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基础长边方向弯矩

1?2G???M??a12??2l?a'??pmax?p??p?pl????max12?A????1?2?3.7?2.2?1.5?20?1.35????1.52???2?2.2?0.4??363.4?357.1??363.4?357.1?2.2????123.7?2.2?????0.1875??4.8?639.5?13.86??572.6kN?m基础短边方向弯矩

M????12G?2p?p??l?a1??2b?b'??min?max?48A??12?3.7?2.2?1.5?20?1.35? 2???2.2?0.4???2?3.7?0.7???363.4?347.8??483.7?2.2??1??3.24?8.1?630.2?344.6kN?m48

14.某钢筋混凝土条形基础,基础高度h?0.35m,基础宽度b?2m,基础埋深

d?2m,混凝土墙厚0.24m,作用在基础顶面竖向力Fk?220kNm,Mk?20kN?m,基础为C20混凝土,HPB235钢筋,试计算基础底板配筋。

【解】

pmax??A?220?2?2?20??1.352?Fk?Gk??1.35?Mk?1.35W 20?1.35??202.5?40.5?243kPa1?1.0?226pmin?202.5?40.5?162kPa

bbc20.24????0.88m 2222墙边地基反力设计值 b1?p?162??2?0.88??2?243?162??207.4kPa

条形基础l?a'?1.0m,混凝土墙体,a1?b1

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M???12?2G???a1??2l?a'??pmax?p??p?pl??max??12?A???1?2?2?2?20?1.35??? ?0.882???2?1?1??243?207.4??243?207.4?1.0????122?????0.0645??3?342.4?35.6??68.6kN?mm(1)按照《混凝土结构设计规范》(GBJ10-89)规定配筋:

As?M,fy?210Nmm2,h0?0.35?0.04?0.31m

0.9h0fy68.6?106As??1170.8mm2,配6Φ16@170(As?1205mm2)

0.9?310?210沿墙体长度方向配置Φ8@250分布钢筋。 (2)按照《混凝土结构设计规范》(GB50010-2002)规定配筋: ①混凝土极限压应变

?cu?0.0033??fcu,k?50??10?5?0.0033??20?50??10?5?0.0036

?cu?0.0033,取?cu?0.0033 ②相对界限受压区高度

?b?1??1fy,?1?0.8,Es?2.1?105Nmm2

?cuEs0.8?0.60

2101?0.0033?2.1?105③受压区高度

?b?x?h0?h02?2M,h0?310mm,b?l?2000mm ?1fcb?1?1.0,M?68.6kN?mm,fc?9.6?103kNm2

2?68.6?106x?310?310??11.7mm??bh0?0.6?310?186mm

1.0?9.6?20002④配筋面积

As??1fcbxfy?1.0?9.6?2000?11.7?1069.7mm2,配6Φ16@170(As?1205mm2)

210⑤验算最小配筋率??

As1205??0.17%,满足最小配筋率0.15%。 bh2000?35046

15.某钢筋混凝土条形基础,基础埋深d?2m,混凝土墙厚0.4m,作用在基础顶面竖向力Fk?400kNm,Mk?30kN?mm,基础为C15混凝土,HPB235钢筋,地下水位在地面下2m,土层分布为:0~5m粉土,??19kNm3,?k?24?,

ck?20kPa,fak?200kPa,IL?0.75,Es?7.5MPa;5m以下为淤泥质黏土,

??17kNm3,?k?18?,ck?8kPa,fak?95kPa,IL?1.1,e0?1.29,Es?2.5MPa;试回答下面问题:

(1)已知粉土质量密度?s?2.7gcm3,w?27%,求e; (2)修正后持力层地基承载力特征值; (3)计算基础宽度;

(4)按土的抗剪强度指标计算持力层地基承载力特征值; (5)验算软弱下卧层地基承载力; (6)验算基底压力;

(7)计算基础板最大剪力; (8)计算地基沉降的附加应力; (9)判断软弱下卧层的压缩性; (10)计算底板最大弯矩。

【解】(1)?s??sg?2.7?9.81?26.5kNm3

e??s?1?0.01??26.5??1?0.01?27??1??1?1.77?1?0.77

?19(2)fa?fak??b?(b?3)??d?m(d?0.5)

设条形基础宽度b?3m,e?0.77,IL?0.75,均小于0.85

?b?0.3,?d?1.6

fa?200?1.6?19?(2?0.5)?245.6kPa

(3) b?Fk400??1.95m,取b?2m

fa??Gd245.6?20?2Mk30??0.0625m?0.033b?0.033?2?0.066m,可

Fk?Gk400?2?2?20(4)偏心距e?以根据土的抗剪强度指标计算持力层地基承载力特征值。

fa?Mb?b?Md?md?Mcck

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查表,?k?24?时,Mb?0.80,Md?3.87,Mc?6.45

fa?0.80??19?10??2?3.87?19?2?6.45?20?290.5kPa

(5)pz?pcz?faz

pcz?19?2?(19?10)?3?38?27?65kPa

pz?b(pk?pc)

(b?2ztan?)pc?19?2?38kPa,pk?Fk?Gk400?2?2?20??240kPa A2Es1Es2?7.52.5?3,zb?32?1.5,地基压力扩散角??23°。

pz?2?(240?38)404??88.8kPa

(2?2?3tan23?)4.55pz?pcz?88.8?65?153.8kPa

faz?fak??d?m(d?0.5),淤泥质粘土?d?1.0

faz?95?1.0?19?2?(19?10)?3?(2.0?3.0?0.5)?95?58.5?153.5kPa

(2?3)pz?pcz?153.8kPa?faz?153.5kPa,基本满足要求。

(6)pk?pkmaxFk?Gk400?2?2?20??240kPa?fa?245.6kPa,满足 A2F?GkMk30?k??240??285kPa

1AW?1?226pkmax?285kPa?1.2fa?1.2?245.6?294.7kPa,满足

(7)基础底面净反力

Fk?1.35Mk?1.35?AW 400?1.3530?1.35???270?60.8?330.8kPa12?1.0?226pjmax?pjmin?270?60.8?209.2kPa

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