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MC LC
PHN I...........................................................................................3THIT K S B............................................................................3
CHNG I..................................................................................5GII THIU CHUNG.....................................................................5
I. NHIM V THIT K.................................................................................5II. C IM A HNH................................................................................5
III. C IM A CHT..............................................................................5IV. QUY TRNH THIT K V CC NGUYN TC CHUNG..................51. Quy trnh thit k:............................................................52. Cc thng s k thut c bn:...........................................6
V. CC PHNG N K THUT..................................................................6CHNG II.................................................................................7PHNG N S B I CU DM GIN N BTCT......................7MT CT CH I...........................................................................7
I.GII THIU CHUNG V CC PHNG N :..........................................7I.1. S b tr chung ca cu:............................................7I.2. M t tm tt kt cu phn trn.....................................7I.3. M t tm tt kt cu phn di:...................................8I.4. Mt cu v cc cng trnh ph khc:.............................8I.5 . Vt liu ch to. ........................................................10I.6 . Bin php thi cng. ....................................................12
CHNG III..............................................................................15PHNG N S B II CU DM GIN N BTCT...................15MT CT SUPER T.....................................................................15
I.GII THIU CHUNG V CC PHNG N :........................................15
I.1. S b tr chung ca cu:..........................................15I.2. M t tm tt kt cu phn trn...................................15I.3. M t tm tt kt cu phn di:.................................16I.4. Mt cu v cc cng trnh ph khc:...........................16I.5 . Vt liu ch to...........................................................18I.6 . Bin php thi cng. ....................................................20
CHNG III..............................................................................23SO SNH LA CHN PHNG N............................................23V Kinh T................................................................................23
V K Thut..............................................................................23V M Quan..............................................................................23
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9.2.Tnh duyt theo TTGH Cng ................................10610.1.Xc nh sc khng ct danh nh............................11110.2.Tnh duyt lc ct theo TTGH cng ....................11310.3.Tnh duyt ct thp doc chu xon...........................114
XI.TNH DUYT BN MT CU V DM NGANG.............................11711.1.Phng php tnh ton bn mt cu........................11711.2.Xc nh ni lc bn mt cu do tnh ti..................11711.3.Xc nh ni lc do hot ti v ngi i b:..............12011.4.Vt liu thit k cho bn mt cu:...........................12211.5.Tnh ton ct thp chu lc:......................................123
PHN I
THIT K S B
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2. Cc thng s k thut c bn:
2.1. Quy m cng trnh:
- Cu vnh cu.
2.2. Ti trng thit k:
- Ti trng thit k : HL93.
- Ti trng ngi i b.
2.3. Kh cu:
Cu c chiu rng nh sau:
- B rng mt ng xe chy: 7m (khng k lan can, g chn bnh, g chn
xe)
- L ngi i b: 2x2m
2.4. Kh thng xe
- Kh thng xe 22.05 x 4.75m
V. CC PHNG N K THUT.
Phng n xy dng cu da trn cc nguyn tc sau:
- Tho mn cc yu cu v tuyn v cu.
- Phng n phi c tnh kh thi.
- Ph hp vi cc iu kin t nhin ti v tr xy dng cu (c bit l iukin a cht, mt bng).
- Tn dng c nguyn liu v cng ngh c sn trong nc.
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CHNG II
PHNG N S B I CU DM GIN N BTCT
MT CT CH I
I.GII THIU CHUNG V CC PHNG N :
I.1. S b tr chung ca cu:
Da vo mt bng cu, kh thng xe, kh vt ng v mt ct a cht,ngi thit k a ra phng n cu dm gin n BTCT mt ct ch I vi s cu nh sau: cu gm 11 nhp vi chiu di cc nhp b tr: 11 x 33.050m
Chiu di ton cu Ltc = 373.55 m
I.2. M t tm tt kt cu phn trn.
I.2.1. Chiu cao dm.
Chiu cao dm c chn theo yu cu sau:
+ Trng lng thp ca dn ch nh.
+ Bo m tnh khng thng xe v vt ng st.
+ m bo m quan v ph hp vi cnh quan khu vc xy cu.
Theo kinh nghim th i vi cu t, chiu cao dm c chn nh sau:
h = (8
1:
25
1)Lnhp = (
8
1:
25
1)x 33 =1.32 4.125 (m).
Ta chn chiu cao dm h = 1.65m.
I.2.2. S dm v khong cch tim cc dm:
S dm v khong cch tim cc dm ca cu do s ln xe quyt nh. Chn 9dm v khong cch cc dm 2.250m. Phn hng tnh n mp lan can 1.35m.
I.2.3. dc dc cu.
Do tnh m quan cng trnh v trnh p t u cu qu ln cu c dcthay i nh sau:
Dc 4% cc nhp: 1, 2, 3, 4 (ln cu); 8, 9, 10, 11 (xung cu).
Dc 0% cc nhp: 5 ,6,7
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I.3. M t tm tt kt cu phn di:
Hai m cu cu to ging nhau, theo kiu m ch U bng BTCT. Thn m ttrn mng cc b thp BTCT Cc cc l cc khoan nhi c ng knh cc
1200mm.Tr: 10 tr ca cu c mt ct ging nhau, l loi tr thn hp , thn tr t
trn h mng cc b thp. Cc cc l cc khoan nhi c ng knh cc
1200mm.
I.4. Mt cu v cc cng trnh ph khc:
Bn mt cu BTCT dy 180mm.
Lp b tng to dc thay i t 0 110mm.
Lp ph mt cu xe chy dy 74 mm. Bao gm phng nc 4 mm ,lp btng nha 70 mm.
Mt cu c dc ngang 2.0 %.
H thng thot nc dng ng gang b tr dc cu thot nc xung gmcu.
Khe co gin loi 50mm.
Gi cu dng gi con ln BTCT.
Lan can trn cu dng lan can thp.
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I.5 . Vt liu ch to.
Vt liu dng cho kt cu phn trn
- B tng dm
+ Cp B tng :M400
+ Cng chu nn tui 28 ngy :fc = 40 Mpa
+ Trng lng ring ca b tng :c = 24.5 kN/m3
+ Modul n hi ca b tng :Es= 32980 Mpa+ H s Poisson : = 0.2
+ H s gin n nhit :c = 1.08E-05 1/oC
- B tng bn
+ Cp B tng :M350
+ Cng chu nn tui 28 ngy :fc = 35 Mpa
+ Trng lng ring ca b tng :c = 24.5 kN/m3
+ Modul n hi ca b tng :Es= 30849.7 Mpa
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+ H s Poisson : = 0.2
+ H s gin n nhit :c = 1.08E-05 1/oC
- Thp
a. Thp d ng lc : Dng loi tao xon 7 si loiA416 cp 270 t chng thp
+ ng knh danh nh 1 tao :15.2 mm
+ Din tch danh nh mt tao cp :138.7 mm2
+ Trng lng danh nh 1 tao :0.1089kN/m
+ M un n hi :Ep=1975000 Mpa
+ Cng chu ko :f pu=1860 Mpa+ Gii hn chy :f py=1670 Mpa
+ Lc ko t nh nht 1 tao cp :260.4 kN
b. Thp thng
+ Cng chu ko :fu=620 Mpa
+ M un n hi:Es=200000 Mpa
+ Gii hn chy :fy=420 Mpa
Vt liu dng cho kt cu phn di
- B tng
+ Cp B tng :M300
+ Cng chu nn tui 28 ngy :fc = 50 Mpa
+ Trng lng ring ca b tng :c = 24.5 kN/m3
+ Modul n hi ca b tng :Es= 28561.32Mpa
+ H s Poisson : = 0.2
+ H s gin n nhit :c = 1.08E-05 1/oC
- Thp
+ Cng chu ko :fu=6200 Mpa
+ M un n hi :Es=200000 Mpa
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+ Gii hn chy:fy=420 Mpa
I.6 . Bin php thi cng.BIN PHP THI CNG M
bc 1: thi cng cc khoan nhi
- san i mt bng thi cng.
- xc nh tim cc ngoi hin trng.
- ng ng vch thp.
- khoan cc bng my khoan.
- thau ra l khoan, h ng thp.
- b tng cc.
- rt ng vch thp.
bc 2: o h mng
- o h mng tr.
- ri lp dm m dy 30 mm.
- b tng to phng dy 10 mm.
- x l u cc.
- th nghim cc.
bc 3: thi cng b cc, thn v tng cnh m
- lp t vn khun v ct thp b cc.
- b tng.
- lp t gio ct thp thi cng tng thn v tng cnh.
- lp t vn khun v ct thp.- b tng.
- di di thit b thi cng v vt t.
- dn dp khu vc thi cng.
BIN PHP THI CNG TR
bc 1: thi cng cc khoan nhi
- san i mt bng thi cng.
- xc nh tim cc ngoi hin trng.
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- ng ng vch thp.
- khoan cc bng my khoan.
- thau ra l khoan, h ng thp.
- b tng cc.- rt ng vch thp.
bc 2: o h mng
- o h mng tr.
- ri lp dm m dy 30 mm.
- b tng to phng dy 10 mm.
- x l u cc.
- th nghim cc.bc 3: thi cng b v thn tr
- lp t vn khun v ct thp b tr.
- b tng.
- lp t gio ct thp thi cng cho tr.
- lp t vn khun v ct thp.
- b tng.
bc 4: thi cng x m tr
- lp t gio ct thp thi cng.
- lp t vn khun v ct thp x m tr.
- b tng.
- di di thit b thi cng v vt t.
- dn dp khu vc thi cng.
BIN PHP THI CNG KT CU NHPbc 1: thi cng ng v chun b thit b lao dm
- chun b mt bng , vt t thit b xe my, lp t. ng di chuyn dm.
- lp t gi 3 chn v ng vn chuyn.
- vn chuyn dm super t t bi c ra u cu.
bc 2: lao dm super t nhp 1
- di chuyn gi lao dm ra kt cu tr 1.
- lao dm vo v tr nhp 1 v h dm xung gi.- c nh dm lao cc dm tip theo.
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CHNG III
PHNG N S B II CU DM GIN N BTCT
MT CT SUPER T
I.GII THIU CHUNG V CC PHNG N :
I.1. S b tr chung ca cu:
Da vo mt bng cu, kh thng xe, kh vt ng v mt ct a cht,ngi thit k a ra phng n cu dm gin n BTCT mt ct super T vi s cu nh sau: cu gm 9 nhp vi chiu di cc nhp b tr: 39.2 + 7 x 40 + 39.2m.
Chiu di ton cu Ltc = 368.4 m.
I.2. M t tm tt kt cu phn trn.
I.2.1. Chiu cao dm.
Chiu cao dm c chn theo yu cu sau:
+ Trng lng thp ca dn ch nh.
+ Bo m tnh khng thng xe v vt ng st.
+ m bo m quan v ph hp vi cnh quan khu vc xy cu.
Theo kinh nghim th i vi cu t, chiu cao dm c chn nh sau:
h = (8
1:
25
1)Lnhp = (
8
1:
25
1)x 40 =1.6 5 (m).
Ta chn chiu cao dm h = 1.75m.
I.2.2. S dm v khong cch tim cc dm:
S dm v khong cch tim cc dm ca cu do s ln xe quyt nh. Chn 9dm v khong cch cc dm 2.30m. Phn hng tnh n mp lan can 1.25m.
I.2.3. dc dc cu.
Do tnh m quan cng trnh v trnh p t u cu qu ln cu c dc
thay i nh sau: Dc 4% cc nhp: 1, 2, 3 (ln cu);7, 8, 9(xung cu).
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Dc 0% cc nhp: 4, 5 ,6
I.3. M t tm tt kt cu phn di:
Hai m cu cu to ging nhau, theo kiu m ch U bng BTCT. Thn m t
trn mng cc b thp BTCT Cc cc l cc khoan nhi c ng knh cc 1200mm.
Tr: 10 tr ca cu c mt ct ging nhau, l loi tr thn hp , thn tr t
trn h mng cc b thp. Cc cc l cc khoan nhi c ng knh cc 1200(mm).
I.4. Mt cu v cc cng trnh ph khc:
Bn mt cu BTCT dy 180mm.
Lp b tng to dc thay i t 0 110mm.
Lp ph mt cu xe chy dy 74 mm. Bao gm phng nc 4 mm ,lp btng nha 70 mm.
Mt cu c dc ngang 2.0 %.
H thng thot nc dng ng gang b tr dc cu thot nc xung gmcu.
Ton cu c 10 khe co gin loi 50mm.
Gi cu dng gi con ln BTCT.
Lan can trn cu dng lan can thp.
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I.5 . Vt liu ch to
Vt liu dng cho kt cu phn trn
- B tng dm
+ Cp B tng :M500
+ Cng chu nn tui 28 ngy :fc = 50 Mpa
+ Trng lng ring ca b tng :c = 24.5 kN/m3
+ Modul n hi ca b tng :Es=36872.5 Mpa
+ H s Poisson : = 0.2
+ H s gin n nhit :c = 1.08E-05 1/oC
- B tng bn
+ Cp B tng :M350
+ Cng chu nn tui 28 ngy :fc = 35 Mpa
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+ Trng lng ring ca b tng :c = 24.5 kN/m3
+ Modul n hi ca b tng :Es= 30849.7 Mpa
+ H s Poisson : = 0.2
+ H s gin n nhit :c = 1.08E-05
1/o
C- Thp
a. Thp d ng lc : Dng loi tao xon 7 si loiA416 cp 270 t chng thp
+ ng knh danh nh 1 tao :15.2 mm
+ Din tch danh nh mt tao cp :138.7 mm2
+ Trng lng danh nh 1 tao :0.1089kN/m+ M un n hi :Ep=1975000 Mpa
+ Cng chu ko :f pu=1860 Mpa
+ Gii hn chy :f py=1670 Mpa
+ Lc ko t nh nht 1 tao cp :260.4 kN
b. Thp thng
+ Cng chu ko :fu=620 Mpa
+ M un n hi :Es=200000 Mpa
+ Gii hn chy:fy=420 Mpa
Vt liu dng cho kt cu phn di
- B tng
+ Cp B tng :M300
+ Cng chu nn tui 28 ngy :fc = 50 Mpa
+ Trng lng ring ca b tng :c = 24.5 kN/m3
+ Modul n hi ca b tng :Es= 28561.32Mpa
+ H s Poisson : = 0.2
+ H s gin n nhit :c = 1.08E-05 1/oC
- Thp+ Cng chu ko :fu=6200 Mpa
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+ M un n hi:Es=200000 Mpa
+ Gii hn chy:fy=420 Mpa
I.6 . Bin php thi cng.
BIN PHP THI CNG M
bc 1: thi cng cc khoan nhi
- san i mt bng thi cng.
- xc nh tim cc ngoi hin trng.
- ng ng vch thp.
- khoan cc bng my khoan.
- thau ra l khoan, h ng thp.
- b tng cc.
- rt ng vch thp.
bc 2: o h mng
- o h mng tr.
- ri lp dm m dy 30 mm.
- b tng to phng dy 10 mm.
- x l u cc.
- th nghim cc.
bc 3: thi cng b cc, thn v tng cnh m
- lp t vn khun v ct thp b cc.
- b tng.
- lp t gio ct thp thi cng tng thn v tng cnh.- lp t vn khun v ct thp.
- b tng.
- di di thit b thi cng v vt t.
- dn dp khu vc thi cng.
BIN PHP THI CNG TR
bc 1: thi cng cc khoan nhi
- san i mt bng thi cng.
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- xc nh tim cc ngoi hin trng.
- ng ng vch thp.
- khoan cc bng my khoan.
- thau ra l khoan, h ng thp.- b tng cc.
- rt ng vch thp.
bc 2: o h mng
- o h mng tr.
- ri lp dm m dy 30 mm.
- b tng to phng dy 10 mm.
- x l u cc.- th nghim cc.
bc 3: thi cng b v thn tr
- lp t vn khun v ct thp b tr.
- b tng.
- lp t gio ct thp thi cng cho tr.
- lp t vn khun v ct thp.
- b tng.
bc 4: thi cng x m tr
- lp t gio ct thp thi cng.
- lp t vn khun v ct thp x m tr.
- b tng.
- di di thit b thi cng v vt t.
- dn dp khu vc thi cng.BIN PHP THI CNG KT CU NHP
bc 1: thi cng ng v chun b thit b lao dm
- chun b mt bng , vt t thit b xe my, lp t. ng di chuyn dm.
- lp t gi 3 chn v ng vn chuyn.
- vn chuyn dm super t t bi c ra u cu.
bc 2: lao dm super t nhp 1
- di chuyn gi lao dm ra kt cu tr 1.- lao dm vo v tr nhp 1 v h dm xung gi.
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- c nh dm lao cc dm tip theo.
- sau khi lao dm xong tin hnh thi cng cc lin kt, dm ngang.
bc 3: lao dm ti cc nhp khc- y gi 3 chn ti v tr nhp 2.
- di chuyn dm n u nhp 2.
- lao dm vo v tr.
- h dm xung gi ca nhp 2.
- sau khi thi cng nhp 2 tin hnh thi cng cc nhp tip theo.
bc 4: hon thin cu
- di di thit b thi cng v vt t.- b tng bn mt cu, g chn, lan can.
- thi cng lp ph mt cu.
- lp t khe co gin.
- lp t h thng thot nc.
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CHNGIII
SO SNH LA CHN PHNG N
la chn mt phng n tt nht ta phi tin hnh so snh chng v cc chtiu Kinh T K Thut M Quan v iu kin duy tu bo dng cng nh khaithc kh nng ca chng.
V Kinh T
a. Cu dm Super T
Nh tnh ton trong phn thit k s b ta c :
Khi lng btng ca 5 dm ch trong 1 nhp : 30.0898 m3
Khi lng btng ca dm ngang trong 1 nhp : 35,2m
b. Cu dm mt ct ch I
Do ch l phng n s b nn khng tnh chi tit phn ny
V K Thut
a. Cu dm Super T
Thi cng kt cu nhp bng gi 3 chn : u im ca n l thi cng rt an tonnhng cng knh v kh phc tp.
Bn cnh dm cng to sn cng tc cho cc cng vic di mt cu ngay saukhi dm c t vo v tr, to s an ton cho cng nhn.
b. Cu dm mt ct ch I
V M Quan
a. Cu dm Super T
Dm c dng mt y dng dm hp vi t gc cnh nn c xem nh tngng vi cc dm hp hay bn c l c ti ch ang c a chung. y ccnhp v x m lin tc to hiu qu cao v m quan.
Ngoi ra n cn c nhiu u im v kh nng s dng nh : giao thng m
thun, trnh c ting n.b. Cu dm mt ct ch I
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V duy tu bo dng
a. Cu dm Super T
Ngy nay chng ta hiu BTCT v BTCT d ng lc l loi vt liu l tng ckh nng cnh tranh vi thp trong lnh vc xy dng ni chung v trong xy dngcu ni ring. Tnh bn c hc ca btng tuy c thua km thp nhng btng lic kh nng chu mi tt, c kh nng chng li cc tc ng ca mi trng tthn thp nn trong qu trnh s dng trnh c chi ph duy tu bo dng
b. Cu dm mt ct ch I
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PHN II
THIT K K THUT
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I. S LI U THIT K
Chiu di ton dm L = 38.3 m
Khong cch t u dm n
tim gi a = 0.35 mKhu tnh ton Ltt = L-2a 37.6 m
Ti trng thit k Hoat ti HL - 93
Ti trng ngi 3Kpa
Mt xe chy: B1 = 6.8 m
L ngi i B2 = 1.9 m
Lan can B3 = 0.5 m
Di phn cch B4 = 0.2 mTng b rng cu B=B1+2.B2+2.B3+2.B4 B = 12 m
Dng kt cu nh p Cu dm
Dng mt ct Super T
Vt liu kt cu BTCT d ng lc
Cng ngh ch to Cng trc
Cp b tng D m chu f'c1= 50 Mpa
Bn mt cu f'c2= 32 Mpa
T trng b tng c= 2450 kg/m3
Loi ct th p DUL: tao thpTao 7
si xon ng knh Dps = 15.2 mm
Cng chu ko tiu chun fpu = 1860 Mpa
Thp thng G60 f u = 620 Mpafy = 420 Mpa
Quy trnh thit k 22TCVN 272-05
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II. THIT K CU TAO
2.1. La chn kch thc mt ct ngang cu
S lng dm ch 12 12 6 4.82.5 2 2 2.5
B Bn n nm m
chn n= 5 dm
Khong cch gia hai dm ch S = 2300 mm
L ngi i ng mc vi mt cu phn xe chy, v c ngn cch vi nhaubng g phn cch
B tr dm ngang ti cc v tr gi cu: 2 mt ct
S lng dm ngang: Nn=(Nb-1).2 Nn= 8 dm
Phn cnh hng: Sk=(B-(Nb-1)S-0.15*2)/2 Sk= 1.25 mChiu dy trung bnh ca bn hf= 18 cm
L p BT atphal t1= 70 mm
Lp phng nc t2= 4 mm
2.2. Thit k dm ch
on ct khc: Lck= 800 mm
on dm c: Ldac= 1200 mm
Chn chiu cao dm ch H
38.30.045 0.045 38.3 1.72 2.13
18 18
L L H x H m H m
Chn H = 1.75m = 1750mm
2.2.1.Mt ct ngang dm trn gi
Chiu cao dm SuperT: H'= 800 mm
B rng bu dm di b'1= 890 mm
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Chiu cao cnh dm h1= 75 mm
Chiu cao vt trn h2= 75 mm
B rng vt trn b8= 100 mm
B rng cnh dm b2= 1220 mm
2.2.2.Mt ct ngang dm c
Chiu cao dm Super T H= 1750 mm
B rng bu dm di b1= 700 mm
Chiu cao cnh dm h1= 75 mm
Chiu cao vt trn h2= 75 mm
B rng vt trn b8= 100 mm
B rng cnh dm b2= 2260 mm
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Vh= 50 mm
Vb= 25 mm
2.3. Cu to dm ngang
Chiu cao dm ngang: Hnd=H'= 800 mm
B dy dm ngang: tdn=Lck= 800 mm
Chiu di dm ngang: a'dn= 1080 mm
a_dn= 1410 mm
B rng vt trn avdn= 100 mm
Cao vt trn: hvdn= 75 mmDin tch mt ct dm ngang: Adn 0.6278 m2
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III. TINH C TRNG HNH HOC DM SUPER-T
Xt mt ct c trng gm:
Mt ct gi : x0= 0 m
Mt ct ti ch thay i tit din: x1= 1.44 mM ct khng dnh bm 1: x2= 3 m
Mt ct khng dnh bm 2: x3= 6 m
Mt ct L/2: x4=Ltt/218.8 m
Trong qu trnh tnh ton ta s qui i tit din Sper-T v tit din n gin hn thun tin cho vic tnh ton
3.1. Mt ct trn gi x0
3.1.1.Tit din nguyn khi:Tnim X Y Xi - Xi+1 Yi + Yi+1 Yi*Yi+1 Fi Yi Ji
0 0 0
1 -0.61 0 0.61 0 0 0 0 0
2 -0.61 -0.075 0 -0.075 0 0 0 0
3 -0.51 -0.15 -0.1 -0.225 0.01125 0.0225 -0.0039 0.00063
4 -0.445 -0.8 -0.065 -0.95 0.12 0.06175 -0.0509 0.040915 0.445 -0.8 -0.89 -1.6 0.64 1.424 -1.7088 1.82272
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6 0.51 -0.15 -0.065 -0.95 0.12 0.06175 -0.0509 0.04091
7 0.61 -0.075 -0.1 -0.225 0.01125 0.0225 -0.0039 0.00063
8 0.61 0 0 -0.075 0 0 0 0
9 0 0 0.61 0 0 0 0 010 0 0 0 0 0 0 0 0
SUM 1.5925 -1.8184 1.9058
Din tch mt ct :F= 0.796 m2
Ta trng tm mt ct so vi th trn dm
0.381 m
Ta trng tm mt ct so vi th di dm
Yc= 0.419 m3
M men tnh ca mt ct i vi trc y dm
Sx= 0.124 m3
M men qun tnh ca mt ct i vi trc X :
Jx = 0.159 m4
M men qun tnh i vi trc trung ha :
Jthx = 0.043 m4
Din tch ca tit din lin hp:
2 1 20 0 . . 0.796 0.8 2.3 0.18 1.1272
cbang g f
cdam
EA A S h x x m
E= + = + =
Trong :
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T trng ca b tng 32450 /c kg m =
Cng chu nn ca b tng lm bn mt cu fc = 32 MPa
Mun n hi ca bn mt cu:
1.5 ' 1.50.043 0.043.2450 . 32 29498cban c cb E f MPa= = =
Cng chu nn ca b tng lm dm fc = 32 MPa
Mun n hi ca bn mt cu:
1.5 ' 1.50.043 0.043.2450 . 50 36872.5cdam c cd E f MPa= = =
294980.8
36872.5cban
cdam
E
E= =
3.2. Mt ct ti ch thay i tit din X1:3.2.1. Tit din nguyn khi:
BNG TNH CC C TRNG CHO DM
Tn im X Y Xi - Xi+1 Yi + Yi+1 Yi*Yi+1 Fi Yi Ji
0 0
1 -1.13 0 1.13 0 0 0 0 0
2 -1.13 -0.075 0 -0.075 0 0 0 0
3 -0.61-0.075 -0.52 -0.15 0.00563 0.078 -0.0088 0.00088
4 -0.51 -0.15 -0.1 -0.225 0.01125 0.0225 -0.0039 0.00063
5 -0.35 -1.75 -0.16 -1.9 0.2625 0.304 -0.5356 0.937846 0.35 -1.75 -0.7 -3.5 3.0625 2.45 -6.4313 15.0063
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7 0.51 -0.15 -0.16 -1.9 0.2625 0.304 -0.5356 0.93784
8 0.61-0.075 -0.1 -0.225 0.01125 0.0225 -0.0039 0.00063
9 1.13-0.075 -0.52 -0.15 0.00563 0.078 -0.0088 0.00088
10 1.13 0 0 -0.075 0 0 0 011 0 0 1.13 0 0 0 0 0
SUM 3.259 -7.5279 16.885
Din tch mt ct :
F= 1.630 m2
Ta trng tm mt ct :
0.770 m
Ta trng tm mt ct so vi th di dm
Yc= 0.980 m3
M men tnh ca mt ct i vi trc X :
Sx= 1.597 m3
M men qun tnh ca mt ct i vi trc X :
Jx = 1.407 m4
M men qun tnh i vi trc trung ha :Jthx = 0.441 m4
3.2.2.Tit din lin h p
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Din tch tit din lin hp:
2 1 21 1 . . 1.630 0.8 2.3 0.18 1.9612
cbang g f
cdam
EA A S h x x m
E= + = + =
3.3. Mt ct gia nhp X4:3.3.1. Tit din nguyn khi:
BNG TNH CC C TRNG CHO DM
Tn im X Y Xi - Xi+1 Yi + Yi+1 Yi*Yi+1 Fi Yi Ji
0 0 0
1 0 -0.075 0 -0.075 0 0 0 02 0.52 -0.075 -0.52 -0.15 0.00563 0.078 -0.0088 0.00088
3 0.62 -0.15 -0.1 -0.225 0.01125 0.0225 -0.0039 0.00063
4 0.78 -1.75 -0.16 -1.9 0.2625 0.304 -0.5356 0.93784
5 1.48 -1.75 -0.7 -3.5 3.0625 2.45 -6.4313 15.0063
6 1.64 -0.15 -0.16 -1.9 0.2625 0.304 -0.5356 0.93784
7 1.74 -0.075 -0.1 -0.225 0.01125 0.0225 -0.0039 0.00063
8 2.26 -0.075 -0.52 -0.15 0.00563 0.078 -0.0088 0.000889 2.26 0 0 -0.075 0 0 0 0
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10 1.54 0 0.72 0 0 0 0 0
11 1.54 -0.05 0 -0.05 0 0 0 0
12 1.515 -0.05 0.025 -0.1 0.0025 -0.0025 0.00019 -1E-05
13 1.515 -0.3 0 -0.35 0.015 0 0 014 1.4275 -1.175 0.0875 -1.475 0.3525 -0.1291 0.15952 -0.1898
15 1.345 -1.475 0.0825 -2.65 1.73313 -0.2186 0.43637 -0.7775
16 1.13 -1.525 0.215 -3 2.24938 -0.645 1.45138 -2.9033
17 0.915 -1.475 0.215 -3 2.24938 -0.645 1.45138 -2.9033
18 0.8325 -1.175 0.0825 -2.65 1.73313 -0.2186 0.43637 -0.7775
19 0.745 -0.3 0.0875 -1.475 0.3525 -0.1291 0.15952 -0.1898
20 0.745 -0.05 0 -0.35 0.015 0 0 021 0.72 -0.05 0.025 -0.1 0.0025 -0.0025 0.00019 -1E-05
22 0.72 0 0 -0.05 0 0 0 0
23 0 0 0.72 0 0 0 0 0
SUM 1.26863 -3.4329 9.14374
Din tch mt ct :F= 0.6343125000 m2
Ta trng tm mt ct :
0.9020100503 m
Ta trng tm mt ct so vi th di dm
Yc= 0.848 m3
M men tnh ca mt ct i vi trc y dm
Sx= 0.538 m3
M men qun tnh ca mt ct i vi trc X :
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Jx = 0.7619781445 m4
M men qun tnh i vi trc trung ha :
Jthx = 0.2458874567 m4
3.3.2.Tit din lin h p
Din tch tit din lin hp:
2 1
4 4 . . 0.6343 0.8 2.3 0.18 0.9655cban
g g f
cdam
E A A S h x x
E= + = + =
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IV. TNH H S PHN B TAI TRONG
4.1. H s phn lnS ln xe thit k:
11
1
1
73.5
2 6 7 2
1 6lan
BB m
m
m B m n
B m
< =
4.2. H s phn b hot ti theo ln i vi mmen4.2.1.H s phn b hot ti i vi mmen trong cc dm gia
Kim tra h s phn b tha mn tiu chun 22TCN - 272 -05 i vi phm vi p
dng ca phng php tra bngGA TR(mm) PHAM VI P DNG KIM TRA
S= 2300 1800 3500( )S mm THA
H= 1750 450 1700( ) H mm KHNG THA
Ltt= 37600 6000 73000( )tt L mm THA
Nb= 5 3bN THA
Kt lun: Khng nm trong phm vi p dng
=> Phi dng phng php n by
4.2.1.1.Phng php on by:
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* Vi 1 xe ti thit k
' '
1 2
900 2300 900.1 .1 0.6087
2300
Sy y
S
= = = =
. '
1 2
1 1( . ) . ( ) 1.2 (0.6087) 2 0.7304
2 2
SI
M LL lanm g m y y x x x = + = =
* Vi 2 xe ti thit k
'
3
1800 2300 1800.1 .1 0.2174
2300
Sy
S
= = =
'
4 1y =
'
5
1200 2300 1200.1 .1 0.4783
2300
Sy
S
= = =
'
6 0y =
. ' ' '
3 4 5 6
1 1( . ) . ( ) 1 (0.2174 1 0.3913 0) 0.42394 4
MI
M LL lanm g m y y y y x x = + + + = + + + =
( )( . ) max ( . ) , ( . ) max(0.7304,0.4239) 0.7304I SI MI M LL M LL M LLm g m g m g = = =
* Vi ti trng ln:
Thin v an ton coi ti trng ln theo phng ngang cu l ti trng t p trung.
( . ) 1.2 1 1.2IM la nm g = =
4.2.1.2.La chn h s phn b mmen thit k
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( )( . ) max ( . ) , ( . ) 0.7304I IM M M LLm g m g m g = =
( )( . ) max ( . ) , ( ) 1.2I IM lan M M lanm g m g mg = =
4.2.2.H s phn b hot ti i vi mmen ca dm bin
4.2.2.1.Vi 1 lan thit k: s dng phng php on by tnh
Xt cho xe ti thit k v xe hai trc: v khong cch ca hai trc bnh xe theophng ngang ca hai loi xe ny l nh nhau nn c chung mt h s phn b ngang
Phng trnh tung ng nh hng: ( )dbx
y xS
=
Mt ln thit k => h s ln m = 1.2
* Vi xe ti thit k:
4 3 25150 600 2300 1250 150 200 500 1900 600
0.21742300
kS S B B ByS
+ + + + = = =
6 0y =
4 51 1 1
( . ) . 1.2 ( ) 1.2 (0.2174 0) 0.13042 2 2
SE
M LL im g m y y y x x = = + = + =
* Vi ti trng ngi i
31
150 2300 1250 150 500 1.39132300
kS S ByS
+ + + + = = =
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3 23
150 2300 1250 150 500 19000.5652
2300
kS S B ByS
+ + + + = = =
B2=1.9m
221 2 1.9( ) (1.3913 0.5652) 1.8587
2 2M PL B y y x m = + = + =
2
1.2( . ) . 1.8587 1.1739
1.9SEM PL M PL
mm g x
B = = =
* Vi ti trng ln:
3 2 43
150 2300 1250 150 200 500 19000.4783
2300
kS S B B ByS
+ + + + = = =
3 2 43150
2
2300 1250 150 200 500 19000.4783 263.065
2
kM L an S S B B B y
x
+ + =
+ + = =
1.2( . ) . 263.065 0.1052
3000 3000
SE
M lan M Lam
mm g x = = =
4.2.2.2. Vi hai hoc nhiu lan thit k:
Khong cch t tim dm bin n mp trong b va hoc lan can chn xe
4 3 2150 1250 150 200 500 1900 1200
e kd S B B B mm= + = + =
* Phm vi p dng:
GA TR(mm) PHM VI P DNG KIM TRA
S= 2300 1800 3500( )S mm THA
de= -1200 0 1400( )ed mm KHNG THA
Kt lun: Khng nm trong phm vi p dng=> Khng s dng cng thc tra bng
4.2.2.3.La chn h s phn b mmen thit k
Ta s dng h s phn b ngang trong trng hp 1 ln theo phng php n by thit k
( . ) 0.1304SEM LLm g =
( . ) 1.1739
SE
M PLm g
=
( . ) 0.1052SEM L AN
m g =
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4.3. H s phn b hot ti theo ln i vi lc ct4.3.1.H s phn b hot ti i vi lc ct trong cc dm gia
Kim tra h s phn b tha mn tiu chun 22TCN - 272 -05 i vi phm vi pdng ca phng php tra bng
GA TR(mm) PHM VI P DNG KIM TRA
S= 2300 1800 3500( )S mm THA
H= 1750 450 1700( ) H mm KHNG THA
Ltt= 37600 6000 73000( )tt L mm THA
Nb= 5 3bN THA
Kt lun: Khng nm trong phm vi p dng
=> Phi dng phng php n by
4.3.1.2.Phng php on by:
* Vi 1 xe ti thit k
' '
1 2
900 2300 900.1 .1 0.6087
2300
Sy y
S
= = = =
. '
1 2
1 1( . ) . ( ) 1.2 (0.6087) 2 0.7304
2 2
SI
M LL lanm g m y y x x x = + = =
* Vi 2 xe ti thit k
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'
3
1800 2300 1800.1 .1 0.2174
2300
Sy
S
= = =
'
4 1y =
'5 1200 2300 1200.1 .1 0.4783
2300Sy
S = = =
'
6 0y =
. ' ' '
3 4 5 6
1 1( . ) . ( ) 1 (0.2174 1 0.3913 0) 0.4239
4 4
MI
M LL lanm g m y y y y x x = + + + = + + + =
( )( . ) max ( . ) , ( . ) max(0.7304,0.4239) 0.7304I SI MI M LL M LL M LLm g m g m g = = =
* Vi ti trng ln:
Thin v an ton coi ti trng ln theo phng ngang cu l ti trng t p trung.
( . ) 1.2 1 1.2IM l anm g = =
4.3.1.3.La chn h s phn b lc ct thit k
( )( . ) max ( . ) , ( . ) 0.7304I IM M M LLm g m g m g = =
( )( . ) max ( . ) , ( ) 1.2I IM lan M M lanm g m g mg = =
4.3.2.H s phn b hot ti i vi lc ct ca dm bin4.3.2.1.Vi 1 lan thit k: s dng phng php on by tnh
Xt cho xe ti thit k v xe hai trc: v khong cch ca hai trc bnh xe theophng ngang ca hai loi xe ny l nh nhau nn c chung mt h s phn b ngang
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Phng trnh tung ng nh hng: ( )dbx
y xS
=
Mt ln thit k => h s ln m = 1.2
* Vi xe ti thit k:
4 3 25
150 600
2300 1250 150 200 500 1900 6000.2174
2300
kS S B B B
yS
+ + =
+ + = =
6 0y =
4 5
1 1 1( . ) . 1.2 ( ) 1.2 (0.2174 0) 0.1304
2 2 2
SE
M LL im g m y y y x x = = + = + =
* Vi ti trng ngi i
31
150 2300 1250 150 5001.3913
2300
kS S ByS
+ + + + = = =
3 23
150 2300 1250 150 500 19000.5652
2300
kS S B ByS
+ + + + = = =
B2=1.9m
221 2
1.9( ) (1.3913 0.5652) 1.8587
2 2
M PL
B y y x m = + = + =
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2
1.2( . ) . 1.8587 1.1739
1.9SEM PL M PL
mm g x
B = = =
* Vi ti trng ln:
3 2 43
150 2300 1250 150 200 500 1900 0.47832300
kS S B B ByS
+ + + + = = =
3 2 43
150 2300 1250 150 200 500 19000.4783 263.065
2 2
kM L an
S S B B By x
+ + + + = = =
1.2( . ) . 263.065 0.1052
3000 3000
SE
M lan M Lam
mm g x = = =
4.3.2.2. Vi hai hoc nhiu lan thit k:
Khong cch t tim dm bin n mp trong b va hoc lan can chn xe de
4 3 2150 1250 150 200 500 1900 1200e kd S B B B mm= + = + =
* Phm vi p dng:
GA TR(mm) PHM VI P DNG KIM TRA
S= 2300 1800 3500( )S mm THA
de= -1200 0 1400( )ed mm KHNG THA
Kt lun: Khng nm trong phm vi p dng
=> Khng s dng cng thc tra bng
4.3.2.3.La chn h s phn b lc ct thit k
Ta s dng h s phn b ngang trong trng hp 1 ln theo phng php n by thit k
( . ) 0.1304SEM LLm g =
( . ) 1.1739SEM PLm g =
( . ) 0.1052SEM L AN m g =
BNG TNG HP H S PHN B NGANG
V tr Ni lc Ti trng xe Ti trng ln Ti trng ngi
Dm gia
Mmen 0.73 1.2 0
Lc ct 0.73 1.2 0
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Dm binMmen 0.13 0.105 1.174
Lc ct 0.13 0.105 1.174
4.4. H s iu chinh t trngH s deo 1D = i vi cc b phn v lin kt thng thng
H s d tha 1R = i vi mc d tha thng thng
H s quan trng 1.05I = Cu thit k l quan trng
H s iu chnh ca ti trng
. . 1.05 0.95 D R I
= = >
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V. XC NH NI LC TA I CC MT CT C TRNG
5.1. Tnh ti tc dng ln 1 dm chu.5.1.1.Tnh ti dm ch
+ Xt on dm ct khc:
Din tch tit din 1 20 0.796gA m=
T trng b tng dm ch 32450 /c kN m =
Trng lng on dm 10 0. . .2 2450 0.796 0.8 2 3120.32d c g ck DC A L x x x kg = = =
+ Xt on dm c:
Din tch tit din1 2
1 1.630gA m=
Trng lng on dm 11 1. . .2 2450 1.630 1.2 2 9584.4d c g dac DC A L x x x= = =
+ Xt on dm cn li:
Din tch tit din 1 24 0.6343gA m=
Trng lng on dm
[ ] ( )1 4. . ( ).2 2450 0.6343 38.3 0.8 1.2 2 53303.4d c g ck dacDC A L L L x x x kg + = = + =
5.1.2.Tnh ti vch ngn
Vi dm c chiu di 38.3 m, ta dng 2 vch ng mng dy 15cm chia dm lm3 khoang, mi khoang di 12.67m. Vch ngn ny c tc dng tng n nh khi culp dm.
S vch ngn: 2
B dy vch ngn: 0.15m
Din tch vch ngn theo phng dc cu
1 1 2
1 4 7. 1.630 0.6343 0.05 0.82 0.9547vn g g h A A A V b x m= = =
Trng lng vch ngn
. . . 2450 0.9547 0.15 2 701.7vn c vn vn vn
DC A t N x x x kg = = =
+ Tnh ti dm ch coi l ti trng ri u sut chiu di dm
0 1 3120.32 9584.4 53303.4 701.7 1741.7738.3
d d d vndc
DC DC DC DC DC
L
+ + + + + += = =
5.1.3.Tnh ti bn mt cu
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
+ Dm gia:
Din tch tit din 2. 2.3 0.18 0.414bmg f A S h x m= = =
T trng b tng bn mt cu 32450 /c kg m =
Trng lng bn mt cu dm gia
. 2450 0.414 1014.3 /bmg c bmg DC A x kg m= = =
+ Dm bin :
Din tch tit din
22.3. ( 1.25) 0.18 0.4322 2bmb k f S
A S h x m = + = + =
Trng lng bn mt cu dm gia
. 2450 0.432 1058.4 /bmb c bmb DC A kg m= = =
5.1.4.Tnh ti dm ngang
Din tch mt ct dm ngang And = 0.6278 m2
Trng lng dm ngang
( . ). 0.6278 0.8 8. 2450 51.404 /
. 5 38.3
dn dn ndn c
b
A t N x xDC x kg m
N L x
= = =
5.1.5.Tnh ti vn khun lp ghp
Chiu di vn khun theo phng ngang 7 0.82vkb b m= =
B dy vn khun 0.05vk hh V m= = =0.05m
Trng lng vn khun trn 1m di dc cu
. . 2450 0.82 0.05 100.45 /vk c vk vk DC b h x x kg m= = =
5.1.6.Tnh ti lan can5.1.6.1. Lan can c tay vn
Phn thp c trng lng
16 /tlc
DC kg m=
Chiu cao lan can 0.3lch m=
Phn b tng 3. . 0.5 0.3 2450 367.5 /btlc lc c DC B h x x kg m= = =
Trng lng lan can c tay vn 16 367.5 383.5 /lc tlc btlc DC DC DC kg m= + = + =
* Trng lng lan can do dm bin chu
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
2300 12501.5435
` 2300
klcb
S Sy
S
+ += = =
. 383.5 1.5435 591.93 /lcb lc lcb DC DC y x kg m= = =
0lcgy =
. 0lcg lc lcg DC DC y= =
5.1.6.2. G chn
Chiu cao g chn 300 0.3gch mm m= =
Trng lng g chn . 2450 0.3 0.2 147 / gc c gc DC A x x kg m= = =
5.1.7. Tnh ti lp ph m t cu v tin ch cng cng (DW)
*Lp ph
Chiu dy lp b tng Asphal 1 70 0.07t mm m= =
T trng b tng Asphal 1 2250 /kg m =
*Lp phong nc
Chiu dy lp phng nc 2 4mm 0.004mt = =
T trng lp phng nc 32 1800kg / m =
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
Tng trng lng lp ph mt cu
1 1 2 2( ) (0.07 2250 0.004 1800) 2.3 378.81kg / m
ti DW xt xt xS x x x = + = + =
*Trng lng tin ch v trang thit b trn cu
0 /lp DW kg m=
378.81 /ti lp DW DW DW kg m= + = =378.81kg/m
* Trng lng DW do dm bin chu
3 4. 0.152
lp
b k ti
DW S DW S B B DW
S
= + + +
378.81 2300(1250 150 200 500 ) 0 304.695kg/m
2300 2= + + + =
* Trng lng DW do dm gia chu
378.81 /g DW DW kg m= =
5.1.8.Tng cng tnh ti tc dng ln cc dm dc ch
5.1.8.1.Dm gia
+ Giai on cha lin hp bn mt cu
0 1 1741.74 /d d d vndc DC DC DC DC
DC kg mL
+ + += =
+ Giai on khai thc: bn mt cu:
2889.58 / g dc bmg dn lcg vk vn DC DC DC DC DC DC DC kg = + + + + + =
378.81 /g DW DW kg m= =
5.1.8.2.Dm binGiai on cha lin hp bn mt cu
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
0 1 1741.74 /d d d vndc DC DC DC DC
DC kg mL
+ + += =
Giai on khai thc: bn mt cu:
3672.42 /b dc bmb dn lcb gc vk vn
DC DC DC DC DC DC DC DC kg m= + + + + + + =
304.695 /b DW kg m=
5.2. Hot ti HL935.2.1.Xe ti thit k:
Xe ti thit k: gm trc trc nng 35KN, hai trc sau mi trc nng 145KN,khong cch gia hai trc trc l 4300mm, khong cch hai trc sau thay i t 4300 -9000mm sao cho gy ra ni lc ln nht, Theo phng ngang khong cch gia hai bnh
xe l 1800mm
5.2.2.Xe hai trc thit k:
Xe hai trc thit k gm c hai trc, mi trc nng 110K, khong cch gia haitrc khng i l 1200mm, theo phng ngang khong cch gia hai bnh xe l 1800mm
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
5.2.3.Ti trng ln:
Ti trng ln bao gm ti trng ri u 9.3N/m. xp theo phng dc cu , theophng ngang cu ti trng ln phn b theo chiu rng 3000mm, ti trng ln c th xdch theo phng ngang gy ra ni lc ln nht.
5.2.4. Ti trng ngi i b:
L ti trng phn b c qui nh ln l 3E-03 Mpa.
5.2.5. Ti trng xung kch:
L ti trng ln vo ti trng xe 3 trc hay xe 2 trc ly bng 25% ti trng ca mixe.
5.3. ng nh hng mmen v lc ct ti cc mt ct c trng
5.3.1. Xc nh cc mt ct c trng
Mt ct gi: x0= 0 m
Mt ct ti ch thay i tit din: x1= 1.65 m
M ct khng dnh bm 1: x2= 3 m
Mt ct khng dnh bm 2: x3= 6 m
Mt ct L/2: x4=Ltt/2 18.8 m
5.3.2.Xc nh ng nh hng ni lc ti cc mt ct
5.3.2.1.Phng trinh ng anh hng
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
* Phng trnh ng nh hng mmen ti mt ct Xknh sau:+ Trn on X = 0 -> Xk
1( , ) .tt k
M ktt
L Xf X X X
L
=
+ Trn on x = Xk-> Ltt
2 ( , ) ( )k
M k tt tt
Xf X X L X
L=
+ Din tch ng nh hng mmen ti Xk
1( , )
2Mk tt M k k L y X X =
* Phng trnh ng nh hng lc ct
+ Trn on X=0 -> Xk
1( , )V ktt
X f X X
L
=
+ Trn on x = Xk-> Ltt
2 ( , ) 1V ktt
X f X X
L=
+ Din tch ng nh hng lc ct ti Xk
Phn h dng:
2
1( ). ( , )
2
Vkd tt k V k k L X f X X =
Phn h m:
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1
1. ( , )
2Vka k V k k
X f X X =
Tng din tch h:
Vk Vkd Vka
= +5.3.2.2.Ti mt ct c trng th 0 ( mt ct gi)
X = Xk= X0 = 0m
1
0
37.6 0( , ) ( , ) . 0 0
37.6
1( , ) 0
2
tt kM k k M k
tt
M tt M k k
L Xy X X f X X X x
L
L y X X
= = = =
= =
1
0 2
( , ) ( , ) 1
1 1( ). ( , ) (37.6 0) 18.8
2 2
tt
V k k V k tt
V tt k V k k
L X
y X X f X X L
L X f X X x
+
+
= = =
= = =
1 0
1( , ) ( , ) 0 0.0 0
2V k k V k V tt
Xy X X f X X x
L
= = = =
5.3.2.3.Ti mt ct c trng th 1
X = Xk= X1 =1.44m
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
1
0
37.6 1.44( , ) ( , ) . 1.44 1.385
37.6
1 1( , ) 37.6 1.385 26.035
2 2
tt kM k k M k
tt
M tt M k k
L Xy X X f X X X x
L
L y X X x x
= = = =
= = =
1
0 2
1.44 37.6 1.44( , ) ( , ) 0.962
37.6
1 1( ). ( , ) (37.6 1.44) 0.962 17.388
2 2
ttV k k V k
tt
V tt k V k k
L y X X f X X
L
L X f X X x x
+
+
= = = =
= = =
1 0
1.44 1( , ) ( , ) 0.0383 1.44 ( 0.0383) 0.0276
37.6 2V k k V k V tt
Xy X X f X X x x
L
= = = = = =
5.3.2.4.Ti mt ct c trng th 2
X = Xk= X2 =3 m
1
0
37.6 3( , ) ( , ) . 3 2.76
37.6
1 1( , ) 37.6 2.761 51.9
2 2
tt kM k k M k
tt
M tt M k k
L X y X X f X X X x
L
L y X X x x
= = = =
= = =
1
0 2
3 37.6 3( , ) ( , ) 0.920
37.6
1 1
( ). ( , ) (37.6 3) 0.920 15.922 2
ttV k k V k
tt
V tt k V k k
Ly X X f X X
L
L X f X X x x
+
+
= = = =
= = =
1 0
3 1( , ) ( , ) 0.0798 3 ( 0.0798) 0.1197
37.6 2V k k V k V tt
Xy X X f X X x x
L
= = = = = =
5.3.2.4.Ti mt ct c trng th 3
X = Xk= X3 =6 m
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
1
0
37.6 6( , ) ( , ) . 6 5.04
37.6
1 1( , ) 37.6 5.04 94.8
2 2
tt kM k k M k
tt
M tt M k k
L Xy X X f X X X x
L
L y X X x x
= = = =
= = =
1
0 2
3 37.6 6( , ) ( , ) 0.840
37.6
1 1( ). ( , ) (37.6 6) 0.840 13.28
2 2
ttV k k V k
tt
V tt k V k k
Ly X X f X X
L
L X f X X x x
+
+
= = = =
= = =
1 0
6 1( , ) ( , ) 0.1596 6 ( 0.1596) 0.479
37.6 2V k k V k V tt
Xy X X f X X x x
L
= = = = = =
5.3.2.4.Ti mt ct c trng th 4
X = Xk= X4 =18.8 m
1
0
37.6 18.8( , ) ( , ) . 18.8 9.4
37.6
1 1( , ) 37.6 9.4 176.72
2 2
tt kM k k M k
tt
M tt M k k
L X y X X f X X X x
L
L y X X x x
= = = =
= = =
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
1
0 2
18.8 37.6 18.8( , ) ( , ) 0.5
37.6
1 1( ). ( , ) (37.6 18.8) 0.5 4.7
2 2
ttV k k V k
tt
V tt k V k k
Ly X X f X X
L
L X f X X x x
+
+
= = = =
= = =
1 018.8 1( , ) ( , ) 0.5 18.8 ( 0.5) 4.7
37.6 2V k k V k V tt
Xy X X f X X x xL
= = = = = =
5.3.3.Bng tng h p din tch ng nh hng ti cc mt ct c trng
M
V + V V
X0 0.00 0.00 18.8 18.80
X1 26.04 -0.03 17.39 17.36X2 51.90 -0.12 15.92 15.80
X3 94.80 -0.48 13.28 12.80
X4 176.72 -4.70 4.7 0.00
5.4. Tnh ni lc do tnh ti tc dng ln dm gia v dm binCng thc tnh l ly gi tr ti trng nhn vi din tch ng nh hng ti mt ct
ang xt. Ta tnh ni lc ti cc mt ct nh sau:5.4.1.Mmen do tnh ti tc dng ln dm bin
5.4.1.1.Giai on cha ban b tng
. . ( . ) DCdc dc M M DC g kN m=
Mt ct M dcDC . . ( . ) DCdc dc M M DC g kN m=
X0 0.00 1741.74 0
X1 26.04 1741.74 440.25
X2 51.90 1741.74 877.46
X3 94.80 1741.74 1602.77
X4 176.72 1741.74 2987.77
5.4.1.2.Giai on a ban b tng
. . ( . )
. . ( . )
DCb b M
DWb b M
M DC g kN m
M DW g kN m
==
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
Mt ct M bDC bDW DCbM DWbM
X0 0.00 2889.58 378.81 0.00 0.00
X1 26.04 2889.58 378.81 738.15 96.77
X2 51.90 2889.58 378.81 1471.20 192.87X3 94.80 2889.58 378.81 2687.27 352.29
X4 176.72 2889.58 378.81 5009.44 656.71
5.4.2.Mmen do tnh ti tc dng ln dm gia
5.4.2.1.Giai on cha ban b tng
. . ( / ) DCdc dc M M DC g kN m=
Mt ct M dcDC . . ( / ) DCdc dc M M DC g kN m=
X0 0 1741.74 0.00
X1 26.04 1741.74 440.25
X2 51.9 1741.74 877.46
X3 94.8 1741.74 1602.77
X4 176.72 1741.74 2987.77
5.4.2.2.Giai on a ban b tng
. . ( / )
. . ( / )
DCb g M
DWb g M
M DC g kN m
M DW g kN m
=
=
Mt ct M gDC gDW DCgM DWgM
X0 0.00 2889.58 378.81 0.00 0.00
X1 26.04 2889.58 378.81 738.15 96.77
X2 51.90 2889.58 378.81 1471.20 192.87
X3 94.80 2889.58 378.81 2687.27 352.29
X4 176.72 2889.58 378.81 5009.44 656.71
5.4.3.Lc ct do tnh ti tc dng ln dm bin
5.4.3.1.Giai on cha ban b tng. . ( . ) DCdc dc V V DC g kN m=
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
Mt ct V dcDC DCdcV
X0 18.80 1741.74 317.85
X1 17.36 1741.74 293.50
X2 15.80 1741.74 267.13X3 12.80 1741.74 216.41
X4 0.00 1741.74 0.00
5.4.3.2.Giai on a ban b tng
. . ( . )
. . ( . )
DCb b M
DWb b M
M DC g V kN m
M DW g V kN m
=
=
Mt ct V bDC bDW DCbV DWbV
X0 18.80 3672.42 304.70 677.30 56.19
X1 17.36 3672.42 304.70 625.42 51.89
X2 15.80 3672.42 304.70 569.22 47.23
X3 12.80 3672.42 304.70 461.14 38.26
X4 0.00 3672.42 304.70 0.00 0.00
5.4.4.Lc ct do tnh ti tc dng ln dm gia
5.4.4.1.Giai on cha ban b tng
. . ( / ) DCdc dc V V DC g kN m=
Mt ct V dcDC DCdcV
X0 18.80 1741.74 317.85
X1 17.36 1741.74 293.50
X2 15.80 1741.74 267.13
X3 12.80 1741.74 216.41
X4 0.00 1741.74 0.00
5.4.2.2.Giai on a ban b tng
. . ( / )
. . ( / )
DCb g M
DWb g M
M DC g V kN m
M DW g V kN m==
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
Mt ct V gDC gDW DCgV DWgV
X0 18.80 2871.26 378.81 529.54 69.86
X1 17.36 2871.26 378.81 488.98 64.51
X2 15.80 2871.26 378.81 445.04 58.71X3 12.80 2871.26 378.81 360.54 47.57
X4 0.00 2871.26 378.81 0.00 0.00
5.4.5.Bng tng h p ni lc
+ i vi dm bin
Dm bin
Giai on 1 Giai on 2
DCdcM DCdcV DCbM DWbM DCbV DWbV
X0 0.00 317.85 0.00 0.00 677.30 56.19
X1 440.25 293.50 938.13 77.84 625.42 51.89
X2 877.46 267.13 1869.77 155.13 569.22 47.23
X3 1602.77 216.41 3415.30 283.36 461.14 38.26
X4 2987.77 0.00 6366.59 528.23 0.00 0.00
+ i vi dm gia
Dm binGiai on 1 Giai on 2
DCdcM
DCdcV DCgM DWgM DCgV DWgV
X0 0.00 317.85 0.00 0.00 529.54 69.86
X1 440.25 293.50 738.15 96.77 488.98 64.51
X2 877.46 267.13 1471.20 192.87 445.04 58.71
X3 1602.77 216.41 2687.27 352.29 360.54 47.57
X4 2987.77 0.00 5009.44 656.71 0.00 0.00
5.5. Ni lc do hot ti tc dng ln dm gia v dm bin:5.5.1.Mmen do hot ti HL93 v PL tc dng ti cc mt ct dm
i vi cc mt ct c trng trong phm vi t gi n Ltt/2 ta xt 2 trng hpxp xe bt li nht ln ng nh hng mmen. Ni lc do xe thit k s ly gi tr max
ca 2 trng hp trn.
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
* S tnh: S tnh ca dm ch l dm gin n nn khong cch gia cc trc caxe ti thit k (Truck) u ly = 4.3m
* Cch xp ti ln ng nh hng:
+TH1: Xp xe sao cho hp lc ca cc trc xe v trc xe gn nht cch u tung lnnht ca ng nh hng.Vi xe Truck ta c phng trnh xc nh khong cch u
x/2 4.3(x+4.3)+145.x=145.(4.3-x)
=>x/2 = 0.7275m
+TH2: Xp cc trc v cng 1 pha ng nh hng c din tch ln nht sao cho trctrc hoc trc sau ca xe trng vi tung ln nht ca ng nh hng.
* s xp ti
5.5.1.1.S xp tai ln ng anh hng cho tng mt ct c trng
Cng thc xc nh
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
2
1
1 1 1 11 2 5
2 2 2 2
1 3 4
1 2
( , ) ( )
( , )
( , ) . 0
35 . 145 . 145 .
145 . 145 . 35 .
( , )
kM k tt k
tt
M k
tt kM k k
tt
Truck M M M
Truck M M M
Truck Truck Truck
X f X X L X X X L
L y X X
L X f X X X X X
L
M kN Y kN Y kN Y
M kN Y kN Y kN Y
M Max M M
=
=
=
= + += + +
=
Mt ct1
1MY1
2MY1
5MY1
TruckM 2
1MY2
3MY2
4MY2
TruckM
TruckM
X0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
X1 0.00 0.69 1.25 280.32 1.38 1.22 1.06 414.67 414.67
X2 0.00 2.09 2.48 662.18 2.76 2.42 2.07 823.44 823.44
X3 0.82 4.43 4.47 1319.63 5.04 4.36 3.67 1491.30 1491.30
X4 6.89 9.04 7.61 2655.27 9.40 7.25 5.10 2592.75 2655.27
5.5.1.2.Mmen do xe 2 truc thit k (Tandem) gy ra ti cc mt ct c trng
Cng thc xc nh
2
1
1 1 1
3 4
2 2 2
1 2
1 2
( , ) ( )
( , )
( , ) . 0
110 . 110 .
110 . 110 .
( , )
kM k tt k
tt
M k
tt kM k k
tt
Tandem M M
Tandem M M
Tandem Tandem Tandem
X f X X L X X X L
L y X X L X
f X X X X X L
M kN Y kN Y
M kN Y kN Y
M Max M M
=
=
=
= +
= +
=
Mt ct 1
3MY 1
4MY 1
TandemM 2
1MY 2 2MY
2
TandemM TandemM
X0 0.00 0.00 0.00 0.00 0.00 0.00 0.00X1 0.81 1.36 238.67 1.38 1.34 299.61 299.61
X2 2.21 2.71 541.34 2.76 2.66 596.81 596.81
X3 4.54 4.95 1043.36 5.04 4.85 1088.30 1088.30
X4 9.10 9.10 2002.00 9.40 8.80 2002.00 2002.00
Vy mmen ti cc mt ct do xe thit k gy ra ( , ) . xetk Truck TandemM Max M M kN m=
BNG TNG HP
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
Mt ct TruckM TandemM xetkM
X0 0.00 0.00 0.00
X1 414.67 299.61 414.67
X2 823.44 596.81 823.44X3 1491.30 1088.30 1491.30
X4 2655.27 2002.00 2655.27
5.5.1.4.Mmen gy ra do tai trng lan.
Theo 3.6.1.2.4, ti trng ln ri u sut chiu di v c ln nh sau:
9,3 /lanq KN m=
Mmen do ti trng ln gy ra ti cc mt ct c xc nh bng phng phpng nh hng : Nhn gi tr qlan vi din tch ng nh hng.
. ( . )lan lan M M q kN m=
Gi tr din tch ng nh hng mmen ti cc mt ct c trng c tnh mc 5.3.2
BNG TNG HP
Mt ct 2( )
Mm lanq lanM
X0 0.00 9.30 0.00
X1 26.04 9.30 242.17
X2 51.90 9.30 482.67
X3 94.80 9.30 881.64
X4 176.72 9.30 1643.50
5.5.1.5.Mmen do tai trng ngi i gy ra dm bin
Coi nh dm bin chu ton b ti trng ngi i
2
2
3000 3 /
. . ( . ) PL M
PL Pa KN m
M PL B kN m
= ==
BNG TNG HP
Mt ct 2
( )M m PL PLM
X0 0.00 3 0.00X1 26.04 3 148.43
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
X2 51.90 3 295.83
X3 94.80 3 540.36
X4 176.72 3 1007.30
5.5.1.6. T h p mmen do hot tai (a nhn h s phn b m.g)
H s xung kch: 1 + IM = 1+ 0.25 =1.25
* Ti cc mt ct ca dm bin
( . ) .(1 ). ( . ) . ( . ) . ( . )SE SE SE LLb M LL xetk M lan lan M PL PLM m g IM M m g M m g M kN m = + + +
Trong
( . ) 0.1304SE
M LLm g =
( . ) 0.105SEM lanm g =
( . ) 1.174SEM PLm g =
BNG TNG HP
Mt ct (1 )IM+ ( . ) .SE
M LL xetk m g M ( . ) .SE
M lan lanm g M ( . ) .
SE
M PL PLm g M ( . )LLbM kN m
X0 1.25 0.0000 0.0000 0.0000 0.0000
X1 1.25 54.0874 25.4807 174.2416 267.3316X2 1.25 107.4058 50.7853 347.2787 532.3213
X3 1.25 194.5178 92.7639 634.3357 970.2469
X4 1.25 346.3394 172.9244 1182.4873 1788.3360
* Ti cc mt ct ca dm gia
( . ) .(1 ). ( . ) . ( . )I I
LLg M LL xetk M lan lanM m g IM M m g M kN m = + +
Trong
( . ) 0.73IM LLm g =
( . ) ) 1.2IM lanm g =
BNG TNG HP
Mt ct (1 )IM+ ( . ) .I
M LL xetk m g M ( . ) .I
M lan lanm g M ( . )LLgM kN m
X0 1.25 0.0000 0.0000 0.0000
X1 1.25 302.8895 290.6064 669.2183
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X2 1.25 601.4722 579.2040 1331.0443
X3 1.25 1089.2997 1057.9680 2419.5926
X4 1.25 1939.5007 1972.1952 4396.5711
5.5.2.Lc ct do hot ti HL93 v PL tc dng ti cc mt ct dm
i vi cc mt ct c trng trong phm vi t gi n Ltt/2 trng hp xp xe btli nht ln ng nh hng lc ct ca mt ct th hin trong hnh v sau:
5.5.2.1.S xp tai ln ng anh hng cho tng mt ct c trng5.5.2.2.Lc ct do xe tai thit k (Truck) gy ra ti cc mt ct c trng
Cng thc xc nh
1
2
1 3 4
( , ) 0
( , )
( , ) 1
145 . 145 . 35 .
V k k
tt
V k
V k k tt
tt
Truck V V V
X f X X X X
L y X X
X f X X X X L
L
V kN Y kN Y kN Y
=
==
= + +
Mt ct 1VY 3VY 4VY ( )TruckV kN
X0 1.0000 0.8856 0.7713 300.4075
X1 0.9617 0.8473 0.7330 287.9600
X2 0.9202 0.8059 0.6915 274.4870
X3 0.8404 0.7261 0.6117 248.5520
X4 0.5000 0.3856 0.2713 137.9075
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5.5.2.3.Lc ct do xe 2 truc thit k (Truck) gy ra ti cc mt ct c trng
1
2
1 2
( , ) 0
( , )
( , ) 1
110 . 110 .
V k k
tt
V k
V k k tt
tt
Tandem V M
X f X X X X
L y X X
X
f X X X X LL
V kN Y kN Y
=
=
=
= +
Mt ct 1VY 2VY ( )TandemY kN
X0 1.0000 0.9681 216.4910
X1 0.9617 0.9298 208.0650
X2 0.9202 0.8883 198.9350
X3 0.8404 0.8085 181.3790X4 0.5000 0.4681 106.4910
Vy lc ct ti cc mt ct do xe thit k gy ra ( , ) xetk Truck TandemV Max V V kN =
BNG TNG HP
Mt ct TruckV TandemV ( )xetkV kN
X0 300.408 216.491 300.4075
X1 287.96 208.065 287.9600
X2 274.487 198.935 274.4870
X3 248.552 181.379 248.5520
X4 137.908 106.491 137.9075
5.5.2.4.Lc ct gy ra do tai trng lanLc ct do ti trng ln gy ra ti cc mt ct ti cc mt ct xc nh bng phng
php ng nh hng , nhn gi tr qlan vi din tch ng nh hng phn dng(i vi mt ct t gi tri n Ltt/2:
Gi tr din tch ng nh hng lc ct phn din tch dng ti cc mt ct ctrng c tnh trn mc 5.3.2
9,3 /
.
lan
lan lan V
q KN m
V q
+
=
=BNG TNG HP
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Mt ct ( )V m+ lanq ( )lanV kN
X0 18.8 9.3 174.8400
X1 17.39 9.3 161.7270X2 15.92 9.3 148.0560
X3 13.28 9.3 123.5040
X4 4.7 9.3 43.7100
5.5.2.5.Lc ct do tai trng ngi i gy ra dm bin
Coi nh dm bin chu ton b ti trng ngi i
2
2
3000 3 /
. . ( ) PL V
PL Pa KN m
V PL B kN += =
=
BNG TNG HP
Mt ct ( )V m+
PL ( )PLV kN
X0 18.8 3 107.1600
X1 17.39 3 99.1230
X2 15.92 3 90.7440X3 13.28 3 75.6960
X4 4.7 3 26.7900
5.5.2.6. T h p lc ct do hot tai (a nhn h s phn b m.g)
H s xung kch: 1 + IM = 1+ 0.25 = 1.25
* Ti cc mt ct ca dm bin
( ) .(1 ). ( . ) . ( . ) .SE SE E LLb V LL xetk V Lan lan V PL PLV mg IM V m g V m g V = + + +
BNG TNG HP
Mt ct (1 )IM+ ( . ) .SE
V L L xetk m g V ( . ) .SE
V lan lanm g V ( . ) .
SE
V PL PLm g V ( )LLbV kN
X0 1.25 39.1836 18.3962 125.7965 193.1722
X1 1.25 37.5600 17.0165 116.3618 180.3283
X2 1.25 35.8027 15.5781 106.5256 166.8571
X3 1.25 32.4198 12.9948 88.8605 142.3801X4 1.25 17.9879 4.5991 31.4491 58.5331
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* Ti cc mt ct ca dm gia
( ) .(1 ). ( . ) .I I LLg V LL xetk V Lan lanV mg IM V m g V = + +
BNG TNG HP
Mt ct (1 )IM+ ( . ) .I
V L L xetk m g V ( . ) .I
V lan lanm g V ( )LLgV kN
X0 1.25 219.4281 209.8080 484.0931
X1 1.25 210.3360 194.0724 456.9924
X2 1.25 200.4949 177.6672 428.2858
X3 1.25 181.5510 148.2048 375.1436
X4 1.25 100.7324 52.4520 178.3675
5.6. T hp ti trng ti cc mt ct c trng5.6.1.Th p ni lc theo cc TTGH ti cc mt ct dm gia
5.6.2.1.Trng thi gii hn cng I.
* Mmen: n v kN.m
1 .(1.75 1.25 1.5 )uCD g LLg DCg DWg M M M M = + +
Mt ct LLgM DCgM DWgM 1uCD gM
X0 1.05 0.00 0.00 0.00 0.000
X1 1.05 669.22 738.15 96.77 2350.923
X2 1.05 1331.04 1471.20 192.87 4680.514
X3 1.05 2419.59 2687.27 352.29 8527.900
X4 1.05 4396.57 5009.44 656.71 15687.908
* Lc ct: n v kN
1 .(1.75 1.25 1.5 )uCD g LLg DCg DWg V V V V = + +
Mt ct LLgV DCgV DWgV 1 ( )uCD gV kN
X0 1.05 484.09 529.54 69.86 1694.5718
X1 1.05 456.99 488.98 64.51 1583.113
X2 1.05 428.29 445.04 58.71 1463.5584
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X3 1.05 375.14 360.54 47.57 1237.4579
X4 1.05 178.37 0.00 0.00 327.7503
5.6.1.2.Trng thi gii hn cng II.* Mmen:n v kN.m
2 .(0. 1.25 1.5 )uCD g LLg DCg DWg M M M M = + +
Mt ct LLgM DCgM DWgM 2uCD g M
X0 1.05 0.00 0.00 0.00 0.0000
X1 1.05 669.22 738.15 96.77 1121.2346
X2 1.05 1331.04 1471.20 192.87 2234.7203X3 1.05 2419.59 2687.27 352.29 4081.8986
X4 1.05 4396.57 5009.44 656.71 7609.2083
* Lc ct: n v kN
2 .(0. 1.25 1.5 )uCD g LLg DCg DWg V V V V = + +
5.6.1.3.Trngthi gii hncng III.
* Mmen: n vkN.m
3 .(1.35. 1.25 1.5uCD g LLg DCg DWg M M M M = + +
Mt ct LLgM DCgM DWgM 3uCD gM
X0 1.05 0.00 0.00 0.00 0.0000
X1 1.05 669.22 738.15 96.77 2069.8516
X2 1.05 1331.04 1471.20 192.87 4121.4755
X3 1.05 2419.59 2687.27 352.29 7511.6711
X4 1.05 4396.57 5009.44 656.71 13841.3478
* Lc ct: n v kN
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Mt ct LLgV DCgV DWgV 2 ( )uCD g V kN
X0 1.05 484.09 529.54 69.86 805.0508
X1 1.05 456.99 488.98 64.51 743.3895
X2 1.05 428.29 445.04 58.71 676.5833
X3 1.05 375.14 360.54 47.57 548.1315
X4 1.05 178.37 0.00 0.00 0.0000
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3 .(1.35. 1.25 1.5 )uCD g LLg DCg DWg V V V V = + +
Mt ct LLgV DCgV DWgV 3 ( )uCD gV kN
X0 1.05 484.09 529.54 69.86 1491.2527
X1 1.05 456.99 488.98 64.51 1391.1762
X2 1.05 428.29 445.04 58.71 1283.6784
X3 1.05 375.14 360.54 47.57 1079.8976
X4 1.05 178.37 0.00 0.00 252.8359
5.6.1.4.Trng thi gii hn s dung.
* Mmen: n v kN.m
.(1. 1 1. )uSDg LLg DCg DWg M M M M = + +
Mt ct LLgM DCgM DWgM uSDgM
X0 1.05 0.00 0.00 0.00 0.0000
X1 1.05 669.22 738.15 96.77 1579.3452
X2 1.05 1331.04 1471.20 192.87 3144.8700
X3 1.05 2419.59 2687.27 352.29 5732.1102
X4 1.05 4396.57 5009.44 656.71 10565.8572
* Lc ct: n v kN
.(1.. 1. 1. )uSDg LLg DCg DWg V V V V = + +
Mt ct LLgV DCgV DWgV ( )uSDgV kN
X0 1.05 484.09 529.54 69.86 1137.6678
X1 1.05 456.99 488.98 64.51 1061.0065
X2 1.05 428.29 445.04 58.71 978.6376
X3 1.05 375.14 360.54 47.57 822.4163
X4 1.05 178.37 0.00 0.00 187.2859
5.6.2.Th p ni lc theo cc TTGH ti cc mt ct ca dm bin
5.6.1.1.Trng thi gii hn cng I.
* Mmen: n v kN.m
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1 .(1.75 1.25 1.5 )uCD b LLb DCb DWbM M M M = + +
Mt ct LLbM DCbM DWbM 1uCD bM
X0 1.05 0.00 0.00 0.00 0.0000
X1 1.05 267.33 938.13 77.84 1845.1154X2 1.05 532.32 1869.77 155.13 3676.5433
X3 1.05 970.25 3415.30 283.36 6711.7019
X4 1.05 1788.34 6366.59 528.23 12474.1790
* Lc ct: n v kN
1 .(1.75 1.25 1.5 )uCD b LLb DCb DWbV V V V = + +
Mt ct LLbV DCbV DWbV 1 ( )uCD bV kN
X0 1.05 193.17 677.30 56.19 1332.4094
X1 1.05 180.33 625.42 51.89 1233.9438
X2 1.05 166.86 569.22 47.23 1128.0884
X3 1.05 142.38 461.14 38.26 927.1292
X4 1.05 58.53 0.00 0.00 107.5546
5.6.1.2.Trng thi gii hn cng II.* Mmen:n v kN.m
2 .(0. 1.25 1.5 )uCD b LLb DCb DWbM M M M = + +
Mt ct LLbM DCbM DWbM 2uCD bM
X0 1.05 0.00 0.00 0.00 0.0000
X1 1.05 267.33 938.13 77.84 1353.8936
X2 1.05 532.32 1869.77 155.13 2698.4029
X3 1.05 970.25 3415.30 283.36 4928.8733
X4 1.05 1788.34 6366.59 528.23 9188.1116
* Lc ct: n v kN
2.(0. 1.25 1.5 )
uCD g LLb DCb DWbV V V V = + +
Mt ct LLbV DCbV DWbV 2 ( )uCD bV kN
X0 1.05 193.17 677.30 56.19 977.4555
X1 1.05 180.33 625.42 51.89 902.5905
X2 1.05 166.86 569.22 47.23 821.4885
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X3 1.05 142.38 461.14 38.26 665.5058
X4 1.05 58.53 0.00 0.00 0.0000
5.6.1.3.Trng thi gii hn cng III.
* Mmen: n v kN.m
3 .(1.35. 1.25 1.5 )uCD g LLb DCb DWbM M M M = + +
Mt ct LLbM DCbM DWbM 3uCD bM
X0 1.05 0.00 0.00 0.00 0.0000
X1 1.05 267.33 938.13 77.84 1732.8362
X2 1.05 532.32 1869.77 155.13 3452.9683
X3 1.05 970.25 3415.30 283.36 6304.1982
X4 1.05 1788.34 6366.59 528.23 11723.0779
* Lc ct: n v kN
3 .(1.35. 1.25 1.5 )uCD g LLb DCb DWbV V V V = + +
Mt ct LLbV DCbV DWbV 3 ( )uCD bV kN
X0 1.05 193.17 677.30 56.19 1251.2771
X1 1.05 180.33 625.42 51.89 1158.2059
X2 1.05 166.86 569.22 47.23 1058.0084
X3 1.05 142.38 461.14 38.26 867.3295
X4 1.05 58.53 0.00 0.00 82.9707
5.6.1.4.Trng thi gii hn s dung.
* Mmen: n v kN.m
.(1. 1 1. )uSDb LLb DCb DWbM M M M = + +
Mt ct LLbM DCbM DWbM uSDbM
X0 1.05 0.00 0.00 0.00 0.0000
X1 1.05 267.33 938.13 77.84 1347.4667
X2 1.05 532.32 1869.77 155.13 2685.0824
X3 1.05 970.25 3415.30 283.36 4902.3522X4 1.05 1788.34 6366.59 528.23 9117.3138
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* Lc ct: n v kN
.(1.. 1. 1. )uSDb LLb DCb DWb
V V V V = + +
Mt ct LLbV DCbV DWbV ( )uSDbV kN
X0 1.05 193.17 677.30 56.19 972.9953
X1 1.05 180.33 625.42 51.89 900.5202
X2 1.05 166.86 569.22 47.23 822.4725
X3 1.05 142.38 461.14 38.26 673.8691
X4 1.05 58.53 0.00 0.00 61.4598
Ta c bng tng hp ni lc tnh ton ca dm bin v dm gia:
DM BIN DM GIA
1max( )uCD bM 1max( )uCD gM
12474.1790 15687.908
9117.3138 10565.8572
Cn c vo cc gi tr ni lc tnh ton th dm gia l dm bt li hn nn ta s chondm gia l dm tnh duyt.
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)max( 2buCDM )max( 3buCDM )max( uSDgM
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VI. TNH TON V B TRI CT THEP
6.1. Tnh ton din tich ct thp+ Dng loi tao c t chng thp Dps= 15.2mm tiu chun ASTM A416 Grade 270.
+ Loi tao thp DL: Tao thp c t chng thp
+ Cng tiu chun:91.86 10pu f x Pa=
+ H s qui i ng sut: 1 0.9 =
+ Cp ca thp: 270
+ Gii hn chy(TCN 5.9.4.4.1): 30.9 1.674 10 py pu f xf x MPa= =
+ ng sut trong thp DL khi kch( TCN 5.9.3.1):3
0.75 1.395 10 pj pu f xf x MPa= =
+ Din tch 1 tao cp: 21 140ps A mm=
+ Mdun n hi cp: 197000p E MPa=
+ Btong dm cp: '1 50c f MPa=
+ Mmen tnh ton: 1 1max( , ) 15687.90kN.u uCD g uCD bM M M= =
+i vi cu kin BTCT chu un v chu ko DL th h s sc khng: 1 =
+ :psA Din tch mt ct ngang ct thp DL:
+ :psgA Din tch mt ct ngang ct thp DL tnh theo kinh nghim.
C th tnh gn ng din tch ct thp theo cng thc kinh nghim sau:
33 2
9
15687.9 106.3 10
0.85 0.9 0.85 1.86 10 0.9 1.75
upsg
pu
M x A x m
f H x x x x
= = =
ps psg A A
S tao cp DL cn thit theo cng thc trn l:
630045
.1 140
psg
cg
ps
An
A= = =
Ta chn 45cn = tao thp Dps=15.2mm
6.2. B tr ct thp DL6.2.1.B tr ct thp DL ti mt ct ngang dm
* Mt ct ti gi: X0
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* Mt ct kim ton ct: X1
* Mt ct khng dnh bm 1: X2
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* Mt ct khng dnh bm 2: X3
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* Mt ct gia nhp L/2: X4
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6.2.2.B tr ct thp theo phng dc dm
Theo phng dc cu ct thp DL c ko thng, trnh xut hin ng sut
ko gy nt th trn do lc d ng lc, v tr dm gim ng sut ko ta s thit kcc oncp khng dnh bm vi b tng c v tr i xng vi tim dm.
S tao thp dnh dm ti cc mt ct nh sau:
Hng Ta X0 X1 X2 X3 X4
A 75 0 7 9 11 11
B 125 0 9 11 13 13
C 175 0 9 11 13 13
D 225 0 2 4 4 4E 275 0 2 2 2 2
F 1690 2 2 2 2 2
Ta cc nhm ct thp d ng lc tnh n y dm c n v l mm
Tnh ta trng tm ct thp DL ti cc mt ct:
*Mt ct ti gi: X0 = 0m
+ Din tch ct thp DL bu dm ti mt ct:
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5
ps0 ps.1 0
1
A = (A . ) 140 (0 0 0 0 0) 0 ps ii
n x=
= + + + + =
+ Din tch ct thp DL th trn dm ti mt ct:
2
ps' ps.1A 2.A 2 140 280 x mm= = =
+ Ta trng tm cc ct thp DL bu dm ti gi( tnh n y dm):
5
01
0 5
01
( . )0
psi ps ii
ps
ps ii
y nC
n
=
=
= =
+ Ta trng tm cc ct thp DLth trn:
20' ' 60 800 60 740psC H mm= = =
Khong cch t ta trng tm ct thp n th trn dm SuperT:
0 0
0' 0'
' 800mm 0.8m
' 60mm 0.06m
ps ps
ps Cps
d H C
d H
= = =
= = =
Trong :
yps:ta tao cp DL
nps : S tao cp ti cc hng*Mt ct duyt lc ct: X1 = 1.44m
+ Din tch ct thp DL bu dm ti mt ct:
52
ps1 ps.1 11
A = (A . ) 140 (7 9 9 2 2) 4060ps ii
n x mm=
= + + + + =
+ Ta trng tm cc ct thp DL bu dm ti gi( tnh n y dm):
5
11
1 5
1
1
( . ) 75 7 125 9 175 9 225 2 275 2145.69 0.1457
7 9 9 2 2
psi ps ii
ps
ps i
i
y n x x x x xC mm m
n
=
=
+ + + += = = =+ + + +
Khong cch t ta trng tm ct thp n th trn dm SuperT:
1 1 1.75 0.1457 1.6043 ps psd H C m= = =
*Mt khng dnh bm 1: X2 = 3 m
+ Din tch ct thp DL bu dm ti mt ct:
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52
ps2 ps.1 21
A = (A . ) 140 (9 11 11 2 2) 4900ps ii
n x mm=
= + + + + =
+ Ta trng tm cc ct thp DL bu dm ti gi( tnh n y dm):
5
21
2 5
21
( . ) 75 9 125 11 175 11 225 2 275 2142.14 0.1421
9 11 11 2 2
psi ps ii
ps
ps ii
y n x x x x xC mm m
n
=
=
+ + + += = = =
+ + + +
Khong cch t ta trng tm ct thp n th trn dm SuperT:
2 21.75 0.1421 1.6079
ps psd H C m= = =
*Mt khng dnh bm 2: X3 = 6 m
+ Din tch ct thp DL bu dm ti mt ct:5
2
ps3 ps.1 3
1
A = (A . ) 140 (11 13 13 4 2) 6020 ps ii
n x mm=
= + + + + =
+ Ta trng tm cc ct thp DL bu dm ti gi( tnh n y dm):
5
3
13 5
3
1
( . )75 11 125 13 175 13 225 4 275 2
143.6 0.143611 13 13 4 2
psi ps i
ips
ps i
i
y n x x x x x
C mm
n
=
=
+ + + += = = =
+ + + +
Khong cch t ta trng tm ct thp n th trn dm SuperT:
3 31.75 0.1436 1.6064
ps psd H C m= = =
*Mt gia nhp L/2: X4 = 18.8 m
+ Din tch ct thp DL bu dm ti mt ct:
52
ps4 ps.1 41
A = (A . ) 140 (11 13 13 4 2) 6020ps ii
n x mm=
= + + + + =
+ Ta trng tm cc ct thp DL bu dm ti gi( tnh n y dm):
5
41
4 5
41
( . )75 11 125 13 175 13 225 4 275 2
143.6 0.143611 13 13 4 2
psi ps ii
ps
ps ii
y nx x x x x
C mm mn
=
=
+ + + += = = =
+ + + +
Khong cch t ta trng tm ct thp n th trn dm SuperT:
4 4 1.75 0.1436 1.6064 ps psd H C m= = =
BNG TNG HP
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Mt ct Apsi (mm2) Dpsi (mm) dpi (m)
X0 0 0.000 0.800
X1 4060 145.690 1.604
X2 5180 146.622 1.603X3 6020 143.605 1.606
X4 6020 143.605 1.606
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VII. DC TRNG HNH HOC CUA CC MT CT DM
7.1. Chiu rng c hiu ca dmchiu rng c hiu ca dm gia ly gi tr nh nht trong cc gi tr sau:
- 1/4 chiu di nhpbch1= Ltt/4 = 37.6/4 = 9.4 m
-12 ln b dy trung bnh ca bn cng gi tr ln hn trong hai gi tr b rng sn dmv na b rng bn trn dm super T
bch2=12.hf+ max(bw,b6)
-khong cc trung bnh gia cc dm: bch3= S
lp bng tng hp (m)
X bch1 bch2 bch3 bch
0 9.4 3.05 2.3 2.3
1 9.4 3.29 2.3 2.3
2 9.4 2.88 2.3 2.3
3 9.4 2.88 2.3 2.3
4 9.4 2.88 2.3 2.3
M un n hi ca b tng:Ecdam=36872.5Mpa
M un n hi ca thp:Eps=197000Mpa
H s qui i thp sang b tngps
cdam
En 5.343
E= =
T l m un n hi gia bn mt cu v dm ch:cban
cdam
En ' 0.800
E= =
7.2. t trng hnh hc mt ct dm Super T giai on I.7.2.1.Mt ct ti gi X0
*Din tch mt ct: Ag01=0.796m2
*Ta trng tm mt ct so vi th di dm: Yc0=0.419 m3
*M men tnh ca mt ct i vi trc y dm: Sx0=0.334 m3
*M men qun tnh ca mt ct i vi trc X: Jx0 = 0.159 m4
*M men qun tnh i vi trc trung ha :Jthx0 = 0.043 m4
*Din tch mt ct quy i
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n Tt Nghip Trng i Hc Giao Thng Vn Ti
Ag0. = A0g1 + (n-1).Aps0 = 0.79625 + (5.343 - 1).0 = 0.796m2
* Mmen tnh ca ct thp cng cao i vi trc trung ha ca mt ct nguyn
Sx-x0 = (n1-1).Aps0.(Dps0-(H-Yc)) = 0 m3
* Khong cch t trng tm ca tit din c ct thp n trng tm ca tit din nguynZ0 = Sx-x0/Ag0 = 0 m3
* Khong cch t trng tm ca tit din c ct thp n th trn dm
ytg0 = H-ybg0 = 0.381m
* Khong cch t trng tm ca tit din c ct thp n th di dm
ybg0 = Yc-Z0 = 0.419 m
* Mmen qun tnh tnh so vi trc trung ha ca mt ct c ct thpIg0.= Ig0 + Z02*Ag0 = 0.043 m4
7.2.2.Mt ct ti on khc X1
*Din tch mt ct : F1 = 1.63 m2
*Ta trng tm mt ct so vi th di dm Yc1 = 0.98 m3
*M men tnh ca mt ct i vi trc X: Sx1 = 1.597 m3
*M men qun tnh ca mt ct i vi trc X : Jx1 = 1.407 m4
*M men qun tnh i vi trc trung ha: Jthx1 = 0.441 m4
*Din tch mt ct quy i
Ag1. = A0g1 + (n-1).Aps1= 1.63 + (5.343 - 1).0.00406 = 1.648 m2
* Mmen tnh ca m ct thp i vi trc trung ha ca mt ct nguyn
Sx-x1= (n - 1).Aps1.(dps1 - (H - Yc1)) = (5.343-1)x0.00406x(1.604 (1.75 0.98))
= 0.0147 m3
* Khong cch t trng tm ca m ct thp n trng tm ca tit din
Z1=Sx-x1/A0g1 = 0.0089 m
* Khong cch t trng tm ca tit din c ct thp n th trn dm
ytg1= H - ybg1= 0.780 m
* Khong cch t trng tm ca tit din c ct thp n th di dm
ybg1 = Yc1-Z1=0.98 0.0089 = 0.970 m
* Mmen qun tnh tnh so vi trc trung ha ca mt ct c ct thp
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Ig1.= Ig1 + Z12xA0g1+ Aps1x(Yc1 - (H - dp1 - Z1 ))2
= 0.441+ 0.00892x1.63 + 0.00406x(0.98 (1.75 1.604 0.0089))2 = 0.444 m4
7.2.3.Mt ct khng dnh bm 1: X2
*Din tch mt ct : F2 = 0.6343 m2
*Ta trng tm mt ct so vi th di dm Yc2 = 0.848 m3
*M men tnh ca mt ct i vi trc X: Sx2 = 0.538 m3
*M men qun tnh ca mt ct i vi trc X : Jx2 = 0.7620 m4
*M men qun tnh i vi trc trung ha: Jthx1 = 0.2459 m4
*Din tch mt ct quy i
Ag2. = A0g2 + (n-1).Aps2= 0.6343 + (5.343 - 1).0.00518 = 0.6568 m2
* Mmen tnh ca m ct thp i vi trc trung ha ca mt ct nguyn
Sx-x2= (n - 1).Aps2.(dps2 - (H - Yc2)) =
= (5.343-1)x0.00518x(1.603 (1.75 0.848)) = 0.0158 m3
* Khong cch t trng tm ca m ct thp n trng tm ca tit din
Z2=Sx-x2/A0g2 = 0.0158/0.6568 = 0.0240 m
* Khong cch t trng tm ca tit din c ct thp n th trn dm
ytg2= H - ybg2= 1.75 0.824 = 0.926 m
* Khong cch t trng tm ca tit din c ct thp n th di dm
ybg2 = Yc2-Z2= 0.848 0.0240 =0.824 m
* Mmen qun tnh tnh so vi trc trung ha ca mt ct c ct thp
Ig2.= Ig2 + Z22xA0g2+ Aps2x(Yc2 - (H - dp2 Z2 ))2
= 0.2459+ 0.02402x0.6343 + 0.00518x(0.848 (1.75 1.603 0.0240))2
= 0.2506 m4
7.2.4.Mt ct khng dnh bm 2: X3
*Din tch mt ct : F3 = 0.6343 m2
*Ta trng tm mt ct so vi th di dm Yc3 = 0.848 m3
*M men tnh ca mt ct i vi trc X: Sx3 = 0.538 m3
*M men qun tnh ca mt ct i vi trc X : Jx3 = 0.7620 m4
*M men qun tnh i vi trc trung ha: Jthx3 = 0.2459 m4
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*Din tch mt ct quy i
Ag3 = A0g3 + (n-1).Aps3= 0.6343 + (5.343 - 1).0.00602 = 0.6604 m2
* Mmen tnh ca m ct thp i vi trc trung ha ca mt ct nguyn
Sx-x3= (n - 1).Aps3.(dps3 - (H - Yc3)) == (5.343-1)x0.00602x(1.606 (1.75 0.848)) = 0.0184 m3
* Khong cch t trng tm ca m ct thp n trng tm ca tit din
Z3=Sx-x3/A0g3 = 0.0185/0.6604 = 0.0279 m
* Khong cch t trng tm ca tit din c ct thp n th trn dm
ytg3= H - ybg3= 1.75 0.824 = 0.9299 m
* Khong cch t trng tm ca tit din c ct thp n th di dmybg3 = Yc3-Z3= 0.848 0.0279 = 0.8201 m
* Mmen qun tnh tnh so vi trc trung ha ca mt ct c ct thp
Ig3.= Ig3 + Z32xA0g3+ Aps3x(Yc3 - (H - dp3 Z3 ))2
= 0.2459+ 0.02792x0.6343 + 0.00602x(0.848 (1.75 1.606 0.0279))2
= 0.2514 m4
7.2.4.Mt ct gia nhp: X4
*Din tch mt ct : F4 = 0.6343 m2
*Ta trng tm mt ct so vi th di dm Yc4 = 0.848 m3
*M men tnh ca mt ct i vi trc X: Sx4 = 0.538 m3
*M men qun tnh ca mt ct i vi trc X : Jx4 = 0.7620 m4
*M men qun tnh i vi trc trung ha: Jthx4 = 0.2459 m4
*Din tch mt ct quy i
Ag4. = A0g4 + (n-1).Aps4= 0.6343 + (5.343 - 1).0.00602 = 0.6604 m2
* Mmen tnh ca m ct thp i vi trc trung ha ca mt ct nguyn
Sx-x4= (n - 1).Aps4.(dps4 - (H - Yc4)) =
= (5.343-1)x0.00602x(1.606 (1.75 0.848)) = 0.0184 m3
* Khong cch t trng tm ca m ct thp n trng tm ca tit din
Z4=Sx-x4/A0g4 = 0.0185/0.6604 = 0.0279 m
* Khong cch t trng tm ca tit din c ct thp n th trn dm
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ytg4= H - ybg4= 1.75 0.824 = 0.9299 m
* Khong cch t trng tm ca tit din c ct thp n th di dm
ybg4 = Yc4-Z4= 0.848 0.0279 = 0.8201 m
* Mmen qun tnh tnh so vi trc trung ha ca mt ct c ct thpIg4.= Ig4 + Z32xA0g3+ Aps3x(Yc3 - (H - dp3 Z3 ))2
= 0.2459+ 0.02792x0.6343 + 0.00602x(0.848 (1.75 1.606 0.0279))2
= 0.2514 m4
7.3. t trng hnh hc mt ct dm Super T giai on II.7.3.1.Mt ct ti gi X0
* Din tch nguyn khi
A.g0 = A2g0+ (n-1).Aps0 = 1.12745 + (5.343 - 1).0 = 1.12745 m2
* Mmen tnh ca tit din i vi TTH I-I (ca giai on 1)
SI-I0 = n'.S.hf(hf/2+ytg0) = 0.8x2.3x0.18(0.09+0.381) = 0.156m3
* Khong cch t trng tm dm n trng tm tit din lin hp.
Z.0 = SI-I0/A.g0 = 0.156/1.12745 = 0.138 m
* Khong cch t trng tm ca tit din lin hp n y dm
y.bg0 = ybg0 + Z.0= 0.419 + 0.138 = 0.557 m
* Khong cch t trng tm tit din lin hp n mp trn dm
y.tg0 = H_ - y.bg0 = 0.8 - 0.557 = 0.243 m
* Mmen qun tnh
2 3 2
. 0 0. 0. .0 . 0
2 3 2 4
1'( . . .( ) )12 2
1 0.18
0.043 0.796 0.138 0.8 ( 2.3 0.18 2.3 0.18 (0.243 )) 0.09612 2
f
g g f f tg
h I I Ag xZ n S h S h y
x x x x x x m
= + + + + =
+ + + + =
7.3.2.Mt ct ti mt ct tnh duyt lc ct X1
* Din tch nguyn khi
A.g1 = A2g1+ (n-1).Aps1 = 1.961 + (5.343 - 1).0.00406 = 1.9786 m2
* Mmen tnh ca tit din i vi TTH I-I (ca giai on 1)
SI-I1 = n'.S.hf(hf/2+ytg1) = 0.8x2.3x0.18(0.09+0.7789) = 0.2878 m3
* Khong cch t trng tm dm n trng tm tit din lin hp.Z.1 = SI-I1/A.g1 = 0.2878/1.9786 = 0.146 m
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* Khong cch t trng tm ca tit din lin hp n y dm
y.bg1 = ybg1 + Z.1= 0.9711 + 0.146= 1.1171 m
* Khong cch t trng tm tit din lin hp n mp trn dm
y.tg1 = H - y.bg1 = 1.75 1.1171 = 0.6329 m* Mmen qun tnh
2 3 2. 1 1. 1. .1 . 1
2 3 2 4
1'( . . .( ) )12 2
1 0.180.444 1.6471 0.146 0.8 ( 2.3 0.18 2.3 0.18 (0.6329 )) 0.653
12 2
fg g f f tg
hI I Ag xZ n S h S h y
x x x x x x m
= + + + + =
+ + + + =
7.3.3.Mt ct khng dnh bm 1: X2
* Din tch nguyn khiA.g2 = A2g2+ (n-1).Aps2 = 0.966 + (5.343 - 1)x0.00518 = 0.9890 m2
* Mmen tnh ca tit din i vi TTH I-I (ca giai on 1)
SI-I2 = n'.S.hf(hf/2+ytg2) = 0.8x2.3x0.18(0.09+0.9260) = 0.3365 m3
* Khong cch t trng tm dm n trng tm tit din lin hp.
Z.2 = SI-I1/A.g2 = 0.3365/0.9890 = 0.340 m
* Khong cch t trng tm ca tit din lin hp n y dm
y.bg2= ybg2 + Z.2= 0.8240 + 0.340= 1.164 m
* Khong cch t trng tm tit din lin hp n mp trn dm
y.tg2 = H_ - y.bg2 = 1.75 1.164 = 0.586 m
* Mmen qun tnh
2 3 2. 2 2. 2. .2 . 2
2 3 2 4
1'( . . .( ) )12 2
1 0.180.2506 0.6568 0.340 0.8 ( 2.3 0.18 2.3 0.18 (0.586 )) 0.47912 2
fg g f f tg
hI I Ag xZ n S h S h y
x x x x x x m
= + + + + =
+ + + + =
7.3.4.Mt ct khng dnh bm 2: X3
* Din tch nguyn khi
A.g3 = A2g3+ (n-1).Aps3 = 0.966 + (5.343 - 1)x0.00602 = 0.9921 m2
* Mmen tnh ca tit din i vi TTH I-I (ca giai on 1)
SI-I3 = n'.S.hf(hf/2+ytg3) = 0.8x2.3x0.18(0.09+0.9299) = 0.3378 m3
* Khong cch t trng tm dm n trng tm tit din lin hp.
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Z.3 = SI-I1/A.g3 = 0.3378/0.9921 = 0.3405 m
* Khong cch t trng tm ca tit din lin hp n y dm
y.bg3= ybg3+ Z.3= 0.8201 + 0.3405= 1.1606 m
* Khong cch t trng tm tit din lin hp n mp trn dmy.tg3 = H - y.bg3 = 1.75 1.1606 = 0.5849 m
* Mmen qun tnh
2 3 2. 3 3. 3. .3 . 3
2 3
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