A pile driven into ground using a hammer question

[paul9619]

A pile driven into ground using a hammer question!

Hi all,

I have been attempting the following question, could you check my answers /method please :)

A pile is driven into the ground by a hammer of mass 500Kg's dropped from a height of 4 meters. The mass of the pile is 40Kg's and the average resistance of the ground penetration is 45K Newtons.

1. Calculate the velocity and Kinetic energy of the hammer prior to the impact with the pile.

I used the formula v^2=u^2+2as to give me 8.8589 m/s. For the KE I used 1/2mv^2 giving me 19620 Joules.

2. Calculate the velocity and Kinetic energy of the hammer/pile combination immediately after the impact.

By using the momentum of pile and hammer after impact = the momentum before impact. (Let v1 = Common velocity after impact)

I got (500KG + 40KG)v1 = 500KG x 8.8589m/s (velocity before)
By transposing I get v1 = 8.203m/s.

Now for kinetic energy I used 1/2mv^2 again but this time I used the total mass of pile & hammer and the comman velocity after impact to give me 18168J.

3. Calculate the distance for which the pile penetrates the ground??

I am really struggling with this question, cannot seem to find a formula to link the distance driven into the ground?? Can anyone help?

 

[radou]

...
3. Calculate the distance for which the pile penetrates the ground??

I am really struggling with this question, cannot seem to find a formula to link the distance driven into the ground?? Can anyone help?

Since you are given the average resistance of the ground penetration, the pile will keep sliding into the ground until the resistance force and the force which drives the pile into the ground are in equilibrium.

 

[paul9619]

Since you are given the average resistance of the ground penetration, the pile will keep sliding into the ground until the resistance force and the force which drives the pile into the ground are in equilibrium.

and now in english

 

[radou]

and now in english

Use the fact that the work of all the forces (gravity and resistance) equals the change of kinetic energy.

 

[paul9619]

Sorry still unsure where the distance comes into it. I must be having one of my thick days

 

[junior_J]

work = force x distance = change in energy

 

[radou]

Sorry still unsure where the distance comes into it. I must be having one of my thick days

Something like that: (G - R)*s = change in kinetic energy. (G is the weight, R is the force of resistance, and s is the displacement.)

 

[lostkid]

So, is the answer 3.6600 cm?

 

[Ghan]

I sure can help you. It is a civil engineering question.

Firstly, using 500 kg hammer dropped from a whoping 4m on a measly 40kg pile will really put too much undue stress on you pile.

Secon you need to consider the efficiency of the hammer. The initial momentum does not equal the final momentum in this case.

You need to is use this equation to work out the pile displacement.

Load ,p=45kN=(0.8*500*9.81*4)/(s) and solve for s, which is the pile displacement.

0.8 is the assumed efficiency of the hammer. It may be even lower, depending on the type and the setup.

Hope it helps

Hi all,

I have been attempting the following question, could you check my answers /method please :)

A pile is driven into the ground by a hammer of mass 500Kg's dropped from a height of 4 meters. The mass of the pile is 40Kg's and the average resistance of the ground penetration is 45K Newtons.

1. Calculate the velocity and Kinetic energy of the hammer prior to the impact with the pile.

I used the formula v^2=u^2+2as to give me 8.8589 m/s. For the KE I used 1/2mv^2 giving me 19620 Joules.

2. Calculate the velocity and Kinetic energy of the hammer/pile combination immediately after the impact.

By using the momentum of pile and hammer after impact = the momentum before impact. (Let v1 = Common velocity after impact)

I got (500KG + 40KG)v1 = 500KG x 8.8589m/s (velocity before)
By transposing I get v1 = 8.203m/s.

Now for kinetic energy I used 1/2mv^2 again but this time I used the total mass of pile & hammer and the comman velocity after impact to give me 18168J.

3. Calculate the distance for which the pile penetrates the ground??

I am really struggling with this question, cannot seem to find a formula to link the distance driven into the ground?? Can anyone help?

 

[technician]

I would use work done by friction force = change in energy.
I would take the total change in energy to be the PE of the hammer (it moves a vertical distance = 4 + d where d is penetration) plus the loss in PE of the stake (it moves d).
The work done by friction is F x d and this should give an equation which can be solved for d ( I got 0.49m)

 

[Ghan]

0.49m is too much.


There a few ways to do this. Please bear with me and I will give you my answer as soon as I am free.

I would use work done by friction force = change in energy.
I would take the total change in energy to be the PE of the hammer (it moves a vertical distance = 4 + d where d is penetration) plus the loss in PE of the stake (it moves d).
The work done by friction is F x d and this should give an equation which can be solved for d ( I got 0.49m)