Mc lc

Phn I:

lp bo co u t xy dng tuyn ng

Trng i hc dn lp Hi Phng B mn Xy dng n tt nghip Ngnh: Xy dng cu ng

Svth: Nguyn Tun Anh Mssv: 100102 Lp: C1001Trang: 3

Chng 1:Gii thiu chung

1. Tn cng trnh: D n u t xy dung tuyn ng T1 T2 thuc Xa Bun Tng Ln Huyn Krong Nang Tnh k Lk.2. a im xy dng:Huyn Krong Nang - tnh k Lk3. Ch u t v ngun vn u t:Ch u t l UBND tnh k Lk y quyn cho s Giao Thng Cng Chnh tnh k Lk thc hin. Trn c s u thu hn ch tuyn chn nh thu c kh nng v nng lc, my mc, thit b, nhn lc v p ng k thut yu cu v cht lng v tin thi cng.Ngun vn xy dng cng trnh do ngn sch nh nc cp. bn cnh c s h tr ca ngun vn ODA.4. K hoch u t:D kin nh nc u t tp trung trong vng 6 thng, bt u u t t thng 9/2010 n thng 3/2011. V trong thi gian 15 nm k t khi xy dng xong, mi nm nh nc cp cho 5% kinh ph xy dng duy tu, bo dng tuyn.5. Tnh kh thi XDCT: nh gi s cn thit phi u t xy dng tuyn ng T1 T2 cn xem xt trn nhiu kha cnh c bit l cho s phc v cho s pht trin kinh t x hi nhm cc mc ch chnh nh sau:* X Bun Tng Ln l x thuc huyn min ni Krng nng nm phang bc tnh k Lk , c din tch t nhin l 621km2 .V ranh gii hnh chnh th Tam Phong gip :Pha Bc gip x Tam Bnh v Thng Nht v Tam HpPha Ty gip x Tam inPha Nam gip Giang Tin V Giang ThnhPha ng gip vi x Tam Hp

Huyn Bun Tng Ln thuc Tnh k Lk l huyn nm phia ng bc tnh k Lk thnh lp ngy 9-11-1987 tch ra t tnh Krong Buk vi tim nng pht trin chnh l cy cng nghip .c bit, l cy trng mi nhn nh c ph cao su v lua nc l vng khu trng im v pht trin cy trng cng nghip v vy nu tin hnh xy dng tuyn ng ny s gip tng trng kinh t v ci thin kh nng trao i hng ha cho c vng Pht huy trit tim nng, ngun lc ca khu vc, khai thc c hiu qu cc ngun lc t bn ngoi. Trong nhng trng hp cn thit phc v cho chnh tr, an ninh, quc phng.Theo s liu iu tra lu lng xe thit k nm u tin khi a d n vo khai thc l: 900 xe/ng.. Vi thnh phn dng xe: Xe con: 44% Xe bus nh: 20% Xe ti nh;23% Xe ti trung: 4% Xe ti nng loi 1: 3%-Xe ti nng loi 2;4%-Xe ti nng loai 3;2% H s tng xe: 6 %.Nh vy lng vn chuyn gia 2 im T1- T2 l kh ln vi hin trng mng li giao thng trong vng khng th p ng yu cu vn chuyn. Chnh v vy, vic xy dng tuyn ng T1- T2 l hon ton cn thit. Gp phn vo vic hon thin mng li giao thng trong khu vc, gp phn vo vic pht trin kinh t x hi a phng v pht trin cc khu cng nghip ch bin, dch v ...6. Tnh php l u t xy dng:Cn c vo:- Quy hoch tng th mng li giao thng ca tnh k Lk.

Quyt nh u t ca UBND tnh Dk Lk s 3769/Q-UBND. K hoch v u t v pht trin theo cc nh hng v quy hoch ca UBND huyn Bun Tng Ln. Mt s vn bn php l c lin quan khc. H s kt qu kho st ca vng (h s v kho st a cht thu vn, h s qun l ng c, ..vv..) Cn c v mt k thut: Tiu chun thit k ng t TCVN 4054 - 05. Quy phm thit k o ng mm (22TCN - 211 -06). Quy trnh kho st xy dng (22TCN - 27 - 84). Quy trnh kho st thu vn (22TCN - 220 - 95) ca b GTVT Lut bo hiu ng b 22TCN 237- 01Ngoi ra cn c tham kho cc quy trnh quy phm c lin quan khc.7. c im khu vc tuyn ng i qua:c im v iu kin t nhin a hnh ni thp c cao t 500m => 1.090m, cc x Bun Tng Ln chim khong 12% din tch t nhin, vng ni c dc >350, a hnh i chim khong 70% din tch c cao 20-500m,i sp xp thnh dng bt p v cu to bi lc nguyn, phn b theo hng ng Ty, dc t 12-350, a hnh thung lng chim khong 8% thng hp, dc vi cu to ch V,t c hnh ch U a hnh ng bng chim 10% din tch, y l din tch t nng nghip trng la l ch yu ca Huyn7.2. c im a hnh : Tuyn i qua khu vc a hnh tng i phc tp c dc ln v c a hnh chia ct mnh. Chnh cao ca hai ng ng mc l 5m.

dc trung bnh ca sn dc khong 19,6%c im v KT-VH-XH huyn Bun Tng Ln Cn c vo c im a hnh ca huyn mc tiu pht trin kinh t ca vng nm 2010 l : Tip tc i mi mt cch su sc ton cnh ca cc ngnh,cc cp tp trung s dng c hiu qu mi ngun lc,khai thc tim nng v tr a l,ti nguyn, y mnh nh hng Cng nghip ho hin i ho Thc hin c cu kinh t : Cng nghip-dch v-nng nghip tip tc y nhanh c cu tong ngnh theo tng trng kinh t gn vi bo v mi trng sinh thi. Kt hp cht ch gia cc tng trng kinh t vi vic gii quyt tt cc lnh vc x hi.Gi vng n nh chnh tr,trt t an ton x hi to th v lc cho huyn Krong nang pht trin.Phn u nm 2010 Bun Tng Ln tr thnh vng kinh t trng im ca Dak Lakc im a cht thu vn: a cht khu vc kh n nh t b phong ho, khng c hin tng nt n, khng b st n. t nn ch yu l t bazan , a cht lng sng v cc sui chnh ni chung n nh . Cao mc nc ngm y tng i thp, cao l -3.7m, cp thot nc nhanh chng, trong vng c 1 dng sui hnh thnh dng chy r rng c lu lng tng i ln v cc sui nhnh tp trung nc v dng sui ny.c im mi trng: y l khu vc rt t b nhim v t b nh hng xu ca con ngi, trong vng tuyn c kh nng i qua c 1 phn l t trng trt. Do khi xy dng tuyn ng phi ch khng ph v cnh quan thin nhin, chim nhiu din tch t canh tc ca ngi dn v ph hoi cng trnh xung quanh.c im iu kin vt liu v iu kin thi cng: Cc ngun cung cp nguyn vt liu p ng vic xy dng ng c ly vn chuyn < 5km. n v thi cng c y nng lc my mc, thit b p ng nhu cu v cht lng v tin xy dng cng trnh. C kh nng tn dng nguyn vt liu a phng trong khu vc tuyn i qua c m cp phi

dm vi tr lng tng i ln v theo s liu kho st s b th thy cc it gn c th p nn ng c. Phm vi t cc m n phm vi cng trnh t 500m n 1000m.c im iu kin kh hu: Tuyn nm trong khu vc kh hu nhit i - gi ma,ma h nng m ma nhiu. t gi. Ma kh lnh kh hanh,t ma gi ch yu l gi ng bc, nhit khng kh trung bnh hng nm khong 210C.Lng ma hng nm khong 1700 - 2400 mm, Ch c 2 ma l ma ma v ma kh8. nh gi vic xy dng tuyn ng:Tuyn c xy dng trn nn a cht n nh nhng l khu vc i ni cao v dy c nn khi thi cng phi ch m bo dc thit k. n v lp d n thit k: S Giao Thng Cng Chnh tnh Dak Lak. n v gim st thi cng: n v thi cng:

Chng 2:Xc nh cp hng ngv cc ch tiu k thut ca ng$1. Xc nh cp hng ng:1.Da vo ngha v tm quan trng ca tuyn ngTuyn ng thit k t im T1 n T2 thuc vng quy hoch ca tnh DakLak, tuyn ng ny c ngha rt quan trng i vi s pht trin kinh t x hi ca tnh.Con ng ny ni lin 2 vng kinh t trng im ca tnh k Lk .V vy ta s chn cp k thut ca ng l cp III, thit k cho min ni.2. Xc nh cp hng ng da theo lu lng xeQuy i lu lng xe ra xe con:(Bng 2.1.1)

LL( N1)Xe ti nng 3Xe ti nng 2Xe ti nng 1

Xe ti trung

Xe ti nhXe bus nh

Xe con

hstx(q)

203543242320446

h s qd(ai)332.52.52.52.51

Xe quy i

32

16

24

32

184

160

352

Nqd(N1)=Ni .ai1392

(H s quy i tra mc 3.3.2/ TCVN 4054-05)Theo tiu chun thit k ng t TCVN 4054-05 (mc 3.4.2), phn cp k thut ng t theo lu lng xe thit k (xcq/ngy m): >3000 th chnng cp III.Cn c vo cc yu t trn ta s chn cp k thut ca ng l cp III, tc thit k 60Km/h (a hnh ni).

$2. Xc nh cc ch tiu k thut theo quy phm1. Cn c theo cp hng xc nh ta xc nh c ch tiu k thut theo tiu chun hin hnh (TCVN 4050-2005) nh sau: (Bng 2.2.1)Cc ch tiu k thutTr s

Chiu rng ti thiu cc b phn trn MCN cho a hnh vng ni (bng 7-T11)

Tc thit k (km/h)60

S ln xe ginh cho xe c gii (ln)2

Chiu rng 1 ln xe (m)3

Chiu rng phn xe dnh cho xe c gii (m)6

Chiu rng ti thiu ca l ng (m)1.5 (gia c 1m)

Chiu rng ca nn ng (m)9

Tm nhn ti thiu khi xe chy trn ng (Bng 10- T19)

Tm nhn hm xe (S1), m75

Tm nhn trc xe ngc chiu (S2), m150

Tm nhn vt xe, m350

Bn knh ng cong nm ti thiu (Bng 11- T19)

Bn knh ng cong nm ti thiu gii hn (m)125

Bn knh ng cong nm ti thiu thng thng (m)250

Bn knh ng cong nm ti thiu khng siu cao(m)1500

dc siu cao (isc) v chiu di on ni siu cao (Bng 14- T22)

R (m)iscL(m)

125 1750.07 0.0670 60

175 2500.05 0.0455 50

250 15000.03 0.0250

dc dc ln nht (Bng 15- T23)

dc dc ln nht (%)7

Chiu di ti thiu i dc (Bng 17- T23)

Chiu di ti thiu i dc (m)150 (100)

Bn knh ti thiu ca ng cong ng li v lm (Bng 19- T24)

Bn knh ng cong ng li (m)

2500

Ti thiu gii hn

Ti thiu thng thng4000

Bn knh ng cong ng lm (m)

1000

Ti thiu gii hn

Ti thiu thng thng1500

Chiu di ng cong ng ti thiu (m)50

Dc ngang mt ng (%)2

Dc ngang l ng (phn l gia c) (%)2

Dc ngang l ng (phn l t) (%)6

$3. Tnh ton ch tiu k thut theo cng thc l thuyt[ 1 ]1. Tnh ton tm nhn xe chy:Tm nhn dng xe:

LpShlo

S1

Tnh cho t cn hm kp dng xe trc chng ngi vt(Bang1.3.1)

Vtktl1= V(m/s).t(s)Sh =KV 2

l0 (m)S1=l1+Sh+l0 (m)Ghi ch

TTXe tt(km/h)Ki(s)3,6(m)254(i)(m)

1Xe con601,20,00,5116,667341060,67

2Xe ti601,40,00,5116,66739,681066,35chn

3Xe Bus601.20.00.5116.667341060.67

[ 1]_ Ni dung tnh ton phn ny thc hin theo y/c n TN trong nh trngl1: qung ng ng vi thi gian phn ng tm l Sh : chiu di hm xelo : c ly an tonV: vn tc xe chy (km/h) K: h s s dng phanh: h s bmi: khi tnh tm nhn

S tnh tm nhn S2Tm nhn 2 chiu:

LpLpShShLoS1S1Tnh cho 2 xe ngc chiu trn cng 1 ln xe.

(Bang1.3.2)

Vtktl1=V(m/s) .t(s)ST1+ST2 =KV 2 .l0 (m)S2= 2l1+ST1+ST2+l0(m)Ghi ch

TTXe tt(km/h)Ki(s)1,8(m)127( 2i2 )(m)

1Xe con601,20,00,5133,3368,0310111

2Xe ti601,40,00,5133,3379,3710123chn

3Xe Bus601.20.00.5133.3368.0310111

Tm nhn vt xe:S tnh tm nhn vt xe

Tnh tm nhn vt xe:Tm nhn vt xe c xc nh theo cng thc (s tay tk ng T1/168). y ta tnh cho xe con vt xe ti(bng 1.3.3)

TTXe ttKV(km/h)l0

S4(m)Ghi ch

1Xe con1,280100,5248,71

2Xe ti1,460100,5263,19chn

3Xe Bus1.280100.5248.71

V 2KV (V

V )KV2lVV

4S1

1 1 2

2 o

1 . 1

3

(V1

V2 ).3,6

254

254

V1V2V1

Theo tiu chun :V1 > V2 =20km/h (i vi ng cp III)Trng hp ny c p dng khi trng hp nguy him nht xy ra V3 = V2=VTK= 60Km/h

2. dc dc ln nht cho php imax:imax c tnh theo 2 iu kin: iu kin m bo sc ko (sc ko phi ln hn sc cn - k cn xe chuyn ng):Df + iimax = D fD: nhn t ng lc ca xe (gi tr lc ko trn 1 n v trng lng, thng s ny do nh sx cung cp) iu kin m bo sc bm (sc ko phi nh hn sc bm, nu khng xe s trt - k xe chuyn ng)

D D'

GK . G

Pwi' G

max

D' f

Gk: trng lng bnh xe c trc ch ng G: trng lng xe.Gi trtnh trong kin bt li ca ng (mt ng trn trt:= 0,2) PW: Lc cn khng kh.

K.F.V213Pw(m/s)

Sau khi tnh ton 2 iu kin trn ta so snh v ly tr s nh hnTnh dc dc ln nht theo iu kin sc ko ln hn sc cn:Vi vn tc thit k l 60km/h. D tnh phn kt cu mt ng s lm bng b tng nha. Ta c:f: h s lc cn ln trng hp lp xe cng v tt th vi mt ng b tng nha, b tng xi mng, thm nhp nha f = 0,02 => f = 0,02V: tc tnh ton km/h. Kt qu tnh ton c th hin bng sau:Da vo biu ng lc hnh 3.2.13 v 3.2.14 s tay thit k ng t ta tin hnh tnh ton c cho bng:(Bng 2.3.1)

Loi xeXe conXetinngXetinngXetinng

loi 1 (2trc)loi 2 (2trc)loi 3 (2trc)

Vtt km/h60606060

f0,020,020,020,02

D0,130,0350,0330,048

imax(%)111.51.32.8

2.2 Tnh dc dc ln nht theo iu kin sc ko nh hn sc bm.Trong trng hp ny ta tnh ton cho cc xe trong thnh phn xe

GK . GPw GD' f va D'ib max

KF(V2Vg2 )13Trong :Pw: sc cn khng kh PW

V: tc thit k km/h, V = 60km/hVg: vn tc gi khi thit k ly Vg = 0(m/s) F: Din tch cn gi ca xe 0,8.B.H(m2) K: H s cn khng kh;

(Bng 2.2.2)

Loi xeKF, m2

Xe con0.025-0.0351.5-2.6

Xe ti0.06-0.073.0-6.0

Xe bus nho0.04-0.061.5-2.6

: h s bm dc ly trong iu kin bt li l mt ng m t,bn ly= 0,2GK: trng lng trc ch ng (kg).Gk= (0,5 0,55) G i vi xe con Gk =(0,65-0,7) G i vi xe ti G: trng lng ton b xe (kg).(Bng 2.2.3)

Xe conXe ti nng loi 1(2trc)Xe ti nng loi 2(2trc)Xeti nng loi 3(2trc)

K0.030.050.060.07

F2.6356

V60606060

Pw21.641.583.1116.4

Gk93762019633

G1875954014820

D'0.090.1210.122

i'max7%10.1%10.2%

Theo TCVN 4054-05 vi ng III, tc thit k V = 60km/h th imax = 0,07, cng vi kt qu va c (chn gi tr nh hn) hn na khi thit k cn phi cn nhc nh hng gia dc dc v khi lng o p tng thmkh nng vn hnh ca xe, ta s dng id5% vi chiu di ti thiu i dcc quy nh trong quy trnh l 150m, ti a l 500m.3. Tnh bn knh ti thiu ng cong nm khi c siu cao:

V2127(iSC )Rmin SC

Trong :V: vn tc tnh ton V= 60km/h: h s lc ngang = 0,16iSC: dc siu cao max 0,07

602127(0,160,07)123.24(m)Rmin SC

4. Tnh bn knh ti thiu ng cong nm khi khng c siu cao:

V2127(in )Rmin 0SC

: h s p lc ngang khi khng lm siu cao ly

602127(0, 080, 02)283.46(m)= 0,08 (hnh khch khng c cm gic khi i vo ng cong) in: dc ngang mt ng in = 0,02

Rmin 0SC

5. Tnh bn knh thng thng:Thay iv iSC ng thi s dng cng thc. V 2R

127(

iSC )

Bn knh thng thng(Bng 2.2.4)

isc%R(m)

=0.150.140.130.120.110.100.090.08

7%128.88134.98141.73149.19157.49166.74177.17188.97

6%134.98141.73149.19157.48166.74177.17188.98202.47

5%141.73149.19157.48166.74177.17188.98202.47218.05

4%149.19157.48166.74177.17188.98202.47218.05236.22

3%157.49166.74177.17188.98202.47218.05236.22257.70

2%166.74177.17188.98202.47218.05236.22257.70283.46

6. Tnh bn knh ti thiu m bo tm nhn ban m:

30.S1Rb.min

Trong :S1: tm nhn 1 chiu: gc chiu n pha= 2o

30.7521125(m)Rb.min

Khi R < 1125(m) th khc phc bng cch chiu sng hoc lm bin bo cho li xe bit.7. Chiu di ti thiu ca ng cong chuyn tip & b tr siu cao:

ng cong chuyn tip c tc dng dn hng bnh xe chy vo ng cong v c tc dng hn ch s xut hin t ngt ca lc ly tm khi xe chy vo ng cong, ci thin iu kin xe chy vo ng cong.ng cong chuyn tip.

3Xc nh theo cng thc: L

V(m)

Trong :

CT47RI

V: tc xe chy V = 60km/hI: tng gia tc ly tm trong ng cong chuyn tip, I = 0,5m/s2 R: bn knh ng cong trn c bn

B(iSC2.iphin )Chiu di on vut ni siu cao

LSC

( m rng phn xe chy = 0)Trong :B: l chiu rng mt ng B = 6 miph: dc ph thm mp ngoi ly iph = 1% p dng cho ng vng ni c Vtt =20 40km/h, vi cc cp ng khc iph= 0,5% (theo tiu chun nc ta quy nh).iSC: dc siu cao thay i trong khong 0,02 - 0,07

Chiu di ng cong chuyn tip v on vut ni siu cao(Bng 2.2.5)

Rtt(m)125-150150-175175-200200-250250-300400400-1000

isc0.070.060.050.040.030.020.02

in0.020.020.020.020.020.020.02

Lct(m)73.53-61.361.3-52.552.5-45.945.9-36.836.8-30.622.9822.98-9.19

Lctchon75625346372322

Lsc(m)54484236302424

Ltc(m)70605550505050

Lmax(m)847260505050111

(Theo TCVN4054-05, chiu di ng cong chuyn tip v chiu di on ni vut siu cao khng c nh hn Ltc v vi ng c tc thit k >60km/h th cn b tr ng cong chuyn tip) n gin, ng cong chuyn tip v on vut ni siu cao b tr trng nhau, do phi ly gi tr ln nht trong 2 on .on thng chm

L1L22on thng chm gia 2 on ng cong nm ngc chiu theo TCVN 4054-05 phi m bo b tr cc on ng cong chuyn tip v on ni siu cao.

Lmax

Tnh on thng chm(Bng 2.2.6)

Rtt(m)Rtt(m)125 150150 175175 200200 250250 300400

125 150847872676767

150 175787266616161

175 200726660555555

200 250676155505050

250 300676155505050

400676155505050

8. m rng phn xe chy trn ng cong nm E:Khi xe chy ng cong nm trc bnh xe chuyn ng trn qu o ring chiu phn ng ln hn do phi m rng ng cong.Ta tnh cho kh xe di nht trong thnh phn xe, dng xe c Lxe : 12.0 (m)L20,1Vng c 2 ln xe m rng E tnh nh sau:EA RR

Trong :LA: l khong cch t mi xe n trc sau cng ca xe R: bn knh ng cong nmV: l vn tc tnh tonTheo quy nh trong TCVN 4054-05, khi bn knh ng cong nm 250m th phi m rng phn xe chy, phn xe chy phi m rng theo quy nh trong bng 3-8 (TK t T1-T53).( Bng 2.2.7)

Dng xeBn knh ng cong nm, R (m)

250200200150150100

Xe con0,40,60,8

Xe ti0,60,70,9

9. Xc nh bn knh ti thiu ng cong ng:Bn knh ng cong ng li ti thiu:Bn knh ti thiu c tnh vi iu kin m bo tm nhn 1 chiuS2R1 2d1

d1: chiu cao mt ngi li xe so vi mt ng, d1 = 1,2m S1: Tm nhn 1 chiu; S1 =75m

Rli

752

2343.75(m)

min

2.1,2

Bn knh ng cong ng lm ti thiu:c tnh 2 iu kin.- Theo iu kin gi tr vt ti cho php ca l xo nhp xe v khng gy cm gic kh chu cho hnh khc.

2lmV

602

Rmin

6, 56, 5

553.84(m)

Theo iu kin m bo tm nhn ban m

S 2602 I 2(0, 675.sin 2o )559.45(m)2(hS .sin)1Rlmmin

Trong :h: chiu cao n pha h = 0,6m: gc chn ca n pha= 2o(Ghi ch: hin nay gc m ca n pha rt ln => s liu tnh tonch l ti thiu gii hn cui cng)10. Tnh b rng ln xe:10.1 Tnh b rng phn xe chy B:Khi tnh b rng phn xe chy ta tnh theo s xp xe nh hnh v trong c ba trng hp theo cng thc sau:

Trong :

B = bcxy2

b: chiu rng ph b (m) c: c ly 2 bnh xe (m)x: c ly t sn thng xe n ln xe bn cnh ngc chiu X = 0,5 + 0,005Vy: khong cch t gia vt bnh xe n mp phn xe chy y = 0,5 + 0,005VV: tc xe chy vi iu kin bnh thng (km/h)Tnh ton c tin hnh theo s xp xe cho 2 xe ti chy ngc chiu

Xe ti c b rng ph b l 2,5mb1 = b2 = 2,5m c1 = c2 = 1,96m

Xe ti t tc 60km/h

x = 0,5 + 0,005 . 60 = 0,8(m)y = 0,5 + 0,005 . 60 = 0,8(m)

2,51,9620,80,83.83mVy trong iu kin bnh thng ta c

b1 = b2 =

Vy trng hp ny b rng phn xe chy l:B=b1 + b2 = 3,83 x 2 = 7,66 (m)Tnh ton cho trng hp xe ti vi xe con Xe con c chiu rng ph b 1,8mb1=1,8 m c1=1,3 mXe ti c chiu rng ph b 2,5mb2=2,5m c2=1,96m

Vi xe con : B1= x+y+

b2c12

=0,8+0,8+

2, 5 1, 32=3,5 (m)

Vi xe ti :B2=x+y+b2=0,8+0,8+2,5= 4,1(m)Vy trng hp ny b rng phn xe chy l:B=B1+B2= 3,5+4,1=7,6 (m)

B rng l ng ti thiu (Bl):Theo TCVN 4054-05 vi ng cp III a hnh ni b rng l ng l 2x1,5(m).B rng nn ng ti thiu (Bn).B rng nn ng = b rng phn xe chy + b rng l ngBnn = (2x3) + (2x1,5) = 9(m)

11. Tnh s ln xe cn thit:S ln xe cn thit theo TCVN 4054-05 c tnh theo cng thc:

Ngcdz.Nlthnlxe

Trong :nlxe: l s ln xe yu cu, c ly trn theo qui trnhN gc: l lu lng xe thit k gi cao im c tnh n gin theo cng thc sau:N gc = (0,100,12) . Ntbn (xe q/h)

Theo tnh ton trn th nm th 15:Ntbn =3367 (xe con q/ng) => N gc =354,9426xe q/ngy mNlth :Nng lc thng hnh thc t. Trng hp khng c di phn cch v t chy chung vi xe th s Nlth = 1000(xe q/h)

4260, 77.10000.553Z l h s s dng nng lc thng hnh c ly bng 0,85 vi ng cp III

Vy nlxe =

V tnh cho 2 ln xe nn khi n = 0,553 ly trn li n = 1 c ngha l ng c 2 ln xe ngc chiu. dc ngangTa d nh lm mt ng BTN, theo quy trnh 4054-05 ta ly dc ngang l 2%Phn l ng gia c ly chiu rng 1m, dc ngang 2%.Phn l t (khng gia c) ly chiu rng 0,5m, dc ngang 6%. Ta c bng tng hp cc ch tiu k thut nh sau: (Trang bn)

Bng tng hp cc ch tiu k thut

(Bng 2.2.8)

S TTCc ch tiu k thtn vTheo tnh tonThe tiu chunChn Thit k

1Cp hng ngIIIIII

2Vn tc thit Kkm/h6060

3B rng 1 ln xem3,8333

4B rng mt ngm7,6666

5B rng nn ngm999

6S ln xeln0.55322

7Bn knh ng cong nm minm128,84125125

8Bn knh khng siu caom283,4615001500

9Tm nhn 1 chium66,47575

10Tm nhn 2 chium123150150

11Tm nhn vt xem263.19350350

12Bn knh ng cong ng lm minm559,4515001500

13Bn knh ng con ng li minm2343,7525002500

14 dc dc ln nht%07070

15 dc ngang mt ng%02020

16 dc ngang l ng%06060

III.Kt lun:Sau khi tnh ton v nh gi ta s ly kt qu ca bng tra theo tiu chun (TCVN4054-2005) lm c s tnh ton cho nhng phn tip theo.

Chng 3:ni dung Thit k tuyn trn bnh I.Vch phng n tuyn trn bnh :1. Ti liu thit k: Bn a hnh t l 1:10000 c H=5m on tuyn thit k nm gia 2 im T1- T2, thuc huyn Bun Tng Ln, tnh Dak Lak. S ha bnh v a v t l 1:10000 thit k trn Nova3.02. i tuyn:Da vo dng a hnh ca tuyn T1- T2 ta nhn thy s phi s dng 2 kiunh tuyn c bn l kiu g b v kiu ng dn hng tuyn tin hnh vch tuyn.i vi on dc, ta i tuyn theo bc Compa.H . 1 (cm)ittitt= (imax- ip)= 7%-1%=6%ip: l dc nng siu cao,vi Vtk=60 km/h th ip=1%Bng tnh bc compa.Bng 3.1.1

ttItt (%)H(m)(cm)

165100000,833

+ Da vo cch i tuyn nh trn, kt hp cc tiu chun k thut tnh ton v chn la ta c th vch c 2 phng n tuyn sau:Phng n I:

Phng n ny vt o ti cao +650m, sau i tuyn hon ton dc , s dng cc ng cong nm vi bn knh ln, chiu di ton tuyn l 6914m.Phng n II:

Phng n ny i bm st vi khu vc dn c thuc huyn Bun Tng Ln, phn u tuyn nm trn sn mt nh ni c cao 495.84,vt o ti cao +608.27m v i xung sn ni. Do c im i tuyn ca phng n ny g b nn i cn ch gii hn bc compa,s dng ng cong nm ln m bo cho xe chy an ton, thun li. Nhng tuyn c chiu di ln hn tuyn ca phng n 1 l:7078.98m So snh s b cc phng n tuyn.Bng so snh s b cc phng n tuyn.Bng 3.1.2

Ch tiu so snhPhng n

III

Chiu di tuyn69147078

S ng cong nm68

S ng cong c Rmin00

S cng trnh cng1011

Bng trn th hin cc yu t dng so snh la chn phng n tuyn.

II. Thit k tuyn:1. Cm cc tim ngCc cc im u, cui (T1, T2), cc l trnh (H1,2 , K1,2), cc cng (C1,2), cc a hnh, cc ng cong (T,TC,P),2. Cm cc ng cong nm:

TTC

Cc yu t ca ng cong nm:T=R.(tg/2)

Krad.R

0 . .R 180

PRR

R 1 Cos/ 2

Cos/ 2D = 2T-K

Cos/ 2

Trong :T: chiu di tip tuyn P: phn co: gc ngotK: chiu di ng cong R: bn knh ng congThit k cc phng n tuyn chn & cm cc cc phng n xem bnh thit k c s 2 tuyn.

Chng 4:Tnh ton thy vnV Xc nh khu cng

I. Tnh ton thy vn:Thit k cng trnh thot nc nhm trnh nc trn, nc ngp trn ng gy xi mn mt ng, thit k thot nc cn nhm bo v s n nh ca nn ng trnh ng trn t, gy bt li cho xe chy.Khi thit k phi xc nh c v tr t, lu lng nc chy qua cng trnh, t chn khu , chiu di cho thch hp. Lu lng ny ph thuc voa hnh ni tuyn i qua.T iu kin tnh ton thy vn ta xc nh khu cng l mt trong nhng iu kin thit k ng .1. Khoanh lu vc Xc nh v tr l trnh cn lm cng tc thot nc . Vch ng phn thu v t thu phn chia lu vc v cng trnh . Ni cc ng phn thu v t thu d phn chia lu vc cng trnh . Xc nh din tch lu vc . Vi lu lng nh th dn cng v bn cnh bng knh thot nc hoc dng cng cu to 0,75m.2. Tnh ton thy vnKhu vc m tuyn i qua x Bun Tng Ln,Huyn Krong nng, tnh k Lk, thuc vng V (Cc lu vc huyn Krong nng - Ph lc 12 TK ng t tp 3).Cn c vo tiu chun k thut ca tuyn ng vi Vtt = 60 km/h ta xcnh c tn xut l tnh ton cho cu cng l P = 2% (TCVN 4054 - 05 ) tra bng ph lc 15 (TK ng t tp 3/ 257) c H2% = 421 mm.Da vo bnh tuyn ta tin hnh khoanh lu vc cho tng v tr cng sdng rnh bin thot nc v v tr cng (din tch lu vc c th hin trn

bnh ). Tnh ton theo Tiu chun 22 TCN 220-95. Cng thc tnh lu lng thit k ln nht theo tn sut xut hin ca l theo c dng sau:QP% =. Ap.. Hp.. FTrong : F: Din tch lu vc ( km2) Ap: Module dng chy nh l (Xc nh theo ph lc 3/ S tay TK ng t T2) ng vi tn sut thit k trong iu kin cha xt n nh hng ca ao h, ph thuc vols, ts v vng ma. HP: Lu lng ma ngy ng vi tn sut l thit k p% : H s dng chy l (xc nh theo bng 9- 6/TK ng t tp 3/175 hoc ph lc 6/ S tay TK ng t T2), ph thuc vo loi t, din tch lu vc, lng ma. : H s trit gim do h ao v m ly (bng 9-5 sch TK ng t tp 3 hoc bng 7.2.6/ S tay TK ng t T2) ts: Thi gian tp trung nc sn dc lu vc ph thuc vo c trng a

sdmo thu vn bsd : Chiu di trung bnh sn dc lu vc (m) mls : H s nhm lng sui (m=11) isd: dc lng sui ()

ls:c trng a mo lng sui

1/ 4.(.I1/ 3)1/ 4lsls Fm.HP 00ls =

1000.Lc

b0,6sdsdI0,3 .m)0,4.( ..Hsdsdp00

- bsd: chiu di trung bnh ca sn dc lu vc

bsd

F

1,8(

liL)

Trong :l ch tnh cc sui c chiu di > 0,75 chiu rng trung bnh ca lu vc. Vi lu vc c hai mi dc B = F/2LVi lu vc c mt mi dc B = F/LL: l tng chiu di sui chnh (km)(cc tr s tra bng u ly trong "Thit k ng t - Cng trnh vt sng, Tp 3 - Nguyn Xun Trc NXB gi,o dc 1998".Isd : dc lng sui (%0). li : Chiu di sui nhnhSau khi xc nh c tt c cc h s trn (xem thm ph lc 4), thay vo cng thc Q, xc nh c lu lng Qmax.Chn h s nhm msd=0,15Bng 4.1.1:Tnh ton thy vn - lu lng cc cng

Phng n tuyn 1:

sstCngF(km2)L(km)ilsisdlstsApQ2%

1C10.0430.6555350.95201350.0450.58

2C20.0230.1634280.95131700.0540.68

3C30.0221.1855420.95191500.0520.74

4C40.0250.1835280.95212100.0530.67

5C50.0170.4646320.95151400.0430.47

6C60.0120.19871390.96282600.0610.37

7C70.0320.10547590.96141200.0540.635

8C80.0150.17689560.96232100.0520.29

9C90.0950.11579310.96171650.0470.164

10C100.0160.11669530.96161620.0490.288

Phng n tuyn 2:

CongFLBsdIsdilsPilsTsdAplQ

C10.0460.650.03925046201900.0570.4346

C20.0870.820.05893630141200.05980.8623

C30.0540.740.04054742161350.06120.5478

C40.0590.160.30922947181600.06260.6122

C50.1070.580.10253642171500.06191.0978

C60.1130.560.11214036201900.0571.0676

C70.1790.620.16044049191800.06331.8781

C80.0860.660.07244138161350.06120.8724

C90.1120.580.10732843161360.0571.0581

C100.1360.830.09213024151250.06051.3638

C110.1530.670.11923235171300.0721.762

II. La chn khu cng La chn cng ta da trn cc nguyn tc sau: Phi da vo lu lng Qtt v Q kh nng thot nc ca cng. Xem xt yu t mi trng, m bo khng xy ra hin tng trn ngp ph hoi mi trng m bo thi cng d dng chn khu cng tng i ging nhau trn mt on tuyn. Chn tt c cc cng l cng trn BTCT khng p c ming loi thng Tnh ton cao khng ch nn ng: Hn= max Khng ch nc dng H1_ Khng ch chu lc H2_ Khng ch thit k kt cu o ng H3 H1= Hd + 0,5 (Hd = Cao y cng +hd)H2= Cao nh cng +0,5

)H3=H + (0,3-0,5) +hm (H= Cao y +

Sau khi tnh ton c lu lng ca tng cng tra theo ph lc 16 - Thit kng t T3- GSTS KH Nguyn Xun Trc- NXB GD 1998. v chn cng theo bng di y:

Bng 4.2.1:Chn khu cc cngPhng n tuyn 1

Vi triloi cngch chyQs lngHVD

KM0+100trn loi 1khng p0.434610.561.621m

KM0+650trn loi 1khng p0.862310.791.961

KM2+500trn loi 1khng p0.547810.61.71

KM3+550trn loi 1khng p0.612210.621.721

KM4trn loi 1khng p10.97810.942.21

KM4+450trn loi 1khng p1.067610.912.011

KM4+642trn loi 1khng p1.878111.142.421.25

KM5+700trn loi 1khng p0.872410.862.081

Km6+300trn loi 1khng p1.058110.961.61

Km6+750trn loi 1khng p1.363810.992.21.25

Phng n tuyn 2:

Vi triloi cngch chyQs lngHVD

KM0+150trn loi 1khng p0.434610.561.621m

KM0+350trn loi 1khng p0.862310.791.961

KM0+600trn loi 1khng p0.547810.61.71

KM1+400trn loi 1khng p0.612210.621.721

KM1+600trn loi 1khng p10.97810.942.21

KM2+450trn loi 1khng p1.067610.912.011

KM3+600trn loi 1khng p1.878111.142.421.25

KM4trn loi 1khng p0.872410.862.081

Km4+850trn loi 1khng p1.058110.961.61

Km6+250trn loi 1khng p1.363810.992.21.25

Km6+850trn loi 1khng p1.363810.762.051.2

Cao khng ch

sttCongCDTNCDDCHdH1H2H3Hn

1C1717.88717.580.56718.64719.18719.41719.41

2C2730.05729.750.79731.04731.35731.58731.58

3C3725.93725.630.6726.73727.23727.46727.46

4C4723.32723.020.62724.14724.62724.85724.85

5C5716.45716.150.94717.59717.75717.98717.98

6C6713.33713.030.91714.44714.63714.86714.86

7C7705.21704.911.14706.55706.76706.99706.99

8C8703.7703.40.86704.76705705.23705.23

9C9694.08693.780.95695.23695.38695.61695.61

10C10686.19685.890.99687.38687.74687.97687.97

Chng 5:Thit k trc dc & trc ngang

I. Nguyn tc, c s v s liu thit k1. Nguyn tcng c thit k trn cc nguyn tc:+ Bm st a hnh.+ Nng cao iu kin chy xe.+ Tho mn cc im khng ch v nhiu im mong mun, kt hp hi ho gia Bnh -Trc dc-Trc ngang.+Da vo iu kin a cht v thu vn ca khu vc phm vi nh hng ca n tuyn ng i qua.2. C s thit kTCVN4054-05.Bn ng ng mc t l 1/10000, H = 5m trn th hin bnh tuyn.Trc dc ng en v cc s liu khc.3. S liu thit kCc s liu v a cht thu vn, a hnh. Cc im khng ch, im mong mun. S liu v dc dc ti thiu v ti a.II. Trnh t thit kPhn trc dc t nhin thnh cc c trng v a hnh thng qua dc sn dc t nhin xc nh cao o p kinh t.Xc nh cc im khng ch trn trc dc: im u tuyn, cui tuyn, v tr cng,...Xc nh cc im mong mun trn trc dc: im o p kinh t, cao o p m bo iu kin thi cng c gii, trc ngang ch L,...Thit k ng .

III. Thit k ng Sau khi c cc im khng ch (cao im u tuyn, cui tuyn, im khng ch qua cu cng) v im mong mun, trn ng cao t nhin, tin hnh thit k ng .Sau khi thit k xong ng , tin hnh tnh ton cc cao o p, cao thit k ti tt c cc cc.IV. B tr ng cong ngTheo quy phm, i vi ng cp III, ti nhng ch i dc trn ng m hiu i s gia 2 dc 1% cn phi tin hnh b tr ng cong ng .Bn b tr ng cong ng xem thm bn v

l~omBn knh ng cong ng lm minRmin

= 1500m

liBn knh ng cong ng li minR min

=2500 m

Cc yu t ng cong ng c xc nh theo cc cng thc sau:K = R (i1 - i2) (m)

T = R

i1i22

(m)

Trong :

P = T

22R

(m)

i (%): dc dc (ln dc ly du (+), xung dc ly du (-) K : Chiu di ng cong (m)T : Tip tuyn ng cong (m) P : Phn c(m)V. Thit k trc ngang & tnh khi lng o pCc nguyn tc thit k mt ct ngang:Trong qu trnh thit k bnh v trc dc phi m bo nhng nguyn tc ca vic thit k cnh quan ng, tc l phi phi hp hi ha gia bnh , trc dc v trc ngang.

Phi tnh ton thit k c th mt ct ngang cho tng on tuyn c a hnh khc nhau.ng vi mi s thay i ca a hnh c cc kch thc v cch b tr lng, rnh thot nc, cng trnh phng h khc nhau. Chiu rng mt ngB = 6 (m). Chiu rng l ng2x1,5 = 3 (m). Mt ng b tng p phan c dc ngang 2%, dc l t l 6%. Mi dc ta luy nn p 1:1,5. Mi dc ta luy nn o 1 : 1. nhng on c ng cong, ty thuc vo bn knh ng cong nm m c m rng khc nhau. Rnh bin thit k theo cu to, su 0,4m, b rng y: 0,4m. Thit k trc ngang phi m bo n nh mi dc, xc nh cc on tuyn cn c cc gii php c bit.Trc ngang in hnh c th hin trn bn v.2. Tnh ton khi lng o p n gin m vn m bo chnh xc cn thit p dng phng php sau: Chia tuyn thnh cc on nh vi cc im chia l cc cc a hnh, ccng cong, im xuyn, cc H100, Km. Trong cc on gi thit mt t l bng phng, khi lng o hocp nh hnh lng tr. V ta tnh c din tch o p theo cng thc sau:ii+12

Fo tb = (F o + F

o )/2(m )

Fp tb= (Fi p + Fi+1 p)/2(m2)Vo = Fo tb .Li-i+1 (m3)

V= F. L(m )3pp tbi-i+1Sau khi tnh ton ta c din tch nh sau:33Phng n 1: So=96537.4 m ;Sp=76783.89 mPhng n 2: So=47941.12 m3;Sp=96684.14 m3

Chng 6:Thit k kt cu o ng

I. o ng v cc yu cu thit ko ng l cng trnh xy dng trn nn ng bng nhiu tng lp vt liu c cng v cng ln hn so vi nn ng phc v cho xe chy, chu tc ng trc tip ca xe chy v cc yu t thin nhin (ma, gi, bin i nhit ). Nh vy m bo cho xe chy an ton, m thun, kinh t v t c nhng ch tiu khai thc-vn doanh th vic thit k v xy dng ong phi t c nhng yu cu c bn sau:+ o ng phi c cng chung tc l trong qu trnh khai thc, s dng o ng khng xut hin bin dng thng ng, bin dng trt, bin dng co, dn do chu ko un hoc do nhit . Hn na cng o ng phi t thay i theo thi tit kh hu trong sut thi k khai thc tc l phi n nh cng .+ Mt ng phi m bo c bng phng nht nh gim sc cn ln, gim sc khi xe chy, do nng cao c tc xe chy, gim tiu hao nhin liu v h gi thnh vn ti.+ B mt o ng phi c nhm cn thit nng cao h s bm gia bnh xe v mt ng to iu kin tt cho xe chy an ton, m thun vi tc cao. Yu cu ny ph thuc ch yu vo vic chn lp trn mt ca kt cu o ng.+Mt ng phi c sc chu bo mn tt v t sinh bi do xe c ph hoi v di tc dng ca kh hu thi tit l nhng yu cu c bn ca kt cu o ng, ty theo iu kin thc t, ngha ca ng m la chn kt cu o ng cho ph hp tha mn mc khc nhau nhng yu cu ni trn.Cc nguyn tc khi thit k kt cu o ng:+ m bo v mt c hc v kinh t.+ m bo v mt duy tu bo dng.+ m bo cht lng xe chy an ton, m thun, kinh t.

II. Tnh ton kt cu o ng1. Cc thng s tnh tona cht thy vn:t ni tuyn ng i qua thuc loi t bazan, cc c trng tnh ton nh sau:t nn thuc loi 1 (lun kh ro) c: E0 = 44Mpa, C = 0.031 (Mpa),= 120,a=w =0.60 ( m tng i)wnh

Ti trng tnh ton tiu chun:Ti trng tnh ton tiu chun theo quy nh TCVN 4054 i vi kt cuo ng mm l trc xe c ti trng 100Mpa, c p lc l 6.0 daN/cm2 v tc dng trn din tch vt bnh xe c ng knh 33 cm.Lu lng xe tnh tonLu lng xe tnh ton trong kt cu o ng mm l s t c quy i v loi t c ti trng tnh ton tiu chun thng qua mt ct ngang ca ng trong 1 ngy m cui thi k khai thc ( nm tng lai tnh ton): 15 nm k t khi a ng vo khai thc.Thnh phn v lu lng xe

Loi xeThnh phn (%)

Xe ti nng 34

Xe ti nng 23

Xe ti nng 12

Xe ti trung4

Xe ti nh23

Xe bus nh20

Xe con44

T l tng trng xe hng nm : q = 6% Quy lut tng xe hng nm:Nt = N0 (1+q)t-0Trong :q: h s tng trng hng nm

Nt: lu lng xe chy nm th t N0: lu lng xe nm th 15

NN 15

800

Nott

333.8( xe / ngd )

(1q)t

(1q)15

(10.06)15

Bng 6.2.1:Lu lng xe ca cc nm tnh ton

Nm

Lai xe

xe conxe bus nhxe ti nhxe ti trungXe ti nng 1Xe ti nng 2Xe ti nng 3

Tphn %44%20%23%4%3%2%4%

(1+q)t

11.07157.1571.4382.15.14.310.1757.1414.28

21.1516976.77488.315.3511.517.6715.35

31.22179.1881.4493.6616.312.218.1416.3

41.31192.487.45100.5717.513.118.7417.5

51.40205.6293.46107.4818.6914.029.34618.69

61.50220.3100.14115.1620.0315.0210.0120.03

71.61236.46107.48123.6021.516.1210.7521.5

81.72252.62114.82132.0522.9617.2211.522.96

91.83268.77122.17140.524.4318.3212.2124.43

101.97289.33131.51151.2426.319.713.1526.3

112.10308.43140.2161.2528.0421.0314.0228.04

122.25330.462150.21172.7430.0422.5315.0230.04

132.41353.96160.7185.0232.224.1316.0932.2

142.58378.9172.24198.0734.4525.817.2234.45

152.76405.36184.25211.936.8527.6318.4236.85

Bng 6.2.2:D bo thnh phn giao thng nm usau khi a ng vo khai thc s dng

Loi xeTrng lng trc Pi (kN)S trc sauS bnh ca mi cm bnh trc sauKhong cch gia cc trc sau (m)

Trc trcTrc sau

Xe con18.0018.001Cm bnh n

Xe bus nh26.4045.201

Cm bnh i

Ti trung25.8069.601Cm bnh i

Ti nh18.0056.001Cm bnh i

Ti nng loi 1

48.20

100.00

1

Cm bnh i

Ti nng loi 2

45.20

94.20

2

Cm bnh i

3.0

Bng 6.2.3:Bng tnh s trc xe quy i v s trc tiu chun 100 KN

Loi xePi(kN)C1C2niC1.C2.ni.(qi/100)^4.4

Ti trungTrc trc25.816.4323.63

Trc sau69.6113230

Ti nhTrc trc1816.41844.94

Trc sau561118472.38

Ti nngTrc trc48.216.42433.47

Trc sau100112496

Ti nngTrc trc45.416.42426.1

Trc sau902.2124138.56

Ti nngTrc trc23.116.4241.75

Trc sau73.2212455.12

Ntt=C1.C2.ni.(pi/100)^4.4=434.95

C1=1+1.2x(m-1), m L s trc xeC2=6.4 cho cc trc trc v C2=1 cho cc trc sau loi mi cm bnh c 2 bnh (cm bnh i)* Tnh s trc xe tnh ton tiu chun trn 1 ln xe NttNtt =Ntk x flV ng thit k c 2 ln xe khng c di phn cch nn ly f=0.55 .Vy:Ntt =434.94 x 0.55=239.2225 (trc/ln.ngy m)Tnh s trc xe tiu chun tch lu trong thi hn thit k, t l tng trng q=7%

Ne[(1

q)tq

1]* 365* Ntt

Bng 6.2.4:Bng tnh lu lng xe cc nm tnh ton

Nm151015

Lu lng xe Ntt(trc/lnng)99.81133.58178.86239.222

S trc xe tiu chun tch lu (trc)0.0178x1060.13x1060.433x1061.062x106

Theo tiu chun ngnh o ng mm - cc yu cu v ch dn thit k 22TCN 211-2006 (T39). Tr s m un n hi uc xc nh theo bng ph lc III.Bng 6.2.5:Bng xc nh m un n hi yu cu ca cc nm

Nm tnh tonNttCp mtngEyc (Mpa)Eyc min(Mpa)Echon (Mpa)

199.81A2120.56120120.56

5133.58A1151.42130130

A2126.42120126.42

10178.86A1157.14130157.14

A2132.14120132.14

15239.222A1162.35130162.35

Eyc:Mun n hi yu cu ph thuc s trc xe tnh ton Ntt v ph thuc loi tng ca kt cu o ng thit k.Emin: Mun n hi ti thiu ph thuc ti trng tnh ton, cp o ng, lu lng xe tnh ton(bng3-5 TCVN 4054-2005)Echon: Mun n hi chn tnh ton Echn= max(Eyc, Emin)

dvV l ng min ni cp III nn ta chn tin cy l 0.9=> K dc dc

= 1,1

Vy Ech=Kdv

x Eyc=162.35x1.1=178.585(Mpa)

Bng 6.2.6:Bng cc c trng ca vt liu kt cu o ng

Bng 1.6.7

STT

Tn vt liuE (Mpa)

Rn (Mpa)

C (Mpa)

()

Tnh ko un (100)Tnh vng (300)Tnh trt (600)

1BTN cht ht mn18004203002.8

2BTN cht ht th16003502502.0

3Cp phi dm loi I300300300

4Cp phi dm loi II250250250

6Cp phi si cui2002002000.03842

Nn tDat do bazan440.03112

Tra trong TCN thit k o ng mm 22TCN 211-062. Nguyn tc cu to Thit k kt cu o ng theo nguyn tc thit k tng th nn mtng, kt cu mt ng phi kn v n nh nhit. Phi tn dng ti a vt liu a phng, vn dng kinh nghim v xy dng khai thc ng trong iu kin a phng.- Kt cu o ng phi ph hp vi thi cng c gii v cng tc bo dngng. Kt cu o ng phi cng , n nh, chu bo mn tt di tc dng ca ti trng xe chy v kh hu. Cc vt liu trong kt cu phi c cng gim dn t trn xung di ph hp vi trng thi phn b ng sut gim gi thnh.- Kt cu khng c qu nhiu lp gy phc tp cho dy chuyn cng ngh thi cng.3. Phng n u t tp trung (15 nm).C s la chnPhng n u t tp trung 1 ln l phng n cn mt lng vn ban u ln c th lm con ng t tiu chun vi tui th 15 nm (bng tui th

lp mt sau mt ln i tu). Do yu cu thit k ng l ni hai trung tm kinh t, chnh tr vn ho ln, ng cp III c Vtt= 60(km/h) cho nn ta dng mtng cp cao A1 c lp mt B tng nha vi thi gian s dng l 15 nm.S b la chn kt cu o ngTun theo nguyn tc thit k tng th nn mt ng, tn dng nguyn vt liu a phng la chn kt cu o ng; do vng tuyn i qua l vngi ni, l ni c nhiu m vt liu ang c khai thc s dng nh , cp phi dm, cp phi si cui ct, xi mng...Theo tiu chun ngnh 22TCN 211-06 Ne> 2.106 th b dy ti thiu tng

1emt cp cao A = 10cm,da v N tt=2,843.106

.Kt hp vi E yc

nn la chn kt

chcu o ng cho ton tuyn T1-T2 nh sau Phng n I:BTN cht ht mn4cmE1 = 420 (Mpa)

BTN cht ht th6 cmE2 = 350 (Mpa)

CPDD loi IE3 = 300 (Mpa)

CP si cuiE4 = 220 (Mpa)

t nnE0 = 44(Mpa)

Phng n II:

BTN cht ht mn 4cm5cmE1 = 420 (Mpa)

BTN cht ht th 6 cm7 cmE2 = 350 (Mpa)

CPDD loi IE3 = 300 (Mpa)

CPDD loi IIE4 = 250 (Mpa)

t nnE0 = 44 (Mpa)

Kt cu ng hp l l kt cu tho mn cc yu cu v kinh t v k thut. Vic la chn kt cu trn c s cc lp vt liu t tin c chiu dy nh ti thiu, cc lp vt liu r tin hn s c iu chnh sao cho tho mn iu kin v Eyc . Cng vic ny c tin hnh nh sau :Ln lt i h nhiu lp v h hai lp xc nh mun n hi cho lp mtng. Ta c:

Ech = 178.585(Mpa)

BTN cht ht mn4cmE1 = 420 (Mpa)

BTN cht ht th6 cmE2 = 350 (Mpa)

Lp 3E3 = 300 (Mpa)

Lp 4E4 = 220(Mpa)

Nn bazanE0 = 44(Mpa)

i 2 lp BTN v 1 lph14

D33

178.5854200.425.Ech E1

= 0.12

Tra ton hnh 3-1.tiu chun nghnh 22TCN211-06

Ech1 E1

0.39

Ech1

= 163.8(Mpa)

h2D6330.18

Ech1E2163.83500.468

Tra ton hnh 3-1.tiu chun nghnh 22TCN211 06

Ech2 E2

0.41

Ech2

131.2 (Mpa)

chn c kt cu hp l ta s dng cch tnh lp cc ch s H3 v H4 . Kt qu tnh ton c bng sau :

Bng 6.2.7:Chiu dy cc lp phng n I

Gii phph3Ech2 E3H3

DEch3 E3Ech3Ech3 E4EoE4H4DH4H4chn

1140.4370.4240.31930.420.20.7825.7426

2150.4370.4550.29870.3950.20.6922.7723

3160.4370.4850.27810.3680.20.5919.4720

Tng t nh trn ta tnh cho phng n 2:Bng 6.2.8:Chiu dy cc lp phng n II

Gii phph3Ech 2E3H 3DEch3E3Ech3Ech3E4 EoE4H 4DH4H4chn

1140.4710.4240.31930.3720.1760.7524.7525

2150.4710.4550.29870.3480.1760.65621.6822

3160.4710.4850.27810.3240.1760.620.7921

S dng n gi xy dng c bn so snh gi thnh xy dng ban u cho cc gii php ca tng phng n kt cu o ng sau tm gii php c chi ph nh nht. Ta c bng gi thnh vt liu nh sau:Tn vt liun gi (ngn ng/m3)

Cp phi dm loi I145.000

Cp phi dm loi II135.000

Cp phi si i120.000

Ta c kt qu nh sau :Bng 6.2.9:Gi thnh kt cu (ngn ng/m3)Phng n I:

Gii phph3 (cm)Gi thnh ()h4 (cm)Gi thnh ()Tng

11420.3002639.60059.900

21521.7502337.20058.950

31623.2002034.80058.000

Phng n II:

Gii phph3 (cm)Gi thnh ()h4 (cm)Gi thnh ()Tng

11420.3002541.85062150

21521.7502237.80059.550

31623.2002133.75056.950

Kt lun: Qua so snh gi thnh xy dng mi phng n ta thy gii php 3 ca phng n II l phng n c gi thnh xy dng nh nht nn gii php 3 ca phng n II c la chn. Vy y cng chnh l kt cu c la chn tnh ton kim tra.Ta c kt cu o ng phng n chn:Bng 6.2.10:Kt cu o ng phng n u t tp trung

Lp kt cuE yc= 181.94(Mpa)hiEi

BTN cht ht mn4420

BTN cht ht th6350

CPD loi I16300

CPD loi II21250

Nn t st: Enn t = 44Mpa

Kt cu o ng phng n u t tp trungKim tra kt cu theo tiu chun vng n hi:- Theo tiu chun vng n hi, kt cu o ng mm c xem l cng khi tr s mdun n hi chung ca c kt cu ln hn tr s mun

chyccn hi yu cu: E> Ex K dv

(chn tin cy thit k l 0.9 =>Kcd

dv=1.1).

Bnng: Chn h s cng v vng ph thuc tin cy

tin cy0,980,950,900,850,80

H s K dvc1,291,171,101,061,02

Tr s Ech ca c kt cu c tnh theo ton hnh 3-1. xc nh tr s mdun n hi chung ca h nhiu lp ta phi chuyn v h hai lp bng cch i hai lp mt t di ln trn theo cng thc:1Kt1/ 3 3Etb = E4 [1K]

Trong :t =

E3 ; K = h3E4h4

Bng 6.2.11:Xc nh Etbi

Lp kt cuE1(Mpa)th1(cm)kHtbE'tb(Mpa)

Cp phi dm loi II2502132250

Cp phi dm loi I3001.2160.7637323

B tng nha ht th3501.0860.16243327

B tng nha ht mn4201.28440.09348335

H+ T s D

481.4533

nn tr s Etb ca kt cu c nhn thm h s iu

chnh= 1.186 (tra bng 3-6/42. 22TCN 211-06)

=EEtttbtb

= 1.19x335= 397(Mpa)

H+ T cc t s D

601.4533;

Eo

Etttb

44397

0.11

Tra ton hnh 3-1 ta c:

Ta c:

Ech Etb

0.464

Ech

= 0.464x397= 184.208 (Mpa)

cdEch > Eyc x KdvV s trc xe tnh ton trn 1 ngy m trn 1 ln xe l 125.87 nn tra bng (ni suy gia Ntt=100 v Ntt=200) tm c Eyc=156.64 Mpa(ln hn Eyc ti thiu vingcpIIItheobng3-5(tiuchun211-06)l140)dovylyEyc=156.64Mpa kim ton.ng cp III 2 ln xe nn theo bng 3-3, chn tin cy thit k l 0.9,do vydvdv

theo bng 3-2 xc nh c Kcd

=1.1 v Eyc x K

cd=156.64 x1.1=172.304

cdVy Ech = 184.208(Mpa) > Eyc x Kdv

= 181.94 (Mpa)

Kt lun: Kt cu chn m bo iu kin v vng n hi.Kim tra cng kt cu theo tiu chun chu ct trt trong nn t m bo khng pht sinh bin dng do trong nn t, cu to kt cuo ng phi m bo iu kin sau:Ctt

Trong :

ax + av

K tr cd

+ ax: l ng sut ct hot ng ln nht do ti trng xe gy ra trong nn t ti thi im ang xt (Mpa)+av: l ng sut ct ch ng do trng lng bn thn kt cu mt ng gy ra trong nn t (Mpa)+ Ctt: lc dnh tnh ton ca t nn hoc vt liu km dnh (Mpa) trng thi m , cht tnh ton.+Kcdtr: l h s cng v chu ct trt c chn tu thuc tin cytr

thit k (0,9), tra bng 3-7 ta c Kcda. Tnh Etb ca c 5 lp kt cu- Vic i tng v h 2 lp

= 0,94

1Etb = E2 [1

Kt1/ 3K

]3 ; Trong : t =

E1 ; K =E2

h1 h2

Bng 6.2.12:Bng xc nh Etb ca 2 lp mng

Lp vt liuEiHiKtEtbiHtbi

Cp phi dm loi I300160.761.2268.7948

Cp phi dm loi II25021

- Xt t s iu chnh = f(H/D=53/33=1.6) nn = 1.186 Do vy: Etb = 1.186x268.79= 318.78 (Mpa)b. Xc nh ng sut ct hot ng do ti trng bnh xe tiu chun gy ra trong

Etb Eo318.78447.245nn t Tax

H1.45 ;D

E1 E2

Tra biu hnh 3-3.22TCN211- 06 (Trang46), vi gc ni ma st ca t

nn = 12o ta tra c

Tax = 0.0243. V p lc trn mt ng ca bnh xe tiuP

chun tnh ton p = 6daN/cm2 = 0.6 MpaTax=0.0243 x 0.6 = 0.0146 (Mpa)c. Xc nh ng sut ct hot ng do trng lng bn thn cc lp kt cu ong gy ra trong nn t,vi gc ni ma st ca t nn = 12o ta tra c Tav:Tra ton hnh 3 - 4 ta c Tav = 0.00085(Mpa)d. Xc nh tr s Ctt theo (3 - 8)Ctt = C x K1 x K2x K3C: l lc dnh ca nn t st C = 0,031 (Mpa)K1: l h s xt n kh nng chng ct trt di tc dng ca ti trng trng phc, K1=0,6K2: l h s an ton xt n s lm vic khng ng nht ca kt cu, Vi Ntt = 310 < 1000(trc/ln,ng), ta c K2 = 0.8K3: h s gia tng sc chng ct trt ca t hoc vt liu km dnh trongiu kin chng lm vic trong kt cu khc vi mu th. K3 = 1.5 Ctt = 0.031 x 0.6 x 0.8 x 1.5 = 0.0149 (Mpa)

ng cp III, tin cy = 0.9. tra bng 3-7:

Kcd

0.94e. Kim tra iu kin tnh ton theo theo tiu chun chu ct trt trong nn t Tax + Tav= 0.0146+0.00085= 0.0155(Mpa)

Ctttr

= 0.0149 =0.0158 (Mpa)

K cd

0.94

Kt qu kim tra cho thy 0.0155 < 0.0158 => Nn t nn c m boTnh kim tra cng kt cu theo tiu chun chu ko un trong cc lp BTN v cp phi dma. Tnh ng sut ko ln nht lp y cc lp BTN theo cng thc:* i vi BTN lp di:ku=ku x P xkbedTrong :p: p lc bnh ca ti trng trc tnh tonkb: h s xt n c im phn b ng xut trong kt cu o ng di tc dng ca ti trng tnh ton l bnh I => kb= 0.85ku: ng sut ko un n v1600 7 1800 51683.3

h1=12 cm; E1=

57(Mpa)

Tr s Etb ca 2 lp CPD I v CPD II c Etb = 268.79 (Mpa) vi b dy lp ny l H = 37 cm.Tr s ny cn phi xt n tr s iu chnh

Vi

H = 37 = 1.12 Tra bng 3-6 c = 1.135

D

End

33Edctb = 268.79x1.135 = 305.08(Mpa)

Echm0.456dcEtb440.144

Vi

dc

Etb

305.08

, tra ton 3-1, ta xc nh c

=> Echm = 139.12(Mpa)Tmku y lp BTN lp di bng cch tra ton 3-5

H1D10330.303;E1Echm1683.3139.1212.1

Kt qu tra ton c=1.86 v vi p=6(daN/cm2) ta c :ku =1.84x0.6x0.85=0.938(Mpa)*i vi BTN lp trn:H1= 4 cm ; E1= 1800(Mpa)Tr s Etb ca 4 lp di n c xc nh phn trn

1Etb = E2 [1

Kt1/ 3]3K

;Trong :t =

E1 ; K =E2

h1 h2

Lp vt liuEiHiKTEtbiHtbi

BTN cht ht th160060.1625.95374.1143

Cp phi dm loi I300160.761.20268.7937

Cp phi dm loi II2502121

HXt n h s iu chnh = f( D

431.30333

) = 1.17

Etb

dc=1.17x374.11= 437.71 (Mpa)

p dng ton hnh 3-1 tm Echm y ca lp BTN ht nh:

Vi

H431.303 V

Enendat44437.710.1dcEtbD33

Tra ton 3-1 ta c

Echm Etb dc

= 0.45

Vy Echm = 0.45x437.71= 196.97(Mpa)Tmku y lp BTN lp trn bng cch tra ton hnh 3-5 vi

E1Echm1800196.97H1D4339.140.12 ;

Tra ton ta c:ku = 2.15 vi p = 0.6 (Mpa)ku = 2.15 x0.6 x0.85 = 1.098 (Mpa)b. Kim tra theo tiu chun chu ko un y cc lp BTN* Xc nh cng chu ko un tnh ton ca lp BTN theo:

ku

RttkuR cd ku

(1.1)

Trong :

Rttku:cng chu ko un tnh tonRcdku: cng chu ko un c la chnRkutt=k1 x k2 x RkuTrong :K1: h s xt n suy gim cng do vt liu b mi (i vi VL BTN th)11.1111.11

NK1=0.22E

(1.062*106 )0.22 =0.525

K2: h s xt n suy gim nhit theo thi gian k2=1 Vy cng ko un tnh ton ca lp BTN lp di lRkutt = 0.525 x 1.0 x 2.0=1.05 (Mpa)

V lp trn l :

Rku

tt = 0.525x1.0x 2.8=1.47 (Mpa)dc

*Kim ton iu kin theo biu thc (1.1) vi h s Kkubng 3-7 cho trng hp ng cp III ng vi tin cy 0.9 Vi lp BTN lp di:

= 0.94 ly theo

ku = 0.938(Mpa) < 1.050.94

= 1.12(Mpa)

Vi lp BTN lp trn:ku = 1.098(daN/cm2) < 1.470.94

= 1.56(Mpa)

Vy kt cu d kin t c iu kin v cng i vi c 2 lp BTN.Kim tra trt ca lp b tng nha.ax + av [ ] = KxCTrong :+ ax: l ng sut ct hot ng ln nht do ti trng xe gy ra trong nn t ti thi im ang xt (Mpa)+ av: l ng sut ct ch ng do trng lng bn thn kt cu mt ng gy ra trong nn t (Mpa), kim tra trt ca lp b tng nha th khng tnh av v lp ny nm trn cng ca o ng (xem nh av = 0)

+ C: lc dnh tnh ton ca b tng nha C = 0.3 Mpa+K: l h s tng hp K = 1.6- i hai lp b tng nha v mt lp:

Lp vt liuEiHiKtEtbiHtbi

BTN cht ht mn45040.751.28388.7410

BTN cht ht th3506

- i hai lp CPD v mt lp:

Lp vt liuEiHiKtEtbiHtbi

CPD loi I300160.761.2268.7937

CPD loi II25021

HD37331.12Ta c: Etbi = 268.79(Mpa);

H371.12

Xt n h s iu chnh = f( D33

EoEtbmHD373344305.080.1441.12 vEtbm = 268.79x1.135= 305.08 (Mpa)

T:

) = 1.135

Ech.m

0.456

Tra ton 3-1 ta c:

Etbm=> Ech.m = 139.12(Mpa)

10330.303T Etb = 268.79 (Mpa); Ech.m = 139.12(Mpa)

Ta c:

Etb Ech.m

268.791.93 v H139.12D

Tax

Tra ton 3-13/101TCTK ng t ta xc nh c:P=> Tax= 0.35 x 0.6 = 0.21 (Mpa)Tax= 0.21 (Mpa) < [ ] = KxC = 0.48 (Mpa)Vy lp b tng nha m bo iu kin chng trtKt lun

= 0.35

Cc kt qu kim ton tnh ton trn cho thy kt cu d kin m boc tt c cc iu kin v cng .

Chng 7:lun chng kinh t - k thut

so snh la chn phng n tuynI. nh gi cc phng n v cht lng s dng Tnh ton cc phng n tuyn da trn hai ch tiu :+) Mc an ton xe chy+) Kh nng thng xe ca tuyn. Xc nh h s tai nn tng hp

14Ki1H s tai nn tng hp c xc nh theo cng thc sau : Ktn =Vi Ki l cc h s tai nn ring bit, l t s tai nn xy ra trn mt on tuyn no ( c cc yu t tuyn xc nh ) vi s tai nn xy ra trn mt on tuyn no chn lm chun.+) K1 : h s xt n nh hng ca lu lng xe chy y K1 = 0.786.+) K2 : h s xt n b rng phn xe chy v cu to l ng K2 = 1.35.+) K3 : h s c xt n nh hng ca b rng l ng K3 = 1.4+) K4 : h s xt n s thay i dc dc ca tng on ng.+) K5 : h s xt n nh hng ca ng cong nm.+) K6 : h s xt n nh hng ca tm nhn thc t c th trn ng K6=1+) K7 : h s xt n nh hng ca b rng phn xe chy ca cu thng qua hiu s chnh lch gia kh cu v b rng xe chy trn ng K7 = 1.+) K8 : h s xt n nh hng ca chiu di on thng K8 = 1.+) K9 : h s xt n nh hng ca lu lng ch giao nhau K9=1.5+) K10 : h s xt n nh hng ca hnh thc giao nhau K10 = 1.5.+) K11 : h s xt n nh hng ca tm nhn thc t m bo ti ch giao nhau cng mc c ng nhnh K11 = 1.+) K12: h s xt n nh hng ca s ln xe trn ng xe chy K12 = 1.

+) K13 : h s xt n nh hng ca khong cch t nh ca ti phn xe chy K13 = 2.5.+) K14 : h s xt n nh hng ca bm ca mt ng v tnh trng mt ng K14 = 1Tin hnh phn on cng dc dc, cng ng cong nm cacc phng n tuyn. Sau xc nh h s tai nn ca hai phng n :KtnPaII= 6.79 Ktn PaI = 5.84

II. nh gi cc phng n tuyn theo nhm ch tiu v kinh t v xy dng.1. Lp tng mc u t.Bng tng hp khi lng v khi ton chi ph xy lp

TTHng mcn vn giKhi lngThnh tin

Tuyn ITuyn IITuyn ITuyn II

I, Chi ph xy dng nn ng (KXDnn)

1Dn mt bngm25001639501760408197500088020000

2o bp/m34000076783.8947941.1230713556001917644800

3o i/m35000019753.5109876755000

4Chuynt np/m345000048743.0202193457950

5Lu lnm2500086893.593301.2434467500466506000

Tng45754736004665628750

II, Chi ph xy dng mt ng (KXDmt)

1Cc lpkm4.345724.3745868470858306892785597

III, Thot nc (Kcng)

1CngCi850000122550000085000000

D = 0.75m3050

2CngCi11000004688000000132000000

D=1.0m80120

3CngCi1370000328220000043840000

D=1.25m6032

Tng195700000260840000

Gi tr khi ton1161825943011819254350

Bng tng mc u t

TTHng mcDin giiThnh tin

Tuyn ITuyn II

1Gi tr khi ton xy lp trc thuA1161825943011819254350

2Gi tr khi ton xy lp sau thuA' = 1,1A1280085370013001179790

3Chi ph khc:B

Kho st a hnh, a cht1%A116182594.3118192543.5

Chi ph thit k c s0,5%A58091297.1559096271.75

Thm nh thit k c s0,02A2323651.8862363850.87

Kho st thit k k thut1%A116182594.3118192543.5

Chi ph thit k k thut1%A116182594.3118192543.5

Qun l d n4%A464730377.2472770174

Chi ph gii phng mt bng50,00081975000008802000000

B90711931099690807927

4D phng phC = 10%(A' + B)21872046812269198772

5Tng mc u tD = (A' + B + C)2405925149024961186490

2. Ch tiu tng hp.Ch tiu so snh s b.

Ch tiuSo snhnh gi

Pa1Pa2Pa1Pa2

Chiu di tuyn (km)69147078+

S cng1011+

S cong ng115+

S cong nm68

Bn knh cong nm min (m)5001000+

Bn knh cong ng li min (m)25002500

Bn knh cong ng lm min (m)25002500

Bn knh cong nm trung bnh (m)400450+

Bn knh cong ng trung bnh (m)55004750+

dc dc trung bnh (%)1.641.86+

dc dc min (%)0.50.1+

dc dc max (%)3.654.10+

Phng n chn

Ch tiu kinh t.Tng chi ph xy dng v khai thc quy i:Tng chi ph xy dng v khai thc quy i c xc nh theo cng thc

Pq =

Etc

.K qd

tss

Ctxt t

- cl t

Trong :

Eqd

t 1 (1

Eqd )

(1Eqd )

Etc: H s hiu qu kinh t tng i tiu chun i vi ngnh giao thng vn ti hin nay ly Etc = 0,12.Eqd: H s tiu chun qui i cc chi ph b ra cc thi gian khc nhau Eq = 0,08Kqd : Chi ph tp trung tng t quy i v nm gc

Ctx : Chi ph thng xuyn hng nmtss : Thi hn so snh phng n tuyn (Tss =15 nm)cl : Gi tr cng trnh cn li sau nm th t.Tnh ton cc chi ph tp trung trong qu trnh khai thc Ktrt.

itrtKtrtn(1 E )trt1qdKqd = K0 +

Trong :K0 : Chi ph xy dng ban u ca cc cng trnh trn tuyn. Ktr.t: Chi ph trung tu nm t.T nm th nht n nm th 15 c 2 ln trung tu(nm th 5 v nm th 10) Ta c chi ph xy dng o ng cho mi phng n l: Phng n tuyn 1:K0I =24059251490 (ng/tuyn) Phng n tuyn 2:IIK0= 24961186490 (ng/tuyn)Chi ph trung tu ca mi phng n tuyn nh sau:

tKtr PAI =

Ktrt10.08 ttrt

100, 051 240592514900, 051 240592514901403438957 (ng/tuyn)

K=PAIItrt

(1 0.08)5

Ktrt

1 0, 08

10.07 ttrt

100, 051 249611864900, 051 249611864901456051180 (ng/tuyn)

(1 0.08)5

1 0, 08

K0KtrtPAKqd

Tuyn I24,059,250,4901,403,438,95725,462,689,450

Tuyn II24,961,186,4901,456,051,18026,417,237,670

Tnh ton gi tr cng trnh cn lai sau nm th t:CL

cl = (Knn x

100

1550+ Kcng x

15 )x0.7

10050

Kx 100 15nn100Kx 50 15cng50

cl

Tuyn I3,889,152,560136,990,0002,818,299,792

Tuyn II3,965,784,438182,588,0002,903,860,707

Xc nh chi ph thng xuyn hng nm Ctx.DT + C VC + C HK + C TN (/nm)

Trong :

Ctxt = Ctttt

CtDT : Chi ph duy tu bo dng hng nm cho cc cng trnh trnng(mt ng, cu cng, rnh, ta luy...) CtVC : Chi ph vn ti hng nmCtHK : Chi ph tng ng v tn tht cho nn KTQD do hnh khchb mt thi gian trn ng.CtTN : Chi ph tng ng v tn tht cho nn KTQD do tai nn giao thng xy ra hng nm trn ng.a. Tnh CtDT.

0CDT = 0.0055x(K0XDM + K XDC

) Ta c:

Phng n IPhng n II

38,735,322.0738,628,840.78

b. Tnh CtVC:CtVC = Qt.S.LL: chiu di tuynQt = 365. . .G.Nt (T)G: Lng vn chuyn hng ho trn ng nm th t: 3.96=0.9 h s ph thuc vo ti trng

=0.65 h s s dng hnh trnhQt = 365x0.65x0.9x3.96xNt = 845.56xNt (T)S: chi ph vn ti 1T.km hng ho (/T.km)

S= Pbd

+ Pcdd

(/T.km)

. .G

. .G.V

Pc:chi ph c nh trung bnh trong 1 gi cho t (/xe km)Pbd xNi

NPc=i

G: l tI trng TB ca t cc loi G=

Gi. xNi Ni

(tn/ xe)

Loi xeThnh phnTi trngGtb

(%)(T)(T)

Ti nh102.5

4.4

Ti trung44

Ti nng87

Pb: chi ph bin i cho 1 km hnh trnh ca t (/xe.km) Pb=K. . a.r =1 x 2.7 x 0.3x17000=13770(/xe.km)Trong K: h s xt n nh hng ca iu kin ng vi a hnh min ni k=1 : L t s gia chi ph bin i so vi chi ph nhin liu =2.7a=0.3 (lt /xe .km) lng tiu hao nhin liu trung bnh ca c 2 tuyn ) r : gi nhin liu r=17000 (/l)V=0.7Vkt (Vkt l vn tc k thut ,Vkt=30 km/h- Tra theo bng 5.2 Tr125- Thit k ng t tp 4)Pcd+d:Chi ph c nh trung bnh trong mt gi cho t (/xe.h)c xc nh theo cc nh mc x nghip vn ti t hoc tnh theo cng thc:Pcd+d = 12% Pbd= 0.12x13770 = 1652.4

Chi ph vn ti S:

S= 137701652.4 +=5377.16

0.65 0.9 4.4S = 6220.88 (/1T.km)

0.65 0.9 4.4 21

P/a tuynL (km)S (/1T.km)QtCtVC

Tuyn I69145377.16845.56xNt25,181,180.06xNt

Tuyn II70785377.16845.56xNt26,807,410.47xNt

c. Tnh CtHK:

VttC HK = 365N xe con Lc

tchoc

.HcxC

Trong :

tN c: l lu lng xe con trong nm t (xe/ng.)L : chiu di hnh trnh chuyn tr hnh khch (km) Vc: tc khai thc (dng xe) ca xe con (km/h)chtc : thi gian ch i trung bnh ca hnh khch i xe con (gi).Hc: s hnh khch trung bnh trn mt xe conC: tn tht trung bnh cho nn kinh t quc dn do hnh khch tiu ph thi gian trn xe, khng tham gia sn xut ly =7.000(/gi)Phng n tuyn I:

CtHK = 365Ntxe con

5.4800 40

.4x7000

= 1400140x NtPhng n tuyn II:

xe con

CHKt= 365Nt

xe con 5.89640

0 .4x7000

= 1506428x Ntxe cond. Tnh Ctc xe:

Ctx = 0e. Tnh Ctainm :Ctn = 365x10-6 (LixaxCixmixNt) Trong :Ci: tn tht trung bnh cho mt v tai nn = 8(tr/1v.tn) a: s tai nn xy ra trong 100tr.xe/1kma = 0.009xk2tainan - 0.27ktainan + 34.5a1 = 0.009x7.862 - 0.27x7.86 + 34.5 = 32.93 a2=0.009x9.262- 0.27x9.26+ 34.5 = 32.77mi: h s tng hp xt n mc trm trng ca v tai nn = 3.98 mi= m1.m2m11 l xt tng nh hngca iu kinng n tn thtdo mt v tai nn gy ra v xc nh theo bng 5-5 TKD4/tr 131 Phng n tuyn I:

tnC= 365x10-6

(5.4x32.93x8.000.000x3.98xNt) = 2066576.2xNt (/tuyn)

Phng n tuyn II:Ctn = 365x10-6 (5.75898x32.77x8.000.000x3.98xNt) =2193248.9xNt (/tuyn)Ta c bng tnh tng chi ph thng xuyn hng nm (xem phu lc 5)

Phng n IPhng n II

629,875,543,493667,825,760,533

- Ch tiu kinh t:

Pt =

EtcE

xKq +

15Ctxt1 (1Eqd )t- cl t

qd(1

Eqd )

Phngn E tc xKEqqd15Ctx(1E )tt 1qd

cl(1E )tqdPq

Tuyn I38,194,034,180330,856,523,41024,123,177,006344,927,380,650

Tuyn II39,625,856,510350,780,000,00024,855,533,834365,550,322,710

Kt lun: T cc ch tiu trn ta chn phng n I thit k k thut - thi cng.III.nh gi phng n tuyn qua cc ch tiu: NPV; IRr; BCR;THV:(Gi phng n nguyn trng l G, phng n mi l M)1. Cc thng s v ng c( theo kt qu iu tra)Chiu di tuyn: Lc = (1.2-1.3) LI =(1.2-1.3)x5486= 6480 (m) Mt ng dm Chi ph tp trung: V ta gi thit ng c l ng dm nn thi gian trung tu l 3 nm, i tu l 5 nm

tCTt = 20% CT

ca ng mi

= 0.2x0.42x24059250490= 2020977041 ()Tt

CtTt = 28% Ct

ca ng mi

= 0.28x14703438957=392962908 () Chi ph thng xuyn hng nm qui i v thi im hin ti:

tCtxt = CtDT + CtVC + C HKChi ph vn chuyn : CtVC

+ CtTN

(/nm)

CtVC = 1.3(CtVC)M =1.3x25181180.06xNt ()Chi ph hnh khch : CtHK

CHK = Lg x [C HK tLmtTX

] = 1.2x1400140x Nt

xe con

Chi ph tc xe: Ct

Trong :

CtTX =

Qt'*D *Ttx * r

288

()

Qt= 0.1xQt = 0.1x845.56x Nt (T)Ttx =0.5 ( thng)D l gi tr trung bnh ca mt tn hng : 2 triu/1 tn r l sut li nhun kinh t ; r =0.12Ta c :

CTXt=35231.67x NtChi ph do tai nn : CtTN

TNTN

CtTN =1.3x[ Ct

]M Ct

=1.3x2066576.2xNt

Chi ph duy tu sa cha hng nm: CtDTCtDT = 45%( CtDT)M=0.45x38735322.07= 17430894.93 ()Vy chi ph thng xuyn qui i v hin ti l:

15Ctxt1 (1Eqd )t826167620490

=(10.08)15

= 260,442,490,700()

2. Tng li ch cho d n ng, v tng chi ph xy dng ng trong thi gian so snh (n) quy v nm gc:Tng li ch:

B=Bt

tss=

(CVCHK

C[tt

TXTN

CCtt

+K ] -

(1r)t1

(1r)t0 G

tss

(CTN

C HK

CVC

CTx )

tsscl(1r)t1[] + tttt M

1(1

r)t

Bng tnh ton cc thng s ca ng c v ng mi: Xem ph lc 7

Ta c: B =83,192,980,079Tng chi ph xy dng ng:

C=Ct

=[K

CDT+t

TrDT

CCtt]

CDT[t

TrDT

CCtt]

(1r)t0

(1r)tG

(1r)tM

Bng tng chi ph ca tuyn ng c v mi nh sau xem trong ph lc 8Ta c:

C= 181,656,238,200 100,407,562,410= 81,248,675,8003.nh gi phng n tuyn qua ch s hiu s thu chi c qui v thi im hin ti ( NPV):

Bt(1r)tCt(1r)tNPV = B- C =-=

= 83,192,980,079- 81,248,675,800=1,944,304,270()Ta thy NPV > 0Phng n la chn l phng n ng gi.4. nh gi phng n tuyn qua ch tiu sut thu li ni ti ( IRR):

tssBt(1IRR )tCt(1IRR)t11tss= 0

Vic xc nh tr s IRR kh phc tp. nhanh chng xc nh c IRR ta c th s dng phng php gn ng bng cch ni suy hay ngoi suy tuyn tnh theo cng thc ton hc:u tin gi thit sut thu li ni ti IRR = IRR1, sao cho NPV1>0 Sau gi thit IRR=IRR2.

Tr s IRR c ni suy gn ng theo cng thc sau:

IRR=IRR1 +

IRR2NPV1

IRR1/ NPV 2 /

* NPV1

-Gi nh IRR1 = r= 12%NPV1= 60,558,182,480> 0

-Gi nh IRR2= 15%NPV2=

tss

1 (1

Bt-

tssCt(1IRR 2)t1IRR 2)t

Ta c bng tnh tng li ch (xem ph lc 9) v tng chi ph (xem ph lc 10)

tnh NPV2 , da vo bng ph lc 9 v 10 ta tnh c: Tng li ch:B= 66,007,436,805 ()Tng chi ph:C=16,105,200,597 ()NPV2= B- C=66,007,436,805- 16,105,200,597= 49,902,236,208 ()

Ta c :

0.150.1260,558,182,48049,902,236,208IRR=0.12+

x60,558,182,480= 0.14=14%

Ta thy IRR > r. Vy d n u t xy dng ng l ng gi.

5.nh gi phng n tuyn qua ch tiu t s thu chi (BCR):

nBt(1r)t1nCt(1r)t1BCR= B =:C

Trong : r = 0.12. Da vo kt qu tnh ton ca bng trn ta c:

BCR=76,694,194,223: 16,136,011,743= 4.75Ta thy BCR >1. Vy d n xy dng ng l ng gi nn u t.

6.Xc nh thi gian hon vn ca d n:

TC)

Nc ta qui nh vi d n ly r= 12%, th thi gian hon vn tiu chun (Thv l 8.4 nm:Thi gian hon vn c xc nh theo cng thc:Thv=T-tT: Tng thi giant: thi gian cha khai thcT xc nh theo phng trnh

TBtCt(1IRR)tt 0NPV ==0

T ta i tm T tha mn vi iu kin trn,tc l tm T sao cho NPV=0 Tnh ton ta c T= 8.6 nmGi s thi gian khai thc tuyn ng l 1.5 nm => Thv=8.6-1.5=7.1 nm1

Vi r= 12% vi quy nh ca nh nc th Thv t

Lx: Chiu di xn t. Lx = q/L.h (m)L = 3.03(m): Chiu di li ih = 0.1(m): Chiu su xn tLx = 1.368/3.03x0.1 = 4.51(m) Vx: Tc xn t. Vx = 20m/phLc: C ly vn chuyn t. Lc = 20(m)Vc: Tc vn chuyn t. Vc = 50m/phLl: Chiu di li li: Ll = Lx + Lc =4.51+20=24.51(m)Vl: Tc li li. Vl = 60m/ph

4.512024.51(3 2 1)20506060tq: Thi gian chuyn hng. tq = 3(s) tq: Thi gian nng h li i. th = 1(s) tq: Thi gian i s. tq = 2(s).1.134( phut)

Thay vo cng thc tnh nng sut trn ta c nng sut my i vn chuyn ngang o b p l:

N = 60.T.Kt .q.kdt.kr

60x7 x0.75x1.368x11.134x1.2

316.67 (m3/ca)

2. Thi cng vn chuyn dc o b p bng my i D271AKhi thi cng vn chuyn dc o b p vi c ly L < 100m th thi cng vn chuyn bng my i t hiu qu cao nht do kh nng vn chuyn ca n. C th c ly vn chuyn ln n 120 (140) ta dng i vn chuyn vn t hiu qu cao.

Qu trnh cng ngh thi cngBng 3.3

STTCng ngh thi cngYu cu my mc

1o t nn o v vn chuyn ti v tr pMy i D271A

2Ri v san t theo chiu dy cha ln pMy i D271A

3Ti nc t m tt nht( nu cn)Xe DM10

4Lu nn p 6ln/im V = 3km/hLu D400A

5Hon thin cc ch ni tip gia cc onMy i D271A

6m ln mt nn ngLu D400A

2. Thi cng nn ng bng my o + t .Qu trnh cng ngh thi cng

STTCng ngh thi cngYu cu my mc

1o t nn oMy o ED-4321

2Ri v san t theo chiu dy cha ln pMy i D271A

3Ti nc t m tt nht( nu cn)Xe DM10

4Lu nn p 6ln/im V=3km/hLu D400A

5Hon thin cc ch ni tip gia cc onMy i D271A

6m ln mt nn ngLu D400A

Chn my o ED-4321 dung tch gu 0.4m3 c ns tnh theo cng thc sau :

thN8x3600.q.K . Kc

(m3/ca)

Trong :q = 0.4 m3 _ Dung tch guKc _ H s cha y gu Kc = 1.2

Kr T

Kr _ H s ri rc ca t Kr = 1.15T_ Thi gian lm vic trong mt chu k o ca my (s) : T = 17 (s) Kt _ H s s dng thi gian ca my Kt=0.7Kt qu tnh c nng sut ca my o l : N = 494.98 (m3/ca)

Chn t Huynai vn chuyn t:S lng xe vn chuyn cn thit phi bo m nng sut lm vic ca myo , c th tnh theo cng thc sau:

Kd.t't. .Kxn(xe)

Trong :Kd - H s s dng thi gian ca my o, ly Kd= 0.7 Kx - H s s dng thi gian ca xe t Kx= 0.9

QKrqKct - Thi gian ca mt chu k o t t = 15 (s) - S gu y c mt thng xeQ - Ti trng xe : Q = 10 (Tn)Kr - H s ri rc ca t : Kr = 1.15 V - Dung tch gu : V=0.4 (m3)- Dung trng ca t :=1.8T/m3 Kc - H s cha y gu : Kc=1.2t' - Thi gian ca 1 chu k vn chuyn t ca t: t' = 30 pht = 1800 giyThay s ta c :

n0,7.1800 15.10.1,15.0,9

1,8.0,4.1,2

7 (xe)

4. Thi cng vn chuyn t t m p vo nn p bng t Maz503Qu trnh cng ngh thi cngBng 3.4

STTCng ngh thi cngYu cu my mc

1VC t t ni khc n nn p t Maz503

2Ti nc t m tt nht( nu cn)Xe DM10

3Hon thin ch ni tip gia cc onMy i D271A

4m ln mt nn ngLu D400A

Bng tnh ton khi lng cng tc thi cng nn cho tng on

Bin php thi cngon Ion II

VC ngangMy thi cngMy iMy i

Khi lng756.44263.54

C ly vn chuyn1212

Nng sut316.67316.67

S ca2.41

VC dc o b p< 100mMy thi cngMy iMy i

Khi lng752.772539.63

C ly vn chuyn83.4298.12

Nng sut316.67316.67

S ca2.48.02

VC dc o bp >100mMy thi cngt + my xct + my xc

Khi lng18025.5613433.77

C ly vn chuyn180.64147.78

Nng sut494.98494.98

S ca36.4127.1

VC t m vMy thi cngt + my xct + my xc

Khi lng28200.2861966.51

C ly vn chuyn10001000

Nng sut494.98494.98

S ca57125.2

Chng 4:Thi cng chi tit mt ng

I. tnh hnh chungMt ng l 1 b phn quan trng ca cng trnh,n chim 70-80% chi ph xy dng ng v nh hng ln n cht lng khai thc tuyn.Do vy vn thit k thi cng mt ng phi c quan tm mt cch thch ng,phi thi cng mt ng ng ch tiu k thut yu cu a ra thi cng.1. Kt cu mt ng oc chn thi cng l:

BTN ht mn4cm

BTN ht th6cm

CPDD loi I16cm

CPDDloi II21cm

2. iu kin thi cng:Nhn chung iu kin thi cng thun li, CP dm loi I v loi II c khai thc t m trong vng c ly vn chuyn trung bnh 5 KmMy mc nhn lc: C y my mc cn thit,cng nhn c trnh tin hnh thi cngII. Tin thi cng chungCn c vo on tuyn thi cng ta thy on tuyn thi cng li dng con tuyn trc hon thnh do khng phi lm thm ng ph,mt khc m vt liu cng nh phn xng x nghip ph tr u c nm pha u tuyn nn chn hng thi cng t u tuyn l hp l.Phng php t chc thi cng.Kh nng cung cp my mc v thit b y , phc v trong qu trnh thi cng, din thi cng va phi cho nn kin ngh s dng phng php thi cng tun t thi cng mt ng. Chia mt ng lm 2 giai on thi cng.+ Giai on I : Thi cng nn v 2 lp mng CPD.+ Giai on II: thi cng 2 lp mt B Tng Nha.

Ch : Sau khi thi cng xong giai on I phi c bin php bo v lp mt CPD cm khng cho xe c i li, m bo thot nc mt ng tt. Tnh ton tc dy chuyn giai on I:Do yu cu v thi gian s dng nn cng trnh mt ng phi hon thnh trong thi gian ngn nht.Do tc dy chuyn c tnh theo cng thcVL

Trong :

min

Tt kt

L: chiu di tuyn thi cng L= 5472(m) T = min(T1,T2)T1 = TL-tiT2 = TL-tiTL: Thi gian thi cng d kin theo lch TL = 31(ngy)ti : S ngy ngh do nh hng ca thi tit xu. D kin 3ngy T1 = 31-3 = 28(ngy)ti : Tng s ngy ngh l.(3 ngy)=> T1 = 31-3 = 28(ngy)=> Tmin = 28 ngyTkt: Thi gian khai trin dy chuyn , Tkt = 2 ngy5472

Vmin I =

(282)

(m/ngy). Chn VI = 240 (m/ngy)

TtktLTnh tc dy chuyn giai on II: VminII =

Trong :L: chiu di tuyn thi cng L = 5472(m) T = min(T1,T2)T1= TL-tiT2 = TL-tiTL: Thi gian thi cng d kin theo lch TL = 20(ngy)

ti : S ngy ngh do nh hng ca thi tit xu. D kin 3 ngy T1 = 20 - 3 = 17(ngy)ti : Tng s ngy ngh l.(2 ngy)=> T1 = 20-2 = 18(ngy)=> Tmin= 17 ngyTkt: Thi gian khai trin dy chuyn Tkt = 1 (ngy)5472

=>VminII = 17 1

337.5 (m/ngy). Chn VII = 360(m/ngy)

III. Qu trnh cng ngh thi cng mt ng1.Thi cng mt ng giai on I . 1.1.Thi cng o khun o ng Qu trnh thi cng khun o ng Bng 4.3.1STTTrnh t thi cngYu cu mymc

1o khun o ng bng my san t hnhD144

2Lu lng ng bng lu nng bnh thp 4 ln/im; V = 2km/hDU8A

Khi lng t o khun o ng l: V = B.h.L.K1.K2.K3 (m3)Trong :+ V: Khi lng o khun o ng (m3)+ B: B rng mt ngB = 6 (m)+ h: Chiu dy ton b kt cu o ngh = 0.6 m+ L: Chiu di on thi cngL = 240 m+ K1: H s m rng ng congK1= 1.05+ K2: H s ln pK2= 1+ K3: H s ri viK3= 1Vy: V = 6x0,6x240x1,05x1x1 = 907.2 (m3)

Tnh ton nng sut o khun o ng:

N =

Trong :

60.T.F.L.Ktt

(m3/ca)

+ T: Thi gian lm vic mt ca T = 8h+ F: Din tch o: F = B.h =6 0,6 = 3.60 (m2)+ t: Thi gian lm vic mt chu k.

nt =2.LxVx

ncnsVcVs

2.t' nx

ncns

t: Thi gian quay u t =1 pht (bao gm c nng, h li san, quay u v sang s)nx= 5; nc = 2; ns = 1; Vx = Vc= Vs = 80 m/pht (4,8Km/h) Vy nng sut my san l:N=60.8.3, 60.240.0.855508 (m3/ca)2.240.( 521 )2.1.(52 1)808080Bng 4.3.2 :Bng khi lng cng tc v s ca my o khun o ng

TTTrnh t cng vicLoi myn vKhi lngNng sutS ca my

1o khun o ng bng my san t hnhD144M3907.255080.165

2Lu lng ng bng lu nng bnh thp 4 ln/im; V = 2km/hD400Km0.240.4410.544

Thi cng lp cp phi dm loi IIDo lp cp phi dm lai II dy 32 cm nn ta t chc thi cng thnh 2 lp (thi cng hai ln).Gi thit lp cp phi dm lai II l lp cp phi tt nht c vn chuyn n v tr thi cng cch 5km.

Bng 4.3.3 :Qu trnh cng ngh thi cng lp cp phi dm loi II

STTQu trnh cng nghYu cu my mc

1Vn chuyn v dI CPD loi II-lp di theo chiu dy cha ln pMAZ 503+EB22

2Lu s b bng lu nh 4 ln/imSau bt lu rung 6 ln/im;V=2km/hLu nh D469A

3Lu ln cht bng lu nng 10 ln/im; V =3 Km/hLu nng TS280

4Vn chuyn v dI CPD loi II-lp trn theo chiu dy tra ln pMAZ 503+EB22

5Lu s b bng lu nh 4 ln/im;Sau bt lu rung 6 ln/im, V = 2 Km/hLu nh D469A

6Lu ln cht bng lu nng 10 ln/im; V = 3 Km/hLu nng TS280

xc nh c bin ch i thi cng lp cp phi dm loi II ,ta xc nh khi lng cng tc v nng sut ca cc loi myTnh ton khi lng vt liu cho cp phi dm loi II ly theo MCB 1999 BXD c:H1 =16(cm) l 14.55 m3/100m2

2H =16(cm) l 14.55 m3/100m2Khi lng cp phi dm cho on 240 m, mt ng 6 m l: VH1 = 6x14,55x2,4=209.52(m3)VH2 = 6x14.55x2,4 = 209.52(m3) tin cho vic tnh ton sau ny, trc tin ta tnh nng sut lu, vn chuyn v nng sut san.a. Nng sut lu: lu ln ta dng lu nng bnh thp D400 v lu nh bnh thp D469A (S lu b tr nh hnh v trong bn v thi cng mt ng).

Khi lu lng ng v lp mng ta s dung s lu lng ng, cn khi lu ln lp mt ta s dng s lu mt ng.Nng sut lu tnh theo cng thc:

Rlu= L

T.Kt .L 0,01.L

Trong :

.N.V

T: Thi gian lm vic 1 ca (T = 8 gi)Kt: H s s dng thi gian ca lu khi m nn mt ng.Kt=0.8 L: Chiu di thao tc ca lu khi tin hnh m nn L=0.20(Km). (L=200m =0,20 Km chiu di dy chuyn).V: Tc lu khi lm vic (Km/h). N: Tng s hnh trnh m lu phi i.Nyc

N = Nck.Nht =

Nhtn

Nyc: S ln tc dng m nn mt ng t cht cn thit. N: S ln tc dng m nn sau mt chu k (xc nh t s lu).Nht: S hnh trnh lu phi thc hin trong mt chu k (xc nh t s lu).

:H s xt n nh hng do lu chy khng chnh xc (= 1,2).Bng 4.3.4:Bng tnh nng sut lu

Loi luCng vicNycnNhtNV(Km/h)Plu (Km/ca)

D469Lu nh mng ng8283220.33

TS280Lu nng lp mngng2021010030.264

D400Lu nng bnh thp42102030.66

b. Nng sut vn chuyn v di cp phi:Dng xe MAZ-503 trng ti l 7 tn

Pvc =

P.T.K t .K tt

lV1lV2t(Tn/ca)

Trong :P: Trng ti xe 7 (Tn)T: Thi gian lm vic 1 ca (T = 8 gi) Kt: H s s dng thi gian Kt = 0,8 Ktt: H s s dng ti trng Ktt = 1,0L : C ly vn chuyn l = 5 KmT : Thi gian xc vt liu v quay xe, xp vt liu bng xe xc, thi gian xp l 6 pht, thi gian l 4 phtV1: Vn tc xe khi c hng chy trn ng tm V1 = 20 Km/hV2: Vn tc xe khi khng c hng chy trn ng tmV2 = 30 Km/h7.8.0,8.1

5564203060Vy: Pvc =

=76.8 (Tn)

Dung trng ca cp phi dm sau khi ln p l:2,4(T/m3) H s m nn cp phi l:1,5

Vy dung trng cp phi trc khi nn p l:

2.4

1.5

1.6 (T/m3)

Vy nng sut ca xe Maz 503 vn chuyn cp phi l:

76.8

1.6

48 (m3/ca)

Ta c bng th hin khi lng cng tc c ca my thi cng lp cp phi dm loi II (xem bng 4.3.5 trang bn)

Bng khi lng cng tc v ca my thi cng lp cp phi dm loi II

STTQu trnh cng nghLoi myKhi lngn vNng sutS ca my

1Vn chuyn v ri cp phi dm loi II lp diMAZ 503+EB22209.52m3484.365

2Lu s b bng lu nh 4 ln/im; Sau bt lu rung 6 ln/im;V = 2 Km/hD469A0.24km0.330.72

3Lu ln cht bng lu nng 10 ln/im; V = 3 m/hTS2800.24km0.2640.90

4Vn chuyn v ri cp phi dm loi II lp trnMAZ 503+EB22209.52m3484.365

5Lu s b bng lu nh 4 ln/im; V = 2 Km/h,Sau bt lu rung 6 ln/im

D469A

0.24

km

0.33

0.72

6Lu ln cht bng lu nng 10 ln/im; V = 3 m/hTS2800.24km0.2640.90

Bng 4.3.6:Bng t hp i my thi cng lp cp phi dm loi II

STTTn myHiu myS my cn thit

1Xe vn chuyn cp phiMAZ - 50315

2My diEB221

3Lu nh bnh thpD469A2

4Lu nng bnh lpTS2802

5Lu nng bnh thpD4003

Thi cng lp cp phi dm loi I:Bng 4.3.7:Bng qu trnh cng ngh thi cng lp cp phi dm loi I

STTQu trnh cng nghYu cu my

1Vn chuyn v ri cp phi dmMAZ 503+my ri EB22

2Lu s b bng lu nh 4 ln/im,Sau bt lu rung 8 ln/im; V=2 Km/hD469A

3Lu ln bng lu nng 10 ln/im; V= 4 Km/hTS280

4Lu ln cht bng lu nng 4 ln/im; V=3 km/hDU8A

xc nh c bin ch i thi cng lp cp phi dm loi I ,ta xcnh khi lng cng tc v nng sut ca cc loi myTnh ton khi lng vt liu cho cp phi dm loi I ly theo MCB 1999 BXD c: H = 16 (cm)l:14.65/100 (m2)Khi lng cp phi dm cho on 240 m, mt ng 8.0m l:V = 8.0x14.65x2,4 = 281.28 (m3) tin cho vic tnh ton sau ny, trc tin ta tnh nng sut lu, vn chuyn v nng sut san.a, Nng sut lu: lu ln ta dng lu nng bnh thp D400 v lu nh bnh thp D469A, lu bnh lp TS280 (S lu b tr nh hnh v trong bn v thi cng mt ng).Nng sut lu tnh theo cng thc:

Rlu= L

T.Kt .L0,01.L

Trong :

.N.V

T: Thi gian lm vic 1 ca (T = 8 gi)Kt: H s s dng thi gian ca lu khi m nn mt ng.L: Chiu di thao tc ca lu khi tin hnh m nn L = 0.24 (Km). (L = 240m = 0,24 Km chiu di dy chuyn).V: Tc lu khi lm vic (Km/h).

N: Tng s hnh trnh m lu phi i.N yc

N = Nck.Nht =

Nhtn

:Nyc: S ln tc dng m nn mt ng t cht cn thit. N: S ln tc dng m nn sau mt chu k (xc nh t s lu). Nht: S hnh trnh lu thc hin trong 1 chu k (xc nh t s lu).H s xt n nh hng do lu chy khng chnh xc (= 1,2).Bng 4.3.8:Bng tnh nng sut lu

Loi luCng vicNycnNhtNV(Km/h)Plu (Km/ca)

D469Lu nh mng ng82104020.53

TS280Lu nng bnh lp20288040.35

DU8ALu nng bnh thp42122430.66

b. Nng sut vn chuyn cp phi:Dng xe MAZ-503 trng ti l 7 tn

lV1lV2tP.T.K t .K tt

Pvc =

(Tn/ca)

Trong :P: Trng ti xe 7 (Tn)T: Thi gian lm vic 1 ca (T = 8 gi) Kt: H s s dng thi gian Kt = 0,8 Ktt: H s s dng ti trng Ktt = 1,0L : C ly vn chuyn l = 5 KmT : Thi gian xc vt liu v quay xe, xp vt liu bng xe xc, thi gian xp l 6 pht, thi gian l 4 phtV1: Vn tc xe khi c hng chy trn ng tm V1 = 20 Km/hV2: Vn tc xe khi khng c hng chy trn ng tmV2 = 30 Km/h

Vy: Pvc =

7.8.0,8.1

=76.8 (Tn)

5564203060Dung trng ca cp phi dm sau khi ln p l:2,4(T/m3) H s m nn cp phi l:1,5

Vy dung trng cp phi trc khi nn p l:

2.4

1.5

1.6 (T/m3)

Vy nng sut ca xe Maz 503 vn chuyn cp phi l:

Bng 4.3.9:

76.8

1.6

48 (m3/ca)

Bng khi lng cng tc v ca my thi cng lp cp phi dm loi I

STTQu trnh cng nghLoi myKhi lngn vNng sutS ca my

1Vn chuyn v rI cp phi dm loi IMAZ 503+EB22281.28m3485.86

2Lu s b bng lu nh 4 ln/im, V=2 Km/hD469A0.24km0.530.45

3Lu ln bng lu nng10 ln/im; V= 4 Km/hTS2800.24km0.350.68

4Lu ln cht bng lu D400 4 ln/im; V=3 km/hDU8A0.24km0.660.36

Bng 4.3.10:Bng t hp i my thi cng lp CP D loi I

STTTn myHiu myS my cn thit

1Xe vn chuyn cp phiMAZ - 50315

2My riEB221

3Lu nh bnh thpD469A2

4Lu nng bnh lpTS2802

5Lu nng bnh thpDU8A3

2. Thi cng mt ng giai on ii .Thi cng lp mt ng BTN ht thCc lp BTN c thi cng theo phng php ri nng, vt liu c vn chuyn t trm trn v vi c ly trung bnh l 3 Km v c ri bng my ri D150B Bng 4.3.11: Bng qu trnh cng ngh thi cng v yu cu my mc

STTQu trnh cng ngh thi cngYu cu mymc

2Vn chuyn BTN cht ht thXe MAZ - 503

3Ri hn hp BTN cht ht thD150B

4Lu bng lu nh lp BTN 4 ln/im; V =2 km/hD469A

5Lu bng lu nng bnh lp lp BTN 10 ln/im; V = 4 km/hTS280

6Lu bng lu nng lp BTN 4 ln/im; V = 3 km/hDU8A

Khi lng BTN ht th cn thit theo MXD c bn BXD vi lp BTN dy 7 cm:10,51(T/100m2)Khi lng cho on di 360 m, b rng 8 m l: V=8.10.51.3,6=302.69(T)Nng sut lu ln BTN :S dng lu nh bnh st D469A,lu lp TS 280,lu nng bnh thp DU8A,v thi cng BTN l thi cng theo tng vt ri nn nng sut lu c th c tnh theo cng thc kinh nghim,khi tnh ton nng sut lu theo cng thc kinh nghim ta c kt qu ging nh nng sut lu tnh theo s luBng 4.3.12:Bng tnh nng sut lu

Loi luCng vicNycnNhtNV(Km/h)Plu(Km/ca)

D469Lu nh bnh thp42122420.44

TS280Lu nng bnh lp10284040.352

DU8ALu nng bnh thp62123630.264

Nng sut vn chuyn BTN:xe t Maz 503:Dng xe MAZ-503 trng ti l 7 tnP.T.K t .K tt

lV1lV2tPvc =

(Tn/ca)

Trong :P: Trng ti xe 7 (Tn)T: Thi gian lm vic 1 ca (T = 8 gi) Kt: H s s dng thi gian Kt = 0,8 Ktt: H s s dng ti trng Ktt = 1,0 L : C ly vn chuyn l = 3 KmT : Thi gian xc vt liu v quay xe, xp vt liu bng xe xc, thi gian xp l 6 pht, thi gian l 4 phtV1: Vn tc xe khi c hng chy trn ng tm V1 = 20 Km/hV2: Vn tc xe khi khng c hng chy trn ng tmV2 = 30 Km/h7.8.0,8.1

3364203060Vy: Pvc =

=106,7 (Tn)

Dung trng ca BTN cha ln p l:2,2(T/m3) H s m nn cp phi l:1,5Vy nng sut ca xe Maz 503 vn chuyn BTN l:

106.71.5

71.13 (m3/ca)

Lng nha dnh bm (0.5 kg/m2): 360.8.0,5 = 1440(Kg)=1.44(T)Theo bng (7-2) sch Xy Dng Mt ng ta c nng sut ca xe ti nha D164 l: 30 (T/ca)

Bng 4.3.13:Bng khi lng cng tc v ca my thi cng lp BTN ht th

STTQu trnh cng nghLoi myKhi lngn vNng sutS ca

1Ti nha dnh bm(0.5 lt/m2)D164A1.44T300.046

2Vn chuyn v ri BTN ht thXe Maz 503 +D150B302.69T71.134.255

3Lu bng lu nh 4 ln/im; V =2 km/hD469A0.36Km0.440.795

4Lu bng lu lp 10 ln/im; V = 4 km/hTS2800.36Km0.3520.994

5Lu l phng 6 ln/im; V = 3 km/hDU8A0.36Km0.2641.3257

5. Thi cng lp mt ng BTN ht mnCc lp BTN c thi cng theo phng php ri nng, vt liu c vn chuyn t trm trn v vi c ly trung bnh l 3 Km v c ri bng my ri D150B Bng 4.3.14:Bng qu trnh cng ngh thi cng v yu cu my mcSTTQu trnh cng ngh thi cngYu cu mymc

1Vn chuyn BTNXe MAZ - 503

2Ri hn hp BTND150B

3Lu bng lu nh lp BTN 4 ln/im; V =2 km/hD469A

4Lu bng lu nng bnh lp lp BTN 10 ln/im; V = 4 km/hTS280

5Lu bng lu nng lp BTN 6 ln/im; V = 3 km/hDU8A

Khi lng BTN ht mn cn thit theo MXD c bn BXD vi lp BTN dy 5 cm:9.70(T/100m2)Khi lng cho on di 360 m,b rng 8 m l: V=8x9.70x3,6=271.6(T)

Nng sut lu ln BTN: S dng lu nh bnh st D469A,lu lp TS 280,lu nng bnh thp DU8A,v thi cng BTN l thi cng theo tng vt ri nn nng sut lu c th c tnh theo cng thc kinh nghim, khi tnh ton nng sut lu theo cng thc kinh nghim ta c kt qu ging nh nng sut lu tnh theo s lu

Loi luCng vicNycnNhtNV(Km/h)Plu(Km/ca)

D469Lu nh bnh thp42122220.44

TS280Lu nng bnh lp10284040.352

DU8ALu nng bnh thp62123630.264

Nng sut vn chuyn BTN:xe t Maz 503:Dng xe MAZ-503 trng ti l 7 tn

lV1lV2tP.T.K t .K tt

Pvc =

(Tn/ca)

Trong :P: Trng ti xe 7 (Tn)T: Thi gian lm vic 1 ca (T = 8 gi) Kt: H s s dng thi gian Kt = 0,8 Ktt: H s s dng ti trng Ktt = 1,0L : C ly vn chuyn l = 3 KmT : Thi gian xc vt liu v quay xe, xp vt liu bng xe xc, thi gian xp l 6 pht, thi gian l 4 phtV1: Vn tc xe khi c hng chy trn ng tm V1 = 20 Km/hV2: Vn tc xe khi khng c hng chy trn ng tmV2 = 30 Km/h7.8.0,8.1

3364203060Vy: Pvc =

=106,7 (Tn)

Dung trng ca BTN cha ln p l:2,2(T/m3) H s m nn cp phi l:1,5

Vy nng sut ca xe Maz 503 vn chuyn BTN l:

106.71.5

71.13 (m3/ca)

Bng 4.3.15:Bng khi lng cng tc v ca my thi cng lp BTN ht mn

STTQu trnh cng nghLoi myKhi lngn vNng sutS ca

1Vn chuyn v ri BTND164A271.6T71.133.81

2Lu bng lu nh 4 ln/im; V =2 km/hD469A0.36Km0.440.795

3Lu bng lu lp 10 ln/im; V = 4 km/hTS2800.36Km0.3520.994

4Lu l phng 6 ln/im; V = 3 km/hDU8A0.36km0.2641.325

Bng tng hp qu trnh cng ngh thi cng o ng giai on I

TTQu trnh cng nghLoi myKhi lngn vNng sutS ca

1o khun o ng bng my san t hnhD144831.6m355080.165

2Lu nng bnh thp4 ln/im; V = 2km/hD4000.24km0.4410.451

3Vn chuyn v ri cp phi dm loi II- lp1MAZ 503+EB22

209.52

m3

48

4.365

4Lu s b bng lu nh 4 ln/im; bt lu rung 6 ln/im;V = 2 Km/h

D469A

0.24

km

0.33

0.72

5Lu ln cht bng lu nng 10 ln/im; V = 3 m/h

TS280

0.24

km

0.264

0.90

6Vn chuyn v rI cp phi dm loai II- lp2MAZ 503+EB22

209.52

m3

48

4.365

7Lu s b bng lu nh 4ln/im;bt lu rung 6 ln/im; V = 2 Km/h

D469A

0.24

km

0.33

0.72

8Lu ln cht bng lu nng10 ln/im;V=3 km/h

TS280

0.24

km

0.264

0.90

9Vn chuyn v ri cp phi dm loi IMAZ 503+EB22281.28m3485.86

10Lu s b bng lu nh 4 ln/im, V=2 Km/hD469A0.24km0.530.45

11Lu ln bng lu nng 16 ln/im; V= 4 Km/hTS2800.24km0.350.68

12Lu ln cht bng luD400 4 ln/im; V=3 km/h

DU8A

0.24

km

0.66

0.36

Bng tng hp qu trnh cng ngh thi cng o ng giai on II

14Ti nha dnh bm(0.5 lt/m2)D164A1.44T300.048

15Vn chuyn v ri BTN ht thXe Maz 503+D150B

294.28

T

71.13

4.137

16Lu bng lu nh 4 ln/im; V =2 km/hD469A0.36Km0.440.795

17Lu bng lu lp 10 ln/im; V = 4 km/hTS2800.36Km0.3520.994

18Lu l phng 6 ln/im; V = 3 km/hDU8A0.36Km0.2641.3257

19Vn chuyn v ri BTND164A271.6T71.133.818

20Lu bng lu nh 4 ln/im; V =2 km/hD469A0.36Km0.440.795

21Lu bng lu lp 10 ln/im; V = 4 km/hTS2800.36Km0.3520.994

Svth: Nguyn Tun Anh Mssv: 100102 Lp: C1001Trang: 109

22Lu l phng 6 ln/im; V = 3 km/hDU8A0.36km0.2641.3257

Tnh ton la chon s my v thi gian thi cng giai on I

STTQu trnh cng nghLoi myS ca myS myS ca thi cngS gi thi cng

1o khun o ng bng my san t hnhD1440.16510.1651.32

2Lu nng bnh thp 4 ln/im; V = 2km/hD4000.54130.1811.451

3Vn chuyn v ri cp phi dm loi II- lp1MAZ 503+EB22

4.128

150.2912.328

4Lu s b bng lu nh 4 ln/im; bt lu rung 6 ln/im;V = 2 Km/h

D469A

0.72

20.362.88

5Lu ln cht bng lu nng 10 ln/im; V = 3 m/h

TS280

0.90

20.453.6

6Vn chuyn v ri cp phi dm loai II- lp2MAZ 503+EB22

4.128

150.2912.328

7Lu s b bng lu nh 4 ln/im;bt lu rung 6 ln/im; V = 2 Km/h

D469A

0.72

20.362.88

8Lu ln cht bng lu lp 10 ln/im;V=3 km/hTS2800.9020.3753

9Vn chuyn v ri cpMAZ 5.86150.393.125

phi dm loi I503+EB22

10Lu s b bng lu nh 4 ln/im, V=2 Km/hD469A0.44520.2251.8

11Lu ln bng lu lp 10 ln/im; V= 4 Km/hTS2800.6820.342.72

12Lu ln cht bng lu nng 4 ln/im; V=3 km/h

DU8A

0.36

1

0.4

3.024

Tnh ton la chon s my v thi gian thi cng giai on II

14Ti nha dnh bm(0.5 lt/m2)D164A0.04810.0460.368

15Vn chuyn v ri BTN ht thXe Maz 503+D150B4.137150,2763.54

16Lu bng lu nh 4 ln/im; V =2 km/h

D469A

0.795

20.3983.87

17Lu bng lu lp 10 ln/im; V = 4 km/h

TS280

0.994

20.4973.976

18Lu l phng 6 ln/im; V = 3 km/h

DU8A

1.3257

30.44193.73

19Vn chuyn v ri BTN ht mnD164A3.818150.2553.46

20Lu bng lu nh 4 ln/im; V =2 km/h

D469A

0.795

20.3983.87

21Lu bng lu lp 10 ln/im; V = 4 km/h

TS280

0.994

20.4973.976

22Lu l phng 6 ln/im; V = 3 km/h

DU8A

1.3257

30.4423.73

3. Thnh lp i thi cng mt ng:+ 1 my ri D150B+ 15 t MAZ 503+ 2 lu nh bnh thp D469A+ 2 lu nng bnh lp TS 280

+ 3 lu nng bnh thp D400+ 3 lu nng bnh thp DU8A+ 1 xe ti nha D164A+ 15 cng nhn

Phn III: Thit k k thut

on tuyn t km0+00 - km1+00 (Trong phn thit k s b )

CHNG 1 : NHNG VN CHUNG

1. Tn d n : D n xy dng tuyn T1 T2.2. a im : Huyn Krong nng tnh k Lk3. Ch u t : UBND tnh k Lk u quyn cho S giao thng cng chnh tnh k Lk thc hin.4. T chc t vn : BQLDA tnh k Lk5. Giai on thc hin : Thit k k thut.Nhim v c giao : Thit k k thut 1Km

I) Nhng cn c thit k Cn c vo bo co nghin cu kh thi (thit k s b) c duyt caon tuyn t Km0+00 Km6 + 914 Cn c vo cc quyt nh, iu l v.v... Cn c vo cc kt qu iu tra kho st ngoi hin trngII) Nhng yu cu chung i vi thit k k thut Tt c cc cng trnh phi c thit k hp l tng ng vi yu cu giao thng v iu kin t nhin khu vc i qua. Ton b thit k v tng phn phi c lun chng kinh t k thut ph hp vi thit k s b c duyt. m bo cht lng cng trnh, ph hp vi iu kin thi cng, khai thc. Phi ph hp vi thit k s b c duyt.

Cc ti liu phi y , r rng theo ng cc quy nh hin hnh.III. Tnh hnh chung ca on tuyn:on tuyn t KM1+750KM2+850 nm trong phn thit k s b c duyt. Tnh hnh chung ca on tuyn v c bn khng sai khc so vi thit k s b c trnh by. Nhn chung iu kin khu vc thun li cho vic thit k thi cng

CHNG 2 : THIT K TUYN TRN BNH

I) Nguyn tc thit k:1) Nhng cn c thit k.Cn c vo bnh t l 1/1000 ng ng mc chnh nhau 1m, a hnh& a vt c th hin mt cch kh chi tit so vi thc t.Cn c vo cc tiu chun k thut tnh ton da vo quy trnh, quy phm thit k thc hin trong thit k s b.Vo cc nguyn tc khi thit k bnh nu trong phn thit k s b.2) Nhng nguyn tc thit k.Ch phi hp cc yu t ca tuyn trn trc dc, trc ngang v cc yu t quang hc ca tuyn m bo s u n, un ln ca tuyn trong khng gian.Tuyn c b tr, chnh tuyn cho ph hp hn so vi thit k s b m bo yu cu k thut, cht lng gi thnh.Ti cc v tr chuyn hng ca tuyn phi b tr ng cong trn, trn ccng cong ny phi b tr cc cc T, TC, P V c b tr siu cao, chuyn tip theo tiu chun k thut tnh ton.Tin hnh di cc : Cc Km, cc H, v cc cc chi tit, cc cc chi tit th c 20 m ri mt cc, ngoi ra cn ri cc ti cc v tr a hnh thay i, cng trnh vt sng nh cu, cng, nn li dng cc cc ng cong b tr cc cc chi tit trong ng cong.Bng cm cc chi tit xem ph lcII) Nguyn tc thit k1) .Cc yu t ch yu ca ng cong trn theo .Gc chuyn hng RChiu di tip tuyn T = Rtg /2

Chiu di ng cong trnK =

Phn cP= R(

1- 1)cos

180

2 Vi nhng gc chuyn hng nh th R ly theo quy trnh.Trn on tuyn t k thut c 1 ng cong nm, c b tr vi nhng bn knh hp l ph hp vi iu kin a hnh, cc s liu tnh ton c th trong bng

Bng cc yu t ng cong

P=RxSTT1 R T=Rtg2162.30K=1800nhL trnhGc ngotR(m)( 1 cos21.561)3001621P1Km1+150600158.512) c im khi xe chy trong ng cong trn.Khi xe chy t ng thng vo ng cong v khi xe chy trong ng cong th xe chu nhng iu kin bt li hn so vi khi xe chy trn ng thng, nhng iu kin bt li l: Bn knh ng cong t +chuyn bng R . Khi xe chy trong ng cong xe phi chu thm lc ly tm, lc ny nm ngang, trn mt phng thng gc vi trc chuyn ng, hng ra ngoi ng2

cong v c gi tr t 0 khi bt u vo trong ng cong v t ti C = GVgRvo trong ng cong.

khi

Gi tr trung gian: C =

GV 2

gpTrong C : L lc ly tmG : L trng lng ca xe V : Vn tc xe chyp : Bn knh ng cong ti ni tnh ton R : Bn knh ng cong nm.Lc ly tm c tc dng xu, c th gy lt xe, gy trt ngang, lm cho vic iu khin xe kh khn, gy kh chu cho hnh khch, gy h hng hng ho .Lc ly tm cng ln khi tc xe chy cng nhanh v khi bn knh cong cng nh. Trong cc ng cong c bn knh nh lc ngang gy ra bin dng ngang ca lp xe lm tiu hao nhin liu nhiu hn, xm lp cng chng hao mn hn. Xe chy trong ng cong yu cu c b rng ln hn phn xe chy trnng thng th xe mi chy c bnh thng. Xe chy trong ng cong d b cn tr tm nhn, nht l khi xe chy trong ng cong nh on ng o. Tm nhn ban m ca xe b hn ch vn pha ca xe ch chiu thng trn mt on ngn hn.

Chnh v vy trong chng ny s trnh by phn thit k nhng bin php cu to ci thin nhng iu kin bt li trn sau khi b tr ng cong trn c bn trn bnh , cho xe c th chy an ton, vi tc mong mun, ci thin iu kin iu kin lm vic ca ngi li v iu kin l hnh ca hnh khch.III) B tr ng cong chuyn tipNh trnh by trn khi xe chy t ng thng vo ng cong th xe chu nhng iu kin bt li : Bn knh t +chuyn bng R.

Lc ly tm t ch bng 0 t ti

GV 2.

gR- Gchp thnh gia trc bnh trc v trc xe t ch bng khng (trnng thng) ti ch bng(trn ng cong).Nhng thay i t ngt gy cm gic kh chu cho li xe v hnh khch, i khi khng th thc hin ngay c, v vy m bo c s chuyn bin iu ho cn phi c mt ng cong chuyn tip gia ng thng vng cong trn.ng cong chuyn tip c dng y l ng cong Clothoide. Chiu di ng cong chuyn tip c xc nh theo cng thc :V 3

Trong

Lct =

47IR

R - Bn knh ng cong trn.V -Tc tnh ton xe chy (km/h), ng vi cp ng tnh ton V = 60km/h.I - tng gia tc ly tm I = 0.5.+ Vi ng cong trn nh 1.V = 60 km/h;I = 0,5 ;R = 250 m.603

=> Lct =

= 18.38(m).47.0, 5.500

Lnsc = isc*B/insc =0.03*6/0.01 =18m;Theo quy nh ca quy trnh th chiu di ng cong chuyn tip, on ni siu cao, on ni m rng trong ng cong c b tr trng nhau.Vi ng cong trn vic chn chiu di ng cong chuyn tip cn ph thuc vo chiu di on ni siu cao.

IV) B tr siu cao gim gi tr lc ngang khi xe chy trong ng cong c th c cc bin php sau:Chn bn knh R ln. Gim tc xe chy.Cu to siu cao: Lm mt ng mt mi, v pha bng ng cong v nng dc ngang ln trong ng cong.Nhn chung trong nhiu trng hp hai iu kin u b khng ch bi u kin a hnh v iu kin tin nghi xe chy. Vy ch cn iu kin th 3 l bin php hp l nht.H s lc ngang :

2= V+ igRn1) dc siu cao dc siu cao c tc dng lm gim lc ngang nhng khng phi l khng c gii hn. Gii hn ln nht ca dc siu cao l xe khng b trt khi mt ng b trn, gi tr nh nht ca siu cao l khng nh hn dc ngang mt ng ( dc ny ly ph thuc vo vt liu lm mt ng, ly bng 2% ng vi mt ng BTN cp cao)Vi bn knh ng cong nm chn v da vo quy nh ca quy trnh la chn ng vi Vtt = 60 Km/h.- nh P1 c : R = 500 isc = 2%.2. Cu to on ni siu cao.on ni siu cao c b tr vi mc ch chuyn ho mt cch iu ho t trc ngang thng thng (hai mi vi dc ti thiu thot nc ) sang trc ngang c bit c siu cao (trc ngang mt mi ). iscin(B)2ipChiu di on ni siu cao:( Vi phng php quay quanh tim).

Lsc =

Trong Lsc: Chiu di on ni siu cao . isc : dc siu cao.in : dc ngang mt, in= 2%B: B rng mt ng phn xe chy (gm c l gia c) B = 8 m. : m rng phn xe chy trong ng cong.

Vi ng cong c bn knh R =500 m, theo tiu chun 4054-05 th xe chy thun li v m bo tm l hnh khch khi chay trn ng cong th nn ly m rng bng 0.iP : dc dc ph tnh bng phn trm (%), ly theo quy nh iP = 0.5%Bng tnh ton Lnsc

S TT1nh ng congP1isc(%)2Lsc (m)32Theo quy nh ca quy trnh th chiu di ng cong chuyn tip von ni siu cao c b tr trng nhau v vy chiu di on chuyn tip hay ni siu cao phi cn c vo chiu di ln trong hai chiu di v theo quynh cu tiu chunBng gi tr chiu di on chuyn tip hay ni siu cao

STTnh ng congLtt (m)Ltc (m)La chn

1P1325050

- Kim tra dc dc ca on ni siu cao: m bo dc dc theo mp ngoi ca phn xe chy khng vt qu dc dc cho php ti a i vi ng thit k. Ta kim tra dc dc caon ni siu cao.Xc nh dc dc theo mp ngoi phn xe chy im:im = i + iPTrong : i dc dc theo tim ng trn on cong .iP dc dc ph thm trn on ni siu cao c xc nh theo s .

Svth: Nguyn Tun Anh Mssv: 100102 Lp: C1001Trang: 119

B.isc L8, 6 0, 02500.344%+ ng vi ng cong nh P1: nm trong on i dc c imax = 0,02ip

im= 1,2% + 0,516% = 1,72%m bo nh hn dc dc cho php imax = 7%

B.isc L8, 6 * 0, 02500,516%+ ng vi ng cong nh 3: imax = 4.8%ip

im= 1,3% + 0,516% = 1.816%m bo nh hn dc dc cho php imax= 7%. Chuyn tip t trc ngang hai mi sang trc ngang mt mi trn on ni siu cao.Vic chuyn t trc ngang mt mi sang trc ngang hai mi c b tr siu cao c thc hin theo trnh t sau:

i=1.03%

ip=SO O NNG SIU CAO QUANH TIM

ghi ch

1Tim ng

23Mp ng phn xe chy pha lng ng cong Mp ng phn xe chy pha bng ng cong

5467Mp phn m rng pha lng ng cong Mp phn m rng pha bng ng cong Mp l ng pha lng ng congMp l ng pha bng ng cong

V) Trnh t tnh ton v cm ng cong chuyn tip

Phng trnh ng cong chuyn tip Clothoide l phng trnh c chuyn sang h to Descarte c dng

x = s -

y =

S 540A4 S 36 A2

tin cho vic tnh ton v kim tra ta c th da vo bng tnh sn tnh ton.1) Trnh t tnh ton v cm ng cong chuyn tip. Xc nh cc yu t ca ng cong tng ng vi cc yu t ca ng cong trn trong bng tnh trn. T chiu di ng cong chuyn tip xc nh c thng s ng cong

R.L A

ng cong nh P1:A =500x50

= 158.11(m). chn A= 158.11 (m)

nh P1 : R = 250 mR/3 =