Download Calculating the impact force of a mass on an elastic

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.