TRNG I HC GTVT TP.HCMKHOA CONG TRNH

GV. NGUYEN THAGV. NGUYEN THANH NH AATT

CNG TRNH TRN T YU

1. Mc ch v ngha mn hc

2. Ni dung mn hc: Gm 6 chng

3. Hnh thc nh gi mn hc: Thi trc nghim

4. Ti liu tham kho

M U

Chng 1 : c im v tnh cht c bn ca t t yu

Chng 2 : Trng thi ti hn ca t st c kt thng v t ri

Chng 3 : Cac bien phap x l ket cau tren NY

Chng 4 : Cc dng m hnh nn v ng dng

Chng 5 : Cc gii php x l v gia c nn t yu

Chng 6 : t c ct

Chng 7: Mng su

NI DUNG MN HC

1.1 Khi nim v t yu

CHNG 1: C IM V TNH CHT C BN CA T YU

Da vo cc ch tiu vt l:Dung trng: H s rng: m:

Da vo cc ch tiu c hc:Modun bin dng: Gc ma st trong:Lc dnh C:

Da vo cng nn n qu t th nghim nn n.t rt yu: t yu:

)/(17 3mkN10 e

(%)40W

)/(5000 20 mkNE 010

)/(10 2mkNC

)/(25 2mkNqu )/(50 2mkNqu

1.2 c im ca t yu

1.2.1 c im v s phn b t yu khu vc thnh ph H Ch Minh

1.2.2. c im v s phn b t yu khu vc ng bng sng Cu Long.

1.2.3 Cc loi t khc cng khng thun li cho xy dng nh sau:

HUYN BNH CHNH

T. TY NINH

Hnh 1.1: Phn b t TP. HCM v khu vc ln cn

- Vng A: Cc loi gc J3-K1 - Vng B: St, st pha ct Ct pha st - Vng C: St nho, bn st, Bn ct pha st, Bn st pha ct

T. BNH DNG

T. NG NAI

T. LONG AN

T. LONG AN

C-V

H. CN GI

C-II

H. NHA BE

B-IQ. TH C

A

B-II

C-I

TP. HCM

B-II

C-III

C-III

C-III

C-III

C-IV

H. HC MNB-II

B-II

H. C CHI

B-I - Khu vc t tt, thun li cho xy dng: mt phn Q1, Q3, mt phn Q9, Q10, mt phn Q12, Q11, Tn Bnh, G Vp, C Chi, Th c.- Khu vc t yu, khng thun li cho vic xy dng: mt phn Q1, Q2, Q4, Q5, Q6, Q7, Q8 , mt phn Q9, Bnh Thnh, Nh B, Bnh Chnh, Cn Gi.

Phn b t yu BSCL

- t ct mn bo ha nc, t ct ri

- t hu c v than bn

- t ln t (ln st)

- t trng n

1.2.3 Cc loi t khc cng khng thun li cho xy dng nh sau:

1.3 Tnh cht ca t yu1.3.1 Tnh bin dng ca t- Th nghim nn c kt (oedometer):

My nn nn c kt

Th nghim nn c kt (oedometer)

Lc tc dng thng qua cc qu

Mu t

bt

Dao vng

ng h o chuyn v

M hnh nn mu t

e0

e1

p2 p1

e2

ng cong nn ln

p

M

M2

a tan

p

S

h

Quan h gia h s rng v lc tc dng

H s nn ln: m2/kN (cm2/kG).

dpdea =

12

21

12

12tanppee

ppeea

==

1

1,1

=

nn

nnnn PP

eea

H s nn ln tng i ao (h s nn th tch mv) (m2/kN)

11 eaam ov +==

PCa cv

435,0=P = (Ptrc + Psau)/2

Biu quan h e-P

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

Ap lc nen P (kG/cm2)

H

e

s

o

r

o

n

g

e

( )11

,1,1 1

+

= nn

nnnn eh

he

( )00

1 eh

he +=

en = e0 e

Tnh h s rng ng vi mi cp p lc

en = en-1 en-1,n

Biu quan h e-logP (nn v d ti)

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0.1 1.0 10.0Ap lc nen P (kG/cm2)

Pressure

H

e

s

o

r

o

n

g

e

V

o

i

d

R

a

t

i

o

0.4 4.0

e4.0

e0.4

Ch s nn Cc

==

peCc log

1

1

loglog

=

nn

nn

ppee

1

1

loglog

=nn

nn

ppee

0,20,4log0,2log0,4log

0,40,20,40,2 eeeeCc=

=

Biu quan h e-logP (nn v d ti)

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0.1 1.0 10.0Ap lc nen P (kG/cm2)

Pressure

H

e

s

o

r

o

n

g

e

V

o

i

d

R

a

t

i

o

0.4 4.0

e4.0

e0.4

Ch s n Cs (Cr)

peC rs log

=

1

)1()(

loglog

=

nn

nrnr

ppee

1

)()1(

loglog

=

nn

nrnr

ppee

0,20,4log0,2log0,4log

)0,4()0,2()0,4()0,2( rrrrs

eeeeC

==

Biu quan h e-p: nn, d ti v nn li

logp'

NG NEN

NG NEN LAI

NG N

e

p'

e

NG NEN

NG N

NG NEN LAI

Phng php 1 xc nh Pc

p lc tin c kt Pc

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0.1 1.0 10.0Ap lc nen P (kG/cm2)

Pressure

H

e

s

o

r

o

n

g

e

V

o

i

d

R

a

t

i

o

1

2

Pc

3

4

A

Phng php 2 xc nh Pc

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0.1 1.0 10.0Ap lc nen P (kG/cm2)

Pressure

H

e

s

o

r

o

n

g

e

V

o

i

d

R

a

t

i

o

pcPc

12

-T s tin c kt OCR (overconsolidation ratio):

pc : p lc tin c kt

p : ng sut hu hiu hin ti theo phng ng (ng sut bn thn)

OCR = 1 : t c kt thng (NC)

OCR < 1 : t km c kt

OCR > 1 : t c kt trc (OC)

ppOCR c=

Xc nh h s c kt cv theo pp logt

H s c kt cv Phng php logt (Casagrandes method)

0.80

1.20

1.60

2.00

2.400.1 1 10 100 1000 10000

Thi gian (phut)Time (min)

S

o

o

c

b

i

e

n

d

a

n

g

(

m

m

)

D

e

f

o

r

m

a

t

i

o

n

d

i

a

l

r

e

a

d

i

n

g

(

m

m

)

D0

D50

D100

t50

21000

50DDD +=

50

2197,0t

Hcv =

( )22

1 1 nn HHH +=

11 eack wv +=

Xc nh h s c kt cv theo pp cn t

Phng php cn t (Taylors method)

12.4

12.8

13.2

13.6

14

14.4

14.8

0 2 4 6 8 10 12 14 16Cn t [ph]

S

c

b

i

n

d

n

g

[

m

m

]

t90

D90

90

2848,0t

Hcv =

x1,15x

1 2

D0

Modul tng bin dng ca t E (kN/m2)

- Xc nh modul bin dng t th nghim nn c kt

nn

nnn a

eE,1

1),1(

1

+=

= 121

2

- Theo kinh nghim th thng ly EBN = (2 6) ETNTr s m khi h s rng e bng Loi t

0,45 0,55 0,65 0,75 0,85 0,95 1,05 Ct pha st 4 4 3,5 3 2 St pha ct 5 5 4,5 4 3 2,5 2

St 6 6 5,5 5,5 4,5

Xc nh ln n nh

ii

iin

ih

eeeS1

21

1 1+=

=

iioi

n

ihpaS =

=1

iii

in

ihp

ES =

=

1

Ngoi ra cn c cc cng thc tnh ln da vo ng nn ln e-logp.

Cho t c kt thng

heeS

01+=

[ ]ooc pppCe log)log( +=

+

+= ooc

ppp

ehCS log

1 0

+

+= = oi ioin

i i

ic

ppp

ehCS log

11 0

Cho t c kt trc nng (po + p pc)[ ]oos pppCe log)log( +=

+

+= oo

o

s

ppp

ehCS log

1

Cho t c kt trc nh (po + p pc)

+

+++= co

o

c

o

c

o

s

ppp

ehC

pp

ehCS log

1log

1

Poi : ng sut hu hiu trung bnh ban u ca lp th i (ng sut bn thn poi = tb= p1)pi = i : Gia tng /s thng ng ca lp th i (/s gy ln)e0 : h s rng ng vi thi im trc khi xy dng cng trnh, tc ng vi ng sut bn thn poi

Cc iu kin cn bng n nh: < s : t trng thi n nh = s : t trng thi cn bng gii hn > s : khng xy ra trong t v t b ph

hoi trc khi t n ng sut .

s = tan + cc

t dnh

s = tant ct

s = c

c

t st thun ty

Cc dng ca ng sc chng ct theo cc loi t

s = tan + c s = tan + c1.3.3 Sc chng ct ca t

Vng trn ng sut Mohr

s = tan + c

c

13

o

M

a

b

2

Bn knh

3

12

3

1,

x=1x=3 (13)/2

(1+3)/2

Vng trn ng sut Mohr

2cos223131 ++= 2sin2

31 =

* Theo QPVN (TCXD 45-70, 45-78) : khu vc bin dng do l b/4

- Pgh = R (Rtc RII)(45-70)

hgchbg

Pgh ++++= )cot25,0(2/cot

cg

ghg

bg

Pgh 2/cotcot1

2/cot2/cot25,0

++

++++=

)*( cDhBbAmRtc ++= )*(21 cDhBbA

kmmRtc

II ++= (45-78)

1.3.4 Kh nng chu ti ca t yu

* Theo Prandtl , = 0

4.3.2.2 Phng php tnh da trn gi thuyt cn bng gii hn im

gcegchPgh cotsin1

sin1)cot( tan ++=

* Theo Terzaghi- Mng bng: Pgh = 0,5 N b + Nq h + Nc c- Mng trn, bk R: Pgh = 0,6 N R + Nq h + 1,3 Nc c- Mng vung cnh b: Pgh = 0,4 N b + Nq h + 1,3 Nc c

N , Nq , Nc : cc h s ph thuc vo

- Th nghim ct trc tip (Direct shear test)- Th nghim nn 3 trc (Triaxial compression test: Undrained Unconsolidated, Undrained Consolidated, Drained Consolidated).- Th nghim nn n (Unconfined compression test)- Th nghim xuyn (ng) tiu chun (SPT)- Th nghim xuyn tnh (CPT)- Th nghim ct cnh (Vane test)

4.2.3 Cc phng php th nghim xc nh sc chng ct ca t

My ct trc tip (my c)

* Th nghim ct trc tip (Direct shear test)

My ct trc tip

* Th nghim ct trc tip (Direct shear test)

T

Tht c nh

Tht di ng

- Ct 3 mu t (dy 30 cm) cho 3 ln th nghim vi 3 cp ti trng khc nhau- Cho my ct vi tc 1 mm/min n khi no mu b ph hoi; ghi li gi tr () ng vi lc ng h o ng lc ngang t gi tr max.

Quan h lc ct v p lc thng ng

- Xc nh gi tr c v bng phng php hnh hc

(kN/m2)

(kN/m2)

s = tan + c

c

- V biu quan h gia (kG/cm2) v (kG/cm2)

- Xc nh gi tr c v bng phng php bnh phng cc tiu

( )2

11

2

111tan

=

==

===n

ii

n

ii

n

ii

n

ii

n

iii

n

n

( )2

11

2

111

2

1

=

==

====n

ii

n

ii

n

iii

n

ii

n

ii

n

ii

nc

- Xc nh gi tr c v bng hm LINEST trong Excel

tan=LINEST(1:3,1:3,1)=DEGREES(ATAN(tan))c=IF ((1/3)*(( 1+2+3)-tan(1+2+3))>0,(1/3)*((1+2+3)-tan(1+2+3)),0)Chuyn kt qu thp phn ca sang gi tr Pht => =((-INT())*60 + pht => =CONCATENATE(ROUND(,0),o,ROUND(pht,0),)

Kt qu tnh ton c v bng Excel

0

20

40

60

80

100

0 20 40 60 80 100 120 140 160

Ap lc thang ng (kPa)

L

c

c

a

t

(k

P

a

)

K E T Q U A tg = 0.3992 = 22 46'C = 5.003 kPa

+ Ct (nn) nhanh khng c kt / UndrainedUnconsolidated (UU): Gi tr cuu v uu+ Ct (nn) nhanh c kt / UndrainedConsolidated (CU): Gi tr ccu & cu ; c v v p lc nc l rng u + Ct (nn) chm c kt / Drained Consolidated (CD): Gi tr c v

* Th nghim nn 3 trc (Triaxial Compression Test)

My nn ba trc

Mu t trong bung nn

Thit b gt mu

S th nghim nn ba trc

1 2 3 4

1 2 3

4

ng du

Bm to p lc bung

7

85

6

9

10

a

bc

e

d

34

- Van 1: dng thot nc khi c kt v n c ni vi ng y mu. - Van 2: c cc tc dng sau:+ Dng cp nc t bnh nc vo bung.+ Dng to o lc bung v kha gi p lc bung khi thc hin cng ngh bm nhi bng bm quay tay+ Trong giai on c kt, th nc trong mu thot ra, lm mu co li. T lng nc trong bung gim, v khi nc s tng du chy xung, qua ng b, ri ng a qua van 2 vo bung.+ ng a c tc dng gn vo van 34 cp nc lm bo ha nc trong cc van 3, van 4 v ng di y b mu, ng ni vi cap (m ca mu)- Van 3, van 4: + 2 van ny c ng li trong giai an c kt+ Khi tin hnh giai an ct 3 trc, ta s m 2 van 3 v 4, ng thi kha van s 3 li.+ Van 3 : o p lc nc l rng pha trn mu+ Van 4 : o c p lc nc l rng pha di mu.+ Hai van ny gp chung thnh p lc nc l rng van 34. T ni ra u dy in tr o p lc nc l rng (trung bnh) ca mu trong qu trnh ct 3 trc khng cho thot nc

Biu quan h ng sut lch v bin dng

0

10

20

30

40

50

60

70

0 2 4 6 8 10 12 14 16 18 20

Bin dng %

n

g

s

u

t

l

c

h

(

1

-

3

)

k

P

a

* Th nghim UU

Th nghim UU thc hin vi thi gian nhanh, khong 10-15 pht. lch ng sut = 1 3 tng nhanh vmu t khng kp thot nc, khng o p lc nc l rng uf nn kt qu chbiu th theo ng sut tng. Th nghim UU thch hp cho loi t st bo ha nc v sc chng ct ca t ph thuc vo cu cn unh.

Biu cc vng Mohr

0

20

40

60

0 20 40 60 80 100 120 140 160 180 200

ng sut chnh (1+3)/2 kPa

n

g

s

u

t

c

t

(

1

-

3

)

/

2

k

P

a

* Th nghim CU

Th nghim CU thc hin sau khi cho mu ckt di p lc bung (ngang) ng hng nc thot ra hon ton. Tin hnh tng p lc ng 1 ng thi o p lc nc l rng uf. Kt qu xc nh c thng s sc chng ct hu hiu (c, ) v thng s tng (ccu , cu ).

* Th nghim CU

0

50

100

150

200

0 2 4 6 8 10 12 14 16 18 20Bin dng %

n

g

s

u

t

l

c

h

(

1

-

3

)

k

P

a

Biu quan h ng sut lch v bin dng

Quan h gia p lc nc l rng v bin dng

05

101520253035

0 2 4 6 8 10 12 14 16 18 20

Bin dng %

p

l

c

n

c

l

r

n

g

k

P

a

Biu cc vng Mohr

0102030405060708090

100

0 40 80 120 160 200 240 280

ng sut chnh (1+3)/2 kPa

n

g

s

u

t

c

t

(

1

-

3

)

/

2

k

P

a

* Th nghim CD

020406080

100120140160

0 40 80 120 160 200 240 280 320 360 400 440 480

ng sut chnh (1+3)/2 kPa

n

g

s

u

t

c

t

(

1

-

3

)

/

2

k

P

a

Biu cc vng Mohr

Th nghim CD thc hin sau khi cho mu c kt di p lc bung (ngang) ng hng nc thot ra hon ton. Tin hnh tng p lc ng 1 vi tc chm m bo p lc nc l rng khng thay i. Kt qu xc nh c thng s sc chng ct hu hiu (c, ).

Phng php gii tch ton hc (pp bnh phng cc tiu) xc nh c, trong th nghim 3 trc

sin

cot23131 =++

gc

++

+=2

4522

45231 oo tgctg

ba += 31

+=2

452 otga

+=2

452 otgcb

oaartg 902 = abc

2=

2

13

1

23

1 131

131

=

nn

n nn

n

na

2

13

1

23

1 1313

11

1

23

=

nn

n nnn

nb

* Th nghim nn n (Unconfined Compression Test)

- Mu t c dng hnh trc, chiu cao bng 2 ln ng knh, c nn thng ng khng c p lc xung quanh. Sc chu nn n (1 trc) l p lc nn ln mu lc b trt, qu. - Sc chng ct khng thot nc hay lc dnh khng thot nc cu = qu/2. Gc ma st trong u = 00 . Thnghim ph hp vi t st bo ha hon ton (u = 00).

Vng Mohr trong th nghim nn n

u=0

qu

max=cu

* Th nghim xuyn tnh CPT (Cone Penetration Test)

- Da vo sc khng xuyn qc , xc nh gc ma st trong ca t ct

() suqc (105 Pa) 2 m 5 m v su hn

10 28 2620 30 2840 32 3070 34 32120 36 34200 38 36300 40 38

- Da vo sc khng xuyn qc , xc nh lc dnh khng thot nc ca t st

: ng sut bn thn ca t nn ti im ang xtA : din tch mi xuyn (10 cm2)

Aqc cu

=

t ri

* Th nghim xuyn (ng) tiu chun SPT (Standard Penetration Test)

N (SPT) Trng thi Gc ma st trong< 4 Rt ri < 300

4 10 Ri 300 35011 30 Cht va 350 40031 50 Cht 400 450

> 50 Rt cht > 450

t dnh

N (SPT) Trng thi Sc chu nn n qu(bar-kG/cm2)< 2 Rt mm (nho) < 0,2

2 4 Mm (do nho) 0,2 0,55 8 Rn va (do mm) 0,5 1

9 15 Rn (do cng) 1 216 30 Rt rn (na cng) 2 4

> 30 Cng > 4

> 50 Rt cng

* Th nghim ct cnh ch thp (Shear Vane Test)

dddhdM xoay 32

42

2 +=

+==

hdhd

Mcs xoayuu

31

2

2

- o moment tc ng t trc xoay M, khi mu t b trt th:

- Sc chng ct khng thot nc:

Bi tp chng 1

2.1 Cc tnh cht trong th nghim nn 3 trc

CHNG 2: TRNG THI TI HN

H 2.1 Ph hoi gin (t cng)H 2.2 Ph hoi chy do

H 2.3 Ph hoi ca t qu yu

S thay i din tch v th tch :

L

P

- Din tch mt ct ngang ca mu t thay i theo ti trng nn nh sau :

0

0

1

1

hh

VV

AA o

=

- Nu th nghim khng thot nc V = 0

0

0

1hh

AA =

=0hh gi l bin dng tng i.

Vng trn ng sut Mohr

s = tan + c

c

13

o

M

a

b

2.2 Phn tch ng sut da vo vng trn Mohr

2'' 31 +

2'' 31

2

Bn knh

3

12

3

1,

x=1x=3 (13)/2

(1+3)/2

Vng trn ng sut Mohr

2cos223131 += 2sin2

31 =

- Khi vng trn tng ng c xy dng vi cc ng sut hu hiu: lch ng sut: q = 1 3Bt bin ng sut: s = 1/2 (1 + 3 )

t = 1/2 (1 - 3 )- Khi vng trn tng ng c xy dng vi cc ng sut tng:ng sut tng: 1 = 1 + u

3 = 3 + u lch ng sut: q = qBt bin ng sut: s = s + u

t = t

2.3 L trnh ng sut (ng ng sut) stress path trong th nghim nn 3 trc

2.3.1 L trnh ng sut trong h trc (1/ 3 ), 1/3

1

1

3 3 3

1 1/

3 /

ESP : ng ng sut c hiu (effective stress

path)

TSP : ng ng sut tng

(total stress path)

2.3.2 L trnh ng sut trong h trc t/s( t/s)

s = 1/2 (1 + 3) t = 1/2 (1 3)

'

CSL

ng ng sut khi tng ti c thot nc

CSL : Critical state line

Cc ng ng sut tng v c hiu khi tng ti khng thot nc

3 1 13

CSL

2.3.3 L trnh ng sut trong h trc q/ p (q/p)

313

CSL

Cc ng ng sut trong trc ta q/p

- ng sut trung bnh : p = 1/3(1 + 2 + 3 )= 1/3(1 + 23 )

- lch ng sut: q = (1 - 3 )p = p + ufq = q

- Khi tng 1 th ng tng ng sut (TSP) l C -> SD c dc 1/3- Khi mu t khng thot nc trong lc ch tng 1, p lc nc l rng tng t 0 ln uf v ng ng sut c hiu ESP l C -> SU.- ng bao ph hoi hay ng ng sut cc hn cth xc nh tng ng vi cc gi tr q v p ti lc ph hoi: qf = M pf

- Quan h gia M v gc ma st trong tng ng xc nh bi ng bao ph hoi Mohr-Coulomb hay ng CSL; t vng trn Mohr, khi c = 0

)(21

)(21

'sin'3

'1

'3

'1

+

= 'sin1'sin1

'1

'3

+=

)2(31

)('3

'1

'3

'1

'

'

+==

f

f

pq

M

'sin3'sin6

)'sin22'sin1()'sin1'sin1(3

'sin1)'sin1(2

)'sin1'sin1(3

'1

'1

'1

'1

'1

'1

=++++=

+++

=M

MM+= 6

3'sin

- Theo l trnh ko: 3 > 1 do gi nguyn 3 gim 1

'1

'3

'1

'3'sin

+=

''sin3'sin6

3'2

3'

32'

32'

3'

'sin pqqp

qqpqp

qpqp

+

=+

=

+

+

+

=

'sin3'sin6*

+=M q = M*p

*

*

63'sin

MM

=

- Theo l trnh nn: 1 > 3 do gi nguyn 1 gim 3iu kin cn bng Mohr-Coulomb l:

'cot'2'sin '

3'1

'3

'1

gc++=

'cot'23

'32'

3'

32'

'sin

gcqpqp

qpqp

+

++

+=

( ) ( )'cot2''cot'2''sin3

'sin6 gcMpMgcpq +=+=

PT ng ti hn CSL ca t dnh: q = M (p+ccotg)- ngha ca ng CSL: Dng nh gi s n nh ca 1 im trong t nn da vo ng l trnh ng sut khi ly mu t em v phng xc nh cc ng sut 1 & 3 . Nu nhng im SU, SD nm di ng CSL th mu t n nh trong nn, ngc li im s b ph hoi .

2.4 L thuyt trng thi gii hn

2.4.1 t vn :2.4.2 L thuyt trng thi gii hn2.4.3 ng trng thi gii hn (CSL) v cc ng ng sut khi cht ti trn nn t st c kt thng (NC) trong cc h trc p/ q ; p/ v v Ln p/v

- Phng trnh ng ng sut ti hn ( CSL)

H 2.10a, h trc q/p: q = M pH 2.10c, h trc v/Lnp: 'ln fpv =

: gi tr th tch ring v trn ng CSL ti p = 1kN/m2

Cc ng ng sut trong h ta p/ q ; p/ v v Ln p/v

31CS

L

3

- Phng trnh ng c kt thng (NCL):H 2.10c, h trc v/Lnp: 'ln pNv =

- Hai ng NCL v CSL song song nhau nn bng nhau

vLnp f

=' Vf ep

='

- Vy pt ng c kt thng NCL trong h trc p/q :

)exp('' vMMpq ==

(v = 1 + e), (vc = 1 + ec : do), (vf = 1 + ef : ph hoi)v: th tch ring)

L trnh cc ng ng sut (TN CU) trong h ta p/ q/ v

L trnh cc ng ng sut (TN CD) trong h ta p/ q/ v

2.4.4 Cc mt gii hn khng b ko, mt Hvoslev vmt Roscoe

1

T

S

C

q/ qe

p/ pe3

O

M 1

Mt RoscoeMt Hvorslev

Mt khng chu ko

3=0

H1

g

Cc mt bin trng thi ti hn

p

v

N

vk

ng nn: v = N-Lnp

ng n: v = vk

NCL

CSL

SL

111

Ln

= dc ng nn = dc ng n (h ta Lnp/v) = cs/2,3

- Mt gii hn khng b ko (OT): q = 3 p l mt gii hn v t khng b ko

V

e

- Mt Hvoslev (TS): q = H p + (M H) exp[( -V)/]l mt ng vi mu t c cng h s rng vi mt Roscoe nhng h s OCR > 2,5 (t c kt trc)- Phng trnh ng Hvorlev c dng:

'exp' hpvNgq +

= - Ti S, im giao vi mt Roscoe, phng trnh mt Hvorslev c dng :

( ) 'exp' hpvhMq +

=

S ba chiu ca ton b mt bin trng thi ti hn

q p

v

S

T

v

v

T

S

S

N

N

N

T

SS: ng trng thi ti hnNN: ng c kt thng VVTT: Mt gii hn khng b ko TTSS: Mt Hvorslev SSNN: Mt Roscoe

2.4.6 bn sc chng ct ca ct v c trng bin dng

O

Ct ri

n

g

s

u

t

n

h

Ct cht

-V Ct ri

Ct cht

+V

Co ngt (gim)

N (tng)

n

g

s

u

t

c

c

h

n

3.1 M hnh nn bin dng cc b (cho t yu)

CHNG 3: CC DNG M HNH NN

M hnh nn 1 thng s

3.1.1 M hnh nn 1 thng s: Cz

h

=

D

f

N

h

=

D

f

N

s

Cz = f (z,F,t)

( )0

021

PP

FbaCC z

++=

Theo Vesic: ( )2001 = bECz

Theo Terzaghi: - i vi t ri

2

3,0 23,0

+=b

mbCC mzz

- i vi t dnhb

mCC mzz3,0

3,0=

Quan h P-S th nghim bn nn hin trng

S

0

S

P P

Vi Cz 0.3m l h s nn khi th nghim bn nn hin trng (Cz = P/S, bn nn c ng knh = 0,3m)

SPCk z ==

3.1.2 M hnh nn 2 thng s: Cz v Cx

S

N

H

P(x) = Cx Px = H/F = F

HCx

- Nu F > 50 m2Cx = 0,7 Cz

- Nu F 50 m2 00

)(217,0PP

FbaCCx

++=

3.1.3 M hnh nn 3 thng s: Cz ,Cx v C

JMC =

S

N

H M

- Nu F > 50 m2

Cx = 0,7 Cz- Nu F 50 m2

J: moment qun tnh ca mng

( )0

0321

PP

FbaCC

++=

3.2 Cc m hnh lu bin

3.2.1 nh ngha: L cc m hnh din t s tng quan gia ng sut (hoc lc Q) v bin dng (hoc l)

an hoi

(Q)

0

(Q)

(l)

an hoi

deo

0 (l)

trt

(Q)

0

Prandtl

(Q)

(l)

Saint - Vernant

Vat the deo cng

0 (l)

an - deoc

c

(Q)

0 (l)

an - deo

tang tien

c

cVat lieu don at - nen mong Kim loai - Ket cau thep

Quan h gia ng sut v bin dng

3.2.2 Cc m hnh lu bin c bn

a) M hnh n hi (l xo = clastic spring)

hoac

=E.

0

E,K E,K

(l)Q

(nen hay keo)

M hnh n hi

Phng trnh trng thi: = E

hay Q = E l

b) M hnh nht (ng nhn = Dash pot): L m hnh xt n tnh nht ca vt liu, c xt n thi gian.

M hnh nhtPhng trnh trng thi:

=

.

0 d/dt

dtd = =

c) M hnh do (ngm trt): L m hnh xt n tnh do ca vt liu

M hnh nht

Q K (trt, chy)Q < K (l = 0)

0 = K

l Q()

3.2.3 Cc m hnh n - nht tuyn tnh

M hnh Kelvin

= E + = E =

E

a) M hnh Kelvin: Da trn th nghim n hi, thnghim nht xy ra ng thi (mc song song, i = const; i = f(t) )

E== += E

M hnh Maxwell

= E = = E +

b) M hnh Maxwell: Dng nghin cu s chng ng sut (M hnh mc ni tip, i = const; i = f(t).)

E

3.2.4 Cc m hnh n - do

M hnh n-do; mc ni tip

Lc:

Q = QE = QKChuyn v:

q = l = qE + qK

a) Mc ni tip

K

Q()QE

E

QK

M hnh n-do; mc song song

Lc:

Q = QE + QKChuyn v:

q = l = qE = qK

b) Mc song song:

K

Q() QEE

QK

3.2.5 Cc m hnh n - nht - do

M hnh n-nht-do

K

E0

E

E1

K

E2

3.3 Cc dng m hnh lu bin khc tnh ton nn mng

Mt s m hnh lu bin

E

Terzaghi Biot

E1

E2 GibonSchiffman

Taylor

XDDD - CN C - TL

(t TP.HCM v BSCL)

Bi tp chng 3

4.1 Khi nim v mng ccCHNG 7: MNG SU

Nn ca mng cc

H cc

i cc- Mng cc: Mng su- i cc:- H cc:

4.2.1 Theo vt liu cc

4.2 Phn loi mng cc

4.2.2 Theo kh nng chu ti4.2.3 Theo chiu su t i

4.2.4 Theo c tnh chu lc

4.3 Cu to cc b tng ct thp

D

L

Ct thp dc

Ct thp ai

1-1,5D 150

1000 Mc cu, 16

6 a1001000

6 a100 20,1m

D L

A-A

Hp ni cc

AA

Mi thpMi hn

on u cc

NI CC

Hnh 3.6 Cu to chi tit cc v ni cc

hh

THEP HOP AU COC TL : 1/10

350350

8x350x180

1

8

0

=8mm 11

334x180x8

350x350x8

10

9

320

3 - 3

230x130x10

(CHIEU CAO NG HAN h=10mm) TY LE 1/10CHI TIET BAN THEP AU COC

9

11

250x250x8

320

10

Li thep 6

LI THEP AU COC TL : 1/10

5850

5

8

5

0

300x300x10

4 - 4TL :1/10

COC CBT-1

350x350x89

COC CBT-212

CHI TIET B NOI COC CBT-1 & CBT-2TY LE :1/10

200x200x12 12

CHI TIET MUI COCTL: 1/10

4181

203

MC 2-2TL: 1/10

HAN CHUM AU

120

CHI TIET COC BETONG CBT1

3

126a50

6a100 126a200

26 1

TL : 1/20

218

21818

1

116a100

4 3 li thep han 6a50 loai B

126a50

1 li thep han 6a50

Ban thep au coc

loai A

1 li thep han 6A50

3 li thep han 6a50 loai B

Ban thep au coc

loai A

120

CHI TIET COC BETONG CBT2

3

146a50

116a100

6 2 6

TL : 1/20

136a200

218

218 18

6

4

126a100

3 li thep han 6a50 loai B

146a50

loai A

Ban thep au coc

1 li thep han 6a50

4.4 Trnh t tnh ton mng cc:

1. D liu tnh ton

- D liu bi ton v cc c tnh ca mng cc

- S liu ti trng (tnh ton)

- Chn vt liu lm mng: mc BT, cng thp, tit din v chiu di cc (cm vo t tt > 1,5 m), on neo ngm trong i cc (on ngm + p u cc 0,5 0,6m); chn ct thp dc trong cc: vRa .

S tnh ton mng cc

Qs

Qp

4

Ntt

Htt

Mtt

2. Kim tra mng cc lm vic i thp

E H

2

21

fap DbK

FSK

H

bKFSK

HD

ap

f

2

Df 0,7 hminbHh

22

45tan 0min

=

Kp = tan2 (450 + /2)Ka = tan2 (450 - /2)FS = 3 (p lc sau i

cha t trng thi b ng)

b : cnh ca y i theo phng vung gc vi H

3. Xc nh sc chu ti ca cc Pc- Theo vt liu lm cc

Qa = (Rb Ab + Ra Aa)

v = 2 v = 0,7 v = 0,5

u cc ngm trong i v mi cc nm trong t mm

u cc ngm trong i v mi cc ta trong t cng hoc

u cc ngm trongi v mi cc ngmtrong

* Cc khoan nhi, cc barrette, cc ng nhi btngQa = (Ru Ab + Ran Aa)

Ru : cng tnh ton ca b tng Ru = R/4,5; Ru 6 MPa: khi btng di nc, bnRu = R/4; Ru 7MPa: khi btng trong h khoan khR : mc thit k ca b tngRan : cng tnh ton cho php ca ct thp < 28mm, Ran = Rc/1,5; Ran 220 MPa.

- Theo iu kin t nn:+ Theo ch tiu c hc

p

pp

s

ss

p

p

s

sa FS

qAFS

fAFSQ

FSQQ +=+=

FSs : h s an ton cho thnh phn ma st bn; 1,5 2,0FSp h s an ton cho sc chng di mi cc; 2,0 3,0FS : h s an ton chung, chn 2 3

FSqAfA

FSQQ

FSQQ ppsspsua

+=+==

Thnh phn chu ti do ma st xung quanh cc Qsfs = ca + h tana

= ca + Ks v tanaca , a = c, : cc ng, p btng ct thpca , a = 0,7[c, ] : cc thp (bng 3.28/213).Ks = K0 = 1 - sin (t)Ks = 1,4 K0 (khi t nn b nn cht do ng cc)

== 1sK OCRKs )sin1( =

Thnh phn sc chu mi ca t di mi cc Qp* Phng php Terzaghi:qp = 1,3 c Nc + h Nq + 0,6 rp N (rp: b/knh cc trn)qp = 1,3 c Nc + h Nq + 0,4 d N (d: cnh cc)Nc , Nq , N : h s sc chu ti, xc nh theo Terzaghi, bng 3.5/174. Df = v* Phng php Meyerhof:

qp = c Nc + q NqNc, Nq : xc nh t biu 3.28/178* TCXD 205-1998:

qp = c Nc + v Nq + d N

+ Theo ch tiu vt lQa = km (Rp Ap + u fsi li) (21-86)km = 0,7 : cc chu nn; km = 0,4 : cc chu nnQtc = mR qp Ap + u mf fsi li (205-1998)

kQQ tca = k =1,4 1,75

=> Chn Pc = min (Pvl ; Pn)

mR , mf : h s iu kin lm vic ca t mi cc m bn hng cc, bng 3.18/201.

Rp : sc chu ti n v din tch ca t di mi cc, bng 3.19/201.

fsi : lc ma st n v gia t v cc, bng 3.20/202

Qtc = m (mR qp Ap + u mf fsi li) (205-1998)* Cc khoan nhi, barrette:

. t dnh, qp tra bng 3.25/204

. t ri, qp c tnh

qp = 0,75 ( dp Ak0 + L Bk0): cc nhi, cc barrette, cc ng ly nhn.

qp = ( dp Ak0 + L Bk0): cc ng gi nguyn nhn : trng lng ring ca t di mi cc : trng lng ring ca t nm trn mi ccCc h s , , Ak0, Bk0 tra bng 3.24/204.

N : S SPT: S SPT trung bnh trong khong 1d di mi cc v

4d trn mi cc. Nu > 60, khi tnh ton ly = 60; nu >50 th trong cng thc ly = 50.

Nc : gi tr trung bnh SPT trong lp t ri.Ns : gi tr trung bnh SPT trong lp t dnh.Ap : din tch tit din mi ccLc : Chiu di cc nm trong lp t ri (m).Ls : Chiu di cc nm trong lp t dnh (m). : Chu vi tit din cc (m).Wp : Hiu s gia trng lng cc v trng lng t b cc

thay th

+ Theo th nghim SPT (TCXD 195 )

N

Qu = qp Ap + fs As

+ Theo th nghim CPT

N

qp: cng chu mi cc hn ca t mi cc c xc nh

cqsc khng xuyn trung bnh ly trong khong 3d pha trn v 3d pha di mi cc

fs : Cng ma st gia t v cc c suy t sc khng mi chiu su tng ng

i

cisi

qf =

=> Sc chu ti ca cc cui cng s ly theo kt qu thnghim nn tnh hin trng.

ccp qkq =

4. Chn s lng cc v b tr cc

=> b tr cc khong (3 6)d, cu to i c mp i cch mp cc ngoi 100 150mm.

= 1,2 1,6c

tt

c

tt

PQN

PN

n +==

5. Kim tra sc chu ti ca cc (lc tc dng ln cc)

++= 2

i

maxttx

2i

maxtty

tt

max yyM

xxM

nNP

++= 2

i

ittx

2i

itty

tt

)y,x( yyM

xxM

nNP

Pmax Pc (Qa)Pmin PnPmin 0

- Kim tra sc chu ti ca cc lm vic trong nhm. H s nhm :

Pnh = nc Pc > Ntt + Q

n1 : s hng ccn2 : s cc trong 1 hngd : ng knh hoc cnh ccs : khong cch gia cc cc

+=21

1221

90)1()1(1

nnnnnn

=sdarctg [deg]

6. Kim tra ng sut di mi cc (mng khi qui c)Fqu = Lqu Bqu

= [(L - 2x) + 2 lc tan] [(B - 2y) + 2 lc tan]

y

tcy

x

tcx

qu

tcqu

minmax/ WM

WM

FN =

=

qu

Lqu

qu

Bqu

qu

tcqu

Le

Be

FN 66

1minmax/e = M/N = (M0 + H h )/N

qu

tcqu

tb FN= )DcBhAb(

kmmR II

*qu

tc

21IItb ++=

max 1,2 RII min 0

7. Kim tra ln ca mng cc

S Sgh = 8 cm

hp tbgl =gl

zgl pk=

ii

iin

i

n

ii he

eeSS1

21

11 1+==

==

iioi

n

ihpaS =

=1ii

oi

in

ihp

ES =

=

1

7. Kim tra chuyn v ngang ca cc- Tnh ton cc chu ti trng ngang - Kim tra chuyn v ngang cho php

H Png (Png : sc chu ti ngang ca cc

301000 lEJ

P ngng= [T]

ng = 1 cm: chuyn v ngang ti u cho php EJ : cng ca cc = 0,65 : khi cc ng trong t st = 1,2 : khi cc ng trong t ctl 0,7 d ; d [cm]: cnh hay ng knh cc.

9. Kim tra iu kin xuyn thng ca iPxt PcxPxt = phn lc ca nhng cc nm ngoi thp

xuyn pha nguy him nhtPcx = 0,75 Rk Sthp xuyn

10. Xc nh ni lc v b tr ct thp- Tnh moment: dm conxn, ngm ti mp ct, lc tc dng ln dm l phn lc u cc.

00 9,0 hRM

hRM

Fa

g

a

ga =

11. Mt s vn thi cng cc- Tnh mc cu vn chuyn v thi cng cc

L

0,207L 0,207L0,586L

Mmax = 0,0214 qL2

0,293L

Mmax = 0,043 qL2

- Nu cc ng th chn ba ng E 25 Pc

5+E

qQ

- Thc t chn my p ti trng gp 2 ln Ptt ca cc.- Tnh chi thit k, etk 2 mm

k: h/s ng nht vt liu = 0,7; m: h/s k lm vic = 0,91; PS : sc chu ti cc n theo k t nn; Ap: din tch tit din ngang cc; q: trng lng cc; Q: trng lng ba (thng chn = 11,25Q); h: chiu cao ri ba; n: h s = 15 kG/cm2 cho cc BTCT, = 10 kG/cm2 cho cc g khng m.

- chi thc t l ln trung bnh ca 10 nht ba cui cng.

qQqQ

AnPmk

P

hQAnmke

pSS

ptk +

+

+

= 2,01

4.5 Cc chu ti trng ngang(Theo TCXDVN 205-1998)

M0yH0

y (kN/m2)

z

L

z

S lm vic ca cc chu ti trng ngang

S tc ng ca moment v ti ngang ln cc

H

M N

n

0 y0

z

l

H0=1

HH H M

z

M0=1 MH

M M

z

N

H

l

l0

l

- p lc tnh ton z [T/m2]:

++= 13 012 01010 DIE

HCIE

MBAyzK

bbdbbdbde

bdz

- Moment un Mz [Tm]:

30

3030302 DHCMBIEAIyEM

bdbbdbbdz ++=

- Lc ct Qz [T]

4040402

403 DHCMBIEAIyEQ bdbbdbbdz ++=

ze : chiu su tnh i, ze = bd zle : chiu di cc trong t tnh i, le = bd lbd : h s bin dng, bc : chiu rng qui c ca cc, d 0,8 m => bc = d + 1 m; d < 0,8 m => bc = 1,5d + 0,5 m (TCXD 205-1998)

5IE

Kb

b

cbd =

- Chuyn v ngang HH , HM , -MH , MM do cc ng lc n v

03

1 AIEbbd

HH =

02

1 BIEbbd

HMMH ==

01 C

IEbbdMM =

A0 , B0 , C0 , D0 tra bng 4.2/250- Moment un v lc ct ca cc ti z = 0 (mt t)

H0 = HM0 = M + H l0

- Chuyn v ngang y0 v gc xoay 0 ti z = 0 (mt t)y0 = H0 HH +M0 HM0 = H0 MH +M0 MM- Chuyn v ngang ca cc cao trnh t lc hay y i

IEMl

IEHlly

bbn 23

20

30

000 +++=

- Gc xoay ca cc cao trnh t lc hay y i

IEMl

IEHl

bb

020

0 2++=

* n nh nn xung quanh cc

( )IIvI

zy ctg +

,21 cos

4

vp

vp

MnMMM

++=2

v : ng sut hu hiu theo phng ng ti su zI : trng lng ring tnh ton ca tcI , I : lc dnh v gc ma st trong tnh ton ca t : h s = 0,6 cho cc nhi v cc ng, = 0,3 cho cc cc

cn li1 : h s = 1 cho mi trng hp; tr ct chn t,

chn nc = 0,72 : hs xt n t l nh hng ca phn ti trng

thng xuyn trong tng tiMp : moment do ti thng xuynMv : moment do ti tm thin = 2,5, tr:n = 4 cho mng bngn = cng trnh quan trng, le < 2,5 ly n = 4; le > 2,5 ly

n = 2,5

4.6 Ma st m 4.6.1 Hin tng ma st m

- Khi t nn ln xung ko cc ln theo s to ra lc ma st m tc dng ln cc. - Lc ma st m ny c chiu i xung lm tng lc tc dng ln cc vlm gim kh nng chu ti ca cc.

fs > 0

z

N

fs > 0

fs < 0Vng t gy ra ma st m

Qp Hin tng ma st m

4.6.2 Cc nguyn nhn gy ra hin tng ma st m

- p ph ti ln nn t sau khi ng cc

- Cht ph ti ln nn nh khi s dng mng cc

- Cc i qu lp t yu l than bn m t nn cn trong giai on ln (tc ln ca nn t ln hn tc ln ca cc)

- Khai thc hoc h mc nc ngm.

4.6.3 Tnh ton ma st m- Tnh ton ln ca t nn

ii

iin

i

n

ii he

eeSS1

21

11 1+==

==ii

i

in

ihp

ES =

=

1

- Xc nh chiu su nh hng z (gy ra ma st m)

)1(s

p

SS

hz =h: b dy lp t yuSp : ln ca ccSs : ln ca nn

- Tnh lc ma st m (fs < 0)

QNSF = As fs = U z fs

4.6.4 Cc bin php ngn nga ma st m v chng ma ma st m

- Khng cht ph ti ln nn c mng cc

- Khng san lp nn sau khi ng cc (Nu san lp nn th phi tnh thi gian c kt ca t nn di tc dng ca ti san lp ln ca t nn khng gy nh hng ma st m ln cc)

- Khng khai thc, h mc nc ngm

- Dng h sn v cc b tng ct thp gim ti chng ma st m

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