A well foundation of 6 m external diameter and of 5 m internal diameter is sunk to a depth of 15 m in a deep deposit of sand. If the average N value of sand is 20, the load that the well can carry by bearing alone will be nearly
- A52 MNCorrect
- B44 MN
- C38 MN
- D29 MN
Solution
Concept
End bearing of a well foundation from SPT
Formula
Qb = qb × Abase
Step-by-step solution
1. Area of the annular base = (π/4)(6² − 5²) = 8.64 m². 2. End bearing pressure in sand from SPT ≈ 300 N kPa = 300 × 20 = 6000 kPa. 3. Qb = 6000 × 8.64 = 51 800 kN ≈ 52 MN.
Explanation
When skin friction is neglected the load is carried by end bearing on the ring-shaped base of the well. The unit end bearing of dense sand is related to the SPT value.
Why the other options are wrong
The other loads correspond to lower unit pressures (5.1, 4.4 and 3.4 MPa) that do not follow from a simple relation with N = 20.
Shortcut
Ring area × end bearing (≈ 300 N kPa).
Common mistake
Using the full circular area of the well.
Correct answer: A) 52 MN