Pile driver - Conservation of enegy

In summary: I feel like I might actually be able to solve this!In summary, In attempting to solve a problem, the individual used two different methods to calculate the outcome. The first method used the conservation of momentum principle and arrived at a result of 27513N. The second method used energy and arrived at a result of 1310.44J. If the problem statement is assumed to be that the pile driver has a velocity before it impacts the pile, then the first method is the most accurate.
  • #1
darkminos
8
0
Hi,

I did some calculations but somehow I feel it's not 100% correct. Could someone have a look please.

Many Thanks

"Task"
A pile driver of mass mh has struck a pile of mass mp and driven the pile D meters into the ground, if the velocity of the strike is V, determine the force due to the resistance in the ground by the following methods:
1. D’Alembers Principle
2. Conservation of energy

V = 3.62m/s
mh = 100kg
mp = 200kg
D = 0.08m
g = 9.81 m/s^2



SOLUTION:


v^2 = u^2 + 2as

where
v = initial velocity
u = final velocity
a = acceleration
s = displacement

transposed the formula to get the acceleration.

Deceleration
a = v^2 - u^2 / 2s
a = 3.62^2 - 0 / 2 * 0.08
a = -81.9 m/s^2

F = ma
F = (200 + 100) * 81.9
F = 24570N

Ground Resistance
R = 300 * 9.81 + 24570
R = 27513N

For the second part I'm trying to work out the Potential and Kinetic energy but then i have no idea what to do next...


K.E = 1/2 mv^2
K.E = 1/2 * 200 * 3.62^2
K.E = 1310.44J

Height of fall
1/2mv^2 = mgh

Transposed to make h the subject

h= 1/2*mv^2 / mg
h = (300 * 3.62^2) / 2+ (200 * 9.81)
h = ~1m


P.E = mgh
P.E = 200 * 9.81 * 1
P.E = 1962J
 
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  • #2
Since the pile has mass, I am unsure from the problem if the given 'strike' velocity is the same as the initial velocity of the driver-pile system as they start their decent into the ground. If it is, I agree with your solution in the first part for R= 27,513 N. To solve using energy methods, remember to include the work done by the ground resistance force (delta KE plus delta PE + W =0). And check the value you are using for the mass.
 
  • #3
PhanthomJay said:
Since the pile has mass, I am unsure from the problem if the given 'strike' velocity is the same as the initial velocity of the driver-pile system as they start their decent into the ground. If it is, I agree with your solution in the first part for R= 27,513 N. To solve using energy methods, remember to include the work done by the ground resistance force (delta KE plus delta PE + W =0). And check the value you are using for the mass.

That’s the conclusion which I came to early today :) and it works out... now how to delete this post, as there are people on my course that won't do the work just Google it and copy the work that took me 3 days to do!

Thanks for the help!
 
  • #4
I guess I can delete my post if you delete yours, but a good teacher or TA or Prof will sort out the 'cheaters' from those who have spent hours of work on the problem. But we don't really have an answer here yet; we've assumed the 'strike' velocity to be the same as the initial velocity of the driver/pile system, but if the problem defines 'strike' velocity as the speed of the driver head just before impact, then you need to first apply conservation of momentum principles to determine the initial speed of the driver/pile system as if the collision of the two were totally inelastic (that is, they join together during impact and proceed downward into the ground as one unit during the 0.08m decent). I think the problem statement is not clear.
 
  • #5
I will actually gamble at this stage and stick with the first answer of 27513N i really hope this is right...
 
  • #6
I would have assumed that the "velocity of the strike" was the speed of the pile driver just before it collided with the pile.
 
  • #7
Doc Al said:
I would have assumed that the "velocity of the strike" was the speed of the pile driver just before it collided with the pile.
I'm inclined to agree.
 
  • #8
damn it... back to work than...
 
  • #9
ok I redone my calculations and now the results from the first method don't match the results of the second... Another remark from my side - I don’t expect this task to by "super duper" complex. While if I apply the numbers as assumed at the beginning it all works out with only a difference of 0.27.
 
  • #10
You've done something wrong. In your first attempt , when using the initial velocity of 3.62, you arrived at the same answer for R using Newton's laws versus Energy methods. Now in your new attempt, you need to calculate the initial velocity based on the conservation of momentum principle, then the steps after that to solve for R by either method, are the same. Only V_initial changes.
 
  • #11
That does look ok now, If u assume that the velocity given is the velocity before strike :) I can always argue the case that the problem was not stated clear enough! But for the first time actually now I feel like this is it.

Still waiting for comments before I delete this post!

Many Thanks
Dark
 
  • #12
darkminos said:
mementum before impact = (100+200)*3.62 = 362Ns
you mean of course (100)(3.62) = 362 N-s
TA-DA!??
Looks good, so I guess its OK to sound the trumpet, on the assumption that Doc Als' assumption, which I agree with, is correct. What sort of a gambler are you?
 
  • #13
PhanthomJay said:
you mean of course (100)(3.62) = 362 N-sLooks good, so I guess its OK to sound the trumpet, on the assumption that Doc Als' assumption, which I agree with, is correct. What sort of a gambler are you?

Thanks for showing me that "(100)(3.62) = 362 N-s".
I'm the type of gambler that will make one paper for each assumption (my initial and Doc Als'), handle them in + an A4 page with an explanation why there are two papers :)
 
  • #14
If you deleted your original post, you would only have had one paper to submit, probably the wrong one. Just something to think about.:wink:
 
  • #15
yup! you right, good that I didn't now how :) ...

Thanks for the help!
 

1. What is a pile driver?

A pile driver is a large machine used in construction to drive piles (long, slender columns) into the ground to provide support for buildings, bridges, or other structures. It typically consists of a heavy hammer, called a ram, that is lifted and then dropped onto the pile, driving it into the ground.

2. How does a pile driver conserve energy?

A pile driver conserves energy through the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred or converted. In this case, the potential energy of the lifted ram is converted into kinetic energy when it is dropped onto the pile, driving it into the ground. This energy transfer allows the pile driver to efficiently and effectively drive piles into the ground without using excessive amounts of energy.

3. What factors affect the conservation of energy in a pile driver?

The conservation of energy in a pile driver is affected by several factors, including the weight and height of the ram, the type and condition of the pile, and the surface and soil conditions where the pile is being driven. Additionally, the efficiency of the pile driver itself, including the design and materials used, can also impact energy conservation.

4. Are there any alternative methods for driving piles that conserve energy?

Yes, there are alternative methods for driving piles that also conserve energy. Some examples include using hydraulic or pneumatic systems to drive the piles, as well as using vibratory methods that use high-frequency vibrations to drive the piles into the ground. These methods often require less energy and can also be more efficient and precise than traditional pile driving methods.

5. How important is energy conservation in pile driving?

Energy conservation is a crucial aspect of pile driving, not only to reduce the cost and environmental impact of construction projects, but also to ensure the safety and stability of the structures being built. By conserving energy and using efficient pile driving methods, we can minimize the risk of damage to the surrounding environment and structures, as well as reduce the overall cost and time of construction projects.

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