Upload
vanle
View
7
Download
0
Embed Size (px)
DESCRIPTION
tkt
Citation preview
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 =