BT chuong 3-4-5

Bi tp tng hp TS. Thanh Hi

1

Bi tp chng 3:

Bi 1: Cho mt mng b c kch thc LxB=20x10m t trn mt t t nhin

chu p lc gy ln 200 kPa. Nn t di y mng l t st c gi tr h s nn

ln th tch l mv = 5x10-5

m2/kN, h s thm ng kv = 1.10

-8 cm/s. Cho bit chiu

dy chu nn ca nn t st di y mng l 8m v bn di 8m ny l nn t

ct cht xem nh khng ln. Khi tnh ton ln chiu dy nn t st c chia

ra thnh 4 lp, mi lp c chiu dy 2m. Cho bit thm w=10 kN/m3 .

a) (1,0 im) Hy xc nh ln c kt ca nn t st di mng b theo

phng php tng lp phn t.

b) (1,0 im) Hy xc nh ln ca nn t st sau 3 thng (90 ngy) k t lc

nn t st chu p lc gy ln ca mng b. Cho bit mi quan h gia c kt

%U v nhn t thi gian Tv c th c xc nh theo biu thc sau:

Khi U60% 1,781 0,933log(100 %)VT U

Bng 1: Gi tr K0 cho ti hnh ch nht phn b u

z/B L/B

1.0 1.5 2.0 3.0

0.0 1.000 1.000 1.000 1.000

0.2 0.960 0.973 0.976 0.977

0.4 0.800 0.854 0.870 0.878

0.6 0.606 0.694 0.727 0.748

0.8 0.449 0.546 0.593 0.627

1.0 0.336 0.428 0.481 0.525

1.2 0.257 0.339 0.392 0.443

1.4 0.201 0.272 0.322 0.377

1.6 0.160 0.221 0.267 0.322

1.8 0.131 0.183 0.224 0.278

2.0 0.108 0.153 0.190 0.241

Giai:

a/Tnh ln bng phng php tng phn t (chia lm 4 lp,mi lp c chiu

dy 2m)

Tnh ln ti tm mng theo bng sau:

Mi lp c hi=2m

p=K0pgl


2

su

lp t

(m)

im

gia lp

su

im

gia (m)

z/B K0 p Si=mvphi

0-2 1 1 0.1 0.988 197.6 0.01976

2-4 2 3 0.3 0.923 184.6 0.01846

4-6 3 5 0.5 0.7985 159.7 0.01597

6-8 4 7 0.7 0.66 132 0.0132

S=0.067m

Vy ln S=6.7cm

b/ ln ca nn t st sau 3 thng (90 ngy)

Nn t thot nc 2 phng nn H=8/2=4m

H s c kt Cv:

Nhn t thi gian Tv:

c kt trung bnh U, gi s U


3

Mt cng trnh t p dng hnh thang m hnh v

bn di din t cng trnh c: chiu cao p 5m

a=10m, b=3m, trng lng ring t p =19

kN/m3. Cho anpha=10

ng sut thng ng trong nn t do ti hnh

thang c tnh theo cng thc:

])([ 121

bp

z

- tnh ng sut z do ti ngoi v trng lng bn thn ti M (giao gia trc tm ti v y lp st), sau khi ln c

kt hon tt.

- tinh ln ca lp st theo phng php mt lp phn t. - tnh thi gian ln t 90% ln n nh.

Giai:

1/ng sut do ti trng ngoi v trng lng bn thn gy ra ti im M

(giao gia trc tm ti v y lp st)

ng sut do ti trng bn thn:

ng sut do ti ngoi hnh thang sau khi c kt:

ng sut tng cng:

2/ ln ca lp st theo phng php mt lp phn t

Xt ti im gia ca lp st (z=3m)

ng sut do ti trng bn thn trc khi c ti:

(0.8694) ng sut do ti ngoi hnh thang gy ra:

ng sut tng cng do ti bn thn v ti trng ngoi:

(0.826)

ln ca lp st:

(13.8cm)

3/Tnh thi gian ln t 90% ln n nh


4

C th xem biu ng sut gia tng do ti trng ngoi gn ng l dng

hnh thang nh hnh v:

Ti nh lp t: 1= 0; 2 = /2 nn

Ti y lp t l im M

c kt tnh theo cng thc

Trong

Theo U0-2=0.9 v t ta c:

T suy ra TV = 0.8392

H s nn ln a:

3

12

21 1047.0244.115

826.0869.0

x

pp

eea (m

2/kN)

H s c kt:

6

3

8

1 1098.3101047.0

)869.01(10)1(

xxxa

ekC

w

v

(m2/s)

Thi gian t c kt 90%


5

Tv=2H

tCv => v

v

C

HTt

2

= 66

2

1059.71098.3

68392.0x

(s)=88 days

Bi 3: Cho s ti trng phn b u kn khp b mt nn t st bo ha nc c b

dy 12m. Trng lng ring bo ha ca t sat = 18 kN/m3. H s thm kv = 1.10

-6

cm/s. Mc nc ngm nm ngay ti mt t, ly w = 10 kN/m3. Kt qu th nghim nn

c kt ca mu t nh sau:

p lc nn p (kN/m2) 0 25 50 100 200 400

H s rng e 1,50 1,42 1,37 1,25 1,16 1,05

1/ Tnh ln n nh ca lp t st do ti trng ngoi gy nn:

Gii:

1) Tnh ln n nh ca nn t Tnh ln theo lp phn t nh sau:

Xt ti im gia ca lp st, im M (z=6m)

S tnh nh sau:

Nn cng khng thm

12m Lp t st bo ha nc

MNN

8kN/m3*12m=96kPa

M 48kPa

Ct, cao 5m, c =20kN/m3

Nn cng khng thm

p = 100 kN/m2

12m Lp t st bo ha nc

Bin thot nc MNN


6

Trong : p1=bt, M = (18-10)*6 =48 kN/m

2 e1= 1,374

p2= p1 + p = 48 + 100 =148 kN/m2 e2= 1,207

ln n nh:

mhe

eeS 845,012

374,11

207,1374,1

1 1

21

=84,5cm

Nhn: "mode/3" ri nhp "2" (nu ni suy tuyn tnh), ri nhp x, y vo cc ct, sau thot (AC).

Nhp "gi tr cn ni suy/shift/1/7 (reg)" ri chn 5 (nu mun

tm y) hoc chn 4 (nu mun tm x), Ni suy cc hm khc tng t (sau khi bm mode/3 c cc dng

tng ng)

2) Xc nh h s c kt Cv [m

2/s] ca lp t st:

H s thm kv = 1.10-6

cm/s =10-8

m/s

H s nn ln a:

001672,048148

207,1374,1

12

21

pp

eea (m

2/kN)

68

1 10419,110001672,0

)374,11(10)1(

xxa

ekC

w

v

(m2/s)

3) Xc nh ln St ca lp t st sau thi gian 6 thng (180 ngy) k t ngy gia ti:

Thi gian gia ti t=180 ngy= 180*86400= 1,55x107 (s)

Tv= 153,012

1055,110419,12

76

2

H

tCv

Tra bng hoc tnh t cng thc, khi U


7

4) Gi s lp st trn c thot nc theo c hai bin (bn trn v bn di), xc nh thi gian t (ngy) lp st t c mc c kt Uv = 80%

Khi c kt U>60% 1,781 0,933log(100 %)VT U (Casagrande) Suy ra Tv= 1,781-0,933log(100-80) = 0,567.

Tv=2)

2(H

tCv => v

v

C

HT

t

2)2

(

= 76

2

1043,110419,1

6567.0x

(s)=166 days

5) Xc nh ch s nn Cc v ch s c kt trc OCR cho bit p lc tin c kt t th nghim ca mu t ti su 6m l 60 kN/m

2.

365,0)2log(

05,116,1

200

400log

200log400log

400200400200

eeeeCc

T s tin c kt (h s c kt trc) OCR (overconsolidation ratio):

25,148

60

0

p

pOCR c

Tnh ti su 10m, cha c kt

07580

60

0

p

pOCR c

Trong :

pc : p lc tin c kt

p0 : ng sut hu hiu hin ti theo phng ng (p0 = bt = h: ng sut bn

thn)

Bi tp chng 4:

Bi 1: Mt mng bng rng b = 3,0m, di L=20m, vi tng ti tp trung N=6000kN, y

mng su Df=1,5m. Mc nc ngm su 1,0m so y mng. t nn trn

mc nc ngm c trng lng th tch = 19 kN/m3, v t di mc nc ngm c

trng lng th tch bo ha sat = 20.0 kN/m3, lc dnh c = 16kN/ m

2, = 20

o. Cho bit

h s n hng (h s Poisson): = 0,3.


8

a. Tnh gc lch ng sut ti im A c to (x = 0, z = 3.0m) b. Tnh gc lch ng sut ti im B c to (x = 1,5m; z = 3,0m) c. Kim tra s n nh ca im A v B

GII

p lc tiu chun ti y mng bng

pgl =

5,1)22(203

6000)( ftb D

F

N 133 kN/m2

a. Tnh gc lch ng sut ti im A c to (x = 0, z = 3.0m):

2

222

max

2

)cot2(

4)(sinsin

gcxz

xzxz

- Tnh ng sut z=z(p)+v(bt)

- z (p)=kz.p

0

1

b

x

b

z

=> kz=0.55 => z (p)=0.55x133= 73.15(kN/m2)

- v(bt)=ihi=19.0x2.5+(20-10)x2=67.5(kN/m2)

z=73.15+67.5=140,65(kN/m2)

- Tnh ng sut x=x(p)+x(bt)

+ x (p)=kx.p

0

1

b

x

b

z

=> kx=0.04 => x (p)=0.04x133=5,32(kN/m2)

+x(bt)=v

1.0m

B

x

z

A

N=6000kN

Df=1,5m

3m


9

1=0.428 => h=0.428x67.5=28.89(kN/m

2)

x=5,32+28.89=34,21(kN/m2)

- xz=0;

T suy ra: sin2max=0.164 => max=23

05340 Ti im A: max>: mt n nh

b. Tnh gc lch ng sut ti im B c to (x = 1,5m; z = 3,0m) Tng t:


- z (q)=kz.p

5,0

1

b

x

b

z

=> kz=0.41 => z (q)=0.41x133=54,53(kN/m2)

- v(bt)=ihi=19.0x2.5+(20-10)x2=67.5(kN/m2)

z=54,53+67.5=122,03(kN/m2)


+ x (p)=kx.p

5,0

1

b

x

b

z

=> kx=0.09 => x (p)=0.09x133=11,97(kN/m2)

+x(bt)=v

1=0.428 => h=0.428x67.5=28.89(kN/m

2)

x=11,97+28.89=40,86(kN/m2)

ng sut tip, tra k= 0,16

- xz=0,16x133=21,28 kPa;

- x=40,86kN/m2

- z=122,03kN/m2

=> sin2max=0,133 => max=21

026

c. Kim tra s n nh ca im A v B

Ti im A: max>: mt n nh


10

Ti im B: max>: mt n nh

Bi 2. Cho mt mng bng c b rng B = 2m chu p lc ti y mng l p=80kN/m2.

su chn mng Df = 1,5m. Mng c t trong nn t st pha ct c cc c trng:

Trng lng ring t nhin = 18 kN/m3, trng lng ring bo ha sat = 19 kN/m

3, h

s Poisson l 0,3. Gc ma st trong =18o. Lc dnh c=20kPa. Cho trng lng ring ca

nc w = 10 kN/m3. Mc nc ngm nm ngay ti y mng. Cc im A, B c ta

trong hnh v so vi y mng.

Kim tra n nh ti A v B

1) Tnh gc lnh ng sut ti A (x=0; z=1m)

2

222

max

2

)cot2(

4)(sinsin

gcxz

xzxz


- z (p)=kz.p

0

1

b

x

b

z

=> kz=0.82 => z (p)=0.82x80= 65,6(kN/m2)

- v(bt)=ihi=18.0x1,5+(19-10)x1=36 (kN/m2)

z=65,6+36=101,6 (kN/m2)


+ x (p)=kx.p

0

1

b

x

b

z

=> kx=0.18 => x (p)=0.18x133=14,4(kN/m2)

=18kN/m3

sat=19kN/m3

Df=1,5m

B (1,0; 3,0m)

A (0; 1m)

p=80kN/m2


11

+x(bt)=v

1=0.428 => h=0.428x36=15,408(kN/m

2)

x=14,4 + 15,408=29,808(kN/m2)

- xz=0;

T suy ra: sin2max=0.079 => max=16

0232 Ti im A: max max=14

01124 Ti im B: max


12

a. Tnh :

2

11

2

111tan

n

i

i

n

i

i

n

i

i

n

i

i

n

i

ii

n

n

= 26001400003

600262583003

x

xx=0.295 => =16

026

b. Tnh c:

2

11

2

111

2

1

n

i

i

n

i

i

n

i

ii

n

i

i

n

i

i

n

i

i

n

c

=2

2

6001400003

5830060014000262

x

xx=28.33kN/m

2

Bi 4: Mt mng n hnh ch nht c kch thc 2,0m3,0m, c su chn mng 2m,

trn nn t c cc thng s sau: mc nc ngm su 1m, trng lng ring di

mc nc ngm sat=20 kN/m3; trng lng ring trn mc nc ngm =18,5kN/m

3,

gc ma st trong ca t =180, lc dnh c =10kN/m

2. Cho dung trng ca nc

W10kN/m3, trng lng ring trung bnh ca t v mng trn y mng l tb = 22

kN/m3.

a. Xc nh sc chu ti ca t nn di y mng (kN/m2) theo TCVN, (cho

1m ).

b. Xc nh sc chu ti ca t nn di y mng (kN/m2) theo Terzaghi, cho h

s an ton theo pp Terzaghi, k = 2.

c. Nu mc nc ngm nm ti y mng, xc nh sc chu ti ca t nn di

y mng (kN/m2).

d. Trong trng hp mc nc ngm nm ti y mng, mng trn chu mt ti

trng dc trc l Ntc =600kN. t nn bn di y mng c tho iu kin n

nh khng?


13

)*( cDDBbAmR ftc

GII

a. Xc nh sc chu ti ca t nn di y mng (kN/m2) theo TCVN, (cho

1m ).

)*( cDDBbAmR ftc

Trong : 1m ,

=180 tra bng c : A=0.4313; B = 2.7252; D=5.3095; b=2.0m

Rtc =1[0.4313x2.0x (20-10)+2.7252x[18.5x1.0+(20.0-10.0)x1.0]+5.3095x10.0]

= 139.38(kN/m2)

Xc nh c bmin=2,0885m

b. Xc nh sc chu ti ca t nn di y mng (kN/m2) theo Terzaghi, cho

h s an ton theo pp Terzaghi, FS = 2.

pgh = 0,4 N b + Nq * Df + 1,3 Nc c

=180 tra bng c : N =5; Nq = 6.042; Nc =15.517

pgh= 0.4x5x(20-10)x2.0+ 6.042x[18.5x1.0+(20.0- 10)x1.0]+1.3x15.517x10.0=

= 413.92(kN/m2).

Sc chu ti cho php: R=2

92.413

FS

pgh=206.96(kN/m

2).

c. Nu mc nc ngm nm ti y mng, xc nh sc chu ti ca t nn

di y mng (kN/m2).

N=600kN

Df=2,0m

3m

2m

=18,5 kN/m3

B

p=N/F+tbDf

= 600/6 + 22*2=144kN/m2

1,0m

sat=20 kN/m3

Rtc =1[0.4313x2.0x (20-10)+2.7252x[18.5x1.0m+(20.0-10.0)x1.0m]+5.3095x10.0]

= 139.38(kN/m2)


14

Rtc=1[0.4313x2.0x (20-10)+2.7252x[18.5x2.0]+5.3095x10.0]

= 162.55(kN/m2)

Nu mc nc ngm nm ngay mt t, th Rtc s nh nht

Rtc=1[0.4313x2.0x (20-10)+2.7252x[(20-10)*2m]+5.3095x10 = 116,23(kN/m2)

d. Trong trng hp mc nc ngm nm ti y mng, mng trn chu mt ti

trng dc trc l Ntc =600kN. t nn bn di y mng c tho iu kin

n nh khng?

22232

600x

xD

F

Np ftb

tc

tb =144.0 kN/m2

ptb


15

Bi tp chng 5:

Bi 1: Mt lng tng chn trn lng, thng ng, cao 4m, mt t nm ngang, chu ti

phn b u p=21kPa kn khp trn mt t sau lng tng.

t p gm 2 lp: lp trn l ct dy 2m, lp

di l st dy 2m, cc c trng ghi trong

hnh.

- Tnh p lc ch ng tc ng ln lng tng ti nh tng, y lp ct, mt trn

lp st v y lp st.

- Tnh tng p lc ch ng tc ng ln tng v moment tc ng do p lc ch

ng i vi im A chn tng.

Gii:

Xc nh cng p lc ch ng lp 1.

- H s p lc ch ng ca lp 1:

Ka1 = tg2(45-1/2) = tg

2(45-30/2) = 0.33

Cng p lc ch ng ti cc v tr

nh tng:

PaC = pKa1 2c 1aK = 21*0.33 2*0*0.57 = 6.99 kN/m

2

y lp ct:

PaB1

= pKa1 + c*h1*Ka1 2c 1aK = 21*0.333 + 18*2*0.33 2*0*0.57 = 18.98

kN/m2.



Ka2 = tg2(45-2/2) = tg

2(45-20/2) = 0.49


p=21kPa

Ct; c=18kN/m3

1=30o; c1=0 kPa

st; s=20kN/m3

2=20o; c2=25kPa

A

p=21kPa

Ct; c=18kN/m3

1=30o; c1=0 kPa

st; s=20kN/m3

2=20o; c2=25kPa

A

B1

C

B2

Ea1

Ea2


16

Mt trn lp st:`

PaB2

= (p+c*h1)*Ka`2 2c 2aK =(21+18*2)*0.49 2*25*0.7 = -7,07kN/m2

y lp st

PaA = pKa2 + (c*h1 +s*h2)*Ka2 2c 2aK =[21 + (18*2+20*2)]*0.49

2*25*0.7 = 12,53 kN/m2.

Tng p lc ch ng tc dng ln lng tng

p lc ch ng v im t ca lp 1: Ea1 = (6.99+18.98)*2/2 = 25.97 kN/m

im t = mH

ba

ba846.0

3

2

98.1899.6

98.1899.62

3

2

so vi im y lp ct

So vi im A, y1= 2+0.846 = 2.846m

p lc ch ng v im t ca lp 2:

Tm v tr ca im M so vi A theo s sau:

Ta c t l AMBMAT

BN

AM

BM564.0

53.12

07.7

V AM+BM =2m AM = 1,28m

Ea2 = 1/2(1.28x12.53) = 8.02 kN/m

im t so vi im A, y1=1/3 (1.28)= 0.43 m

Tng p lc ch ng = Ea1+Ea2 = 25.97 + 8.02 =33.99kN/m

Moment ti A = 25.97x2.84 + 8.02x0.43 = 77.20kN

A

M

Ea2

N B

7.07kPa

12.53kPa

2m

y2 T

a=6.99kPa

b=18.98kPa

H=2m


17

Bi 2: Tng t bi 1, mc nc ngm cch mt t 1m, cho dung trng bo ha ca

lp ct v lp st ln lt l 20kN/m3, v 21kN/m

3. Tnh v v biu p lc ch

ng,v tnh moment tc dng ln chn tng?

Xc nh cng p lc ch ng lp 1. - H s p lc ch ng ca lp 1:

Ka1 = tg2(45-1/2) = tg

2(45-30/2) = 0.33


nh tng: Pa

C = pKa1 = 21*0.33 = 6.99 kN/m

2

im D: mt nc ngm

PaD = (p + c*1m)Ka1=[21 + 18*1m]*0.3= 12,87 kN/m

2.

y lp ct:

PaB1

= (p + c*h1*)Ka1 2c 1aK =[21 + 18*1m+ (20-10)*1m]*0.3= 16,17 kN/m2.



Ka2 = tg2(45-2/2) = tg

2(45-20/2) = 0.49


Mt trn lp st:`

PaB2

= (p+c*h1)*Ka`2 2c 2aK =(21+18*1m+10*1m)*0.49 2*25*0.7 = -10.99 kN/m2

y lp st

PaA = pKa2 + (c*h1 +s*h2)*Ka2 2c 2aK =[21 + (18*1+10*1+11*2)]*0.49

2*25*0.7 = -0.21 kN/m2.

p=21kPa

Ct; c=18kN/m3

1=30o; c1=0 kPa

st; s=20kN/m3

2=20o; c2=25kPa

A

2m

2m

1m

C

B

D

12,87

1=30o; c1=0 kPa

st; s=20kN/m3

2=20o; c2=25kPa

A

B1

C

B2

Ea

Ea2: khng

c lc

D

A

D

16,17

30

12,87

26,17

30

10

Ea1, CD

Ea1, B1D

Ea2


18

Bi 3: Mt tng chn t bng BTCT cao 8m, t sau lng tng gm 2 lp c cc c

trng nh hnh v. Ti trng sau lng tng phn b kn u khp c ln q = 100

kN/m2. t trc lng tng cao 3m, c cc c trng ging nh lp t s 2 sau lng

tng.

Gi thit tng thng ng, trn lng, t sau lng tng nm ngang. Mc nc ngm

nm rt su. B qua phn p lc tc dng ln mt hng ca mng tng chn.

Cu 1) Xc nh ln (kN/m) v im t (m) (cch chn tng C) ca tng p lc

ch ng (trn 1m tng) lp 1 (on AB) tc dng ln thn tng.


Ka1 = tg2(45-1/2) = tg

2(45-25/2) = 0.406


nh tng:

PaA = qKa1 2c 1aK = 100*0.406 2*12*0.637 = 25,3 kN/m

2

y lp 1:

PaB = qKa1 + 1*h1*Ka1 2c 1aK = 100*0.406 + 18*4*0.406 2*12*0.637 = 54,54 kN/m

2.

4m

4m

Lp 1:

= 18kN/m3

= 250

c = 12 kN/m2

Lp 2:

= 19kN/m3

=280

c = 0

q =100kN/m2

A

B

C

= 19kN/m3

=280

c = 0

3m


19


Ea1 = (25,3+54,54)*4/2 = 159,68 kN/m

im t = mH

ba

ba76.1

3

4

52,543,25

52,543,252

3

2

so vi im y lp 1

So vi im C, y1= 4+1,76 = 5,76m

Cu 2) Xc nh ln (kN/m) v im t (m) (cch chn tng C) ca tng p lc ch

ng (trn 1m tng) lp 2 (on BC) tc dng ln thn tng Xc nh cng p lc ch ng lp 2.


Ka2 = tg2(45-2/2) = tg

2(45-28/2) = 0.361


Mt trn lp 2

PaB = (1*h1+q)Ka2 2c 2aK = (18x4+100)*0.361 -2*0*0,6= 62,1kN/m

2

y lp 2

PaC = qKa2 + (1*h1 +2*h2)*Ka2 2c 2aK =(100 + 18*4+19*4)*0.361 2*0*0.7

=89,53 kN/m2.

p lc ch ng v im t ca lp 2: Ea2 = (62,1+89,53)*4/2 = 303,3 kN/m

im t = mH

ba

ba88,1

3

4

53,891,62

53,891,622

3

2

so vi im y lp 2

Vy y2= 1,88 m

Cu 3) Tnh tng p lc ngang (kN/m) tc dng ln ton thn tng (trn 1m tng),

(bao gm p lc ch ng v b ng)

Tng p lc ch ng Ea = Ea1 + Ea2 = 159,7 =303,3 =463 kN/m

Tng p lc b ng Ep

- H s p lc b ng ca lp t trc tng

Kp = tg2(45+ /2) = tg

2(45+ 28/2) = 2,77


Mt trn lp t trc tng Pp = 0

A

B

25,3kN/m2

62,1kN/m2

Ea1=159,64kN/m 4m

4m

C

54,54kN/m2

89,53kN/m2

Ea2=303,3kN/m

y1

y2

89,53kN/m2

Ep=236,8kN/m

yp


20

y lp t trc tng Pp

C = Kp**h=2,77*19*3 = 157,9 kN/m

2

Tng p lc b ng Ep = *(157,9)*3= 236,8 kN/m

im t yp= 1/3*H=1/3*3 =1m so vi im C

Vy tng p lc tc dng = Ea-Ep = 463 236,8 = 226,2 kN/m

Cu 4) Kim tra lt ti chn tng (im C)

15,088,1*3,30376,5*7,159

1*8,236

lat

giu

M

M< [FS]=3 nn tng b lt

H s an ton cho php l FS=3


21

Bi 3 : (30 pht) Mt tng chn t bng BTCT cao 8m, t sau lng tng gm

2 lp c cc c trng nh hnh v. Ti trng sau lng tng phn b kn u khp

c ln q = 100 kN/m2. t trc lng tng cao 3m, c cc c trng ging

nh lp t s 2 sau lng tng. Mc nc ngm nm cch mt t 2m.


Ka1 = tg2(45-1/2) = tg

2(45-25/2) = 0.406


nh tng:

PaA = qKa1 2c 1aK = 100*0.406 2*12*0.637 = 25,3 kN/m

2

y lp 1:

PaB = (q+ 1*2m+1*2m)*Ka1 2c 1aK = (100+ 18*2+9*2)*0.406 2*12*0.637 =

47,23kN/m2.


Ka2 = tg2(45-2/2) = tg

2(45-28/2) = 0.361


Mt trn lp 2, ti lp 2, c2=0

PaB = (1*2m+1*2m+q)Ka2 2c2 2aK = (100+ 18*2+9*2)*0.361 = 55,59kN/m

2

y lp 2

PaC = qKa2 + (1*h1 +2*h2)*Ka2 2c 2aK =(100 + 18*4+19*4)*0.361 2*0*0.7

=89,53 kN/m2.

4m

4m

Lp 1:

1 = 18kN/m3

sat,1 = 19kN/m3

'1 = 9kN/m3

= 250

c = 12 kN/m2

Lp 2:

sat,2 = 20kN/m3

'2 = 10kN/m3

=280

c = 0

q =100kN/m2

A

B

C

= 19kN/m3

=280

c = 0

3m

2m


22


Ea1 = (25,3+54,52)*4/2 = 159,7 kN/m

im t = mH

ba

ba76.1

3

4

52,543,25

52,543,252

3

2

so vi im y lp 1

So vi im C, y1= 4+1,76 = 5,76m

p lc ch ng v im t ca lp 2: Ea2 = (62,1+89,53)*4/2 = 303,3 kN/m

im t = mH

ba

ba88,1

3

4

53,891,62

53,891,622

3

2

so vi im y lp 2

Vy y2= 1,88 m

Cu 3) Tnh tng p lc ngang (kN/m) tc dng ln ton thn tng (trn 1m tng),

(bao gm p lc ch ng v b ng)

Tng p lc ch ng Ea = Ea1 + Ea2 = 159,7 =303,3 =463 kN/m

Tng p lc b ng Ep

- H s p lc b ng ca lp t trc tng

Kp = tg2(45+ /2) = tg

2(45+ 28/2) = 2,77


Mt trn lp t trc tng Pp = 0

y lp t trc tng Pp

C = Kp**h=2,77*19*3 = 157,9 kN/m

2

Tng p lc b ng Ep = *(157,9)*3= 236,8 kN/m

im t yp= 1/3*H=1/3*3 =1m so vi im C

Vy tng p lc tc dng = Ea-Ep = 463 236,8 = 226,2 kN/m

Cu 4) Kim tra lt ti chn tng (im C)

15,088,1*3,30376,5*7,159

1*8,236

lat

giu

M

M< [FS]=3 nn tng b lt

H s an ton cho php l FS=3

A

B

25,3kN/m2

62,1kN/m2

Ea1=159,7kN/m 4m

4m

C

54,52kN/m2

89,53kN/m2

Ea2=303,3kN/m

y1

y2

89,53kN/m2

Ep=236,8kN/m

yp


23

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BT chuong 3-4-5