xem ebook1.pdf

TRNG I HC THU LI B MN A K THUT

---------------------

BI GING

C HC T

H ni 7/2011

Mt chy do

n hi E, , c

C

MC LC

CHNG 1. TNH CHT VT L CA T

1.1. Cc pha hp thnh t v tc dng tng h gia chng ............................................................. 1 I. Pha rn (Ht t). ........................................................................................................... 1 II. Pha lng (Nc trong t) ............................................................................................ 6 III. Pha kh trong t ......................................................................................................... 9

1.2. Cc ch tiu tnh cht vt l v trng thi vt l ca t ........................................................... 9 I. Cc ch tiu tnh cht vt l ca t. .............................................................................. 9 II. Cc ch tiu trng thi vt l ca t .......................................................................... 14

1.3. Phn loi t .............................................................................................................. 17 I. Mc ch ...................................................................................................................... 17 II. Gii thiu mt s tiu chun phn loi t in hnh ................................................. 18

CHNG 2. TNH CHT C HC CA T ................................................................... 27 2.1. Tnh thm nc ca t ............................................................................................. 27

I. Khi nim dng thm trong t ................................................................................... 27 II. nh lut Darcy .......................................................................................................... 30 III. H s thm v phng php xc nh ....................................................................... 32

2.2. Tnh p co v bin dng ca t ................................................................................ 35 I. Khi nim tnh p co v bin dng ca t .................................................................. 35 II. Quan h gia bin thin th tch (V) v h s rng (e) ............................................ 36 III. Th nghim p co khng n hng v nh lut p co ............................................... 36 IV. Xc nh cc c trng bin dng ca t ................................................................ 44 V. C kt ca t dnh bo ha nc v s chuyn ha ng sut trong qu trnh c kt thm ................................................................................................................................. 47 VI. Nhn t nh hng n tnh p co v bin dng ca t .......................................... 50

2.3. Cng chng ct ca t ...................................................................................... 50 I. Khi nim v cng chng ct ca t ................................................................... 50 II. Th nghim ct trc tip v nh lut Coulomb.......................................................... 52 III. Tiu chun ph hoi Mohr - Coulomb ...................................................................... 56 IV. Th nghim ba trc .................................................................................................... 63 V. Cng chng ct ca t ct ................................................................................. 66 VI. Cng chng ct ca t st ................................................................................ 73

2.4. Tnh m cht ca t ................................................................................................ 81 I. ngha thc t v Mc ch ca m cht t............................................................ 81 II. Nguyn l m cht .................................................................................................... 81 III. Cc nhn t nh hng n tnh m cht ca t ................................................... 87

CHNG 3. XC NH NG SUT TRONG T ................................................................... 89 3.1. Cc loi ng sut trong t v cc gi thit c bn tnh ton ............................... 89

I. Cc loi ng sut trong t: ......................................................................................... 89 II. Cc gi thit tnh ton: ........................................................................................... 89

3.2. Xc nh ng sut bn thn ....................................................................................... 90 I. ng sut bn thn trong nn t: ................................................................................. 90 II. ng sut bn thn trong cng trnh t: ..................................................................... 91

3.3. Xc nh p sut y mng........................................................................................ 94 I. Khi nim: ................................................................................................................... 94 II. Xc nh p sut y mng (cho mng cng) ............................................................ 95

3.4. ng sut tng thm trong nn cng trnh .................................................................. 98 I. Hai bi ton c bn: ..................................................................................................... 98 II. ng sut tng thm trong nn ng cht khi mt nn chu ti trng phn b trn din tch hnh ch nht:......................................................................................................... 102 III. ng sut tng thm trong nn ng cht bi ton phng: .................................... 116 IV. Mt s phng php xc nh ng sut tng thm: ................................................ 124

CHNG 4. SC CHU TI CA NN T ................................................................... 139 4.1. M u ..................................................................................................................... 139 4.2. Cc hnh thc mt n nh ca nn khi chu ti ...................................................... 139

I. Th nghim bn nn chu ti trng thng ng .......................................................... 139 II. Cc hnh thc ph hoi nn ...................................................................................... 139

4.3. L thuyt sc chu ti ca Terzaghi ......................................................................... 141 I. Cc gi thit ............................................................................................................... 141 II. Cng thc tnh ton .................................................................................................. 141

4.4. H s an ton ........................................................................................................... 143 4.5. Phng trnh sc chu ti tng qut ......................................................................... 144

I. Khi qut .................................................................................................................... 144 II. Phng trnh tng qut ca Mayerhof...................................................................... 144 III. Tnh sc chu ti ca nn trong trng hp ti trng lch tm ............................... 148

4.6. Cc phng php tnh sc chu ti ca nn t theo tiu chun Vit Nam ............. 150 I. Cc giai on lm vic ca t nn ........................................................................... 150 II. Cc phng php xc nh sc chu ti ca nn ...................................................... 151 III. Xc nh sc chu ti ca nn da vo s pht trin ca vng bin dng do. ....... 151 IV. Xc nh sc chu ti theo ti trng ph hoi (phng php Evdokimov) ............. 155

CHNG 5. LC T LN TNG CHN .................................................................... 163 5.1. M u ..................................................................................................................... 163 5.2. Cc loi p lc t tc dng ln tng chn t v iu kin sn sinh ra chng .... 165 5.3. Xc nh p lc t tnh ........................................................................................... 167

I. Trng hp lng tng thng ng, mt t nm ngang .......................................... 167 II. Trng hp lng tng chn v mt t p nghing ............................................. 168

5.4. Tnh ton p lc t tnh theo l thuyt ca Rankine .............................................. 169 I. Nguyn l tnh ton ................................................................................................... 169 II. Cc gi thit c bn .................................................................................................. 170 III. Xc nh p lc t ch ng.................................................................................. 170 IV. Xc nh p lc t b ng .................................................................................... 173 V. Tnh ton p lc t trong mt s trng hp ......................................................... 174

5.5. Tnh ton p lc t theo l lun ca Coulomb ....................................................... 179 I. Cc gi thit c bn ................................................................................................... 179 II. Nguyn l tnh ton .................................................................................................. 179 III. Xc nh p lc t ch ng.................................................................................. 179 IV. Xc nh p lc t b ng .................................................................................... 183

CHNG 6. XC NH LN CA NN CNG TRNH .......................................... 188 6.1. M u ..................................................................................................................... 188 6.2. XC NH LN C KT N NH. ............................................................ 188

I. Tnh ton ln c kt mt hng. .......................................................................... 188 II. Tnh ton ln c kt c xt n bin dng hng. ................................................. 195

6.3. Xc nh ln c kt theo thi gian ...................................................................... 200 I. L thuyt c kt thm ca Terzaghi. ......................................................................... 200 II. Tnh ln theo thi gian. ....................................................................................... 204

LI NI U

C hc t l mn hc c s k thut nhm trang b cho sinh vin kin thc c bn v t phc v cho mc ch xy dng. Gip cho sinh vin c kin thc tip cn chuyn mn lnh vc Nn Mng ni ring v Cng trnh xy dng ni chung. Mc d l mn mn c s c dy t lu trong cc trng khi k thut xy dng, c mt h thng gio trnh v sch tham kho kh hon chnh nhng vn phi thng xuyn cp nht, chnh sa p ng tt nht nhu cu hc tp, nghin cu, ng dng ca sinh vin cng nh cn b k thut xy dng.

B mn a k thut -Trng i hc Thu li bin son cun Bi ging C hc t vi mc tiu sao cho st vi chng trnh ging dy, cung cp c hu ht kin thc c bn ca mn hc v cp nht c thng tin t cc sch chuyn ngnh mi dch ca nc ngoi. Ni dung ca cun Bi ging ny c bn da trn ni dung ca cun C hc t do GS.TSKH Cao Vn Ch v bn dch cun: Gii thiu a k thut ca Hotz v Kovacs

Tham gia bin son gm cc thy c trong b mn:

ThS Hong Vit Hng vit chng 1

PGS.TS Nguyn Hu Thi vit chng 2

ThS Mc Th Ngc vit chng 3

PGS.TS Nguyn Hng Nam vit chng 4

ThS Phm Huy Dng vit chng 5

GVC.ThS Nguyn Vit Quang vit chng 6

Cun bi ging c tinh gin ni dung theo phng chm c bn, hin i c k tha kin thc v kinh nghim ca cc lp thy c ging dy ti b mn. Mc d tp th bin son rt c gng nhng khng th trnh c cc sai st. Rt mong nhn c s gp ca cc bn sinh vin v c gi cun bi ging ngy cng hon chnh hn.

Cc tc gi

CHNG 1. TNH CHT VT L CA T

1.1. Cc pha hp thnh t v tc dng tng h gia chng

t l sn phm ca s phong ha gc thnh cc ht t, cc ht t t sp xp to thnh khung ct t c nhiu l rng, trong cc l rng c cha nc v khng kh. Nh vy t gm 3 thnh phn vt cht:

Ht t ( pha rn).

Nc trong t ( pha lng).

Kh trong t ( pha kh).

Tnh cht ca t c xc nh bi cc yu t:

Tnh cht ca cc pha hp thnh t.

T l v s lng gia cc pha.

Cc tc dng a phn t, tc dng ha l, tc dng c hc gia cc pha vi nhau v gia cc nhm ht.

I- Pha rn (Ht t). Pha rn ca t bao gm cc ht t ( ht khong vt) c kch thc khc nhau chim phn ln

th tch khi t, to thnh khung ct t. C ba yu t ca pha rn nh hng n nhng tnh cht ca t l: Thnh phn khong vt ca ht t, kch thc ht t, hnh dng ht t. Sau y s i phn tch c im chi tit ca ba yu t ny.

1. Thnh phn khong vt ht t Khong vt c nh ngha l nhng n cht hay hp cht ha hc trong t nhin, hnh thnh

v tn ti trong v tri t hay trn mt t trong nhng iu kin a cht nht nh.

Hin nay khoa hc tm c khong 2800 khong vt trong c khong 50 loi khong vt to thnh t .

Cc c tnh ca khong vt c trnh by cc sch chuyn ngnh k thut a cht cng trnh.

V t l sn phm ca s phong ha gc do vy thnh phn khong vt ca t ph thuc ch yu vo thnh phn gc v tc dng phong ha . Tc dng phong ha khc nhau s sn sinh cc khong vt khc nhau ngay c khi tc dng phong ha trn cng mt loi gc.

Thnh phn khong vt ca ht t c th chia thnh ba loi: Khong vt nguyn sinh, khong vt th sinh ( hai loi ny l khong vt v c), v cht ha hp hu c.

Khong vt nguyn sinh thng gp l fenpt, thch anh v mica. Cc ht t c thnh phn khong vt nguyn sinh thng c kch thc ln, ln hn 0,005 mm.

Cc khong vt th sinh chia lm hai loi:

Khong vt khng ho tan trong nc, thng gp l kaolint, ilit v monmorilont, chng l thnh phn ch yu ca cc ht st trong t nn cn gi l khong vt st.

Khong vt ho tan trong nc thng gp l canxit, dolomt, mica trng, thch cao, mui m v.v ....

Cc khong vt th sinh thng c kch thc rt nh, nh hn 0,005 mm.

Cht ho hp hu c l sn phm c to ra t di tch thc vt v ng vt, giai on ph hu hon ton, sn phm ny c gi l mn hu c.

nh hng ca thnh phn khong vt n cc tnh cht ca t c th thy:

Vi t c kch thc ht ln: Thnh phn khong vt khng nh hng nhiu n tnh cht ca t.

Vi t c kch thc ht nh: Thnh phn khong vt nh hng rt nhiu n tnh cht ca t v chng nh hng n hot tnh b mt ht t dn n nh hng ti lp nc kt hp mt ngoi ht t. Cc tnh cht nh hng ny s c phn tch k hn trong phn phn tch s hnh thnh lp nc kt hp mt ngoi.

2. Thnh phn cp phi ht 2.1. Cc khi nim

Nhm ht: L tp hp cc ht trong t c kch thc nm trong mt phm vi nht nh.

Cp phi ht: Lng cha tng i ca cc nhm ht trong t tnh bng phn trm tng lng t kh.

2.2. Biu th cp phi ht ca t

Ly t v, sy kh, gi nh, lm th nghim phn tch ht. Mc ch ca th nghim phn tch ht l xc nh phm vi kch c ht trong t v phn trm cc ht ca mi nhm kch c. C hai phng php thng dng th nghim phn tch ht l phng php sng (ry) c thc hin vi t ht th v phng php t trng k ( phng php lng) c thc hin vi t ht mn.

a) Phng php sng

Vi cc ht c ng knh d > 0,1mm s s dng phng php sng. Phng php ny s dng mt h thng cc sng c kch thc mt sng khc nhau v thng c gi l b ry tiu chun. Cc tiu chun ban hnh khc nhau th c s chnh lch i cht v kch c mt sng. Chng hn theo tiu chun M ( US Standard) th quy nh sng s 4 c ng knh mt sng l 4,76mm, sng s 10 c ng knh 2mm vv. Nhng nguyn l chung khi phn tch ht th khng thay i.

Hnh 1.1: H thng ry tiu chun

phn tch ht theo phng php sng, mu t c sy kh sau gi nh lm ti mu t bng ci s - chy cao su. t sau khi gi ti c vo h thng sng v lc u. Cc ht ln ng cc sng bn trn, cc ht nh hn ng ln lt cc sng pha di. Cc nhm ht ng trn cc sng s c cn xc nh cp phi ht.

b) Phng php t trng k

Vi cc ht c ng knh nh d 0,1mm dng phng php t trng k phn tch ht (Lu tiu chun M th qui nh d 0,074mm). Phng php ny da trn nh lut Stokes, cc ht c ng knh khc nhau khi lng chm trong nc s lng ng vi cc tc khc nhau. nh lut Stokes a ra vn tc lng chm ca ht hnh cu:

v =

18ws d2 (1.1)

Trong :

v: Vn tc ca ht hnh cu lng ng trong cht lng.

d: ng knh ht.

s : Trng lng ring ht.

w: Trng lng ring ca nc.

: nht ca cht lng.

Nu bit khong cch lng chm l (h) v thi gian chm lng l (t) th s tnh c:

thv = (1.2)

T suy ra ng knh ht ( kt hp vi cng thc 1.1).

Sau khi lm th nghim phn tch ht, biu din cp phi ht bng ng cong cp phi.

Hnh 1.2: Phng php t trng k

V d 1.1:

C 300gam t kh, sau khi cho vo ry xc nh c khi lng ring trn mi ry nh sau:

Cc ht c ng knh d 0,1mm c khi lng l 60g, tip tc lm th nghim t trng k xc nh c:

0,05 mm < d 0,1mm => 30g

0,01 mm < d 0,05mm => 15g

0,005 mm < d 0,01mm => 9g

d 0,005mm => 6g

ng knh ry Khi lng

2mm 15g

1mm 15g

0,5mm 30g

0,25mm 60g

0,1mm 120g

y hng 60g

Lp bng tnh X%

d (mm) >2 21 10,5 0,5 0,25 0,25 0,1 0,1 0,05 0,05 0,01 0,01 0,005

V h s cp phi Cc:

1060

230

c D.DDC = (1.4)

D60 l ng knh ca ht, cc ht c ng knh nh hn v bng n chim 60% khi lng t kh.

D10 l ng knh ca ht, cc ht c ng knh nh hn v bng n chim 10% khi lng t kh. D10 thng c gi l ng knh hiu qu ca t.

D30 l ng knh ca ht, cc ht c ng knh nh hn v bng n chim 30% khi lng t kh.

H s Cu cng ln th ng cong cp phi cng xoi, t khng u ht ( cp phi tt). Ngc li nu h s Cu nh th t u ht ( Cp phi xu).

Thng thng t c gi l cp phi tt vi Cu > 46 v Cc t 13. Nu Cu =1 th t c gi l cp phi xu.

3. Hnh dng ht t t c kch thc ht ln ( Cc loi ct, cui, si, dm) thng cc ht c dng hnh khi,

hnh cu trn nhn, hnh gc cnh. Cc hnh dng ny nh hng nhiu n tnh cht ca t, c bit l tnh chng ct.

t c kch thc ht nh ( Cc loi set) thng cc ht c dng hnh phin, hnh kim. Cc hnh dng ny t nh hng n tnh cht ca t.

Hnh 1.4: Hnh dng ht t

II- Pha lng (Nc trong t) t trong t nhin lun tn ti mt lng nc nht nh v nhng dng khc nhau. Nc tc

dng mnh vi nhng ht khong vt, c bit l nhng ht nh c kch thc ht keo to nn hot tnh b mt ht t.

Theo quan im xy dng, nc trong t c phn thnh cc loi:

Nc trong ht khong vt Nc ht bm

Nc kt hp mt ngoi ht t Nc mng mng kt hp mnh

Nc mng mng kt hp yu

Nc t do Nc mao dn

Nc trng lc

4. c im tng loi nc trong t 4.1. Nc trong ht khong vt

L loi nc trong mng tinh th ca ht khong vt, n tn ti di dng phn t nc H2O hoc dng ion H+, OH. Loi nc ny ch c th tch khi khong vt khi nhit cao (>105C). Theo quan im xy dng loi nc ny c coi l mt b phn ca ht khong vt.

4.2. Nc kt hp mt ngoi ht t

Nc kt hp mt ngoi tn ti di tc dng ca lc ht in trng nn cc phn t nc v nhng ion dng b ht vo b mt ht t v c sp xp mt cch cht ch, c nh hng. Cng cch xa b mt ht t, lc ht cng yu nn s sp xp km cht ch v thiu qui tc hn. Do vy tnh cht ca nc kt hp mt ngoi rt khc so vi tnh cht ca nc thng thng.

Nc ht bm: C tnh cht gn vi th rn, khng c kh nng di chuyn, khng truyn p lc thy tnh, t trng khong 1,5. nhit -78oC nc ht bm mi ng bng. Khi t st ch cha nc ht bm s trng thi rn.

Nc mng mng kt hp mnh: C kh nng di chuyn theo hng bt k t ch mng nc dy sang ch mng nc mng, nhng s di chuyn khng lin quan n tc dng ca trng lc. Khi t st cha nc kt hp mnh, t s trng thi na rn.

Nc mng mng kt hp yu: C tnh cht gn ging vi nc thng thng. Khi t st cha lp nc mng mng kt hp yu v kt cu t b ph hoi th t th hin tnh do.

4.3. Nc t do

L loi nc nm ngoi lc ht in trng v chia lm 2 loi: Nc mao dn v nc trng lc.

a) Nc mao dn:

L loi nc b ko ln trong cc ng dn nh trong t, bn trn mc nc ngm, do sc cng b mt ca nc. Hin tng ny c th m t v gii thch tng t hin tng mao dn trong ng thy tinh nh.

Hnh 1.5: Hin tng mao dn v lc mao dn ti mt phn cch

cao mao dn c th xc nh t iu kin cn bng gia tng sc cng b mt ( Cn gi l lc nn mao dn) v tng trng lng ca ct nc dng ln trong ng:

k2

w h.4d.cos.d..T = (1.5)

=>

= cos.dT4h

wk (1.6)

Trong

hk: cao mao dn.

w: Trng lng ring ca nc.

d: ng knh ng thy tinh.

T: Sc cng b mt, ly gn ng T = 0,075.10-3KN/m.

: Gc nghing ca sc cng b mt vi thnh ng

T iu kin cn bng (a) c th rt ra tr s p lc mao dn Pk

Pk = = cosdT4h. kw (1.7)

p lc mao dn pk c tc dng nh mt lc dnh kt nu cht cc ht t vo nhau. iu ny tri ngc vi bn thn p lc nc l rng trong t v c th coi p lc mao dn l p lc nc l rng m.

Pk = w.hk = -uw (1.8)

Trong xy dng cn ch hin tng mao dn, cao mao dn v tc dng ln ca nc mao dn. Nc mao dn s lm cho t m t khin sc chu ti ca nn v tnh n nh ca mi dc gim. i vi nhng cng trnh nn v tr thp gn mc nc ngm cn ch hin tng mao dn.

b) Nc trng lc

Nc trng lc tn ti trong cc l rng ca t, chu s chi phi ca trng lc v tun theo nh lut Darcy. Cn quan tm n cc vn sau y ca nc trng lc.

Kh nng ha tan v phn gii ca nc.

nh hng ca p lc thy tnh i vi t v cng trnh.

nh hng ca lc thm.

Pha kh trong t Nu cc l rng ca t khng cha y nc th kh ( thng l khng kh) s chim nhng ch

cn li. Cn c nh hng ca kh i vi tnh cht c hc ca t, c th phn th kh trong t thnh hai loi:

Loi thng vi kh quyn.

Loi khng thng vi kh quyn.

Kh thng vi kh quyn khng c nh hng g ng k i vi tnh cht ca t, khi m cht kh ny s thot ra ngoi.

Kh khng thng vi kh quyn ( Bc kh Ti kh) thng thy trong cc loi t st. Loi kh ny c nhiu nh hng n tnh cht ca t, c bit l tnh thm v tnh m cht ca t.

1.2. Cc ch tiu tnh cht vt l v trng thi vt l ca t

Cc ch tiu tnh cht vt l ca t. t l sn phm ca s phong ha gc v gm 3 pha vt cht: pha rn, pha lng, pha kh.

Tnh cht vt l ca t ph thuc vo tnh cht ca 3 pha vt cht v t l v s lng gia 3 pha vt cht ny.

biu th nh lng t phn ca 3 pha vt cht hp thnh t, ngi ta thng dng s 3 th minh ha:

Hnh 1.6: S 3 pha vt cht to thnh t

Cc ch tiu tnh cht vt l ca t c th c chia lm hai nhm

1. Nhm cc ch tiu tnh cht vt l trc tip L cc ch tiu c xc nh trc tip t th nghim trong phng. Cc ch tiu ny bao gm:

1.1. Khi lng ring t nhin ca t : k hiu

T

T

VM

= (kg/m3) (1.9)

Mt: Khi lng tng cng ca mu t (kg).

Vt: Th tch tng cng ca mu t (m3).

Cch xc nh khi lng ring t nhin ca t: Dng mt dao vng c th tch V (cm3) v khi lng M1 (gam). C nhiu loi dao vng vi cc kch c khc nhau, dao vng ct cho t ht mn thng c kch thc b hn dao vo dng cho cc loi t sn si. Dng dao vng nay ct mu t ang cn xc nh khi lng ring. Ct sao cho t ngp y dao vng, sau gt bng hai mt ca mu t v cn mu. Xc nh c khi lng c t v dao l M2(gam). Theo nh ngha nu trn ta c:

V

MM 12 = (g/cm3) (1.10)

1.2. m ca t: W

m ca t l t s gia khi lng nc v khi lng ht trong mt mu t.

s

W

MM

W = (%) (1.11)

Cch xc nh m ca t: Dng mu t c kt cu khng cn nguyn dng nhng m phi cn nguyn vn. Cn mu v xc nh c khi lng M1 ca mu. Mang mu sy kh trong iu kin t sy nhit 105oC v thi gian sy trong khong 8gi, y ht nc ra khi mu. Sau khi sy kh cn xc nh c khi lng M2

100.M

MMW2

21 = (%) (1.12)

1.3. T trng ht t: Gs

ws

ss .V

MG

= (1.13)

Cch xc nh: Cn ht kh xc nh Ms, cho ht t kh vo nc xc nh th tch ht nh th tch nc dng ln trong bnh. Theo nh ngha trn s xc nh c Gs, w l khi lng ring ca nc v c gi tr bng 1000kg/m3.

2. Nhm cc ch tiu gin tip Cc ch tiu gin tip l cc ch tiu tnh c thng qua cc ch tiu trc tip bng cc cng

thc tnh i hoc cc lin h thng qua m hnh ba pha vt cht.

Cc ch tiu gin tip bao gm

2.1. Khi lng ring kh ca t: k hiu d

t

sd V

M= (1.14)

Cch xc nh: T cng thc nh ngha trn, c th chng minh c mt trong cc cng thc sau y

wd +

=1

(1.15)

v w l cc ch tiu tnh cht vt l trc tip, xc nh t th nghim trong phng, t y tnh ra c d.

d l ch tiu nh gi cht ca t p.

2.2. Khi lng ring ht s

s

ss V

M= (1.16)

Cch xc nh: T cng thc nh ngha, chng minh c: s =Gs. w. (1.17)

2.3. H s rng ca t (e)

s

v

VVe = (1.18)

Cch xc nh: T cng thc nh ngha, chng minh c cng thc:

1ed

s

= (1.19)

2.4. rng ca t (n)

(%)100VVn

t

o = (1.20)

Cch xc nh: T cng thc nh ngha, chng minh c cng thc

een+

=1

(1.21)

2.5. Khi lng ring bo ha (sat)

L khi lng ring ca t khi cc l rng trong t cha y nc.

t

wssat V

MM += (1.22)

wM l khi lng nc cha y trong l rng ca t. T cng thc trn c th chng minh c cng thc xc nh sat

wdt

wvd

t

w

t

s

t

wssat .nV

.VV

MVM

VMM

+=

+=

+=+

= (1.23)

2.6. Khi lng ring y ni ()

L khi lng ring ca t khi b ngp trong nc, cc ht t b y ni bi lc y acsimet.

t

sws

VV.M

= (1.24)


e1).1G( ws

+

= (1.25)

2.7. bo ha ca t (S)

o

w

VVS = (1.26)


ew.GS s= (1.27)

Cc bi tp v d:

V d 1.1:

Xut pht t cng thc nh ngha ca khi lng ring y ni, chng minh cng thc sau:

e1).1G( ws

+

=

Bi gii

Cng thc nh ngha ca khi lng ring y ni l:

t

sws

VV.M

= = sv

swss

VVV.V.

+

Chia c t v mu s cho Vs ta c: e1ws

+

=

Vi s = Gs. w ta c ( )

e11G ws

+

= (pcm)

V d 1.2:

Mt lai t c cc ch tiu tnh cht vt l trc tip c xc nh l:

= 1760 kg/m3; w = 10%; s = 2700 kg/m3

Yu cu: Dng s 3 pha, xc nh cc ch tiu d , e, n, S v sat.

Bi gii

Vi cc bi tp dng s 3pha xc nh cc ch tiu tnh cht vt l ca t th phi thc hin in tt c cc i lng trn s 3pha. Cc ch tiu s c tnh sau theo cng thc nh ngha.

Hnh 1.7: V d s dng s 3 pha xc nh cc ch tiu ca t

Gi thit Vt = 1 (m3), th c Mt = .Vt = 1760 (kg)

Vi W =10%, nn Mw = 0,1Ms

Nn Mt = 0,1Ms + Ms = 1760 (kg)

=> Ms = 1600 (kg)

Mw = 160 (kg)

Vi s = 2700 kg/m3; Ms = 1600 kg => Vs = 0,593 m3.

Vi Mw = 160 kg; w = 1000 kg/m3 => Vw = 0,16 m3.

Vi Vt = 1 m3 ; Vw = 0,16 m3; Vs = 0,593 m3 => Va = 0,247 m3.

Vv = Va + Vw = 0,407 m3.

Nh vy vic in y cc i lng trong s 3 pha hon thnh.

Tnh ton cc ch tiu:

16001

1600VM

t

sd === (kg/m

3).

686,0593,0

160,0247,0V

VVVVe

s

wa

s

v =+

=+

== .

%7,401000,1

160,0247,0100V

VVVVn

t

wa

t

v =+

=+

== .

%3,39100160,0247,0

160,1100VV

VVVS

wa

w

v

w =+

=+

== .

Th tch (m3)

Khi lng (Mg)

20701

1600)160247(V

MM

t

wssat =

++=

+= (kg/m3).

V d 1.3:

Mt lai t c khi lng ring = 1700 kg/m3.

Yu cu: tnh trng lng ring ca t .

Bi gii

Trng lng ring ca t (k hiu ) c tnh theo nh lut 2 ca Newton:

= .g = 17009,81 = 16677 (N/m3) hay 16,677 (KN/m3).

Cc ch tiu trng thi vt l ca t

3. Mc ch Trong xy dng, nu ch cn c cc ch tiu vt l nu trn th cha th c c nhn bit

y v mt loi t no . Nhng nu ni t trng thi xp hay cht, do mm, chy hay rn th s b nh gi c loi t no dng cho xy dng s tt hn.

4. Trng thi v cc ch tiu trng thi vt l ca t ri t ri l nhng loi t c kch thc ht ln, chng hn theo TCXD Vit Nam th cc ht c

kch thc ln l nhng ht c ng knh t (0,05mm 200mm). Cc loi t ny ri rc khng c tnh dnh. V d cc lai t ct.

Trng thi ca t ri c th phn tch ra trng thi cht v trng thi m.

4.1. Cc ch tiu nh gi cht

a) Dng ch tiu h s rng e nh gi cht

Theo nh ngha

s

v

VVe =

Tnh c cc gi tr ca e, so snh ch tiu qui nh ca qui phm bit c trng thi ca loi t ri ang xt.

b) Dng ch tiu cht tng i Dr

minmax

omaxr ee

eeD

=

Trong : Dr; cht tng i.

emax: H s rng ca loi t ri ang xt trng thi xp nht.

emin: H s rng ca loi t ri ang xt trng thi cht nht.

eo: H s rng ca t trng thi t nhin.

xc nh emax ta c:

1e mind

smax

=

mind c xc nh khi loi t ri ti xp nht. C th tm tt th nghim nh sau: Cn mt

khi lng ct kh M1(gam), lng ct ny vo trong ng nghim c sn mt cnh khoy. Ko cch khuy ln lm ti ct, c th tch ct V(cm3) trong ng nghim ta c

VM1min

d = (g/cm3)

xc nh emin, tc l xc nh h s rng ca t trng thi cht nht, ta c

1e maxd

smin

=

maxd c xc nh ng vi t ri trng thi cht nht. xc nh c thng s ny, dng

mt ci hnh tr bng ng c th tch V(cm3) v ct kh vo m cht, cn lng ct kh ny s c khi lng M1(gam).

VM1max

d = (g/cm3)

Thay maxd vo cng thc trn s tnh c emin.

Nhn xt:

Khi eo = emin th Dr = 1, t trng thi cht nht.

Khi eo = emax th Dr = 0, t trng thi xp nht.

Trong khong Dr bin i gi tr t 0 n 1 th hin c s thay i trng thi cht ca t ri t xp nht n cht nht. V vy trong khong 0-1 c th phn ra 3 mc nh gi cht

t ct cht 1> Dr > 0,67 t ct cht va 0,67 Dr 0,33 t ct xp 0 < Dr < 0,33

4.2. Ch tiu nh gi m ca t ri

Dng bo ha S nh ga m ca t ri

S > 0,8 t bo ha 0,5 < S 0,8 t m S 0,5 t hi m

5. Trng thi v cc ch tiu trng thi vt l ca t dnh 5.1. Trng thi vt l ca t dnh

t dnh thng cha phn ln nhng ht c kch thc ht keo, do trng thi vt l ca loi t ny khng nhng ch c quan h ti lng cha tng i gia cc th trong t m cn c quan h ti tc dng gia cc ht t v nc.

t dnh thng c cc trng thi sau: Rn, na rn, do, chy.

i vi t dnh, ch tiu cht v m khng th tch ri v khi m tng th th tch ca t cng tng ln, ng thi s thay i m cng quyt nh n s thay i trng thi ca t dnh.

Kt qu cc th nghim khi thay i m ca t dnh.

Hnh 1.8: Kt qu th nghim khi thay i m ca t dnh

Biu biu din kt qu th nghim ny cho thy khi m tng th t dnh chuyn dn trng thi t rn, sang na rn, sang do, sang chy v ngc li.

5.2. Gii hn Atterberg v ch s do

Gii hn Atterberg l nhng m qu khi t chuyn t trng thi ny sang trng thi khc. m qu ny c nh khoa hc Thy in l Atterberg pht hin ra vo nm 1911.

Atterberg phn bit ba loi m qu l gii hn chy, k hiu LL, gii hn do k hiu PL v gii hn co k hiu l SL. Cn mt s m qu na l m gii hn dnh ( Sticky limit) khng trnh by trong phn ny m ch yu s dng trong cc mc ch dng t khc.

Cc m qu ca t c xc nh trong phng th nghim. C th tm tt qui trnh xc nh cc m gii hn LL v PL nh sau:

xc nh LL ngi ta dng th nghim th chy Vaxiliep hoc th nghim ca Casagrande.

Hnh 1-9 l thit b th nghim ca Casagrande:

Hnh 1.9: Thit b th nghim ca Casagrande.

lm th nghim ny, mu t c ch b vi m sao cho gn vi m gii hn chy. Cho t vo trong bt ca thit b, dng dao ct rnh ct mt rnh ngn i t trong bt. Quay tay quay bt t g xung b 25 ln. Sau ln g th 25 rnh ct khp li ch cn mt on khong 13mm. Mang t i xc nh m v m ny l m gii hn chy.

xc nh PL ngi ta dng th nghim ln t thnh dy t. t c ch b ti m gn vi m gii hn do. Ln t thnh dy t c ng knh khong 3mm v t thnh tng on di t 3mm n 10mm th mang i xc nh m, ta c PL. Trong trng hp dy t c ng knh nh hn 3mm hoc dnh nhiu vo bn ln l t qu t, vt qu m gii hn do. Trong trng ng knh dy t ln hn th t qu kh. Vi nhng trng hp ny u phi lm th nghim li.

Khi m ca t bin thin trong phm vi PL v LL th t th hin tnh do. Tnh do l mt c trng quan trng ca t dnh v ngi ta thng dng ch s do, k hiu PI biu th phm vi do

PI = LL PL (1.28)

5.3. Ch tiu nh gi trng thi ca t dnh

Dng ch s st LI nh gi trng thi ca t dnh

PLLLPLwLI

= (1.29)

Trong : W: m ca t trng thi t nhin.

LI > 1 t trng thi chy. 0 LI 1 t trng thi do. LI < 0 t trng thi rn.

V d 1.4:

Th nghim mt loi t cho kt qu: Gii hn chy LL = 30; ch s do PI = 20; t trng Gs=2,7. Cho bit khi t gii hn chy th xem nh n trng thi bo ha nc.

Yu cu: Xc nh h s rng v rng ca t gii hn chy.

Bi gii

Ta c: eWG

S s.

=

Khi t bo ha: S=1 e = Gs.W = 2,7 0,3 = 0,81

rng: 448,081,01

81,01

=+

=+

=e

en

1.3. Phn loi t

Mc ch Phn loi dng

Lm c s la chn phng php nghin cu. c phng php s dng ng n cc loi t vi mc ch xy dng. Thng nht tn gi cho t.

Gii thiu mt s tiu chun phn loi t in hnh

1. H thng phn loi t thng nht USCS ( Unified Soil Classipication System)

H thng phn loi t ny do U.S.Burean of Reclamation a vao nm 1952. Hin nay c dng rt ph bin trn th gii. Theo cch phn loi ny t c chia lm 3 nhm chnh l t ht th, t ht mn v t hu c.

t ht th li chia thnh t cui si (G) v ct (S) W = Cp phi tt, M = Ht mn khng do, P = Cp phi xu, C = Ht mn do. t ht mn chia thnh bi (M) v st ( C), bi hu c hoc st hu c Do L Do thp nu LL < 50% Do H Do cao nu LL > 50% Tiu chun phn loi s b + Cui si nu hn 50% ht ng li trn sng s 4 (4,76mm).

+ Ct nu hn 50% ht lt qua sng 4 v hn 50% ht ng li trn sng 200 (0,074mm).

+ Bi st nu hn 50% ht lt qua sng 200 (0,074mm). Biu phn loi t theo USCS (ASTM D 2487).

Hnh 1.10: Biu phn loi t theo USCS (ASTM-D2487)

1.1. i vi Cui si v Ct

Nu c hn 5% lt qua sng s 200 (0,074mm) th c th l GW, GP, SW, SP

Trong

G l cui ( Gravel) W l cp phi tt (Well) P l cp phi xu ( Poor)

V d: GW c l cui si c cp phi tt, cn kt hp vi h s Cu, Cc phn bit.

Nu c hn 12% lt qua sng s 200 (0,074mm) th c th l GC, GM, SM, SC

Trong M l bi, C l st.

Cn da vo ch s do biu do phn bit.

1.2. i vi t ht mn

LL < 50% ML, CL, OL. LL >50% MH, CH, OH.

V d 1.5:

Lm th nghim vi mt loi t c phn trm ht lt qua sng nh sau:

Sng s 4 = 92% Sng s 10 = 81% LL = 48 Sng s 40 = 78% PI = 32 Sng s 200 = 65%

Bi gii

1) C hn 50% ht lt qua sng s 200 ( 0,074mm) vy y l t ht mn v c th l ML, CL, OL, MH, CH, OH.

2) Da vo biu do ta c LL v PI nm vng CL

Vy kt lun t th nghim l CL, t st c tnh do thp.

2. H thng phn loi t theo AASHTO H thng phn loi t AASHTO co U.S.Breau of Public Roads ngh. C 7 nhm phn loi

chnh t A-1 n A-7. Nhng loi t nm trong nhm l nhng loi t c tnh cht tng t nh nhau. Nhm t A-1 n A-3 l nhm ht th. T A-4 n A-7 l nhm ht mn.

S phn loi theo AASHTO da trn kt qu phn tch ht qua cc sng s 200, 40, 10 v ca cc th nghim chy - do. S khc bit gia cc nhm ht t A-1 n A-7 th hin bng ch s GI

GI = (F-35)[0,2+0,005(LL-40)] +0,01(F-15)(PI-10)

Trong F = % lt qua sng 200.

Cht lng chung ca cc lp t thng qua ch s nhm nh sau:

GI = 0 Cp phi rt tt. GI = 01 Cp phi tt. GI = 24 Cp phi trung bnh.

GI = 59 Cp phi xu. GI = 1020 Cp phi rt xu.

Hnh 1.11: Biu phn loi t theo AASHTO

V d 1.6:

Phn tch loi t theo tiu chun AASHTO cho 1 loi t, bit LL = 39% v PI = 19%. Phn trm ht lt qua sng

No10 (2mm) =41%

No40 (0,425mm) =29,5%

No10 (0,074mm) =21%

Bi gii

T biu phn loi t theo AASHTO th nhm A-2-6 ph hp hn c ( da vo ch s do PI chn).

Tnh GI

GI = 0,01(F-15)(PI-10) = 0,01(21-15)(19-10) = 0,54.

Vy t thuc nhm A-2-6 v c cp phi tt.

3. H thng phn loi t theo TCVN. Vit Nam hin nay ang tn ti hai tiu chun phn loi t l TCXD 45-78 v TCVN 5447-

1993. Tiu chun xy dng 45-78 c phn phn loi t v chia ra thnh t dnh v t ri. Cc gio trnh hin hnh phn ln trnh by theo TCXD 45-78 phn loi t.

TCVN 5447-1993 v c bn tng t nh tiu chun phn loi t thng nht USCS. Tuy nhin, tiu chun ny cha c dng ph bin. Sau y s trnh by mt s im chnh ca tiu chun TCXD 45-78.

3.1. Phn loi t theo TCXD 45-78.

3.1.1. Phn loi t dnh theo TCVN 45-78

t dnh c phn theo ch s do Ip ( hoc A):

Ip=WL-Wp

Vi quy nh:

- Phn nhm ht theo bng 1.1. (Bng tra sinh vin)

- Dng b ry tiu chun ca Lin X.

- Gii hn chy WL c xc nh theo phng php Vaxiliev vi t ch b, ht qua ry 0,1mm.

Ty thuc ch s do, t dnh c phn theo bng 3.1.

Bng 1.1

Ch thch bng:

a) Khi t dnh c cha nhng ht > 2mm th tn t trong bng c lm r nh sau:

- Nu lng cha t 12-25% khi lng th thm t c

ct c cui (dm), c si (sn).

st c cui (dm), c si (sn).

St c cui (dm), c si (sn).

- Nu lng cha t 25-50% khi lng th thm t pha

ct pha cui (dm), pha si (sn).

st pha cui (dm), pha si (sn).

St pha cui (dm), pha si (sn).

b) Khi t cha trn 50% khi lng nhng ht > 2mm th t c xp vo loi t ht th (bng phn loi t ri).

c) t dnh cn bao gm: t bn, t ln t v t trng n. t ln t v t trng n c xp vo loi t c bit.

Mi loi t dnh cn c lm sng t v kh nng chu lc thng qua st ca t ghi trong bng 3.2.

Tn t dnh Ch s do (A)

ct 1

Bng 1.2

st ca t dnh st tng i B

t ct

cng B < 0

do 0 B 1

lng B > 1

t st v st

cng B < 0

na cng 0 B 0,25

do cng 0,25 B 0,50

do mm 0,50 B 0,75

do nho 0,75 B 1,0

lng B>1

3.1.2. Phn loi t ri theo TCXD 45-78 (nn nh v cng trnh)

t ri c phn thnh: t ht th v t ct. Mi loi c phn thnh tng loi theo ch dn ca bng 3.3.

Bng 1.3

Tn t** Ch tiu phn loi*

t ht th

ln, tng Lng cha ht ln hn 200mm trn 50%

cui, dm Lng cha ht ln hn 10mm trn 50%

t si, sn Lng cha ht ln hn 2mm trn 50%

t ct

t ct ln si Lng cha ht ln hn 20mm trn 25%

t ct th Lng cha ht ln hn 0,5mm trn 50%

t ct va Lng cha ht ln hn 0,25mm trn 50%

t ct nh Lng cha ht ln hn 0,10mm bng v trn 75% (75%)

t ct mn ( ct bi) Lng cha ht ln hn 0,10mm di 75% (

Bng 1.4

bo ha S* Mc m

O < Sr 0,50 m t

0,50 < Sr 0,80 Rt m

0,80 < Sr 1,0 No nc (bo ha nc)

cht ca t c xc nh theo h s rng e ca mu t nguyn dng (bng 3.5) hoc cht tng i D (bng 3.6).

Bng 1.5. Phn loi cht ca t ri theo h s rng

Loi t cht ca t

Cht Cht va Xp

Ct cha sn, ct to v ct va e < 0,55 0,55 e 0,70 e > 0,70

Ct nh e < 0,60 0,60 e 0,75 e > 0,75

Ct mn (ct bi) e < 0,60 0,60 e 0,80 e > 0,80

Bng 1.6. Phn loi cht ca t ri theo cht tng i D

cht tng i D cht ca t

1 D > 0,66 Cht

0,66 D > 0,33 Cht va

0,33 > D 0 Xp

Bi tp chng I

1) Mt mu t ly t mt tng t nm di tng nc di t c m = 44%, c t trng = 2,7. Hy tm: h s rng , rng n, trng lng ring t nhin (cng l trng lng ring bo ha v t di mc nc ngm), trng lng ring kh v trng lng ring y ni ca t .

2) Mt mu t c trng lng ring t nhin =18kN/m3, m =25%, t trng =2,7. Hy xc nh trng lng ring kh, h s rng v bo ha ca t .

3) Cho bit 1m3 ct kh nng 16,5 kN (k=16,5kN/m3). Cho ct bo ha nc. Bit t trng ca ct =2,65. Hy xc nh h s rng v m ca ct .

4) Mt loi t c trng lng ring t nhin =17kN/m3 vi m =15%. Tnh m ca t sau khi thm vo 1m3 t 120 lt nc.

5) Mt loi t c trng lng ring t nhin =17kN/m3 vi m =15%. Xc nh trng lng ring ca t khi c m l 25%. (Cho bit th tch ca t khng i khi m ca n thay i, ngha l cho bit t c k=const).

6) Th nghim mt loi t cho kt qu: Gii hn do =30, ch s do A=20, t trng =2,7. Hy xc nh h s rng v rng ca t gii hn chy. Cho bit khi t gii hn chy th xem nh n bo ha nc.

7) Lm th nghim hai loi t thy chng c cng gii hn chy (ch=40% v d=25%). Nhng loi t th nht v loi t th hai c m t nhin ln lt l =45% v =20%. Hy xc nh tn t v trng thi ca hai loi t . Loi no dng lm nn tt hn?

8) Phn tch ht mt lng ct kh khi lng 300g. Cn lng ht ry ng knh 0,5mm l 120g, lng ht ry ng knh 0,25mm l 90g. Xc nh tn loi ct ? Cng loi ct em lm th nghim nhn c max=1,2 v min=0,7. Hy xc nh trng thi t nhin ca loi ct . Cho h s rng trng thi t nhin ca n l =0,9.

9) Xut pht t nh ngha chng minh cc cng thc sau:

+

=

1)1(n

n ; 1)1(

+

=

n

CHNG 2. TNH CHT C HC CA T Nh trnh by trong Chng I, t c cc c im c bn c tm tt nh sau:

Tnh ri, rng.

Cng lin kt gia cc ht nh hn nhiu ln cng bn thn ht t.

t thng gm 2, 3 pha (Rn, Lng, Kh), tc dng v nh hng ln nhau

Di tc dng ca ti trng , tnh rng ca t thay i v do cc tnh cht c hc ca t thay i theo.

Tt c nhng c im nu trn to cho t nhng tnh cht c hc in hnh, c th phn bit r rt vi cc vt rn lin tc nh b tng, thp:

Tnh thm nc,

Tnh p co v bin dng kt cu pha,

Tnh chng trt ma st.

2.1. Tnh thm nc ca t

I. Khi nim dng thm trong t t gm cc ht phn tn khong rng gia chng lin thng vi nhau nn nc c th chy t

do bn trong khi t.

Trong mi trng rng nh vy, nc s chy t vng c p lc cao ti vng c p lc thp. V vy, c th nh ngha tnh thm ca t l kh nng ca t cho nc i qua.

Dng thm c th l n nh hoc khng n nh, tng ng vi cc iu kin l hng s hoc bin i theo thi gian. Trong a K Thut, dng thm sinh ra trong trng ng sut l dng khng n nh trong mi trng c l rng thay i theo thi gian.

Dng chy cng c th c phn loi thnh mt-, hai- hay ba-chiu. Dng thm trong a k thut thng c gi s l mt- hoc hai-chiu v iu ny l ph hp vi hu ht cc vn thc t.

Trong a k thut, ti cc mc p lc thng thng c th b qua cc thay i khi lng ring, nn dng chy ca nc trong t c coi nh khng nn c.

Dng chy c th l chy tng, hoc chy ri. Trng thi qu tn ti gia dng chy tng v dng chy ri.

Trong hu ht cc loi t, dng chy c vn tc rt nh nn c th coi l dng chy tng. Do vy t hnh 2.1, v t l vi i :

v = ki (2.1)

Phng trnh ny chnh l nh lut Darcy (s xt k hn phn sau)

Hnh 2.1: Cc vng dng chy tng v dng chy ri (theo Taylor 1948)

Phng trnh Bernoulli di dng nng lng ca mt n v trng lng (cho dng chy n nh khng nn c) (Thy lc hc):

=++=++ 22

22

11

21

22z

gpp

gvz

gpp

gv

ww

constant total head (2.2)

Theo phng trnh ny: nng lung tng (hay ct nc tng) ca h l tng ca ct nc vn tc v2/2g, ct nc p lc p/wg v ct nc th z .

Ty thuc vo dng chy trong cc ng, knh h hoc qua mi trng rng s tn ti cc tn tht ct nc hoc tn tht nng lng i km vi dng chy, hf.

fww

hzgp

pg

vzgp

pg

v+++=++ 2

222

11

21

22 (2.3)

trong :

v1, v2 - vn tc ti mt ct 1 v 2

g - gia tc trng trng

w - khi lng ring cht lng (nc)

p1, p2 - p lc ti mt ct 1 v 2

z1, z2 - cao mt ct 1 v 2 so vi mt chun

Trong phng trnh Bernoulli p dng gii cc bi ton thm cho t:

p1 v p2 = p lc gy ra ct nc p lc = p lc nc l rng u, v vy ct nc p lc tng c th vit li:

zug

vH

w

++=2

2

(2.4)

Do t c kt cu ht, dng thm chu sc cn ln nn v thng qu nh, v vy c th b qua ct nc vn tc:

zuHw

+=

(2.4)

Vi:

H = h + z

h = H1 - H2

i = h/L

Nh vy, dng thm sinh ra trong t khng ch do chnh ct nc trng trng, m quan trng l do chnh ct nc p lc.

Hnh 2.2

thy r hn bn cht ca dng thm trong t, cng nh iu kin p dng nguyn l dng chy (cng thc 2.4, 2.5) cho t, ta cng cn phn bit dng thm thc v dng thm khng thc:

Trong cc phng trnh nu trn chng ta s dng din tch ton b mt ct ngang trong khi r rng nc khng th chy xuyn qua cc ht rn m ch qua cc l rng gia cc ht t.

Vy ti sao ta khng s dng phn din tch rng v tnh tc thm da trn din tch rng ?

Vi mt chiu rng n v ca mu trong hnh 2.3, chng ta c th d dng tnh din tch phn rng qua cng thc h s rng:

v v

s s

V AeV A

= =

Hnh 2.3: Tc thm v tc b mt trong dng chy u (theo Taylor 1948)

Vn tc vo va v vn tc ra vd trong hnh 2.3 u bng v = q/A . Do vy v trong quan h ny l vn tc mt, i lng khng thc nhng thun tin trong k thut.

Vn tc thm thc vs, l vn tc thc ca dng nc chy qua cc l rng. Chng ta c:

v = nvs

Do 0% n 100%, vn tc thm thc lun ln hn vn tc b mt (vn tc ra). Nh vy, H s rng hay rng ca t nh hng n dng chy ca nc qua n v do nh hng n gi tr h s thm ca mt loi t (k).

Hnh 2.4: M hnh ca tc thm v tc b mt ca dng chy

(dng chy vung gc vi trang giy)

nh lut Darcy Hn mt trm nm trc, k s thy lc ngi Php tn l Darcy (DArcy, 1856) thng qua cc

th nghim ch ra rng vn tc cht lng trong ct sch t l vi gradien thy lc, PT (2.1):

v = ki (2.1)

v vA V nA V

= =

a d s vq v A v A vA v A= = = =

p dng nh lut bo ton khi lng (trong c hc cht lng) cho dng chy n nh khng nn c, chuyn thnh phng trnh lin tc:

q = v1A1 = v2A2 = constant (2.6)

T (2.1), (2.6), nh lut Darcy thng c vit l (Hnh 2.3):

ALhkkiAvAq === (2.7)

Trong :

q - lu lng thm trong n v thi gian qua mt ct A

(n v: th tch/thi gian, m3/s)

v1,v2 - vn tc ti mt ct 1 v 2

A1, A2 - din tch mt ct 1 v 2

k - h s thm Darcy, hoc l h s thm. k c n v ca vn tc [m/s, cm/s] (bi i khng c th nguyn)

Mt s yu t nh hng n h s thm (k):

ng knh hiu qu, mc khng u ht

Tnh phc tp ca hnh dng l rng v cc ng chy qua cc l rng

bo ha S

Cc c tnh ca cht lng; nht, i lng ph thuc vo nhit , v t trng.

Phm vi thch dng ca nh lut Darcy:

Cc th nghim thn trng cho thy l PT 2.7 (hoc 2.1) ng cho mt phm vi rng cc loi t khc nhau, c bit l t ct sch.

Vi si rt sch v khi p bng cp phi h, dng thm c th l ri v nh lut Darcy khng c gi tr.

Vi cc t mn (t st) khi gradien thy lc rt thp, mi quan h gia v v i l phi tuyn (Hnh 2.5).

v = k2(i-io) vi i i1 (2.8a)

v = k1i n vi i < i1 (2.8b)

Vi t st Thy in in hnh s m n c gi tr trung bnh vo khong 1.5.

Tuy nhin khng c s nht tr hon ton vi khi nim c ch ra trong Hnh 2.5. on cong ca ng v~i thc t khng n nh, kh xc nh. Hin nay tha nhn ng v~i ko di ct ti io ( dc thy lc ban u) cho t dnh:

v = k(i io) (2.8c)

Hnh 2.5: lch so vi nh lut Darcy c quan st trong t st Thy in (theo Hansbo 1960)

H s thm v phng php xc nh H s thm rt cn thit cho vic thit k cc cng trnh k thut c s xut hin ca dng thm.

C th xc nh k bng cc th nghim trong phng v hin trng.

Th nghim trong phng: S dng thit b my o thm trong cc th nghim

- ct nc khng i (hnh 2.6a)

- ct nc gim dn (hnh 2.6b)

Th nghim hin trng: Thng s dng thit b bm trong cc th nghim

- ct nc khng i

- ct nc gim dn

Trong khun kh mn hc ch trnh by cc th nghim trong phng.

1. Th nghim ct nc khng i: Th tch nc Q thu nhn c trong thi gian t l

Q = Avt

Theo nh lut Darcy, Lhkikv ==

T rt ra: tAh

LQk = (2.9)

Trong :

Q - tng th tch nc thot ra (m3) trong thi gian t (s)

A - din tch mt ct ngang ca mu t (m2)

Hnh 2.6,a

V d 2.1:

Mu t hnh tr trn, ng knh 7.3 cm v di 16.8 cm, c th nghim vi thit b o thm ct nc khng i. Ct nc 75 cm c duy tr trong sut thi gian th nghim. Sau 1 pht th nghim, thu c tng cng 945.7 g nc. Nhit l 20oC. H s rng ca t l 0.43.

Yu cu: Tnh h s thm theo cm/s.

Bi gii:

Din tch mt ct ngang ca mu t:

222

cm9.41)3.7(44

===DA

T phng trnh 2.9, thay s liu tm k:

scmscmcm

cmcmhAtQLk /08,0

min/60min19,41758,167,945

2

3

=

==

2. Th nghim ct nc gim dn:

Vn tc gim trong ng o p l: dtdhv =

Lu lng chy vo mu t l: dtdhaqin =

T nh lut Darcy (phng trnh 2.7), lu lng chy ra l:

ALhkkiAqout ==

Theo phng trnh lin tc 2.6, qin = qout nn:

ALhk

dtdha =

Hnh 2.6,b

Phn ly bin s v tch phn hai v phng trnh trn cc gii hn s c:

1 2

2 1

h t

h t

dh Aa k dth L

=

v nhn c: 2

1lnhh

tAaLk

= (2.10a)

y, t = t2 - t1 . Theo log10 ta c: 2

110log3,2 h

htA

aLk

= (2.10b)

Trong : a - din tch ng o p

A, L - din tch v chiu di mu t

t - thi gian ct nc trong ng o p gim t h1 n h2

V d 2.2:

Th nghim ct nc gim dn c tin hnh trong phng vi t ct ln si xm nht (SW) v thu c cc d liu sau (nhit nc l 20oC):

a = 6.25 cm2 h1 = 160.2 cm A = 10.73 cm2 h2 = 80.1 cm L = 16.28 cm t = 90 s

Cho ct nc gim t h1 n h2

Yu cu: Tnh h s thm theo cm/s.

Bi gii:

S dng phng trnh 2.10b ta c:

( )o

6.25 16.28 160.22.3 log10.73 90 80.1

0.07 cm/s 20

k

C

=

=

Ch : nu nhit nc khc 20oC khi phi s dng cc h s chuyn i tnh ng nht ca nc.

Cc nhn t nh hng n kt qu th nghim thm trong phng:

S tn ti cc bc kh lm t khng hon ton bo ha, bo ha < 100%

S dch chuyn ca cc ht mn trong mu th nghim

S thay i nhit , c bit trong cc th nghim vi thi gian di

Cu trc t nhin ca mu t th nghim kh c m bo

xt n mt cch tng i chnh xc bn cht hay thay i v tnh khng ng nht theo t nhin ca cc loi t trm tch v kh khn cng nh nhng hn ch ca cc th nghim trong phng, nn tin hnh cc th nghim bm hin trng o h s thm trung bnh ton b khu vc.

H s thm cng c th xc nh da trn cc th nghim nn mt hng trong phng hoc th nghim ba trc.

C th xc nh h s thm nh cc cng thc thc nghim - cc cng thc 7.10, 7.11, 7.12; hoc cc bng gi tr k cho cc loi t khc nhau, v d nh hnh 7.6 (Casagrande 1938). - (Xem trong Gii thiu KT, W.D. Kovas)

2.2. Tnh p co v bin dng ca t

Khi nim tnh p co v bin dng ca t Gi s bin dng ca lp t chu nn ch theo mt hng, nh trng hp bin dng gy

ra bi ti trng thng ng trn mt vng t rng.

Hnh 2.7

Xt phn t t ti su z, chu nn mt hng:

bin dng theo phng z: z 0

bin dng theo phng x: x = 0

bin dng theo phng y: y = 0

Khi chu ti trng, t b p co, bin dng ca t sinh ra l do th tch l rng thay i. Nh vy, Bin thin th tch ca t chnh l do th tch rng thu hp, V Vv

Quan h gia bin thin th tch (V) v h s rng (e) Xt bi ton: Mt khi t c th tch ban u V1, h s rng e1. Hy tnh bin thin th

tch V khi h s rng l e2. (vi e1 > e2).

Tnh th tch ht t Vs1 c trong V1: Vs1 = V1m1 = 1

1 11

eV

+

Tnh th tch ht t Vs2 c trong V2: Vs2 = V2m2 = 2

2 11e

V+

Nh bit, bin thin th tch khi t l do th tch rng thu hp li (th tch ht khng i), ngha l:

Vs2 = Vs1 2

2 11e

V+

= 1

1 11e

V+

1

212 11

eeVV

++

=

V = V1 - V2 1

211 1 e

eeVV+

= , e1 - e2 = e , l bin thin h s rng

ee

VV +

=1

1

1 (2.11a)

eV = (2.11b)

Nh vy, Bin thin th tch ca t t l bc nht vi bin thin h s rng

Cng c th vit (2.11a) di dng bin dng th tch tng i:

11 ee

VV

v +

=

= (2.11c)

vi zyxv ++= (theo l thuyt n hi) (2.11d)

Th nghim p co khng n hng v nh lut p co

1. Th nghim p co khng n hng m phng p co mt hng trong phng th nghim, thng nn mu t bng thit b nn

khng n hng hay nn c kt. (Hai dng thit b nn khng n hng c th hin hnh 2.8).

Mt mu t nguyn dng, i biu cho mt phn t t b nn trong nn, c ct gt to mu cn thn v t vo hp nn.

Hnh 2.8: S thit b th nghim nn khng n hng (a) hp nn di ng (b) hp nn c nh

(theo Hi cc k s qun i M).

Hp nn thnh cng khng cho php bin dng ngang xy ra. Trn v di mu t c lt thm khi chu nn th nc thot ra. Thng thng thm nh mu c ng knh nh hn ng knh ca hp nn cng khong 0,5 mm, n khng th ko r dc theo thnh khi ti trng tc dng. T s gia ng knh v chiu cao mu trong khong t 2,5 n 5 v cc ng knh ny thng ph thuc vo ng knh ca mu nguyn dng khi th nghim.

Bin php lm gim ma st thnh bng cch dng hp nn bng s hoc cc cht bi trn xung quanh.

Th nghim hp nn di ng: qu trnh nn din ra c hai mt mu th nghim. C th thy rng ma st hp nn th nghim ny nh hn th nghim hp nn c nh

Th nghim hp nn c nh: t ch chuyn v xung. u im c bn ca th nghim hp nn c nh l c th o hoc kim sot nc thot ra t thm y. Cng c th kt hp tin hnh th nghim thm bng hp nn.

Tin hnh th nghim thit lp quan h gia ti trng v bin dng theo mu th nghim p co khng n hng:

Ti trng tc dng ln mu tng dn tng cp (c th tng ti bng h thng tay n c hc hoc bng kh nn).

Vi mi cp ti trng tc dng, ch cho mu t ln n nh v p lc nc l rng d trong mu xp x v khng. ng sut cui cng hay ng sut cn bng c gi l ng sut hiu qu.

Qu trnh ny c lp li cho n khi s im d liu th hin ng cong quan h bin dng ~ ng sut (s ~ vc )

T Ct. (2.11c): 11 e

eVV

v +

=

= Hnh 2.9

+

= ++i

iii

eee

Hs

111

Hs

eee iiii1

1 )1(+

+ += (2.11e)

=

= +H

sVV i

v1

Hsi

ii1

1+

+ = (2.11f)

Da vo cc Ct. (2.11e v 2.11f) kt hp ng quan h thc nghim (s ~ vc ), ta c th xy dng cc ng quan h (e ~ vc) v ( ~ vc) .

Mc tiu ca th nghim c kt l:

m phng s p co ca t di tc dng ca ti trng ngoi cho.

Xc nh thng s mun ca t khi nn khng n hng.

D on ln ca cc lp t hin trng bng cch nh gi cc c trng nn ca mu nguyn dng tiu biu.

Hai phng php biu din d liu ti trng-bin dng c th hin Hnh 2.10:

Hnh 2.10: Hai cch th hin d liu th nghim c kt: a) Phn trm c kt (hay bin

dng) v% vi ng sut hiu qu vc ; b) H s rng e vi ng sut hiu qu vc

(Th nghim vi t bn ti vnh San Francisco su -7,3m).

C hai th ny u cho thy t l vt liu bin dng tng bn, c ngha l gi tr mun (tc thi) tng khi ng sut tng. Quan h ng sut-bin dng th hin hnh 2.10 l hon ton phi tuyn.

Hnh 2.11: Th hin d liu th nghim c kt trn h trc bn logarit (cng s liu vi Hnh 2.10): a) Phn trm c kt (hay bin dng) v% vi log ng sut hiu qu vc ;

b) H s rng e vi log ng sut hiu qu vc .

C th thy c hai th u c hai on gn nh thng ni tip vi ng cong chuyn tip trn. ng sut ti im chuyn tip hay l im gy xut hin ng cong th hin hnh 2.11 ch ra gi tr ng sut lp ph thng ng ln nht m mu t ny chu trong qu kh. Gi tr ny c hiu l gi tr ng sut c kt trc p. i khi cng dng k hiu pc hay vm, ch m vit di biu th p lc qu kh ln nht.

H s qu c kt OCR:

H s qu c kt l t s gia ng sut c kt trc v ng sut nn hiu qu hin ti theo phng ng:

'

'

vo

pOCR

= (2.12)

p = p lc c kt trc

vo = p lc lp ph thng ng hin ti

OCR=1, ngha l p = vo t c kt bnh thng (NC)

OCR>1, ngha l p > vo t qu c kt (OC)

OCR

Cc bc thao tc ca Casagrande nh sau:

1) Chn bng mt mt im c bn knh cong nh nht (hoc cong nht) ca ng cong c kt (im A trn hnh 2.12).

2) T im A k ng nm ngang.

3) T im A k ng tip tuyn vi ng cong c kt.

4) K ng phn gic ca gc c to bi bc 2 v 3.

5) Ko di on ng thng ca ng cong nn nguyn sinh cho n khi ct ng phn gic to bc 4. Giao im ny cho ta tr s ng sut c kt trc ( im B trn hnh 2.12).

Mt phng php n gin hn nh gi tr s ng sut c kt trc c mt s k s s dng: - Ko di hai on thng ca ng cong c kt, im giao nhau ca chng cho p lc c kt trc c th ng nht (im C trn hnh 2.12).

Nu trn hnh 2.12 th tr s ln nht p c th s l gi tr ti im D, tr s nh nht p c th l tr s ti im E, l giao im ca ng cong nn nguyn sinh vi ng nm ngang k t tr s eo.

Hnh 2.12: Phng php Casagrande(1936) xc nh ng sut qu c kt, trn th ch ra

c tr s nh nht, c th xy ra nht v ln nht c th ca ng sut c kt trc.

2. nh lut p co khng n hng v cc c trng p co ca t Khi kt qu th nghim c biu th theo h s rng: (hnh 2.10b)

dc ca ng cong (e vc ) ti im bt k c xc nh bng tr s o hm ti im :

vv

adde

iv

=

,

(2.13)

Nhn xt:

- Khi v,i nh, av ln t d p co

- Khi v,i ln, av nh t kh p co

Nh vy, av biu th mc p co ca t, gi l h s p co (h s nn); n v thng

l [m2/kN].

Trong nhiu trng hp, phm vi thay i ca v khng ln (100300 kPa) c th coi l on thng. V vy ta c th vit li (2.11) di dng gn ng:

vv

ae =

(2.14)

T ta c: vvae = (2.15)

vi: e = e i - e i+1 (2.16a)

v = v,i+1 - v,i (2.16b)

Pht biu nh lut p co ca t:

Khi bin thin p lc nn khng ln th bin thin h s rng t l bc nht vi bin thin p lc. nh lut p co c th hin bng cng thc 2.13.

Khi kt qu th nghim c biu th theo bin dng (hnh 8.4a) th dc ca ng cong nn ln c gi l h s bin thin th tch, mv, tc l:

o

v

v

v

v

vv e

addm

+=

==1,,

(2.17)

Trong , v l bin dng ng. Trong nn mt hng, v = e/(1+eo).

Khi kt qu th nghim c biu din bng quan h (e log v), hnh 2.11b, th dc ca ng cong p co nguyn sinh c gi l ch s nn Cc

,1

,2

21,1

,2

21,

logloglog)(logeeee

ddeC

vc

=

=

= (2.18)

Ch : Cc khng th nguyn

Khi kt qu th nghim biu th bng (v % log v), Hnh 2.11a, dc ca ng cong nn nguyn sinh c gi l ch s nn ci bin Cc (i khi, t s nn):

,1

,2log

vcC

= (2.19)

Quan h gia ch s nn ci bin Cc v ch s nn Cc c biu din bng:

o

cc e

CC+

=1

(2.20)

Ch s nn li Cr l dc trung bnh ca phn nn li ca ng cong (elogvc), Hnh 2.14 - Cr c nh ngha tng t Cc , PT (2.18).

Nu kt qu th nghim c v bng quan h (v logvc) th dc ca ng cong nn li c gi l ch s nn li ci bin Cr (i khi gi l t s nn li).

Lin h Cr v Cr :

o

rr e

CC+

=1

(2.21)

Hnh2.1

Hnh 2.14: ng cong e ~ log (p lc) minh ha qu trnh trm tch, ly mu (d ti) v c kt li trong thit b th nghim c kt

V d 2.3:

Th nghim nn khng n hng trong phng th nghim mt mu t vi 1=100 KN/m2 th xc nh c h s rng e1=1,2; vi 2=200KN/m2 th xc nh c h s rng e2=1,1.

Yu cu: Tnh h s p co (h s nn ln) ca loi t ny.

Bi gii

H s p co ca loi t ny l:

001,0100200

1,12,1

12

21 =

=

=eeav (m

2/kN)

3. Th nghim bn nn ti hin trng v nguyn l bin dng tuyn tnh 3.1. Th nghim bn nn

(bn nn ch chu ti trng thng ng)

Khi p < pIgh , bin dng ng l ch yu, do Vv thu hp; quan h /s v b/d trong nn l tuyn tnh. cui giai on I (p = pIgh) bin dng do xut hin u tin ti hai mp bn nn pht trin thnh vng do (su khong B)

Khi p > pIgh , vng do pht trin theo p tng, quan h S~p trong nn l phi tuyn. Khi p pIIgh b/d do chim u th, cong cng ln.

khi p = pIIgh , vng do pht trin hon ton, khi nn trng thi Cn bng Gii hn. Tng mt lng p rt nh, nn b ph hoi trt (p tri).

Hnh 2.15

3.2. Nguyn l bin dng tuyn tnh

Theo kt qu th nghim bn nn:

Khi p < pIgh , bin dng ca t ch yu do th tch rng (Vv) thu hp; quan h (S p) c dng gn thng, c th coi l bc nht. T c th a ra nguyn l bin dng tuyn tnh:

Khi ti trng tc dng khng ln (p < pIgh ), quan h (S p) c dng gn thng th c th xem t nh vt liu bin dng tuyn tnh, v quan h gia ln v p lc ln nn l bc nht.

Nguyn l bin dng tuyn tnh c th vn dng cho t nh sau:

So snh c tnh bin dng ca 2 vt liu (xem Hnh v):

Thp (vt liu n hi) v

t (vt liu ri)

C s tng t v hnh thc, quan h gia bin dng v p lc l bc nht. V th, trong giai on bin dng tuyn tnh, ta c th vn dng cc biu thc ca l thuyt n hi tnh cho t, trc ht l nh lut Hooke lin h bin dng vi ng sut.

Vi gi thit phn t t l n hi ng hng, ngha l Ex = Ey = Ez = Eo , th nh lut Hooke c th hin di dng cc biu thc (2.20):

x = 1/Eo [x - o(y + z)]

y = 1/Eo [y - o(z + x)] (2.22)

z = 1/Eo [z - o(x + y)]

trong ,

Eo v o tng ng l mun bin dng v h s n hng ca t.

x , y , z l cc ng sut php tc dng ln phn t t theo cc phng x, y, z .

x, y , z l cc bin dng tng i ca phn t t theo cc phng x, y, z .

Xc nh cc c trng bin dng ca t

4. Xc nh h s n hng, o. Vi phn t t chu nn mt hng

(n hng t do):

0

0

=

==

yx

yx

z

y

z

xo

== (2.23)

5. Xc nh h s p lc hng, Ko.

Hnh 2.16

Hnh 2.18

Hnh 2.17

Hnh 2.19

Vi phn t t chu nn khng n hng:

0

0

==

=

yx

yx

z

y

z

xoK

== (2.24)

Quan h gia h s p lc hng Ko v h s n hng o :

V mu t b nn khng n hng, t (2.24), x = y = 0 v (2.22), x =1/Eo [x - o(y + z)] = 0, thu c:

x = o(y + z) = oy + oz

zo

oyx

==1

o

o

z

xoK

==

1 (2.25)

6. Xc nh mun bin dng, Eo. M un Eo - l mt c trng bin dng quan trng ca t, c ngha tng t mun

n hi Ee, nhng khc v bn cht:

Ee biu th tnh n hi ca t

Eo biu th tnh bin dng ca t, bao gm bin dng d (khng hi phc) v bin dng n hi (hi phc), trong bin dng d l ch yu.

6.1. Xc nh Eo t th nghim nn khng n hng

Mu t b nn trong iu kin khng n hng:

T PT (2.11c), 11 e

eVV

v +

=

= zvv ea

VV

11+=

= (a)

Theo LT n hi: )(21 zyxo

ozyxv E

++=++= (b)

theo PT (2.25), zo

oyx

==

1, thay vo (b), nhn c:

o

z

ov E

o

=

12

12

(c)

cn bng (c) vi (a), v t

=

o

o

12

12

(2.26)

cui cng nhn c: v

o aeE 11+= (2.27)

V d 2.4:

Th nghim nn t ti hin trng mt h o vi mt bn nn trn c din tch bn nn F=5000 cm2 (d=2 /F =79,8cm), kt qu bng sau:

P(kN/m2) 0 100 150 200 250 300 350 400

S(mm) 0 8 12 20 32 65 100 150

Cng loi t , khi th nghim p co khng n hng trong phng th nghim mt mu t c chiu cao ho=2,54cm cho kt qu sau:

P(kN/m2) 0 100 200 300 400

S(mm) 0 1,24 1,71 2,10 2,35

Yu cu:

a) Tnh cc h s rng e i v v quan h e-p trong th nghim p co khng n hng, cho bit h s rng ban u ca t eo=0,814.

b) Tnh m uyn bin dng ca t theo kt qu ca hai phng php th nghim ng vi cng cp ti trng thay i t p=0 n p=100kN/m2. So snh kt qu v cho nhn xt.

Bi gii

a) Tnh cc h s rng ei

T cng thc ei = eo -(1+ eo)o

i

HS

Vi p=100KN/m 2ta c:

725,04,25

24,1)814,01(814,0100 =+=e

692,04,25

71,1)814,01(814,0200 =+=e

664,04,25

1,2)814,01(814,0300 =+=e

646,04,25

35,2)814,01(814,0400 =+=e

V ng quan h e-p:

b) Tnh E trong khong p=0 n p=100 KN/m2

562,163000089,0

814,018,01

10

00 =

+=

+=

ae

E

Vi cc bi tp dng s 3pha xc nh cc ch tiu tnh cht vt l ca t th phi thc hin

Vi cc bi tp dng s 3pha xc nh cc ch tiu tnh cht vt l ca t th phi thc hin

6.2. Xc nh Eo t th nghim bn nn

Khi p pIgh , l thuyt n hi chng minh c ln ca mt bn nn trn t trn mt bn khng gian bin dng tuyn tnh:

dP

ES

o

o

21 =

SP

dE oo

21 = (2.28a)

Nu bn nn vung, c th ly ng knh tng ng theo cng thc: Fd 2= , vi

F l din tch y bn nn vung.

Lu : Biu thc trn tnh cho mi trng bn khng gian v hn n hi. Tuy nhin i vi t th phm vi nh hng ca P khng ra v cng m hu hn, v th khi dng cn phi hiu chnh bng cch thm mt h s thc nghim vo cng thc, mo < 1.

SP

dmE ooo

21 = (2.28b)

C kt ca t dnh bo ha nc v s chuyn ha ng sut trong qu trnh c kt thm

7. Khi nim v tnh p co ca t bo ha nc Xt trng hp bin dng ca lp t chu nn mt hng (Hnh v):

Khi chu ti trng, t b p co bi:

- Bin dng ca cc ht t

- Nc v kh trong l rng ca t b p co

- Nc v kh b p thot ra khi l rng

Tuy nhin, di tc dng ca ti trng thc t, p co ca bn thn cc ht khong vt l rt nh v thng b qua. i vi t bo ho hon ton (gi thit bo ho l 100%), tnh

Hnh 2.20

00089,0100

725,0814,0

12

100010 =

=

= ppeea

nn ln ca nc trong l rng cng c b qua. V th, yu t lm thay i th tch ca t trm tch chnh l s thot nc l rng trong t. Khi nc trong t thot ra th bn thn cc ht t t sp xp li n v tr n nh hn v khi t tr nn cht hn. Khi th tch t gim th s dn n ln b mt nn.

S sp xp li cc ht t v s p co ca t cng ph thuc vo cng ca khung ct t v l hm ca kt cu t. Kt cu t li ph thuc vo lch s a cht v trm tch ca t.

S p co trong t Ri v t Dnh cng c nhng khc nhau, cn phn bit.

7.1. i vi t Ri (p co mt hng):

S p co (bin dng) din ra trong thi gian rt ngn bi v t th nghim l t ht th thot nc tt. Nc v kh d dng thot ra khi l rng ca t.

Trong thc t, vi t ct th qu trnh p co xy ra ngay trong khi xy dng v phn ln qu trnh ln kt thc sau khi xy dng xong cng trnh.

S ln din ra nhanh, thm ch ln tng kh nh ca cc lp t ht ri, c th s bt li cho cng trnh c bit nhy cm vi s ln nhanh.

7.2. i vi t Dnh bo ha nc (p co mt hng):

V kh nng thot nc trong t st kh nh, nn qu trnh p co ca t st c nh gi bng tc thot nc khi l rng ca t.

Qu trnh ny gi l

Qu trnh c kt, v l quan h ng sut - bin dng - thi gian.

Qu trnh ln c th ko di hng thng, hng nm thm ch hng chc nm. y l s khc bit c bn v duy nht gia nn ca t ri v c kt ca t dnh:

Nn ca t ct xy ra tc thi,

C kt l qu trnh ph thuc thi gian. S khc nhau v tc ln ph thuc vo s khc nhau v tnh thm ca t.

M hnh C kt thm Terzaghi:

t dnh bo ha nc gm 2 pha:

Pha Rn: gm cc ht t, to thnh khung kt cu (ct t).

Pha Lng: gm nc chim y th tch rng trong t ( bo ha S=1)

Hai pha di tc dng ca p lc s c nhng phn ng khc nhau: Phn p lc truyn cho pha rn, lm t bin dng, gi l ng sut hiu qu ()

Phn p lc truyn cho pha lng (nc), khng lm bin dng t, m ch to nn ct nc v gy ra s thm trong t (lm nc thot ra ngoi mu t), gi l p lc nc l rng hoc p lc trung ha (u).

Hnh2.21

Qu trnh nc thot ra, l rng thu hp v t cht li, l qu trnh, u tng v gim. Nh vy, nu gi ng sut tng l , ta c:

= + u (2.29)

C kt ca t st c gii thch d dng bng m hnh ca Terzaghi (1923): Mt pt-tng P chu ti trng ng v nn mt l xo t trong mt bnh ng y nc.

L xo tng trng cho ct t,

Nc trong bnh tng trng cho nc trong l rng ca t.

Van V t trn nh pt-tng tng trng cho kch thc l rng ca t.

Hnh 2.22: M hnh Terzaghi

Hnh 2.22a: Cn bng p lc xy ra khi van V m nhng khng c nc thot ra ngoi. Trng hp ny m phng mt lp t bn di cn bng vi trng lng cc lp t pha trn n (lp ph)

Hnh 2.22b: Lp t chu thm mt gia lng . Ban u, van V cha kp m, ton b ti trng truyn cho nc trong bnh (v nc khng chu nn v cha th thot ra), lc ny cha c s chuyn ng ca pt-tng, v ng h ch u= . p lc nc l rng u c gi l p lc thu tnh d v n l gia lng ca p lc thu tnh ban u uo.

Sau , m van V v cho nc thot dn dn ra khi bnh di p lc d ban u u. Theo thi gian, nc thot dn ra, p lc nc gim v ti trng chuyn dn sang l xo, l xo b nn li do ti trng tng ng: < ; u <

Hnh 2.22c: Khi t cn bng, khng c nc thot ra thm na, p lc nc l rng li t trng thi cn bng thu tnh v l xo t trng thi cn bng vi ti trng tc dng: u = 0 ; =

Nh vy, M hnh l xo m phng c qu trnh c kt xy ra trong t dnh hin trng v trong phng khi chu ti trng nh sau:

Lc u (t = 0), nc cha kp thot ra, ton b ti trng ngoi c chuyn thnh p lc nc l rng d hoc p lc thu tnh d. V th ti thi im ban u khng c s thay i v ng sut hiu qu trong t (l xo cha bin dng, t cha b nn).

Dn dn (0< t < T) nc thot ra di tc dng ca chnh lch p lc, ct t b nn li v tip nhn ti trng, ng sut hiu qu tng ln (u < v, v < v). Qu trnh nn ca l xo m phng qu trnh nn ca ct t. (T= thi gian nc d thot ra ht).

Cui cng (t = T), nc d thot ra ht, p lc thu tnh d s bng khng v p lc nc l rng li tr li p lc thy tnh nh khi cha tc dng ti trng (u = 0, v = v). L xo b nn hon ton (t c kt hon ton)

Kt lun:

Qu trnh chuyn ha ng sut trong MH c kt thm m t qu trnh chuyn ha ng sut trong t dnh bo ha nc.

C th ni, qu trnh c kt ca t dnh bo ha nc v mt c hc l qu trnh chuyn ha t p lc nc l rng d sang ng sut hiu qu.

Nhn t nh hng n tnh p co v bin dng ca t

8. Cc nhn t ch quan

Lin kt kt cu ca t: nu lin kt kt cu b ph hoi th tnh p co v bin dng ca t ln, nu lin kt kt cu cha b ph hoi th tnh p co v bin dng ca t s b hn.

Loi t khc nhau th tnh p co v bin dng ca t s khc nhau. t dnh ni chung bin dng ln hn t ri, c bit l ct to v ct si c bin dng rt b.

cht ban u ca t c nh hng n tnh p co v bin dng. V d t ri c cht ban u cng ln th tnh p co v bin dng cng nh.

9. Cc nhn t khch quan

nh hng do cch tc dng ti trng: tnh cht nn ln v bin dng ca t rt khc nhau khi tng ti, d ti v nn li.

nh hng do tc gia ti: Vi cng gi tr ti trng nh nhau, nu tc gia ti cng ln, bin dng s cng ln. Trong thc t xy dng dng bin php gia ti vi tc chm hn ch ln tng nhanh.

nh hng ca ti trng ng: Ti trng ng lm cho t ct c nn cht mnh.

2.3. Cng chng ct ca t

I. Khi nim v cng chng ct ca t

Nu ti trng hoc ng sut () trong nn, mi dc t hoc t p sau tng chn tng cho n khi bin dng vt qu mc cho php, cc khi t bt u dch trt ln nhau theo mt mt trt th c th ni rng t trong nn, trong mi dc hoc sau tng chn b ph hoi. Trong trng hp ny s lin quan n bn ca t hay l ng sut ln nht hoc gii hn m vt liu c th chu c (f). Trong a k thut, ngi ta thng quan tm n cng chng

ct ca t v, trong phn ln cc vn trong nn mng v khi thi cng t, s ph hoi thng xy ra khi ng sut ct tc dng vt qu mc cho php, > f .

l ng sut ct do ti trng

f hoc o l cng chng ct ca t

Hnh 2.23

Nh vy cng chng ct f ca t l nhn t ch yu quyt nh i vi s n nh ca khi t (nn, t p) v an ton ca cng trnh.

Cng chng ct f c hiu l: lc chng trt ln nht trn mt n v din tch ti mt trt khi khi t ny trt ln khi t kia.

Mt trt ch c th i qua cc im tip xc gia cc ht v khng th ct qua cc ht (v cng lin kt

Di y s ln lt trnh by nh lut c bn v cng chng ct, iu kin ng sut gii hn, phng php xc nh cc c trng cng chng ct v mt s vn c lin quan.

Th nghim ct trc tip v nh lut Coulomb

1. L thuyt ph hoi Mohr Mohr (1900) a ra mt tiu chun ph

hoi cho cc vt liu thc theo ng cho rng vt liu b ph hoi khi ng sut ct trn mt phng ph hoi t n mt hm duy nht no ca ng sut php trn mt , ngha l (xem Hnh 2.31):

( )ffff f = (2.40) Trong l ng sut ct v l ng

sut php. Ch s f u tin lin quan n mt phng chu tc dng ca ng sut (trong trng hp ny l mt ph hoi) v ch s f th hai ngha l ti lc ph hoi.

Tha nhn tn ti mt mt ph hoi. Gi thit ny ph hp vi cc loi t, v nhiu loi vt liu khc.

Nu bit cc thnh phn ng sut ti thi im ph hoi, ta c th dng c mt vng trn Mohr c trng cho trng thi ng sut ca phn t ny. Bng cch:

Tin hnh th nghim n ph hoi i vi mt s mu cng loi, v dng cc vng trn Mohr tng ng vi mi mu (phn t). Ch rng trong c hc t thun tin, ch v na pha trn ca cc vng trn Mohr (xem Hnh 2.32).

Do cc vng trn Mohr c xc nh ti thi im ph hoi, ta hon ton c th tm c ng bao gii hn (hoc ph hoi) ca ng sut ct. ng bao ny c gi l ng bao ph hoi Mohr, cho bit mi quan h hm s gia ng sut ct v ng sut php ti thi im ph hoi (Pt.2.40).

Hnh2.25: (a) Tiu chun ph hoi Mohr; (b) phn t ti thi im ph hoi, cho bit cc ng sut chnh v cc ng sut trn mt ph hoi.

Hnh 2.26: Vng trn Mohr ti thi im ph hoi xc nh ng bao ph hoi Mohr

Vi vng Mohr (v d A) nm di ng bao ph hoi Mohr, c trng cho iu kin n nh.

Khi vng trn Mohr tip xc vi ng bao ph hoi, th hin tng ph hoi xut hin

Khng tn ti nhng vng trn nm pha trn ng bao ph hoi Mohr (nh B). Vt liu s b ph hoi trc khi t n trng thi ng sut .

Nu vi mi loi vt liu xc nh, ng bao ph hoi ny l duy nht th im tip xc ca ng bao ph hoi cho ta cc iu kin ng sut trn mt ph hoi ti thi im ph hoi. Ta c th xc nh gc ca mt ph hoi t im tip xc ca vng trn Mohr v ng bao ph hoi Mohr.

Gi thit ph hoi Mohr cho rng: im tip xc ca ng bao ph hoi vi vng trn Mohr ti thi im ph hoi s cho ta gc nghing ca mt ph hoi, f.

Nu gi thit ph hoi Mohr l hp l, th mt ph hoi cng s to thnh mt gc -f, nh trn Hnh.2.33a. Thc t, do iu kin ng sut nh v y ca mu th nghim l khng ng nht v bn thn mu th nghim cng khng phi l ng nht hon ton, thng l nguyn nhn gy ra mt mt ph hoi n trong mu th nghim.

Nu coi tt c l ng nht v ng sut phn b u trn mu th nghim, th s hnh thnh nhiu mt ph hoi vi cc gc lin hp l f , nh trn Hnh.2.33c.

2. Th nghim ct trc tip Nm 1773, C.A.Coulomb tin hnh th nghim ct i vi t ct.

My ct trc tip (Hnh 2.25 a): V c bn, gm mt hp ng mu th nghim hay hp ct, c chia lm hai na theo phng ngang. Mt na c gi c nh, na cn li c th b y hoc ko theo phng ngang. Ti trng thng ng P c t vo mu t trong hp nn thng qua mt tm cng chu lc. Lc ct, chuyn v ngang v chuyn v ng c o trong sut qu trnh th nghim. Chia lc y ngang v ti trng ng cho din tch danh ngha ca mu, ta c ng sut ct cng nh ng sut php trn mt ph hoi.

Ch : to cc mu t cng loi v cng cht e c th s dng ng n tng ng trn Hnh 2.25 b . Mt ph hoi bt buc l mt ngang tip xc gia hai na ca hp nn.

Kt qu th nghim ct trc tip vi t ct: Th hin trn Hnh 2.26.

Hnh 2.26a biu th quan h gia /s ct v bin dng trt tng ng vi cc ng sut php n1 < n2 < n3. Hnh 2.26b l quan h chuyn v ng vi : u tin H hoc V gim nh (p co), tip theo H hoc V tng ln (n ra); khi ng sut php tng, t cng hn s hn ch tnh n ca mu t trong khi ct

Hnh 2.29: Kt qu th nghim ct trc tip vi t ct cht

Hnh 2.26c th hin Phng trnh ng Coulomb vi t ri:

tgf = (2.30)

Trong :

- gc ma st trong ca ct

f - l cng chng ct ca t ct

- ng sut php tc dng trn mt ct

Hnh 2.28

Kt qu th nghim ct trc tip vi t st: Th hin trn Hnh 2.27.

ng quan h f ~ giao vi trc tung ti im c ta = c,

ctgf += (2.31)

v c l cc thng s bn ca t, chng khng phi l cc c tnh c hu ca vt liu; ngc li chng ph thuc vo cc iu kin khi tin hnh th nghim.

- gc ma st trong

c - lc dnh n v

f - l cng chng ct ca t dnh

- ng sut php tc dng trn mt ct

Hnh 2.30: Kt qu th nghim ct trc tip vi t st

3. nh lut coulomb (1776) Kt qu ca rt nhiu th nghim chng minh rng: Biu cng chng ct ca t

ri l mt ng rt thng, Pt (2.30).

i vi t st, khi p lc khng ln (< 700 kPa) th tt c cc im th nghim nm chnh xc trn ng thng, Pt. (2.31).

nh lut Coulomb v cng chng ct ca t:

Cng chng ct ca t ri l lc ma st, t l bc nht vi ng sut php,

tgf = (2.30)

Cng chng ct ca t dnh l hm s bc nht ca ng sut php, v gm 2 phn:

Lc ma st, t l bc nht vi ng sut nn, tg

Lc dnh n v c, khng ph thuc ng sut nn

ctgf += (2.31)

i vi t ri (t ct) c = 0, do c th coi biu thc (2.30) l trng hp c bit ca biu thc (2.31).

Tiu chun ph hoi Mohr - Coulomb

4. ng sut ti mt im v vng Mohr ng sut 4.1. Lp cc cng thc v biu din trng thi ng sut ti mt im bng vng trn Mohr ng sut

t l vt th ri, rng, do c th p dng l thuyt v ng sut ti mt im trong vt th lin tc th cn phi coi ng sut trong t l lc trn mi din tch n v, theo din tch xt n l ton b mt ct ngang hay din tch k thut. Din tch ny bao gm c din tch tip xc gia cc ht v cc l rng.

Gi s mt khi t chu tc dng ca mt nhm lc F1, F2, , Fn, nh minh ha trn Hnh 2.28. Ti thi im tnh ton, gi thit rng cc lc tc dng trong mt mt phng hai chiu.

C th phn tch cc lc ny ra thnh cc lc thnh phn trn mt phn t nh ti im bt k trong khi t (v d nh ti im O). Cc lc thnh phn c chiu l php tuyn v tip tuyn trn mt mt phng i qua im O to vi phng ngang mt gc nh trn Hnh 2.29, l hnh nh phng to ca mt phn t nh ti im O.

Quy c du:

cc lc v ng sut (ng sut php) gy nn l dng.

cc lc ct dng gy mmen theo chiu kim ng h quanh im pha ngoi phn t

Gc quay tng ng cng c coi nh dng.

Cc quy c ny ngc vi cc quy c gi nh thng thng trong c hc kt cu.

Gi thit khong cch AC = 1 n v, v chiu dy vung gc vi mt phng trang giy = 1 n v th kch thc cc cnh khc th hin trn Hnh 2.29.

Tng hnh chiu cc lc theo cc phng ngang v ng l:

Fh = H Tcos Nsin = 0 (2.32a)

Fv = V + Tsin Ncos = 0 (2.32b)

0sincossin = x (2.33a)

0cossincos = y (2.33b)

Hnh2.3

Hnh2.32: Phn tch cc lc trong Hnh 10.1 thnh cc thnh phn trn mt phn t nh ti im O. Cc quy c du c th hin nh hnh nh pha trn.

Gii h phng trnh (2.33), thu c:

2cos22cossin 22 yxyxyx

+

+== (2.34)

2sin2cossin)( yxyx

== (2.35)

Nu ly bnh phng hai v sau cng hai phng trnh ny, s c phng trnh ca mt vng trn vi bn knh (x - y)/2, c tm ti im [(x + y)/2, 0]. Vng trn v ln h trc ~ , c gi l vng trn Mohr ng sut (Mohr, 1887), c trng cho trng thi ng sut ti mt im khi cn bng.

V x v y l cc ng sut chnh (x = 1, y = 3), nn:

2cos223131 +

+= (2.36)

2sin231 = (2.37)

C th x/ (1, 3) theo cc thnh phn /s (x, y, zx):

22

31 22 xy

xyxy

+

+= (2.38)

Gc nghing ca ng sut 1 vi phng ng (phng ng sut y), c th xc nh theo cc cng thc:

x

xytg

=

1 hoc

xy

xytg

=

22 (2.39)

C th biu din v theo cc thnh phn /s (x, y, zx) bng cch thay 1, 3 trong (2.38) vo cc Ptr. (2.36, 2.37).

4.2. Cc bi ton tnh ng sut thng gp:

Tnh ng sut php , v ng sut ct trn mt phng nghing gc bt k khi bit cc thnh phn ng sut chnh (1, 3). C hai trng hp xy ra:

Trng hp x v y thuc cc mt phng chnh

Cc ta (, ) trn Hnh. 2.30b c th c xc nh bng cc Pt. 2.36 v 2.37. Cng t cc phng trnh ny, thy rng ta ca tm vng trn Mohr l [(1 + 3)/2, 0], vi bn knh (1 - 3)/2.

Trng hp tng qut, trong x v y khng thuc cc mt phng chnh:

Ta d dng c th suy ra cc phng trnh cho trng hp tng qut ny. Chng han, c th biu din v theo cc thnh phn /s (x, y, zx) bng cch thay 1, 3 trong (2.38) vo cc Ptr. (2.36, 2.37).

Tuy nhin, trong thc t s dng phng php gii tch i lc kh phc tp, khi c th dng phng php gii. Phng php gii da trn mt im duy nht trn vng trn Mohr gi l cc hay gc ca cc mt phng.

c tnh ca im cc l, bt k ng thng no v qua im cc s ct vng trn Mohr ti mt im, im ny cho bit trng thi ng sut trn mt phng nghing cng phng trong khng gian vi ng thng .

Hnh 2.33: (a) phn t lc cn bng; (b) Vng trn Mohr ng sut.

Khi nim ny c ngha l nu bit trng thi ng sut, v trn mt phng trong khng gian, ta c th v mt ng thng song song vi mt phng q```ua cc``` ta v trn vng trn Mohr. im cc y chnh l giao im ca ng thng vi vng trn Mohr.

Khi im cc c xc nh, th c th tm c cc thnh phn ng sut trn bt c mt phng no, n gin bng cch v mt ng thng t im cc song song vi mt phng; cc ta ca im giao vi vng Mohr chnh l cc thnh phn ng sut trn mt phng .

V d 2.5: (Trng hp x v y thuc cc mt phng chnh)

Cho bit: Cc thnh phn ng sut trn mt phn t nh trn hnh

Yu cu: Xc nh ng sut php v ng sut ct trn mt phng nghing gc =35o so vi mt phng quy chiu nm ngang.

Hnh Vd. 2.5

Bi gii:

1) V vng trn Mohr theo t l thch hp (xem Hnh. Vd. 2.1b).

- Tm ca vng trn =

kPa322

12522

31 =+

=+

- Bn knh ca vng trn =

kPa202

12522

31 =

=

2) Xc nh gc ca cc mt phng hay im cc: - Dng mt phng nm ngang chu tc dng ca 1, c trng thi s l im A. V mt ng thng qua A v song song vi mt phng 1 s tm c im cc P c ta (3 , 0)

ng thng i qua im cc P nghing gc = 35o so vi mt phng nm ngang s song song vi mt phng trn phn t trong Hnh Vd.10.1a, y cng l mt phng m trn ta cn tnh ng sut php v ng sut ct. Giao im l im C trn Hnh Vd.10.1b, vi = 39 kPa v = 18.6 kPa.

C th kim tra li cc kt qu ny bng cch s dng cc Pt. 2.36 v 2.37. Ch rng l dng v im C xut hin p hn trn trc honh. Do chiu ca trn mt phng nghing gc 350 c xc nh nh trn Hnh Vd.2.1.c, d, n i din cho phn nh v y ca phn t cho (u l dng).

V d 2.6: (Trng hp x v y khng thuc cc mt phng chnh)

Cho bit: Cng xt phn t trn vi cc thnh phn ng sut nh Hnh Vd.2.1a, nhng lc ny phn t xoay mt gc 200 so vi phng ngang, nh trn Hnh Vd.2.5.

Yu cu: Nh trong V d 2.5, xc nh ng sut php v ng sut ct trn mt nghing gc 350 so vi mt y ca phn t.

Bi gii:

1) V vng trn Mohr (Hnh Vd.2.7)

2) Xc nh im cc ca vng trn. Tng t nh trong v d trc, nu li bt u vi mt mt ng sut chnh ln nht, mt ny nghing gc 20o so vi mt ngang. Bt u t im A, ti giao im ca ng thng ny v vng trn Mohr, xc nh im cc P ca vng trn.

Hnh vd.2.7

3) By gi tm cc thnh phn ng sut trn mt phng nghing gc 350 so vi y ca phn t. T ng AP, m gc 350 c cng hng nh trn phn t, cc thnh phn ng sut trn mt phng c xc nh bi giao im ca ng thng vi vng trn Mohr (trong trng hp ny l im C). T ta ca C theo t l ta xc nh v . Ch rng cc ng sut ny cng tng t nh trn Hnh Vd. 2.1. iu xy ra l do khng c g thay i, ngoi tr hng ca phn t trong khng gian.

5. Tiu chun ph hoi Mohr - Coulomb Phn trn ta cp nh lut Coulomb v cng chng ct ca t v th hin bng Pt. Coulomb (2.31):

ctgf += (2.31)

Cng bit Tiu chun ph hoi Mohr, Pt. (2.40):

( )ffff f = (2.40) T 2 Phng trnh trn, ta c nhn xt quan trng sau: /s ct tng nhng khng th vt qu f (ngha l lun lun f ).

Cng nh vy, lun lun c ff , m trng thi /s ti phn t l duy nht, do c th kt hp phng trnh Coulomb, Pt.2.31, vi tiu chun ph hoi Mohr, Pt. 2.40, bng cch:

Ly xp x ng cong ny bng mt ng thng i qua mt s gi tr ng sut cho trc; sau lp c phng trnh ca ng thng theo cc thng s bn Coulomb, ngha l cho ff = f

T Tiu chun ph hoi Mohr Coulomb c th vit nh sau:

ctgffff += (2.41)

Tiu chun (2.41) c vit di dng /s php ff v /s tip ff trn mt trt. y l tiu chun n gin, d p dng, n c rt nhiu u im ring bit khi so snh vi cc tiu chun ph hoi khc. N l tiu chun ph hoi duy nht, d on c cc ng sut trn mt ph hoi ti thi im ph hoi, v do cc khi t c quan st n khi ph hoi trn cc mt rt khc bit, ta c th d on c trng thi ng sut trn cc mt trt tim nng.

6. Vn dng tiu chun ph hoi M-C T tiu chun M-C ta cn lu cc vn sau:

6.1. H s an ton ca t:

Mt phn t t c xc nh cc ng sut chnh, c tr s nh hn cc ng sut gy ra ph hoi. Trng thi ng sut nh th c th c biu th bng vng trn Mohr nh trn Hnh.10.9a. Trong trng hp ny f l sc khng ct c huy ng trn mt ph hoi tim nng, v ff l cng chng ct vn c (ng sut ct trn mt ph hoi ti thi im ph hoi). V vn cha t ti mc ph hoi, vn cn li bn d tr, nn c th nh ngha v h s an ton ca vt liu. nh sau:

H s an ton (FS) = ff (vn c)

f (tc dng)

Nu tng cc ng sut cho n khi xut hin ph hoi, th vng trn Mohr s tin ti tip xc vi ng bao ph hoi Mohr. Theo cc gi thit ph hoi Mohr, s ph hoi xut hin trn mt phng nghing gc f , vi ng sut ct trn mt l ff .

6.2. Gc ca mt ph hoi v cc ng sut trn mt ph hoi:

C th chng minh gc ca mt ph hoi so vi mt ng sut chnh ln nht:

245 += of (2.42)

Hnh 2.34

Cc ng sut php v tip trn mt trt:

ff 2cos22

3131 ++

=

(2.43)

ff 2sin2

31 = (2.44)

6.3. nh gi trng thi ng sut ti mt im:

Hnh 2.35 ch ra cc trng thi ng sut (c biu th bng vng Mohr) khi dng tiu chun ph hoi Mohr-Coulomb:

Hnh (a): Trng thi ng sut trc khi ph hoi;

Hnh (b): Trng thi ng sut ti thi im ph hoi. Ni chung, ff khng phi l max v nh hn max :

1f 3fmax ff2

= >

Hnh (c): ng bao ph hoi Mohr ca loi vt liu thun dnh ( = 0), l ng nm

ngang. S ph hoi v mt l thuyt xy ra trn mt 450, v

2

231

31max

fff

fff

+=

==

Hnh 2.35

6.4. Tiu chun ph hoi MohrCoulomb biu th qua cc ng sut chnh:

Kho st Hnh 2.34, ch rng: DR

=sin

cos

2

2sin31

31

cff

ff

++

=

ctg2sin

31

31

cffff

++

= (2.45)

C th bin i (2.45) a v dng thng dng:

++

+=

245tan2

245tan231

cff (2.46)

Cc Ptr. 2.45 v 2.46 l tiu chun ph hoi Mohr-Coulomb biu th qua cc thnh phn ng sut chnh 1f v 3f.

Nu c = 0 ff

ff

31

31sin

+

= (2.47)

Sp xp li (10.13), ta c:

sin1sin1

3

1

+

=f

f hoc

+=

245tan2

3

1 o

f

f

(2.48)

Ly nghch o (10.14):

sin1sin1

1

3

+

=f

f hoc

=

245tan2

1

3 o

f

f

(2.49)

Cc phng trnh 2.48 v 2.49 c gi l cc quan h nghing v nghing (hay dc) ln nht ca ng bao ph hoi Mohr xut hin khi c bng khng.

Cc ta ca im tip xc ca ng bao ph hoi v vng trn Mohr (ff, ff ) l cc ng sut trn mt phng (f = 45o+/2) vi nghing ln nht trong phn t t, (xem Hnh 2.35b) . Ni cch khc, t s ff /ff l ln nht trn mt phng ny. Mt phng vi ng sut ct ln nht ( = 45o) c nghing nh hn gi tr ln nht v

ff

ff

3crit :

Cc vng trn Morh ng vi thi im ph hoi trong trng hp ny c th hin trong Hnh 2.44a. Vng trn t nt E th hin cc ng sut hiu qu, vng trn lin nt T biu din cc ng sut tng. V cng thc 2.29, = + u lun duy tr, nn hai vng trn tch ri nhau bi gi tr p lc l rng d u hnh thnh trong qu trnh ct mu. Do th tch c xu hng gim, hnh thnh p lc l rng dng (gia tng), dn ti ng sut hiu qu gim (xem H. 2.43). V vy, trong v d ny, p lc l rng d u = 3c 3f = 3c 3crit. lch ng sut chnh ti thi im mu ph hu (1-3)f c tnh theo cng thc 2.52 khi p lc ng hng ti thi im ph hu l 3crit:

= 1)(max3

1331

critf (2.52)

Hnh 2.44: Cc vng Mohr ca th nghim nn ba trc khng thot nc v thot nc: (a) trng hp 3c > 3crit; (b) trng hp 3c < 3crit .

Cng nh vy, nu tin hnh th nghim thot nc vi p lc ng hng bng vi 3c ti im C, th sc khng ct thot nc s ln hn rt nhiu so vi sc khng ct khng thot nc v vy vng trn Mohr ca n phi tip xc vi ng bao ph hoi Mohr ng sut hiu qu. Kch thc tng i ca hai vng Morh ng sut hiu qu th hin trong hnh 2.44a.

Trng hp 3c < 3crit :

ng x ca t s khc nu th nghim vi cp p lc ng hng hiu qu nh hn p lc ng hng hiu qu ti hn 3crit. T biu trong Hnh 2.43, c th cho rng mu t c xu hng n ra. V thc t khng cho php mu t tng th tch, p lc l rng m hnh thnh lm tng ng sut hiu qu t im D (A) ti im H (3crit). Do , nh trong mu trc, ng sut hiu qu gii hn l p lc ng hng hiu qu ti hn 3crit.

Vng Mohr i din cho trng hp 3c < 3crit th hin trong hnh 2.44b. Th nghim ct khng thot nc xut pht p lc ng hng hiu qu 3c, im A, v v p lc nc l rng m, p lc ng hng hiu qu tng cho ti khi xy ra ph hu mu ti im H. Lu rng cc vng Mohr ng sut hiu qu E lc ph hoi trong hnh 2.44a, v b c cng kch thc bi v, vi h s rng ec , ng sut hiu qu khi mu b ph hu 3crit l nh nhau.

Nu ng sut hiu qu v h s rng ca cc mu nh nhau th chng phi c cng sc khng nn, 1f 3f ; do cc vng Mohr c cng ng knh. Lu rng vng Mohr ng sut tng T, khi mu b ph hoi, cng c cng kch thc vi vng Mohr ng sut hiu qu v lch ng sut (1 3)f l nh nhau cho c vng Mohr T v E; v vng Mohr T nm bn tri vng Mohr E. Trng hp ny ngc li vi Hnh 2.44a. (ng ph hoi ca vng Mohr ng sut tng b b qua hnh ny c n gin.) Cng cn lu rng, vng Mohr cho trng hp ct thot nc v c bn nh hn vng Morh ng sut hiu qu trong trng hp ct khng thot nc. Nh cp, vng Mohr xut pht t 3c, v phi tip xc vi ng bao ph hoi vng Mohr ng sut hiu qu. V h s rng sau c kt ec cho tt c cc th nghim l nh nhau nh trong hnh 2.44, tt c cc vng Mohr ng sut hiu qa phi tip xc vi ng bao ph hoi vng Mohr ng sut hiu qu.

Nhng chnh tho lun v c th hin trong hnh 2.44 c tng kt trong bng 2-1. nghin cu ton din cc c tnh khng ct khng thot nc ca t ct c th tham kho Seed v Lee (1

xem ebook1.pdf

Documents