00:02
In this problem, we have a hammer of a given mass of a pile driver that falls from a certain height onto the top of a pile and drives it into the ground a certain distance.
00:14
Assuming a perfectly elastic, or sorry, a perfectly plastic impact, we want to determine the average resistance of the ground to penetration.
00:24
So let's look at the velocity of the hammer at impact.
00:34
So as you can see from our diagram, the hammer starts from rest and reaches some maximum velocity just before it impacts the pile.
00:46
So we'll use the conservation of energy to calculate this velocity.
00:58
Now the initial kinetic energy of the hammer is zero.
01:00
So we'll call that t1.
01:02
T1 is equal to zero.
01:04
The potential energy of the hammer vh is equal to its mass.
01:10
X x times 1 .2 the hammer falls 1 .2 meters before striking the pile and so this is 650 kg the mass of the hammer times g 9 .81 times 1 .2 meters and so the gravitational potential energy initially since this is just for the hammer we'll call it v1 v1 is equal to 7 ,652 joules just before the hammer strikes the pile it has kinetic energy a half m vh squared and so again we can substitute our values in here this is 650 over 2 which is the mass times vh squared so that's simply 325 times the speed that we reach to calculate vh squared and the final potential energy, gravitational potential energy is zero since we take the datum at the top of the pile.
02:46
So now if we apply the conservation of energy, we know that the kinetic plus the gravitational potential energy before the hammer falls must equal to its kinetic and potential energies at its point just about the pile.
03:07
So initially the kinetic energy is 0, v1, 7 ,652, and this is equal to the final kinetic energy of the hammer 325 vh squared.
03:25
And so if we rearrange this equation, you can find the speed of the hammer just before the impact vh to be 4 .852 meters per second.
03:39
So that's the velocity of the hammer before the impact.
03:46
Now let's look at the velocity of the pile after impact.
04:03
So in this case we'll use the conservation of momentum.
04:09
Now the ground reaction and the weights are non -impulsive.
04:14
So we have the mass of the hammer plus the speed at which the hammer strikes the pile.
04:25
Mhvh is equal to the mass of the hammer.
04:28
Of the hammer plus the mass of the pile mh plus mp and they move off with a common velocity which we'll call v prime and so if we rearrange this we can find v prime to be the mass of the hammer times the speed of the hammer so that's mh vh over the mass of the hammer and the pile together mh plus mp and so if we thought to chew our values into this into this equation we get this to be 650 over 650 plus 140 multiplied by the speed that the hammer strikes the pile for 0 .852 we get the common velocity of the hammer and pile after the impact to be 3 .992 meters per second...