The Pile Driver?

Question

Watching piles being driven into the ground is an interesting sight -- the huge machine, the chug and puff and hiss of the engine and then the bang of the hammer. Suppose the hammer and the pile each weigh one ton. Then suppose the hammer drops on the pile from a height of two feet and the impact drives the pile one inch into the earth. What would be the average force of the pile on the ground as it penetrates that one inch?

a) One ton
b) Two tons
c) Twelve tons
d) Thirteen tons
e) Fourteen tons

Odu83 – There is a reason 6 tons is not an answer choice.

“Dr. H” aka Tom Flood/Thomas Flood/Floodtl/(Ω)Jack/ (Ω)flood IR/(Ω)floodtl/(Ω)flood/Ineedattentionftl/et al. – And they say imitation is the sincerest form of compliment. On the bright side, I don’t have to live with you! :)

Alexey V – Please. Try again.

Kirchwey – More like it but not there yet. And thank you for the um compliment. Since you liked it…

Said Einstein, "I have an equation,"
"Which some might call Rabelaisian:"
"Let P be viginity,"
"Approaching infinity,"
"And let U be a constant, persuasion."

"Now, if P over U be inverted,"
"And the squareroot of U be inserted,"
"X times over P,"
"The result, Q.E.D."
"Is a relative." Einstein asserted.

The answer is e. There are two foot tons of kinetic energy in the hammer when it bangs the pile. Since the hammer and the pile have equal masses, half of that energy is converted to heat by the impact (recall my question SMUSH AGAIN). That leaves one foot ton of kinetic energy to drive the pile one inch. Now one foot ton is equal to twelve inch tons. That is, a force of one ton pushing one foot does as much work as a force of twelve tons pushing one inch. So does this mean the pile exerts a force of twelve tons on the earth? Even more! The pile pushes on the ground with a weight of one ton even before the hammer hits it, and after the impact the hammer which also weighs one ton momentarily rests on the pile. So there are two more tons -- fourteen in all. Does that mean fourteen inch tons of work were done on the earth? Yes, the two extra inch tons came from the weight of the pile and hammer as they went down that last inch. Tricky? A little.

That's why engineers earn as much per hour as plumbers -- or do they?

Answer 1

Evidently this is an inelastic collision, since otherwise Alexey's answer would be OK and there would be a 24 ton answer choice. (Actual pile drivers work mostly inelastically; they bounce but only a little.) In the inelastic case, with equal masses, half the kinetic energy is lost and both the hammer and the pile move at v/2 after impact.
odu almost had it, but since both the hammer and the pile decelerate from v/2 to 0 in 1 inch, F = 2 tons * 193/32 = 12 tons.
P.S. Asker's poetic, um, talent is duly noted.
EDIT: Boffo profundo! OK, another try. Forces for vertical and horizontal pile-driving have to differ, so add the static weight of the stack for a total of 14 tons.

Answer 2

The Benefits of the hydraulic static pile driver

· Free of noise, vibration and pollution; suitable for operations in urban areas or country sites
· Able to install a wide range of concrete pile sizes from 250mm x 250mm to 550mm x 550mm and spun piles from 250mm to 600mm
diameter
· Easy movement : X-Y direction and rotation
· Wide base, low center of gravity and stable; suitable for various types of ground condition
· Easily dismantled for transportation
· Pile jacking load is applied continuously and evenly, without hammering and vibration
· Able to handle individual cast pile length exceeding 12m, thus reducing the number of welding joint
· No heavy reinforcement needed for the pile as compared to a hammer-driven system
· Excellent piling accuracy
· Hassle-free operation and maintenance

Answer 3

All the potential energy of hammer mgH is spend on work to penetrate the ground Fh, so mgH=Fh and F=mg(H/h), so it is largen then the weight of hammer in H/h=24 times. So average force would be 24 tons and none of the answers is correct.

Answer 4

F=m*a

The driver will have kinetic energy of
.5*m*v^2=m*g*h where h is 2 feet, g is 32 ft/s^2

From that we can calculate v at impact.

The driver decellerates in 1 inch at an average velcocity of v/2 from above

so the time to decellerate is
0=v/2-a*t

and

1/12=v/2*t-.5*a*t^2

we can now solve for t and a. Then we calculate F as m*a.

1/12=v/2*t-.5*v*t/2
1/(3*v)=t
v^2=128
v=11.3 ft/s
t=0.03 seconds

a=(11.3*0.03/2-1/12)*2/(.03^2)
a=191.5 ft/sec^2

F=1ton*191.5/32
6 tons

j

Answer 5

Ahahahahaha!!!