习题解答
习题3－2 某过江隧道底面宽度为33m，隧道A、B段下的土层分布依次为：A段，粉质粘土，软塑，厚度2m，Es＝4.2MPa，其下为基岩；B段，粘土，硬塑，厚度12m，Es= 18.4MPa，其下为基岩。试分别计算A、B段的地基基床系数，并比较计算结果。 〔解〕本题属薄压缩层地基，可按式（10-52）计算。 A段： k?Es?4200?2100kN/m3
Ah2B段： k?18400?1533kN/m3
B12比较上述计算结果可知，并非土越硬，其基床系数就越大。基床系数不仅与土的软硬有关，更与地基可压缩土层的厚度有关。
习题3－3 如图10-13中承受集中荷载的 钢筋混凝土条形基础的抗弯刚度EI＝2×106
kN·m2，梁长l＝10m，底面宽度b＝2m，基床 系数k＝4199kN/m3，试计算基础中点C的挠 度、弯矩和基底净反力。
〔解〕 图10-13
??4kb44199?2?1
??0.18m4EI4?2?106?x?0.18?5?0.9 ?l?0.18?10?1.8
查相关函数表，得Ax=0.57120，Bx=0.31848，Cx=-0.06574，Dx=0.25273，Al=0.12342，Cl=-0.19853，Dl=-0.03765，El=4.61834，Fl=-1.52865。
(1)计算外荷载在无限长梁相应于A、B两截面上所产生的弯矩和剪力Ma、Va、Mb、Vb 由式（10-47）及式（10-50）得：
Ma?FMCx?Dx 4?21000100 ????0.06574???0.25273??103.9kN?m4?0.182FMCx?Dx 4?21000100 ????0.06574???0.25273??78.7kN?m4?0.182 Mb? Va?FM?Dx?Ax 221000100?0.18 ??0.25273??0.57120?121.2kN22 Va??FM?Dx?Ax 221000100?0.18 ???0.25273??0.57120?131.5kN22(2)计算梁端边界条件力
FA=(El+FlDl)Va+λ(El-FlAl)Ma-(Fl+ElDl)Vb+λ(Fl-ElAl)Mb =(4.61834+1.52865×0.03756)×121.2
+0.18×(4.61834+1.52865×0.12342)×(-103.9) -(-1.52865-4.61834×0.03756)×(-131.5)
+ 0.18×(-1.52865-4.61834×0.12342)×(-78.7)
=282.7kN
FB=(Fl+ElDl) Va+λ(Fl-ElAl) Ma-(El+FlDl)Vb+λ(El-FlAl)Mb =(-1.52865-4.61834×0.03756)×121.2
+0.18×(-1.52865-4.61834×0.12342)×(-103.9) -(4.61834+1.52865×0.03756)×(-131.5)
+ 0.18×(4.61834+1.52865×0.12342)×(-78.7)
=379.8kN
VaV??El?FlDl?Ma??Fl?ElCl?b??Fl?ElDl?Mb2?2? 121.2 ???4.61834?1.52865?0.19853??2?0.18 MA???El?FlCl???4.61834?1.52865?0.03756????103.9? ???1.52865?4.61834?0.19853???131.5
2?0.18 ???1.52865?4.61834?0.03756????78.7?
=-396.7kN·m
VaV??Fl?ElDl?Ma??El?FlCl?b??El?FlDl?Mb2?2? 121.2 ???1.52865?4.61834?0.19853??2?0.18 MB??Fl?ElCl????1.52865?4.61834?0.03756????103.9? ??4.61834?1.52865?0.19853???131.5
2?0.18 ??4.61834?1.52865?0.03756????78.7?
=756.8kN·m
(3) 计算基础中点C的挠度、弯矩和基底净反力
(FA?FB)?(MA?MB)?2F?wC?Ax?Bx?2kbkb2kb
?282.7＋379.8??0.18(?396.7?756.8)?0.1821000?0.18 ??0.57120??0.31848?2?4199?24199?22?4199?2 ?0.0134m?13.4mm