Soil Mechanics Laboratory_KTL.pdf

7/24/2019 Soil Mechanics Laboratory_KTL.pdf

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H KIN TRC TP HCMKHOA XY DNG B MN NN MNG

Trn ng Kim Lam

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SOIL MECHANICS LABORATORY

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CHAPTER 0

SOIL MECHANIC LABORATORYTH NGHIM C T

GII THIU MN HC

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TIEU CHUAN KHAO SAT A KY THUAT

CHAPTER 0GII THIU MN HC

0.1. V SAO PHI NGHIN CU TH NGHIM C T

0.2. V TR KHO ST A C TRONG XY DNG

0.3. QUY TRNH XY DNG H S KHO ST

0.4. PHN LOI TH NGHIM C T

0.5. KT QU T C CA TH NGHIM C T

0.6. TI LIU THAM KHO TH NGHIM C T

0.7. QUY TRNH NH GI CHT LNG MN HC TH NGHIM C T

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V Cng Ng - Nguyn Thi; Th nghim t hin trng v ng dng trong phn tch nn mng.

V.. Lmtaze (bn dch): Phng php nghin cu tnh cht c l ca t phng th nghim;

Tiu chun v kho st v o c xy dng.

Tuyn tp Tiu chun xy dng ca Vit Nam; Tp XI t xy dng.

Nguyn Uyn; Thc tp & bi tp a cht cng trnh;

Chu Ngc n. C hc t. NXB HQG TP.HCM 2009;

L Qu An v cng s. C hc t. NXB GD H Ni 1995;

Cao Vn Ch v cng s. C hc t. NXB XD H Ni 2003;

Bi Anh nh. C hc t. NXB XD H Ni 2004;

T c Thnh v cng s. C hc t. NXB XD H Ni 2002;

R. Whitlow. C hc t. NXB GD H Ni 1999;[email protected]


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T. W. Lambe et al. Soil mechanics. John Wiley, 1979;

R. F. Craig. Soil mechanics. London, 1997;

J. Atkinson. An Introduction to the mechanics of soil and foundations. Mcgraw Hill, 1993;

B. M. Das. Soil mechanics laboratory manual. Engineering press, 1996.

P. P. Raj. Geotechnical Engineering. McGraw Hill, 1995;

D. P. Coduto. Geotechnical Engineering Principles and Practices. Prentice Hall, 1998;

R. L. Schiffman. One dimensional consolidation of saturated clay layers nonlinear finite strain primary

consolidatin. University of Colorado, 1984;

B. M. Das. Principles of geotechnical engineering. Pws, 1995.

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CHAPTER 1

CHNG M U

TH NGHIM C T

SOIL MECHANIC LABORATORY

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CHNG M U

CHAPTER 11.1. TNG QUAN CNG TC KHO ST A CHT CNG TRNH

Mc ch v nhim v ca kho st a cht cng trnh

Cc giai on kho st a cht cng trnh

Nguyn l chung ca kho st a cht cng trnh

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1.1.1. MC CH V NHIM V CA KHO ST A CHT CNG TRNH

CHNG M U

CHAPTER 1

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1.1.1. MC CH V NHIM V CA KHO ST A CHT CNG TRNHNn ng b st l (trt) do t nn mt n nh . Tc l b

ph hoi ct (hay b ph hoi v cng ).

CHNG M U

CHAPTER 1

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CHNG M U

CHAPTER 11.1.1. MC CH V NHIM V CA KHO ST A CHT CNG TRNH

Nn mng cngtrnh b trt do t

nn mt n nh .

Tc l b ph hoi

ct (hay b ph

hoi v cng ).

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1.1.1. MC CH V NHIM V CA KHO ST A CHT CNG TRNH

Cung cp y , chnh xc v ton din cc iu kin CCT ca nn t trong

khu vc xy dng.

Kho st CCT c nhng nhim v sau:

CHNG M U

CHAPTER 1

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1.1.2. CC GIAI ON KHO ST A CHT CNG TRNH

Giai on iu tra ban u

Giai on kho st

Giai on kho st b sung phc v thi cng

CHNG M U

CHAPTER 1

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1.1.2. CC GIAI ON KHO ST A CHT CNG TRNH

Giai on iu tra ban u

Giai on kho st Giai on kho st s b

Giai on kho st chi tit

CHNG M U

CHAPTER 1

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1.1.3. NGUYN L CHUNG CA KHO ST A CHT CNG TRNH

Nguyn l k tha

Nguyn l giai on

Nguyn l kt hp

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CHAPTER 1

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1.2. CNG TC KHO ST NGOI TRI

o v a cht cng trnh

Phng php thm d a vt l

Phng php khoan, o thm d a cht cng trnh

Cc phng php th nghim a cht cng trnh hin trng

CHNG M U

CHAPTER 1

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1.2.3. PHNG PHP KHOAN O, THM D A CHT CNG TRNH

1

2

3

4

CHNG M U

CHAPTER 1

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Khi lng khoan o thm d

Cch b tr im khoan o thm d

su khoan o thm d

Phn loi khoan o thm d

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CHAPTER 1

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CHNG M U

CHAPTER 1

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o thm d

Ging o

Hm thm d

Ho thm d

Khoan thm d

CHNG M U

CHAPTER 1

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Khoan th cng

CHNG M U

CHAPTER 1

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Khoan gung xon

Khoan p cp

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CHAPTER 1

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CHNG M U

CHAPTER 11.2.3. PHNG PHP KHOAN O, THM D A CHT CNG TRNH

Khoan xoay my

u im

Nhc im

Cu to

Cng tc ly mu

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CHNG M U

CHAPTER 11.2.3. PHNG PHP KHOAN O, THM D A CHT CNG TRNH

Cng tc ly mu

Mu khng nguyn dng

Mu nguyn dng

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1.2.4. CC PP TH NGHIM A CHT CNG TRNH HIN TRNG

Cc th nghim a cht cng trnh

Cc th nghim a cht thy vn

Cc cng tc quan trc a cht cng trnh

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CHAPTER 1

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1.2.4. CC PP TH NGHIM A CHT CNG TRNH HIN TRNG

CHNG M U

CHAPTER 1

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1.3. CNG TC TRONG PHNG

Th nghim trong phng cc ch tiu c l

Tng hp, chnh l thng k kt qu th nghim trong phng

ngha v ni dung bo co CCT

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1.3.1. TH NGHIM TRONG PHNG CC CH TIU C L

CHNG M U

CHAPTER 1

Nhm th nghim xc nh cc ch tiu vt l

Nhm th nghim xc nh cc ch tiu c hc

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1.3.2. TNG HP, CHNH L THNG K KT QU TH NGHIM TRONG PHNG

CHNG M U

CHAPTER 1

Phn loi cc tr cc c trng ca t

Phm vi s dng cc c trng

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CHNG M U

CHAPTER 1

Trnh t thng k s liu a cht

3. Loi tr sai s th nghim

max 0

min 0

i CM

i CM

A A

A A

g s

g s

-

-

Mu no khng tha Loi Lm li t Bc 1.

Nu tha ht Bc 4

0: h s tiu chun thng k, ly theo s lng mu th nghim.

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CHNG M U

CHAPTER 1


3. Loi tr sai s th nghimn 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0 2.07 2.18 2.27 2.35 2.41 2.47 2.52 2.56 2.6 2.64 2.67 2.7 2.73 2.75 2.78

n 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

0 2.8 2.82 2.84 2.86 2.88 2.9 2.91 2.93 2.94 2.96 2.97 2.98 3.0 3.01 3.02

n 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

03.03 3.04 3.05 3.06 3.07 3.08 3.09 3.1 3.11 3.12 3.13 3.14 3.14 3.15 3.15

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CHNG M U

CHAPTER 1


4. Xc nh li s ton phng trung bnh2

1

1( )1

n

i

iA A nns - -

5. Xc nh h s bin ng:

S

A

H s bin ng tra bng ph thuc vo loi c trng

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CHNG M U

CHAPTER 1


5. Xc nh h s bin ng:c trng ca t H s bin ng

T trng ht 0.01

Trng lng ring 0.05

m t nhin 0.15

Gii hn Atterberg 0.15Module bin dng 0.30

Ch tiu sc chng ct 0.30

Cng chu nn 1 trc 0.40 [email protected]


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CHNG M U

CHAPTER 1


6. Tr tiu chun ca cc ch tiu 1

n

i

i

A

A n

Ring tr tiu chun v lc dnh n v v gc ma st trong xc nh theo cng

thc sau:

-

-

n n

ii

n

i

n

i

n

ii

n

n

tg

1

2

1

2

111

ss

tsts

j

-

n n

ii tgn

c1 1

1 sjt

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CHNG M U

CHAPTER 1


6. Tr tiu chun ca cc ch tiu 1

n

itc

i

A

A n

Ring tr tiu chun v lc dnh n v v gc ma st trong xc nh theo cng

thc sau:

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CHNG M U

CHAPTER 1


6. Tr tnh ton ca cc ch tiu

tc

tt

d

AAk

Trong kd l h s an ton v t.

kd = 1, vi tt c cc ch tiu tr lc dnh, gc ma st trong, dung trng, cng nn n

kd = 1/(1) vi cc ch tiu lc dnh, gc ma st trong, dung trng, cng nn n

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CHNG M U

CHAPTER 1


6. Tr tnh ton ca cc ch tiu i vi c v tgj: = t.

i vi dung trng v cng nn n:.t

n

Xc sut tin cy ca tr tnh ton cc c trng ca t c ly bng:

= 0,95 khi tnh nn theo sc chu ti; = 0,85 khi tnh nn theo bin dng.

Khi tnh ton theo sc chu ti th tr tnh ton ca cc c trng j, c v g k hiu l jI, cI v gI; cn tnh

theo bin dng th k hiu l jII, cII v gII. [email protected]


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CHNG M U

CHAPTER 1

S bc t do (n -1) ivi R

nv , (n - 2) i vi

c v j

H s t

ng vi xc suttin cy

0,85 0,952 1,34 2,923 1,25 2,354 1,19 2,135 1,16 2,016 1,13 1,947 1,12 1,908 1,11 1,869 1,10 1,8310 1,10 1,8111 1,09 1,8012 1,08 1,7813 1,08 1,77

S bc t do (n -1)i vi R

nv , (n - 2)

i vi c v j

H s t

ng vi xcsut tin cy

0,85 0,9514 1,08 1,7615 1,07 1,7516 1,07 1,7517 1,07 1,7418 1,07 1,7319 1,07 1,7320 1,06 1,7225 1,06 1,7130 1,05 1,7040 1,05 1,6860 1,05 1,67

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CHNG M UCHAPTER 1

Sai s ton phng trung bnh

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1.3.3. NGHA V NI DUNG BO CO CCT

ngha

Ni dung

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CHNG M UCHAPTER 1


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CHNG M UCHAPTER 1


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CHNG M UCHAPTER 1


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CHNG M UCHAPTER 1

1.4. LA CHN TH NGHIM PHC V CNG TC THIT K

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CHNG M UCHAPTER 1


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CHNG M UCHAPTER 1


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BI TP 1

1.3. BI TP

CHNG M UCHAPTER 1

Mt lon t m c khi lng 68,3g, sau khi sy kh khi lng cn li l 44,8g. Khi lng ca lon l

11,4g. Xc nh m ca mu t.

Dao vng s dng xc nh khi lng th tch ca mu t trn c chiu cao h=2cm, ng knh

trong d=6,3cm. Khi lng ca dao vng v t ( c ct v gt phng); 187,9g. Khi lng ca dao

vng: 89,8g. Xc nh khi lng th tch mu t.

Khi lng ring ht ca mu t l s=2,653g/cm3. Xem khi lng ring ca nc w=1g/cm3. Xc

nh khi lng th tch t kh, h s rng, rng, khi lng th tch y ni v bo ha ca mu

t. [email protected]


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CHNG M UCHAPTER 1

BI TP 2

1.3. BI TP

T th nghim xc nh gii hn

nho (bng phng php chm cuCasagrance) v gii hn do, s liu

ghi nhn c th hin bng sau.

Xc nh gii hn nho (c th s

dng biu trong ti liu ny), gii

hn do ca mu t. Tnh st ca

mu t, bit rng m ca mu t

W=67,3%.

Gii hn nho Gii hn do

Khi lng tm v lon

20,8 22,2 20,9 20,5 21,9

Khi lng tkh v lon

16,0 17,1 16,0 18,1 19,0

Lon s A15 A28 A93 B39 C45Khi lng lon 7,2 8,3 7,9 8,0 7,0

S ln ri N 36 26 17 m 54.54% 57.95% 60.49% 23.76% 24.17%

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CHNG M UCHAPTER 1

BI TP 3

1.3. BI TP

Gii hn Atterberg (gii hn nho bng phng php Casagrande) ca hai mu t c xc nht kt qu th nghim nh sau:

Mu 1: WL=75,3%, WP=37,7%

Mu 2: WL=41,3%, WP=18,9%

Xc nh tn t theo biu ng A.

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CHNG M UCHAPTER 1

BI TP 4

1.3. BI TP Th nghim nn n trc mt mu t c chiu cao ban u l h=7,6cm, ng knh

d=3,8cm. Gi tr ti trng tc dng ln mu ng vi bin dng cho theo bng sau.

V ng cong quan h ng sut bin dng, xc nh module bin dng, cng

sc khng nn n.

Titrng

(N)0 21,98 28,18 32,69 33,82 36,08 38,33 40,58 41,71 43,97 45,09 47,35 49,60 51,86 54,11 55,24 55,24 53,55 51,29

Bindngdc

trcrh

(0,01mm)

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900

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CHNG M UCHAPTER 1

BI TP 5

1.3. BI TP

Th nghim nn khng n hng (nn c kt) mt mu t c chiu cao ban u h0=2,0cm, h s rng ban

u e0-2,635. S c ng h bin dng n nh di cc cp p lc khi gia ti v d ti cho theo bng:

Thigian

(pht)0,1 0,25 0,5 1 2 4 5 8 15 30 1h 2h 3h 6h 12h

ln

(mm)

0,347 0,351 0,355 0,362 0,3708 0,385 0,4032 0,421 0,441 0,455 0,466 0,4725 0,479 0,486 0,493

V ng cong quan h bin dng v log(t), t xc nh h s c kt Cv v h s thm kv mu t.

Chiu cao ban u ca mu t 2cm, h s rng ban u e=2,[email protected]


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CHNG M UCHAPTER 1

BI TP 6

1.3. BI TP

Th nghim nn khng n hng (nn c kt) mt mu t c chiu cao ban u h0=2,0cm, h s rng ban

u e0=0,681. S c ng h bin dng n nh di cc cp p lc 1,0 v 2,0 kG/cm2 tng ng l 72

v 97 (0,01mm). Bit rng h s Poisson ca mu t =0,3. Xc nh gi tr module bin dng ca mu

t.

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CHNG M UCHAPTER 1

BI TP 7

1.3. BI TP

Th nghim nn ln mt mu t trong phng. Khi ng sut ng hng tng t 100kPa ln 200kPath th tch gim 0,1%. Gi s tng ng sut t 200kPa ln 300kPa, bin dng th tch s bng, nh hn

hay ln hn gi tr 0,1%? Ti sao?

Kt qu th nghim trong phng cho h s rng ca lp bn st dy 20m l e=2,345, Khi c lng

ln bng ng cong nn ln t kt qu th nghim trong phng th h s rng ban u c chn thno?

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CHNG M UCHAPTER 1

BI TP 8

1.3. BI TP

Cn c vo ng cong nn ln, nh gi v s thay i module bin dng ca t nn theo su.

T chng minh rng phm vi nn ln ca vng nn l c gii hn.

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CHNG M UCHAPTER 1

BI TP 9

1.3. BI TP

Kt qu ct trc tip hai mu t cho theo bng, xc nh gi tr c trng sc chng ct, nhn xt nh

gi v loi t.Mu 1.

Mu 2.

ng sut php (kG/cm2) ng sut tip (kG/cm2)0,25 0,1240,5 0,1511 0,210

ng sut php (kG/cm2) ng sut tip (kG/cm2)

1,0 0,606

2,0 1,187

3,0 1,[email protected]


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CHNG M UCHAPTER 1

BI TP 10

1.3. BI TP

STT Cng nn n qu (kG/cm2)1 0,192 0,223 0,184 0,275 0,17

6 0,267 0,218 0,28

Gi tr cng nn n trc t kt qu th nghim cc mu t trong mt lp theo

bng sau. Xc nh gi tr tnh ton.

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CHNG M UCHAPTER 1

BI TP 11

1.3. BI TP

Th nghim 3 1 - 31 30 31,42 70 32,7

Tin hnh th nghim nn ba trc mt mu t theo s UU. p lc bung c gi khng i trongsut qu trnh th nghim. Giai on k tip l nn lch trc trong iu kin khng thot nc cho n khi

mu t ph hoi. Kt qu nn lch trc khi ph hoi cho theo bng sau vi cc gi tr ng sut vi n v

kPa:

Xc nh sc chng ct ca mu t.

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CHNG M UCHAPTER 1

BI TP 12

1.3. BI TP

Skempton ngh quan h gia tng p lc nc l rng v ng sut xc nh theo biu thc

sau:u=B[3+A1-3]u - gia tng p lc nc l rng khi nn khng thot nc.A, B - Cc h s p lc nc l rng.viB=u3 VA=u1

Tin hnh nn 3 trc mt mu t bo ha nc hon ton theo s CU. Trc tin l gia tngp lc bung v cho mu t c kt n khi p lc nc l rng phn tn ht hon ton. p lc bung

c gi khng i trong sut qu trnh th nghim. Giai on k tip l nn lch trc trogn iu kinkhng thot nc cho n khi mu ph hoi v c o p lc nc l rng. Kt qu nn lch trc khi phhoi cho theo bng sau vi cc gi tr ng sut vi n v kPa:Xem nc l rng khng chu nn, nhn xt nh gi cc gi tr h s p lc nc l rng A v B.

Th nghim 3 1 - 3 u1 20 41 8,22 40 70 11,93 60 98 15,6

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CHNG M UCHAPTER 1

BI TP 13

1.3. BI TP

Th nghim nn n trc mt mu ct kt hnh lng tr trn c ng knh d0=47mm, chiu cao h0=93mm

cho theo bng, H s vng lc R=575 N/div. Tnh v v ng quan h ng sut - bin dng. Xc nh gi

tr module bin dng trung bnh. Nhn xt.

S c ng h bin dng

dc trc(div.=x10-2mm) 0 9 16 23 27 31 36 41 45 49 52 56 61 65 70 76

S c ng h vng lc DR(div.)

0 34 45 55 65 75 85 95 105 115 125 135 145 155 165 175

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CHNG M UCHAPTER 1

BI TP 14

1.3. BI TP

Gi tr p lc trn ng h khi nn n trc mt mu hnh lng tr trn ng knh d=54mm im ph hoi l 326kG/cm2. Din tch pistong Apis =13.35 cm2. Xc nh cng sc khng nn

n qu (kG/cm2, kN/m2).

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CHNG M UCHAPTER 1

BI TP 15

1.3. BI TP

STT qu (kG/cm2) STT qu (kG/cm2)

1 0.099 7 0.217

2 0.174 8 0.223

3 0.186 9 0.258

4 0.167 10 0.202

5 0.229 11 0.214

Xc nh gi tr tnh ton ca cng khng nn n ca lp t bn st t cc s liu th nghim.

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CHNG M UCHAPTER 1

BI TP 16

1.3. BI TP STT Module bin dng E0 (kG/cm2)

1 66.7

2 61.2

3 68.54 60.1

5 57.8

6 69.3

7 64.4

8 67.2

9 58.5

10 72.6

11 66.9

Gi tr module bin dng t kt qu th

nghim cc mu t trong mt lp theo bng sau.

Xc nh tr tnh ton vi xc sut tin cy=0.85.

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CHNG M UCHAPTER 1

BI TP 17

1.3. BI TP

Mt ct Granite to bi cc khi c kch thc 1mx1mx1m. Vi bn nn n trc qu=200Mpa = 2000 x 10.2 = 20.400 T/m2 v khi lng ring =2.6 T/m3. Module n hi Ee = 50.000 Mpa. Xc

nh chiu cao ti a ca ct sao cho ct vn n nh (bin dng xem nh khng ng k v khng

xt nh hng ca ti trng ngang). Khi bin dng ca ct do tc dng ca trng lng bn thn

s l bao nhiu?

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CHAPTER 2

PHN LOI T THEO MC CH XY

DNG

TH NGHIM C TSOIL MECHANIC LABORATORY

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CHAPTER 2 PHN LOI T THEO MC CH XY DNG

2.1. CC NGUYN TC PHN LOI T

Phi m t mt s thut ng xc nh, ngn gn nhng y cho ngi s

dng

Cc lp v ph lp c xc nh r theo cc thng s tng i d o lng

nh lng

Cc lp v ph lp phi tp hp c cc loi t c cc c trng tnh cht xy

dng tng t li vi nhau

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cng nacng

t ri t dnh t cbit


2.2. PHN LOI T TNG QUT

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2.3. TIU CHUN PHN LOI T VIT NAM

1. nh ngha mt s thut ng theo TCVN2. Phn loi t xy dng Vit Nam

3. Phn loi t ti hin trng


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2.3.1. NH NGHA MT S THUT NG THEO TCVN

t xy dng

tng

Cui v dm

Si v sn

Ht ct

Ht bi

Ht st

Ht mn

Ht th

t hu c

t ht mn

t cui si

t ct

t bi

t st

t ri

t dnh


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2.3.2. PHN LOI T XY DNG VIT NAM 5949 - 1993

Sinh vin t nghin cu


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2.3.3. PHN LOI T THEO TCVN 9362:2012Loi t hn ln v t ct Phn b ca ht theo ln tnh bng phn trm trng lng

ca t hong kh1 2

A. t hn ln

t tng ln (khi c ht sc cnh gi l a khi)t cui (khi c ht sc cnh gi I t dm)

t si (khi c ht sc cnh gi l t sn)

Trng lng ca cc ht ln hn 200 mm chim trn 50 %Trng lng cc ht ln hn 10 mm chim trn 50 %Trng lng cc ht ln hn 2 mm chim trn 50 %

B. t ct

Ct si

Ct th

Ct th va

Ct mn

Ct bi

Trng lng cc ht ln hn 2 mm chim trn 25 %

Trng lng cc ht ln hn 0,5 mm chim trn 50 %

Trng lng cc ht ln hn 0,25 mm chim trn 50 %

Trng lng cc ht ln hn 0,1 mm chim trn 75 % hoc hn

Trng lng ht ln hn 0,1 mm chim di 75 %CH THCH: nh tn t theo Bng 2 phi cng dn phn trm hm lng ht ca t nghin cu: Bt u t ccht ln hn 200 mm, sau l cc ht ln hn 10 mm, tip n l cc ht ln hn 2 mm ... Tn t ly theo ch tiuu tin c tha mn trong th t tn gi Bng 2.


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Tn t hn ln v t ct theo no nc

no nc, G

t m 0 < G 0,5

m 0,5 < G 0,8

No nc 0,8 < G 1

2.3.3. PHN LOI T THEO TCVN 9362:2012


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Loi ct cht ca ctCht Cht va Ri

A. Theo h s rng (e)

Ct si th v th vaCt mn

Ct bi

e < 0,55e < 0,6

e < 0,6

0,55 e 0,70,6 e 0,75

0,6 e 0,8

e > 0,7e > 0,75

e > 0,8

B. Theo sc khng xuyn cn pt (MPa) khi xuyn tnhCt th v th va (khng ph thuc m)

Ct mn (khng ph thuc m)

Ct bi:

a) t m v m

b) No nc

pt > 15

pt >12

pt > 10

pt > 7

15 pt 5

12 pt 4

10 pt 3

7 pt 2

pt < 5

pt < 4

pt < 3

pt < 2

C. Theo sc khng xuyn cn quy c p (MPa) khi xuyn ngCt th v th va (khng ph thuc m) p > 11 11 p 3 p < 3Ct mn:

a) t m v m

b) No nc p > 8,5 8,5 p 2 p < 2Ct bi t m v m p > 8,5 8,5 p 2 p < 2CH THCH:

1. Khng cho php dng xuyn ng xc nh cht ca ct bi no nc.

2. Khi xuyn t, dng hnh nn c gc nh l 60 v ng knh l 36 mm xuyn tnh v 74 mm xuyn [email protected]


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Loi t st Ch s do IP

-ct 0,01 Ip 0,07-st 0,07 Ip 0,17St Ip > 0,17CH THCH:

1. Khi trong t st c nhng ht ln hn 2 mm th thm vo tn gi Bng 6 t c cui (cdm) hoc c si (c sn) nu lng cha cc ht tng ng chim 15 % n 25 % theotrng lng v t cui (dm) hoc si (sn) nu cc ht ny cha trong t t 25 % n50 % theo trng lng.2. Khi loi ht ln hn 2 mm chim trn 50 % trng lng t th c xp vo t hn ln(xem 3.2).



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Tn t st theo ch s st Ch s st Is-ct:

- Cng Is 1-st v st:- Cng Is < 0- Na cng 0 Is 0,25- Do cng 0,25 Is 0,50

- Do mm 0,50 Is 0,75- Do nho 0,75 Is 1- Nho Is > 1



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Tnh cht Ht th Ht mn Hu c

Cc loi t

tng Bn Than bn

Cui St

CtDng ht Trn n gc cnh Bng Th

Kch thc ht Th Mn -

rng Thp Cao Cao

Tnh thm Cao Km n khng thm Thay i

Lc dnh gia cc ht Cao Thp Khng c n thp

Tnh do Khng Thp n cao Thp n trung bnhTnh p co Rt thp Trung bnh n rt cao Thng rt cao

Tc c kt Tc thi Trung bnh n chm Trung bnh n nhanh

2.4. PHN LOI T TI HIN TRNG


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2.5. PHN LOI T THEO TIU CHUN KHC

Phn loi theo BS 5930:1999

Phn loi theo ASTM D2487

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CHAPTER 3

CC TNH CHT VT L C BN CAT


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CHAPTER 3 CC TNH CHT VT L C BN CA T

3.1. M HNH T

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3.2. CC CH TIU TNH CHT C BN NH NGHA THUT NG

PHN LOI V CC TNH CHT C L CA T CHAPTER 5

Water content (w) is the ratio of the weight of water to the weight of solids.

Void ratio (e) is the ratio of the volume of void space to the volume of solids.

Porosity (n) is the ratio of the volume of void to the total volume of soil.

Degree of saturation (S) is the ratio of the volume of water to the volume of void.

Bulk unit weight () is the weight density, that is, the weight of a soil per unit volume.

Saturated unit weight (sat) is the weight of a saturated soil per unit volume.

Dry unit weight (d) is the weight of a dry soil per unit volume.

Effective unit weight () is the weight of soil solids in a submerged soil per unit volume.

Specific volume (V) is the volume of soil per unit volume of solids Specific gravity (Gs) is the ratio of the weight of the soil solids to the weight of water of equal

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Khong kh

Nc

Hat ran

Va

Vw

Vs

VvV

Wa

Ww

Ws

WWv

Pha Khoi lng Trong lng The tch

Qte VN Qte VN Qte VN

Ran(Soil) Ms Qh Ws Qh Vs Vh

Rong(Void) Mv Qr Wv Qr Vv VrNc Mw Qn Ww Qn Vw VnKh (Air) Ma Qk Wa Qk Va VkTong M Q W Q V V

3.2. CC CH TIU TNH CHT C BN K HIU


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Ten goiKy hieu

Ten goiKy hieu

Qte VN Qte VNDung trong t nhien , w o am w w

Dung trong kho d k o bao hoa S G

Dung trong bao hoa sat bh He so rong e e

Dung trong ay noi n o rong n n

Trong lng rieng hat s h o cha khong kh Av Av

Ty trong hat Gs D Dung trong nc w n

3.2. CC CH TIU TNH CHT C BN K HIU


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Ten Cong thc Ten Cong thc

Dung trongt nhien

o am

Dung trongkho

o baohoa

Dung trongbao hoa

He so rong

Dung trongay noi

o rong

Trong lngrieng hat

o chakhong kh

Ty tronghat

The tchrieng phan

VW

g

VWs

dg

VV.W vws

satg

g

VV.W

' sws g-

g

s

ss V

Wg

wss /G gg

s

w

WW

w

v

wr V

VS

s

v

VV

e

VV

n v

3.2. CC CH TIU TNH CHT C BN CNG THC THEO NH NGHA


1)w1.(.G

e ws -g

g

1eds -gg

e

G.wS sre1

en

1d wgg a

v

v

Va

V

Air

WaterSoil Solid

Va

Vw

Vs

VvV

Wa=0

Ww

Ws

WWw

Volume Weight

' sat wg g g -'

s

VV

V

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3.3. MI QUAN H GIA CC CH TIU TH TCH RN N V

Air

Water

Soil Solid

Va

Vw

Vs

VvV

Wa=0

Ww

Ws

WWw

Volume Weight

AirWater

Soil Solid

Sr.e

1

eV =1+e

Wa

Sr.e.gw=w.Gs.gw

Gs.gw

(1+w).Gs.gw

Volume Weight


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3.3. MI QUAN H GIA CC CH TIU TH TCH N V

Air

Water

Soil Solid

Sr.n

1-n

n1

Gsgw(1-n)

Gsgw(1-n)(1+w)

av.n

wGsgw(1-n)

Volume Weight

Air

Water

Soil Solid

Va

Vw

Vs

VvV

Wa=0

Ww

Ws

WWw

Volume Weight


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5.2.3. MI QUAN H GIA CC CH TIU

Moist unit weight () Dry unit weight (d)Saturated unit weight (sat)

w, Gs, e , w Gs, e

S, Gs, e Gs, e Gs, n

S, Gs, w S, Gs, w Gs, wsat

w, Gs, n Gs, n e, wsat

S, Gs, n e, w, S n, wsat

sat, e e, d

sat, n S, d

sat, Gs wsat, d

( )11

s ww G

e

g

( )

1s wG Se

e

g

( )1

1

s w

s

w G

wG

S

g

( )( )1 1s wG n wg -

( )1s w wG n nS g g-

1 w

g

1s wG

e

g

( )1s wG ng -

1

s w

s

G

wG

S

g

( )1weS

e w

g

1w

sat

e

e

gg -

sat wng g-

( )( )1

sat w s

s

G

G

g g-

-

( )1

s wG e

e

g

( )1s wG n n g -

( )11

sat s w

sat s

w G

w G

g

1

1sat

w

sat

we

w eg

1 satw

sat

wn

wg

1

d w

e

e

g g

11 d w

sGg g

-

( )1d sat wg [email protected]


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3.4. CC CH TIU TRNG THI


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3.4.1. CHT TNG I (T RI)

Dr(%) 0 15 15 50 50 70 70 85 85 100

Trang thai Rat ri RiChat trung

bnhChat Rat chat

Bng trng thi t ri (theo Braja. M. Das) theo cht tng i


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Bang trang thai at ri (theo tieu chuan Lien Xo)

D o chat cua at

0 D 0.33 Xop (ti)0.33 D 0.66 Chat va0.66 D 1 Chat

Loai ato chat

Chat Chat va Xop

Cat lan san, cat to, cat va e < 0.55 0.55 e 0.7 e > 0.7

Cat nho e < 0.6 0.6 e 0.75 e > 0.75

Cat bui e < 0.6 0.6 e 0.8 e > 0.8



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Bang trang thai at ri c xac nh theo th nghiem xuyen tieu chuan

SPT Standard Penetration Test (theo tieu chuan Lien Xo)

Tr so N D o chat cua at

1 4 0.2 Rat xop

5 9 0.2 0.33 Xop

10 29 0.33 0.66 Chat va

30 50 0.66 1 Chat> 50 > 1 Rat chat



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at hat tho (at ri)

Tr so N Trang thai cua at0 4 Rat xop4 10 Xop10 30 Chat va30 50 Chat

> 50 Rat chat

Bang trang thai at ri c xac nh theo th nghiem xuyen tieu chuan

SPT Standard Penetration Test (theo Peck va cong s Hoa Ky)



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3.4.2. GII HN ATTERBERG (T DNH)

PLLLPI -

Ch s do

Ch s nho

PLLL

PLwLI

--

Loai at wL Ipat cat wL < 35 Ip < 15

at bot wL > 20 5 < Ip < 25

at set wL > 30 Ip > 15

Ch so deo Ip Ten at

Ip < 1 at ri1 < Ip < 7 A cat7 < Ip < 17 A set

Ip > 17 Set


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Ten at Ch so long IL Trang thai cua at

at a catIL < 0 Cng

0 IL 1 DeoIL > 1 Long

at a set va set

IL 0 Cng0 < IL 0.25 Na cng

0.25 < IL 0.5 Deo cng

0.5 < IL 0.75 Deo mem0.75 < IL 1 Deo chay

1 < IL Long

3.4.2. GII HN ATTERBERG (T DNH)


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Tnh lng nc sch cn iu ch va st bentonite t 1 tn bt st c m 10%, t trng ht Gs = 2.75. Gi thit trng lng ring ca va stg = 11.5 kN/m3.

BI TP 13.5. BI TP


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wsat

wsat

n

nwa

gg

g

-)

)11

()sd

wsatwbgg

g -

Chng minh khi trng thi bo ha, m ca t bng:

BI TP 23.5. BI TP


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Mt mu t c dung trng t nhin l 16.5 kN/m3, w = 1 5 % , Gs = 2.70. Yu cu tnh:

1. Dung trng kh,gd.

2. rng, n.

3. bo ho, S

4. Khi lng nc cn thm vo mu t trng thi bo ho.

BI TP 33.5. BI TP


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Mt mu t c dung trng kh 17.5kN/m3, m w = 4%, t trng ht Gs = 2.65. Xc nh

trng lng nc cn thm vo c bo ho S = 95%.

BI TP 43.5. BI TP


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BI TP 53.5. BI TP


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BI TP 63.5. BI TP


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BI TP 73.5. BI TP


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BI TP 83.5. BI TP


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BI TP 93.5. BI TP


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BI TP 103.5. BI TP


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CHAPTER 4

NHM TH NGHIM CC CH TIU TNHCHT


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CHAPTER 4 NHM TH NGHIM CC CH TIU TNH CHT

4.1. TH NGHIM XC NH M

- Sinh vin t nghin cu.

- TCVN 4196:2012

- ASTM D2216

VIDEO TH NGHIM XC NH M CA T M HNH

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4.2. TH NGHIM XC NH TRNG LNG RING


- TCVN 4202:2012

- ASTM D7263


VIDEO TH NGHIM XC NH TRNG LNG RING CA T M HNH

VIDEO SAND CONE METHOD - I

VIDEO SAND CONE METHOD - [email protected]


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4.3. TH NGHIM XC NH T TRNG HT

- Sinh vin t nghin cu

- TCVN 4195:2012

- ASTM D854.


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4.4. TH NGHIM XC NH THNH PHN HT RY SNG


- TCVN 4198:2012

- ASTM D422

- BS 1377 Part II


VIDEO TH NGHIM XC NH THNH PHN HT CA T M HNH

VIDEO TH NGHIM XC NH THNH PHN HT CA T THC T

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10

60

D

DC

u

1060

2

30)(

DD

DC

g



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CHAPTER 5

NHM TH NGHIM CC CH TIUTRNG THI


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CHAPTER 5 NHM TH NGHIM CC CH TIU TRNG THI

5.1. TH NGHIM XC NH GII HN DO CA T

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Phng php xc nh gii hn chy kiu Va-xi-li-ep

Phng php xc nh gii hn chy theo phng php Casagrande

5.2. TH NGHIM XC NH GII HN CHY CA T


VIDEO TH NGHIM XC NH GII HN CHY CA T M HNH

VIDEO TH NGHIM XC NH GII HN ATTERBERG CA T THC T I

VIDEO TH NGHIM XC NH GII HN ATTERBERG CA T THC T II

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CHAPTER 6

NHM TH NGHIM M CHT T


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CHAPTER 6 NHM TH NGHIM M CHT T


- TCVN 4201:2012, 22 TCN 332:06

- ASTM D698, D1557, D1883, D4429

- BS 1377 Part IV

VIDEO TH NGHIM M CHT T M HNH

VIDEO TH NGHIM M CHT T THC T I

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CHAPTER 7

BIN DNG CA T


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CHAPTER 7 BIN DNG CA T

7.1. CONSOLIDATION THEORY

- ln l g?

- nh hng n cng trnh?

- Cc loi ln?

Si

t

S

t100

Sc

SsSt

t

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**** Assumptions for Consolidation theory

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**** Some Key Words

- Degree of Consolidation: U%

- Coefficient of Consolidation: Cv

- Time Factor: Tv

%100*01

1

uu

uu

S

S

U

wt

-

-

2H

t

v

C

vT

wo

z

w

z

va

k

a

ekC

gg

.

)1( )(%v

TfU

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**** Some Key Words

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Si

t

S

t100

Sc

SsSt

t

**** Phases of Consolidation

- Initial compression.

- Primary consolidation.

- Secondary compression.

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**** Curve - fitting procedures

- The log-time method.

- The square root time method

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**** The log time method

( )242 U%

vT

v

T.)

vf(TU%

)/(*197.0

50

22

50

50 smmt

hh

t

TC

v

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**** The square root time method

vv

v T

T.)f(TU% 128.12

)/(*848.0

90

22

90

90 smmt

hh

t

TC

v

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**** Relative merits of the two methods

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7.2. TH NGHIM NN T C IM BIN DNG CA T KHI B NN

- Th nghim hin trng

- Bn nn hin trng

- Th nghim trong phng

- Th nghim nn khng n hng

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7.2.1. TH NGHIM BN NN HIN TRNG

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VIDEO TH NGHIM BN NN HIN TRNG

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Sd

pO

ShS

p

A

BC

D

S

F

G S = Sd+ Sh

pO p

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P,

S

Pult,sult

k (cat tuyen)

k (tiep tuyen)

)()()/( 3

mSMPamMNk s

21

12

ssSk

-

-

sss

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- u im.

- Nhc im

- Cc nghin cu xc nh h s k sinh vin t nghin cu.

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7.2.2. TH NGHIM NN KHNG N HNG

**** Tiu chun p dng

- TCVN 4200:2012.- ASTM D2435, D4186

- BS 1377 Part VI

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**** Dng c th nghim:

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1 2

34

5

6

7

7

8

b

1 2

3

4

5

6

7

a

5. ming kim loi tn lc

6. chuyn v k

1. mu t

2. vng cha mu

3. bt

4.

7. lc tc ng

**** Cu to hp nn + mu:

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**** Quy nh chung th nghim:

- Quy nh v bo ha mu

- Quy nh v mu t th nghim.

- Quy nh v ti trng th nghim.

- Quy nh v cp ti trng

- Quy nh v thi gian th nghim.

- Quy nh v dng c th nghim

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**** Cng tc chun b mu:

- i vi mu nguyn dng

- i vi mu khng nguyn dng

**** Quy trnh th nghim:

VIDEO TH NGHIM NN KHNG N HNG

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t

e1

e3

e

t1 t2

e2

p1

t

p

t1 t2

p2

p3101

.

.

0 -

D--

-

sh

hh

sh

h

sh

shh

sh

vh

shA

vhA

sV

vV

e

Tthai cuoi

eo

ef

t

Tthai au

Toc o nhanh

Toc o cham

e

eo

ef

po pf Log p

Tthai au

Tthai cuoi

e

eo

ef

po pf p

Tthai au

Tthai cuoi

e

FG

Wh

sw

ss ..g

11 0 -D-D--s

mn

s

ii

hhhh

hhe

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eo

e1

p1 p4 p

e2e3

e4

e

p2 p3 p1 Log p

e

e1e2

p2 p3 p4

e3

e4


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How To Draw Void RatioPressure Plots

Step 1: s sss w s w

W MH

AG AGg

Step 2: v sH H H -

Step 3: 0.

.v v v

s s s

V H A H e

V H A H

Step 4: 11 s

He

H

DD

Step 5:1 0 1

22 1 1 1

2

e e e

He e e e

H

- D

D - D - [email protected]


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BI TP 1

Following are the results of a laboratory

consolidation test on a soil specimen obtainedfrom the field: Dry mass of specimen = 128 g,

height of specimen at the beginning of the test =

2.54 cm, Gs = 2.75, and area of the specimen =

30.68 cm2.

Make necessary calculations and draw an e

versus logs curve.


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7.2.3. C IM BIN DNG CA NN T KHI B NN

Sd

pO

ShS

p

A

BC

D

S

F

G

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7.2.5. CC NHN T NH HNG N BIN DNG LN CA T

- cht ban u ca t

- Tnh kt cu ca t

- Lch s chu nn ca t

- Tnh hnh tng ti

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7.3. CC L THUYT BIN DNG LN

L thuyt nn bin dng tuyn tnh


Sp

Se

S (m)

O p (kN/m2)

13

4

2

L thuyt bin dng n hi cc b

L thuyt tng bin dng n hi

L thuyt hn hp

L thuyt tng qut

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7.4. CC C TRNG BIN DNG CA T

3. H s nn tng i ao H s nn th tch mv

4. Module bin dng E0 (kg/cm2):

)1('1 11 e

e

e

ama vo

D

D-

s

)

1

21(

1 20

-

-vm

E

'

11

sD

e

va

HS 'sDv

mHS

s

0

2

0 1

21

'

E

phH

E

S

-

-D

Cc c trng bin dng ca t ph thuc vo cc cp ti [email protected]

CHAPTER


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5. p lc tin c kt

s2

e1

e2

s1 Log p

Tthai au

Tthai cuoi

e

s1 Log p

e

e1e2

s2 s3 s4

e3

e4

Cs

Cc

sp

Ngun gc ca qu trnh c kt trc- Sinh vin t nghin cu.

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CHAPTER 7


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5. p lc tin c kt

ng nen nguyen thuy

trong ieu kien t nhienng n do quatrnh lay mau

ng nen do th nghiemco ket trong phong cua maunguyen dang

ng nen lai do th

nghiem co ket trong phong

ng n dogiam tai

sp

A

logsvc

e

D

C

B

F

E

O

ng co ket cuamau xao tron

Phng php xc nh p lc tin c kt

Phng php Casagrande


Hiu th no v p lc tin c kt

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CHAPTER 7


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5. p lc tin c kt

Hiu th no v p lc tin c kt

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CHAPTER 7


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s1 s2 logs

e

e1

e2ng nen

nguyen thuy

ng n va

nen lai

s1 s2 sp logs

eer1

er2

ng nennguyen thuy

ng nen lai

6. Ch s nn Cc, ch s n Cs

,1

,2

21s lglg eeC s-s -

,1

,2

21

loglog)(log sss -

-

DD-

eee

Cc

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7. T s tin c kt OCR

svo ng sut hiu qu do ti trng bn thn cc lp t

bn trn hin hu gy ra

* OCR < 1 t di c kt (UC)

* OCR = 1 t c kt thng (NC)

* OCR > 1 t c kt trc (OC)

'

vo

'p

OCR

s

s

p'1 p'2 Log p

e

e1

e2

Cs

Cc

sp

OC NC

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CHAPTER 7


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7.6.2. D BO LN BNG CCH DNG KT QU BI TON NN MT CHIU

c. Phng php tng phn t

Ap lc ay mong u nho e vung bien dang deo trong nen khong phat trien qua ln Coi nenlam viec trong gii han an hoi

Tnh ap lc gay lun: pgl= p g.Df

Xac nh chieu day vung nen lun Ha:

sp 0.2sbt vi at co Module bien dang E5MPa

sp 0.1sbt vi at co Module bien dang E5MPa

Chia vung nen lun thanh cac phan to co chieu day nho hn 0.25 be rong mong coi S trong

cac phan to thay oi khong ang ke.Nen nhieu lp: Mat phan chia cac lp at trung mat chia cac phan to

Khi so lp n=64, th ket qua tnh lunso lp [email protected]

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Tnh toan va ve bieu o S hu hieu do trong lng ban than va tai trong gay lun gay ra tai gia cac lp

phan to:sbt(i)= gi.hi = p1i he so rong e1icua cac phan to at trang thai ban au, khi cha ganh chu cong trnh.

sp(i)= k.(p g.Df) S do tai trong ngoai gay ra

p2i=sbt+sp S trong at sau khi ganh cong trnh he so rong e2icua cac phan to at.

S dung cac cong thc tnh lun cua bai toan 1 chieu tnh bien dang ng cua cac pha n to (S dung bieu oe p, e logp)

o lun cua mong la tong cac o bien dang ng cua cac phan to

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Dpi =sgl(i)

pgl

15

p1i =sbt(i)

p2i

Ha

eo

e1

p1 p4 p

e2e3e4

e

p2 p3

p1i p2i

e2i

e1i

p1 Log p

e

e1e2

p2 p3 p4

e3

e4

p1i

e1i

p2i

e2

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pgl

15

15

Vung nenlun khikhong xetay noi

Vung nenlun khi coxet ay noi

sbt khi khongxet ay noi

sbt khi coxet ay noi Nen xet co lc ay noi vi tat ca cac

loai at nam di [email protected]



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7.6.2. D BO LN THEO L THUYT BN KHNG GIAN N HI


7.6.3. D BO LN THEO PP LP TNG NG


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7.6.4. D BO LN THEO THI GIAN

a. c kt Ut

- c kt Ut l g?

t

tt

SSU

- ln ti thi im t?

ttt SUS .

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b. Phng php thc hnh

-

-

-

-

...e251

e91

e8

1U v

2

v

2

v

2

T4

25T

49

T4

2t

2H

tCT vv

wo

z

w

zv

a

k

.a

)e1(kC

g

g

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b. Phng php thc hnh

2H

tCT vv

wo

z

w

zv

a

k

.a

)e1(kC

g

g

-- ---

...125

1271321 4

254942

222

vvv TTT

t eeeU

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b. Phng php thc hnh

2H

tCT vv

wo

z

w

zv

a

k

.a

)e1(kC

g

g

--

--...

3

21

9

121

161 4

94

2

22vv TT

t eeU

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b. Phng php thc hnh

Tv(0-1)= Tv(0)+(Tv(1)- Tv(0))I0-1 Tv(0-2)= Tv(2)+(Tv(0) Tv(2))[email protected]



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b. Phng php thc hnh

Trng hp 0-1 Trng hp 0-2

I0-1 I0-20

0.10.20.30.40.50.6

0.70.80.9

1.00

00.840.690.560.460.360.27

0.190.120.060.00

11.523456

789

12

10.830.710.550.450.390.3

0.250.20.170.13

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172/275

7.7. BI TP

BI TP 2

set

B

C

DE

F

A

G

3m

6m

cat

a

im A B C D E F G su (m) 0 1 2 3 4 5 6

u (KPa) 0 13.4 23.22 26.82 23.22 13.4 0


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7.7. BI TP

BI TP 3


Cho 1 nn t nh hnh v. Trn mt v y lp st l lp ct thot nc tt. Lp st bo ha v c

m W = 30%, Gs = 2.7, a = 0.002cm2/N, kv = 2*10-9cm/s. Biu ng sut do ti cng trnh gy ra

trong lp st nh hnh v. Tnh thi gian lp st t c kt l 96%

2m

cat

cat

set

18KPa

10KPa

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7.7. BI TP

BI TP 4


Cho 1 nn t st dy 6m c kt thm 1 chiu nh hnh v. Lp c m e0 = 0.8, cv = 144x103

cm2/nm, a = 0.0025cm2/N, kv = 2*10-9cm/s. Biu ng sut do ti cng trnh gy ra trong lp stnh hnh v. Tnh thi gian lp st c kt gn xong

6m

cat

set

300KPa

120KPa

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7.7. BI TP

BI TP 5


Nn t c p t gia ti trc cho lp st nh hnh v. Sau 1 thng gia ti ln ca lp st

l 100mm. Sau 2 thng gia ti ln ca lp st l 139.4mm

8m

cat

cat

set

100KPa

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7.7. BI TP

BI TP 7

CHAPTER 7

Ti 1 nn t do iu kin cht ti gy ra p lc nc l rng thng d ban u trong lp st nh

hnh v. Tnh c kt trung bnh ca lp st ng vi Tv = 0.3. ui = 15- 1.5z-2sin(z/6)

6m

cat

cat

set

15KPa

6KPa

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7.7. BI TP

BI TP 8

CHAPTER 7

Cho 1 nn t nh hnh v. Lp st bo ha nc c a0

= 0.045cm2/KG, lp ct c a0 = 0.0085cm2/KG1. Tnh c kt ca t nn ti cc im c su tnh

t mt t ln lt l 2m, 4m, 6m ti thi im c

kt trung bnh ca lp st l 50%:

2. Tnh ln ca t nn ti thi im ni trn bit tng

gc l khng ln:

q=120KPa

4m

6m

cat

cat

set

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7.7. BI TP

BI TP 9

CHAPTER 7

Cho 1 nn t c lp st bo ha nc dy 6m nm trn 1 tng . Trn mt c ti phn b u khp

l q. Ta ly 1 mu t st c chiu cao 2cm em lm TN nn c kt vi ti tc dng l q. Sau 10 pht

mu t t c kt l 50%.

1. Tnh thi gian c kt cn thit lp st t c kt l 50%, xem tng l nt n v thot nc

tt:

2. Tnh thi gian c kt cn thit lp st t c kt l 80%, xem tng l khng thot nc:

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180/275

7.7. BI TP

BI TP 12

CHAPTER 7

p lc nn (N/cm2) 0 5 10 20 40

Chiu cao mu khi n nh (mm) 20 19.49 19.13 18.78 18.58

Kt qu nn khng n hng mt mu t bo ha cho trong bng sau:

Yu cu:- V ng cong p co (e - p)

- Xc nh h s p co a ng vi ti trng p = 15 N/cm2

- Chp nhn gi thit t l vt liu n hi vi h s poisson = 0.3, hy xc nh mdun n hi (E) ca t t

h s p co (a) ni trn.

Cho bit t trng ht ca t Gs = 2.72, m ca mu sau khi th nghim xong W = 30.51%.

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CHAPTER 8

SC CHNG CT CA T

SOIL MECHANIC LABORATORY

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CHAPTER 8 SC CHNG CT CA T


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CHAPTER 8

8.1. TNG QUAN V TNH BN V SC CHNG CT CA T

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a b c

N

WE

Mt trt

Mt trt

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8.2. L THUYT SC CHNG CT CA T

Sc chng ct ca t bao gm hai thnh phn: ma st gia cc ht v lc dnh

gia cc ht t.

8.2.1. SC CHNG CT CA T RI

B A

Lc php tuyn

Lc tip tuyn

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B A

Lc php tuyn

Lc tip tuynd

DC

Lc php tuyn

Lc tip tuyn

[email protected]



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189/275


Hien tng hoa long

Vi at ri bao hoa nc khi chu cat the tch lo rong giam. Neu nc thoat ra khong kp

th ap lc phap tuyen truyen len nc lo rong khi chu cat nhanh co the mat i sc

chong cat cua no hien tng hoa long.

Vi at cat chat co he so rong nho hn he so rong gii han khi b cat nhanh, nc lo

rong thoat ra khong kp c bu vao phan the tch lo rong gia tang nen khong co hientng ap lc phap tuyen truyen vao lo rong, do o, hien tng hoa long khong xay ra.

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8.2.2. SC CHNG CT CA T DNH

Lc ct

Lc php tuyn

A

Lc ct

Lc php tuyn

0

Thnh phn dnh

B

C

A

Thnh phn ma st

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8.3. NH LUT MORH - COULOMB

t=s = .tgj + c

Sc chng ct ca t ti mt im trn mt mt phng l mt hm tuyn tnh theo ng sutphp tuyn trn mt

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ng bao chng ct Mohr

ng Mohr - Coulomb

c j

ng bao chng ct [email protected]


193/275



194/275


**** Nhng iu chnh b sung sau MORH - COULOMB

Cac nghien cu sau Morh Coulomb cho thay cac thong so chong cat c, jcon phu thuoc vao nhieuyeu to khac nh: trang thai S ban au, o am cua at, ieu ien thoat nc, ieu kien th

nghiem,Terzaghi:

Sc chong cat phu thuoc vao S phap hu hieu ch khong phai la S phap tong, v ch co phanS hu hieu mi phat sinh ma sat.

s S hu hieu

j goc ma sat noi tai cua at

c la lc dnh

''' ctgs js

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8.4. CC TH NGHIM XC NH THNG S CHNG CT CA T

**** Nhng kha cnh sc chng ct ca t cn quan tm

1. The shear resistance of free-draining non-cohesive soils (i.e. sands and gravels), which is virtually

independent of time.2. The drained shear strength of cohesive soils, which depends upon the rate of displacement being slow

enough to permit full drainage to take place during shear.

3. The long-term or residual drained shear strength of soils such as over-consolidated clays, for which a

slow rate of displacement and a large displacement movement are required.

4. The shear strength of very soft cohesive soils under undrained conditions, i.e. in which shearing is

applied relatively [email protected]



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Nhom cac th nghiem trong phong:

Nhom cac th nghiem hien trng:

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* Cc th nghim xc nh sc chng ct ca t u c hai giai on c bn:

* Giai on 1: Tc ng h lc ln mu tao trng thi S tng t nh mu t

th nm t nhin (tr th nhim nn n)* Giai on 2: Tc ng thay i p lc ln mu tng ng vi mu t hot

ng khi xy dng cng trnh

*UU

*CD*CU


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8.4.1. CC TH NGHIM TRONG PHNG

8.4.1.1. TH NGHIM CT TRC TIP

**** Tiu chun p dng

- TCVN 4199:2012.

- ASTM D3090

- BS 1377 Part VIII

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8.4.1. CC TH NGHIM TRONG PHNG

8.4.1.1. TH NGHIM CT TRC TIPHop cat Vong ng bien

Chuyen v ke

Tht di ong

Tht co nhMau at [email protected]



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[email protected]



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**** Phn loi:

1. Quick test: is applicable to free-draining soils (i.e. sands), whether dry of fully saturated. Drained

condition apply in this test.2. Slow drained shear test applies to soils that are not free-draining and therefore require more time to

allow drainage to take place. In order to carry out a slow drained shear test, provision is made for the

specimen to be consolidated before shearing and for further drainage to take place during shear at a

suitably slow rate of displacement, so that the consolidated drained shear strength parameters can

be determined.

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203/275


**** Application of normal stress

** Loading yoke only

kgmL

m

and

kgL

m

or

mkNL

m

h)

9810.

(

9810.

/1000..81.9

2

1

2

2

2

-

s

s

s

** Loading yoke only

( ) )

( )a

cWW

a

bWWWW

or

cWWbWaWWaW

jbh

jbh

..

....

21

21

(kg/cm2)

m(kg)

0.25 6.75

0.5 12.5

0.75 20.25

1 25

1.5 37.5

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**** How to calculate shear stress

2

2/1000*

.mkN

L

RCRt

VIDEO TH NGHIM CT TRC TIP M HNH

VIDEO TH NGHIM CT TRC TIP THC T I

VIDEO TH NGHIM CT TRC TIP M HNH II

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-40

-20

0

20

40

60

80

100

120

140

160

180

0 100 200 300 400 500 600

chuyen v ngang (10-2 mm)

chuyenvng(10-2m

m)

(kPa)- Ct cht

Ct ri

Ct ri

Ct cht

nh

im un

**** Shear Strength of Dry and Saturated Sand

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**** Shear Strength of Clay

**Undrained Test on Clays: not available

**Drained Tests on Clays and Silts

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**** Residual Shear Strength

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**** Effect of Stress History

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209/275


**** Applications of Shear Strength Parameters

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**** Limitations of the Shearbox Test

SELF

STU

DY

SELFSTU

DY

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**** Advantages of the Shearbox Test

SELF

STUDY

SELFSTU

DY

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Chiu cao: 2.00 cmTit din: 31.172 cm2

H s vng: 0.00442 KG/cm2/div.

p lc nn S c Lc ct, tKG/cm2 ln nht KG/cm2

0.50 112.0 0.4951.00 132.0 0.5831.50 167.0 0.738

j= 14040 c= 0.362 KG/[email protected]



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8.4.1.2. TH NGHIM NN BA TRC

**** Tiu chun p dng

- TCVN 8868:2011.

- ASTM D2850, D4767, D7181

- BS 1377 Part VIII

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[email protected]



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Th nghim nn ba trc l th nghim tin cy nht xc nh cc c trng sc chng ct

ca t, v:

* M t xc thc nht mu t trong nhng iu kin chu ti trong t nn khi gnh cc

loi cng trnh khc nhau.

* C th m phng cc iu kin thot nc khc nhau ca t nn.

* C th xc nh ng thi cc c trng bin dng ca t nn ng thi vi cc ch tiu

sc chng ct

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BC 1 BC 2 BC 4BC 3 [email protected]



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Ct c kt thot nc (CD)

100max

tbHV fo

e

3

1

3

3

=

1-

3

+

Bien dang doctruc, DLLc ng, P

Ap lcbuong, 3

Ap lc nclo rong, u=0

The tch thayoi,DV

Mangcao su

Mau at

a tham

Van

Piston

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8.4.1.2. TH NGHIM NN BA TRC - CD

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s1-s3Cat chat, set cng (OC)

+DV(n)

-DV(nen)

ea %

ea %

Cat chat, set cng (OC)

Cat ri, set mem (NC)

Cat ri, set mem (NC)


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Cong trnh ng suat chong cat CDThi cong cong trnh at ap tng lp thatcham tren nen set mem

ng suat chong cat tai cho co thoat ncsau khi a co ket

ap at vi s tham qua thng trc ng suat chong cat cua loi ap co thoatnc, sau khi a co ket

Mai doc t nhien hoac mai ao cua homong

ng suat chong cat tai cho co thoat nc,sau khi a co ket

D/ Mong nong tren nen set sau khi lun on

nh

ng suat chong cat tai cho co thoat nc,

sau khi a co ket


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Ct c kt khng thot nc (CU)Bien dang doc

truc, DLLc ng, P

Ap lcbuong, 3

Ap lc nclo rong, u

The tch thayoi,DV=0

Mangcao su

Mau at

a tham

Van

Piston

3

1

3

3

=

1-

3

+

[email protected]



223/275

at co ket thng (NC)

t

Vong tron MohrS hu hieu

s3 s1

c= 0

j > jCD

8.4.1.2. TH NGHIM NN BA TRC - CU

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NC


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225/275

at qua co ket (OC)

t

Vong tron MohrS hu hieu

s3 s1

c > 0

j > jCD


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OC


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Th d 8.1.

Kt qu th nghim nn ba trc c kt khng thot nc mt mu t dnh c kt thng

cho trong bng sau:

p lc trong bung nn: sc = s3 (kPa) 100 200 300 lch ng sut cc hn: (s1f-s3) (kPa) 137 210 283p lc nc l rng khi mu b trt: uf(kPa) 28 86 147

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229/275


Th d 8.2.

Kt qu th nghim nn ba trc c kt khng thot nc mt mu t dnh c kt trc cho

trong bng sau:

p lc trong bung nn: sc = s3 (kPa) 100 250 400 lch ng sut cc hn: (s1f-s3) (kPa) 340 410 474p lc nc l rng khi mu b trt: uf(kPa) -42 64 177

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p lc trong bung nn: sc = s3 (kPa) 100 250 400 lch ng sut cc hn: (s1f-s3) (kPa) 340 410 474p lc nc l rng khi mu b trt: uf(kPa) -42 64 177

Th d 8.2.

0

50

100

150

200

250

0 200 400 600 800 1000

jcu =110

ccu = 121(kPa)

j = 270

c= 31(kPa)

t (t) (kPa)

s (s) (kPa)[email protected]



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H s A v B ca Skempton:

Theo Skempton, trong th nghiem nen 3 truc khong thoat nc:

Du = B[Ds3+ A(Ds1-Ds3)]

3sDD ouB - HS ap lc nc lo rong trong qua trnh nen ang hng,

vi at bao hoa va khong chu nen: B=1

- He so ap lc nc lo rong trong qua trnh ap S lech

31

1

ss D-DD

u

AAB

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Gia tr he so BLoai at S=100% S=99%

Cac loai set mem co ket thng 0,9998 0,986 cac loai set va at bot am chat cac loai set co ket trc nhe

0,9988 0,93

cac loai set cng co ket trc manh phan ln cac loai cat

0,9877 0,51

cac loai cat rat chat cac loai set rat cng ap lc chen bo ngang ln

0,913 0,10

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Loai at set Afat set rat nhay T +0,75 en +1,50

at set co ket thng T +0,5 en +1at a set am chat T +0,25 en +0,75at set co ket trc nhe T 0 en +0,5Hon hp a at set am chat T 0,25 en +0,25at set co ket trc manh T 0,5 en 0Theo Skempton 1954

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Th d 8.3.

Mt mu t trong th nghim nn ba trc, trc tin c kt vi p lc ng hng

trong hp nn l 600kPa, qu trnh thot nc c dng li khi p lc nc l rng

trong mu l 300kPa, cc vi thot nc c ng li v tng p lc trong hp nn

n 720 kPa, p lc nc l rng tng n 415 kPa. Sau , tng lc dc trc to

mt ng sut lch (D 1 -D 3) l 550 kPa trong khi vn gi nguyn gi tr p lc trong

hp nn, p nc l rng cui cng t n 562 kPa. Tnh cc h s p lc nc lrng A, B.

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Cong trnh va giai oan lam viec tngng

ng suat chong cat CU

A/ ap lp at tiep theo sau khi a chonen co ket hoan toan vi lp ap trc

cua cong trnh at ap, hoac gia tai tngcap trong gia tai trc

ng suat chong cat tai cho khong thoatnc, ngay sau khi ap lp sau va nen a

co ket hoan toan bi tai cua lp ap trc

B/ Xa nc ho cha nhanh. Loi khongtham cua ap cha kp thoat nc trongqua trnh thao ap

ng suat chong cat loi ap khong thoatnc, trong luc xa nhanh ho cha, sau khiloi a co ket hoan toan trong s thamthng trc

C/ Thi cong nhanh cong trnh at ap trensn doc ng suat chong cat ngay sau khi gia taikhong thoat nc, nen a co ket hoantoan bi trong lng ban than trc o


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Ct khng c kt khng thot nc (UU)

s3 = sc

s1= scuc= uo+ sc

=

s1= -uo

+ s3= -uo

s3 = sc

s1= sc + Ds1uc= uo+sc+ADs1

=

1 = -uo- AD 1+D 1

+ 3=- uo-AD 1

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8.4.1.2. TH NGHIM NN BA TRC - UU

j

t

s, s

s3=gz

s13 s21fs11fs

23

tmax = cu ju = 0s = cUUjUU = 0

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Cong trnh va giai oan lam viec tngng

ng suat chong cat UU

A/ Thi cong nhanh cong trnh at ap tren

lp set mem

ng suat chong cat ngay sau khi gia tai

khong thoat nc, nen set mem cha coket hoan toan bi trong lng ban than(NC)

B/ ap at kch thc ln thi cong nhanh,o am cua loi khong kp thay oi.

ng suat chong cat khong thoat nc cualoi set am chat cua ap

C/ Mong nong thi cong nhanh tren nenset Sc chu tai phu thuoc ng suat chong catkhong thoat nc

8.4.1.2. TH NGHIM NN BA TRC - UU

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8.4.1.2. TH NGHIM NN BA TRC NHN XT

jCD>jCU>jUU

cCD > cCU > cUU

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8.4.1.3. TH NGHIM NN N (UNCONFINED COMPRESSION TEST)

**** Tiu chun p dng

- ASTM D2166

- BS 1377 Part VII

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8.4.1.3. TH NGHIM NN N (UNCONFINED COMPRESSION TEST)

s3 = sc

s1= sc + Ds1uc= uo+sc+ADs1

=

1 = -uo- AD 1+D 1

+

3=- u

o-AD

1qu

t

s = cu= 0.5qucu = 0.5quju = 0

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8.4.3. CCH LA CHN CH TIU CNG CHNG CT

i vi t ct, khi ly c cc mu nguyn dng, gi tr gc ma st xc nh bng th

nghim trn my ct trc tip.

i vi t st, khi khng cho php ly mu nguyn dng, th ch tiu chng ct ca n cth xc nh bng my ct cnh ti hin trng. Nu t cho php ly mu nguyn dng th

gi tr cc tham s chng ct c th xc nh bng my nn ba trc hoc ct trc tip.

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8.4.4. CC YU T NH HNG N CNG CHNG CT CA T

Thnh phn khong, hnh dng v cp phi ht

p lc c hiu theo phng php tuyn

cht ban u ca t ct

m

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CHAPTER 9

TH NGHIM SPT

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CHAPTER 9 TH NGHIM XUYN TIU CHUN


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9.1. NGUYN L TH NGHIM

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9 2 G C G


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9.2. DNG C V TRNH T TH NGHIM

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9.2. DNG C V TRNH T TH NGHIM - VIDEO

VIDEO TH NGHIM SPT M HNH

VIDEO TH NGHIM SPT THC T I

VIDEO TH NGHIM SPT THC T II

VIDEO TH NGHIM SPT THC T III

VIDEO TH NGHIM SPT THC T [email protected]


9 2 DNG C V TRNH T

Documents

Soil Mechanics Laboratory_KTL.pdf