39《地基处理2》换填垫层法课件5.04.pdf

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    3  ! $#%()*+,-./ 1.8x10 kg1 m3 20;=,?@AB/280kpa( 1 EC DFGH/K A LI B NJ C PJ D ABC ?M R SOU!V4.2.1  LIQPJTMW( ZXY[\.] ^X_`,LIH( cA 2 ƒ KPa EC 20!Jabd,efg7AhiB l mn a DFBH/jkK A 80 B 105 C 115 D 121 R qUV5.2.4 MosprJη =0tη =1 b d ƒ =ƒ +η γ (d-0.5) a ak d m =80+1x17x(2-0.5) =105.5 KPa B 105 vu%wc ( 3 ) 2m \,LI EC yx.z{ }20|~ m €%‚,Hƒ„…l m nDFBH/jkK A 0.3 B 0.9 C 1.2 D 1.5 R SOU!V4.2.3  ‡o\z=2m x.b=3.6m z/b=2/3.6=0.560.5 0 ˆU!V‰4.2.2θ=30 b‹ b+2ztanθ 20†Šx. Œ 0 ~€x.= ztanθ=2xtan30 =1.15m0.3m C 1.2 vu%wc ( 4 3 Ž EC;= 20!,@ P kpa AB l m nDFBH/jkK Z A 130 B 150 C 170 D 190 R SOU!V4.2.2  P =b P -P / b+2zxtgθ l m l m Z k c ‡oP =280kpa k P =17x2=34kpa c P =3.6 280 -34 / 3.6+2x2xtan300 =149.9 kpa Z l m l m B 150 vu%wc ( ‘’“ 1 ”•–—˜™š›œžŸ ”¡¢£¤¥¦§ 2 P -P θ