TKMH: Kt cu tu

1.Nhim v thit kThit k kt cu tu hng kh hot ng vng bin khng hn ch c cc thng s c bn sau:Chiu di tu: L= 113,8 m.Chiu rng tu : B= 17 m.Chiu cao mn: D= 10,3 m.Chiu chm tu: d= 6,8 m.H s bo th tch: Cb= 0,77.Vn tc thit k:v = 13 hl/h.Vng hot ng : Bin khng hn ch.2. La chn quy phm:+) Quy phm phn cp v ng tu bin v thp QCVN21- 2010/BGTVT - kt cu thn tu v trang thit b (tu c chiu di t 90m tr ln).3.La chn vt liu tnh ton:

Thp ng tu cp A ( Theo quy phm phn cp v ng tu QCVN21-7A : 2010/BGTVT ) c gii hn chy = 235 MPa.4. La chn h thng kt cu:a, Vng gia tu: (Khoang hng) - Dn y kt cu h thng dc. - Dn mn kt cu h thng ngang. - Dn boong c 2 boong Boong thi tit kt cu h thng hn hp: T thanh quy dc ming hm hng n mn kt cu h thng dc. Gia hai thanh quy dc kt cu h thng ngang. Boong xp hng kt cu h thng ngang - Dn vch kt cu vch phng gm np ng, c np ng khe.b, Vng mi: - Dn y kt cu h thng ngang. - Dn mn kt cu h thng ngang. - Dn boong kt cu h thng ngang. - Dn vch kt cu vch phng gm np nm, sng ng.c, Vng ui: - Dn y kt cu h thng ngang. - Dn mn kt cu h thng ngang. - Dn boong kt cu h thng ngang. - Dn vch kt cu vch phng gm np nm, sng ng.d, Vng bung my: - Dn y kt cu h thng ngang. - Dn mn kt cu h thng ngang. - Dn boong kt cu h thng ngang. - Dn vch kt cu vch phng gm np ng, c np ng khe5. Phn khoang:5.1. Khong sn: (iu 5.2)Yu cu: Khong sn chun: S = 2.L + 450 = 2.113,8 + 450 = 677,6 mm. Khong cch chun ca c cu dm dc mn: S = 2.L + 550 = 2.113,8 + 550 = 777,7 mm. Khong sn khoang mi, khoang ui Sf v Sa 610 mm. Khong sn trong on cch ng vung gc mi 0,2L ti vch mi 700 mm. Chn khong sn sai khc vi khong sn chun khng qu 250 mm.Chn khong sn tho mn yu cu: Khong sn khoang ui : Sa = 600 mm. Khong sn khoang mi : Sf = 600 mm. Khong sn t vch ui n vch mi: S= 650mm. Khong cch cc c cu dc: a=710mm .5.2. Phn khoang (iu 11.1 v 11.2 tp 1).a, Chiu di khoang mimax(5%L+3;8%L) Lmi min (5%L;10m)9,104m Lmi min (5,69;10m) = 5,69 mb, Chiu di khoang ui 8%L L 5%L 9,104 m L 5,69 Chn L = 8,4mc, Chiu di khoang my15%L LKM 10%L17,07 m LKM 11,38 mChn LKM = 14,3 md, Chiu di khoang hngLKH 30m. Tra bng 2A/11.1 qui phm vi 102 L 123 (m) S lng vch ti thiu l 6 vche,Chiu cao y iTheo Quy phm 2A/4.2.2 th chiu cao y i c xc nh nh sau: d0 = B/16= 1,0625 mTrong :B= 17 (m) l chiu rng tu.Vy ta chn chiu cao y i l 1,2 (m) S phn khoang theo chiu di:

STTVngT snn snKhong snChiu di

1Khoang ui014600mm8,4m

2Khoang my1436650mm14,3m

3Khoang hng 13668650mm20,8m

3Khoang hng 268100650mm20,8m

5Khoang hng 3100132650mm20,8m

6Khoang hng 4132164650mm20,8m

7Khoang mi164V mi600mm7,9m

8Chiu cao y iTnh : 1,05mChn : 1,2 m

S phn khoang theo chiu rng:Khong cch cc c cu dc vng khoang hng: a = 710 (mm) Bmingkhoang hng 0,7.B = 0,7.17 = 11,9 (m) Chn Bmingkhoang hng = 11,36 (m);Khong cch t sng ph n mn: 2,82 (m);Khong cch t sng ph n sng ph hoc sng ph n sng chnh: 2,84m

Hnh 5.2: Phn khoang theo chiu rng.6. Kt cu vng khoang hng:6.1.Kt cu dn vch:6.1.1B tr kt cu: (iu 11.2)S kt cu

1: np ng khe. 2: np ng thng. 3: Boong 2.

6.1.2.Tnh ton kt cu:a.Tn vch: (iu 11.2.1 v 11.2.2)Theo iu 11.2.1, chiu dy tn vch khng nh hn tr s tnh theo cng thc : t = 3,2.S. 1 +2,5 (mm) S - khong cch np, chn S = 0,71 m h - khong cch thng ng o t cnh di ca tm tn vch n boong vch ng tm tu (m) ( h 3,4m )+ Tm tn th nht - tn di cng Theo iu 11.2.2, mc 1, chiu dy di di cng ca tn vch t nht phi ln hn 1mm so vi chiu dy tnh ton theo cng thc 11.2.1=> Chiu dy t1 = 3,2.S.1 + 2,5 +1 (mm) h1 = D + B/50-d0 = 10,3 + 17/50-1,2 = 9,44 m t1 = 10,628mmChn t1 = 12 mm Theo iu 11.2.2, mc 2, on c y i, di di cng tn vch t nht phi ln n 610mm cao hn mt tn y trn, v theo iu 4.2.2, chiu cao tit din sng chnh khng nh hn B/16, tr trng hp c ng kim chp nhn c bitChiu rng b1 B/16 + 610 = 16800/16+610 = 1672,5 mmChn b1 = 2000 mm = 2mTa c bng kch thc cc tm tn nh sau:

STTht Chiu dy (mm) Chiu rng (m)

Tm th nht9,4410,48122

Tm th hai7,448,69102

Tm th ba5,447,7882

Tm th t3,46,7782

Tm th nm 3,46,7781,44

b.Np thng vch: (iu 11.2.3) Np thng vch di boong 2:(Khoang chnh)Theo iu 11.2.3 Mmen chng un k c mp km xc nh theo biu thc: W = 2,8CShl2 =275,93 (cm3) Trong :+C: h s ph thuc lin kt mt np ( bng 2A/11.2 ). C = 0,8+l: nhp np (m) l = 5 (m).+S: khong cch np gia cng cho vch (m). S = 0,71(m) +h: khong cch thng ng t trung im np n nh ca boong vch o ti tm tu (m). h = l1/2 + l2 + B/50 = 6,94 (m) Chn thp lm np vch c qui cch : thp ch L khng u cnh. Mp km: b = min(0,2l;S) = min(1000;710) =710 (mm). Chn chiu dy ca tm tn mp km: t = 10 (mm). Vy kch thc mp km l: bt = 71010 mm.Bng chn thp:

7101011018012Kt lun: C cu tha mn qui phm.Vy chn np vch c qui cch: L18011012 (mm).

Np thng ni boong ( trong on t boong 2 n boong 1):Theo iu 11.2.3 Mmen chng un k c mp km xc nh theo biu thc: W = 2,8CShl2 =79,562 (cm3) Trong :+C: h s ph thuc lin kt mt np ( bng 2A/11.2 ). C = 0,8+l: nhp np (m) l = 4,1 (m).+S: khong cch np gia cng cho vch (m). S = 0,71 (m)+h: khong cch thng ng t trung im np n nh ca boong vch o ti tm tu (m).+h = ( l2 /2 + B/50)/2 =2,22 (m) V h< 6 (m) nn ta c h= 1,2 + 0,8*2,22 = 2,976 (m)Chn thp lm np vch c qui cch : thp ch L khng u cnh. Mp km: b = min(0,2.l;S) = min(820;710) = 710 (mm). Chn chiu dy ca tm tn mp km: t = 8 (mm). Vy kch thc mp km l: bt = 7108 mm.Bng chn thp:

7108701108Kt lun: C cu tha mn qui phm.Vy chn np vch c qui cch: L110708 (mm)c. Np khe vch:Np khe vch di boong 2(khoang chnh)Theo iu 11.2.3 thi m un chng un (W) cua tit din np phai khng nho hn tri s sau y : W= 2,8CShl2 =1103,737 (cm3 ) Trong : C: h s ph thuc lin kt mt np ( bng 2A/11.2 ). C = 0,8l: nhp np : l = 5 (m).S: chiu rng ca vng m sng phi (m). S = 2,84 (m)h : khong cch thng ng o t trung im ca l ca sng ng hoc o t trung im ca S ca sng nm n nh boong vch ng tm tu (m). Nu khong cch thng ng nh hn 6m th h ly bng 1,2m cng vi 0,8 ln khong cch thng ng thc.h = l1/2 + l2 + B/50 = 6,94 (m)* Chn thp lm sng vch c qui cch : thp ch T . Mp km:b = min(S; l/5) = min(2840; 1000) =1000 (mm).Chn chiu dy ca tm tn mp km: s = 8 (mm).Vy kch thc mp km l: bs = 10008 mm.Chn chiu cao bn thnh : h = 400 (mm)=>kch thc ca bn thnh l : h tt = 400 12 mm+Chiu dy bn cnh : Chn t = 12 (mm)=> kch thc bn cnh l : bc tc = 150 12 mm* Bang chon thep

Kim tra iu kin np vch di sng boong

81240015010001251252012250160156012+Z0: m dun chng un yu cu ca np Z0 = 1103,737 (cm3) +Z: m un chng un thc ca np Z= 1205,92 (cm3)+C=17,7+A: din tch tit din np k c mp km A=146 cm2+W: ti dc tm np. W= h.b.S= 28,7.2,6.2,84= 209,7 (kN)

Suy ra : = 17,64 =17,7Kt lun: C cu tha mn qui phm.Vy chn np vch c qui cch: T (mm).

Np khe ni boong (trong on t boong 2 n boong 1):Theo iu 11.2.6 thi m un chng un (W) cua tit din np phai khng nho hn tri s sau y : W= 2,8CShl2 =332,8 (cm3 ) Trong : C: h s ph thuc lin kt mt np ( bng 2A/11.2 ). C = 0,8l: nhp np : l = 4,1 (m).S: chiu rng ca vng m sng phi (m). S = 2,84 (m)h : khong cch thng ng o t trung im ca l ca sng ng hoc o t trung im ca S ca sng nm n nh boong vch ng tm tu (m). Nu khong cch thng ng nh hn 6m th h ly bng 1,2m cng vi 0,8 ln khong cch thng ng thc.h = l/2+B/50 = 2,39 (m) h= 1,2+ 0,8.2,39= 3,112 (m)* Chn thp lm sng vch c qui cch : thp ch T . +Mp km:b = min(S; l/5) = min(2840; 820) =820 (mm).Chn chiu dy ca tm tn mp km: s = 8 (mm). =>Vy kch thc mp km l: bs = 8208 mm. +Chn chiu cao bn thnh : h = 270 (mm)=> kch thc bn thnh l : h tt = 270 8 mm +Chiu dy bn cnh : Chn t = 6 (mm)=> Vy kch thc ca bn cnh l : bc tc = 100 8 mmBng tra thp :

Kim tra iu kin np vch di sng boong

Z0: m dun chng un yu cu ca npZ0 =332,8 (cm3 )

88270100820125125201225016015608Z: m un chng un thc ca npZ= 380,913(cm3 )C=17,7A: din tch tit din np k c mp km :A=95,2 cm2W: ti dc tm np.W =S.h.b= 28,7.2,6.2,84=209,7(kN)

Suy ra = 17,62=17,7Kt lun: C cu tha mn qui phm.Vy chn np vch c qui cch: T mm6.1.3.Lin kt: (iu 1.1.14 v 1.1.15)

M lin kt ca np vch : lm = = 625 mm (Vi l=5,0m)

chn m c quy cch :

M lin kt ca np vch khoang ni boong : lm = = 512,5 mm (l=4,1m) chn m c quy cch : 6.2. Kt cu dn y:6.2.1B tr kt cu: (Chng 4) Dn y vng khoang hng c b tr kt cu theo s :

1: sng ph 3: dm dc dy (np gia cng) 2: ngang 4: sng chnh. 6.2.2Tn y ngoi: (iu 14.2 v 14.3)1. Tn y: (iu 14.3.1)* Theo iu 14.3.4, tu kt cu theo h thng dc chiu dy tn y ngoi khng nh hn tr s tnh theo biu thc sau: tmin = = 10,667 mm

=11,926 (mm) Trong :+C1 : H s ph thuc vo chiu di tu. C C1 = 1 khi L < 230 m

+: H s ph thuc vo h thng kt cu

Trong :

.: T s ca m un chng un ca tit din ngang thn tu tnh theo l thuyt chia cho m un chng un thc ca tit din thn tu tnh vi y. V ta xt on ngoi 0.3L k t mi tu (ly X = 0.3L).=> C2 = 4,101 > 3,78+S = 0,71 (m) : Khong cch gia cc dm dc y. +d = 6,8 (m) : Chiu chm tu.+L' = 113,8 (m) : Chiu di tu.+h1 = 0 : Chiu cao ct p => Chn t = 12 (mm)2.Tn gia y: (iu 14.2.1)*Chiu rng tn gia y:Theo iu 14.2.1.1, trn sut chiu di ca tu ,chiu rng ca di tn gia y (b) khng c nh hn tr s tnh theo cng thc sau y : b= 2L+1000 =1227,6(mm) =>chn b=1500 (mm)*Chiu dy tn gia yTheo iu 14.2.1.2, chiu dy ca di tn gia y phi ln hn 2 (mm) so vi chiu dy tn y on gia tu => t=14 (mm) Kch thc di tn sng nm: 1500x14 mm.3: Tn hng: (iu 14.3.5)*Chiu dy ca tn hng:Theo 14.3.5, chiu dy ca di tn hng (t) on gia tu phi khng nh hn tr s tnh theo cng thc sau y, tuy nhin cng khng nh hn chiu dy di tn y k vi n :

= 10,28 (mm).Trong : +R = 1,1 (m): bn knh cong hng (o trn tuyn hnh )+a = -0,32 (m) khong cch t cnh di cung hng n dm dc tng ng gn nht +b = -0,1 (m): khong cch t cnh trn cung hng n dm dc tng ng gn nht .Vy (a+b) < 0 ta ly (a+b) = 0 +L = 113,8 (m): chiu di tu.+l = 2,6 (m): khong cch gia cc ngang c .+d = 6,8 (m): chiu chm thit k.Do chiu dy tn y ngoi t= 12 mm , chiu dy tn y trn t = 14 mm =>nn chn t = 14 (mm)6.2.3Tn y trong: (iu 4.5)1. Tn y trong:Theo iu 4.5.1, mc 1, Quy phm phn cp v ng tu bin v thp TCVN 21- 2A: 2010 chiu dy tn y trn khng nh hn tr s tnh theo cc cng thc sau :

t1= = 6,1 (mm)

t2 = = 11,696 (mm)Trong :+d0 = 1,2 (m): Chiu cao tit din sng chnh.+d = 6,8 (m): Chiu chm tu.+B = 17(m): Chiu rng tu.+S = 0,71(m): Khong cch c cu gia cng+h : Khong cch thng ng t mt tn y trn ti boong thp nht o tm tu. h = D + B/50 d0 = 9,44 (m)+C: H s ly theo qui nhC B/lh = 17/20,8= 0,817 => 0,8Vy chn chiu dy sng chnh t = 14 (mm).b, M gia cng cho sng chnh y (iu 4.2.4.2)Yu cu: Khong cch khng qu 1,75m, ta b tr hai khong sn 1,42 (m)M trong hng yu cu chiu dy tng ln 1,5mm so vi tr s tnh theo cng thc 4.2.4.2

(0,6 + 2,5) = (0,6. + 2,5) = 8,9 mm.Chn chiu dy m trong hng l 12mm, cnh t do ca m phi c gia cng thch ng. Nu do hnh dng ca tu m m hng qu di th phi t thanh thp gc b sung dc trn cnh cc m hoc phi dng bin php thch hp khc.c,Np gia cng cho sng chnh y (iu 4.2.4.1)Theo iu 4.2.4.1, do cc m ngang lin kt sng vi dm dc gn nht cch nhau 1,42 > 1,25 (m) nn ta gn thm cc np b sung+Chiu dy ca np t = tsngchnh = 14 + Chiu cao tit din np d 0,08d0 = 0,096. Chn d = 0,1 Vy chn np : hnp x tnp = 100 x 14 vt mp hai u.2.Sng ph y a.tnh chn sng ph y (iu 4.2.3) Theo iu 4.2.3, Chiu dy tit din sng ph y phi khng nh hn tr s tnh theo cng thc sau, ly tr s no ln hn :

t1 = C1 (2.6 - 0.17) + 2.5 =11,23 (mm) t2 = C1'd0 + 2.5 = 11,26 (mm)=>Vy chn chiu dy sng ph t = 14 (mm).Trong : +S = 2,84 (m) : Khong cch t sng ph ang xt n cc sng ph k cn. +d0 = 1.2 (m) : Chiu cao tit din sng ph. +d1 : Chiu cao l khot ti im ang xt, d d chn d = 0,5 (m) +lH = 20,8(m): Chiu di khoang hng. +x = 0,45lH = 9,36 (m): Khong cch theo chiu dc t trung im ca lH ca mi khoang n im ang xt. +y = 3 (m) : Khong cch theo phng ngang t tm tu ti sng dc. +C - H s cho theo cng thc ; ph thuc vo t s Vi 0,817 C1 = = 0,0212 +d = 6,8 (m) Chiu chm tu +B = 17 (m) Chiu rng tu. +S1 = 1.2 (m) Khong cch gia cc np t ti sng ph +C - H s ph thuc vo t s Vi = 1. Tra bng 2A/4.1 ta c C1 = 7,3b,Np gia cng cho sng ph y Np gia cng cho sng ph l thanh thp dt c chiu dy bng chiu dy ca tm sng ph ; c chiu cao khng nh hn 0,08d. Chn : +Chiu dy np gia cng cho sng ph l t = 12 +Chiu cao tit din l 0,1 Vy chn np : hnp x tnp = 100 x14 vt mp hai u.c,L khot trn sng phTrn sng ph on k vch ngang khng t l khot.Trong phm vi 10% chiu di khoang k t mi u khoang, ng knh l khot gim trng lng trn sng ph khng ln hn 1/3 chiu cao tit din sng. Chn ng knh l khot l 0,3 (m)6.2.7. ngang c:(iu 4.3)1) Yu cu b tr:Theo iu 4.3.1.1, ngang c phi c t cch nhau xa khng qu 3,5m Khong cch gia cc ngang l l = 2,6 < 3,5 tha mn2)Chiu dy ngang:

Theo iu 4.3.2, chiu dy ca ngang c khng c nh hn tr s tnh theo cng thc sau, ly tr s no ln hn :

t1 = C2 () + 2.5 = 11,501(mm)

t2 = 8.6 + 2.5 = 12,2118 (mm)Trong :+ B' = 16,5 (m) =(17+17.0,1-2.R): Khong cch gia cc ng nh m hng o mt tn y trn on gia tu +B'' = 16,5 (m): Khong cch gia cc ng nh m hng o mt tn y trn ti v tr ca ngang c +S = 2,6 (m) : Khong cch gia cc ngang c +y = 8,25 (m) :Khong cch t ng tm tu n im kho st theo phng ngang tu, vi y ly y = +d0 = 1,2 (m): Chiu cao tit din ngang y ti im ang xt. + d1 = 0,5 (m): Chiu cao l khot ti tit din kho st + C2 : H s ly theo bng 2A/4.2. Vi 0,817 C2 = 0,022

+ C : H s ly theo bng 2A/4.3. Vi = = 1 C = 9 +H - H s ty thuc vo vic c gia cng bi thng hay khng khi khot l, vi trng hp ngang c c l khot v l nh c gia cng bi thng , H = 1 (v d1/S1 = 0,5/1,2 = 0,416kch thc ca mp km l : b x t = 820 x 12 mm

*Khoang hng 1,2, : h1 = C. (C=1,18- tra bng 2A/8.2) => h1 = 15,102 kN/m2 --> Z = 810,532 cm3Bng chn thp

121233012082012 Kt lun: C cu tha mn qui phm.Vy chn sn khe c qui cch: T

* Khoang hng 3h4 = 32,32 kN/m2 => Z = 1326,346 cm3

121235025082012Kt lun: C cu tha mn qui phm.Vy chn sn khe c qui cch: T

* Khoang hng 4 h4 = 50,382 kN/m2 => Z = 1867,445 cm3Bng chn thp

121246023082012Kt lun: C cu tha mn qui phm.Vy chn sn khe c qui cch: T

C,Sn cng xon trong h thng cng xon:c tnh chn trong mc dn boong.6.3.4Lin kt: Sn thng lin kt vi x ngang boong 2 v dn y bng m c mp b.Chiu di m: lm l/8 = 5000/8 = 625 (mm). Chn : lm = 650 (mm). Tra bng 2A/1.3 ta chn m c kch thc: 650 x 650 x 9 ; Chiu rng mp l 60 mm. Sn thng lin kt vi x ngang boong 2 v boong 2 bng m c mp b.Chiu di m: lm l/8 = 4100/8 = 512,5 (mm). Chn : lm = 550 (mm). Tra bng 2A/1.3 ta chn m c kch thc: 550 x 550 x 8,5 ; Chiu rng mp l 55 mm. Sn khe lin kt vi sng ngang boong 2 v boong 2 bng m c mp b.Chiu di m: lm l/8 = 4100/8 = 512,5 (mm). Chn : lm = 512,5 (mm). Tra bng 2A/1.3 ta chn m c kch thc: 550 x 500 x 8,5 ; Chiu rng mp l 55mm. Sn khe lin kt vi sng ngang boong 2 v dn y bng m c mp b.Chiu di m: lm l/8 = 5000/8 = 625 (mm). Chn lm = 625 (mm). Tra bng 2A/1.3 ta chn m c kch thc: 650x 650 x 9; Chiu rng mp l 60 mm.6.4Kt cu dn boong:6.4.1B tr kt cu:+Tu thit k c hai boong , chiu cao boong di l 5 m (tnh t tn y di ) v chiu cao boong trn l 4,1 m (tnh t tn boong di) .+Boong di kt cu h thng ngang.+Boong trn : Vng t hai thanh quy dc ra mn kt cu h thng dc. Vng trong hai thanh quy dc kt cu h thng ngang.

15600x568015600x5680boong diboong trn 12365412374561: Ct chng 3: Thanh quy dc 5: X ngang boong khe2: Thanh quy ngang 4: X ngang boong thng 6: X ngang cong xon7: 6.4.2 Tn boong:(iu 15.3)A,Ti trng tc dng ln boong thi tit: Ti trng boong h (kN/m2) phi khng nh hn tr s tnh theo cng thc sau: h = max(h* ; hmin; 12,8)Trong : +h* = a(bf y) Cb = 0,77: H s bo th tch. f : H s tnh theo cng thc sau

= = 7,363 y : Khong cch thng ng t ng nc trng ti thit k n boong chu thi tit o mn. y phi c o mi tu cho on boong pha trc ca 0,15L tnh t mi tu , c o 0,15L tnh t mi tu cho on boong t 0,3L n 0,15L tnh t mi tu , c o sn gia cho on boong t 0,3L tnh t mi tu n 0,2L tnh t ui tu v c o ui tu cho on boong pha sau ca 0,2L tnh t ui tu. Ly chung y cho vng t mi tu n 0,2 L tnh t ui theo kch thc o ti sn gia y= 3,1. Khu vc sau 0,2L tnh t ui do c thng tng nn y= 6,148=>H s a, b, h* c xc nh theo bng 2A/8.1

DngV tr ca boongabh*

X boong Tn boongCtSng boongX boong Tn boongCtSng boong

1 pha trc ca 0,15L tnh t mi tu11,774,97,351,35280,6833,58850,382

2T 0,15L n 0,3L tnh t mi tu9,4513,95,91,16551,7721,36332,32

3T 0,3L tnh t mi tu n 0,2L tnh t ui tu5,5262,252,25 3,45123,559,69,6 14,70

4 pha sau ca 0,2L tnh t ui tu7,843,254,91,12816,917,0110,57

+Tr s ca hmin c tnh theo bng sau :DngV tr ca boongChmin

X boong, Tn boongCt, Sng boongX boong, Tn boongCt, Sng boong

1 v 2Pha trc ca 0,3L tnh t mi tu4,21,3753,75317,533

3T 0,3L tnh t mi tu n 0,2L tnh t ui tu2,051,1826,23615,102

4Pha sau ca 0,2L tnh t ui tu2,951,4731,4715,68

Boong thng 2 trn boong mn kh1,950,6920,87,36

=>Vy ti trng boong h = max(h* ; hmin,12,8) c xc nh theo bng sau DngV tr ca boongKhoang hng(sn)h

X boong Tn boongCtSng boong

1 pha trc ca 0,15L tnh t mi tuKhoang hng480,6833,58850,382

2T 0,15L n 0,3L tnh t mi tuKhoang hng 353,75321,26332,32

3T 0,3L tnh t mi tu n 0,2L tnh t ui tu

Khoang hng 1,226,23615,10215,102

4 pha sau ca 0,2L tnh t ui tu31,4715,6815,68

B,Ti trng tc dng cc c cu trn boong chu lc:-i vi boong xp hng :Theo quy phm 2-A/8.2.1-1, ti trng h (kN/m2) i vi boong hng chnh dung xp hng ha thng thng c tnh:h = 7h2= 28,7 (kN/m2)Trong h2 = 4,1 m :chiu cao t boong xp hng n cnh trn ca thnh ming khoang boong chu thi titC,Chiu dy tn boong: ( 15.3.1 qui phm )1.Vi boong xp hngChiu dy tn boong: t = 1,25.C.S. + 2,5 = 6,852 (mm)Trong : +C l h s xc nh theo cng thc: C = 0,905 + = 0,905 + = 1 +L = 113,8 (m) v L < 230(m) nn ta ly L = 230 (m). +S = 0,65 (m): Khong cch gia cc x ngang boong. +h = 28,7 (KN/m2) l ti trng hng ha. =>Chn t = 7 mm2.Vi boong chu thi tit Chiu dy tn boong ngoi vng ng ming l khot on gia tu c x dc boong: t1 = 1,47.C.S. + 2,5 (mm) Chiu dy tn boong trong vng ng ming l khot on gia tu c x ngang boong t2 = 1,25.C.S. + 2,5 Trong +h : l ti trng boong tnh theo bng trn. .Khoang hng 1,2 : h = 26,236 (KN/m2). =>T ta c: t1 = 7,39 mm , t2 = 6,661 mm .Khoang hng 3 : h = 53,753 (KN/m2). =>T ta c : t1 = 9,505 mm , t2 = 8,456 mm .Khoang hng 4 : h = 80,68 (KN/m2). =>T ta c : t1 = 11,08 mm , t2 = 9,798 mmVy ta chn t1 = t2 = 9 mm i vi khoang hng 1,2 t1 = t2 = 11 mm i vi khoang hng 3 t1 = t2 = 12 mm i vi khoang hng 46.4.3X boong: (iu 8.3)A, Boong xp hnga,X ngang boong (iu 8.3.3.2)- M un chng un tit din x ngang boong phi khng nh hn tr s tnh theo cng thc sau: W = 0,43Shl2 =63,784 (cm3)Trong :+S = 0,65 (m): Khong cch gia cc x ngang boong.+h = 28,7 (kN/m2): Ti trng boong.+l = 2,84 (m): Khong cch gia cc sng dc boong.- Chn thp lm x dc boong c qui cch thp ch L khng u cnh. Mp km: tmk =8 (mm) bmk = min( l/5;S ) = min( 560 ; 650 ) = 560 (mm)Vy kch thc ca mp km l : bmk x tmk = 560 x 7 mm

5608631008Kt lun: C cu tha mn qui phm.Vy chn x ngang boong c qui cch: L100638

b,Sng dc boong- M un chng un tit din sng dc boong khng nh hn tr s tnh theo cng thc sau: Z = 0,484.l.( l.b.h+k.w ) = 266,681cm3) Trong : +h = 28,7 (kN/m2) l ti boong.+l = 2,6 m: nhp ca sng boong+b = 2,84 m : chiu rng trung bnh din tch boong c bi sng.+w= :ti trng boong c bi ct ni boong (kN)

+k= = 0 .a =2,6 (m) khong cch t ct ti vch-M men qun tnh ca tit din phi khng nh hn tr s tnh theo cng thc sau: I = 4,2.Z.l = 2912,157 (cm4) -Chiu dy bn cnh ca x ngang boong khe phi khng nh hn chiu dy ca bn thnh.+ Chiu dy ca bn thnh theo 10.2.3 qui phm: tt = 10S1+2,5 = 5,2 (mm). .S1 : Khong cch gia cc np gia cng bn thnh hoc chiu cao tit din bn thnh ly gi tr nh hn. S1 = 0,27 (m). =>Chn tt = 8 (mm)+ Chiu cao tit din bn thnh khng nh hn tr s tnh theo cng thc sau y : d0 = 2,5d1 = 2,5.0,105 = 0,2625 (m)=>kch thc ca bn thnh l : =bt x tt = 270 x 6 mm - Chiu rng bn cnh khng nh hn tr s tnh theo cng thc sau: b = 85,4 = 99,226 mm tc = 8 mm=>Kch thc ca bn cnh l : bc x tc = 100 x 8mmChn thp lm sng dc boong c qui cch thp ch T. Mp km: tmk = 8 (mm) bmk = min(l/5;b) = min( 520;2250 ) = 520 (mm)=>Kch thc ca mp km l : bmk x tmk = 520 x 8 mmBng chn thp:

1058882701000000005200Kt lun: C cu tha mn qui phm.Vy chn x ngang boong khe c qui cch: T

c,Sng dc boong lm thanh quy dc ming hm hng - M un chng un tit din sng dc boong khng nh hn tr s tnh theo cng thc sau: W = 0,484.l.( l.b.h+k.w ) = 2629,25 (cm3) Trong :+l = 5,2 (m): Khong cch gia cc x ngang cng xon.+b = 7,1 (m): Chiu rng trung bnh din tch boong bi thanh quy dc.+k: H s k = 12(1 - )2 = 0 ( ).+w: Ti trng boong c bi ct ni boong.+h = 28,7 (KN/m2) l ti trng tc dng ln sng boong. M men qun tnh ca tit din x ngang boong khe phi khng nh hn tr s tnh theo cng thc sau: I = 4,2.Z.l =57422,82 (cm4)Chiu dy bn thnh sng ngang khng nh hn : t = 10S1 + 2,5 = 8 (mm). .S1 l chiu cao tit din bn thnhNgoi ra i vi cc khoang hng 4 chiu dy bn thnh cn phi tha mn khng nh hn tr s trn v tr s tnh theo cng thc sau, ly tr s no ln hn : t = + 2,5 = 10,914 +d0 l chiu cao tit din bn thnh: d0 = 0,55m Chn thp lm x ngang boong khe c qui cch thp ch T. Mp km: tmk = 8 (mm) bmk = min( l/10;b/2) = min( 520;3500 ) = 520 (mm)Bng chn thp:

Ta cng thy J = 77119,92 (cm4) > I = 57422,82 (cm4) 550x 13520 x 8300x 13 Kt lun: C cu tha mn qui phm.d,Sng ngang boong lm thanh quy ngang - M un chng un tit din sng ngang boong khng nh hn tr s tnh theo cng thc sau: Z = 0,484.l.( l.b.h+k.w ) = 2548,35 (cm3) Trong :+l = 8,5 (m): Chiu di nhp thanh quy ngang.+b = 2,6 (m): Chiu rng trung bnh din tch boong bi thanh quy ngang. +k: H s k = 12(1 - )2 = 0 ( ). +w: Ti trng boong c bi ct ni boong.+h = 28,7 (KN/m2) l ti trng tc dng ln sng boong. -M men qun tnh ca tit din x ngang boong khe phi khng nh hn tr s tnh theo cng thc sau: I = 4,2.Z.l =89905,8 (cm4) -Chiu dy bn thnh sng ngang khng nh hn : t = 10S1 + 2,5 = 8 (mm). Trong S1 l chiu cao tit din bn thnh S1 =0,55 m -Ngoi ra i vi cc khoang hng 1 v 4 chiu dy bn thnh cn phi tha mn khng nh hn tr s trn v tr s tnh theo cng thc sau, ly tr s no ln hn : t = + 2,5 = 7,548mmTrong : d0 = 0,55 (m): l chiu cao tit din bn thnhChn thp Mp km: tmk = 8 (mm) bmk = min(l/10;b/2) = min( 840;4550 ) = 840 (mm)Bng chn thp:

Ta cng thy J = 88239,9(cm4) > I = 89905,8 (cm4)

550 x 12840 x 8300 x 12Kt lun: C cu tha mn qui phm.e,Sng ngang boong - M un chng un tit din sng ngang boong khng nh hn tr s tnh theo cng thc sau: Z = 0,484 l( lbh+kw ) = 287,195 (cm3) Trong :+l = 2,84 (m): Chiu di nhp sng ngang boong.+b = 2,6 (m): Chiu rng trung bnh din tch boong bi sng ngang boong.+h = 28,7 (KN/m2) : Ti trng boong -Momen qun tnh tit din ca sng ngang boong: I = 4,2.Z.l = 3329,84 (cm4 )Chn thp Mp km: tmk = 8 (mm) bmk = min(l/5;b) = min( 560;2600 ) = 560(mm)Bng chn thp:

Kt lun: C cu tha mn qui phm.

882501005608Vy chn sng vch c qui cch: T

f,H x ngang cng xon*Bao gm x ngang cng xon v sn kho x ngang cng xon* X ngang cong xon ( theo mc 2A/5.5.1 )-Chiu cao tit din x ngang cng xon o nh trong ca m mt ngoi : d0 = l0/5 = (2820/2)/5 = 282 mm-Chiu cao tit din x ngang cng xon o mt trong ca x c th gim bng 0,5 chiu cao ti nh trong ca m mt ngoi -Din tch tit din bn mp ca x ngang cng xon ti mt trong c th gim bng 0,6 din tch tit din bn mp ti nh trong ca m mt ngoi-Chiu dy bn thnh x ngang cng xon phi khng nh hn : t = max ( t1; t2 ) = 16,89 mm t1 = 0,0095 + 2,5 = 17,72 mm t2 = 5,8 + 2,5 = 13,285 mm -Mun chng un tit din x ngang cng xon ti nh trong ca m mt ngoi : Z = 7,1Sl0(b1h1 + b2h2 ) = 9790,74 cm3S =5,2 m : khong cch gia cc x ngang cng xonl0 = 1,41 m :Khong cch nm ngang t mt trong ca x cng xon ti nh trong ca m mt ngoib1 = 2 m : khong cch theo phng nm ngang t mt trong x ngang cng xon n nh trong ca m mt ngoib2 = 5,7 m : na chiu rng ming khoangh1 = 28,7 kN/m2 h2 = (28,7 + 1,41) = 30,11 kN/m2 : ti trng tc dng ln sng ngang boong v ln np ming khoangdc = 0,65 m : chiu cao tit din ang xt ca x ngang cng xon Mp km : chiu rng : b = min (l0/5,S) = min ( 280 ; 5200 ) Chn b = 280 mm , Chiu dy : t = 8 mmBng chn thp

140 x83 x 650x 181320 x 181580 x8Vy c cu tha mn quy phm Sn kho x ngang cng xon: ( 5.5.2 qui phm )-Chiu cao tit din sn kho khng nh hn: = 625 (mm) vi l = 5000 (mm) l chiu di sn kho k c cc chiu di ca cc lin kt u sn. -M un chng un tit din sn kho phi khng nh hn tr s sau: W = 7,1C1Sl1 b1h1+b2h2 = 10909,682 (cm3) -Chiu dy bn thnh: t = max (t1;t2 ) = 10,7

t1=0,0095+2,5= 8,607(mm)

t2 = 5,8 +2,5 = 10,858 (mm) =>Chn tt = 11 (mm), tc =12(mm)Trong :+C1 = 0,15 + 0,5.(0,5b1 h1 +b2h2)/(b1h1 + b2h2 )= 0,65+ S = 5,2 (m): Khong cch gia cc sn cng xon.+ l1 = 2,4 (m): Khong cch theo phng ngang t mt trong x ngang cng xon n cnh trong ca sn kho.+ b1 = 2 (m) , b2 = 5,7 (m) , h1 = 28,7 ; h2 = 30,11 (kN/m2): Nh tnh cho x ngang cng xon.+dw = 0,7(m) Chiu cao tit din sn ang xt.+C2 = 0,9 v c sn khe ni boong cng x ngang cng xon boong trn t ln nh ca sn khe trong khoang.- Chn mp km: + Chiu dy ca mp km l t = 12 mm. + Chiu rng ca mp km l b = min ( l/5, S) = (1000, 5200) =1000 mm=>Vy mp km c qui cch bt = 100012 (mm).Bng chn thp

Vy c cu tha mn quy phm

500 x123x700x111320x11 Lin kt x ngang cng xon vi sn kheX ngang cng xon v sn khe n phi c lin kt chc chn vi nhau bng mBn knh cong cnh t do ca m : r = 700 mmChiu dy m : t = 16 mmM c gia cng bng cc npDin tch tit din bn mp cnh t do ca m 1420x16 , bn mp ca m c lin kt vi bn mp ca x ngang cng xon v vi bn mp ca sn khe B, Boong thi tita,X dc boong*i vi khoang hng 1 v 2 - M un chng un tit din x dc boong phi khng nh hn tr s tnh theo cng thc sau: W = 1,14Shl2 = 143,551(cm3)Trong :

52099014010h = C.=26,236 (kN/m2): Ti trng boong.S = 0,71 m : l khong cch gia cc x dc.l = 2,6 m : l khong cch gia cc sng ngang boong.- Chn thp lm x dc boong c qui cch thp ch L khng u cnh. Mp km: tmk = 9 (mm) bmk = min( l/5;S ) = min( 520 ; 700 ) = 520 (mm)Bng chn thp:

Kt lun: C cu tha mn qui phm.Vy chn x dc boong c qui cch: L1409010*i vi khoang hng 3 -M un chng un tit din x dc boong phi khng nh hn tr s tnh theo cng thc sau: W = 1,14Shl2 = 294,112 (cm3) +h = 53,753 (kN/m2): Ti trng boong. +S,l nh trn- Chn thp lm x dc boong c qui cch thp ch L khng u cnh. Mp km: tmk = 11 (mm) bmk = min( l/5;S ) = min( 520 ; 700 ) = 520 (mm)Bng chn thp:

5201112520011

Kt lun: C cu tha mn qui phm.Vy chn x dc boong c qui cch: L20012511*i vi khoang hng 4 :-M un chng un tit din x dc boong phi khng nh hn tr s tnh theo cng thc sau: W = 1,14Shl2 = 441,444 (cm3)+h = 80,68 (kN/m2): Ti trng boong.(Bng trn)+S,l nh trn- Chn thp lm x dc boong c qui cch thp ch L khng u cnh. Mp km: tmk = 12 (mm) bmk = min( l/5;S ) = min( 520 ; 700 ) = 520 (mm)Bng chn thp:

5201212520014Kt lun: C cu tha mn qui phm.Vy chn x dc boong c qui cch: L20012514

b) X ngang boong. (8.4 qui phm)*i vi khoang hng 1 v 2 -M un chng un tit din x ngang boong phi khng nh hn tr s tnh theo cng thc sau: W = 0,43Shl2 = 57,49(cm3)h = 26,236(kN/m2): Ti trng boong.S = 0,65 m : khong cch gia cc x ngang boong.L = 2,8 m : khong cch gia cc sng dc boong. - Chn thp lm x dc boong c qui cch thp ch L khng u cnh. Mp km: tmk = 9 (mm) bmk = min( l/5;S ) = min( 560 ; 650 ) =560 (mm)Bng chn thp:

5609631007 Kt lun: C cu tha mn qui phm.Vy chn x ngang boong c qui cch: L100637.

*i vi khoang hng 3 :-M un chng un tit din x ngang boong phi khng nh hn tr s tnh theo cng thc sau: W = 0,43Shl2 = 117,787 (cm3)h = 53,753 (kN/m2): Ti trng boong.- Chn thp lm x dc boong c qui cch thp ch L khng u cnh. Mp km: tmk = 11 (mm) bmk = min( l/5;S ) = min( 560 ; 650 ) = 560 (mm)Bng chn thp:

560118012510Kt lun: C cu tha mn qui phm.Vy chn x ngang boong c qui cch: L1258010.

*i vi khoang hng 4 :M un chng un tit din x ngang boong phi khng nh hn tr s tnh theo cng thc sau: W = 0,43Shl2 = 190,391 (cm3)h = 80,68 (kN/m2): Ti trng boong.- Chn thp lm x dc boong c qui cch thp ch L khng u cnh. Mp km: tmk = 12 (mm) bmk = min( l/5;S ) = min( 560 ; 700 ) = 560 (mm)Bng chn thp:

560121001609Kt lun: C cu tha mn qui phm.Vy chn x ngang boong c qui cch: L1601009.

c) Sng ngang boong - Chiu dy bn cnh ca x ngang boong khe phi khng nh hn chiu dy ca bn thnh. Chiu dy ca bn thnh theo 10.2.3 qui phm: tt = 10S1+2,5 S1 : Khong cch gia cc np gia cng bn thnh hoc chiu cao tit din bn thnh ly gi tr nh hn. - Chiu rng bn mp khng nh hn tr s tnh theo cng thc sau: b = 85,4 d0: Chiu cao tit din bn thnh.Tha mn d0 > 2,5dlkhot - M un chng un tit din x ngang boong khe khng nh hn tr s tnh theo cng thc sau: Z = 0,484.l( lbh+kw ) Trong : +l = 2,84 (m): Chiu di nhp sng ngang boong. +b = 2,6 (m): Chiu rng trung bnh din tch boong c bi sng ngang boong. +k: H s k = 12(1 - )2 = 0 ( ). +w: Ti trng boong c bi ct ni boong.*i vi khoang hng 4 -- M un chng un tit din x ngang boong khe khng nh hn tr s tnh theo cng thc sau: Z= 504,161 (cm3) +h = 50,382 (kN/m2) l ti trng tc dng ln sng boong- M men qun tnh ca tit din x ngang boong khe phi khng nh hn tr s tnh theo cng thc sau: I = 4,2Zl =5845,443 (cm4)-Chn thp lm x ngang boong khe c qui cch thp ch T. Mp km: tmk = 12 (mm) bmk = min(l/5;b) = min(560 ;2600 ) = 560 (mm)Bng chn thp:

205101210510100560Kt lun: C cu tha mn qui phm.

*i vi khoang hang 3 : - M un chng un tit din x ngang boong khe khng nh hn tr s tnh theo cng thc sau: Z= 332,42 ( cm3 )+ h = 32,32 (kN/m2) l ti trng tc dng ln sng boong - M men qun tnh ca tit din x ngang boong khe phi khng nh hn tr s tnh theo cng thc sau: I = 4,2Zl =3803,415 (cm4) -Chn thp lm x ngang boong khe c qui cch thp ch T. Mp km: tmk = 11 (mm) bmk = min(l/5;b) = min(560 ;2600 ) = 560 (mm)Bng chn thp:

185101110465100560Kt lun: C cu tha mn qui phm.

i vi khoang hng 1,2 - M un chng un tit din x ngang boong khe khng nh hn tr s tnh theo cng thc sau: Z= 151,122 ( cm3 ) + h = 15,102 (kN/m2) l ti trng tc dng ln sng boong - M men qun tnh ca tit din x ngang boong khe phi khng nh hn tr s tnh theo cng thc sau: I = 4,2Zl =1802,589 (cm4)-Chn thp lm x ngang boong khe c qui cch thp ch T. Mp km: tmk = 9 (mm)

1451091036590560 bmk = min(l/5;b) = min(560 ;2600 ) = 560 (mm)

Kt lun: C cu tha mn qui phm.d) Sng ngang boong lm thanh quy ngang ming hm hng-Mun chng un Z tit din ca sng ngang boong : Z = 0,484 l( lbh + kw ) = 4850,701 cm3+l = 8,5 m : chiu di nhp thanh quy ngang+b = 2,6 m : chiu rng trung bnh din tch boong c bi thanh quy ngang +h = 50,382 kN/m2 : ti trng boong ( p dng cho khoang hng 4 c ti trng boong ln nht )+kw = 0 : do ct sng boong trng v tr ct ni boong -Mmen qun tnh I ca tit din sng ngang boong : I = 4,2Zl = 159705,878 cm4 mp km: chiu rng: b = min ( l/5 ,S ) = min ( 1680; 2600 ) =1680mm chiu dy : t = 12 mm Bng chn thp:

Vy c cu tha mn qui phm c qui cch nh hnh v 250 1100 x 12510900250 x 11150 x 111410 x 11200 x 125 x 12 e) Sng dc boong *Trong vng ming hm hng khoang 4 :- Chiu dy bn cnh ca sng dc boong phi khng nh hn chiu dy ca bn thnh. Chiu dy ca bn thnh theo 10.2.3 qui phm: tt = 10S1+2,5 = 6,65 (mm). Chn tt = 8 (mm). Chn tc = 8 (mm)S1 : Khong cch gia cc np gia cng bn thnh hoc chiu cao tit din bn thnh ly gi tr nh hn. S1 = 0,415 (m). - Chiu rng bn mp khng nh hn tr s tnh theo cng thc sau: b =85,4 = 88,441 (mm). d0 = 2,5.d1 (m) ( d1=0,165m.)tha mn d0 > 2.5dlkhot = 2,5.165 = 412,5(mm).- M un chng un tit din sng dc boong khng nh hn tr s tnh theo cng thc sau: W = 0,484 l( lbh+kw ) = 468,150 (cm3) Trong :+l = 2,6 (m): Nhp sng dc boong.+b = 2,84 (m): Chiu rng trung bnh ca din tch boong bi sng dc boong.+kw = 0.+h = 50,382 (kN/m2) l ti trng tc dng ln sng boong. - M men qun tnh ca tit din sng dc boong phi khng nh hn tr s tnh theo cng thc sau: I = CZl = 5112,206 (cm4)C = 4,2: H s i vi cc sng boong trc on gia tu.-Chn thp lm sng dc boong c qui cch thp ch T. Mp km: tmk =12 (mm) bmk = min(l/5;b) = min( 520;2800 ) = 520 (mm)Bng chn thp:

165812841590520Kt lun: C cu tha mn qui phmVy chn sng dc boong c qui cch: T

*Trong vng ming hm hng khoang 1, 2 :- Chiu dy bn cnh ca sng dc boong phi khng nh hn chiu dy ca bn thnh. Chiu dy ca bn thnh theo 10.2.3 qui phm: tt = 10S1+2,5 =5,13 (mm). Chn tt = 6 (mm). => Chn tc = 6 (mm)+S1 : Khong cch gia cc np gia cng bn thnh hoc chiu cao tit din bn thnh ly gi tr nh hn. S1 = 0,265 (m). - Chiu rng bn mp khng nh hn tr s tnh theo cng thc sau: b =85,4 = 70,619 (mm). d0 = 0,263 (m): Chiu cao tit din bn thnh. tha mn d0 > 2.5dlkhot = 2,5.105 = 262,5 (mm).- M un chng un tit din sng dc boong khng nh hn tr s tnh theo cng thc sau: W = 0,484 l( lbh+kw ) = 140,327 (cm3) +l = 2,6 (m): Nhp sng dc boong.+b = 2,84 (m): Chiu rng trung bnh ca din tch boong bi sng dc boong.+kw = 0.+h = 15,102 (kN/m2) l ti trng tc dng ln sng boong. - M men qun tnh ca tit din sng dc boong phi khng nh hn tr s tnh theo cng thc sau: I = CZl = 1532,382 (cm4)+C = 4,2: H s i vi cc sng boong trc on gia tu.=>Chn thp lm sng dc boong c qui cch thp ch T. => Mp km: tmk =9 (mm) bmk = min(l/5;b) = min( 520;2800 ) = 520 (mm)Bng chn thp :

105899826380520Kt lun: C cu tha mn qui phm.Vy chn sng dc boong c qui cch: T

*Trong vng ming hm hng khoang 3 :- Chiu dy bn cnh ca sng dc boong phi khng nh hn chiu dy ca bn thnh. Chiu dy ca bn thnh theo 10.2.3 qui phm: tt = 10S1+2,5 = 5,75mm). Chn tt = 8 (mm). Chn tc = 8 (mm)+S1 : Khong cch gia cc np gia cng bn thnh hoc chiu cao tit din bn thnh ly gi tr nh hn. S1 = 0,325 (m). - Chiu rng bn mp khng nh hn tr s tnh theo cng thc sau: b =85,4 = 78,5 mm + d0 = 0,325m): Chiu cao tit din bn thnh. tha mn d0 > 2,5dlkhot = 2,5.130 = 325 (mm).- M un chng un tit din sng dc boong khng nh hn tr s tnh theo cng thc sau: Z = 0,484 l( lbh+kw ) = 296,088 (cm3) Trong :+l = 2,6 (m): Nhp sng dc boong.+b = 2,84 (m): Chiu rng trung bnh ca din tch boong bi sng dc boong.+kw = 0.+h = 32,32 (kN/m2) l ti trng tc dng ln sng boong. - M men qun tnh ca tit din sng dc boong phi khng nh hn tr s tnh theo cng thc sau: I = CZl = 3233,285 cm4+C = 4,2: H s i vi cc sng boong trc on gia tu.Chn thp lm sng dc boong c qui cch thp ch T. Mp km: tmk =11 (mm) bmk = min(l/5;b) = min( 520;2800 ) = 520 (mm)Bng chn thp:

130811832580520 Kt lun: C cu tha mn qui phm. Vy chn sng dc boong c qui cch: T

f/ Sng dc lm thanh quy ming hm hng: ( 10.2 qui phm )- M un chng un tit din sng dc boong khng nh hn tr s tnh theo cng thc sau: Z = 1,29 l( lbh+kw ) = 12301,833 (cm3) Trong :+l = 5,2 (m): Khong cch gia cc x ngang cng xon.+b = 7 (m): Chiu rng trung bnh ca din tch boong c bi thanh quy dc.+kw = 0+h = 50,382 (kN/m2): Ti trng boong tc dng ln sng boong (Tnh cho khoang hng 4). - M men qun tnh ca tit din sng dc boong phi khng nh hn tr s tnh theo cng thc sau: I = CZl = 102351,253 (cm4)+C = 1,6 : H s i vi cc sng boong nm ngoi vng ming khoang hng. Mp km: tmk = 12 (mm) bmk = min(l/5;b) = min( 1040;7000).Bng chn thp:

250 1040 x 13510900600 x 13250 x 131410 x 13250 x 160 x 20Vy c cu tha mn quy phm6.4.7X ngang cng xon: (iu 5.5)Bao gm x ngang cng xon v sn khe x ngang cng xon:*X ngang cng xon-Chiu cao tit din x ngang cng xon o nh trong ca m mt ngoi: d0 = l0/5 = 1/5 x 1800 = 360 mmChiu cao tit din x ngang cng xon o mt trong ca x c th gim bng 0,5 chiu cao ti nh trong ca m mt ngoi -Din tch tit din bn mp ca x ngang cng xon ti mt trong c th gim bng 0,6 din tch tit din bn mp ti nh trong ca m mt ngoi -Chiu dy bn thnh x ngang cng xon phi khng nh hn : t = max ( t1 ; t2 ) = 27,58 mm t1 = 0,0095 + 2,5 = 27,58 mm t2 = 5,8 + 2,5 = 15,24 mm -Mun chng un tit din x ngang cng xon ti nh trong ca m mt ngoi : Z = 7,1Sl0(b1h1 + b2h2 ) = 21930,62 cm3+S = 5,2 m : khong cch gia cc x ngang cng xon+b1 = 1,9 m : khong cch theo phng nm ngang t mt trong x ngang cng xon n nh trong ca m mt ngoi+b2 = 5,7m : na chiu rng ming khoang+h1 = h2 = 50,382 kN/m2 : ti trng tc dng ln sng ngang boong v ln np ming khoang ( cho khoang hng 4)+dc = 0,65 m : chiu cao tit din ang xt ca x ngang cng xon =>Mp km : chiu rng : b = min (l0/5,S) = min ( 360 ; 5200 ) Chn b = 360 mm Chiu dy : t = 12 mm (bng chiu dy tn boong)

180x123x650x281400x283x205x28Sn kho x ngang cng xon: ( 5.5.2 qui phm ) Chiu cao tit din sn kho khng nh hn: = 512,5 (mm) vi l = 4100 (mm) l chiu di sn kho k c cc chiu di ca cc lin kt u sn. -M un chng un tit din sn kho phi khng nh hn tr s sau: W = 7,1C1Sl1 b1h1+b2h2 = 18610,567 (cm3) -Chiu dy bn thnh: t = max (t1 ;t2 ) = 19,185 mm

t1=0,0095 +2,5= 19,185 (mm)

t2 = 5,8 +2,5 = 15,66 (mm)Chn tt = 20 (mm).Trong : +S = 5,2 (m): Khong cch gia cc sn cng xon. +l1 = 2,35 (m): Khong cch theo phng ngang t mt trong x ngang cng xon n cnh trong ca sn kho. +b1 = 1,9 (m) , b2 = 5,6 (m) , h1 = h2 = 50,382 (kN/m2): Nh tnh cho x ngang cng xon.+dw = 0,7 (m) Chiu cao tit din sn ang xt. +C1 = 0,15 + 0,5 = 0,65+ b1',b2', h1', h2': tng ng l b1 , b2 , h1 ,h2 trong cng thc tnh cho x ngang cng xon. +C2 = 0,65+0,6=1,25: sn khe ni boong Chn mp km :

+ Chiu dy ca mp km l t = 12 mm. + Chiu rng ca mp km l b = min ( l/5, S) = (800, 5200) =800 mm. Vy mp km c qui cch bt = 80012 (mm).Bng chn thp

6.4.8 Ct chng ( theo mc 9.2 ) Ct chng ca hai boong b tr trng nhau ti ch giao nhau gia thanh quy ngang vi mt phng dc tm . ct ni boong c t trc tip ln ct di boong , ct di boong c t ln sng chnh y . nh v chn ct c gn bng tm kp v bng m a. Ct ni boong -chn ct ni boong dng ct ng c ng knh ngoi d = 246 mm chiu dy t ca ct chng : t = 0,022dp + 4,6 = 10,012 mm + dp = 246 mm (=d1) : ng knh ngoi ca ct ng => chn t = 25 mm (d2= 196mm) - din tch thc t ca ct : f = ( d12 d22 ) = 173,485 cm2 -ti trng boong w m ct phi : w = sbh + kw0 = 1692,835kn -din tch tit din ct a theo quy phm : a = = 166,074 cm2 < f +S = 8,5 m : khong cch gia cc trung im 2 nhip sng boong c bi ct chng + b = 2,84 m : khong cch gia trung im 2 nhp k nhau ca x boong m ct + h = 33,588 kn/m2 : ti trng boong ( cho khoang hng 4 c ti trng boong ln nht ) +l = 4,1 m : nhp ca ct chng k0 = =10,917 cm : bn knh qun tnh ti thiu ca tit din ct,-mmen quan tnh tit din ct I = ( d14 d24 ) =10727,011 cm4 =>vy chn ct ng c ng knh ngoi 246 mm , dy 25 mmb. Ct chng boong di -chn ct chng boong di dng ct ng c ng knh ngoi d = 368 mm -chiu dy t ca ct chng : t = 0,022dp + 4,6 = 12,696 mm dp = 368 mm : ng knh ngoi ca ct ng =>chn t = 30 mm -din tch thc t ca ct : f = ( d2 - d2 ) = 318,557 cm2 -din tch ti thiu ca ct theo quy phm : a = = 305,64 cm2 < f -ti trng boong w m ct phi : w = sbh + kw0 = 3021,278 kn + kw0 = 1692,835 kN : ti trng m ct trn phi +s , b : xc nh nh trn +h = 28 kN/m2 : ti trng boong +l = 5 m : nhp ca ct chng +k0 = = 11,997 cm : bn knh qun tnh ti thiu ca tit din ct -mmen quan tnh tit din ct I = ( d4 - d4 ) = 45849,98 cm4: mmen quan tnh tit din ct =>vy chn ct chng boong di c ng knh 368 mm , dy 30 mm6.4.9. Lin kt: Lin kt boong di: X ngang thng lin kt vi sn thng v x dc lin kt vi np ng vch bng m c b mp. Chiu di m: lm > l/8 = 5000/8 = 625 (mm) Chn m c kch thc 650 x 650 x 9; Chiu rng mp l 60 mm . X ngang kho lin kt vi sn kho bng m c b mp Chn m c kch thc 650x650x9; Chiu rng mp l 60 mm . X ngang cng xon lin kt vi sn kho bng m lin Lin kt boong trn: X ngang thng lin kt vi sn thng v x dc lin kt vi np ng vch bng m c b mp. Chiu di m: lm > l/8 = 4100/8 = 512,5 (mm) Chn m c kch thc 550 x 550 x 8,5 ; Chiu rng mp l 55 mm . X ngang kho lin kt vi sn kho bng m ch T 550x550x8,5,chiu rng mp l 55(mm) X ngang cng xon lin kt vi sn kho bng m lin.7. Kt cu vng khoang my:7.1.Kt cu dn vch:7.1.1B tr kt cu: (iu 11.2)

1: Np ng khe 3: Boong 2 2: Np ng 4: Sng dc boong- Dn vch kt cu vch phng gm np ng, c np ng khe v cc sn- V tr np ng, np ng khe ging nh vch ngang trong vng khoang hng7.1.2Tn vch: (iu 11.2.1 v 11.2.2)Ta c bng kch thc cc tm tn nh sau:STT Chiu dy (mm) Chiu rng (m)

Tm th nht 12 2

Tm th hai 10 2

Tm th ba 8 2

Tm th t 8 2

Tm th nm 8 1,440

7.1.3Np vch: (iu 11.2.3)a.Np thng vch: (iu 11.2.3) Np thng vch di boong 2:(Khoang chnh)Theo iu 11.2.3 Mmen chng un k c mp km xc nh theo biu thc: W = 2,8CShl2 =275,93 (cm3) Trong :+C: h s ph thuc lin kt mt np ( bng 2A/11.2 ). C = 0,8+l: nhp np (m) l = 5 (m).+S: khong cch np gia cng cho vch (m). S = 0,71(m) +h: khong cch thng ng t trung im np n nh ca boong vch o ti tm tu (m). h = l1/2 + l2 + B/50 = 6,94 (m) Chn thp lm np vch c qui cch : thp ch L khng u cnh. Mp km: b = min(0,2l;S) = min(1000;710) =710 (mm). Chn chiu dy ca tm tn mp km: t = 10 (mm). Vy kch thc mp km l: bt = 71010 mm.Bng chn thp:

7101011018012

Kt lun: C cu tha mn qui phm.Vy chn np vch c qui cch: L18011012 (mm).

Np thng ni boong ( trong on t boong 2 n boong 1):Theo iu 11.2.3 Mmen chng un k c mp km xc nh theo biu thc: W = 2,8CShl2 =79,562 (cm3) Trong :+C: h s ph thuc lin kt mt np ( bng 2A/11.2 ). C = 0,8+l: nhp np (m) l = 4,1 (m).+S: khong cch np gia cng cho vch (m). S = 0,71 (m)+h: khong cch thng ng t trung im np n nh ca boong vch o ti tm tu (m).+h = ( l2 /2 + B/50)/2 =2,22 (m) V h< 6 (m) nn ta c h= 1,2 + 0,8*2,22 = 2,976 (m)Chn thp lm np vch c qui cch : thp ch L khng u cnh. Mp km: b = min(0,2.l;S) = min(820;710) = 710 (mm). Chn chiu dy ca tm tn mp km: t = 8 (mm). Vy kch thc mp km l: bt = 7108 mm.Bng chn thp:

7108701108Kt lun: C cu tha mn qui phm.Vy chn np vch c qui cch: L110708 (mm)b. Np khe vch:Np khe vch di boong 2(khoang chnh)Theo iu 11.2.3 thi m un chng un (W) cua tit din np phai khng nho hn tri s sau y : W= 2,8CShl2 =1103,737 (cm3 ) Trong : C: h s ph thuc lin kt mt np ( bng 2A/11.2 ). C = 0,8l: nhp np : l = 5 (m).S: chiu rng ca vng m sng phi (m). S = 2,84 (m)h : khong cch thng ng o t trung im ca l ca sng ng hoc o t trung im ca S ca sng nm n nh boong vch ng tm tu (m). Nu khong cch thng ng nh hn 6m th h ly bng 1,2m cng vi 0,8 ln khong cch thng ng thc.h = l1/2 + l2 + B/50 = 6,94 (m)* Chn thp lm sng vch c qui cch : thp ch T . Mp km:b = min(S; l/5) = min(2840; 1000) =1000 (mm).Chn chiu dy ca tm tn mp km: s = 8 (mm).Vy kch thc mp km l: bs = 10008 mm.Chn chiu cao bn thnh : h = 400 (mm)=>thc ca bn thnh l : h tt = 400 12 mmChiu dy bn cnh : Chn t = 12 (mm)=>kch thc bn cnh l : bc tc = 140 12 mm* Bang chon thep

81240014010001251252012250160156012Kim tra iu kin np vch di sng boong

+Z0: m dun chng un yu cu ca np Z0 = 1103,737 (cm3) +Z: m un chng un thc ca np Z= 1205,192 (cm3)+C=17,7+A: din tch tit din np k c mp km A=146 cm2+W: ti dc tm np. W= h.b.S=28,7.2,6.2,84= 209,7 (kN)

Suy ra : = 17,64 =17,7Kt lun: C cu tha mn qui phm.Vy chn np vch c qui cch: T (mm).Np khe ni boong (trong on t boong 2 n boong 1):Theo iu 11.2.6 thi m un chng un (W) cua tit din np phai khng nho hn tri s sau y : W= 2,8CShl2 =332,8 (cm3 ) Trong : C: h s ph thuc lin kt mt np ( bng 2A/11.2 ). C = 0,8l: nhp np : l = 4,1 (m).S: chiu rng ca vng m sng phi (m). S = 2,84 (m)h : khong cch thng ng o t trung im ca l ca sng ng hoc o t trung im ca S ca sng nm n nh boong vch ng tm tu (m). Nu khong cch thng ng nh hn 6m th h ly bng 1,2m cng vi 0,8 ln khong cch thng ng thc.h = l/2+B/50 = 2,39 (m) h= 1,2+ 0,8.2,39= 3,112 (m)* Chn thp lm sng vch c qui cch : thp ch T . +Mp km:b = min(S; l/5) = min(2840; 820) =820 (mm).Chn chiu dy ca tm tn mp km: s = 8 (mm). =>Vy kch thc mp km l: bs = 8208 mm. +Chn chiu cao bn thnh : h = 300 (mm)=> kch thc bn thanh l : h tt = 300 8 mm +Chiu dy bn cnh : Chn t = 8 (mm)=> kch thc ca bn cnh l : bc tc = 110 6 mmBng tra thp :

88270110820125125201225016015608Kim tra iu kin np vch di sng boong

Z0: m dun chng un yu cu ca npZ0 =328,8 (cm3 ) Z: m un chng un thc ca npZ= 380,193(cm3 )C=17,7A: din tch tit din np k c mp km :A=95,2 cm2W: ti dc tm np.W =S.h.b= 28,72,6.2,84. 15,102= 209,7(kN)

Suy ra = 17,62 =17,7Kt lun: C cu tha mn qui phm.Vy chn np vch c qui cch: T mm7.1.4. Lin kt: (iu 1.1.14 v 1.1.15)

M lin kt ca np vch : lm = = 625 mm (Vi l=5,0m)

chn m c quy cch :

M lin kt ca np vch khoang ni boong : lm = = 512,5 mm (l=4,1m) chn m c quy cch : 7.2.Kt cu dn y:7.2.1. B tr kt cu: (Chng 4) Dn y vng khoang my c b tr kt cu theo s :

- Dn y khoang my kt cu y i h thng ngang.- Cc ngang y cch nhau 650 mm.Ta t thm sng ph di thnh dc b my, cch nhau 2000 mm7.2.2. Tn y ngoi: (iu 14.2 v 14.3)*Chiu dy ti thiu: tmin = = = 10,667 mm (iu 14.3.1)* Theo iu 14.3.4, tu kt cu theo h thng ngang, chiu dy tn y ngoi khng nh hn tr s tnh theo biu thc sau:

= 12,446 (mm) Chn t = 13 (mm)Trong :+C1 : H s ph thuc vo chiu di tu. C C1 = 1 khi L < 230 m

+: H s ph thuc vo h thng kt cu

= 4,66

0,9: T s ca m un chng un ca tit din ngang thn tu tnh theo l thuyt chia cho m un chng un thc ca tit din thn tu tnh vi y. x = X/(0,3L) = 0,33 : V ta xt on ngoi 0.3L k t mi tu (ly X = 0,1L).+S = 0,65 (m) : Khong cch gia cc sn ngang. +d = 6,8 (m) : Chiu chm tu.+L' = 113,8 (m) : Chiu di tu.+h1 = 0 kN/m2 : Chiu cao ct p (v theo qui nh 14.3.2/2A quy phm vng ang xt khng vng 0,3L k t mi tu).2.3Tn y trong: (iu 4.5)Theo iu 4.5.1, mc 1, Quy phm phn cp v ng tu bin v thp TCVN 21- 2A: 2010 chiu dy tn y trn khng nh hn tr s tnh theo cc cng thc sau :

t1= = 6,102 (mm)

t2= = 10,488 (mm)Trong :+d0 = 1,2 (m): Chiu cao tit din sng chnh.+d = 6,8 (m): Chiu chm tu.+B = 17(m): Chiu rng tu.+S = 0,65 (m): Khong cch c cu gia cng+h : Khong cch thng ng t mt tn y trn ti boong thp nht o tm tu. h = D + B/50 d0 = 9,44 (m)+C: H s ly theo qui nhC B/lh = 17/20,8= 0,817 => 0,8