Question

A square plate of section 30 cm \(\times\) 30 cm and length 10 m penetrates a deposit of clay having \( C = 5 \) kN/m\(^2\) and the mobilizing factor \( m = 0.8 \). What is the load carried by the pile by skin friction only?

• A. 192 kN
• B. 75 kN
• C. 60 kN
• D. 48 kN

Hint:

- The skin friction of a pile in clay is calculated using \( P = \alpha C A_s \), where \( A_s \) is the shaft surface area. - Mobilization factor \( m \) accounts for partial mobilization of cohesion along the pile shaft.

Answer 1

The Correct Option is C

Step 1: Use the skin friction formula: \[ P = \alpha C A_s \] where, \( C = 5 \) kN/m\(^2\) (cohesion of clay), \( m = 0.8 \) (mobilization factor), \( A_s = {Perimeter} \times {Length} \) \( = 4 \times 0.3 \times 10 = 12 \) m\(^2\). Step 2: Compute the load carried by skin friction: \[ P = 0.8 \times 5 \times 12 \] Step 3: Calculation: \[ P = 60 { kN} \] Thus, the correct answer is (C) 60 kN.