Analytical and numerical modelling of repeated impacts on elastic-strain hardening beams

Abstract

The repeated impact problem of a clamped elastic, strain-hardening plastic beam repeatedly impinged by a rigid heavy wedge with a low velocity at the mid-span is studied analytically and numerically. The beam motion is described using a Single-Degree-of-Freedom mass-spring model, governed by two structural parameters, i.e., structural resistance and the equivalent mass of beam. The explicit expressions of resistance functions of the beam during loading/unloading/reloading process are obtained from a series of simplified nonlinear quasi-static analysis with material strain hardening being taken into account. The equivalent mass of beam is related to the assumed transverse displacement profile of the beam which varies with its elastic–plastic state. Thereafter, the analytical solutions of the repeated impact response of beams made from the elastic-linear hardening (bi-linear) material are well validated by the detailed numerical simulations obtained by using ABAQUS/Explicit. Additional theoretical and numerical investigations with various tangent modulus values reveal that strain hardening can increase the elastic strain energy absorbed by the beam, but it has little influence on the duration of each impact.

Appendix: Determination of the elastic behavior of the clamped beam

The elastic axial force and the elastic bending moment of the beam are given as (Campbell and Charlton 1973):

$$ N = AE\frac{{\pi^{2} }}{16}\left( \frac{W}{L} \right)^{2} $$

(31)

$$ M = \frac{6EI}{{L^{2} }}W $$

(32)

where A = BH is the cross-sectional area of the beam and I = BH³/12 is the section moment of inertia.

Substituting Eqs. (31) and (32) into Eq. (5), we have

$$ R = \frac{24EI}{{L^{3} }}W + AE\frac{{\pi^{2} }}{8}\left( \frac{W}{L} \right)^{3} . $$

(33)

The axial force N in Eq. (31) is determined from the deformation shape function of the beam in the elastic regime:

$$ w\left( x \right) = \frac{W}{2}\left[ {1 + \cos \left( {\frac{\pi x}{L}} \right)} \right]. $$

(34)

Then, the equivalent mass of the beam is related to total kinetic energy of the beam (Sha and Hao 2014):

$$ \frac{1}{2}m_{b}^{e} \dot{W}^{2} = \int_{0}^{L} {\mu \dot{w}^{2} \left( x \right)} $$

(35)

where $\dot{w}\left( x \right)$ is the assumed transverse velocity profile and μ = ρBH is the beam mass per unit length.

Thus, Eq. (35) together with Eq. (34) predicts:

$$ m_{b}^{e} = \frac{3}{4}\mu L. $$

(36)