2} } } \right)r_{1}^{^{\prime}} \sin \omega {\text{d}} \omega ,$$

$$M_{4} = 2\sigma_{t} \int_{{\varphi_{0} }}^{{\frac{\pi }{n}}} {\left[ {r_{1} \cos (\frac{\pi }{2n}) - r_{1}^{^{\prime\prime}} \cos \varphi } \right]} \left( {\frac{{r_{1}^{^{\prime\prime}} \cos \varphi }}{{\cos \frac{\pi }{n}}}\sin \frac{\pi }{n} - r_{1}^{^{\prime\prime}} \sin \varphi } \right)r_{1}^{^{\prime\prime}} \sin \varphi {\text{d}} \varphi ,$$

$$\omega_{1} = \cos^{ - 1} \left( {r_{1} \cos \frac{\pi }{2n}/r_{1}^{^{\prime}} } \right),$$

$$\omega_{2} = \cos^{ - 1} \frac{{r_{1}^{^{\prime\prime}} }}{{r_{1}^{^{\prime}} }},$$

$$\varphi_{0} = \cos^{ - 1} \left( {\frac{{r_{1}^{^{\prime}} \cos \frac{\pi }{n}}}{{r_{1}^{^{\prime\prime}} }}} \right).$$

When $y_{F} > y_{C} > y_{E}$, as shown in Fig. 14b, the total bending moment of all the strips of the tube is

$$M_{p} = n\left( {M_{5} + M_{6} + M_{7} + M_{8} } \right),$$

(A2)

where

$$\begin{gathered} M_{5} = 2\sigma_{t} \int_{0}^{{\frac{\pi }{n}}} {\left( {r_{1}^{^{\prime}} \cos \omega - r_{1} \cos \frac{\pi }{2n}} \right)} \left( {r_{1}^{^{\prime}} \sin \omega } \right)^{2} {\text{d}} \omega , \hfill \\ M_{6} = \sigma_{t} r_{1}^{^{\prime}} \sin \frac{\pi }{n}\left( {r_{1} \cos \frac{\pi }{2n} - r_{1}^{^{\prime}} \cos \frac{\pi }{n}} \right)^{2} , \hfill \\ M_{7} = \sigma_{t} r_{1}^{^{\prime}} \sin \frac{\pi }{n}\left( { - r_{1}^{^{\prime\prime}} + r_{1}^{^{\prime}} \cos \frac{\pi }{2n}} \right)^{2} , \hfill \\ M_{8} = 2\sigma_{t} \int_{0}^{{\frac{\pi }{n}}} {\left( {r_{1} \cos (\frac{\pi }{2n}) - r_{1}^{^{\prime\prime}} \cos \varphi } \right)} \left( {r_{1}^{^{\prime\prime}} \cos \varphi \tan \frac{\pi }{n} - r_{1}^{^{\prime\prime}} \sin \varphi } \right)r_{1}^{^{\prime\prime}} \sin \varphi {\text{d}} \varphi . \hfill \\ \end{gathered}$$

When $y_{C} > y_{F} > y_{E}$, as shown in Fig. 14c, the total bending moment of all the strips of the tube is

$$M_{p} = n\left( {M_{9} + M_{10} + M_{11} + M_{12} } \right),$$

(A3)

where

$$\begin{gathered} M_{9} = 2\sigma_{t} \int_{0}^{{\omega_{1} }} {\left( {r_{1}^{^{\prime}} \cos \omega - r_{1} \cos \frac{\pi }{2n}} \right)} \left( {r_{1}^{^{\prime}} \sin \omega } \right)^{2} {\text{d}} \omega , \hfill \\ M_{10} = 2\sigma_{t} \int_{{\omega_{1} }}^{{\frac{\pi }{n}}} {\left( {r_{1} \cos \frac{\pi }{2n} - r_{1}^{^{\prime}} \cos \omega } \right)} \left( {r_{1}^{^{\prime}} \sin \omega } \right)^{2} {\text{d}} \omega , \hfill \\ M_{11} = \sigma_{t} r_{1}^{^{\prime}} \sin \frac{\pi }{n}\left( { - r_{1}^{^{\prime\prime}} + r_{1}^{^{\prime}} \cos \frac{\pi }{2n}} \right)^{2} , \hfill \\ M_{12} = 2\sigma_{t} \int_{0}^{{\frac{\pi }{n}}} {\left[ {r_{1} \cos \left( {\frac{\pi }{2n}} \right) - r_{1}^{^{\prime\prime}} \cos \varphi } \right]} \left( {r_{1}^{^{\prime\prime}} \cos \varphi \tan \frac{\pi }{n} - r_{1}^{^{\prime\prime}} \sin \varphi } \right)r_{1}^{^{\prime\prime}} \sin \varphi {\text{d}} \varphi . \hfill \\ \end{gathered}$$

When $y_{E} > y_{C} > y_{F}$, as shown in Fig. 14d, the total bending moment of all the strips of the tube is

$$M_{p} = n\left( {M_{13} + M_{14} + M_{15} + M_{16} } \right),$$

(A4)

where

$\begin{gathered} M_{13} = 2\sigma_{t} \int_{0}^{{\omega_{2} }} {\left( {r_{1}^{^{\prime}} \cos \omega - \cos \frac{\pi }{2n}} \right)} \left( {r_{1}^{^{\prime}} \sin \omega } \right)^{2} {\text{d}} \omega , \hfill \\ M_{14} = 2\sigma_{t} \int_{{\omega_{2} }}^{{\omega_{1} }} {\left( {r_{1}^{^{\prime}} \cos \omega - r_{1} \cos \frac{\pi }{2n}} \right)} \left[ {r_{1}^{^{\prime}} \sin \omega - r_{1}^{^{\prime\prime}} \sin \left( {\cos^{ - 1} \frac{{r_{1}^{^{\prime}} \cos \omega }}{{r_{1}^{^{\prime\prime}} }}} \right)} \right]r_{1}^{^{\prime}} \sin \omega {\text{d}} \omega , \hfill \\ M_{15} = 2\sigma_{t} \int_{{\omega_{2} }}^{{\frac{\pi }{n}}} {\left( {r_{1} \cos \frac{\pi }{2n} - r_{1}^{^{\prime}} \cos \omega } \right)} \left[ {r_{1}^{^{\prime}} \sin \omega - r_{1}^{^{\prime\prime}} \sin \left( {\cos^{ - 1} \frac{{r_{1}^{^{\prime}} \cos \omega }}{{r_{1}^{^{\prime\prime}} }}} \right)} \right]r_{1}^{^{\prime}} \sin \omega {\text{d}} \omega , \hfill \\ M_{16} = 2\sigma_{t} \int_{{\varphi_{0} }}^{{\frac{\pi }{n}}} {\left( {r_{1} \cos \frac{\pi }{2n} - r_{1}^{^{\prime\prime}} \cos \varphi } \right)} \left( {r_{1}^{^{\prime\prime}} \cos \varphi \tan \frac{\pi }{n} - r_{1}^{^{\prime\prime}} \sin \varphi } \right)r_{1}^{^{\prime\prime}} \sin \varphi {\text{d}} \varphi . \hfill \\ \end{gathered}$

Appendix 2: Derivation of solving for the curling radius

This appendix is the derivation of solving for the curling radius.

Neglecting the compression of the metal foam, according to the conservation of energy [15],

$$F\dot{w} = M_{p} \frac{{\dot{w}}}{{R + r(1 - \cos \frac{\pi }{2n})}} + n\sigma_{t} H^{2} \dot{w} + \frac{\mu F}{{\mu \cos \alpha + \sin \alpha }}\dot{w}$$

(B1)

or

$$F = \frac{\mu \cos \alpha + \sin \alpha }{{\mu \cos \alpha + \sin \alpha - \mu }}\left[ {\frac{{M_{p} }}{{R + r(1 - \cos \frac{\pi }{2n})}} + n\sigma_{t} H^{2} } \right],$$

(B2)

where $\beta$ is the angle between the beginning of bending and the position of splitting. According to the geometrical relationship [15],

$$\beta = \cos^{ - 1} (1 - \frac{nH}{{2\pi R}}).$$

(B3)

The bending moment at the crack tip angle $\beta$ is

$$M_{P} = NR\sin (\alpha - \beta ) - \mu NR[1 - \cos (\alpha - \beta )]$$

(B4)

as shown in Fig. 5b. Thus,

$$N = \frac{{M_{p} }}{R\sin (\alpha - \beta ) - \mu R[1 - \cos (\alpha - \beta )]}.$$

(B5)

By combining Eqs. (3) and (B5),

$$\frac{F}{\mu \cos \alpha + \sin \alpha } = \frac{{M_{p} }}{R\sin (\alpha - \beta ) - \mu R[1 - \cos (\alpha - \beta )]}.$$

(B6)

Combining (B2) and (B6), the curling radius R can be obtained.