答案解析
ES1/ES2=9/3=3Z/b=3/5=0.6>0.5θ=23°Pc=18×1.4=25.2kpa
Pz=lb(Pk-Pc)/[(b+2ztanθ)(l+2ztanθ)]
=2.5×5.0×(145-25.2)/[(5+2×3tan23°)(2.5+2×3tan23°)]
=39.3kpa
Pcz=18×1.4+(18-10)×3=49.2
Pz+Pcz=39.3+49.2=88.5kpa≈89kpa
faz=fak+ηdfm(h-0.5)=70+1.0×(18×1.4+8×3)×(4.4-0.5)/4.4=113.6kpa≈114kpa