26

2.2. Xc nh kch thc s b y mng 2.2.1 Mng n ch nht Bc 1: Gi thit mt gi tr b rng mng b. Gi tr ban u ny c th chn bt k, vi ngi cha c kinh nghim thit k c th chn trong khong 1m n 3m. Bc 2: Xc nh cng tnh ton trn nn R: C 2 cch xc nh R: tnh theo cc ch tiu c l hoc s dng bng tra. * Xc nh R theo cc ch tiu c l:

R = ' '1 2 II II II II 0tc

m m (Ab B.h. D.c .h )k

+ + (2.1)m1 - h s iu kin lm vic ca nn t, tra Bng 2.1. Lu ct pha l t loi st (t

dnh) do m1 tra theo st IL. m2 - h s iu kin lm vic ca cng trnh trong s tng tc vi nn, tra Bng 2.1. Lu

nh khung l kt cu mm m2 = 1. ktc - h s tin cy.

ktc= 1 Nu cc ch tiu c l c xc nh bng th nghim trc tip i vi t. ktc = 1,1 Nu cc ch tiu tra theo bng ca quy phm.

A,B,D - cc h s khng th nguyn, tra Bng 2.2 theo II hoc tnh theo cng thc gii tch:

II II

0,25Acot g 0,5

= + ,

II II

B 1cot g 0,5

= + + ,

IIII II

cot gDcot g 0,5

= + , (n v ca II l rad)

II II, c , II - tr tnh ton ca trng lng ring hiu qu, lc dnh n v v gc ma st trong ca t ti y mng (II = trng thi gii hn II).

h - su chn mng k t y mng n ct quy hoch (ct san nn), xc nh nh ch dn Hnh 2.8. Trng hp c cc tnh ti phn b tc dng ln mt nn hai bn mng th tr s ca h cn cng thm chiu cao ct t quy i ca ca cc ti trng .

'II - tr tnh ton trung bnh ca trng lng ring hiu qu ca t trong phm vi h:

n

i i' I 1II

h

h= =

ho= h - hth , khi khng c tng hm ho = 0. hth- chiu cao ct t quy i t y mng n mt trn sn tm hm c xt n chnh lch

gia trng lng ring ca vt liu sn so vi 'II : hth = h1+ hs . s'

II

h1- chiu dy lp t t y mng n mt di sn tng hm; hs , s- Chiu dy v trng lng ring ca kt cu sn tng hm

27

Ch : - Trng lng ring hiu qu ca t, ly bng trng lng ring t nhin cho t trn mc

nc ngm, trng lng ring y ni cho t di mc nc ngm. Ring i vi t st cng, na cng (IL 0,25), khng thm nc th ly bng trng lng ring bo ho.

- C th p dng cng thc (2.1) vi mng c hnh dng trn mt bng bt k. i vi mng c dng hnh trn hoc a gic u, tr s "b" ly bng mA , trong Am l din tch y

mng. - Khi chiu su t mng nh hn 1m, tnh ton R theo cng thc (2.1), ly h = 1m, tr

trng hp khi nn l ct bi no nc hoc t loi st c st IL > 0,5, lc ny chiu su t mng ly theo thc t t ct quy hoch.

- h. 'II = q cn gi l p lc hng mt bn mng, c tc dng chng s y tri ca mng. Trng hp p lc hng hai bn mng khc nhau, v d tn nn khng u..., th ly tr s nh hn a vo tnh ton. Mng bin nh khng

- Khi chiu rng ca tng hm ln hn 20m: R = + +'1 2 II td II IItc

m m ( Ab B.h . D.c )k

Bng 2.1 H s m1, m2

a. Mng khng nm trong phm vi tng hm; b. Mng trong phm vi tng hm khi chiu rng tng hm B 20m; c. Tng t khi B > 20m Hnh 2.8 S tnh ton chiu su t mng nh khi xc nh R

hhh

Ct san nn

Ct a hnht nhin

Ct a hnht nhin

Ct san nnCt 0,00

h =

htd

h =

htd

b)

B > 20 m

h =

htd

h1hs

hh

hsh1

h

hoht

d

B< 20 m

a)

c)

Ct san nn

Ct a hnht nhin

Ct san nn

Ct a hnht nhin

Ct a hnht nhin

Ct san nn

Ct a hnht nhin

Ct san nn

28

Loi t H s

m1

H s m2 i vi nh v cng trnh c s kt cu cng vi t s chiu di ca nh (cng trnh) hoc tng n nguyn vi chiu cao L/H bng:

4 1,5 t hn ln c cht nht l ct v t ct khng k t phn v bi 1,4 1,2 1,4 Ct mn: - Kh v t m - No nc

1,3 1,2

1,1 1,1

1,3 1,3

Ct bi: - Kh v t m - No nc

1,2 1,1

1,0 1,0

1,2 1,2

t hn ln, c cht nht l st v t st c st IL 0,5 Nh trn, c st IL > 0,5

1,2 1,1

1,1 1,0

1,1 1,0

Ch thch: 1. S kt cu cng l nhng nh v cng trnh m kt cu ca n c kh nng c bit chu ni lc thm gy ra bi bin dng ca nn, mun th phi dng cc bin php nu iu 3.75 ca TCXD 45-78. 2. i vi nh c s kt cu mm th h s m2 ly bng 1. 3. Khi t s chiu di trn chiu cao ca nh, cng trnh nm gia cc tr s ni trn th h s m2 xc nh bng ni suy.

Bng 2.2 H s A, B, D xc nh cng tnh ton R ca t nn

oII A B D oII A B D

0 0 1,00 3,14 24 0,72 3,87 6,45 2 0,03 1,12 3,32 26 0,84 4,37 6,904 0,06 1,25 3,51 28 0,98 4,93 7,406 0,10 1,39 3,71 30 1,15 5,59 7,95 8 0,14. 1,55 3,93 32 1,34 6,35 8,55 10 0,18 1,73 4,17 34 1,55 7,21 9,2112 0,23 1,94 4,42 36 1,81 8,25 9,9814 0,29 2,17 4,69 38 2,11 9,44 10,80 16 0,36 2,43 5,00 40 2,46 10,84 11,7318 0,43 2,72 5,31 42 2,87 12,50 12,7720 0,51 3,06 5,66 44 3,37 14,48 13,96

29

V d 2.1 Nn t t nhin c chiu dy v ch tiu c l cc lp nh sau:

STT Lp t dy (m) w

(kN/m3)h

(kN/m3)W

(%) WL (%)

WP (%)

II (o)

cII (kPa)

1 Trng trt 0,7 17

2 Ct pha 10 19,1 26,7 26 24 18 20

Mng nng trn nn thin nhin, y mng cos -1,95m. Nh khung b tng ct thp tn nn trong nh cao hn ngoi nh 0,45 m, trng lng ring ca t tn nn tn = 18kN/m3. Xc nh cng tnh ton ca nn cho mng gia c kch thc lxb = 2,2x1,8 m v mng bin c kch thc lxb = 2,0x1,6 m khi mc nc ngm nm di mt t t nhin : a) 1,2 m b) 4 m Gii a) Mc nc ngm nm di mt t t nhin 1,2 m. Xc nh cc ch tiu vt l ca lp ct pha cn cho tnh ton.

- st: PLL P

W W 26 24I 0,33W W 30 24

= = =

- H s rng: hw

(1 W) 26, 7(1 0, 26)e 1 1 0, 76119,1

+ += = =

- Trng lng ring y ni: h ndn26, 7 10 9, 48

1 e 1 0, 761 = = =+ + kN/m

3

Cng tnh ton ca nn:

R = ' '1 2 II II II II 0tc

m m (Ab B.h. D.c .h )k

+ +

h0 = 0 v khng phi mng di tng hm, m1 = 1,2 : y mng t trn ct pha c IL = 0,33 < 0,5 (Tra bng 2.1) m2 = 1,0 : khung b tng ct thp l kt cu mm (Tra bng 2.1) ktc = 1,0 : cc ch tiu c l c xc nh bng th nghim trc tip i vi t II = dn2 9,48 = kN/m3 : t ct pha ti y mng nm di mc nc ngm cII = 20 kPa : y mng t trn nn ct pha II = 18o tra bng 2.2 c: A = 0,43 ; B = 2,72 ; D = 5,31 Mng gia chiu su chn mng 2 bn bng nhau h = 1,95 m

n

i i' tn w1 w 2 dn 2I 1II

h 0, 45 0, 7 0,5 0,3h 1,95

= + + + = =

0,45.17 0,7.17 0,5.19,1 0,3.9,481,95

+ + += = 16,61 kN/m3

R = 1,2.1(0,43.1,8.9,48 2,72.1,95.16,61 5,31.20)1

+ + = 242,36 kPa

30

Mng bin chiu su chn mng bn tri htr = 1,5 m , bn phi hph = 1,95 m h = min(htr , hph) = 1,5 m

n

i i' w1 w2 dn 2I 1II

h 0,7 0,5 0,3h 1,5

= + + = =

0,7.17 0,5.19,1 0,3.9,481,5

+ += = 16,19 kN/m3

R = 1,2.1(0,43.1,6.9,48 2,72.1,5.16,19 5,31.20)1

+ + = 214,83 kPa b) Mc nc ngm nm di mt t t nhin 4 m. II = w 2 19,1 = kN/m3 : t ct pha ti y mng nm trn mc nc ngm Mng gia chiu su chn mng 2 bn bng nhau h = 1,95 m

n

i i' tn w1 w 2I 1II

h 0,45 0,7 0,8h 1,95

= + + = = 0,45.17 0,7.17 0,8.19,1

1,95+ += = 18,09 kN/m3

R = 1,2.1(0,43.1,8.19,1 2,72.1,95.18,09 5,31.20)1

+ + = 260,74 kPa Mng bin chiu su chn mng bn tri htr = 1,5 m , bn phi hph = 1,95 m h = min(htr , hph) = 1,5 m

n

i i' w1 w 2I 1II

h 0,7 0,8h 1,5

= + = =

0,7.17 0,8.19,11,5+= = 18,12 kN/m3

R = 1,2.1(0,43.1,6.19,1 2,72.1,5.18,12 5,31.20)1

+ + = 232,25 kPa * Xc nh R s dng bng tra: C th xc nh R ca t nn di mng c b rng b, chn su h theo cng tnh ton quy c R0 ca t nn ng vi mng c b rng b1=1m, h1=2m tra Bng 2.3, 2.4, 2.5, 2.6 ph thuc trng thi ca t, loi t. - Khi h m2 :

R = R0 1 111 1

b b h h1 kb 2h

++ (2.2)

- Khi h > 2m:

R = R0 '11 2 II 11

b b1 k k . (h h )b

+ + (2.3)

b, h: B rng v chiu su chn mng thc t k1, k2: H s k n nh hng ca b rng mng v su chn mng.

k1= 0,125 cho nn t hn ln v t ct; 0,05 nn ct bi v t st. k2= 0,25 cho nn t hn ln v t ct; 0,2 nn ct pha v st pha; 0,15 nn st.

Ch : Xc nh R s dng bng tra c chnh xc khng cao ch p dng khi khng c s liu tnh theo cc ch tiu c l.

31

Bng 2.3 p lc tnh ton quy c Ro trn t hn ln v t ct (Phm vi s dng xem iu 3.59 TCXD 45-78)

Loi t Ro (kPa)

t hn ln - t cui (dm) ln ct - t si (sn) t nhng mnh vn kt tinh trm tch

600

500 300

t ct Cht Cht va - Ct th khng ph thuc m - Ct th va khng ph thuc m - Ct mn: + t m + m v no nc - Ct bi: + t m + m + no nc

600 500

400 300

300 200 150

500 400

300 200

250 150 100

Bng 2.4 p lc tnh ton quy c Ro trn t st (khng ln t) (Phm vi s dng xem iu 3.46 TCXD 45-78)

Loi t st H s rng e Ro kPa ng vi st ca t IL = 0 IL = 1

Ct pha 0,5 300 300 0,7 250 200

St pha 0,5 300 250 0,7 250 180 1,0 200 100

St 0,5 600 400 0,6 500 300 0,8 300 200 1,10 250 100

Ch thch: i vi t st c cc gi tr trung gian e v IL cho php xc nh tr s Ro bng cch ni suy lc u theo e i vi cc tr IL = 0 v IL = 1 sau theo IL gia cc tr s Ro tm i vi IL = 0 v IL = 1.

32

Bng 2.5 p lc tnh ton quy c Ro trn nn t ln t (Phm vi dng xem iu 4.9 TCXD 45-78)

Loi t Ro (kPa)

t cu trc t nhin tng ng vi dung trng t kh k (kN/m3)

t m cht ng vi dung trng t kh k (kN/m3)

1,35 1,55 1,6 1,7

Ct pha 3

1,5 3,5

1,8 2 2,5

St pha 3,51,8

42

2,5 3

Ch thch: 1. Trong bng 2.5, t s l gi tr Ro thuc t ln t cu trc t nhin c no nc G

0,5 v khi khng c kh nng thm t chng. Mu s l gi tr Ro thuc t nh trn nhng c no nc G 0,8 v t c no nc b khi c kh nng thm t chng.

2. i vi t ln st c cc gi tr k v G trung gian th Ro xc nh bng ni suy.

Bng 2.6 p lc tnh ton quy c Ro trn nn t p n nh (Phm vi dng xem iu 10.6 TCXD 45-78)

Loi t

Ro (kPa) Ct th, ct trung, ct mn x v.v...

Ct bi, t st tro v.v...

ng vi bo ho G G 0,5 G 0,8 G 0,5 G 0,8

t trong lc san nn m cht theo iu 10.8

250 200 180 150

Cc bi thi t v ph liu sn xut sau khi m cht theo iu 10.8

250 200 180 150

Cc bi thi t v ph liu sn xut khng m cht

180 150 120 100

Cc ni t v ph liu sn xut sau khi m cht theo iu 10.8

150 120 120 100

Cc ni t v ph liu sn xut khng m cht

120 100 100 080

Ch thch: 1. Tr s Ro bng 2.6 l ca cc mng c su t mng h1 =2m. Khi su t

mng h< 2m gi tr Ro s gim bng cch nhn vi h s: 112

+= h hkh

.

2. Tr s Ro 2 im sau cng trong bng 2.6 l thucv t rc v ph liu sn xut c cha tp cht hu c khng qu 10%.