Pile driver: impact loss and clay resistance
A pile driver is used to drive a steel pile into stiff clay. The hammer has mass 420 kg and is released from rest 2.0 m above the top of the pile. The pile has mass 180 kg. The collision between hammer and pile is assumed perfectly inelastic, so they move together immediately after impact. The hammer and pile then penetrate the clay a further 0.14 m before coming to rest. The clay exerts a resistance that may be modelled as a constant upward force during this penetration. Air resistance and friction in the guides are negligible. The site engineer is concerned that the hammer is too light for the job and that the pile may not reach the required depth. A second concern is that the energy lost in the hammer–pile impact may be so large that the penetration estimate is unrealistic. The engineer also wants to know how the resistance force would change if the clay were stiffer and the penetration were only 0.10 m. Use g = 9.8 m/s².
1. Find the common speed of hammer and pile immediately after impact and the kinetic energy lost in the collision.
Read the worked answer
Speed of the hammer just before impact:
u = √(2gh) = √(2 × 9.8 × 2.0) = 6.26 m/s.
Momentum conservation for the perfectly inelastic collision:
420 × 6.26 = (420 + 180)V,
so V = 2629/600 = 4.38 m/s.
Initial kinetic energy before impact = ½ × 420 × 6.26² = 8232 J.
Kinetic energy after impact = ½ × 600 × 4.38² = 5762 J.
Energy lost = 8232 − 5762 = 2470 J, about 30% of the initial kinetic energy.
2. Find the constant resistance force exerted by the clay during the 0.14 m penetration.
Read the worked answer
During penetration, the hammer–pile system loses 5762 J of kinetic energy and also loses gravitational potential energy:
600 × 9.8 × 0.14 = 823 J.
Let R be the constant upward resistance. Then
(R − 600g) × 0.14 = 5762,
so
R = 600 × 9.8 + 5762/0.14 = 5880 + 41160 = 47,040 N.
The average resistance is about 47 kN.
3. If the clay were stiffer and penetration were only 0.10 m, calculate the new resistance and comment on whether the engineer should expect the pile to reach the same depth.
Read the worked answer
With 0.10 m penetration,
R = 600 × 9.8 + 5762/0.10 = 5880 + 57,620 = 63,500 N,
so R ≈ 63.5 kN. The resistance is larger because the same post-impact kinetic energy must be absorbed over a shorter distance. The engineer should not expect the same depth in stiffer clay; for a given hammer energy, depth falls as resistance rises. The constant-resistance model is an idealisation, and real clay resistance often increases with depth, so the calculated force is an average rather than a precise local value.