粘性土的弹粘塑性模型及其固结分析应用（讲稿）一种可考虑排水板、多层土和多级荷载的软黏土固結沉降的通用簡單計算方法 A General Simple Method for Calculating Consolidation Settlements of Layered Clayey Soils with Vertical Drains under Staged Loadings

2.2 Calculation of S, for the j-layerExampleLoading from point 1 to point 42=00:40.1=0Case (i):Initially:OCR=0p /o,=1After increased loading:o. ≥0p for0≤z≤HIn this case, we have:z=h+0mit2)dzSy,1-4 = [ 8.,1-4dz =ZI+e.0+0:a0unit22=0logo,E1(og.za)Substituting:3(0g.5g)o0-1=00101.0+2HHamec,/d-e)(0,.a(i) frowPoinr Tto Poinf2andz)fromPoumrIroPomt(oOC0P.0P.HPHHo6=0a4,0=4,H=4.0H(o...E)4(c.....Tilonding /rewC,/d+e) (c.We have:Varmalhdatauloe C,/(I+e,) (ar t))2000.=4.HC=4,0.CaotHIn]ld(1+e.)ln(10)1+e.)ln(10)I1(o.01,0?0:1,0)+0mii2--LHH==HT In[o/ +(o,-0:/0)+m dmin2 Jdz-In[0-4,0(o:4,H-0-4.0)+m-HH==0=016

2.2 Calculation of Sfj for the j-layer 16 Example: Loading from point 1 to point 4 Case (i): Initially: After increased loading: In this case, we have: ' ' 4 0 z zp    for z H ' ' 1 / 1 OCR     zp z ' 4,0 4, 4,0 2 ,1 4 ' 0 ,0 , ,0 2 ' ' 4,0 4, 4,0 2 ,0 , ,0 2 0 0 ( ) ln[ ] ( )ln(10) ( )ln(10) ( ) ln[ ( ) ] ln[ ( ) ] ' '' z H z z H z unit c c f ' '' z o o z1 z1 H z1 unit z H ' ' ' ' '' z z H z unit z1 z1 H z1 unit z z z σ σσ C C H S dz 1e 1e z σ σσ H z z σ σσ dz σ σσ dz H H                          z H        ' 4 2 ,1 4 ,1 4 ' 0 0 2 ( ) zH zH ' c z unit f z ' z z o zp unit C σ S dz log dz 1 e σ                 ,0 , ,0 ( ) '' ' ' z1 z1 z1 H z1 z σσ σ σ H   ,0 , ,0 ( ) '' ' ' zp zp zp H zp z σσ σ σ H   4,0 4, 4,0 ( ) '' ' ' z z zH z z σσ σ σ H   Substituting: We have: 1 ' ' z zp σ σ  z=0 z=h z 4 ' z σ

2.2 Calculation of S, for the j-layerExampleLoading from point 1 to point 42=00.1=0p0-4Case (i):LetusintroducetwonewvariablesZ-0:4,0)+0x=o4.Hunit24.0Hz=hN101,0)+0.m2(O-H=0-1:0ZH[dz =[H / (o-4,H -0=4. ]dx+o+01=0V=01,0=4,0umit2unit2We have:anddz=[H /(o,H-0-1.0)]dy+oyO1,H+Omit20-4,H-=Humit2Through the substitution method:X=0:4,H+Omt2y=O-1,H+Oni12HHCInxdxnydi(1 +e。) ln(10)(O=4,H0a=4.00-LH=1.0y=0:1,0+0mit2X=0:4,0+0mit217

2.2 Calculation of Sfj for the j-layer 17 ' 4,0 4, 4,0 2 ' 1,0 1, 1,0 2 ( ) ( ) ' '' z z H z unit ' '' z z H z unit z x σ σσ H z y σ σσ H            4, 4,0 1, 1,0 [ / ( )] [ / ( )] ' ' zH z ' ' zH z dz H σ σ dx dz H σ σ dy        Let us introduce two new variables: We have: Example: Loading from point 1 to point 4 Case (i): ' ' 0 4,0 2 0 1,0 2 ' ' 4, 2 1, 2 and , ' ' z z unit z z unit ' ' z H z H unit z H z H unit x σ y σ x σ y σ                       ' ' 4, 2 , 2 ' ' 4,0 2 ,0 2 ,1 4 4, 4,0 , ,0 ln ln ( ) ln(10) ( ) ( ) ' ' z H unit z1 H unit ' ' z unit z1 unit x σ y σ c f ' ' '' o z H z z1 H z1 x σ y σ C H H S xdx ydy 1 e σσ σσ                            Through the substitution method: 1 ' ' z zp σ σ  z=0 z=h z 4 ' z σ

2.2 Calculation of S, for the j-layerExampleLoading from point 1 to point 42=00.1=0p0 :4Case (i):[Inxdx=xlnx-x,Jinydy=ylny-ySincez=hWe have:ZHcHAHTUxinxylnyL0+0mair2(I+e.)In(10) [(α:4,H -0.(O-1,H0From all the above equations, we have:HC[(4,H+om2) In(o:4,H +min2)(=-4,H +oum2)Sf.1-4(I +e.)In(10) [(α=4,H -0=4,0)H(0-4.0 +0m2) n(04.0 +0mm2)(0=4.0 +0mm2)mi2) /n(0=, +0m12)(oH+Oa0(01,n +0m2)(:1,o +0mn2)n(o:1,0 +0mn2)(01,o +2)18

2.2 Calculation of Sfj for the j-layer 18 Example: Loading from point 1 to point 4 Case (i): Since , ln ln xdx x x x    ln ln ydy y y y    We have:     ' ' 4, 2 1, 2 ' ' 4,0 2 1,0 2 ,1 4 4, 4,0 , ,0 ln ln ( ) ln(10) ( ) ( ) ' ' z H unit z H unit ' ' z unit z unit x σ y σ c f ' ' '' x σ y σ o z H z z1 H z1 C H H S x xx y yy 1 e σσ σσ                           ''' ,1 4 4, 2 4, 2 4, 2 4, 4,0 ''' ' ' 4,0 2 4,0 2 4,0 2 1, 2 1, 2 , ,0 [( ) ln( ) ( ) ( )ln(10) ( ) (( )ln( ) ( ))] [( ) ln( ( ) c ''' f z H unit z H unit z H unit ' ' o zH z ''' ' ' z unit z unit z unit z H unit z H unit ' ' z1 H z1 C H S σσσ 1 e σ σ H σσσ σ σ σ σ                       ' ''' 1, 2 1,0 2 1,0 2 1,0 2 ) ( ) (( )ln( ) ( ))] ' ''' z H unit z unit z unit z unit       σ σσσ   From all the above equations, we have: 1 ' ' z zp σ σ  z=0 z=h z 4 ' z σ

2.2 Calculation of S, for the j-layerz=0ExampleLoading from point 1 to point 420=40Case (ii):Initially:OCR=0β /,>1After increased loading:o. ≥, for0≤z≤Hz=hIn this case, we have:Z(ntitSt.1-4 = [ 8.,1-4dz =0ldz1+e。I+e5.,+Ouninl+o0mit2Using the same method as (i):HCph +oumn) n(o=4,p + mi)-(cp,h +oumm) -I(o.f1-4==(（1+e,) ln(10) [(p,H-0p.0)H(op,o +omn)In(p.o +omn)(p,o +om[(o1,H +,mt) In(C1, +omm)i(O=1,H-0:1,0)(01, +0min)(1,0 +0min)In(1,0 +oumm)(C1,0 +0minl)))) +cH2) In(c=4,H + 0 um2 ) -(0=4,H1+ou+wmir2(I+e.)In(10) [(o-4,H -7H((04,0 +oumin2)ln(0=4,0 +omn2)(=4,0 +0umn2))[(cp,H +Oumm2) In( p,H +om2)(CP,H-O(op,H+umi2)-(op,0+omir2)In(op,0+oimin2)-(αp.0+oumit2)19

2.2 Calculation of Sfj for the j-layer 19 Example: Loading from point 1 to point 4 Case (ii): Initially: After increased loading: In this case, we have: ' ' 4 0 z zp    for z H ' ' 1 / 1 OCR     zp z Using the same method as ( i): ' ' 1 2 ,1 4 ,1 4 ' ' 0 1 2 [ ( ) ( )] p zh zh ' ' r zp unit c z unit f z ' ' z zz o z1 unit o zp unit C σ C σ S dz log log dz 1 e σ 1 e σ                     ,1 4 , 1 4, 1 , 1 , ,0 ,0 1 ,0 1 ,0 1 1, 1 1, 1 , ,0 [( ) ln( ) ( ) ( )ln(10) ( ) (( )ln( ) ( ))] [( ) ln( ( ) r '' ' ' '' f zp h unit z p unit zp h unit ' ' o zp H zp '' '' '' ' ' ' ' zp unit zp unit zp unit z H unit z H unit ' ' z1 H z1 C H S σσ σ σ σσ 1 e σ σ H σσ σσ σσ σ σ σ σ σ σ                   1, 1 1,0 1 1,0 1 1,0 1  ''' 4, 2 4, 2 4, 2 4, 4,0 ' ' 4,0 2 4,0 2 ) ( ) (( )ln( ) ( ))] [( ) ln( ) ( ) ( )ln(10) ( ) (( )ln( ) ( ' ' '' '' '' z H unit z unit z unit z unit c ''' ' ' z H unit z H unit z H unit o zH z ' ' z unit z unit σ σ σσ σσ σσ C H σσσ 1 e σ σ σσσ                      ' '' 4,0 2 , 2 , 2 , ,0 ' ''' , 2 ,0 2 ,0 2 ,0 2 ))] [( ) ln( ) ( ) ( ) (( )ln( ) ( ))] ' '' z unit zp H unit zp H unit ' ' zp H zp ' ''' zp H unit zp unit zp unit zp unit H σ σ σ σ σ σσσ               1 ' z σ z=0 z=h z ' zp σ 4 ' z σ

2.2 Calculation of S, for the j-layerZ=0ExampleLoading from point 1 to point 4Z=Zp001EDCase (ii):Initially:OCR=p/o,>1-4z=h4≤0for0≤z≤zAfter increased loadingZforz,≤z≤H0:4>0Inthis case,wehave+0unit)dzfor0≤z≤zSimilar to Case ():+0umitl+0+0unit!wit2)ldzforz,≤z≤HI+e。+e-1+Ounirl+Ounit2OPresentedin Case (ii):20

2.2 Calculation of Sfj for the j-layer 20 Example: Loading from point 1 to point 4 Case (iii): Initially: After increased loading: In this case, we have: ' ' 4 ' ' 4 0 z zp p z zp p for z z for z z H            ' ' 1 / 1 OCR     zp z 1 ' z σ z=0 z=h z ' zp σ 4 ' z σ p z z  1 0 1 ,1 4 ,1 4 0 1 2 1 2 ( )0 [ ( ) ( )] p p z z ' ' r z unit ' ' p z H z o z1 unit f z z H ' ' ' ' z zp unit r c z unit '' '' p z z o z1 unit o zp unit C σ σ log dz for z z 1 e σ σ S dz C σ σ C σ σ log log dz for z z H 1 e σ σ 1 e σ σ                                 Similar to Case (i): Presented in Case (ii):