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How do I work out the forces in an impact of a steel ball and a plate?
Structural Calculations
In the special case of a ball or cylinder bouncing off a hard plate the answer is complex.
Not very surprisingly Timoshenko covers this problem in "Theory of Elasticity".
Unfortunately I didn't copy all the right equations down, so I'll just solve the simpler case of a steel ball bouncing off its twin. If you think about that case it is likely that this is also the same as the steel ball bouncing off an infinitely stiff flat plane.
The time of the contact is
t=2.94(1.25*sqrt(2)*pi*rho*(1-nu^2)/E)^0.4*R*v^-0.2 eqn 244
R=0.031 m
v=3.132 m/s
rho=7843 kg m-3
E=210*10^9 N m-2
nu=0.3
t=0.1493 ms
The average force, F during the contact is 2*m*v/t (ie the change of momentum divided by the time)
F=42 kN, ie a little over 4 tons force.
Timoshenko actually gives a direct solution for a ball on a flat plate, and the peak force rather than the average force, but it is spread over two pages. This solution assumes that the contact time is long compared with the period of the lowest modes of vibration.
Not very surprisingly Timoshenko covers this problem in "Theory of Elasticity".
Unfortunately I didn't copy all the right equations down, so I'll just solve the simpler case of a steel ball bouncing off its twin. If you think about that case it is likely that this is also the same as the steel ball bouncing off an infinitely stiff flat plane.
The time of the contact is
t=2.94(1.25*sqrt(2)*pi*rho*(1-nu^2)/E)^0.4*R*v^-0.2 eqn 244
R=0.031 m
v=3.132 m/s
rho=7843 kg m-3
E=210*10^9 N m-2
nu=0.3
t=0.1493 ms
The average force, F during the contact is 2*m*v/t (ie the change of momentum divided by the time)
F=42 kN, ie a little over 4 tons force.
Timoshenko actually gives a direct solution for a ball on a flat plate, and the peak force rather than the average force, but it is spread over two pages. This solution assumes that the contact time is long compared with the period of the lowest modes of vibration.