Example calculation. Solution: For medium clay layer: Su=45 kPa, α=1−(45−25/90)=0.78 For stiff clay layer: Su=85 kPa, α=0.5 Friction capacity: Qfs=0.78×45×(4×0.25)×7+0.5×85×(4×0.25)×3=373.2 kN End bearing capacity: Qb=9 Su Ab=9×85×(0.25×0.25)=47.82 kN ...