14

chng ii

mng nng trn nn thin nhin $1.Khi nim chung: t trc tip ln nn thin nhin. Mng xy trong h mng o trn ( khong di 2-3m). Thi cng n gin. Trong t/ton b qua s lm vic ca t t y mng tr ln. * Tu theo tnh hnh tc dng ca ti trng ta phn thnh: Mng chu ti ng tm Mng chu ti lch tm * Tu theo kh nng chu un ca mng ngi ta chia mng lm 2 loi: Mng cng: L loi mng t hoc khng chu un Mng mm: L loi mng chu un nhiu. Tnh ton 2 loi mng ny hon ton khc nhau. Mng cng ch yu chu nn, mng mm ngoi kh nng chu nn cn c kh nng tip thu /s ko. Vt liu 2 mng cng khc nhau: Mng cng: B tng, b tng hc, , gch. Mng mm: BTCT

* Ngoi ra cn c vo phng php th/cng chia ra: Mng ton khi Mng lp ghp Mng cng tnh ton tng i n gin Mng mm tnh ton phc tp tnh ton nh KC Dm & Bn t trn nn n hi. $.2. Cu to mng phm vi s dng:

1. Mng n: y hnh vung, ch nht, trn. Vt liu: Gch xy, b tng, b tng ct thp.

Do kh nng chu un km nn cc mng: Gch, , B tng cn cu to kch thc thch hp xem n l loi mng cng m khng phi xem xt ti kh nng chu ko do un.

15

*L ln th M cng ln, mng c th b gy theo mt mn. *Do H cng cn phi ln.

Vt liu c cng nh H cng phi cng ln. Thng da vo gc m quy nh mng cng

hay mm (tc l t s H/L hoc h/l). Trn c s kinh nghim KC mng khng xut hin: Vt nt do ng sut ko gy ra th gc m khng c > max nht nh, ngha l t s: H/L i vi ton mng Hoc h/l i vi mi bc khng c nh hn tr s trong bng sau: * Vi mng BT, BT hc, hc:

p lc trung bnh di y mng

p 1,5kg/cm2 p> 1,5kg/cm2 Mc b tng

Loi mng

16

* Vi mng bng BTCT th khng cn khng ch t s H/L m cn c vo kt qu tnh ton theo nguyn tc mng mm xc nh kch thc ca mng v ca ct thp. Nu ch t thp bc cui cng th cc bc bn trn phi c t s h/l >1

h V.liu: Gch: 14cm; 21cm; 28cm... xy: cho 2 lp xy: 35 60cm. B tng hc: h 30cm. + Mng ton khi: - Ct chu lc ca ct c th cm xung n li thp y mng, nhng d thi cng ngi ta thng t thp ch v b tng phn ct cho n cao trnh ct nn ( +0,00) Thp ch ct c ng knh bng hoc ln hn ng knh ct thp dc trong ct, vic ni ct ch vi ct chu lc phi m bo quy nh v ct v ni ct thp, ngha l khi ni buc th trn mt tit din trong phm vi lneo khng c ni > 50% din tch ct thp nu l ct c g v khng > 25% nu l ct trn. - Ct thp di mng theo phng cnh di t di. Vi mng c cnh a 3m cho php c mt na s thanh khng phi ko ra mp mng chiu di = 0,8a - Trong phm vi chiu cao mng cn c t nht 2 ct ai, mt ct ai st y v mt ct ai cch mt trn ca mng 100mm. - Chiu cao chung ca mng v cc bc phi c kim tra v chc thng v chiu cao mng phi neo chc ct dc vo mng mt on khng nh hn lneo - Chiu dy ca mng mp ti thiu phi l cm66 + ; ng knh thanh thp ln nht di y mng tnh bng cm v bng khong 2/5 chiu cao mng b tng khng b chy xung khi thi cng.

500

800

600

1400

2400

2000

400

300

400

100

350

250

5050

100

0.000

-1.400

500

800

600

1400

2400

2000

400

300

400

100

350

250

5050

100

0.000

-1.400

17

+ Mng lp ghp:

18

Y/cu: - Ct n: hc: Chiu su cc hc ak + 0,05m - Ct 2 nhnh: hc 1,5ak

- Chiu su ngm ct vo mng phi 30d; d: ng knh ct thp dc trong ct - Chiu dy thnh cc 200mm; Chiu dy b tng t y cc n y mng

200mm. - B tng mc khng < 200#

2. Mng bng:

Mng c chiu di rt ln so vi chiu rng. Do cu to lin tc ca cng trnh bn trn nh tng nh, tng chn th dng mng bng l ng nhin. Cn di hng ct th nu dng mng n kch thc ln n mc gn nhau th tt nht l dng mng bng.

19

u: - Gim p lc y mng. - Phn b ti tng i u n ln nn. - Nu cng mng c tc dng lm gim chnh lch ln gia cc ct. C th cu to mng bng giao nhau.

Vt liu: gch, , BT hc, BTCT. Mng bng cng gc m c th ly > 2o 3o so vi tr s cho mng n. Vi mng bng di tng khng cn xt n cng ca mng bng theo

phng trc mng Vi mng bng di hng ct phi xt n cng ca mng theo phng dc

trc.

3.Mng b (Bn): Mng bn c kch thc va di va rng.

KC bn trn c th nm gn trn mt bn mng lin tc hoc nhiu bn ghp li vi nhau

Mng bn thng lm bng BTCT mng c kh nng chu un theo 2 phng nn n c dng trong trng hp:

t nn c cng thp. Ti cng trnh ln v phn b khng u.

Mng c tc dng phn b ti trng ln mt nn u hn nn pht huy c ht kh nng lm vic ca t nn. Mng b c th lm dng bn phng hoc bn sn. tng cng chu un ca mng bn c khi ngi ta dng mng bn kiu vm ngc. c im ca mng vm ngc l nu chn c trc vm ca bn thch hp th

20

di tc dng ca ti trng cng trnh v phn lc nn mng ch chu nn dc trc m khng chu un tuy nhin tnh ton phc tp. * Mng hp: L hp rng di ton b cng trnh, n va lm tng hm. C tc dng nh mng bn Nhng c u im hn ch c cng ln, c kh nng phn b ti trng t min gia ra vng bin. Tuy nhin c cu to phc tp v tn thp. Dng hp l cho cc nh cao tng vi nhiu tng hm su. Vt liu mng:

Mng s dng b tng Mc 250. B tng lt mc 100, thp AI hoc AII vi ng knh 10mm

4. Cu to ging mng: Thng cu to nhm tng cng cng trnh gim chnh lch ln

H ging chn theo kinh nghim tu thuc vo Iu kin a cht v li ct

21

$.3. tnh ton thit k mng nng cng: Mng cng l mng c cng ln di tc dng ca ti trng cng trnh mng bin dng nh c th b qua ng sut tip xc di y mng coi l tuyn tnh. C th coi: Mng n di ct, tr; mng bng di tng l mng cng. Ni dung c bn trong thit k l xc nh cc c trng ca mng gm:

Vt liu; su t mng; Kch thc mng. Sao cho tho mn cc yu cu k thut, thi cng v kinh ttrnh t:

Bc 1

Ti liu

- Cng trnh - Nn - Cc tiu chun thit k

Bc 2

Phng n H mng nng

- Nn - Cngmm - n, bng, b ...

Bc 3

Vt liu mng

- #, thp Rn, Rk, Fa... - Lt lp bo v ao

Bc 4

su mng hm (xem phn 6 chng mt) Bc 5

Chn kch thc mng

b h (bng di tng) b l h (n) B L h (b)

- ptc Bc 6 - ttp0 (khng k TLBT)

ng sut di mng - ptcgl = p

tc - 1hm 1 (Dung trng lp 1)

Bc 7

Kim tra kch thc y

1- Kh nng chu ti: trt, lt v iu kin kinh t v bl. 2- Bin dng: SSgh

Bc 8 Kim tra chiu cao mng v Fa

Tnh ton cng vt liu mng: H, Fa. m bo iu kin kinh t.

Bc 9 Cc kim tra khc

Bc 10

Cu to v Bn v

Vd: trt ngang, lt, c kt chm

22

Tnh ton: Xc nh kch thc y mng tho mn 2 iu kin: m bo n nh v sc chu ti ca nn: TTGH1

Bin dng trong phm vi cho php: TTGH2 A. Tnh ton nn mng theo trng thi gii hn I: Phi chn kch thc y mng sao cho ng sut do ti trng tnh ton ti mc y mng khng c vt qu sc chu ti gii hn ca nn. Trn c s xem xt ti liu a cht, Cng trnh ta gi nh kch thc v su t mng: l; b; hm Xc nh kh nng chu ti ca nn: ( Theo nhng phng php hc trong C hc t)

a. i vi nn : R = k.m.R

R: cng gii hn tm thi ca mu chu nn 1 trc trong trng thi bo ho nc. k,m h s ng nht v h s iu kin lm vic ly km =0,5

b. Vi nn t: C mt s phng php sau:

Phng php 1: Dng l thuyt CBGH v nhng kt qu ca li gii na thc nghim, nhng kt qu ny c lp cho nn t ng nht, hoc cng phi c mt lp t ng nht kh dy di y mng (khong 3 ln b rng mng hoc 1 n 1,5 ln b rng ca nh hay cng trnh) thng thng hay dng kt qu ca Terzaghi. Phng php 2: Nn t khng ng nht, gm 2 hoc 3 lp t c cc ch tiu cng khc nhau; ph ti 2 bn mng chnh lch nhau qu 25%; mng t trn mi dc, t di mi dc hoc t trn mt tng t phn b rt dc th trong nhng trng hp ny phi dng phng php gii vi vic gi thit mt trt xc nh sc chu ti ca nn, thng thng hay dng phng php mt trt tr trn. ( cho n gin trong trng hp nn nhiu lp c th lm theo cch gn ng sau: Tnh ton vi lp th nht xong th chuyn sang kim tra lp th hai bng cch to ra mt mng khi quy c trn mt lp hai tnh ton)

23

Phng php 1: Theo Terzaghi cng ti trng gii hn trung bnh trong trng hp:

Bi ton phng nh sau: cqu cNhNbNP ++= .5,0

Bi ton khng gian: ccqqu cNshNsbNsP ....5,0 ++=

Trong : cq NNN ;; cc h s sc chu ti ca nn ph thuc vo: c v tra bng. cq sss ;; : h s hnh dng.

lb

s .2,01= ; ;1=qs lb

s .2,01+=

Nhng nghin cu v sau cho kt qu xc nh c ti trng gii hn cho nhng trng hp ti trng tc dng ln mng c dng phc tp, mt t dc hoc su t mng tng i ln (tham kho ti liu)

Cng ti trng cho php ca nn t c chn: s

u

FPR =

Fs - h s an ton thng ly t 2 - 3. Phng php 2: * Cng trnh xy dng trn nn t yu: + M hnh trt su: mt trt gi nh : - tr trn: ABCDE - hn hp: ABCDE

h s n nh c tm theo cch loi dn: C t + Trng hp n gin: coi ti phn b trn mt din tch quy c no sau kim tra tng t nh mng nng trn nn thin nhin.

B

h

A

C'

C

D

E

mh

h

oN

oQ

Lp t yuy

24

hhm

hy

No

Cng trnh xy dng trn mi dc hoc t nm trn gc c mi nghing:

- kim tra mt trt tr trn trong lp t - kim tra mt trt phng nh trc ti cc mt phn chia t -

1. Ti trng ng tm:

Mng n: iu kin: P tb R Ptb: p lc tiu chun di y mng. R: cng ti trng cho php ca nn t.

lb

GNP

tc

tb

+=

0

tcN 0 : ti trng tiu chun do cng trnh truyn xung ti chn ct G: trng lng mng v t trn mng: G = (tb hm F); (tb=20KN/m3)- dung trng trung bnh ca mng v t trn mng l,b: cnh y mng. F: din tch y mng.

Mng bng: b: b rng mng l: chiu di mng ( ly 1 on no tnh cho tin lu : khng ct ra 1 on ri tnh nh mng n)

mtb

tc

tb hlbN

P += 0

tcN 0 : ti trng tiu chun do cng trnh truyn xung ti chn tng. 2. Ti lch tm: iu kin: P tb R

Pmax < 1,2 R

)661(0minmax, be

le

FN

P batc

=

ea; eb lch tm theo cc cnh y mng *Nu cc iu kin trn ph hp th thi nu cha ph hp chn li l,b,hm v tnh li.

25

C th s b chn kch thc y mng theo cch sau:

+ Vi t s gia 2 cnh mng ch nht bl

= ;

c th chn: = (1+e) (1+2e); ( e= Mo/No) Mo;No : m men v lc dc ti chn ct.

+ Chn kch thc b no , tnh ptb v R + Gii ph/trnh: RPtb = theo b, chn b theo nghim ca ph/trnh. Kim tra b theo chun: RP 2,1max Nu khng tho mn thay b ln hn cho n khi tho mn.

iu kin hp l v kch thc: [ ] [ ]{ }RpR %52,1 max Vn n nh v v tr ca mng: * m bo mng khi b trt theo mt y mng, phi tha mn iu kin sau y:

tttt

d TfN > Trong : ttdN - tng ti trng ng tnh ton ti mc y mng. ttT - tng ti trng ngang tnh ton. f - h s ma st tnh ton gia mng v t nn- tham kho bng:

H s f gia mng xy hoc mng b tng vi t nn.

Tn t f

t st cng t st do t ct t m t ct m t st cng t st do t ct cng t ct do

0,30 0,20 0,55 0,45 0,45 0,25 0,50 0,35 0,75

* m bo mng khng b quay quanh mp A phi tha mn iu kin:

quaygiu MM > Trong :

giuM - m men ca cc lc so vi tm quay ( im A) c xu hng gi mng. quayM - m men ca cc lc so vi im A, c xu hng lm cho mng b quay. Ch : Do trong cc ngnh xy dng cn cha thng nht v vic tnh ton nn theo trng thi gii hn th nht- nn vi cc cng trnh cu ng v thy li c th tham kho cc ti liu chuyn ngnh.

M

26

Bi 1: Xc nh kch thc mng di ct theo iu kin v sc chu ti ca nn, ct tit din (3040) cm2 vi t hp ti trng tiu chun mc mt t No = 45 T; Mo = 3,5Tm v Qott= 1,5 T. Nn t gm 2 lp c cc ch tiu c l c bn nh sau: Lp trn: t lp dy 0,8m, = 1,8T/m3 Lp di: st do cng - trng lng th tch n v = 1,85T/m3 - gc ma st trong = 23o, lc dnh n v c = 2,2 T/m2 - h s an ton ti thiu Fs=2 Bi lm:

1200

1400

tt

Qo

ttNo

tt

Mo

0.000

-1.000

-0.800

su t mng chn s b hm = 1,0m

Chn t s: = a/b: lch tm ca ti trng: e = Mo /No M = Mo+Qo. hm =3,5Tm + 1,5T.1m = 5Tm

e = Mo /No = 3,5/45 = 0,08m = 1+ 2e = 1,16 1,2 ; Chn: = 1,2; kch thc b = 1,2m

a = 1,2 1,2 = 1,4m Xc nh ti trng cho php tc dng ln t

s

u

FPR =

Trong : cccqqqu NcisNqisNbisP ...........5,0 ++= cq NNN ;; : cc h s sc chu ti ca nn ph thuc vo tra bng.

27

cq sss ;; : h s hnh dng

s = 1- 0,2/ = 1- 0,2/1,2 = 0,83 sq = 1 sc = 1+ 0,2/ = 1+0,2/1,2 = 1,17

h s iu chnh nghing ca ti trng: 2

1

=

i 2

21

==

pi

cq ii

(gn ng coi l ti thng ng ( tc l coi = 0) nn ( 1=== cq iii ) Vi = 23o, tra bng ta c N = 7,73 Nq=8,66 Nc= 18,1 { } 2/3,631,18.2,2.1.17,166,8.1.8,1.1.173,7.2,1.85,1.1.83,0.5,0 mTPu =++=

Chn Fs = 2 2/65,31

23,63

mTFPR

s

u===

Tnh pmax; ptb:

2222max /54,4104,1.2,1

5.61.22,1.4,1

45.

.6

.

.6.

.

mTba

M

abMh

baNP yxmtbo =+++=+++= (My=0)

2/8,281.22,1.4,1

45.

.

mThba

NP mtbotb =+=+=

1,2.R= 1,2 . 31,65 = 37,98 T/m2 So snh:

Ptb= 28,8T/m2 < R = 31,65 T/m2

Tuy nhin Pmax = 41,54 > 1,2.R= 1,2.31,65 = 37,98 T/m2

Vy la chn s b b = 1,2m v a = 1,4m l khng t yu cu v mt cng . chn v tnh li.

Bi 2: Cho nn t gm 3 lp :

Lp 1: t lp dy 0,8m; = 1,8T/m3 Lp 2: st do cng, dy 2,2m - trng lng th tch n v = 1,85T/m3 - gc ma st trong = 23o , lc dnh n v c = 2,2 T/m2 Lp 3: t st do nho c cc ch tiu c l nh sau:

- trng lng th tch n v = 1,8T/m3 - gc ma st trong = 5o , lc dnh n v c = 0,8 T/m2

- h s an ton ti thiu Fs=2 Ti trng tiu chun ti mc mt t:

No = 45 T; Mo = 3,5Tm v Qo= 1,5 T. Bi lm

Bc 1: Vic tnh ton kch thc y mng ti mt lp 2 lm tng t nh trn:

Kch thc mng: F= (1,2.1,4)m2

gi s l hp l.

28

Bc 2: Do khng su di y mng c lp t yu nn ta phi kim tra p lc ln b mt lp t yu . To mng khi quy c: b q = b + 2.h*.tg (trong c th ly bng gc ma st trong ca lp 2 ) = 23o h*= 2m ( chiu dy t y mng n b mt lp t yu)

tg = tg 23o = 0,4245 b q = 1,2 + 2. 2. 0,4245 2,9m h q = hm + h* = 1,0 + 2,0 = 3,0m

Tng t: a q = 1,4 + 2.2.0,4245 3,1m

2/8,281.22,1.4,1

45.

.

mThba

NP mtbotb =+=+=

Kim tra p lc ln lp t 3:

Xc nh ng sut trn mt lp t 3: 3** dhhz

bthhz Rmm + +=+=

M

2900

3100

3100

tt

Qo

ttNo

tt

Mo0.000

-1,000

-3.000

1400

-0.800

2323

-3.000

29

233

2211 /51,52,2./85,18,0./8,1.. * mTmmTmmthhbt hhz m =+=+=+=

* . .m

o tb mz h h k p h

= +

=

Trong

- dung trng trung bnh ca t t y mng tr ln:

31 1 2 21 2

. . 1,8.0,8 1,85.0,2 1,81 1,81 /0,8 0,2 1

h h T mh h

+ +

= = = =

+ +

a/b =1,4/1,2 1,2; z/b = 2/1,2 = 1,7 Tra bng ni suy: ko = 0,184

2/97,4)1.81,18,28.(184,0*

mThhz m

==

+=

ng sut trn b mt lp t 3 l: 5,51 + 4,97 = 10,48T/m2 Xc nh cng t nn mt lp 3: Tng t nh trn:

cccqqqu NcisNqisNbisP ...........5,0 ++= Vi = 5o; tra bng ta c N = 1 Nq=1,56 Nc= 6,47

= a q /b q = 3,1m/2,9m = 1,071,1 do : s = 1- 0,2/ = 1- 0,2/1,1 = 0,82 sq = 1 sc = 1+ 0,2/ = 1+ 0,2/1,1 = 1,18

gn ng coi l ti ng ( tc l = 0) nn: 1=== cq iii

3

21

2211 /84,12,28,0

2,2.85,18,0.8,1.. mt

hhhh

=

+

+=

+

+=

Thay s; { } 2/85,1647,6.8,0.1.18,156,1.3.84,1.1.11.9,2.8,1.1.82,0.5,0 mTPu ++=

2/42,8285,16

mTFP

Rs

u===

R = 8,42 T/m2 < 10,48T/ m2 Nh vy kch thc mng trn khng m bo. Ta tng kch thc y mng sau khng cn tnh ton bc 1 na m i tnh ton kim tra nh bc 2 lun. Bi 3: Xc nh s b mng bng di tng dy 40 cm. t nn:

Lp trn: t lp dy 0,8m. = 1,8T/m3 Lp di: st do cng

30

- trng lng th tch n v = 1,85T/m3 - gc ma st trong = 24o, lc dnh n v c= 2,2 T/m2 T hp ti trng tiu chun mc mt t No = 18T/m v Mo= 2,2 Tm/m

h s an ton ti thiu Fs=2 Bi lm: Chn b = 1,4 m; ly ra 1m tnh p lc y mng

2/86,14286,120,1.24,1.1

18.

.

mThba

NP mtb

otb =+=+=+=

Tnh Pmax: 2

22max /59,2173,6286,124,1.16.2,20,1.2

4,1.118

.

6..

.

mTba

Mhba

NP mtb

o=++=++=++=

Xc nh ti trng cho php tc dng ln t

s

u

FPR =

Trong : cqu NcNhNbP ......5,0 ++=

-0.800

400

1000

300

700

1400

tt

Qo

ttNo

tt

Mo

0.000

-1.000

1000

Vi = 24o, tra bng ta c N = 8,97; Nq= 9,6; Nc= 19,3 Tng t nh trn ta c kt qu sau:

{ }2/4,7136,7146,4228,1762,113,19.2,26,9.1.8,197,8.4,1.85,1.5.0

mT

Pu=++

=++=

Chn Fs = 2 2/7,35

24,71

mTFP

Rs

u===

So snh R vi Ptb ta thy Ptb = 14,86 < R=35,7 T/ m2

So snh 1,2.R vi Pmax ta thy Pmax = 21,59

31

B. Tnh ton nn mng theo trng thi gii hn II:

D tnh ln:

C nhiu phng php nu trong C hc t ( thng dng phng php cng ln phn t)

Ni dung: + Xc nh p lc gy ln: mtbgl hpp =

ptb: p lc tiu chun di y mng. hm: chiu su chn mng. : dung trng ca lp t t y mng tr ln. + Tnh v v biu . + Chia chiu su vng nh hng thnh tng lp tnh ln + ln:

i

i

n

SS1==

Tu theo s liu c cung cp m chn cng thc tnh ln:

ii

iii

n

he

eeS1

211

1+

==

p dng cho tr/ hp c kt qu th nghim nn mt chiu.

Trng hp c cc th nghim khc, ln d bo theo m hnh: nn n hi hoc lp n hi

21 ogl const

o

S p bE

= hoc 2

11

1 ( )n

oigl i i

i oi

S p b k kE

=

=

Trong E; Eoi c xc nh t th nghim CPT, SPT, PLT

1. Kim tra chnh lch ln:

Chnh lch ln gia 2 mng n k nhau: Tuyt i: S = S1-S2

Tng i: L

SSS 21 =

S1; S2 - ln tuyt i ca hai mng L: khong cch gia tim hai mng

Chnh lch ln do mng nghing. Chnh lch ln do cng trnh b un.

Ngoi ra cn lu n chnh lch ln thay i theo thi gian- Do t dnh bin dng ko di theo thi gian (c kt). V vy khi thit k cn phi xt n tc bin dng ca mng.

Bi v tr chnh lch ln tnh theo cc tr s n nh cha hn l tr chnh lch ln nht. Do tm chnh lch ln thay i theo thi gian c ngha quan trng.

32

* Cc trng hp nh sau cn xem xt: Cc b phn CT c thi cng xong a vo s dng khng ng thi. t nn c tc c kt chm (Cv < 110

7 cm2/nm), hoc khc nhau mi ni. Chiu dy tng t chu nn di mng khc nhau.

Bi 4: Kim tra iu kin gii hn v ln ca mng la chn trong bi 1 cho bit m uyn bin dng Eo=1500T/m

2, o=0,3. ln cho php ca mng, [S]= 3cm. Nu iu kin bin dng khng tho mn, hy xut phng n x l. Bi lm: Mng c t su 1,0m do ln ch do lp t th 2 gy ra, c th p dng cng thc d bo ln ca nn ng nht.

o

o

E

bpS

)(.. 21 . =

Trong p l ti trng gy ln xc nh theo gi tr tiu chun:

mtbgl hppp .

'==

Trong

- dung trng trung bnh ca t t y mng tr ln:

321

2211 /81,1181,1

2,08,02,0.85,18,0.8,1..

mThh

hh==

+

+=

+

+=

2/86,27286,251.22,1.45,1

45).( mThba

NP mtb

otb =+=+=+=

p = 27,86-1,81.1=26,05T/m2 Vi =1,2 tra bng ta c o= 1,22.

ln ca mng d bo s l:

cmmE

bpSo

o 3,2023,01500

)3,01.(22,1.2,1.05,26)1(... 22==

=

=

ln d bo (S=2,3cm

33

Th nghim nn cho kt qu nh sau: P(kg/cm2) 0 1 2 3 4

H s rng e 0,544 0,360 0,268 0,218 0,205

Lp 2: Ct ht nh 1,8T/m3; qc=50kg/cm2

Bi lm:

2 kg/cm2

Ct ht nh

st pha

P2 =

zP1

Phn t

8

7

6

5

4

3

2

1

0

po =2,4 kg/cm2

1,816 kg/cm2

1,468 kg/cm2

1,204 kg/cm2

0,94 kg/cm2

0,698 kg/cm2

0,648 kg/cm2

0,508 kg/cm2

0,304 kg/cm2

1,204 kg/cm2

1,468 kg/cm2

1,0 kg/cm2

0,8 kg/cm2

1,86 kg/cm2

1,68 kg/cm2

1,5 kg/cm2

1,4 kg/cm2

1,2 kg/cm2

1 kg/cm2

0,8 kg/cm2

0,6 kg/cm2

0,4 kg/cm2

1000

1000

500

1000

1000

1000

1000

1000

Phn t

1000

-7.500

-2.000

0.000

P1 + z

4000

34

* Xc nh p lc gy ln: pgl =po- .hm; pgl=2,4 - 2. 0,2 = 2kg/cm

2 * V biu p lc bn thn ca t v biu ng sut ph thm. Chia nn t ra thnh tng lp phn t vi chiu dy hi b/4. y ta chia: Lp 1: thnh 6 lp phn t vi 5 phn t u hi = 1m; cn lp phn t cui hi =0,5m. Lp 2: thnh 4 lp phn t vi hi= 1m. Tnh p lc bn thn ca t ti cc im 1,2,3.... theo cng thc:

bt= i .( hm + zi) trong : bt - p lc bn thn ca t ti im i i - trng lng n v ca lp t cha im i zi- chiu su k t y mng ti im i hm- su t mng Tnh ng sut ph thm ti cc im 1,2,3.... theo cng thc i= ko.p trong : zi - ng sut ph thm ti im th i p - p lc tnh ln ko - h s ng sut tm mng, ph thuc vo cc t s a/b v z/b

35

Kt qu tnh ton c lp thnh bng nh sau:

Lp im

tnh Zi(m) bt(kg/cm

2) a/b z/b ko zi

I 0 1 2 3 4 5 6

0 1 2 3 4 5 5,5

0,4 0,6 0,8 1,0 1,2 1,4 1,5

2 2 2 2 2 2 2

0 0,25 0,5 0,75 1

1,25 1,375

1 0,908 0,734 0,602 0,470 0,349 0,324

2,0 1,816 1,468 1,204 0,940 0,698 0,648

II 7 8

6,5 7,5

1,68 1,86

2 2

1,625 1,875

0,254 0,152

0,508 0,304

Tnh ln: * Xc nh chiu su vng chu nn. Ta thy chiu su z = 7,5m tng ng vi im 8 th tr s ng sut bn thn bt8 = 1,86 kg/cm2 v tr s ng sut ph thm z8 = 0,304 kg/cm

2 tho mn iu kin: 0,2. bt8 > z8. Do vy, ta ly chiu su vng chu nn Hc = 7,5m. ( Vi E=100kG/cm

2 coi l t tt ly chiu su vng chu nn)

* Tnh ln theo cng thc: in

i

ii he

eeS +

=

1 1

21

1 Cho lp t 1 - t dnh

Trong : S - ln n nh cui cng ca trng tm y mng e1i; e2i h s rng ca t ng vi p1i v p2i

Trong : 2

11

btibtiip

+=

+= ziii pp 12

2

1 zizizi

+=

hi - chiu dy lp t th i

izio

hE

S ..

=

Cho lp t 2 - t ri

Trong : - h s tnh t h s poisson ca t:

=

121

2

C th ly = 0,8

36

21 zizi

zi

+

=

hi - chiu dy lp t th i Eo = . qc Ct ht nh qc= 50kg/cm

2 tra bng chn: =2 Eo = 2 . 50 = 100 kg/cm

2 Kt qu tnh ton c trnh by trong bng sau y:

Tng

hi(m)

p1(kg/cm

2)

P2(kg/cm2)

e1i

e2i i

n

i

iii h

e

eeS +

=

1 1

211

(cm) 1 2 3 4 5 6

1,0 1,0 1,0 1,0 1,0 0,5

0,5 0,7 0,9 1,1 1,3 1,45

2,408 2,342 2,236 2,172 2,119 2,123

0,44 0,4 0,37 0,35 0,33 0,31

0,25 0,246 0,253 0,255 0,260 0,268

13,2 11 8,5 7,0 5,3 1,6

7 8

1,0 1,0

Eo(kg/cm

2)

100 100

z(kg/cm

2)

0,578 0,406

izi

oi hE

S

=

0,46 0,32

Vy ln bng: S = Si = 47,4cm

Bi 6: Kim tra iu kin gii hn v ln ca mng la chn trong bi 3 cho bit m uyn bin dng Eo=1500T/m

2, o=0,3 ln cho php ca mng, [S]= 3cm. Bi lm: Mng c t su 1,0m do ln ch do lp t th 2 gy ra, c th p dng cng thc d bo ln ca nn ng nht.

o

o

E

pbS

)( 21 =

Trong p l ti trng gy ln:

mtbgl hppp .

==

37

2/86,141.24,1.1

18.

.

mThba

NP mtb

o

tb =+=+=

Trong

- dung trng trung bnh ca t t y mng tr ln:

321

2211 /81,1181,1

2,08,02,0.85,18,0.8,1..

mThh

hh==

+

+=

+

+=

p = 14,86 - 1,81.1= 13,05T/m2 Vi mng bng cng tra bng ta c const = 2,12 ln ca mng d bo s l:

cmmE

bpSo

o 3,2023.01500

)3,01.(12,2.4,1.05,13)1(... 22==

=

=

ln d bo (S=2cm

38

Trng hp 2:

* Vi mng n chu ti ng tm:

otbk hbRP 75.0 t tm ra ho H = ho+ a( lp bo v) Rk - cng chu ko tnh ton ca b tng.

ho - chiu cao lm vic ca mng btb - gi tr trung bnh s hc ca chu vi pha trn v pha di ca thp m thng: ( )occtb hbhb 22 ++=

P- lc m thng xc nh theo tnh ton. Phn p lc di mng nm trong phm vi thp m thng ch gy lc p cho thp m khng c tc dng ct b tng theo mt nghing ca thp.

N l lc dc tnh ton tit din chn ct P = N - Ft. p Trong : Ft din tch y ca thp m thng,

)2)(2( ococdt hbhhF ++= p - p lc di mng do ti trng tnh ton gy ra.

PmF

N=

Fm din tch mng.

39

( y b qua trng lng t v mng nm pha trn thp m thng.) * Chiu cao bc cui cng phn con son nm ngoi thp m thng xc nh t iu kin m bo b tng chu ct m khng cn t ct ngang.

018.0 hRcp kd c- vn ca bc di ra ngoi thp m thng

oc hhac = )(5.0 * Vi mng n chu ti lch tm:

Xc nh tng t xt v pha pmax thin v an ton Trong :

octb hbb += ( nu )2 bhb oc + hoc 2bbc +

= ( nu )2 bhb oc >+

dtF Din tch phn gch cho hnh trn.

'ttctdt pFP = 21max'tttt

tt ppp +=

Lu : Nu cu to mng c 2 mi dc th chiu cao mp ngoi h ca mng ly theo ng knh ct thp mng: h 6 + 6 (h v tnh bng cm) - /k thanh thp ln nht trong mng v bng 2/5 chiu cao tnh ton ca mng khi b tng khng b chy xung. Trng hp 3: Tnh bn chu un ca mng tnh ton ct thp y mng: Vi mng n chu ti ng tm:

Xem mng lm vic nh nhng bn con son b ngm tit din chn ct, tit din git cp. Tnh cho c 2 phng, mi phng phi tnh cho cc loi tit din k trn.

A

B

D

C

E

F

G

H

I

K

B1 B

L1

L

1

1

2

2

1 2

1 2

40

V d: Tnh ct thp theo phng cnh a: M men un trn tit din I-I l M1 s do phn phn lc t trong phm vi hnh thang ABCD gy ra, Tng t M2 do EBCF . Tuy nhin n gin tnh ton:

thin an ton M1 BCHG; M2 BCKI M1= 0,125p b (l-hc)

2 M2= 0,125p b (l-a1)

2

Vy: oa

a hRMF

9.0= M ly gi tr M1 tnh fa1; M2

tnh fa2 Hm lng ct thp khng nh hn min i vi cu kin chu un. Vi mng bng: Ct chu lc t theo phng ngang Ct cu to t theo phng dc. Tuy vy nu k n s ln khng u theo phng dc tng do a cht thay i, cng nh khi c khot l ca th ct thp t theo phng dc s phi chu lc. nn khi khi nn phc tp thng cu to thm sn v t ct thp dc trong sn. Vi mng n chu ti lch tm: Cng ging nh i vi mng chu nn ng tm nhng phi thay P bng p tb V d:

Nh hnh bn Fa1 tit din I-I c tnh theo 21max

1ppp tb

+=

Nh hnh bn Fa2 tit din II-II c tnh theo 22max

2ppp tb

+=

Ct thp theo phng cnh b c tnh theo: 2

minmax pppbtb+

=

Nhng ch : + Tnh ton thit k mng bng cng tng t nh mng n. im khc nhau chnh l ti trng xc nh trn 1 n v chiu di mng: No ( Lc / chiu di); Mo ( Lc. Chiu di / chiu di). Cc tnh ton thc hin trn mt n v chiu di mng Ch khng phi l ct ra mt n v chiu di mng.

A

B

D

C

E

F

G

H

I

K

B1 B

L1L

1

1

2

2

1 2

1 2

tt

No tt

Mo

Lng

41

Bi 7: Tnh ton chiu cao mng. Cho mng: kch thc (3.2)m2; hm = 1,2m; ct tit din (20.30) cm

2 Ti trng tnh ton ti mt t:

Nott= 100T Mott= 12Tm Qott= 5T

Dng B tng M# 200; Rn= 90kG/cm

2; Rk= 7,5kG/cm2

Bi lm: Tnh ng sut y mng do ti trng cng trnh gy ra:

200

300

1200

ttQo

ttNo

ttMo

0.000

-1.20045

450900

900

200

900

2000

3001350

3000

500

660

40

660 450

860

1520

240

w

MF

Np

ttott =

minmax

Mtt= Mott + Qott . hm = 12 + 5.1,2 = 18Tm

42

67,163.26.18

2.3100

2minmax ==p

Pmax = 22,7 T/m2

Pmin = 10,7 T/m2

Ptb= 16,7 T/m2

lp bo v a = 4cm gi thit H= 70 cm

vy ho = H- a = 70 - 4 = 66 cm iu kin kim tra: Pt 0,75.Rk. btb. ho Pt lc m thng: gn ng c ly l hp lc phn lc t trong phm vi gch cho

dtdt Fpp

P2

max

* +=

lll

pppp dt

+= )( minmaxmin*

Ft lt.b

mhall ocdt 69,066,023,03

2=

=

=

2* /94,1924,97,103

69,03)7,107,22(7,10 mTp =+=+= 238,12.69,0. mblF dtdt ==

TPdt 4,2938,1.27,2294,19

=

+=

Kh nng chng m thng: Pt 0,75.Rk.btb.ho bc + ho = 0,2 + 0,66 = 0,86 m 0,75.7,5.86.66 = 31928 kG = 32,0 T

So snh: Pt = 29,4 T < kh nng chng m thng = 32,0 T Vy chiu cao gi thit H= 70 cm m bo yu cu v chng m thng. Bi 8: Tnh ton ct thp cho mng trn Bi lm: * Tnh ton ct thp theo phng cnh di:

M men ti mp ct:

43

Mng = Mmax

bapp

M ngngng .2.

2

2max

* +=

( ) ( )a

aapppp ngng

+= minmaxmin*

maa

a cng 35,123,03

2=

=

=

2* /3,176,67,103

35,13)7,107,22(7,10 mTpng =+=

+=

TmM ng 8,212.235,1

.

27,223,17 2

=

+=

Ct thp yu cu:

22 130013,066,0.28000.9,0

8,21..9,0

cmmhR

MF

oa

nga ====

12 thanh 12 a = 180 ( Fa = 13,57cm2) * Tnh ton ct thp theo phng cnh ngn:

M men ti mp ct:

Tmab

pM ngtbng 3,203.29,0

.7,16.2

.

22

===

Ct thp yu cu:

900

3000

2000

200

300

1000

300

660

40

tt

Qo

tt

No

tt

Mo

0.000

-1.000

1350

1

1

2 2

44

22 1200121,066,0.28000.9,0

3,20..9,0

cmmhR

MF

oa

nga ====

min = 0,05% . a . ho = 0,05% . 300 . 66 = 9,9cm2 16 12 a= 200 ( fa=18cm2)

Bi 9: Tnh ton chiu cao mng bng b tng ct thp di tng s liu nh sau: Ti trng tnh ton ti mc mt t:

Nott = 50T/m Mott = 3Tm/m Qott = 1T/m

Tng dy: bt= 30 cm Mng b= 1,4m; hm = 1m; BT 200#; Rn = 90kG/cm

2; Rk = 7,5 kG/cm2

Bi lm:

Tnh ng sut y mng do ti trng cng trnh gy ra:

WM

FN

p ottott =minmax

TmhQMM mottotttt 4113. =+=+= === 2,127,35

4,1.16.4

4,1.150

2minmaxp

1000

300

700

1400

ttQo

ttNo

ttMo0.000

-1.000

1000

300

45

300550

40

260

290

45

2max /9,47 mTp =

2min /5,23 mTp =

2/7,35 mTptb = iu kin kim tra: Pt 0,75. Rk. btb. ho

Theo phng cnh di ca mng ta ct ra 1m tin tnh ton ( Lu : Khng phi ct ra 1m ri sau tnh nh mng n)

Mi tnh ton nh: cng , ng sut vn ca mng bng vi cnh b = 1,4m. gi thit H = 30cm; a = 4cm; ho = 26cm

?* =dtP dtdt Fppp .

2max

* +=

bbbpppp dt+= )( minmaxmin*

mhbbb otdt 29,026,023,04,1

2=

=

=

2* /85,4235,195,234,1

29,04,1)5,239,47(5,23 mTp =+=+=

229,01.29,0 mmmFdt ==

TPdt 2,1329,0.29,4785,42

=

+=

Kh nng chng m thng: Pt 0,75. Rk. btb. ho ; btb:on m- n = 1m 0,75 . 7,5 . 100 . 26 = 14,6 T so snh Pt = 13,2 T < kh nng chng m thng Pct = 14,6 T

Chiu dy mng chn nh trn l hp l. Bi 10 : Tnh ton ct thp cho mng trn vi H = 30cm; a = 4cm

Bi lm Tnh ton ct thp theo phng cnh ngn: M men ti mp tng Mng = Mmax

lbpp

M ngngbng .2.

2

2max

* +=

( ) ( )bbb

pppp ngng

+= minmaxmin*

300

300

700

ttQo

ttNo

ttMo

0.000

-1.000

1000

550

1400

46

mbbb tng 55,02

3,04,12

=

=

=

TmM bng 5,41.255,0

.

25,361,23 2

=

+=

Ct thp yu cu:

22 700068,026,0.28000.9,0

5,4..9,0

cmmhR

MF

oa

bng

a ===

7 12: a = 160; (Fa = 7,92cm2) Tnh ton ct thp theo phng cnh di: Theo phng cnh di cng ca mng ln ct thp c b tr cu to 8 12; a = 200; ( Fa= 9,04cm

2)

$4. mt s trng hp ring:

1. Mng nh c tng hm: su t mng c xc nh nh sau:

3.2 * hhhh +=

Khi x/ hh c xt n s chnh lch gia trng lng th tch vt liu sn so vi trng lng th tch ca t.

d

s

hhh hhh

''' +=

Hoc c th chn gn ng nh sau: *)21

31( hh =

Do c p lc t nn y l trng hp c bit ca mng chu ti trng ngang: ti trng lc gm: p lc t ch ng, p lc t b ng, ti trng tm thi tc dng ln mt t ngoi nh, p lc nc ... (tham kho STTKNM-1)

2. Mng chu ti thng ng lch tm: C th lm nh sau:

a, Sc chu ti ca t di mng: S c tnh tng t nh trnh by trn, nhng b rng mng dng trong tnh ton l:

B = B - 2.e Trong : e l lch tm ca mng.

Lc ta coi rng ti trng tc dng ng tm 1 mng c b rng hu hiu B

e

2e B'

B

47

b, Tnh /s tip xc: Trng hp ti lch tm vt ngoi phm vi y mng (dng phn b /s hnh tam gic

6b

e ) lc c s phn b li /s lc ta phi xc nh

bng phng php cn bng lc trn c s 0min =p , th d nh trng hp lch tm mt phng:

0;2 minmax == pLxNp

Trong : L: chiu di ca mng x: chiu rng hot ng ca khong /s tip xc:

).5,0.(3 BeBx = Nu Bx > l phn b hnh thang. Nu Bx l phn b hnh tam gic. c, Tnh chiu cao v ct thp: Nu dng phn b /s hnh tam gic, lc c s phn b li /s ng sut cng lm tng t nh trn, sau tnh v pha maxp

Trng hp lch tm c 2 phng lm tng t (tham kho ti liu)

3.Tnh ton mng chu ko ln:(Mng bc):

).(. 21 cdtcdtcm TNNnQn ++ n1: h s vt ti i vi lc ko ln P ti nh mng, ly bng 1,2 n2 : h s vt ti i vi lc gi ly bng 0,8.

tcmN : trng lng mng tcdN : trng lng t nm trn cc bc mng

Tc : tng hp lc ca sc chng trt ca t c xc nh nh sau: V biu p lc t trn mt phng thng ng ngha l tnh p lc ngang v ;

sau phn thnh cc phn ring v tm p lc trung bnh trn mi phn tgv.= : gc ma st trong ca t. Vi t ct thay cho ly ta ly h s ma st tnh.

Nd Nd

Nm

Q

H1

H2

1

2

e

x

B

B

48

4. Mng di ct st nhau:

Gi thit mng l cng: Ti trng:

0201 NNN += + Trng lng mng + t ph lp

0201202101 .. MMeNeNM +++= p lc di mng:

mtb

n

oih

F

N

FN

p .1 +=

WMpp +=max W

Mpp =min

Cc bc tnh ton sau ging mng di ct, tng. Khi kim tra chiu cao v ct thp trong mng th cn c vo cu to v tr cc ct m c s tnh ton ph hp.

$5. Tnh ton mng mm:

Mnhp

Mgi

1

1-1

L' L L'nhp

Mgi

1

B

N01

M01

N02

M02

Hm

N

M

49

1. Khi nim :

+ Nn t yu mng n phi m rng ra n gn nhau, nn t bin dng nhiu, cn lm mng lin tc cng ln chu ln khng u.Ta i n gii php dng mng bng, bng giao nhau, hoc b. +Ti trng ngoi v phn lc nn mng b un mng b un li nh hng n s phn b phn lc nn. + Khng xt cng ca mng th tnh ton ch c ngha thc tin cho vic tnh ng sut cn tnh mng sai s s ln. + Tuy nhin n gin tnh ton ta ch xt khi bin dng un ln n mc no

1033

>=hl

EE

t o Eo - m uyn bin dng ca nn

E - m uyn n hi ca vt liu mng l - na chiu di mng h - chiu dy mng

Khi t s hai cnh: 7bl

mng dm; 7bl

mng bn

Bi ton: - Xc nh phn lc nn. - ln ca mng. - Kt hp vi ti trng ngoi tnh c kt cu mng. Cc cch tnh ton:

a, Theo s n gin: Lt ngc mng ln xem n nh 1 dm lin tc, chn ct nh gi ta, ti trng chnh l phn lc nn v xem phn lc nn phn b u vi cng p. Tc l ta b qua bin dng ca bn thn Mng v b qua bin dng ca CT bn trn. Do n ch ng khi: CT bn trn l tuyt i cng phn lc nn phn b u (nh > 9 tng)

50

b, CT+M+N c gn vi nhau v c tc ng qua li vi nhau nn ng n phi xem xt nh 1 th thng nht tnh ton tuy nhin kh khn v mt thut ton nn cha p dng rng ri

c, Hin nay ph bin dng cch ri rc kt cu: tc l tch ring mng v xt s lm vic ng thi ca mng v nn: Tc l xt mt KC c bin dng t trn nn cng c bin dng v xc nh ni lc trong mng Tnh ton kt cu trn nn n hi.

Do : Vic thit k mng bng mm bao gm:

+ Xc nh s b b rng mng tng t mng bng cng vi gi thit phn lc nn phn b u. + La chn s b KC mng ph hp b rng ni trn. + Tnh ton chuyn v mng, phn lc t v ni lc trong mng theo s : Dm trn nn n hi. + Kim tra cc trng thi gii hn ca nn. Nu khng tho mn tng kch thc mng, hoc tng cng ca mng ( sa i KC la chn) v xc nh li ni dung bc 3 + Thit k ct thp cho mng trn c s kt qu bc 3.

2. Tnh ton dm trn nn n hi: Xt mng dm:

Phng trnh trc vng ca dm:

[ ]bpqxd

dEJ xx

x=4

)(4

E.J - cng ca dm; b: b rng dm; : chuyn v ng Vi 2 n x v px ta tm thm 1 ph/trnh na.

51

Quan h: ln ca nn vi p lc y mng )(1)( xx pfS = )(2)( xx SfP =

Th hin c ch lm vic (bin dng) ca nn gi l m hnh nn. Cc m hnh nn t:

+ M hnh nn Winkler

xx sCp .= C: h s t l h s nn. Cn gi l m hnh nn bin dng cc b.

+ M hnh na khng gian bin dng tng th.

Theo But xi nt:

RE

pSo

oyx

pi

2),(

1=

Nhn xt: *M hnh nn Winkler khng phn nh c tnh phn phi ca t th hin t khng h b ln ngoi phm vi t ti. *M hnh BKGH: nh gi qu cao tnh phn phi ca t th hin n nn vn ln. *Thc t t khng phi l vt th n hi tnh phn phi ca n yu hn nhiu so vi vt th n hi nn n ch ln trong phm vi nh no thi.

so snh: ta xem xt mt bin dng ca nn:

theo m hnh Winkler (ng 4) theo m hnh na khng gian n hi (ng2) theo kt qu th nghim (ng 3)

52

Do thiu st ca 2 m hnh trn nn c rt nhiu m hnh khc c nu ra:

* M hnh nn mng. * M hnh nn tm. * M hnh nn n hi vi 2 h s nn. * M hnh lp n hi c chiu dy hu hn * M hnh nn na khng gian n hi c oE bin i theo su...

V vic la chon m hnh nn: T kt qu thc nghim cc tc gi a ra nhng kt lun:

1. ln ca t ngoi phm vi t ti tt rt nhanh: thng thng vng ln khong 0,3 0,5 ng knh tm nn. Nh vy t c tnh phn phi rt yu

2. khi m tng tnh phn phi gim, t bo ho nc tnh phn phi ca n khng ng k

3. mt bin dng ca nn khi xem l bn khng gian n hi tt qu chm so vi thc t quan st.

Do kt lun: t mm dng m hnh winkler hp hn, tuy nhin h s nn C khng c ngha r rng: n khng phi l hng s m n thay i ph thuc cng CT; khong tc dng ca ti trng. Nc ta ng bng sng Hng v Cu long s gp t mm, cha nhiu nc, nc ngm cao nn tnh phn phi ca t yu do nn chn m hnh Winkler 3. Tnh ton mng dm theo m hnh nn bin dng cc b: Phng trnh c bn:

xx CSp =

xx S= CbqS

dSd )(4)(4)(4

4

=+

Trong : xa.= 44EJCb

a = (1/m)

Dm khng ti: 0)(4)(44

=+ S

dSd

Gii tm S tm p

53

a, Tnh dm di v hn trn nn n hi. Hai u dm xa ti trng khng b ln 2 u. 1. Lc P: Bi ton i xng ct i tnh

+++= eCCeCCS )sincos()sincos()( 4321

Xc nh C: H/s da vo /k bin: x () : C1=C2=0 nu khng S s khng bng khng.

Lc : += eCCS )sincos()( 43 (1) Ti x=0 ln s ln nht do :

0=dxdSx

(2)

0=

ddS

o hm (1) ( gc xoay =0) kt hp vi (2) ta c: ( ) ( )[ ] 0sincos 3434 =+ eCCCC 43 CC = (V cos 0 khi x=0)

Gi l Co ta c: += eCS 0)sin(cos)(

Co=? Ti x=0

20PQ =

Ti =a.x=0 0333

42

)( CEJaP

dSd

==

(SBVL)

30 8JaPC =

54

)sin(cos8

3 axaxeEJaPS axx +=

Sau xc nh: Mx; Qx: o hm lin tc

)cos(sin42

2

axaxea

Pdx

SxdEJM axx ==

axeP

dxSxdEJQ axx cos23

3

==

t )sin(cos1 axaxe ax += )cos(sin2 axaxe ax = Ta c: axe ax cos3

= 138

EJa

PSx = 138C

EJaPpx = 24

a

PM x = 32PQx =

: tra bng theo = a. x Smax ti im t ca ti trng P (tc ti x=0) v c tr s:

CbaP

EJaPS

28 3max==

baPp2max

= a

PM4max

= 2maxPQ =

Cc trng hp khc tng t Trong cc gio trnh NM c cc bng tra kt qu tnh ton cho cc trng hp c bn. Lc tp trung, m men tp trung im gia dm di, bn vung hoc trn. C th gii bi ton mng dm, bn chu nhiu ti tp trung bng cch p dng cc bi ton c bn vi nguyn l cng tc dng.

55

4.Tnh ton mng dm theo m hnh nn l bn khng gian BDTT H p/trnh vi phn c bn i vi mng loi dm:

dxxxpE

bS ox

ox

o

ox )ln(

)1(2 2

1)(

2

)(

=

Thng l gii theo phng php gn ng, c th tm thy cc li gii in hnh: + Phng php M.I. Gorbunov Pasadov + Phng php I. A. Ximvulidi + Phng php ca Giemoskin + Phng php phn t hu hn. Trong cc gio trnh nn mng: Nguyn l ca cc phng php ny l chuyn s bi ton dm t trn nn n hi Dm trn cc gi hu hn (dm lin tc) v dng phng php chuyn v trong c hc kt cu gii xc nh phn lc gi.

Phng php phn t hu hn: C th dng cc chng trnh phn mm dng cho vic tnh ton mng loi dm bn: V d chng trnh SAP 90; SAP 2000 vi m hnh Winkler

* Phng php Phn t hu hn ng dng SAP 2000

Bng vic chia nh kt cu mng thnh cc phn t nh v h mng v nn c m hnh ho nh sau: + Kt cu mng: dm c thay th thnh cc phn t frame, bn l cc phn t shell. + Nn:

c thay th bng cc gi n hi spring l cc l so c cng: K=C. ax. bx ( ax; bx l din tch chu ti ca nt ang xt)

56

a, Mng dng dm n:

b, Bng giao nhau:

c, Mng b:

Nh vy c th d dng k n s thay i ca EJ v K (trong mi phn t coi EJ, K l hng s khc nhau)

57

*Cc phng php xc nh cng ca l xo:

+ Tra bng tm h s nn tng ng vi cc loi t( bng tra ph lc) + Cng thc thc nghim: Poulos; Terzaghi; Meyerhof + T kt qu th nghim bn nn.

* ng dng SAP 2000 gii ni lc mng: Bc1: to s hnh hc: frame (mng dm), shell (bn) Bc 2: khai bo s liu u vo: vt liu, tit din, ti trng, t hp nu c Bc 3: gn ti trng, iu kin bin. Bc 4: chia nh kt cu mng thnh cc phn t nh kch thc bng nhau ( kch thc phn t tu thuc vo mt bng mng v yu cu tnh ton cng nh cng chnh xc s lng phn t ln) Bc 5: tnh cng ca gi n hi thay th v gn cho tng nt ch din tch thay th. ttFcK .=

Bc 6: kim tra s liu nhp phn tch ni lc Bc 7: xem kt qu ni lc: + biu (M;Q) ca dm mng, phn lc ca cc gi n hi: Rz v Uz Bc 8: kim tra kt qu: Phn lc nn Rz lun dng ( nn khng chu ko) nu Rz < 0 b lin kt ti cho Uz= 0 phn tch li ni lc. Bc 9: kim tra cng nn v tnh ton ct thp.

*Tnh lp h s nn:

+ Ti sao phi tnh lp: cng K tnh theo cc cng thc thc nghim khng chnh xc tnh lp tm K + Cch tnh lp: Ln 1: tnh K1 theo cc cng thc thc nghim. phn tch kt qu c Rz1, Uz1 nu K ng th Uz1 = S; S - ln ca nn tnh cho mng c kch thc phn t theo phng php cng ln tng lp hc trong C hc t.

58