Question

A single acting steam hammer weighing 2000 kg and falling through a height of 2 m drives a pile an average penetration of 0.5 cm under the last few blows. What will be the allowable load (in tonnes) for the pile?
1. 40
2. 80
3. 88.88
4. 92.44

Answer 1

Correct Answer - Option 3 : 88.88

Given:

Single acting steam hammer

W = weight of hammer = 2000 kg

H = Height of falls of Hammer = 2m

S = Penetration or set per blow = 0.5 cm

For drop hammer and single acting steam Hammer

ultimate load according to Engineering News Record formula

\(Q_{up}=\dfrac{W\cdot H}{S+C}\)

Where,

W = Weight of Hammer = 2000 kg

H = Height of falls of Hammer = 2m = 200 cm

S = Penetration = 0.5 cm

C = Constant

= 2.5 cm for drop Hammer

= 0.25 cm for steam hammer (Single and double)

So,

\(Q_{up}=\dfrac{2000\times 200}{0.5 + 0.25}\)

\(=\dfrac{2000 \times 200}{0.75}\)

= 5,33,333.33 kg

Allowable load

\(Q_{ap}=\dfrac{Q_{up}}{FOS}\)

\(=\dfrac{Q_{up}}{6}\) {In ENR formula Taken FOS = 6}

\(=\dfrac{533333.33}{6}\)

= 88,888.88 kg      {1 ton = 1000 kg}

= 88.88 tonnes