据《建筑桩基技术规范》(JGJ 94-94)计算
ln=0.8lo=0.8×20=16m
σ'=p+γ1z1=60+(18-10)×10/2=100kPa
=δn1σ'1=0.3×100=30kPa<qsk=40kPa
=δn2σ'2=0.4×130=52kPa>qsk=50kPa
取=50kPa