btxm theo qd 3230-qd-bgtvt

NG VNH AI PHA TY

THNH PH NNG

Ngi tnh:L Hng Lp

Kim ton kt cu o ng BTXM Kim tra: Mai Triu Quang

Tiu chun p dng:

1. S liu u vo:

- Cp ng: ng cp III

- Khu vc tuyn i qua Min Nam

- B rng phn xe chy: 7 m

- Sut tng trng trung b nh nm ca cc xe nng: (g r ) 6 %

- Thi hn phc v thit k: (t) 20 nm

- Ti trng trc tiu chun: (P s) 100 kN

- Tr s gradien nhit ln nht (T g) 0.92oC/cm

TNH TON KT CU O NG BTXM

Hng mc:

CNG TRNH:

A IM:

L trnh Ton tuyn

Q S 3230/Q-BGTVT NGY 14-12-2012

lyhonglap-pkt-ECC qt md btxm 3230/qd-bgtvt 14-12-2012 1

- Tr s gradien nhit ln nht (T g) 0.92oC/cm

- H s phn b vch bnh xe theo chiu ngang ( ) 0.58

- S ln trc xe quy i v trc xe tiu chun tch ly trong thi hn thit k: (N e)Trng lng

trcPi (kN)

kpi

= (Pi/Ps)16

N(xe/3000 xe)

n(trc)

ADTT(xe/ngy.2 chiu)

pi(%)

(ki . Pi)

Trc 1 90 0.19 0 0.00 0.000Trc 2 180 12143.95 0 0.00 0.000Trc 1 60 0.00 2900 31.52 0.000Trc 2 120 18.49 2900 31.52 5.828Trc 3 120 18.49 2900 31.52 5.828Trc 1 90 0.19 0 0.00 0.000Trc 2 180 12143.95 0 0.00 0.000Trc 3 140 217.80 0 0.00 0.000Trc 4 140 217.80 0 0.00 0.000Trc 1 90 0.19 100 1.09 0.002Trc 2 180 12143.95 100 1.09 131.999Trc 3 120 18.49 100 1.09 0.201Trc 4 110 4.59 100 1.09 0.050Trc 5 110 4.59 100 1.09 0.050Trc 1 90 0.19 0 0.00 0.000Trc 2 150 656.84 0 0.00 0.000Trc 3 120 18.49 0 0.00 0.000Trc 4 110 4.59 0 0.00 0.000Trc 5 110 4.59 0 0.00 0.000

3000 9200 18 100 143.96

C

D 6

Loi xe

A 0

B 12

0

0

2900

0

100

0

0

Tng cng

E


= 1.93E+07 (ln/ln) (A-3)

Vi = 2.48E+03 (ln trc/ngy.ln) (A-2)

Trong :

- n : Tng s trc n thng qua 3000 xe iu tra (loi xe c 2 trc, 6 bnh tr ln)

- pi: % s trc n c trng lng trc P i trong ph trc xe nng iu tra

- Kpi: H s tnh i cc trc n Pi trong ph trc xe nng v trc xe tnh mi tiu chun P s

- ADTT: S xe nng ngy m trung bnh nm trn ln xe thit k nm u tin a vo khai thc (xe/ngy.ln)

+ H s phn phi lng giao thng cho l n xe thit k = 0.625

+ H s phn phi xe cho mi chiu xe chy = 0.5

+ ADTT = 5.625 (xe/ngy.ln)

* S ln tc dng quy i v trc xe tiu chun P s =100kN tch ly l N e = 1.90E+07 (ln/ln)

=> Quy m giao thng thuc cp Nng

2. D kin kt cu mt ng:

- D kin tng mt BTXM dy hc = 0.26 m

- Cng ko un thit k f r = 5 MPa

- Tr s m un n hi tnh ton E c = 31000 MPa

r

tr

se g

gNN

]1)1[(...365 1

n

iipis pk

nADTTN

1

1 ).().3000

.(


- Tr s m un n hi tnh ton E c = 31000 MPa

- B tng dng ct liu Granit c c= 10 .10-6/oC

- H s poisson ca BTXM mc = 0.15

- S dng tm BTXM c kch thc rng B = 3.5 m di L = 4.8 m

- Lp mng trn bng Cp phi dm gia c XM 5% dy hb = 0.2 m

Eb = 1300 MPa mb = 0.2

- Lp mng di bng Cp phi dm loi 1, Dmax25 dy hsb = 0.18 m

M un n hi Esb= 300 Mpa; msb = 0.35

- nh nn ng: Cp phi t i, K98

Eo = 45 Mpa; dy ho = 0.3 m

- L ng: Cng kt cu vi phn xe chy

3. Kim ton kt cu d kin:

3.1. Tnh m un n hi chung E t ca nn t v mng di bng vt liu ht:

+ M un n hi tng ng ca cc lp vt liu ht E x:

300.00 MPa (8-8)

+ Tng chiu dy cc lp vt liu ht h x:

0.18 m (8-10)

+ H s hi quy li n quan n tng chiu dy cc lp vt liu ht :0.41 (8-9)

+ M un n hi chung E t:

M un n hi 90 ngy tui

= 0.86 + 0.26 * ln(hx) =

Ex= n

1(h2

i*Ei)/n

1.h2

i =

hx= n

1 hi =

r

tr

se g

gNN

]1)1[(...365 1

n

iipis pk

nADTTN

1

1 ).().3000

.(


98.73 MPa (8-7)

3.2. Tnh cng tng i chung ca c kt cu r g :

+ cng un cong tit din ca tm BTXM D c:

46.45 MN.m (8-6)

+ cng un ca tit din lp mng trn c gia c cht lin kt D b:

0.90 MN.m (8-20)

+ cng tng i chung ca c kt cu r g:

0.95 m (8-20)

3.3. Tnhng sut do ti trng trc gy ra :

+ng sut ko un ps do ti trng xe, vi P s = 100 kN

0.00 MPa

1.54 MPa (8-20)

+ng sut ko un ps do ti trng xe, vi Ps = Pmax = 180 kN

0.00 MPa

2.68 MPa (8-20)

+ng sut ko un gy mi do ti trng xe chy ti v tr gia cnh dc tm pr:

3.66 MPa (8-5)

2.45 MPa (8-15)

Vi k r = 0.87 ,kf = Ne0.057 = 2.60 v kc = 1.05

ps =1.45*10-3

* rg0.65

* hc-2

* Ps0.94

/(1 + Db/Dc) =

pm =1.47*10-3

* rg0.7

* hc-2

* Ps0.94

=

rg = 1,21. [(Db + Dc)/Et]1/3

=

ps =1.47*10-3

* rg0.7

* hc-2

* Ps0.94

=

pm =1.45*10-3

* rg0.65

* hc-2

* Ps0.94

/(1 + Db/Dc) =

pr = kr * kf * kc * ps =

pmax = kr * kc * pm =

Et = (Ex/Eo) * Eo=

Dc=Ec.h3

c/[12*(1-2

c)]=

Db=Eb.h3

b/[12*(1-2

b)] =

ChtShttt

tChttShtCL .sin.cos

sin.cos.).

1

1(1

ChtShttt

tChttShtC L .sin.cos

sin.cos.1


Vi k r = 0.87 ,kf = Ne0.057 = 2.60 v kc = 1.05

3.4. Tnhng sut ko un do gradien nhit gy ra :

+ cng tip xc theo chiu dc gia tng mt v tng mng k n

3081.98 Mpa/m (8-23)

+ H s xt n trng thi tip xc gia 2 lp r

0.13 m (8-23)

+ H s lin quan n kt cu tm 2 lp

0.14 (8-23)

+ H s ng sut un vng C L

0.00

0.70 (8-23)

1.69 ; Sht = (et- e

-t)/2 ; Cht = (e

t+ e

-t)/2

+ H s ng sut nhit tng hp B L0.35 (8-18)

+ng sut ko un do gradien nhit gy ra trong tm ti gia cnh dc tm tmax1.28 Mpa (8-17)

+ H s mi nhit k t0.316 Mpa

Trng hp: at = 0.841 bt = 1.323 ct = 0.058

Vi t = L/(3. rg)=

BL = 1,77. e-4,48.hc

.CL - 0,131.(1-CL) =

tmax = c. hc. Ec .Tg. BL/2=

kt = f r /tmax. [at. (tmax/f r )bt - ct] =

kn = 1/2.(hc/Ec + hb/Eb)-1

=

r = {Dc.Db / [(Dc + Db). kn]}1/4

=

= -{(kn. rg4 - Dc). r

3 / [(kn. r

4 - Dc).rg

3]}=

ChtShttt

tChttShtCL .sin.cos

sin.cos.).

1

1(1

ChtShttt

tChttShtC L .sin.cos

sin.cos.1


+ H s mi nhit k t0.313 Mpa (8-19)

Trng hp: at = 0.871 bt = 1.287 ct = 0.071

(at, bt, ct: l cc h s quy hi)

+ng sut ko un do gradien nhit gy mi gia cnh dc tm tr0.41 Mpa (8-16)

3.5. Kim ton iu kin gii hn :

+ Vi ng cp III ta c h s tin cy t. k r = 1.11

Ta c: 4.49 MPa < fr = 5 MPa; (8-1)

V: 4.12 MPa < fr = 5 MPa; (8-2)

* Kt lun: Kt cu chn m bo yu cu

* Kt cu mt ng BTXM c thit k l:

- Tm BTXM c fr =5MPa, dy: Hc = 0.266 m

Trn lp mng: - Cp phi dm gia c XM 5% Hb = 0.2 m

- Cp phi dm loi 1, Dmax25 Hsb = 0.18 m

- nh nn ng: Cp phi t i, K98 Ho = 0.3 m

- L ng: Cng kt cu vi phn xe chy

tr = ktmax

. tmax =

kt = f r /tmax. [at. (tmax/f r )bt - ct] =

r. (pr + tr) =

r. (pmax + tmax) =

lyhonglap-pkt-ECC qt md btxm 3230/qd-bgtvt 14-12-2012 4lyhonglap-pkt-ECC qt md btxm 3230/qd-bgtvt 14-12-2012 4

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