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Radial Inertia Effects in Kolsky Bar Testing of Extra-soft Specimens

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Abstract

Impact responses of extra-soft materials, such as ballistic gelatins and biological tissues, are increasingly in demand. The Kolsky bar is a widely used device to characterize high-rate behavior of materials. When a Kolsky bar is used to determine the dynamic compressive response of an extra-soft specimen, a spike-like feature often appears in the initial portion of the measured stress history. It is important to distinguish whether this spike is an experimental artifact or an intrinsic material response. In this research, we examined this phenomenon using experimental, numerical and analytical methods. The results indicate that the spike is the extra stress from specimen radial inertia during the acceleration stage of the axial deformation. Based on this understanding, remedies in both specimen geometry and loading pulse to minimize the artifact are proposed and verified, and thus capture the intrinsic dynamic behavior of the specimen material.

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Acknowledgements

This research was supported by US Army Research Laboratory (ARL) through a collaborative research agreement with Purdue University. The authors wish to thank Dr. Michael Scheidler of ARL for his efforts in the proofreading of the analytical modeling work.

Author information

Authors and Affiliations

  1. School of Aeronautics and Astronautics and School of Materials Engineering, Purdue University, West Lafayette, IN, 47907, USA

    B. Song (SEM member) & W. W. Chen (SEM member)

  2. School of Aeronautics and Astronautics, Purdue University, West Lafayette, IN, 47907, USA

    Y. Ge

  3. Impact Physics Branch, US Army Research Laboratory, Aberdeen Proving Ground, MD, 21005, USA

    T. Weerasooriya (SEM member)

Authors
  1. B. Song
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  2. Y. Ge
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  3. W. W. Chen
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  4. T. Weerasooriya
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Corresponding author

Correspondence to W. W. Chen.

Appendix: Analysis of Radial Inertia in an Annular Specimen

Appendix: Analysis of Radial Inertia in an Annular Specimen

General Hooke’s law for linear elastic materials is

$$ E_{{\text{s}}} \varepsilon _{r} = \sigma _{r} - \nu {\left( {\sigma _{z} + \sigma _{\theta } } \right)} $$
(19)
$$ E_{{\text{s}}} \varepsilon _{\theta } = \sigma _{\theta } - \nu {\left( {\sigma _{z} + \sigma _{r} } \right)} $$
(20)
$$ E_{{\text{s}}} \varepsilon _{z} = \sigma _{z} - \nu {\left( {\sigma _{\theta } + \sigma _{r} } \right)} $$
(21)

where E s is the Young’s modulus and ν is the Poisson’s ratio. The Hooke’s law was used to convert the strains into stress components. The resultant stress components must satisfy the equation of motion equation (15)

$$ \frac{{\partial \sigma _{r} }} {{\partial r}} = - \rho _{{\text{s}}} {\left[ {\frac{r} {2}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} + \frac{{{\mathop C\limits^{ \cdot \cdot } }_{1} {\left( t \right)}}} {r}} \right]} - \frac{{\sigma _{r} - \sigma _{\theta } }} {r} = - \rho _{{\text{s}}} {\left[ {\frac{r} {2}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} + \frac{{{\mathop C\limits^{ \cdot \cdot } }_{1} {\left( t \right)}}} {r}} \right]} - \frac{{4E_{{\text{s}}} C_{1} {\left( t \right)}}} {{3r^{3} }} $$
(22)

Integration of equation (22) yields

$$ \sigma _{r} = - \rho _{{\text{s}}} {\left[ {\frac{{r^{2} }} {4}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} + \ln {\left( r \right)}{\mathop C\limits^{ \cdot \cdot } }_{1} {\left( t \right)}} \right]} + \frac{{2E_{{\text{s}}} C_{1} {\left( t \right)}}} {{3r^{2} }} + C_{2} {\left( t \right)} $$
(23)

Corresponding stress components in the other two directions from equations (19), (20), (21) are

$$ \sigma _{\theta } = \frac{4} {3}E_{{\text{s}}} \varepsilon _{\theta } + \frac{2} {3}E_{{\text{s}}} \varepsilon _{z} + \sigma _{r} $$
(24)
$${\,\matrix{ {{\sigma _{z} = {4 \over 3}E_{{\rm{s}}} \varepsilon _{z} + {2 \over 3}E_{{\rm{s}}} \varepsilon _{\theta } + \sigma _{r} } \hfill} \cr {{ = {4 \over 3}E_{{\rm{s}}} \varepsilon _{z} + {2 \over 3}E_{{\rm{s}}} {\left[ { - {{C_{1} {\left( t \right)}} \over {r^{2} }} - {1 \over 2}\varepsilon _{z} } \right]} - \rho _{{\rm{s}}} {\left[ {{{r^{2} } \over 4}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} + \ln {\left( r \right)}{\mathop C\limits^{ \cdot \cdot } }_{1} {\left( t \right)}} \right]} + {{2E_{{\rm{s}}} C_{1} {\left( t \right)}} \over {3r^{2} }} + C_{2} {\left( t \right)}} \hfill} \cr {{ = E_{{\rm{s}}} \varepsilon _{z} - \rho _{{\rm{s}}} {\left[ {{{r^{2} } \over 4}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} + \ln {\left( r \right)}{\mathop C\limits^{ \cdot \cdot } }_{1} {\left( t \right)}} \right]} + C_{2} {\left( t \right)}} \hfill} \cr } }$$
(25)

Both the outer and inner hoop surfaces are stress-free, so the boundary conditions are σ r (r = b) = 0 and σ r (r = a) = 0. These boundary conditions and equation (23) lead to

$$ \sigma _{r} = - \rho _{{\text{s}}} {\left[ {\frac{{a^{2} }} {4}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} + \ln {\left( a \right)}{\mathop C\limits^{ \cdot \cdot } }_{1} {\left( t \right)}} \right]} + \frac{{2E_{{\text{s}}} C_{1} {\left( t \right)}}} {{3a^{2} }} + C_{2} {\left( t \right)} = 0 $$
(26)
$$ \sigma _{r} = - \rho _{{\text{s}}} {\left[ {\frac{{b^{2} }} {4}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} + \ln {\left( b \right)}{\mathop C\limits^{ \cdot \cdot } }_{1} {\left( t \right)}} \right]} + \frac{{2E_{{\text{s}}} C_{1} {\left( t \right)}}} {{3b^{2} }} + C_{2} {\left( t \right)} = 0 $$
(27)

Combination of these two equations results in

$$ \rho _{{\text{s}}} {\left[ {\frac{{b^{2} - a^{2} }} {4}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} + \ln {\left( {\frac{b} {a}} \right)}{\mathop C\limits^{ \cdot \cdot } }_{1} {\left( t \right)}} \right]} + \frac{{2{\left( {b^{2} - a^{2} } \right)}E_{{\text{s}}} C_{1} {\left( t \right)}}} {{3a^{2} b^{2} }} + C_{2} {\left( t \right)} = 0 $$
(28)

which has the solution when \( {\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} \) is a constant and t < t m

$$ C_{1} = A\cos \lambda t + B\sin \lambda t + m $$
(29)

where \( \lambda = {\sqrt {\frac{{2E_{s} {\left( {b^{2} - a^{2} } \right)}}} {{3\rho _{s} \ln {\left( {\frac{b} {a}} \right)}b^{2} a^{2} }}} } \).

Because the specimen is initially at rest, the initial displacement and velocity of a material particle must vanish: u| t−0 = 0 and \( \left. {{\mathop u\limits^ \cdot }} \right|_{{t = 0}} = 0 \), which lead to

$$ C_{1} {\left( 0 \right)} = 0,\,{\text{and}}\;{\mathop C\limits^ \cdot }_{1} {\left( {t = 0} \right)} = 0 $$
(30)

Thus \( m = - \frac{{3\rho _{{\text{s}}} a^{2} b^{2} }} {{8E_{{\text{s}}} }}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} = - A \), and B = 0. And we have

$$ C_{1} {\left( t \right)} = m{\left( {1 - \cos \lambda t} \right)} = - \frac{{3\rho _{{\text{s}}} a^{2} b^{2} }} {{8E_{{\text{s}}} }}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} {\left( {1 - \cos \lambda t} \right)} $$
(31)

Substitution of equation (31) into equation (28) results in

$$ C_{2} = {\left( {\rho _{s} \frac{{a^{2} }} {4}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} - \frac{{2E_{s} }} {{3a^{2} }}m} \right)} + {\left( {\rho _{s} \ln {\left( a \right)}\lambda ^{2} + \frac{{2E_{s} }} {{3a^{2} }}} \right)}m\cos \lambda t $$
(32)

Finally, the axial stress in the specimen is found to be

$${\sigma _{z} = E_{s} \varepsilon _{z} + {\left\{ {\rho _{s} {\left( {{{a^{2} + b^{2} - r^{2} } \over 4}} \right)} - {{3\rho _{s} a^{2} b^{2} } \over {8E_{s} }}{\left[ {\rho \ln {\left( {{a \over r}} \right)}\lambda ^{2} + {{2E_{s} } \over {3a^{2} }}} \right]}\cos \lambda t} \right\}}{\mathop \varepsilon \limits^{ \cdot \cdot } }_{z} }$$
(33)

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Song, B., Ge, Y., Chen, W.W. et al. Radial Inertia Effects in Kolsky Bar Testing of Extra-soft Specimens. Exp Mech 47, 659–670 (2007). https://doi.org/10.1007/s11340-006-9017-5

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