Think Fencing Impact Resistance Test.
Let:
m be the mass of the Striker,
D be the diameter of the Striker,
h1 be the initial height of the Striker,
h2 be the rebound height of the Striker,
h3 be the height of the test specimen above the ground,
g be the acceleration due to gravity,
t be the time it takes the Striker to reach the specimen, i.e.
the time it takes to traverse h1,
v(t) be the velocity of the Striker at time t,
x(t) be the position of the Striker at time t,
p1 be the momentum of the Striker on its downwards journey,
p2 be the momentum of the Striker on its upwards, rebound
journey,
p be the change in momentum of the Striker,
F be the impact force exerted on the specimen by the Striker
during the initial collision and
be the total stress the test specimen experiences during the
initial collision.
Newtons Second Law of Motion gives
= () [Minus sign denotes downwards motion] which, upon
integration gives the velocity as
= (2) Furthermore we find the position of the Striker at time,
t, is
= (3)
By setting = and solving for t we can find the fall time of the
Striker. Doing this yields = / ()
So the velocity at the moment of impact is given by substituting
(4) into (2), = () Hence, the momentum of the Striker at impact is
! = () Now, to find the rebound momentum of the Striker let us
consider the kinetic and potential energy of the Striker. At impact
the potential energy is zero while at the peak of the Strikers
rebound journey, the kinetic energy is zero. Since energy is
conserved here, we can equate the potential energy at the peak of
the rebound journey with the kinetic energy at the start of the
rebound journey, and solve for the rebound velocity, and thus
momentum. We get = Thus, the change in momentum is (7)-(6) i.e.
= + Now, the force experienced by the test specimen during a
collision is equal and opposite to that
experienced by the Striker. Using the definition of a force, =
!"!" . Since the test specimen can deform at most h3 during impact
and h3h1 we will assume that the duration of the impact is
approximately twice the time it takes the Striker to travel h3 at
the velocity at time of impact, that is,
= Therefore, dividing (8) by (9) we get the force exerted on the
test specimen during a Think Fencing
Impact Test;
= + () Now, the area of impact is
. So the stress loading, endured by the test specimen during
impact is given by
= = + ()