De Thi Olympic Co Hoc Dat

5/21/2018 De Thi Olympic Co Hoc Dat

1/84

Phan Hng Qun thi Olympic C hc t .

Mc lc

thi nm 1997 Hng dn

thi nm 1998 Hng dn thi nm 1999 Hng dn

thi nm 2000 Hng dn

thi nm 2001(ti HTL) Hng dn

thi nm 2002(ti HGTVT) Hng dn

thi nm 2003(ti HKT) Hng dn

thi nm 2004(ti HXD) Hng dn

thi nm 2005 (ti H T.nguyn) Hng dn thi nm 2006(ti HVKTQS) Hng dn

thi nm 2007(ti HBK) Hng dn

thi nm 2008(ti HTL) Hng dn

thi nm 2009 (ti HHH) Hng dn


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Phan Hng Qun thi Olympic C hc t

thi nm 1997

s 1

Cu 1: Nn trm tch nh trn hnh 1. Lc u mc nc ngm mt t t nhin.Do khai thc mc nc ngm h thp 3m so vi mt t t nhin. bo ha ca

t trn mc nc ngm gim 20%.

- Tnh ng sut hu hiu phn t t nm gia lp st trc v sau khi h nc

ngm;

- T kt qu tnh, nhn xt nh hng ca vic khai thc nc ngm i vi cc

cng trnh th

Hnh 1.

Cu 2: Boussinesq cho kt qu:

z= 53

R

z

2

P3

-

Nhn xt v vic dng kt qu ny tnh ng sut trong nn t- Tnh ng sut ti nhng im c r = 2m (khong cch trn mt bng) cc

su z = 2m; z = 3m; z = 7m v cho nhn xt v kt qu

- mt su no , dng ca ng ng zl g?

-8.0

-6.0

0.0

st : bh= 18 kN/m3

ct th: = 17 kN/m3bh= 20 kN/m

3

t cng, khng thm


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Cu 3: Ngi ta d ct ht th sch t xe xung mt bi vung 20 x 20 (m). Th

nghim cho thy p lc p = 1 kG/cm2sc khng ct ca ct s = 0.68 kG/cm2.

Khi ct to thnh hnh thp nhn.Hy xc nh th tch khi ct.

Cu 4: Biu thc xc nh sc chu ti di hn ca t di mng bng c dng:

pgh=2

1Nb+ Nqq + Ncc

- Vi cng mt ti trng cng trnh ln nn t cho trc, gi nguyn su t

mng, lm th no tng c sc chu ti ca nn ln 1.5 ln?

- C mt mng bng rng 2m, chn su 1m trong ct c trng lng th tch t

l 17 kN/m3; trng lng th tch t bo ha bh= 20 kN/m3. Ch tiu khng

ct ca ct l = 400, h s sc chu ti tng ng N= 100, Nq= 81. Hy xc

nh sc chu ti gii hn ca nn khi:

+ mc nc ngm su 1m

+ mc nc ngm su 5m

Cu 5: Mt mng bng rng 2m, ng sut tip xc p = 200 kPa. Mng chn su

1m trong nn ng nht c cc c trng c l = 18.5 kN/m3; = 200; c = 30kPa.

- Kho st s n nh ca cc phn t t nm trn trc ngang su 1m v

tm im c nguy c mt n nh nht, im an ton nht.

Cho cng thc Michelle tnh ng sut chnh trong bi ton phng c dng

)2sin2(p

3,1 =

Gi thit ng sut php do trng lng bn thn t lun lun bng trnglng ct t nm bn trn im


4/84


s 2( chnh thc)

Cu 1: Nn t ct b ngp nc (hnh 1). thi cng, ngi ta lm tng c v

bm ht nc n l mt t.

Hnh 1.

a, Tnh ng sut trung ha v ng sut hu hiu ti cc im a, b trng thi ban

u. Sau khi c c v bm ht, cc ng sut thay i nh th no?

b, Kim tra xem c hin tng xi (ct chy) khi bm ht khng?Cu 2: Dng kt qu ca Boussinesq:

z= 53

R

z

2

P3

z= 52

R

xz

2

Q3

vi P, Q l lc tp trung tc dng thng ng v nm ngang trn mt bn khng

gian n hi tnh ng sut trong nn t. Cho lc N tc dng trn mt t,nghing 300so vi phng thng ng.

a, Nhn xt v vic dng kt qu ca Boussinesq tnh ng sut trong nn t

b, Tm im c zln nht trn mt c su z = 2m di mt nn t

Mc nc trc khi cv bm ht

a b

3m4m

4m

bh=20kN/m3


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Cu 3: Ngi ta d ct ht th sch t xe xung mt bi vung 20 x 20 (m). Th

nghim cho thy p lc p = 1 kG/cm2sc khng ct ca ct s = 0.68 kG/cm2.

Khi ct, ngi ta ti cho ct m ri to thnh hnh thp nhn. D tnh do mct c lc dnh gi khong (0.05 0.07) kG/cm2

Hy xc nh th tch khi ct.

Cu 4: Biu thc xc nh sc chu ti di hn ca t di mng bng c dng:

pgh=2

1Nb+ Nqq + Ncc

- Vi cng mt ti trng cng trnh ln nn t cho trc, gi nguyn su t

mng, lm th no tng c sc chu ti ca nn ln 1.5 ln?- C mt mng bng rng 2m, chn su 1m trong ct c trng lng th tch t

l 17 kN/m3; trng lng th tch t bo ha bh= 20 kN/m3. Ch tiu khng

ct ca ct l = 400, h s sc chu ti tng ng N= 100, Nq= 81. Hy xc

nh sc chu ti gii hn ca nn khi mc nc ngm mc mt t v vic

thi cng bm ht to ra dng thm c i = 0.2 ngc t di ln.

Cu 5: Mt mng bng rng 2m, ng sut tip xc p = 200 kPa. Mng chn su

1m trong nn ng nht c cc c trng c l = 18.5 kN/m3; = 200; c = 30kPa.

- Kho st s n nh ca cc phn t t nm trn trc ngang su 1m v

tm im c nguy c mt n nh nht, im an ton nht.

Cho cng thc Michelle tnh ng sut chnh trong bi ton phng c dng

)2sin2(p

3,1 =

Gi thit ng sut php do trng lng bn thn t lun lun bng trng

lng ct t nm bn trn im


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thi nm 1998

(Cc bi tp gi n )

Cu 1: Sc khng ct khng thot nc ca t: ngha, th nghim xc nh, vd v trng hp phi dng sc khng ct khng thot nc d bo s lm vic

ca nn cng trnh.

Cu 2: Gi thit ca Wincler l p = ky

p - cng ca ti trng tc dng ln nn

y - ln ca nn

k - h s nn

Nhn xt v vic p dng gi thit ny cho nn t v nhn xt v vic xc nhh s nn k

Cu 3: Tnh th tch biu ng sut ztc dng trn mt phng z = 2m v z =

3.5m do ti trng tc dng l lc tp trung thng ng P = 2000 kN v ti trng

phn b q = 500 kPa trn din tch 2 x 2 (m) tc dng thng gc vi mt gii hn

ca bn khng gian n hi (bi ton Boussinesq)

Cu 4: Cng trnh p vi ti trng d tnh p c t trn mt lp t st yu bo

ha chiu dy H. rt ngn thi gian ln, ngi ta dng cc vt thot nc

thng ng (VTN) v gia ti trc n 2p. Bit c trng ca nn t:

q = 500 kPa P = 2000 kN

z = 2m

z = 3.5m

z


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mv= a0= (e1- e2)/(p2- p1)/(1 + e1) ; Cvv xem nh chng khng thay i cho ton

b lp t trong qu trnh x l.

Chp nhn cc gi thit c s ca l thuyt c kt thm ca Tersaghi v b quabin dng n hi khi d ti.

Sau khi cht ti n thi gian t1th d ti trng d ch cn ti trng cng trnh.

Hy thit lp cng thc tnh (ch cn vit dng, khng cn tnh ra s c th) tng

ln ca cng trnh sau khi x l nn.

Cu 5: Cho hai mng:

- mng ang tn ti (a)

- mng s xy dng (b)

t nn di mng c n nh

Mng mi c ti trng ng tm O l Ptc= 1000 kN.

t nn xem nh bn khng gian bin dng tuyn tnh vi E0= 18000 kPa v 0=0.3.Cc kch thc mt bng cho trn hnh v. Xem mt phng cha hai mng

cng nm mt su t mng l h v l mt phng gii hn ca bn khng

gian n hi (bin dng tuyn tnh).

Kch thc mt bng mi mng l 2 x 2(m)

t0 < t t1: q = 2pt > t1: q = p

Lp y khng thm,

khng ln

H


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Yu cu: Gii thch cc phng c th dng xc nh nghing ca mng (a)

do mng (b) gy ra v tnh gn ng nghing .

Cu 6: Mt tng chn thng ng vi t sau tng l t ri thot nc t do

(hnh 2). Trng lng n v ca t sau lng tng l 18 kN/m3, gc ma st trong

= 300. Trn mt t sau lng tng c ti trng phn b u dc theo chiu di

tng vi cng q = 50 kPa trn b rng b = 2m. Gi thit tng hon ton

khng chuyn v, lng tng nhn v thng ng. Yu cu:- Xc nh tr s p lc t tc dng ln tng chn

2m2m

4m

q = 50 kN/m2

= 300

= 18 kN/m3

2m1m2m

P = 1000 kN


9/84


Cu 7: Xc nh ln ca tng t qua cc thi gian 1 nm, 2 nm v 5 nm

nu nh p lc trn lp t ny l p = 2 kG/cm2, chiu dy lp t h = 5m, h s

nn tng i a0= 0.01 cm

2

/kG, h s thm k = 1*10

-8

cm/s. Cho bit:e-N= e-0.3*1= 0.741

e-9N= e-0.9*1= 0.067

e-0.3*2= 0.549

e-0.3*5= 0.222

Gi thit nc thot ra theo mt hng.

Cu 8: Kt qu nn khng n hng mt mu t bo ha cho trong bng sau:

p lc nn (N/cm ) 0 5 10 20 40Chiu cao mu khi n nh (mm) 20.00 19.49 19.13 18.78 18.58

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 = 2.72, m ca mu sau khi th nghim xong W

= 30.51%

Cu 9: Hnh di y l h mng cng trnh. y h mng cao trnh -4.2m.

Thnh h mng c vy kn bng cc bn c di 8m. Mc nc ngm n nh

cao trnh -0.7m. Bng bin php

bm lin tc s m bo c mc

nc trong h mng thng xuyn

cao trnh y h mng phc v

thi cng.

Hy kim tra n nh chy t

y h mng do dng thm gy ra

0.0

-4.0

-0.7

-4.2

-8.0


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trong hai trng hp:

- t nn l ct th vi t trng ht = 2.60, rng n = 0.3, h s thm k =

1.2*10-4

m/s- t nn gm hai lp: ct th dy 4m trn c tnh cht nh trng hp 1 v

lp di l st c n= 10.8 kN/m3; k = 3.6*10-6m/s

H s an ton chy t yu cu Fs = 2


11/84


thi nm 1999

s 1( chnh thc)

Cu 1: Trn c s no c th kt lun gc ma st trong ca ct kh sch xp xbng gc ngh ca n?

Cu 2: Mt mng bng rng 3.0m truyn ti trng phn b u mc y mng

200 kPa. Nn t t nhin t mt t xuung gm ba lp nh sau: ct ht trung

dy 6m c = 19 kN/m3; st do dy 3m c = 20 kN/m3; ct sch. Hy xc nh

su t mng khng gy ln tng t st vi gi thit ng sut tip xc

y mng gi gi tr thay i. Bit rng th nghim nn mu t nguyn dng ly

t su gia lp st cho p lc tin c kt c= 200 kPa.Cu 3: a tng khu vc bao gm mt lp ct dy 9m nm trn lp st dy 6m

nh hnh v H.1. Mc nc ngm trong t su 3m (k t mt t).

Trng lng th tch n v ca t nh sau:

ct trn mc nc ngm: = 16 kN/m3

ct di mc nc ngm : = 19 kN/m3

st bo ha: = 20 kN/m3

-15.0

-9.0

-6.0

-3.0

A (-8.0)

B (-12.0)


12/84


Do khai thc nc ngm, mc nc trong t h nhanh xung su 6m v n

nh ti . Hy xc nh nh ng sut hu hiu ti cc im A ( su 8m) v

B ( su 12m),Cu 4: Mt lp t st dy 8m nm trn nn cng khng thm nc nh s

A trn hnh H.2. H s rng ban u ca t e0= 1.400; h s nn ln a = 0.144

cm2/kG; h s thm kA. B mt lp st chu ti trng nn phn b u 100 kPa.

Sau 72 ngy k t khi gia ti ln ca nn t ti 24cm.

Hy xc nh thi gian nn t st dy 16m trong s B t ti ln 48cm.

Bit rng h s thm ca t trong s B l kB= 2kA, cc ch tiu c l khc ca

t hai s l nh nhau v khng thay i trong qu trnh c kt.

Cu 5: Th nghim thm ct nc thay i trn mu t ct bi thu c kt qu

nh sau: sau 1 pht mc nc trong ng o din tch tit din 1 cm2 gim t vch90 n vch 45. Mu th nghim c chiu di 16cm, ng knh 4cm.

Hy xc nh h s thm ca t.

16m

8m kA

kB= 2kA

S B

p = 100 kN/m2 p = 100 kN/m2

S A

e0= 1.400a = 0.144 cm2/kg

tng khng thm


13/84


s 2(1999)

Cu 1: Mt ct ngang mt h mng di c dng nh trn hnh H.1. H mng c

bo v bng tng vn c lin tc, cch nc hon ton. Nc trong h mnglun c gi n nh mc y mng nh bm ht lin tc. Hy xc nh ng

sut theo phng ng ti cc im A, B, C, D tn ti trong qu trnh bm ht

nc. Bit rng t nn gm hai lp ct c cc ch tiu c l c bn nh sau:

lp trn dy 10m c = 19 kN/m3; bh= 20 kN/m3; k = 10 m/ngym

lp di c bh= 19 kN/m3; k = 5m/ngym

Cu 2: a tng mt khu vc gm lp ct ht trung dy 8m nm trn lp st yu

dy 6m v kt thc bng lp cui si cha nc c p. Sau mt nm khai thc

nc t tng cui si ct nc p trong tng ny gim thp 3m. Kt qu th

nghim t cho bit:

trng lng th tch n v ca ct trn mc nc ngm l 17 kN/m3; di mc

nc ngm l 19 kN/m3; ca t st l 20 kN/m3; h s c kt ca t st l 1.2

m2/nm; quan h gia h s rng ca t st vi ng sut nn hu hiu (tnh theo

kPa) c m t bi phong trnh e = 0.80 - 0.35lg(/100)

a, Hy d bo ln ca nn do vic khai thc nc ngm gy ra

b, D tnh ln ring ca lp st trong thi gian 3 nm k t khi bt u khai

thc nc

-10.0

MNN

-15.0

-12.5

DC

B

A

O-5.0

0.0


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Cu 3: Mt lp t st dy 8m nm trn tng cng khong thm nc nh trn

hnh H.2. H s rng ban u ca t st e0= 1.400; h s nn ln a = 0.144

cm

2

/kG; h s thm k = 1.2*10

-8

cm/s.B mt lp st chu ti trng phn b u p = 100 kPa. Sau 72 ngy k t khi gia

ti ln ca nn t 24cm.

Hy xc nh thi gian nn t st dy 16m trn s B t ti ln 48cm,

bit rng cc ch tiu c l ca t trn hai s l nh nhau v khng thay i

trong qu trnh c kt ca t.

Cu 4: Hnh v di y m t s xc nh ti trng ngang tc dng ln cc

trong trng hp mng bng c tng cng bng cc BTCT. Mng c b rng

2m, s lng cc trong mng l 2cc/1m di.

16m

8m kA

kB= 2kA

S B

p = 100 kN/m2 p = 100 kN/m2

tng khng thm

e0= 1.400a = 0.144 cm2/kg

S A


15/84


Hy xc nh ti trng ngang ln cc nu bit rng tng ti trng ngang H = 600

kN/m, t nn c = 300, c = 0 v = 18 kN/m3.

Cu 5: Mt tng chn trng lc cao 5m c xy dng chn gi bi thi cc

vt liu ri. Gi s c th b qua ma st gia tng v vt liu thi.

a, Chng minh rng, khi vt liu thi cao u 2m trn nh tng th mt trt

nguy him xc nh theo cc gi thit ca Coulomb vn khng thay i

b, Xc nh p lc vt liu thi ln tng trong trng hp nu bit rng vt

liu thi c = 16 kN/m3

v = 400

N

H h = 5m


16/84


thi nm 2000

Cu 1: Lp st dy 8m nm gia hai lp ct: lp ct trn dy 4m, mc nc ngm

su 2m (hnh 1). Lp ct di cha nc c p, ct nc p trn mt t 6m.Do bm ht nc lp ny, ct nc p h xung 3m sau thi gian ht 6 thng.

Cho bit h s nn th tch ca lp st mv= 0.94*10-3m2/kN, h s c kt Cv= 1.4

m2/nm, 0= 9.81 kN/m3.

a, Tnh ln ca lp st sau 3 nm k t khi bt u bm ht (xem nh thi

im bt u c kt giua thi gian ht nc)

b, Nu c mt lp ct mng thot nc t do nm trn, cch y lp st 2m, th

ln tnh theo cu a, s l bao nhiu?

Cu 2: Th nghim nn khng n hng mt mu t nhn c kt qu bng

di. Yu cu xc nh:

a, H s rng ban u ca mu t th nghim (e0)b, H s rng ca mu sau khi ln di mi cp ti trng (ei)

c, H s nn tng ng vi phm vi ti trng 20 40 N/cm2

6m3m

4m

8m

ct

st

ct

2m

6m st

st

ct


17/84


Cho bit sau khi th nghim xong mu t bo ha nc, W = 30.6%, t trng ht

t = 2.71.

p (N/cm2

) 0 10 20 40 80H (mm) 20.00 19.60 19.34 18.77 18.20

* H l chiu cao ca mu sau khi ln.

Cu 3: Mt mng bng t su 3m trong nn t c mc nc ngm ngang mt

t (xem hnh 2). Mng chu ti trng ng tm P = 1400 kN/m. t nn c cc

ch tiu nh sau:

bh= 21 kN/m3; c = 25 kPa; h s sc chu ti Nc= 20; Nq= 10; N= 7.5.Yu cu xc nh b rng mng hp l v sc chu ti ca nn tng ng h s an

ton Fs = 2.5 trong trng hp thi cng bm ht h nc ngm ngang y h

mng to ra dng thm ngc ln vi Gradient thy lc i = 0.2

Cho php s dng cng thc Terzaghi tnh ti trng gii hn ca nn. Trng

lng ring t nn hai bn mng c

th dng bhhoc n.

Cu 4: Mt tng chn c lng tng nhn, thng ng, chn gi khi t ti

su 10m. Cc c trng ca t sau tng nh sau:

c = 0; = 280; = 18 kN/m3; bh= 19.5 kN/m3

a, Xc nh ln v v tr ca tng p lc ch ng ln tng trong cc iu kin

sau:

- Mc nc ngm di chn tng

- Mc nc ngm ngang mt t

P = 1400 kN

b

3m


18/84


- Mc nc ngm nm gia mt t vi chn tng

b, Gi s tng c b rng y di B, b rng nh b, dung trng vt liu tng

. Vit iu kin n nh chng lt ca tng.

Cu 5: Th nghim thm ct nc thay i trn mu st pha. Buret chia lm

ng o p gim t vch 0cm3n vch 45cm3sau 1 pht th nghim. Ct nc

tnh ban u l 90 cm v cui cng l 45cm. Mu c chiu di 16cm, ng knh

4cm. Hy xc nh h s thm ca t.

(trng lng th tch n v ca nc ly 0= 10 kN/m3)

mc nc ngm

mc nc ngm

mc nc ngm

5m

b

B

5m


19/84


thi nm 2001

Cu 1:

1.

Ti sao khi mu t b ph hoi (hnh 1) mttrt li khng trng vi mt phng c ng

sut ct cc i? Hy chng minh.

2. Trong trng hp no hai mt trng nhau?

Hy gii thch.

Hnh 1

Cu 2:

Dng bin php ph u khp mt lp ct dy 3m c trng lng th tch n v = 16.66 kN/m3 nn trc lp t st

bo ha dy 6m nm trn tng cng nt

n thot nc tt (hnh 2). t st c h s

rng e0 = 1.4, h s nn ln a = 12

cm2/kN, h s thm k = 10-7cm/s

Sau khi ph ct mt thi gian t cng trnh

c khi cng xy dng v khi tr s plc nc l rng do trng lng lp ct

ph gy ra xc nh c nh bng sau.

Hnh 2

im A B C D E F G

su (m) 0 1 2 3 4 5 6

u (kPa) 0 13.40 23.22 26.82 23.22 13.40 0

Yu cu: 1. Xc nh ln ca tng st ti thi im t v c kt Qttng ng.

2. Nu cn i tng st ln xong mi khi cng th thi gian ch i l bao

lu?

1

3

1

3

mt trt

ABCDEF

G

3m

6m

ct h

st bo ha

cng nt n


20/84


Cho bit trng lng n v ca nc 0= 10 kN/m3.

Cu 3:

Hai nn cng trnh A v B u c kt thm mt chiu (hnh 3). Yu cu:1. Xc nh ln cui cng ca mi nn

2. Xc nh thi gian cn thit ln ca mi nn t 7cm.

Cho bit:

- ch tiu c - l ca hai nn ging nhau: e0 = 0.8; a = 0.0025 cm

2/N; Cv =

144*103cm2/nm.

- b qua ln ca lp ct nn B (v qu nh)

-

c kt ca hai trng hp c kt TH-3 v TH-4 tnh theo cng thc

Qt=+

+1

Q)1(Q2 t1t0

Cu 4:

Tng chn kiu bn y rng c mt ct nh hnh 4. t p sau tng l ct c

c = 0; = 400; = 17 kN/m3. t p trc tng l ct c c = 0; = 360; = 17

kN/m3. B qua ma st gia t vi tng. Gc ma st gia nn v bn y l =

300. Yu cu:

1. Xc nh p lc y mng.

12 N/cm212 N/cm2

st6m

30 N/cm2

3.2m

30 N/cm2

st

st1.6m

Hnh 3

ct


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2. Xc nh h s n nh chng trt phng theo mt nn

Cho bit trng lng n v ca vt liu tng = 25 kN/m3.

3.0m

1.75m1.0m

0.3m5.4m

0.4m

Hnh 4


22/84


thi nm 2002

Cu1: Sau mt trn ma, trong mi dc hnh thnh dng thm nh hnh v. Ti R,

ng dng i ra v men theo mt mi dc.1. Hy xc nh gc dc gii hn ca mi trong trng hp

2. Nu yu cu h s an ton Fs = 1.5 th gc mi dc phi l bao nhiu?

Cho bit ct bo ha c trng lng ring = 18 kN/m3, = 300

Cho php dng n= 10 kN/m3

Hnh 1

Cu 2: Mt lp ct dy 8.9m (hnh 2) c h s rng e = 0.5, t trng = 2.67.

Mc nc ngm su 3.9m. Trn mc nc ngm l i bo ba mao dn vi

mc bo ha G = 1. Trn i bo ha mao dn t trng thi kh.

Hy tnh v v biu phn b ng sut tng, ng sut trung ha v ng sut hu

hiu theo chiu su qua cc im ABCD. Cho php dng n= 10 kN/m3

Hnh 2.

R

B

C

D

A

Mc nc ngm

Mc bo ha mao dn

h=2,5m

h=1,4m

h=5m


23/84


Cu 3: Hnh 3 l din tch y mng cng trnh chu ti trng phn b u p = 100

kN/m2. Yu cu tnh ng sut thng ng zdo ti trng p gy ra ti im M

su cch y mng 3m nm trn trc ng qua O.

Cho bit h s ng sut trong nn hai bng sau:H s kc- im gc khi mt nn chu H s K khi mt nn

ti trng thng ng phn b u p chu ti trng tp trung P

trn din tch ch nht

l/bz/b

1 2 2 r/z K r/z K

2 0.084 0.120 0.131 0 0.478 0.4 0.329

3 0.045 0.073 0.087 0.22 0.424 0.42 0.318

5 0.018 0.033 0.044 0.33 0.369 0.44 0.307

Cu 4: Mt cng trnh xy dng trn nn ct ht trung trng thi cht c kp

mt lp st do mm bo ha nc dy 2m. Lp st c cc ch tiu W = 30%, =

M

O1m

1m

p = 100 kN/m2

3m

1m 1m 1m


24/84


2.70, a = 0.002 cm2/N, k = 2.10-9cm/s. Biu ng sut do ti trng cng trnh

gy ra nh hnh 4.

Yu cu:

1. Xc nh thi gian cn thit lp st ln gn xong (tng ng vi Q t =

0.96)

2. Nu gi s di y lp st l lp cng khng thm th thi gian lp st ln

gn xong l bao nhiu?Gi thit biu ng sut khng thay i.

3. Nhn xt cc kt qu tnh ton.

Khi tnh ton cho php b qua ln ca ct cht v qu nh khng ng k.

Cho bit gi tr Qt-N theo bng di y:

NQ0 Q1 Q2

Q0-1 Q0-2

= 0.6 = 0.8 = 1.0 = 2.0

2 0.89 0.86 0.92 0.88 0.89 0.90 0.90

3 0.96 0.95 0.97 0.96 0.96 0.96 0.96

5 0.99 0.99 0.99 0.99 0.99 0.99 0.99Ch thch: a = ng sut ti mt thot nc/ng sut ti mt khng thot nc; Q =

c kt (U); N = nhn t thi gian, N = (2/4).Tv; Tv = Cv.t/h2.

2m

18 N/cm2

10 N/cm2

st do mm

ct cht

ct cht

Hnh 4.


25/84


Cu 5: Mt mng bng chiu rng b = 2m t trn nn t ng nht c cc ch

tiu: = 20 kN/m3, = 300, c = 10 kN/m2. Mng chu ti trng thng ng phn

b u p v ti trng bn q = 30 kN/m2

(hnh 5).

Yu cu:

1. Lp cng thc xc nh ti trng p theo chiu su ln nht Zmaxca vng do

Cho bit phng trnh ng ranh gii phm vi vng do nh sau:

Z =

gcot

cq2

sin

2sinqp

2. Xc nh ti trng p khi vng do c im su nht Zmax trn trc ng i

qua mp mng A.

3. Xc nh su ln nht Zmaxcc i (maxZmax) ca vng do c th t c

v gi tr ti trng tng ng.

p

b = 2m

p = 30 kPa

BA

M

z

Hnh 5.


26/84


thi nm 2003

Cu 1: Hnh 1 l mt ct ngang h mng o su, di trong nn ct c trng

lng ring = 17 kN/m3

v bh= 19 kN/m3

. H mng c bo v bng tngc cch nc hon ton. Nc trong h n nh mc y do bm ht lin tc.

a) Xc nh chiu su H (so vi mt t) tng c m bo cho y h n nh

(khng b y bng) vi h s an ton FS = 1.5;

b) Xc nh ng sut hu hiu ti cc im B v D vi chiu su H trn (ly

gn ng 0= 10 kN/m3).

Cu 2: C hai lp st mm bo ha nc trn nn cng nh trn hnh 2. Ti

trng p trn mt c b rng rt ln so vi chiu dy lp t. Ngi ta quan trc

ln v lun thy 2SA= SB.

a) H s thm ca lp B, kB, phi bng bao nhiu c kt qu quan trc trn

(2SA= SB);b) Nu lp t B nm trn tng cui si th kBbng bao nhiu vn c kt qu

SB= 2SA? Gi tr CvAv CvBkhi y bng bao nhiu?

mt t t nhin

mc nc ngm

mc y hH

D

C

B-10.0

A

-2.0

0.00

tng c

Hnh 1


27/84


Cu 3: Mt mng bng rng 5m chn su 1m, ti trng y mng p = 280 kN/m2.

Nn t c = 20 kN/m3, = 200, c = 25.5 kN/m2.

Chp nhn li gii n hi ca Michelle: 1,3= )2sin2(

p

.

a) Kho st s n nh ca cc im M1(x = 0; z = 1.25m); M2 (x = 0.28; z =

1.25m);

4m

8m

S A S B

a0= 0.045 cm2/kg

kA= 10-8cm/s a0= 0.045 cm

2/kg

kB= ?

1 kg/cm2 1 kg/cm2

Hnh 2

b

p

x

z

hm


28/84


b) Phn tch xc nh v tr tng i ca M1, M2so vi vng bin dng do

pht trin trong nn;

c) Nhn xt phn tch v tnh hp l, xc thc ca vic xc nh vng bin dngdo theo cch lm trn.

Cu 4: Th nghim nn mt chiu bng hp nn (met) trong phng th

nghim. p lc ban u 0.1 kg/cm2 c coi l p lc tip xc. Kt qu th

nghim nh sau:

p lc nn (kg/cm2) 0.25 0.50 1.00 2.00 4.00

H s rng 0.83 0.815 0.75 0.65 0.60

Mu sau c sy kh xc nh trng lng th tch ht, sv h s rng ban

u e0 = 0.85.

a) Trong th nghim ny ngi ta o lng nhng g, lm th no xc nh cc

iv ei;

b) Trnh by kt qu trn th e log

c) Xc nh cc c trng nn ca t trn c s th nghim ny.

Cu 5: Xc nh c trng khng ct ca mt lp t st bo ha bng cch thnghim nn ba trc cho hai mu t ly t lp . Cc mu c cho c kt di

p lc bung 200 v 400 kPa sau chu ti trng dc trc gia tng cho ti khi

ph hoi trong iu kin th tch khng i c o p lc nc l rng. Kt qu t

nghim nh sau:

Mu 3(kPa) (kPa) u (kPa)

1 200 150 140

2 400 300 280

Tm c trng khng ct ca t v nhn xt t ny thuc loi qu c kt hay c

kt bnh thng.


29/84


thi nm 2004

Cu 1: Tng chn cao 9m, lng tng nghing 800so vi phng ngang. t

sau tng l t ct c = 240

, = 20 kN/m3

p nghing 200

(hnh 1). Gc mast gia t p vi tng = 200.

a) Yu cu xc nh tr s tng p lc t ln tng chn vi mt nghing gi

nh BC lm vi phng ngang gc 600.

b) Theo l thuyt ca Coulomb th p lc thuc loi p lc t g? c phi p

lc t ch ng khng? Hy gii thch?

Cu 2: Nn ng p cao 6m vi b rng tnh ton 20m. Trng lng ring t

p = 18 kN/m3. t ngay di khi p l st do mm bo ha nc dy

25m c = 19 kN/m

3

. Kt qu th nghim ct theo hai ch UU v CD chotrong bng sau.

Ch th nghim () c (kPa)UU 0 25CD 10 30

B

C

A 200

600800

9m

Hnh 1.


30/84


Hy nh gi mc n nh tng th ca nn di ti trng p vi h s an

ton Fs = 1.5 i vi hai phng n thi cng p t nh sau:

a) p rt nhanh (ti trng p c xem l gia tc thi ln nn, nc trong tnn khng thot ra c). Nu h s an ton Fs = 1.5 khng m bo th chiu

cao b phn p ti thiu bng bao nhiu?

b) p rt chm (ti trng p tng dn, nc trong nn thot ra c phn ln).

Cho php xc nh h s sc chu ti gii hn ca nn theo cng thc sau y:

Nq= )2/'45(02 +tge tg ; Nc= (Nq-1 ) ctg; N= 1.8(Nq-1)tg;

Trng hp = 0, Nc= (+ 2).

Cu 3: Ngi ta ct v cng l ti trng nn trc p = 100 kPa trn lp st dy

5m, pha di l tng ct kh dy (hnh 2). ln ca tng st sau 1 thng l

100mm; sau 2 thng l 139.4mm. Yu cu:

a) Xc nh ln n nh ca lp st;

b) Xc nh h s thm k ca lp st.

Bi 4: H mng trong t ct c trng lng ring y ni n= 11.2 kN/m3, h

s thm k = 2.3x10-6m/s. y h mng cao trnh -3.0. Di lp ct l ct th

cha nc c p vi ct nc p lc cao n -1.2. H mng c din tch mt bng

7.5 x 35 (m) vy kn bng cc bn c (hnh 3).

5m l t st

tng ct

Hnh 2

= 100 kPa


31/84


Yu cu:

a) Xc nh cong sut ti thiu ca my bm nu bm ht lin tc c th gi

mc nc lun ngang mc y h.

b) Kim tra s n nh ca y h mng (trong iu kin bm ht ni trn) vi h

s an ton K = 2 (dng 0= 9.81 kN/m3).

c) D tnh sau bao lu k t khi ngng bm mc nc s dng cao hn y h

mng 0.5m.Cho php tnh ton vi hai gi thit sau:

- ct nc p ca tng cha nc c p lun khng i

- thi im T bt k, gi tr tn tht ct nc l hng s i vi mi im trn

y h mng.

0.0-1.2

-3.0

-6.4

t ct :n= 11.2 kN/m

3;k = 2.3 x 10-6m/s

ct th cha nc c p

Hnh 3


32/84


thi nm 2005 ( 1)

Bi 1: Mt h mng bng c thi cng trong nn t nh hnh v. Lp t ct

di lp st nng c cha nc c p vi chiu cao ct nc p h = 8m. Lp stxem nh khng thm nc c h s rng e = 0.55, t trng ht = 2.78, m t

nhin W = 15%. Hi:

a) Chiu su h o h c th ln nht l bao nhiu y mng n nh?

b) Xc nh p lc nc l rng ti im N (nm trn trc trng tm ca mng)

sau khi gia ti mc y mng p = 100 kN/m2 vi chiu su t mng hm =

1.5m.

Bi 2: Cho nn t nh hnh v. Ti trng ngoi p = 120 kN/m2tc dng kn khp

b mt. Bit lp t st bo ha nc c h s nn th tch trung bnh a01= 0.045

cm2/kg; ca lp ct l a02= 0.0085 cm2/kg. Yu cu:

a) Tnh p lc nc l rng ti cc im cc su 0; -2; -4 v -6 (m) k t mtt ti thi im c kt ca lp st t 50%;

b) Tnh ln ca nn ti thi im .

3m

N

4m

6m

hm8m

lp 1: st

lp 2: ct

khng thm


33/84


Bi 3: Ngi ta thi cng con ng vo khu cng nghip c b rng mt ng

10m, trng lng ring vt liu lm ng = 21 kN/m3. Gi s mt ct ngang

con ng nh hnh v, t nn ng c cc ch tiu c l nh sau: trng

lng ring tn= 20 kN/m3; gc ma st = 100; lc dnh c = 15 kN/m2. Yu cu:

a) C th p vi chiu cao bao nhiu nn ng n nh vi h s an ton Fs

= 2

b) Vi chiu cao p trn, xc nh cng chng ct 0 trn mt phng

nghing 500so vi phng ngang ti im M nm trung tm ng v cch mt

y mng ng mt on z = 2.5m. Bit ng sut php trn mt nghing xc

nh theo cng thc:

4m

6m

ct trung

st bo ha nc

p = 120 kN/m

lp khng thm, khng ln

+9.0

+10.0 (MTN)

+15.0

b = 10m

= 21 kN/m3

= 20 kN/m3; = 100; c = 15 kN/m2.


34/84


=2

1[(1+ 3) + (1 3) cos 2]

Cc h s sc chu ti tnh theo cng thc sau :

Nq= ( )2/45. 02 +tge tg ; Nc= (Nq 1)ctg; N= 1.8(Nq 1)tg.

Bi 4: Mt mu t st c kt bnh thng c th nghim nn ba trc thot

nc vi p lc bung l 100 kN/m2 v lch ng sut cc hn = 200

kN/m2.

a) Xc nh cc thng s bn ct ca t;

b) Nu trong th nghim mu c c kt di p lc ng hng 200 kN/m2v

giai on gia ti dc trc khng thot nc. Hy xc nh lch ng sut cc

hn nu p lc nc l rng cui cng o c l u = 50 kN/m2.


35/84


thi nm 2005 ( 2)

Bi 1: Tnh lng nc sch cn iu ch va st bentonite t 1 tn bt st c

m 10%, t trng ht = 2.75. Gi thit trng lng ring ca va st = 11.5kN/m3.

Bi 2: Mt con ng rng (gi thit bi ton mt chiu) d kin p cao 6m trn

nn t st bo ha nc dy 7m, bn di lp st l cng khng thot nc.

Bit trng lng ring ca st = 16.5 kN/m3, gc ma st trong c kt-khng

thot nc cu= 160, lc dnh khng thot nc cu= 25 kN/m

2, h s c kt Cv=

3x10-3cm2/s (coi l hng s). Trng lng ring t p d kin = 20 kN/m3v

p lm hai giai on. Yu cu:

a) Giai on mt p nhanh, chiu cao ng l bao nhiu nn n nh vi h

s an ton Fs = 1.5;

b) Chng minh rng c th p thnh hai giai on v hy tnh thi gian ch i

p giai on hai vi gi thit rng lc dnh khng thot nc tng theo qui lut

cu= cutgUh ..21

.

Bi 3: Cho mt tng st dy 8m trn tng khng thm nc. Lp st c cc ch

tiu sau: h s rng e0= 1.4; h s nn a = 14.4 cm2/kN; h s thm k = 1.2 x10-6

cm/s. B mt lp st chu ti u v hn vi cng p = 100 kN/m2. Sau khi gia

ti 72 ngy lp st t ln 24 cm.

a) Lp st c tnh cht nh trn nhng dy 16m v c h s thm k = 2.4x10-6

cm/s sau khi gia ti 72 ngy cng t ln 24cm. iu c ng khng v ti

sao?b) Tnh thi gian t cn thit lp st dy 16m c h s thm k = 2.4 x 10-6cm/s

t ln 48cm.


36/84


Bi 4: Khi thi cng h mng, h nc ngm xung ngang mc y mng

ngi ta s dng cc c v bm ht nh trn hnh. Hi chiu su c h chn vo

lp 2 ti thiu l bao nhiu khng xy ra hin tng chy t (xi ngm) y

h mng vi h s an ton Fs = 2. Nn gm hai lp:

- lp 1: st pha dy h1= 5m, h s thm k1= 1.5 x 10-4cm/s;

- lp 2: st dy v cng, h s

thm k2 = 4 x 10-6 cm/s, trng

lng ring = 19.8 kN/m3.

Gi thit khi bm ht mc nc

ngm ngoi h khng i v -

1.5m so vi mt t.

16m

Tng khng thm

Tng khng thm

8m k = 1.2 10-6cm/s

p = 100 kN/m

k = 2.4 10-6cm/s

p = 100 kN/m2

tng c1.5m

5m

h

lp 1: st pha

lp 2: st


37/84


thi nm 2006 (ti Hc vin KTQS)

Cu 1: Cho lp t c chiu dy H vi h s thm

tng tuyn tnh theo su t gi tr k1( nh) n k2( y lp), k2> k1.

Hy tnh h s thm tng ng ca t khi:

- thm ngang;

- thm ng.

Cu 2: Mt mi dc v hn c dc b

mt = 200. Nn gm 2 lp: lp trn t

ct dy 3m c = 18 kN/m3; lp di t

st.

a. Hy kim tra n nh trt phng theo

b mt lp t st vi h s an ton Fs =

1.5, bit rng t st c c = 10 kN/m2v = 200.

b. Nu mi t ngp nc vi dng thm n nh song song vi mt t th mi

dc c n nh hay khng? Bit rng ct bo ha c bh= 20.5 kN/m3; t st c

= 200; c = 0.

Cu 3: Hnh 3 l mt ct mt h mng o gn b sng c bo v bng tng

c. Din tch h mng F = 500m2c o trong nn ct bi c bh= 20 kN/m3,

h s thm k = 3.6 x 10 -3 m/h. Nc thm t sng qua tng cui si coi nh

khng tn tht.a. Xc nh h s an ton chy t khi mc nc trong h lun cao hn y h

2m;

thm ng

Hthmngang

Hnh 1

3m

200

ct

st

Hnh 2


38/84


b. Trc khi thi cng mng ngi ta dng my bm cng sut 20m3/h bm

ht. Hy xc nh thi gian cn thit bm h nc trong h ti y.

Cu 4: Tng chn t trng lc bng b tng ct thp c bt= 25 kN/m3. y

mng trn nn st pha c = 18 kN/m3, = 220, c = 5 kN/m2(hnh 4). t p l

ct c = 20 kN/m3, = 300. B qua ma st lng tng, gc ma st gia y

mng vi t l = 220.

a. Xc nh chiu cao H ln nht tng khng b trt phng theo y mng;

T NN

1m 1m

MT T P

3m

1m

1m

H

Hnh 4

mc nc trong htctbi

5

2y h o9

t cui si

tng

Hnh 3


39/84


b. lm tng h s an ton chng trt phng theo y mng ngi ta m rng

y mng v pha t p. Hy xc nh b rng y mng ti thiu c Fs =

1.5 vi chiu cao H xc nh theo cu a.Cu 5: Ngi ta p lp t san nn dy 5m c = 18.5 kN/m3 ln nn t st

yu bo ha dy 3m, mc nc ngm ti nh lp nh hnh v.

a. Gia ti nhanh nh trn c m bo iu kin n nh ca nn vi h s an ton

Fs = 1.2 hay khng? Bit rng st yu c lc dnh khng thot nc c u = 23

kN/m2v u= 0;

b. Nu cho thi gian p san nn ko dn 3 thng th sau khi p xong bao lu lp

t st t c kt U = 55%, bit rng h s c kt ca t st Cv= 1.4 m2/nm.

5m t san nn

3m t st yu

t khng thm nc

Hnh 5


40/84


thi nm 2007 (ti Bch khoa)

Cu 1: Ly 3 mu t st bo ha cng mt su. Th nghim nn ba trc c

kt thot nc cho mt mu c kt qu sau: thi im xy ra ph hoi plc bung bng 100 kPa, lch ng sut bng 145 kPa, mt phng ph hoi to

thnh gc 540so vi phng ngang. Th nghim nn mt trc cho mu 2 xc nh

c bn nn mt trc bng 320 kPa. Th nghim nn ba trc khng c kt-

khng thot nc mu 3 vi p lc bung 320 kPa.

a) Xc nh cc ch tiu khng ct ca t

b) Xc nh p lc nc l rng trong mu 2 v mu 3 thi im ph hoi

Cu 2: Ngi ta khai thc mt m t t nhin phc v cho cng trnh p t

Cho bit t t nhin c trng lng ring = 1.8 t/m3, t trng ht = 2.72,

m W = 20%. t khai thc c = 1.4 T/m3, m W = 13%, m m nn tt

nht Wopt = 18% v trng lng ring kh ln nht kmax= 1.72 T/m3.

Gi s t p c m cht n K = 0.9, m c duy tr khng i trong

qu trnh m cht. Hy xc nh:

a) Lng nc ti thm cho t khai thc trc khi m

b) Lng t t nhin cn khai thc ( tnh cho 1m3t c m cht)

Cu 3: xc nh h s thm ca t ngi ta o mt ging ht v hai ging

quan trc nh hnh v. Khi lu lng t trng thi n nh l q th mc nc

trong cc ging quan trc l 2.4 v 1.8m. Bit mc nc ngm ban u cch mt

t 1m v gi thit quan h h s thm ca cc lp t l k2= mk1(m hng s).

a) Hy tnh h s thm ca cc lp t theo lu lng q

b) Nn t l mt lp ng nht v lu lng ht q = 600 lt/pht. Xc nh h sthm ca t.


41/84


Cu 4: Tng c chn t c dng trong thi cng h o chiu di ln c mt

ct ngang nh hnh v. t trc v sau tng l ct c = 18 kN/m3, bh= 21

kN/m3, gc ma st trong = 350. Nc trong h c gi mc y mng cn

nc ngm su 2m k t mt t.

a) Tnh v v biu p lc t ln tng c

b) Tnh h s an ton chng lt (quanh im D) nu sc khng b ng c huy

ng mc 70%.

(Gi thit tng c tuyt i cng v b qua ma st gia t- tng)

q1.8m

50m

15m

4m

6m

2.4m

MNN

t c k1

t c k2= mk1

Tng khng thm nc

y h o

C

MNN

D

6m

3m

B

2m

A

Tng c


42/84


Cu 5: Ngi ta gia ti nn trc x l nn vi p lc p0= 150 kPa. Sau 1 nm

th d ti v xy dng cng trnh c din tch y mng ln, p lc y mng p

=100 kPa. Nn t st dy 6m nm trn tng thm nc tt (h s nn tngi a0= 0.001 m

2/kN; h s thm k = 5x10-10m/s).

a) Tnh chiu dy vng qu nn sau khi d ti

b) Tnh ln n nh sau khi xy dng cng trnh

Cho rng thi gian d ti v xy dng cng trnh l tc thi v b qua bin dng

n ca t.

Ch : Khi tnh ton ly trng lng ring ca nc 0= 10 kN/m3.


43/84


thi nm 2008 (ti i hc thy li)

Cu 1: Nn t gm: lp 1 ct ht nh dy 3m c trng lng ring t nhin g =

18.2 kN/m3, m W = 18%, t trng ht D = 2.62; h s nn ln tng i a0 =8.10-5 kPa-1; lp 2 l st pha dy 1m c h s c kt Cv = 30m2/nm, a0 = 1.10-

4 kPa-1; lp 3 l st dy 4m c h s c kt Cv = 5 m2/nm, a0 = 2.10-4 kPa-1.

Di cng l lp gc khng thm nc. Mc nc ngm ban u ngang mt

t. V nhiu l do, nc ngm h thp cch mt t 1.5m v n nh cng

vi mc nc mao dn dng cao 0.8m trn mc nc ngm.

a) Tnh v v biu ng sut hiu qu sau khi nn c kt hon ton do h

nc ngm.b) Tnh ln n nh ca nn t.

c) Tnh ln n nh ca nn t sau 6 thng cho rng nc ngm h thp tc

thi.

Cu 2: Tng chn c thit k nh hnh v. t sau tng l t dnh c

bn khng ct khng thot nc cu = 26 kPa, trng lng ring g = 19.6 kN/m3.

a) Xc nh chiu dy lp t (DH) ng vi on tng c th ct b tng chn

gi.

b) Xc nh p lc t ln tng chn sau khi ct b on tng chn ni trn v

to mi t c dc 1/1 k t v tr khong c tng chn nh hnh v.

Cu 3: p dng nc trn nn ct c mt ct nh hnh v. p bng BTCT c g

= 25 kN/m3, cao trnh nh p ngang mc nc thng lu (0.00). Pha h lu,

y p c lp st ph dy 1m, chiu di kh ln coi nh khng thm nc c

gbh = 20 kN/m3 v j = 300.a) Tm h s an ton n nh trt theo mt phng y p vi b rng y b =

2m, chiu rng nh b = 1m. Khi tnh ly 70% p lc b ng pha h lu.

b) Tm chiu rng y b c h s an ton Fs = 1.5, khi tnh ly p lc b ng

pha h lu bng 70%.


44/84


c) Vi chiu rng b tnh trong cu trn (b), cho chiu di lp st ph bng 10m,

mc nc h lu bng mt t sau lp ph (-5.00). Tm h s an ton chng trt

nu p v lp ph cng b trt theo mt phng y.Cu 4: a tng lp st pha dy 11m c gbh = 20 kN/m3 nm trn tng ct kt

cha nc co p. Khi o h mng n su 3.5m thy c nc ngm xut hin

v n nh su trong h l 0.5m. o n su 6m v bm ht gi nguyn

mc nc y th thy xut hin chy t y.

a) Xc nh chiu cao ct nc p trong tng ct kt.

b) Xc nh Gradient thy lc ban u ca t.

c) Xc nh chiu su ti a ca h o sao cho trong h khng c nc.


45/84



46/84

Phan Hng Qun Hng dn v p n .

p n k thi nm 1997Cu 1: a) Trc khi h nc ngm, ti gia lp st ng sut hu hiu l

= 6x(20 9.81) + 4 x (18 9.81) = 61.14 + 32.76 = 93.9 kPa

Sau khi h nc ngm:- trng lng ring t trn nc ngm thay i: = 17 + (1 0.2)(20 17) = 19.4 kN/m3;

ng sut hu hiu gia lp st s l : = 3 x 19.4 + 3 x (20 9.81) + 4x(18 9.81) = 58.2 +

30 + 32 = 120.2 kPa

b) Vic h nc ngm lm gia tng ng sut hu hiu trong t ( = 120 93.9 = 24.1) dn

n ln st b mt, do khai thc nc ngm c th dn n h hi cc cng trnh th

trong vng nh hng.

Cu 2: a) Nu nn t khng ng nht, khng ng hng th vic p dng kt qu ca

Boussinesq ch l gn ng. Cc im gn v tr t ti cho kt qu tnh theo Boussinesq qu

ln, khng ph hp vi tnh cht v kh nng tip nhn ca t.

b) p dng cng thc =2/522

3

)(2

3

zr

zP

+ln lt vi r = 2m; z = 2; 5; 7 ta c

(z=2) = 0.007P; (z=5) = 0.013P; (z=7) = 0.008P

t c nhn xt rng ngoi ng tc dng ca P gi tr ztng dn theo su n gi tr

cc i sau gim dn theo su.

c) mt su no bt k cc ng ng zl nhng ng trn tm giao im ca mt

phng z vi phng ca lc tc dng do z= f(r, z) do khi z = const, r = const z= const

do trn ng trn bn knh r z= const.Cu 3: Ct kh c c = 0 do ng ct c dng thp nhn vi gc y bng . Th tch ng

ct V = HB2

3

1

tg= /p = 0.68/1 = 0.68 chiu cao H = Btg2

1= 6.8m

V = 202 x 6.8/3 = 906m3.

Cu 4: a) Khi gi nguyn su chn mng, tng sc chu ti ca nn ta phi tng b rng

mng b = b0+ b

0.5Nb+ Nqhm+ Ncc = 1.5[0.5Nb0+ Nqhm+ Ncc]

hay b = 1.5b0+

N

cNhN cmq +

b) Khi nc ngm cch mt t 1m, t di y mng chu tc dng y ni do trng

lng ring t nhin di mc y thay i dn n pghthay i.

= bh 0= 20 9.81 = 10.19 kN/m3


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Sc chu ti gii hn ca nn:

pgh= 0.5 x 10.19 x 2 x 100 + 17 x 1 x 81 = 2396 kPa.

Trng hp mc nc ngm su 5m tc ngoi phm vi nh hng ca mng:

pgh= 0.5 x 17 x 2 x 100 + 17 x 1 x 81 = 3077 kPa.Cu 5: Gi thit ng sut php do trng lng bn thn t gy ra nh nhau theo mi phng

ta c th tnh c ng sut chnh ti mt im theo cng thc Michell: 1,3 =

)()2sin2( hzmhp

++

Phn tch trn mt na k t trc i xng su z = 1m:

- im a trn trc qua tm: 2= 900

1,3= = 185.5 v 69.9 kPa.

Gc lch ng sut m rng: sin=

tgc /231

31

++

= 0.274

- im b mp mng: 2= 63030 . Tng t, 1,3= 152.7 v 49.3 kPa.

Gc lch sin= 0.281

Kt qu (a) < (b) chng t im b nguy him hn im a.

p n s 2

Cu 1: a) trng thi ban u cc im a v b c ng sut nh nhau:

ng sut tng, = 0Hn+ bhz = 9.81 x 4 + 20 x 3 = 99.24 kPa;

p lc nc, u = 0(Hn+ z) = 9.81 x ( 4 + 3) = 68.67 kPa;

ng sut hu hiu, = u = 99.24 68.67 = 33.57 kPa

Sau khi nc trong h b ht gim, ng sut ti im b thay i: ng sut tng v p lc nc

gim trong khi ng sut hu hiu vn khng thay i.

b) Khi ht nc, dng thm t ngoi vo theo tng c c chiu di ngn nht do c

Gradient ln nht.

Chnh cao ct nc trong v ngoi h l H = 4m; chiu di ng thm L = 8m do

Gradient thm I = 4/8 = 0.5 to ra p lc thm j = i0= 0.5 x 9.81 = 4.9 kN/m3.

Trng lng ring y ni ca t n= bh 0= 10.19 kN/m3

So snh ta thy j < ndo cha c xi xy ra. H s an ton tng ng:

Fs = n/j = 10.19/4.9 = 2.07

Cu 2: a) Nu nn khng tha mn iu kin ng nht v ng hng, vic p dng cng thc

Boussinesq ch l gn ng. c bit ti cc v tr gn im t lc tc dng kt qu tnh ton

cho ng sut rt ln, khng ph hp vi tnh cht v kh nng lm vic ca t.

b) Nhng im c zln nm trn giao tuyn ca mt phng cha N vi mt z = 2: ng OO

trn hnh v.


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Ti trng N phn lm hai thnh phn: lc thng ng P = 0.866N v nm ngang Q = 0.5N.

Thay vo cng thc Boussinesq ta c:

z= 5

2

2

3

5

3

2

3

R

xzQ

R

zP

+ = N(0.413z3

+ 0.238rz

2

)/(z

2

+ r

2

)

5/2

.

Vi z = 2: z= N(3.304 + 0.952r)/(4 + r2)5/2

Ln lt thay cc gi tr r = 0; 1; 2; 3.. ta xc nh c gi tr ztng ng nh sau

r (m) -1 -2 0 1 2 3 4

z(N) 0.042 0.008 0.103 0.076 0.029 0.010 0.004

Kt qu tnh ton chng t ng sut ngay di im t ca lc (cho d nghing 30 0) vn c

gi tr ln nht.

Cu 3: Do nh hng ca m t ct c mt phn tnh dnh. C th qui i tng ng

vi t khng dnh theo biu thc sau: = tg+ c = tg hay tg = tg+ c/

trong = gc ma st tng ng khi c lc dnh;

tg= /= 0.68

Th tch ng ct tnh c s l:

V = B2H/3 = B2.(Btg/2)/3 = 973 1000 m3.

Cu 4: a) Cn phi tng su t mng thm mt on h xc nh theo phng trnh sau:

0.5Nb+ Nq.(h + h) + Nc.c = 1.5[0.5Nb+ Nqh + Nc.c]

hay

h + h = 1.5h +

qN

cc

NbN 5.025.0 +

b) p lc nc tnh v p lc thm lm thay i trng lng ring ca t. Sc chu ti gii

hn ca nn khi c tnh theo cng thc:

pgh= 0.5Nb + Nq.h = 0.5 x 100 x 2 x 8.23 + 81 x 8.23 x 1 = 823 + 666.6 = 1489.6 kPa.

Cu 5: (xem 1)


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p n k thi nm 1998

Cu 1: Sc khng ct ca t, s , l kh nng bn trong ca t chng li ng sut ct do ti

trng ngoi gy ra:

s = tg + c = (s u) tg + c

trong :

u = p lc nc l rng tc l p lc nc xut hin trong trng thi tnh do nc ngm trong

t gy ra hoc do ti trng nga gy ra;

= gi tr trung bnh ng sut tip xc gia cc ht t, nguyn nhn chnh gy ra ma st gia

cc ht t khi dch chuyn do ng sut ct gy ra.

= gc ma st trong gia cc ht t vi nhau;

c = lc dnh gia cc ht t trn mt n v din tch (lc dnh n v - gi tt l lc dnh ca

t)

b) Cc c trng khng ct ca t c th xc nh t th nghim ct phng hoc th nghim

nn ba trc; th nghim ct cnh hin trng hoc cc th nghim gin tip khc, chng hn th

nghim xuyn.

- th nghim ct trc tip cho php xc nh c trng khng ct trong iu kin khng thot

nc cn gi l th nghim ct nhanh khng c kt. T th nghim ta xc nh c cu.

- th nghim nn ba trc cho php xc nh c cc c trng khng thot nc, c kt thot

nc v c kt khng thot nc.

- th nghim ct cnh cho php xc nh bn ct khng thot nc, c u, ca t ti hin

trng.

c) Sc khng ct khng thot nc c dng kim tra sc chu ti ca t yu chng hn

di nn ng p trn t yu.

Cu 2: a) Gii thit ca Winkler ch ng cho cc vt liu khng c tnh phn phi: p = 0 y

= 0 hay ngoi phm vi t ti (p = 0) nn khng b ln. iu ny khng hon ton ng i vi

nn t ni chung. Trong thc t, gia cc ht t c lin kt, d rng rt yu, nh vo lin kt

kt tnh, lin kt keo nht gia cc ht t v c bit quan trng l ma st gi cc ht do

di tc dng ca ti trng ngoi vng t ti vn c ln. Gi thit Winkler n gin v m t

ton hcdo c s dng rng ri trong tnh ton ch nn coi l ph hp t vi cc loi tyu hoc phn tch ngay di phm vi t ti.

b) H s nn k nn xc nh bng th nghim bn nn hin trng:tc dng ti trng p phn b

u ln tm nn cng, o ln ca tm nn, y, ta xc nh c h s nn k =y

p. Thc t

cho thy k ph thuc khng ch vo tnh cht ca t m c vo din tch tm nn, mc ti

trng cng nh phng php gia ti. Ni chung, nn xc nh k ng vi mc ti trng thit

k d kin c hiu chnh theo b rng mng.


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Cu 3: Th tch biu ztrn mt phng nm ngang bt k khng ph thuc su m thc

cht bng gi tr ca ti trng tc dng bn trn theo nguyn l cn bng tnh.

Cu 4: Phn bit hai trng hp

a) ti thi im d ti t1 ln ng sut hu hiu do 2p gy ra trn ton b nn vt qu gi trp: trng hp ny nn qu c kt do sau khi d ti , b qua bin dng n ta c ln

cui cng, S = 1.mv

b) sau thi gian t1ch mt phn nn t qu c kt, ln ca nn s gm 2 phn: ln do ti

trng 2p trong thi gian t1, S1v ln do ti trng p cho n khi n nh, S2: S = S1+ S2= mv(1

+ 2).

S tnh ln cho hai trng hp nh hnh v

Cu 5: c th xc nh c nghing ca mng (a) ta cn phi xc nh c ln tihai im I v K trn mng (a) do ti trng t mng (b) gy ra. C hai cch xc nh nh sau:

a) Dng phng php im gc xc nh v v biu ng sut do ti trng mng (b) gy ra

trn cc trc ng i qua I v K sau tnh ln ti cc im theo phng php cng ln

cc lp phn t nh bnh thng:

S(I, K) = iihE

b) Thay ti trng ln mng (b) bng lc tp trung P t ti trng tm mng. ln ti cc im

I v K xc nh theo cng thc ca Boussinesq:

S(I, K) =ER

P 21

Gc nghing ca mng: =IK

ISKS )()(

H

t1 ln

11

z

2

Ti trng 2p

t1khng lnz

H

Ti trng 2p


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Cu 6: p lc t ln tng do t sau tng v ti trng gy ra. Trng hp tng khng

chuyn v , p lc ln tng l p lc tnh. H s p lc tnh c th xc nh theo cng thc

thc nghim sau:

K0= 1 sin

p lc t ln tng: Et=2

021

HK

p lc tnh do ti trng gy ra c th xc nh theo nguyn l n hi tng ng hai ti trng

i xng nh trn hnh v. Cng ng sut nn theo phng ngang do ti trng ng gy

ra xc nh theo cng thc Flamant:

x= 2.

+ 2222

)(

2

zx

zxp

p lc ln tng l tng p lc trn ton chiu cao tng:

Etp= 2. =

=

=

= +

4

2

4

0

222

2

)(

2x

x

z

z

dxdzzx

zxp

= +

4

2

4

0

222

2

)(

4zx

zdzdxx

p

( )222

224

0

4

0

222 )(2

1

)( xz

xzd

zx

zdz

++

=+ =

4

022

121

+

xz

=)16(

822 +xx

Cui cng, tng p lc ln tng: E = Et+ Etp= 112 kN/m.

Cu 7: ln cui cng ca lp t: S = mvph = 0.01 x 2 x 500 = 10cm

Nhn t thi gian: N = th

Cv

2

2

4

= 10-5Cv.t = 0.3t

trong : Cv= 587

0 1010)1015.3(

= xmk

v

= 3x104cm2/nm

Ti thi im t = 1 nm ta c:

N = 0.3 do U =

+ NN ee 92 9

181

= 0.39

S(1 nm) = 0.39 x 10 = 3.9 cm

Tng t, ti t = 2 nm, U = 0.56, S (2 nm) = 5.6 cm;

ti t = 5 nm, U = 0.82, S (5 nm) = 8.2 cm

Cu 9: H s an ton chng xi xc nh theo cng thc: Fs = n/j

trong :n= trng lng ring y ni ca t di nc ngm, n= bh 0;

j = lc thm n v, j = i0.

a) Trng hp nn ng nht: bh=)1(

)(0e

e

++


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trong , e = h s rng ca t, e =n

n

1

di ng thm: L = (8 0.7) + (8 4.2) = 11.1

Chnh cao ct nc H = 4.2 0.7 = 3.5mH s an ton Fs = 3.67

b) Trng hp nn hai lp:

Gi tn tht qua lp t th nht l H1, chu di ng thm L1= 3.3m; qua lp t th hai

l H2vi chiu di ng thm L2= 7.8m.

Gradient thy lc dng thm qua cc lp t ln lt l I1= H1/L1v I2= H2/L2.

Vn tc dng thm qua cc lp t khng i, v = const ta c phng trnh:

2

2

21

1

1 kL

Hk

L

H

=

H1+ H2= H

Gii theo H2ta c H2= 3.46m do I2= 3.46/7.8 = 0.443

H s an ton Fs = n/(I20) = 10.8/(9.81x0.443) = 2.49


53/84


p n k thi nm 1999 s 01

Cu 1: Gc ngh t nhin ca t ri kh c hiu l gc nghing ca mi t c hnh

thnh mt cch t nhin v t n nh nh ma st gia cc ht t trng thi kh. Mi t

nhn to c thc hin vi gc nghing ln hn gc ngh s t ng iu chnh v gc ngh t

nhin ca n. Trng hp ngc li (gc mi nh hn gc ngh) mi t tip tuc n nh. C

th ni rng h s n nh ca mi t ri kh ng vi gc ngh t nhin bng 1.

Kho st phn t trn mi c gc nghing nh trn hnh v. Phn t chu tc dng ca lc gy

trt T v n nh nh lc ma st F. H s an ton tng ng xc nh theo biu thc:

Fs =F

T

Thay T = Qsinv F = Ntg= Qcostgta c: Fs =

tg

tg

trng thi cn bng gii hn, Fs = 1 ta c tg= tghay gc ngh t nhin ca ct sch, kh

bng gc ma st trong ca t.Cu 2: ng sut gy ln do ti trng t mng bng gy ra su z t gi tr ln nht trn

mt i xng ca ti trng xc nh theo cng thc Michell:

max= )2sin(2(

+p

trong :

2= gc t im ang xt n y mng, b = f(z, hm) ;p = ti trng gy ln, p = p0 ghm;

p0= ti trng tip xc y mng; hm= su t mng; = trng lng ring ca t t y

mng tr ln.

im gia ca lp t st cch mt t mt khong 7.5m. Khi hm thay i, gi tr ng sut gy

ln ti im kho st thay i nh trong bng sau:

hm(m) 0 1 2 3 4 5

p (kPa) 200 181 162 143 124 105

(kPa) 52.0 53.4 55.1 57.2 62.1 64

TN

Q

Q = dVN = QcosT = Qsin


54/84


T kt qu phn tch ta tm c hm 3.8m cho ng sut gy ln ti gia lp t st 60

kPa.

Cu 3: ng sut trong t trc khi h mc nc ngm:

- ti A, trc khi h mc nc ngm:+ ng sut tng = 16 x 3 + 19 x 5 = 143 kPa

+ p lc nc l rng u = 10 x 5 = 50 kPa

+ ng sut hu hiu = u = 143 50 = 93 kPa

- ti B, trc khi h mc nc ngm:

+ ng sut tng = 16 x 3 + 19 x 6 + 20 x 3 = 222 kPa


+ ng sut hu hiu = 222 90 = 132 kPa

Do thay i mc nc ngm ng sut trong t thay i theo thi gian. Hai thi im c trng

cn c kho st:a) Ngay sau khi nc ngm thay i

- ti A, s thot nc nhanh chng lmthay i cc thnh phn ng sut trong t:+ ng sut tng = 16 x 6 + 19 x 2 = 134 kPa


+ ng sut hu hiu = u = 134 20 = 114 kPa.

- ti B, ng sut tng thay i lm thay i p lc nc l rng nhng cha lm thay i

ng sut hu hiu:

+ ng sut tng = 16 x 6 + 19 x 3 + 20 x 3 = 213 kPa

+ ng sut hu hiu cha thay i, = 132 kPa

+ p lc nc l rng thay i u = = 213 132 = 81 kPa

b) Khi thi gian di, nn n nh di s thay i ca mc nc:

ti B:

+ ng sut tng = 213 kPa


+ ng sut hu hiu = u = 213 60 = 153 kPa

Nh vy: ng sut hu hiu ti im A thay i ngay sau khi mc nc ngm trong t b h

thp, Atng mt gi tr = 114 93 = 21 kPa; ng sut hu hiu ti im B khng thayi ngay sau khi mc nc ngm b h. S thay i ng sut hu hiu ti B ch xy theo thi

gian v khi thi gian lu nc l rng trong t th thot ht ra ngoi ng sut hu hiu

ti B mi t ti gi tr n nh cui cng B= 153 kPa vi mc tng = 153 132 = 21

kPa tng t ti im A.

Cu 4: a) Trc ht xt s A: ln cui cng S = =+ pHe

a

10.48m = 48 cm


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c kt sau 72 ngy: U =4824)(

=S

tS= 0.5

t trong s A c kt mt chiu theo s 0 vi chiu di ng thm hA= 8m.

b) i vi s B, tng t ta xc nh c ln cui cng S = 96 cm do c kt khiln t ti 48 cm cng l U = 0.5.

t trong s B c kt theo s 0, nc thot c hai pha do chiu di ng thm

cng l 8m: hB= hA= 8m.

Ta c phng trnh: UA= UB

Thay UA= 1 -vAT

e2

4

2

8

; UB= 1 -vBT

e2

4

2

8

ta c: TvA= TvBhay Bb

vB

A

A

vA th

Ct

h

C22

=

V hA= hB, ta c:A

B

v

A

v

B

vA

vB

B

A

k

k

m

k

m

k

C

C

t

t===

0

0

Theo u bi, kB= 2kAdo tA= 2tBhay tB= 72/2 = 36 ngy.

Thi gian cho t s B ln c 48 cm l 36 ngy.

Cu 5: H s thm ca t xc nh t th nghim thm vi ct nc thay i c xc nh

theo cng thc sau:

k =

21

12 lg)(

3.2h

h

ttA

aL

trong : L = chiu di ng thm (chiu di mu th nghim), L = 16cm;

a = din tch ng cp nc, a = 1 cm2;A = din tch tit din mu, A = 1.56cm2;

t1= thi im bt u th nghim, t1= 0;

t2= thi im kt thc th nghim, t2= 1 pht = 60s;

h1= chiu cao ct nc thi im bt u th nghim, h1= 90 cm;

h2= chiu cao ct nc thi im kt thc th nghim, h2= 45 cm.

k = 2105.1)2lg(6056.121613.2 = xxx

cm/s.


56/84


s 02:

Cu 1: Do nh hng ca dng thm t bn ngoi vo ng sut trong t b thay i ti nhng

v tr c p lc thy ng khc khng.

a) Xc nh Gradient thy lc ca dng thm qua cc lp t:

Cc im cn quan tm nm trn ng dng men theo tng c v cng l ng dng ngn

nht. Gi H l tng tn tht ct nc, ta c H = 7.5m; H1l tn tht ct nc khi thm qua

lp t th nht vi chiu di ng thm L1= 5.0m; H2l tn tht qua lp t th hai vi

chiu di ng thm L2= 7.5m. Cc phng trnh sau y nghim ng:

H1+ H2= H

v1= v2hay k12

22

1

1

L

Hk

L

H =

Suy ra H1= 22

1

1

2 HLL

kk

Kt hp hai phng trnh ta c: 22

1

1

2 HL

L

k

k + H2= 7.5

Thay gi tr L1= 5; L2= 7.5; k1= 10; k2= 5, gii theo H2ta c:

H2= 5.63m v H1= H H2= 1.87m

I1= H1/L1= 1.87/5 = 0.375

I2= H2/L2= 5.63/7.5 = 0.751

b) Tnh ng sut ti cc im A, B, C, D.

- ti A: + ng sut tng = 19 x 5 = 95 kPa;+ p lc nc l rng u = 0

+ ng sut hu hiu = u = 95 kPa.

- ti B:

+ ng sut tng = 95 + 20 x 5 = 195 kPa

+ p lc nc tnh ut= 10 x 5 = 50 kPa

+ p lc thy ng ud= - 0.375 x 10 x 5 = -18.75 kPa

+ ng sut hu hiu = u = 195 50 + 18.75 = 163.75 kPa

- ti C:

+ ng sut tng = 195 + 19 x 5 = 290 kPa


+ p lc thy ng ud= - 0.375 x 10 x 5 - 0.75 x 10 x 5 = - 56.25 kPa

+ ng sut hu hiu = 290 100 + 56.25 = 246.25 kPa

- ti D:

+ ng sut tng = 19 x 5 = 95 kPa


57/84



+ p lc thy ng ud= 0.75 x 10 x 5 = 37.5 kPa

+ ng sut hu hiu = 95 50 37.5 = 7.5 kPa.

Cu 2: a) Vic khai thc nc ngm ch lm thay i trng thi ng sut trong lp st do chlp t st b ln. p lc nc ti nh lp st khng thay i trong khi p lc nc ti y lp

thay i gim ng vi s gim ct nc 3m. ng sut hu hiu trong lp st tng tng ng v

phn b bc nht trong phm vi : ti nh lp = 0; ti y lp = 10 x 3 = 30 kPa.

ln S xc nh theo cng thc:

S =

++ 0

0

0 '

''lg

1

H

e

Cc

trong , 0= ng sut hu hiu gia lp st trc khi thay i;

e0= h s rng trung bnh ca lp st trc khi thay i xc nh ng vi gi tr 0;

= s gia ng sut hu hiu trung bnh, xc nh ng vi gia lp;

Cc= ch s nn ca t st.

S = 0.054m

b) Vic khai thc nc ngm xy ra trong thi gian 1 nm c th coi gn ng nh s gia ti

t ngt thi im gia qu trnh khi phn tch c kt. Trong trng hp ny, thi gian 3 nm

k t khi bt u khia thc nc ch coi nh thi gian sau gia ti l 2.5 nm. Lp t st c

kt theo s 0 vi s thot nc hai pha do chiu di ng thot h = 3m.

Nhn t thi gian Tv= 5.23

2.122

xth

Cv = = 0.333

c kt U = 1 -vT

e 42

2

8

= 1 - 8216.02

8 e

= 0.6436

S(t) = 0.6436x0.054 = 0.035m

ln sau 3 nm khi thc nc ngm l 3.5 cm

Cu 3: (xem s 01)

Cu 4: Khi mng chu ti trng ngang H ton b h thng s dch chuyn theo hng ca ti

trng gay ra p lc t ch ng pha sau mng (cng chiu vi ti trng) v p lc b ng

pha trc mng (ngc chiu vi ti trng). Gi tr p lc t ch ng ton phn (Ecmax) xut

hin khi ch c mt gi tr chuyn dch tng i b so vi gi tr cn thit pht huy p lct b ng. Do ch khi tng gi tr ca ti trng ngoi H vi p lc t ch ng vt qu

gi tr gii hn ca p lc b ng th cc mi phi chu ti.

Trn 1m di mng:

- p lc t ch ng: Ec=20022 518)2/3045(

21

21

xxtgHKc

= = 75 kN/m


58/84


- p lc t b ng: Eb= 20022 518)2/3045(21

21

xxtgHKb

+= = 675 kN/m

Tng ti trng theo hng lc H: Ec+ H = 500 + 75 = 575 kN/m < gi tr p lc t b ng

gii hn Eb= 675 kN/m do cc khng chu ti trng ngang:Hc= 0.

Cu 5: a) Coi phn p cao trn nh tng nh ti trng phn b u cng q = h. Xt p

lc t ng vi mt trt bt k, theo phng php Coulomb ta c quan h gia p lc t vi

cc ti trng tc dng ln lng th trt:

E* = (Q + P))sin(

)sin(

1

+

trong 1= 900 (+ ) = 900;

Q = trng lng ca lng th trt;

P = tng gi tr lc phn b q trn lng th trt, P = q. AC.

t k =Q

Pta c: E* = Q(1 + k)

)sin(

)sin(

1

+

= (1 + k) E.

R rng nu (1 + k) khng ph thuc vo th khi E t gi tr cc i EcmaxE* cng t gi

tr cc i hay ni cch khc mt trt khi c ti trng q khng thay i v tr.

Tht vy, k =AB

q

BCxAB

BCqx

2

2

1 = khng ph thuc vo .

b) p lc t trong trng hp ang xt l p lc t ch ng:

Ec= qKHK cc +2

2

1 = 50.44 kN/m


59/84


p n k thi nm 2000Cu 1: a) Vic h thp mc nc ngm lm gim p lc nc l rng trong t trong khi ng

sut tng khng thay i. ng sut hu hiu trong t tng lm cho nn b ln. S thay i ng

sut trong t th hin trong hnh. ln cui cng ca lp st S = mvh = mvuh = 0.11m

trong : u = mc gim p lc nc l rng trung bnh trong tng st, u = umax/2 .

Sau 3 nm k t khi bt u bm ht, thi gian c kt coi gn ng (theo u bi) l 2.5 nm.

Lp st thot nc hai chiu theo s 0 vi chiu di n thot h = 4m. c kt sau 2.5

nm: U = 1 -vT

e 42

2

8

= 0.53 trong Tv= =th

Cv

20.2188

ln sau 3 nm: S = 0.058 m

b) Lp ct mng chia tng st lm hai u thot nc hai chiu vi chiu di ng thot nc

ln lt l h1= 3m v h2= 1m. c kt sau 2.5 nm nh sau:

U1= 1 -14

2

2

8 vTe

= 0.69 ; U2= 1 -

24

2

2

8 vTe

= 1

trong Tv1= th

Cv

2

1

; Tv2= th

Cv

22

S1= 2mvu1h1= 0.062

S2= 2mvu2h2= 0.048

ln sau 3 nm s l: S = U1S1+ U2S2= 9.1 cm

Cu 2: a) H s rng ca mu sau khi nn xong ng vi cp ti 4= 80 N/cm2

l:e4= w= 0.306x2.71 = 0.8293

H s rng ca ban u ca t: e0=05

055

/1

/

hS

hSe

+

= 1.010

b) H s rng ca t di cc cp ti trng: ei= e0 (1+e0)Si/h0

1= 10 N/cm2, e1= 0.97

...

c) H s nn ca t trong khong thay i ng sut 20 40 N/cm2:

a =2040

4020

ee= 0.0029 cm2/N

trong , e20= h s rng ca t ng vi ti trng = 20 N/cm2, e20= e(=20); e40= tng t,

e40= e(= 40)

Cu 3: t di y mng chu p lc y ni v p lc thy ng cng chiu do trng

lng ring hu hiu = bh 0 i0= 9 kN/m3

Sc chu ti gii hn ca nn: pgh= 0.5Nb + Nqhm+ Ncc (1)


60/84


Ti trng tip xc y mng: ptx= P/b (2)

Gii phng trnh Fs.ptx pghtheo b ta c b 3.67m

B rng mng hp l: b = 3.7m.

Cu 4: a) Mc nc ngm di chn tng: tng chu p lc t ch ng Ecmax=2

21 HK

c

t ti su 2H/3 k t nh tng. Ecmax= 324.9 kN/m

- Mc nc ngm ngang mt t: Tng p lc ln tng bao gm p lc t ch ng tnh vi

nv p lc nc. Emax= 671.5 kN/m. im t khng thay i.

- Mc nc ngm gia tng: Na trn chu p lc t ch ng; na di chu p lc t

ch ng v p lc nc. Emax= 411.5 kN/m t cch chn tng 1.98m

b) iu kin n nh chng lt:

Fs = [ ]FsM

M

l

gi

trong Mgi = m men gi ch yu do trng lng bn thn tng gy ra; M gi = Qt.zQ =

Qt1.zQ1+ Qt2.zQ2;

Qt= trng lng tng, Qt= Qt1+ Qt2= HbB

bH2

+ ;

zQ = khong cch t trng tm Qt n mp mng tng chn, zQ1 =3

)(2

2

bBb + ; zQ2 =

3

)(2 bB ;

Ml= m men gy lt quanh mp mng do p lc t gy ra, Ml= Mmax.zE;

Emax= tng p lc (t + nc) ln tng;

zE= khong cch t im t ca Emaxn chn mng.

Cu 5: k =)(

)/lg(3.2

12

12

ttA

hhaL

= 0.879 cm/pht (1.5x10-2cm/s)


61/84


p nk thi nm 2001

Cu 1: Mt trt l mt c ng sut tip bng sc khng ct ca t tng ng trn mt .

Kho st trng thi ng sut ti mt im trong bi ton phng c biu din qua vng trn

ng sut Morh trn hnh. ng sut tip ln nht xy ra trn mt nghing 450so vi phng ca

ng sut chnh 1:

max= (1- 3)/2v = (1+ 3)/2

Sc khng ct tng ng trn mt

s = tg+ c = [(1+ 3)tg]/2 + c= (1 3)/2cos

s max=

1

cos

1

(1 3)/2

Trong phn ln trng hp >0 hay cos< 1 do hiu (s max) > 0 chng t khng xy ratrt trn mt c tmax hay ni cch khc mt trt khng trng vi mt c ng sut ct ln

nht.

b) Mt trt trng mt c ng sut ct ln nht maxkhi v ch khi hiu (s max) = 0 hay cos

= 1, = 0. t c gc ma st trong = 0 cnc gi l t dnh l tng. Trong thc t ta

gp t c = 0 i vi t st bo ha nc chu lc trong iu kin khng thot nc.

Cu 2: a) ln cui cng ca lp st: S = mvph = 0.15m

trong mv= h s nn th tch, mv=01 e

a

+

p = ti trng t nn p, p = h

h = chiu dy ca lp t st, h = 6m

Ti thi im t ln S(t) = mv(t)

(1+3)/2c

450

max

s

3 1

c

s = tg+ c


62/84


trong (t) = din tch biu ng sut hu hiu trn ton b chiu dy h ca lp t st ti

thi im t, (t) =i

n

i

i

h

hupdztup = =

)()]([(10

.

T kt qu o ta c:

(t) = 200 kNm

S(t) = 0.1m

c kt ca lp t st: U(t) =15.0

1.0)(=

S

tS= 0.67

b) V l thuyt, c kt hon thnh khi U = 1 hay t = . Trong thc t, c th coi hon thnh c

kt khi U 1, chng hn U = 0.99. Khi thi gian tng ng xc nh t biu thc:

U = 1 - Ne2

8

= 0.99 N = 4.395 t = NC

h

v

2

2

4

= 2.54 nm.

trong , Cv=0vm

k = 6.31 m2/nm.

Cu 3: a) Xc nh ln cui cng ca nn:

- Nn s A: SA= mv(A) = 17.5 cm

- Nn s B:SB= mv(B) = 11.2 + 3.2 = 14.4cm

b) Thi gian cn thit ln t 7cm

- Nn A: U(t) = 7/17.5 = 0.4

Nn A c kt theo s kt hp vi = 30/12 = 2.5.

(kPa)50

36.6

26.8

23.2

z(m)

u (kPa)

13.4

23.2

26.8

23.2

13.41

2

3

4

5

6

z(m)


63/84


c kt xc nh theo cng thc: U(t) =

++

1

)()1()(2 10 tUtU

trong : U0(t) = 1 -Ne

2

8

; U1(t) = 1-Ne

3

32

Gii theo N ta c: N = 0.176

Do , t = NC

h

v

2

2

4

= 0.1783 nm = 65 ngy.

- Nn B: rng lp trn thot nc hai chiu vi h = 3.2/2 = 1.6m (theo s 0); phn di

thot nc mt chiu c h = 1.6m (theo s kt hp) do ti cng mt thi im chng c

N nh nhau. Gi ln ca lp trn l S1t= Ut(N)xS1; lp di l S2t= Ud(N)xS2ta c phng

trnh sau vit theo N:

S1t+ S2t= 7 (cm)

hay Ut(N).S1+ Ud(N).S2= 7 (cm)trong ; S1= 11.2 cm; S2= 3.2cm;

Ut(N) =Ne

2

81

; Ud(N) =

+

+

1

)32

1)(1()8

1(232

NNee

= 16.8/12 = 1.4

Gii theo N ta c N = 0.4455

Thay N = tth

Cv 88.13

4 2

2

=

ta c t = 11.7 ngy.

Cu 4: Cc lc tc dng ln tng bao gm: p lc t ch ng sau tng; p lc t b ngtrc tng; trng lng tng v mng tng; trng lng t v ti trng ln bn mng

tng trc v sau tng.

H s p lc t ch ng: Kc= tg2(450 /2) = tg2(250) = 0.2174

H s p lc t b ng: Kb= tg2(450+ /2) = tg2(630) = 3.8518

Kt qu cho trong bng sau (gi tr p lc t b ng ngh ly 30% cho phn tch n nh)

cng vi cnh tay n tng ng so vi mp mng A

Lc P0 P1 P2 P3 P4 Ecd Ecq Eb P E

Tr s(kN/m) 30 9.7 37.5 148.7 70 53.9 47 9.8 295.9 91

z (m) 1.5 .475 1.2 2.125 2.125 1.8 2.7 0.33 1.89

a) p lc ln y mng:

ptb= 295.9/3 = 98.63 kN/m2

M = 53.9 x 1.8 + 47 x 2.7 + 295.9 x (1.5 - 1.89) = 108.5 kNm/m

pmax= ptb+ 6M/b2= 98.63 + 6 x 108.5/9 = 171 kN/m2

pmin= ptb 6M/b2= 26.3 kN/m2


64/84


b) H s n nh chng trt phng:

Fs =E

EPtgb

3.0+= 1.79

Ecd

Ecq

P4

P3P2

P1

P0A

E

q = 40 kN/m2


65/84


p n k thi 2002

Cu 1: a) Xc nh gc dc gii hn

Dng thm tn ti trn khong AR vi ng dng theo mt ngoi ca mi dc c gradient i =sins tc ng ln phn t dV trn bin lc thm = i0. H s n nh ca phn t trn mi

xc nh theo cng thc:

Fs =jdVdV

tgdV

dn

dn

+

sin)(

.cos)(

Vi j = i0= 0sinta c: Fs =

tg

tg

dn

dn

0( +

Gc mi gii hn xc nh ng vi h s an ton Fs = 1 hay

tg=

tgdn

dn

0( + = 0.2566

= 14023.

b) Vi h s an ton theo yu cu [Fs] = 1.5 tng t ta xc nh c = 9043.

Cu 2: Trng lng ring kh: k=e+

1

0 = 17.8 kN/m3

Trng lng ring y ni: n=e+

1

)1( 0 = 11.13 kN/m3

Trng lng ring bo ha: bh= n+ 0= 21.13 kN/m3

ng sut tng:

- ti A: A= 0

- ti B: B= h1= 17.8 x 2.5 = 44.5 kN/m2

- ti C: C= B+ bhh2= 44.5 + 21.13 x 1.4 = 44.5 + 29.58 = 74.1 kN/m2

- ti D: D= C+ bhh3= 74.1 + 21.13 x 5 = 179.8 kN/m2

p lc nc l rng:

dQ.cos

j.dV = 0sin.dV

A

RdV

dQ = ndV

dQ.sin


66/84


- ti A: uA = 0

- ti B: uB= - 0h1= - 10 x 1.4 = - 14 kN/m2

- ti C: uC= 0

- ti D: uD= 0h3= 10 x 5 = 50 kN/m2ng sut hu hiu:

- ti A: A= A uA= 0

- ti B: B= B uB= 44.5 (-14) = 58.5 kN/m2

- ti C: C= C uC= 74.1 0 = 74.1 kN/m2

- ti D: D= D uD= 179.8 50 = 129.8 kN/m2

Biu ng sut c dng nh trn hnh v.

Cu 3: p dng nguyn l cng tc dng, c th xc nh ng sut ti M bng phng php

im tm; phng php im tm kt hp lc tp trung hoc qui ti trng phn b v cc lc

tp trung.

Chia vng t ti thnh 2 phn: hnh vung ( gia) cnh b = 2m v hnh tam gic cn li. ng

sut ti M, M= 1+ 2trong 1v 2tng ng l ng sut do ti trng trn hnh 1 v

hnh 2 gy ra.

1= k01p = 4kcp trong : k01= k0

== 5.1;1b

z

b

l; kc= kc

== 3;1b

z

b

l

2= )(4

10201 kk p = k 2z

Ptrong k02= k0

== 06.1;1b

z

b

l; k = k

= 443.0z

r

Tra bng ta c kc= 0.045; k = 0.307

= 18 + 3.41 = 21.41 kN/m2.

Nu qui ti trng v 5 lc tp trung t ti trng tm cc tam gic ta c Pi= P = 100 kN; r1=

0.67; r2= 1.33

44.5

74.1

179.

z(m)

8.

3.

2.

0 u

2.5 -14

50

z(m)

8.

3.

0

44.5

58.5

74.1

129.8

z(m)

8.9

3.9

2.

0


67/84


= (4k1+ k2) 2zP

= (4 x 0.424 + 0.307) x 100/9 = 22.3 kN/m2

(sai s (22.3 21. 4)/21.4 = 4.2%)

Cu 4: a) Lp t st c kt theo s 0 vi chiu di thot nc h = 1m, thi gian c

kt t ti 96% xc nh t quan h: t = NC

h

v

2

2

4

trong , vi U = 0.96 ta c N = 3 (gii t quan h U = 1 - Ne2

8

= 0.96);

Cv= 00

)1(

a

ek

m

k

v

+

= = 0.586 m2/nm (t bo ha: e = 0.01w= 0.801)

t = 2.07 nm.

b) Nu tng di khng thm nc, chiu di ng thot h = 2m: t = 8.56 nm.

c) Trng hp thot nc hai chiu, thi gian l ln thot nc mt chiu.

Cu 5: a) su ln nht ca vng bin do, zmax, xc nh t iu kin:

0=d

dz

02sin

2cos2==

d

dz- cos2= 1 sin 2= (900 )

Thay vo phng trnh z ta c: zmax=

ctgcq

ctgqp

+

2

Hay p = qctg

zctgcq+

+

++

2

.( max

b) Khi im c zmaxnm trn trc ng qua mp mng, zmax= b.ctg(2) = 1.15m

p = 100 kPa P = 100 kN

2m 1m

1m

1m

1.33m

M

M

3m


68/84


Ti trng tng ng p = 352 kN/m2.

c) Xc nh su ln nht c th c ca vng do (maxzmax) v ti trng tng ng.

su ln nht ng vi im c zmaxnm trn trc ng qua tm mng:

zmax= ctgb2

= 1.732m do tnh c p = 405 kN/m2.

b = 2m

maxzmax

B

2

A

zmax

q = 30 kPap


69/84


p n k thi 2003

Cu 1: a) Fs =0

i

dn = 1.5 i =0

Fs

dn = 0.6

i =)10()2(

8

+=

HHL

H

Gii theo H c H = 12.67m

b) Ti B: = u = 134 kPa trong u bao gm p lc nc tnh v p lc thy ng.

Cu 2: a) 2SA(t) = 2SA.UA(t) = 2mvHAp [1 -NAe

2

8

]

SB(t) = SB.UB(t) = mvHBp[1 -NBe

2

8

] = 2mvHAp[1 -NBe

2

8

]

V 2SA(t) = SB(t) NA= NB

Thay N = th

Cv

2

2

4

ta c:

22B

vB

A

vA

h

C

h

C=

Cv=0vm

ksuy ra

22B

B

A

A

h

k

h

k= hay kB= kA. 2

2

A

B

h

h= 4kA.

b) Trng hp B t trn tng cui si, chiu di thot nc hB= hAdo kB= kA.

Cu 3: a) Xc nh 1,3ti cc im M1v M2vi gi thit 1,3(g) = z, gc 2ln lt l 760

v 87.80. Thay vo cng thc tnh gc lch ng sut ta xc nh c gi tr gc lch . S n

nh ca im ng kho st c nh gi thng qua quan h v : sin< sin- n nh v

ngc li.Ti M1(0; 1.25): 1= 235.8; 3= 74.5

sin =

tgc /231

31

++

= 0.358 > sin= 0.342 M1mt n nh.

Ti M2(0.28; 1.25): 1= 254.5; 3= 89.1

sin= 0.342 = sin M2cn bng gii hn.

b) im M1c sin> sindo thuc bn trong vng cn bng gii hn; im M2c sin=

sin do thuc trn bin vng bin dng do.

c) Vic xc nh vng bin dng do theo phng php ny ni chung cha tha ng v khi

trng thi ng sut ti mt im t ti cn bng gii hn th vic xc nh ng sut ti im

theo nguyn l n hi khng cn hp l na, cc im bn trong vng bin dng do xy ra

tnh hung gc nghing ca tng ng sut ln hn gc ma st trong ca t.

Cu 4: a) Th nghim ny thot tin o ln ca mu di tc dng ca ti trng. Vi mt

vi gi thit n gin ha h s rng ca t ng vi tng cp ti trng c xc nh theo


70/84


cng thc: ei= e0- )1( 00

eh

Si + ; ng sut nn tng ng i= pitrong e0= h s rng ca t

trc khi th nghim (ng vi chiu cao ban u h0 ca mu); Si = ln ca mu di ti

trng pi; h0= chiu cao ban u ca mu; pi= ti trng th nghim cp th i.c) Trn c s kt qu th nghim, c trng bin dng ca t c; Crv Ccc xc nh.

Cu 5: Cc mu c th nghim theo ch CU c o p lc nc khi ph hoi do c th

xc nh c c trng khng ct (cu; ccu) v ( c).

Xc nh cuv ccu:

Mu 1: ng sut khi ph hoi 11= 350; 13= 200

Mu 2: 21= 700; 23= 400 .Lp h phng trnh:

sincu=cucuii

ii

tgc

/231

31

++

vi cc gi tr ng sut khi ph hoi mu ta xc nh c cu

= 15050v ccu= 0.

Xc nh v c:

Mu 1: 11= 11 u = 210; 13= 60

Mu 2: 21= 420; 23= 120

Tng t, gii h phng trnh cn bng gii hn ta c = 33045v c = 0.

Kt qu th nghim cho c = 0 chng t t thuc loi c kt bnh thng.


71/84


p n k thi 2004

Cu 1:

p lc t ln tng ng vi mt trt BC lm gc 600so vi phng ngang c xc nh t

phng trnh c bn:

E = Q

)sin(

)sin(

1

+

= Q sin(360)/sin(960)

Q = trng lng lng th ABC, Q = dt(ABC)x

dt(ABC) = BCxAH2

1

AH = AB sin(400) = )40sin(cos

0

H= 5.87m; BC = AB

)40sin(

)110sin(0

0

= 14m

dt(ABC) = 5.87 x 14 / 2 = 41.09m2 Q = 41.09 x 20 = 822.36 kN/m

E = 822.36 x sin(360)/sin(960) = 486 kN/m

b) Theo l thuyt Coulomb, tr s E trn l p lc ca t ln tng vi mt trt gi nh

BC. Gi tr p lc t ch ng l gi tr ln nht c th c ng vi mt trt nguy him.Vic xc nh Ecmaxdo thc hin trn nguyn tc gii bi ton cc tr theo .

Cu 2: a) Mc n nh tng th xc nh thng qua h s Fs =p

pgh

trong , pgh= sc chu ti gii hn ca nn. Trng hp nn chu ti trng do p nhanh, pgh=

(+ 2)cu= 128.5 kPa;

H

1= 900

= 600

( ) = 360

R

E

Q

C

B

= 600

= 100

= 200A

H = 9m


72/84


p = ti trng tc dng ln nn. Trong trng hp ny l ti trng t p, p = h= 108 kPa

Fs = 128.5/108 = 1.19 < [Fs] = 1.5

Khng m bo an ton khi thi cng p nhanh.

m bo n nh, b phn p c s dng vi chiu cao ti thiu xc nh trn c s: pgh=(+ 2)cu + h = 1.5p

hay h = [1.5p (+ 2)cu]/= 1.86m

b) Vic p chm cho php nc di nn thot ht ra ngoi do bn khng ct ca t

xc nh theo kt qu th nghim CD. Sc chu ti gii hn ca nn xc nh theo cng thc:

pgh= cNqNbN cq ++21

trong : N, Nqv Ncxc nh theo gc ma st trong hu hiu .

Vi = 100:

Nq= etgtg2(450+ /2) = e0.5538tg2(500) = 2.471

N= 1.8(Nq 1)tg = 0.467

Nc= (Nq 1)ctg = 8.342

Trng hp khng c b phn p:

pgh= cNbN c+21

= 338.98 kPa

Fs = 338.98/108 = 3.14 > [Fs] = 1.5

Vy nn m bo n nh khi thc hin phng n p chm.

Cu 3: a) Gi St1v St2 l ln ti thi im t1v t2; Qt1v Qt2l c kt tng ng. Ta c:

S =2

2

1

1

t

t

t

t

Q

S

Q

S=

trong S = ln ca lp st.

do :2

1

2

1

t

t

t

t

Q

Q

S

S=

p dng cho c kt mt chiu trng hp 0 ta c: Q = 1 - Ne2

8

ta c:2

1

2

1

1

1N

N

t

t

ke

ke

S

S

=

trong N2= 2N1v t2= 2t1.

Thay St1= 100mm; St2= 139.4mm ta c: St1/St2= 0.7174

1 ke N1= 0.7174(1 ke -2N1)

Gii theo e-Nta c : e-N= 0.697 do Qt1= 1 k x 0.697 = 0.435

S = St1/Qt1= 100/0.435 = 230mm.

b) H s thm ca t: k = Cvmv0


73/84


Vi t1= 1 thng e-N1= 0.697 hay N1= 0.361 ta c Cv=

t

Nh2

24

= 10.97 m2/nm

mv=pH

S = 4.6 10-4m2/kN k = 0.05 m/nm = 1.67 10-7cm/s

Cu 4: a) Gi lu lng nc thm vo h l q, ta c:

q = kiF= 1.15 m3/h

trong , k = h s thm ca t, F = din tch mt bng y h, i = Gradient thy lc dng

thm.

Vy cng sut bm ti thiu cn la chn phi l 1.15 m3/h.

b) Khi bm ht lin tc vi my bm cng sut q = 1.15 m3/h, gradient dng thm n nh i =

0.53 do h s n nh chng ct chy ca y h:

Fs =1053.0

2.11

0 xi

dn =

= 2.11

Fs = 2.11 > [Fs] = 2.0 y h m bo an ton.

c) Gi h(t) l cao mc nc dng trong h mng ti thi im t (so vi mc so snh l y

lp st) ta c: i(t) =L

thH

)(

trong H = chiu cao ct nc p, H = 5.2m; L = chiu di ng thm, L = 3.0m.

Lng nc vo h sau thi gian dt s l: dQ = ki(t)Fdt

lm cho nc trong h dng ln mt lng dh = dQ/F = ki(t)dt

Ta c phng trnh vi phn:

dt = dhhHk

L

)( - t = ChH

k

L + )ln(

Ti thi im bt u t = 0, h(0) = L do C = )ln( LHk

L

Thay vo phng trnh trn ta c:

t = )ln( LHk

L

)ln( hH

k

L

=

)(ln

thH

LH

k

L

Khi mc nc trong h n 0.5m ta c h = 3.4 + 0.5 = 3.9m:

t =

)9.32.5(

)4.32.5(

ln103.2

4.36x = 4.81x10

5

s = 134 h (5.6 ngy m)


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p n k thi 2005

Bi 1:

a) iu kin n nh: tng ng sut do trng lng t cn bng p lc nc y ni: (6 - h)

0.hw

= 0(1 +w)/(1 + e) = 20.8 kN/m3

hw= 8 h 2.12m. Vy chiu su o ti a l 2.12m

b) Ngay sau khi gia ti ton b ti trng gy ra p lc nc l rng d: u = 1. Tng p lc

nc l rng u = u + u0

u = ( )

2sin2 +p

= 39.7 kN/m2

u0= hw0= 8 x 10 = 80 kN/m2

Nh vy ngay sau khi chu ti, p lc nc l rng ti N: uN= 119.7 kN/m

2

; Khi nn nnh, p lc nc ti N; uN= 80 kN/m2.

Bi 2a) p lc nc l rng d ti th im bt k c th xc nh theo cng thc sau: u(z,

t) =

vT

h

zp

4exp

2sin

4 2

Ti thi im ng vi c kt U = 50%, gii theo Tvta c:

Tv= 502 th

Cv = 0.197

do : u(z, t50) =

197.0.

4exp

2sin

4 2

h

zp = f(z).

thay p = 120 kN/m2; h = 6/2 = 3m ta c: u(z) = 94.sin(0.524z).

ti z = 0: u = 0;

ti z = 2: u = 81.4 kN/m2

ti z = 4: u = 81.3 kN/m2

ti z = 6: u = 0 kN/m2

b) ln ca nn ti thi im bao gm ln ca lp t st t ti c kt U v

ln ca lp ct coi l kt thc: S(t) = S1.U(t) + S2

S1= a01p1h1= 0.324m ;S2= a02p2h2= 0.041m

S(t) = 0.324 x 0.5 + 0.041 = 0.203mBi 3 a) Sc chu ti gii hn ca nn (coi nh mng bng b rng b = 10m, chn su hm=

1m): qu= 0.5Nb+ Nqhm+ Nc.c = 221.23 kN/m2

trong : Nq= 2.47; Nc= 8.34; N= 0.48; = 20 kN/m3; q = 1 x 20 = 20 kN/m3; c = 15 kN/m2

Ti trng do t p p qu/Fs = 110.6 kN/m2

Thay p = .h= 110.6, gii theo hta c h= 5.1m


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b) ng sut chnh ti M: 1,3= ( ) )(2sin2 zhp

++

1= 175.6 kN/m2; 3= 118.8 kN/m

2

Trn mt nghing 500so vi phng ngang: = 142.3 kN/m2Cng chng ct: s= tg+ c = 40 kN/m

2.

Bi 4a) t st c kt bnh thng do c = 0. Khi mu b ph hoi iu kin cn bng gii

hn tha mn: sin = (1 3)/(1+ 3)

trong , 3= 100 kN/m2; 1= 3+ = 300 kN/m

2.

Gii theo sin ta c = 300.

b) ng sut hu hiu 3= 200 50 = 150 kN/m2

Gii theo 1ta c 1= 450 kN/m2do lch ng sut khi mu b ph hoi l = 1

3= 300 kN/m2.


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p n k thi 2006

Cu 1: a) Trng hp thm ng:

Chiu di ng thm l L = H = chiu dy lp t

H s thm ti su z bt k c th vit: k(z) = k1+H

kk 12 z = k1+ mz, trong , m = (k2

k1)/H

Dng thm qua z vi gradient i(z ) =dz

dh

Gi tn tht trn ton b l H, gradient chung l I = H/H. Cc phng trnh sau tha mn:

H = H

dh0

v = I.kt= i(z).k(z) = dzdh [k1+ mz]

hay I.ktmzk

dz

+1= dh

Ly tch phn hai v ta c: I.kt. ( )H

mzkm

01ln1

+ = H

thay I = H/H ta c:

kt.

+

+

0.ln.ln)(

.1 12

112

1

12 H

kkkH

H

kkk

kk

H

H= 1

kt=

1

2

12

lnk

k

kk

b) Trng hp thm ngang: Tn tht ct nc trn chiu di thm nh nhau l nh nhau do

ti cc lp phn t dy dz dng thm xy ra cng gradient I = H/L. Vn tc thm ti su z

l v(z) = I.k(z) = ( )mzkL

H+

1

Tng lu lng thm qua ton b chiu dy H s l:

q = +

H

dzmzkL

H

0

1 )( =

H

mz

zkL

H

0

2

1 2

+

= I.kt.H

Gii theo ktta c : kt=2

21 kk +

Cu 2: Xt cn bng mt lng th c b rng b nh hnh v.


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W = zbcos; N = Wcos; T = Wsin; R = Ntg+ cb

Cc phn lc bn Et = Ep(do mi dc ang trng thi cn bng)

H s an ton chng trt:

Fs =

sincos.cos

sin.cos. 2

z

ctgz

W

bctgW

T

bcNtgt

T

R +=

+=

+=

Thay = 200; = 18 kN/m3; z = 3m; c = 10 kN/m2ta c Fs = 1.576

b) Trng hp mi dc c dng thm song song, p lc nc tc dng ln y lng th (p lc

nc hai thnh bn t cn bng): U = (0.zcos).(bcos) = 0bz.cos2lm gim ma st v gim

mc an ton:

Fs =T

bctgUN

T

R +=

)(=

tg

tg

bh

bh '0 = 0.512

Bi 3: a) H s an ton i vi chy t y h (an ton chng boiling) xc nh theo cng

thc: Fs =0

0

.

i

bh trong , i = Gradient thy lc dng thm ngcln y h khi ht nc.

mc nc 2m trong h, ct nc khng tn tht qua lp cui si ta c :

- chiu di ng thm L = 5m;- tn tht ct nc H = 9 (5+2) = 2m

i = H/L = 0.4

Fs = 10/4 = 2.5

b) Ti thi im t k t khi bm ht mc nc cn li trong h l h(t) ng vi gradient dng

thm vo h i(t) =5

)(45

))(5(9 thth =

+.

3m

b

b

Et

T

W

NEp

N

RU

Et

T

W

NEp

N

R


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Sau thi gian dt, mc nc trong h gim i mt lng dh tng ng lng nc dQ = -F.dh.

Lng nc ny c bm ra ngoi v thm b sung trong khong thi giam . Ta c

phng trnh cn bng lng nc:

dQ = -F.dh = [qb F.k.i(t)]dttrong , qb= lu lng ht ca my bm, qb= 20m

3/h.

dt =

bq

thkF

Fdh

)5

)(4(.

=)(..)5.4(

.5

thkFqkF

dhF

b

=- [ ]

( )hkFqkFhkFqkFd

kb

b

..5.4

..5.45

Tch phn hai v t t = [0; T] 0 v h = [0; 2] trong T = thi gian cn thit bm kh h

mng ta tm c T = 52.86 gi.

Bi 4: a) n nh trt:

p lc t gy trt: Ea=2

2

1HK

a trong Ka= tg

2(450 /2) = 1/3.

Lc chng trt bao gm ma st v lc dnh bm y mng tng:

R = Ntg+ bc trong , N = trng lng tng (BTCT) v t trn tng,

N = [3.1 + 1.(H-1)] bt+ .1.(H-1) = (H-1).45 + 75

R = [45(H-1) + 75].tg+cb

H s an ton trt phng theo y: Fs =2

2

1)3045(

HK

bctgH

E

R

aa

++=

Vi Fs = 1 gii theo H ta c H = 6.67m.

b) Gi s y mng c m rng thm b tha mn iu kin Fs = 1.5 khi trng lngtng v t s thay i:

N = [(3 +b).1+(H-1).1]bt+ (1+b).(H-1).= N + N

Thnh phn gi n nh R = Ntg+ (b + b)c = R + R

trong R = Ntg+ b.c = b[bt+ (H-1)+ c]

Fs =aaa

E

R

E

RR

E

R +=

+= 1

'

Vi H = 6.67 cho Fs =1, gii R/Ea= 0.5 ta c b rngcn m rng b = 1.217m.Bi 5: a) Sc chu ti ca nn t st khi gia ti nhanh: p gh= (+ 2)cu= 118.22 kPa. Ti trng

do san nn: p = h = 92.5 kPa

H s an ton Fs =5.92

22.118=

p

pgh = 1.278

Fs = 1.278 > 1.2 nn n nh.


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b) Lp t st c kt thot nc mt hng theo s 0 do c kt tnh theo cng thc U

= 1 - Ne2

8

= 0.55. Gii theo N c N = 0.5895 do t =v

C

Nh2

24

= 1.537 nm.

Nu coi gn ng nn bt u c kt gia thi gian p (1.5 thng) th thi gian cn thit c kt ca nn t ti 55% s l t = 1.537 0.125 = 1.412 nm.


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p n k thi nm 2007

Cu 1: a) Xc nh cc c trng khng ct ca t

i vi mu th nghim CD cc thnh phn ng sut l ng sut hu hiu:

3= 100

1= 100 + 145 = 245

Gc nghing ca mt ph hoi so vi phng ngang 540trong khi mt phng nm ngang chnh

l mt chnh chu 1do :

= 450+ /2 = 54 = 180.

T iu kin cn bng gii hn: 'sin'/'231

31

=

++

tgc

gii theo c ta c: c = 20.18 kPa.

i vi mu th 2 th nghim thc hin nn mt chiu cho php xc nh bn ct khngthot nc cu=

2

320

2 =u

q=160 kPa.

Ch c hai ch th nghim c thc hin trn 3 mu do ch xc nh c cc c trng

khng ct sau: = 180; c = 20.18 kPa v cu= 160 kPa.

b) Xc nh p lc nc l rng khi th nghim mu th 2 v th 3

Mu s 2: Gi p lc nc l rng khi mu b ph hoi l u2fta c:

1= 1 u2f= 320 u2f; 3= 3 u2f= 0 u2f

Thay vo biu thc cn bng gii hn vi = 180v c = 20.18, gii theo u2fta c u2f= -

295.66 kPaMu th 3 c th nghim theo ch UU vi p lc bung = 200 kPa cho gi tr s gia

ng sut = 320 kPa khi ph hoi ( v theo ch ny ta c cu= 160v u= 0) do ta c 1

= 520 kPa v 3= 200 kPa. Tng t trn, gii theo u ta c p lc nc l rng khi mu b

ph hoi

u = -95.66 kPa.

Bi 2: a) Tnh lng nc ti thm cho 1m3 t khai thc c trc khi m nn:

Gi lng nc thm vo cho 1m3t l Q c th m cht nht, trng lng ring tr

thnh 2= 1+ Q. Trng lng ring kh ca t khng thay i:

k=2

2

1

1

11 ww +=

+

2= 11

2

1

1

w

w

++

Q = 2 1= 11

12

1 w

ww

+

= 0.062 T.

b) Tnh lng t t nhin cn khai thc:

Trng lng ring kh t khi m nn n K = 0.9 l k= 1.55 (T/m3)


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Gi th tch t t nhin cn cho 1m3t sau m cht l V. Trng lng cc ht t khng

thay i hay Vh= V.k= 1.55 m3

k= trng lng ring kh ca t t nhin, k =w+1

=1.5 T/m3

V = 1.55/1.5 = 1.033m3

Bi 3: a) Tnh h s thm ca t nn theo lu lng ht q:

Ti khong cch r k t tm ging, ct nc tng l h(r), dc thy lc i(r) =dr

dhnh nhau

cho c hai lp t (v khng c thm ng).

Lu lng thm qua hnh tr ng knh 2r, chiu cao h:

q = q1+ q2= v1.F1+ v2.F2= k1.i.(h hII)2r + k2.i hII.2r

q = 2r[k1(h h2) + k2h2]dr

dh

trong ,

q1, q2= lu lng thm qua lp t th nht v th hai;

v1, v2= vn tc thm qua tng lp;

F1, F2= din tch t ca dng thm trong t lp 1 v lp 2;

k1, k2= h s thm ca t lp 1 v lp 2, k2= mk1;

hII= chiu dy ca lp t th 2, hII= 4m.

[ ]dhmkhhhkqr

drIIII 11 )(

2+=

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