Click here to load reader

第四章 天然地基上的浅基础

Embed Size (px)

DESCRIPTION

第四章 天然地基上的浅基础. 天然地基上的浅基础设计 Shallow foundation on natural ground GB50007-2002. 1 地基基础分类,荷载和承载力计算 2 基础埋深和基础尺寸确定 3 承载力验算和下卧层验算 4 基础的冲切验算 5 基础的抗震验算 6. 基础的结构设计 7. 算例. 地基基础分类荷载和承载力计算.  D. D. ( 一 ) 天然地基与浅基础.  D. D. 桩基础. 一 地基基础分类. - PowerPoint PPT Presentation

Citation preview

  • Shallow foundation on natural groundGB50007-20021 2 3 4 5 6. 7.

  • () 5m5m

  • 3023b1.5b5m

  • 1 2 ()3 Shallow foundationRigid foundationSpread foundationMat foundationPile foundationIndividual footing(pad foundation) Strip footingCross strip footingBox foundation

  • 2 Rigid foundation ()()Spread foundation,1.1 1.5

  • 3 1: ,Individual footing, pad foundation

  • 2 , :,Strip foundation

  • 3 :,,

  • 4 ,,,,Mat foundation

  • 5

  • 6

  • 2 ():(1),(2),(3)()()

  • 3 :

  • 4 5 F G

  • 6

  • 1 (1) (2)2 3 SGK SQiK SQ1K

  • fakpcrpu
  • fa=Mbb+Md md+Mcck fa : p1/4=NB /2+Nq d+Ncc 2Mb Nb6m6m3m3m

  • ( D50cm, 10cm

  • 1 ,: F

  • L/ H=1~2

  • (1) ,(2) ,

  • 1 ,,2 (1) ()(2) (3) ,

  • ,h1IIh1< 2m h1=2m~4m,h1>4 m () I II III IVh1h1

  • Pingo

  • 1 1 () ,,

  • 1 1928-1929 Casagrande,:45cm,13cm,=8%~12%, 60%~110%, 13cm

  • (2) ,.0oc,-0.5~-30oc

  • :,,+(),,,,

  • 2 1 2 3

  • 3 P32-33,25

    Z0, 1.0m 2.0m 2.5m

  • 4 dmin z0 t dfrZ0 dfr

  • 4 /(%)hwZd (h -z)zzdh

  • 4 dmin > zd hmaxZd hmax

  • 693

  • 5 pk fa

  • 1A*Fk+Gk Gk=Ad, =20KN/m3*fa*Fk+Gk =Afa

  • (2) ,blA, bb0+2htg, l l0+2htgd h+0.1mb, l h=45o (1:1) 30o (1:2)Fll0b0bbth

  • Fhbb0+2htg h:b 60~70cm, b024,37,48cm b 50cm

  • 1A1(2)A=(1.1~1.4) A1 Abl

  • .

  • ,() 1

    b>6m 6m, < 3m 3m fa=Mbb+Md md+Mcck fa :

  • 2 Dbfa(1)

    (2)

    * *(,) *e>b/6, pmin

  • 2 (2)

    * *(,)

    *e>b/6, pmin

  • 3a 0.75b25%*A*l, b, A*

  • () 1 pk=(Fk+Gk)/A 2 c03 p0=pk-c04 Es1/ Es2z/b

  • 5

    2196

  • () 1 5.3.4

  • 2 (1):,

    :,e-p,logp, Es GB 50007-2002 K.0.1-2(2)S=Si s0.025S

  • Ai = p0(izi- I-1 zi-1)si=Ai/Esis=s si

  • (3)S=sSs,,s=1.4~0.2, (1)Es,,(2)p0/fk

  • 3 ,(p0)(p0),-

  • () 1

  • 2 (2)(1)

  • (1) G1.2 G1.35Q1.00.9(2) G1.00.9

  • 2 F1=A1pj[V]=0.7hpftamh0Alamamhp

  • 3 baIIIa1

  • IIIIII

  • Kobe earthquake, 1995()

  • Concrete-frame structure with a mid-story collapse

  • () () 1 pE=sf s=1~1.5(~),8-8 ,

  • 2 (),(1) GE, G FE M FE

  • (2)

  • 3 FEFE=Geq Geq = Ge Geq =0.85 Ge~

  • :. ().12 (1) (2) (3):..:M=logA,

  • Tg ,**,,max

  • 4 bb/45

  • () liquefaction1 :,,(),,,.,,,. u = 3, , C,

  • 2 (1)* Q3(10000),* >10~16%, 7~9* du,dw() : du > 6~9m, dw > 5~8m,

  • 2 (2),N63.5,,ds dw ds>dwcN0

  • N0 7 8 9 6 10 16 8 12 /

  • 30030030080 150 150

  • 120200240 60 170240 60 150 150

  • e=0.1Il=0.2w=0.1n=6()

  • 1

    (A)2.00m(B)1.00m(C)1.40m(D)3.50m

  • 2.(A)180kPa(B)200kPa(C)250kPa(D)150kPa

  • .3.(A)100kPa(B)80kPa(C)95kPa(D)112kPaw=38%;95kPaf 0.83279.0kPa

  • 4.1.51.7(A)220kPa(B)250kPa(C)300kPa(D)175kPafa=Mbb+Md md+Mcck5.2.5 k=18, Mb0.43, Md2.72, Mc=5.31, =9kN/m3, m=18.92kN/m3fa=0.4391.7+2.7218.921.5+5.3125217kPa

  • 5. 1.51.7(A)230kPa(B)210kPa(C)300kPa(D)175kPa

    179.2kPab0.3 d=1.6,

    fa179.2+1.618.921.50.5=209.5kPa

  • 6kPa)(A)50(B)100(C)70(D)8018.91.2+0.31994.7=70.7kPa

  • 7.P0)kPa):(A)155(B)166(C)177(D)200(300+19.61.71.5)/1.7-18.91.5177kPa

  • 8.kPa)(A)72(B)67(C )83(D)56z/b>0.5, 323oz(1771.7)/(1.7+24.7tan23o 53kPa

  • 9.(kPa)(A)100(B)120(C)140(D)15079.0kPa , d=1.1 m=11.4kN/m3fa79.+1.111.46.50.5=154.2kPa

  • 10.s=1.0)(cm):(A)20(B)30(C)10(D)35

  • 1 ABCD

  • 2?ABCD

  • 3 ABCD

  • 45000m5030m5060m4900m105A d=1.0mB d=1.3mC d=1.6mD d=1.45m

  • 5A d=1.0mB d=1.3mC d=1.6mD d=1.45m

  • 6 pe?A B C D

  • 5500