Suppose the 480 lb pile driver that drops 20 ft drives the pile 2 in. every time it strikes the pile. What is the average resisting force of the ground where the pile is driven?
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I am wondering about your instructor, using these obsolite English units, lb, in,
I would do it this way:
energy in the pile=resistive force*distancetmoved
masspile*g*height=F*drivenmove
solve for F, but I would do it in SI units.
To calculate the average resisting force of the ground where the pile is driven, we need to consider the work done by the pile driver. The work done is equal to the product of force and distance.
First, let's convert the weight of the pile driver from pounds to pounds-force (lbf) using the acceleration due to gravity.
Weight of the pile driver = 480 lb
Acceleration due to gravity = 32.2 ft/s² (approximate value)
Weight (in pounds-force) = Weight (in pounds) * Acceleration due to gravity
Weight (in pounds-force) = 480 lb * 32.2 ft/s² ≈ 15,456 lbf
Now, let's calculate the work done by the pile driver.
Work (in foot-pounds) = Force (in pounds-force) * Distance (in feet)
Force (in pounds-force) = Work (in foot-pounds) / Distance (in feet)
Given:
Distance driven per strike = 2 in = 2/12 ft = 1/6 ft
Height the pile is driven = 20 ft
Number of strikes = Height the pile is driven / Distance driven per strike
Number of strikes = 20 ft / (1/6) ft = 120 strikes
Total work done = Work per strike * Number of strikes
Work per strike = Weight * Distance driven per strike
Work per strike = 15,456 lbf * (1/6) ft
Total work done = 15,456 lbf * (1/6) ft * 120 strikes
Now we can calculate the average resisting force.
Average Resisting Force (in pounds-force) = Total work done / Distance driven
Average Resisting Force (in pounds-force) = (15,456 lbf * (1/6) ft * 120 strikes) / 20 ft
Calculating this value will give you the average resisting force of the ground where the pile is driven.