基础工程课后习题答案

作业:

解:

N=920KN, M=276KN.m, V=25kN

Nk=920/1.2=767KN, Mk=276/1.2=230KN.m, Vk=25/1.2=21kN (1)验算基底尺寸

b=3.4m,l=2.2m ,A=7.48㎡ G=20*1.5*7.48=224.4KN

230?21?0.9b3400Mbotk?e0?Nbotk767?224.4=0.251m<6=6=0.567m

(可以)

PkmaxNbotkMbotkNbotk?6e?????1??AWA?b?767?224.4?6?0.251?991.4??1.443 ?1???2.2?3.4?3.4?7.48

=191.2〈1.2f=1.2×200=240kNm (可以)

2

PkminNbotkMbotkNbotk?6e?991.4?6?0.251??????1??1????AWA?b?7.48?3.4?=73.82>0 (可以);

Pk=

Pkmax?Pkmin2=133<200kNm(满足要求)

2该车间是可不作地基变形的二级建筑物,不作地基变形验算

(2)验算基础高度

已假定基础H=0.9m, 如采用锥形基础 荷载设计值作用下的地基净反力

Pj,maxNM920276?25?0.9????kN2 blW2.2?3.412=193.41m?2.2?3.46m2Pj,min?52.57kN

基础抗冲切验算简图如图所示

故冲切破坏荷载:

FL?Pj,maxAl2??bhc??lbc???193.41×??2?2?h0?l??2?2?h??

????????at?abam?2 2??3.40.6??2.20.4????0.85?2.2????0.85????kN =193.41×??22=233.5422??????at?b?0.4ab?bc?2h0?0.4?2?0.85?2.1抗冲切力

6?fah=0.7htm0=0.7×0.99×1.1×10

?0.4?2.1??×kN×?0.85=809.92??〉233.54kN

满足要求

3、基础配筋验算

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