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Calculating the impact force of a mass falling on an elastic structure T. J. Mackin Dept of M&IE The University of Illinois 1206 W. Green St. Urbana, IL 61802 [email protected] References: D. R. H. Jones, Engineering Materials 3, Materials Failure Analysis, Case Studies and Design Implications, Pergamon Press, NY. Consider a mass, M, falling under the influence of gravity from a height, d, onto an elastic structure of cross-sectional area, A and elastic modulus, E, Figure 1. In order for the structure to arrest the mass it must absorb all the energy of the mass. The energy of the falling mass is easily calculated by considering the change in gravitational potential energy. Following impact, the structure deforms by an amount u M d L Figure 1. The falling mass, M, descends an amount d. Once it hits the structure, it continues to descend by an amount u until all the impact energy is absorbed. If one knows the mass of the falling structure, one can calculate the strain required to arrest it. In the present case, we know the system failed, so we estimate the failure strain using a typical value when materials behavior becomes non-linear. The mass free-falls over a distance d and contacts the structure. It then falls by an additional amount, u, after contact with the structure. The deformation of the structure, through the distance u is determined by the elastic deformation of the structure it falls upon. I presume that, once the elastic limit of the structure is exceeded, the structure will fail, likely through buckling, or certainly through plastic deformation. There certainly can be a large amount of energy absorption through plastic deformation, but the forces we will calculate, up to the proportional limit, are so large that collapse is going to be a done deal and we needn’t worry about the exact magnitude of the ensuing deformation and the relative amounts of plastic deformation. So, back to this deflection u: after the mass contacts the structure it deforms the structure by an amount that depends upon the stiffness of the structure. Treat the structure like a spring of stiffness k. Then we know that the force is related to deflection, u, through the spring stiffness using: F = ku (1) If we divide by the cross sectional area, we obtain: F k = u A A (2) Now, upon noting that the deformation of the structure is related to the strain in the structure, we obtain a relation between u, the standing height of the structure, L, and the strain, ε: u = εL (3) Now that we have the deflection of the structure, under the influence of a force, F, we can relate the impact energy to the deformations to extract the impact force, F. In order to arrest the falling mass we must absorb all the energy of that mass. The mass falls through a total drop height of d+u, so the energy that must be absorbed is given by: U = mg (d + u ) (4) Now, replace u with the expression given in (3) to obtain: U = mg (d + ε ⋅ L ) (5) This energy is absorbed through deformation of the structure, approximated up to the elastic limit as: 1 U elastic = σ ⋅ε ⋅ A ⋅ L 2 (6) Where AL is the volume of the structural material. Now, recall by definition, that the stress is related to the force by: σ= F A (7) So, equating (5) and (6) but using the definition (7), we obtain and energy balance equation, as follows: mg (d + ε ⋅ L ) = 1 F ⋅ ε ⋅A⋅L 2 A (8) If we simplify this expression, by eliminating A, multiplying thru by 2, and dividing through by ε and L, we obtain: F = 2 ⋅ mg ( d + 1) ε ⋅L (9) This is the anticipated impact force for strains up to a chosen value. It is reasonable to assume that strains that are at, or very near yield, represent the end of recoverable deformation, so I used a strain value of ε=0.001 in my calculation of the impact force.