n b tng ct thp s ii k.cong ngh - i hc vinh n b tng ct thp s II n b tng ct thp s IIThit k khung ngang nh cng nghip mt tng Thit k khung ngang nh cng nghip mt tngS LIU CHO TRC: S LIU CHO TRC:Nh cng nghip 1 tng lp ghp, ba nhp u nhau,c ca mi thng gi v chiu sng t ti nhp gia. L = 21m, cng cao trnh ray R=6.5m, mi nhp c hai cu trc chy in, ch lm vic nngsc trc Q=20/5 T, dy mc cu cng. Bc ct a = 12 m. a im xy dng ti Bc Ninh. I - LA CHN KCH THC CC CU KIN: I - LA CHN KCH THC CC CU KIN:1. Chn kt cu mi :Vi nhp L=21 m chn kt cu dn b tng ct thp dng hnh thang, dc cc thanh cnh trn i=1/10 , chiu cao gia dn c th chn nh sau:H = (1/7 1/9)L = (21/7 21/9) = (3, 2,33) m.Chn H = 2,8 m. Chiu cao u dn: Hd =H- .2Li = 2,8 - 1 21.10 2 =1,8 m.-Khong cch gia cc mt dn thanh cnh thng ly bng 3m-Khong cch gia cc mt dn thanh cnh h ly bng 6m.Ca mi ch t nhp gia c b tr dc nh, chiu rng ca ca mi ly bng 12m, cao 4 m.Cc lp mi c cu to t trn xung di nh sau: - Hai lp gch l nem k c va lt dy 5 cm. - Lp b tng nh cch nhit dy 12 cm.- Lp b tng chng thm dy 4 cm.- Panen mi l dng panen sn, kch thc 12 3 m cao 45 cm.Tng chiu dy cc lp mi: t = 5 + 12 + 4 + 45 = 66 cm.2. Chn dm cu trc : 2. Chn dm cu trc :- Trc nh v ca nh :Cu trc c sc trc Q = 20T thuc trng hp Q < 30T do trc nh v ca nh c xc nh nhhnh sau: Ct bin: trc nh v trng mp ngoi ct.Ct gia: trc nh v trng vi trc hnh hc ca ct.SVTH : dng Ch linhlp 48k1_ksxd 1 n b tng ct thp s ii k.cong ngh - i hc vinhA B- Nhp cu trc c xc nh : LK = L- 2=21 2.0,75 =19,5 m.- Cu trc lm vic ch nng, sc trc Q=20/5 T , LK =19,5 m tra bng ta c cc ch tiu ca cu trc chy in nh sau:B = 6300 (mm); K = 4400 (mm) ; Hct = 2400(mm); B1 = 260(mm).- p lc bnh xe ln ray:Pmaxtc =22 T. Ptcmin = 4,8 T.-Trng lng xe con: G = 6,0 T-Ton cu trc: G = 33,5 T.-T Ptcmax = 22 TTachn ng ray c chiu cao :Hr = 150 (mm).Trng lng tiu chun ca ng ray:grc = 150 kG/m. - Vi bc ct a =12 m, nhp nh L= 21 m, sc trc Q = 20 T ta chn dm tit din ch I bng b tng ct thp vi cc kch thc c chn theo thit k nh hnh:Hc = 1400 ;b =140 ; bf= 650; hf= 180 ; bf = 340 ; hf =300trng lng dm cu trc : 11,3 T.3. Xc nh chiu cao nh :Ly cao trnh nn nh tng ng vi ctt0.00 xc nh cc kch thc khc.-Cao trnh vai ct: V = R- ( Hr + Hc).R- cao trnh ray cho R = 6,5 m.Hr- chiu cao ray v cc lp m Hr = 0,15 m.Hc- chiu cao dm cu trc Hc = 1,4 m.V = 6,5- ( 0,15 + 1,4) = 4,95.SVTH : dng Ch linhlp 48k1_ksxd 2 n b tng ct thp s ii k.cong ngh - i hc vinh -Cao trnh nh ct:D = R + Hct + a1 Hct- chiu cao cu trc tnh t cao trnh ray n nh xe con, Hct = 2,4 ma1- khe h an ton t nh xe con n mt di kt cu mang lc mi,chn: a1= 0,15m m bo a1 0,1 m D = 6,5 + 2,4 + 0,15 = 9,05 m.- Cao trnh nh mi: M = D + h + hcm + t.Chiu cao kt cu mang lc mi: h = 2,8m.Chiu cao ca mi: hcm = 4mTng chiu dy cc lp mi: t = 0,66 m Cao trnh nh mi nhp th hai c ca mi: M2 = 9,05 + 2,8 + 4 + 0,66 = 16,51 m Cao trnh mi 2 nhp bin khng c ca mi: M1 = 9,05 + 2,8+ 0,66= 12,51 m4. Chn kch thc ct : Trong trng hp s liu ca n: sc nng trc nh hn 30T, nhp nh nh hn 30 m ta chn loi ct c c tit din ch nht. - Chn chiu di cc phn ct:+ Chiu di phn ct trn:Ht = D V = 9,05 4,95 = 4,1( m ).+ Chiu di phn ct di:Hd = V + a2 a2 l khong cch t mp trn ca mng n mt nn. Ta chn a2 = 0,5 m m bo a20,4mHd = 4,95 + 0,5 = 5,45 (m) * Chn tit din ct:- B rng ct b c chn theo thit k nh hnh, thng nht cho ton b phn ct trn v ct di, cho c ct bin ln ct gia, vi a =12 cm ta chn b = 50 cm tho mn mnh ca ct theo phng ngoi mt phng l : l0/b = 1,2 Hd/ b =1,2.5,45/ 0,5 = 13,08 < 30- Xc nh chiu cao tit din phn ct trn:SVTH : dng Ch linhlp 48k1_ksxd 3 n b tng ct thp s ii k.cong ngh - i hc vinh+Ct bin: chn da vo iu kin a4 = - B1 ht >6 ( cm ). Vi a4 l khong cch t mp bn cu trc n mp trong ca ct.Tra bng ta c B1 = 26 (cm); = 75 ( cm ).Chn ht = 40 ( cm ) tho mn iu kin a4 =75 - 40 26 = 9 > 6 (cm).+Ct gia: chn da theo iu kin a4 = - B1 0,5 .ht Chn ht = 60 (cm) tho mn iu kin. a4 =75 0,5 .60 26 = 19 >6 (cm).- Xc nh chiu cao tit din phn ct di:+Ct bin chn hd = 60 (cm) tho m iu kin : hd >Hd / 14 = 5,45/14 = 0,39 m.+Ct gia chn hd = 80 (cm).- Xc nh kch thc vai ct:+ Ct bin : hv = 60 cm ; lv= 40 cm ;gc nghing cnh di ca vai bng 45o.+ Ct gia : hv =60 cm ; lv=60 cm ;gc nghing cnh dica vai bng 45o.- Kim tra mnh ca ct : * Ct bin: + mnh ca ct trn theo phng trong mt phng khung l: 02, 5 2, 5.4,125, 625 300, 4tt tl Hh h 5 cm, tho mn cc quy nh v cu to.Khong cch gia cc ct thp pha t 1 28 + 4 22 l: (50 2,5.2 2,8.1 -2,2.4)/4 =8,35 cm tho mn yu cu cu to. 2. Phn ct di:Tra bng ta c chiu di tnh ton: l0 = 1,5 Hd = 1,5 545 =817,5cm. Kch thc tit din b = 50 cm, h = 60 cm. Gi thit chn a = a = 4 cm, h0 = 60 - 6 = 56 cm; h0 - a = 56 - 4 = 52 cm mnh h = l0/h = 817,5/ 60 = 13,625 > 8 cn xt n un dcT bng t hp ni lc chn ra ba cp nghi ng l nguy him ghi bng sau:K hiu cp ni lcK hiu bng t hpM(T.m)N(T)e1=M/N(m)e0=e1+ea(m)Mdh(Tm)Ndh(T)123IV-13IV-17IV-1840,359-28,958744,6874138,545199,6091210,66560,29130, 1450,2120,31130,16510,23214,9294,9294,929138,545138,545138,545 lch tm tnh ton: e0 = M/N + eavi ea l lch tm ngu nhin, ly bng 2 cm tho mn iu kin:ea( h/30 = 60/30 = 2 cm; Hd/600 =545/600 = 0,9 cm; 1cm)Dng cp ni lc 3 tnh ton sau kim tra cho cc cp cn lia. Tnh vi cp 3 :SVTH : dng Ch linhlp 48k1_ksxd 30 n b tng ct thp s ii k.cong ngh - i hc vinh tnh ton nh hng ca un dc, tm gi thit t = 1,2%, IS = t b h0 (0,5 h a)2 = 0,012 50 56 (0,5 60 4)2 = 22713,6 cm4Ib = b b3/12 = 50 603 /12 = 900000 cm4 Vi cp 3 c e0/h = 23,21/60 = 0,387l = 1+(0, 5 ) 4, 929 138, 545(0, 5 0, 6 0, 04)1(0, 5 ) 44, 6874 210, 6656(0, 5 0, 6 0, 04)dh dhM N h aM N h a+ + ++ + = 1,412 S = 0,110,10,1 0, 387 ++ = 0,326 Ntr =206, 4(lSl x Eb Ib + ES IS) Ntr = 26, 4 0, 326(817, 5 1, 412 300.103 900000 + 200.10422713,6) = 1032 T H s xt n nh hng ca un dc:= 11trNN = 1210, 665611032 = 1,257 ep = 0,4(1,25 h - Rh0) = 0,4(1,25 60 0,563 56) = 17,39 cm e0 = 1,257 23,21 = 29,18 cm > ep=17,39 cm Tnh theo trng hp lch tm ln:e

= e0 + 0,5 h a = 29,18 + 0,5 60 4 = 55,18 cmTnhRh0 = 0,563 56 =31,53 cmChn x= 30 cm . Ta tnh :AS=0'0. . ( 0, 5 )( )bSCN e Rb x h xR h a =210665, 6 55,18 145 50 30 (56 15)3650 (56 4) = 14,26 cm2 Kim tra = 100AS /b ho = 10014,26/(50 56) = 0, 51% vi mnh h = 13,625 c min = 0,05% > min tho mn.Dng AS= 14,26 tnh AS theo cng thc :AS ='. . .b SC SSRb x R A NR+ =145 50 30 3650.14, 26 210665, 63650 + = 16,13 cm2Kim tra t =( AS + AS )/b ho=(16,13 + 14,26 )/(50 x 56) = 0,0109 = 1,09%so vi tr s gi thit l 1,2% l xp x nhau, c th khng cn tnh li. Chn thp: AS: 3 22 + 2 18( 16,49 cm2 ) AS: 3 20 + 2 18( 14,51 cm2 )SVTH : dng Ch linhlp 48k1_ksxd 31 n b tng ct thp s ii k.cong ngh - i hc vinh322218320218216Chn lp bo v ct dc chu lc l abv = 2,5 cm a = .(2, 5 1,1).11, 4 (2, 5 0, 9).5, 093, 5416, 49i iiyFcmF+ + + ;a = (2, 5 1).9, 42 (2, 5 0, 9).5, 093, 4714, 51cm+ + +;b. Kim tra vi cp 1 :V cp 1 c m men cng du vi cp 3 l cp tnh thp nn vi cp 1 c:AS : 3 22 + 2 18(16,49 cm2) ;AS : 3 20 + 2 18( 14,51 cm2 )a = 3,54 cm ; a = 3,47 cm ; h0 = 60 3,54 = 56,46 cm. tnh ton un dc ta tnh li IS vi tng :(AS + AS) = 16,49 +14,51 = 31 cm2IS = (AS + AS) (0,5 h a)2 = 31 (30 3,54)2 =21704 cm4 ; Ib = 900000 cm4Vi cp 1 c e0/h = 31,13/60 = 0,52H s xt n nh hng ca ti trng tc dng di hn :l = 1+(0, 5 ) 4,929 138, 545(0, 5 0, 6 0, 0354)1(0, 5 ) 40,359 138, 545(0, 5 0, 6 0, 0354)dh dhM N h aM N h a+ + ++ + = 1,54H s xt n nh hng ca lch tm e0 :S =0,110,10,1 0, 52 ++ = 0,28Lc dc ti hn:Ntr =206, 4(lSl x Eb Ib + ES IS) = 26, 4 0, 28(817, 5 1, 54 300.103 900000 + 200.10421704) SVTH : dng Ch linhlp 48k1_ksxd 32 n b tng ct thp s ii k.cong ngh - i hc vinh= 885,8 TH s xt n nh hng ca un dc:= 11trNN = 1138, 5451885, 8 = 1,185e

= e0 + 0,5 h a = 1,185.31,13 + 0,5 60 3,54 = 63,35 cmGi thit l nn lch tm ln : 2a =6,94 < x < Rh0 = 0,563 56,46 =31,79 cmXc nh s b x theo cng thc :x = '. ..S S SC SbN R A R AR b+ = 138545 3650 (16, 49 14, 51)145 50+ = 20,1 cmtho mn.Kim tra kh nng chu lc theo cng thc lch tm ln.Tnh Mu =Rb.b.x(h0- 0,5x)+RSC.AS(h0-a) =145.50.20,1 (56,46 - 0,5.20,1)+3650.14,51(56,46 - 3,47) =9569527 kG.cm = 95,695 T.mN.e = 138,545.0,6335 = 87,768 T.mN.e < Mu tit din kh nng chu cp ni lc 1b. Kim tra vi cp 2 :V cp 2 c m men ngc du vi cp 3 l cp tnh thp nn vi cp 2 c:AS : 3 20 + 2 18( 14,51 cm2 ) ;AS : 3 22 + 2 18(16,49 cm2) a = 3,47 cm ; a = 3,54 cm ; h0 = 60 3,54 = 56,53 cm. tnh ton un dc ta tnh li IS vi tng :(AS + AS) = 16,49 +14,51 = 31 cm2IS = (AS + AS) (0,5 h a)2 = 31 (30 3,47)2 =21819 cm4 ; Ib = 900000 cm4Vi cp 2 c e0/h = 16,51/60 = 0,275H s xt n nh hng ca ti trng tc dng di hn : v Mdh ngc chiu vi M nn mang du m.l = 1+(0,5 ) 4,929 138, 545(0,5 0, 6 0, 0347)1(0,5 ) 28, 9587 199, 6091(0,5 0, 6 0, 0347)dh dhM N h aM N h a+ + ++ + = 1,389H s xt n nh hng ca lch tm e0 :S =0,110,10,1 0, 275++ = 0,393Lc dc ti hn:Ntr =206, 4(lSl x Eb Ib + ES IS) SVTH : dng Ch linhlp 48k1_ksxd 33 n b tng ct thp s ii k.cong ngh - i hc vinh = 26, 4 0, 393(817, 5 1, 389 300.103 900000 + 200.10421819) = 1149,5 TH s xt n nh hng ca un dc:= 11trNN = 1199, 609111149, 5 = 1,21e

= e0 + 0,5 h a = 1,21.16,51 + 0,5 60 3,47 = 46,51 cmGi thit l nn lch tm ln : 2a =7,08 < x < Rh0 = 0,563 56,53 =31,83 cmXc nh s b x theo cng thc :x = '. ..S S SC SbN R A R AR b+ = 199609,1 3650 (14, 51 16, 49)145 50+ = 26,54 cmtho mn.Kim tra kh nng chu lc theo cng thc lch tm ln.Tnh Mu =Rb.b.x(h0- 0,5x)+RSC.AS(h0-a) =145.50.26,54 (56,53 - 0,5.26,54)+3650.16,49(56,53 - 3,54) =11513261 kG.cm = 115,13 T.mN.e = 199,6091.0,4651 = 92,84 T.mN.e < Mu tit din kh nng chu cp ni lc 1Kt lun : chn thp nh sau : pha tri chn: 3 22 + 2 18(AS =16,49 cm2 ) pha phi chn: 3 20 + 2 18(AS =14,51 cm2 )+V phnct di khngdi (Hd=5,45m) vni lctit dinIII-III cng khng nh khng th b qua nn ta khng ct ct thp m ko di thp trn sut chiu di phn ct di.+Ct dc cu to: phn ct di c h = 60 cm 50cm nn gia cnh cn c ct dc cu to, khong cch cc ct dc theo phng cnh h l: Sd = (h0 a)/2 = (56,46 3,47)/2 = 26,495 cm, tho mn Sd < 40cm. +Din tch tit din thanh cu to khng b hn 0,0005. b. Sd = 0,0005 x 50 x 26,495 = 0,66 cm2 Dng thp : 16 , AS =2,01 cm2. * Kim tra ct theo phng ngoi mt phng un: Tra bng ta c chiu di tnh ton: l0 = 1,2 Hd = 1,2 5,45 = 6,54 m mnh b = l0/h = 654/50 = 13,08 H s un dc tra bng ph lc (ni suy) c = 0,942SVTH : dng Ch linhlp 48k1_ksxd 34 n b tng ct thp s ii k.cong ngh - i hc vinh Tnh ton kim tra cu kin theo cu kin chu nn ng tmAb = 50 60 = 3000 cm2ASt = (16,49 + 14,51) = 31 cm2 t = 31/ 3000 = 0,01 < 0,03iu kin kim tra: N (Rb Ab + RSC ASt)N chn theo Nmax ly cp ni lc IV-18 ; Nmax = 210,6656 T (Rb Ab + RSC ASt) = 0,942 (145 3000 + 3650 31 )= 516357kG = 516,357 T > 210,6656 TVy ct kh nng chu lc theo phng ngoi mt phng un.VI . TNH TON CT TRC A THEO CC IU KIN KHC : a . Kim tra theo kh nng chu ct : phn ct di, lc ct ln nht xc nh t bng t hp Qmax = 10,494 TK1 Rbt b h0 = 0,6 10,550 56,46 = 17785 kG = 17,78 T, tho mn iu kin:Q < K1 Rbt b h0. B tng kh nng chu lc ct. Ct ai t theo cu to:

max1 1.22 5, 54 45d mmdmm ' ; chn ai 6 min15 15 1, 8 275050d cma b cmcm ' chn a = 25 cmb . Kim tra v nn cc b : nh ct chu lc nn do mi truyn xung :N = Gm + Pm = 116,94 + 12,285 = 129,225 (T) B rng dn mi k ln ct 24cm, b di tnh ton ca on k 26cm. Din tch trc tip chu nn cc b Fcb = 24 x 26 = 624 cm2 din tch tnh ton ca tit din ly i xng qua Fcb tnh c Ft = 50 30 = 1500 cm2H s tng cng c xc nh:mcb = 3cbtFF =36241500 = 1,34 < 2 vi cb = 0,75 th kh nng chu p cc b vai l:SVTH : dng Ch linhlp 48k1_ksxd 35 n b tng ct thp s ii k.cong ngh - i hc vinhcb mcb Rb Fcb = 0,75 1,34 145 624 = 90932,4 kG < N, khng tho mn iu kin v kh nng chu nn cc b. Gia c u ct bng li thp ngang. Dng 4 li thp vung, kch thc li6 6cm, dng thp C-I 6 vi din tch 0,283cm2. Chiu di ca thanh li theo phng cnh b l: l1 = 48 cm, theo phng cnh h l:l2= 38 cm s thanh theo phng cnh b l: n1= 9. s thanh theo phng cnh h l: n2 = 7. Khong cch cc li l: Sl = 12 cm, khong t li l: 3 12 + 2 = 38 cm, m bo khong t li khng di on quy nh i vi thp c g l: 15 d = 15 2,2 = 33 cm.3x120360400202020 20l u i t hp gia c u ctbin24050020 260 20400FcbFt15040020460500Din tch tit din b tng c bao bn trong li:Fl = 46 36 = 1656 cm2 > Ft = 1500 cm2 .T s ct thp ca li tnh theo cng thc : l=(n1 f1 l1+n2 f2 l2)/Fl Sl=(9 0,283 48+ 7 0,283 38)/1565 12= 0,0105 c =1.SlbRR= 0,0105 2250/145 = 0,163k1 = 5 5 0,1632, 981 4, 5. 1 4, 5.0,163cc+ + + +Tnh l = 6244, 5 3, 5. 4, 5 3, 5. 3, 0441500cbtFF Kim tra kh nng chu lc theo cng thc :N (mcb.Rb + k1.1 1. .SlR ).Fcb = (1,34.145 + 2,98.0,0105.2250.3,044).624= 254970 kG= 254,97 TN = 129,225 T < 254,97 T nn m bo kh nng chu lc cc b.SVTH : dng Ch linhlp 48k1_ksxd 36 n b tng ct thp s ii k.cong ngh - i hc vinhc. Tnh ton vai ct :Kch thc v s tnh thp trong vai th hin hnh v : 85,83 600 1000 150 600400 400 45 s t nh vai ctbin 8a150 2 18 2 18 2 18 Chiu cao lm vic h0 = 96 cm, b di vai Lv = 40cm c Lv < 0,9 h0 =86,4 cm nn vai ct thuc kiu cng son ngn.Lc tc dng ln vai:P = Dmax + Gd = 71,42 + 14,41 = 85,83 TKim tra kch thc vai ct theo cc iu kin:P 2,5 Rbt b h0(1)P 1,2 KvRbtb h20/av(2)P = 85,83 T < 2,5Rk b h0 = 2,5 10,5 50 96 = 126000 kG = 126 T tho mn iu kin (1)CutrccchlmvicnngKv=0,75, khongcchtlcPnmpct di SVTH : dng Ch linhlp 48k1_ksxd 37 n b tng ct thp s ii k.cong ngh - i hc vinh av = 75 - 60 = 15 cmP = 85,83 T < 1,2 KvRbtb h20/av = 1,2 0,75 10,5 50 962 / 15 = 870912 kG = 870,912 T tho mn Iu kin(2)* Tnh ct dc:M men un ti tit din mp ct 1-1:M1 = P av = 85,83 0,15 = 12,8745 (T.m)Tnh ct thp vi m men tng 25%Mtt = 1,25 M1 = 1,25 12,8745 = 16,093 (T. m)m =20. .bMRb h = 21609300145 50 96 = 0,024, tra bng c = 0,988AS = 0. .SMRh = 16093000, 988 3650 96 = 4,65 cm2 chn 2 18, AS = 5,09 cm2* Tnh ct ai v ct xin:V P = 85,83 T > Rbt b h0 = 10,5 50 96 = 50400 kG = 50,4 T. v h =100 cm >2,5av = 2,5 15 = 37,5 cm nn trong vai ct dng ct ct ai nm ngang v ct xin.+ Ct ai chn 8, khong cch gia cc ct ai :S 1 1. .100 254 415h cmcm ' khong cch S = 15 cm. + Din tch ct xin ct qua na trn on Lx (Lx = 100,12 cm) khng b hn 0,002 bh0 = 0,002 50 96 = 9,6 cm2, chn AS =10,18 cm2 4 18 t thnh 2 lp. ng knh ct xin tho mn b hn 25 v Lx / 15 = 67 mm(Lx =2 2100 15 += 101,12 cm)* Kim tra p mt ln vai :Dm cu trc lp ghp, lc nn ln nht t 1 dm truyn vo vai l: N = 0,5 Gd + Dmax1Gitr Dmax1doPmaxgyranhngch tnhcho1bndm. Davongnh hng v phn xc nh hot ti ng dm cu trc ta tnh c :Dmax1 = n Pmax(y1+ y3+ y4) = 1,1 22 (1 + 0,842 + 0,475) = 56,07 TSVTH : dng Ch linhlp 48k1_ksxd 38 n b tng ct thp s ii k.cong ngh - i hc vinhN = 0,5 14,41 + 56,07 = 63,275 (T)B rng dm cu trc trong on gi c m rng ra 34cm, on dm gi ln vai 23 cm.Fcb=34 23 = 782 cm2. Din tch tnh ton khi nn cc b l Ft ly theo hnh sau:Ft = 54 23 = 1242 cm2h s tng cng : mcb = 3cbtFF =31242782 = 1,17 < 2 vi cb = 0,75 th kh nng chu p cc b vai l:cb mcb Rb Fcb = 0,75 1,17 145 782 = 99500 kG = 99,5 TvN = 63,275 T < 99,5 T nn tho mn iu kin v kh nng chu p cc bnn ch phi gia c vai ct theo cu to. y ta dng tm thp bn m c chiu dy 1 cm 600100 100230500400340Ft Fcbd. Kim tra ct khi chuyn ch, cu lp :Lcnyct b un, ti trnglybngtrnglngbnthnct nhnvi hs ng lc 1,5 : - on di: g1 = 1,5 0,5 0,6 2,5 = 1,125 (T/ m)SVTH : dng Ch linhlp 48k1_ksxd 39 n b tng ct thp s ii k.cong ngh - i hc vinh - on trn: g2 = 1,5 0,5 0,4 2,5 = 0,75 (T/ m)* Xt cc trng hp bc xp, treo buc chn ra 2 s tnh hnh trang bn.Khi chuynchvbcxp. Ct ct nmtheophngngang, ccimk hoc treo buc cch mt di mt on a1 = 3 m, cch mt trn 1 on a2=3,9 m. M men m ti gi : M1 = 0,5 0,75 3,92 = 5,7 (T . m) M3 = 0,5 1,125 3 2 = 5,06 (T . m)M men dng ln nht on gia phn ct di tm c ti tit din cch gi 1,6 m ti M2 = 3,7 ( T. m)Qua so snh m men v tit din, ta thy cn kim tra vi M1= 5,7 T.m cho phn ct trn v M3 = 5,06 T.m cho phn ct di.Kim tra kh nng chu lc vi tit din nm ngang, h = 50 cm, h0 = 46 cm+ Kim tra cho phn ct trn :M1= 5,7 (T. m), ct thp a vo tnh ton ch ly hai ct ngoi : 1 22 + 1 28 Kim tra theo cng thc Mtd = RS AS (h0 - a') = 3650 (3,801 + 6,158) (46 - 4) = 1526714 kG.cm = 15,3 (T. m) V Mtd = 15,3 (T. m) > M1 = 5,7 (T. m) nn ct kh nng chu lc.+ Kim tra cho phn ct di : M3 = 5,06 (T. m), ct thp : 1 22 + 1 16 +1 20 Kim tra theo cng thc Mtd = RS AS (h0 - a') = 3650 (3,801+2,011+3,142) (46 - 4) = 1372648 kG.cm = 13,73 (T. m) V Mtd = 13,73 (T. m) > M3 = 5,06 (T. m) nn ct kh nng chu lc.* Khi cu lp, lt ct theo phng nm nghing ri mi cu. im cu t ti vai ct, cch mt trn 4,3 m. Chn ct t ln t.M men ln nht phn ct trn, ch tip gip vi vai ct:M4 = 0,5 0,75 4,12 = 6,3 (T. m)Vi tit din ct AS = 26,07 cm2 (3 28+2 22), h = 40 cm, h0 = 36 cm Mtd = RS AS (h0 - a') = 3650 26,07 (36 - 4) SVTH : dng Ch linhlp 48k1_ksxd 40 n b tng ct thp s ii k.cong ngh - i hc vinh= 3044976 kG.cm = 30,45 T. m > M4, vy ct kh nng chu lc.+ phn di m men ln nht tm c cch chn ct 1 on 2 m. M6 = 2,27 (T. m). Tit din c h = 60 cm, h0 = 56 cm ct thp 3 22+2 18 , AS = 16,49 cm2, tnh c:Mtd = RS AS (h0 - a') = 3650 16,49 ( 56 4 ) = 3129803 kG.cm = 31,29 (T. m) > M5 = 2,27 (T. m), nh vy ct kh nng chu lc.* chn ct khi lm vic chu lc tp trung kh ln cn c li thp gia c chn ct. Chn dng 4 li thp ng knh 6 .khong cch gia cc li l 12 cm+ theo phng cnh ngnb = 50 cm dng 9 6chiu di mi thanh l = 58 cm + theo phng cnh dih = 60 cm dng 11 6chiu di mi thanhl= 48 cm 3x120500600202020 20l u i t hp gia c chn ctbi n60020SVTH : dng Ch linhlp 48k1_ksxd 41 n b tng ct thp s ii k.cong ngh - i hc vinhg2=0,75 T/mg1=1,125 T/m3900 3450 30004100 3250 3000M2 =3,7 T.mM3 =5,06 T.m M1 =5,7 T.m1600 3000400500122 12860050012211612041004300 6050g2=0,75 T/mg1=1,125 T/mM4 =6,3 T.mM5 =6,94 T.mM6 =2,27 T.m2000328222500400218322600500SVTH : dng Ch linhlp 48k1_ksxd 42 n b tng ct thp s ii k.cong ngh - i hc vinhVII. TNH TON TIT DIN CT TRC B :Ct trc B chnh dngbnngoi ixng vni lctheo 2chiu xp xnhau nn t ct thp i xng l thun tin v hp l nht.1. Phn ct trn:Chiu di tnh ton l0 = 2,5Ht = 2,5 410 = 1025 cm. Kch thc tit din b = 50cm, h = 60cm. Gi thit chn a = a = 5cm, h0 = 60 5 = 55 cm; h0 - a = 55 - 5 = 50 cm mnh b = l0/h = 1025/60 = 17,08 > 8 cn xt n un dc. lch tm tnh ton: e0 = M/N + eavi ea l lch tm ngu nhin, ly bng 2 cm tho mn iu kin:ea( h/30 = 60/30 = 2 cm; Ht/60 = 410/600 = 0,68 cm; 1cm) tnh ton nh hng ca un dc, tm gi thit t = 2%, tnh m men qun tnh ca tit din :IS = t b h0 (0,5 h a)2 = 0,02 50 55 (30 5)2 =34375 cm4Ib = b b3/12 = 50 603 /12 = 900000 cm4K hiu cp ni lcK hiu bng t hpM(T.m)N(T)e1=M/N(m)e0=e1+ea(m)Mdh(T.m)Ndh(T)123II-16II-17II-1844,582-43,869644,132257,0565257,0565268,1130,17340,1710,16460,19340,1910,18460,3560,3560,356246246246V tnh ct thp i xng nn ch cn tnh ton vi cc cp 1 v 3 l s thomn cp 2a. Tnh ct thp i xng vi cp 1:Vi cp 1 c e0/h = 19,34/60 = 0,322H s xt n nh hng ca ti trng tc dng di hn :

l =1+(0, 5 ) 0, 356 246(0, 5 0, 6 0, 05)1(0, 5 ) 44,582 257, 0565(0, 5 0, 6 0, 05)dh dhM N h aM N h a+ + ++ + = 1,568H s xt n nh hng ca lch tm e0 :S = 0,110,10,1 0, 322 ++ = 0,361Lc dc ti hn:SVTH : dng Ch linhlp 48k1_ksxd 43 n b tng ct thp s ii k.cong ngh - i hc vinhNtr = 206, 4( lSl Eb Jb + Ea Ja) = 26, 4 0, 361(1025 1, 568 300.103 900000 + 200.104 34375 ) = 797,5 TH s xt n nh hng ca un dc:= 11trNN = 1257, 05651797, 5 = 1,476e0 = 1,476 19,34 = 28,55 cme

= e0 + 0,5 h a = 28,55 + 0,5 60 5 = 53,55 cm.Gi thit l nn lch tm ln : 2a x Rh0Xc nh s b chiu cao vng chu nn x:x1 = btNR b = 257056, 5145 50 = 35,46 cm x1 >Rh0 = 0,563.55 = 30,965 cm gi thit khng ng, xy ra l nn lch tm b. Tnh li x theo phng php ng dn:Vi x= x1 , tnh AS v t l AS* :AS* = 00( 0, 5 )( ')SCNe x hR h a+ = 257056, 5(53, 55 0, 5.35, 46 55)3650(55 5)+ = 22,93 cm2x = 001[ 2 . *( 1)].12 . *. .1S SRS SbRN R A hR ARb h+ += 1[257056, 5 2.3650.22, 93( 1)].551 0, 5632.3650.22, 93145.50.551 0, 563+ += 33,26 cm.Tho mn iu kin :Rh0 = 30,965 cm < x Rh0 = 0,563.55 = 30,965 cm gi thit khng ng, xy ra l nn lch tm b. Tnh li x theo phng php ng dn:Vi x= x1 , tnh AS v t l AS* :AS* = 00( 0, 5 )( ')SCNe x hR h a+ = 268113(52, 6 0, 5.36, 98 55)3650(55 5)+ = 23,638 cm2x = 001[ 2 . *( 1)].12 . *. .1S SRS SbRN R A hR ARb h+ += 1[268113 2.3650.23, 638( 1)].551 0, 5632.3650.23, 638145.50.551 0, 563+ += 33,99 cm.Tho mn iu kin :Rh0 = 30,965 cm < x 50 cm nn t ct dc cu to. Dng 216 .3282223282222162. Phn ct di:Chiu di tnh ton l0 = 1,5Hd = 1,5 545 = 817,5 cmKch thc tit din b = 50 cm, h = 80 cm. Gi thit chn a = a = 5cm, h0 = 80 - 5 = 75 cm; h0 - a = 75 - 5 = 70 cm mnh h = l0/h = 817,5/80 = 10,22 > 8 cn xt n un dc. lch tm tnh ton: e0 = M/N + eavi ea l lch tm ngu nhin, ly bng 3cm tho mn iu kin:ea = ( h/30 = 80/30 = 2,7cm; Hd/600 = 0,91 cm; 1cm)tnh ton nhhngcaundc,tm gi thitt=1%,tnh m men qun tnh ca tit din : IS = t b h0 (0,5 h a)2= 0,01 50 75 (0,5 80 5)2 = 45937, 5 cm4SVTH : dng Ch linhlp 48k1_ksxd 46 n b tng ct thp s ii k.cong ngh - i hc vinhIb = b b3/12 = 50 803 /12 = 2133333 cm4T bng t hp ni lc chn ra ba cp nghi ng l nguy him ghi bng sau: K hiu cp ni lcK hiu bng t hpM(T.m)N(T)e1=M/N(m)e0=e1+ea(m)Mdh(T.m)Ndh(T)123IV-14IV-17IV-18-51,434-58,94-58,166281,56353,6806406,51780,18270,16670,14310,21270,19670,1731-0,914-0,914-0,914281,56281,56281,56a. Tnh ct thp i xng vi cp 1:Vi cp 1 c e0/h = 21,27/80 = 0,266H s xt n nh hng ca ti trng tc dng di hn :

l =1+(0, 5 ) 0, 914 281,56(0, 5 0,8 0, 05)1(0, 5 ) 51, 434 281, 56(0, 5 0,8 0, 05)dh dhM N h aM N h a+ + ++ + = 1,6632H s xt n nh hng ca lch tm e0 :S = 0,110,10,1 0, 266 ++ = 0,401Lc dc ti hn:Ntr = 206, 4( lSl Eb Jb + Ea Ja) = 26, 4 0, 401(817, 5 1, 6632 300.103 2133333 + 200.104 45937,5 ) = 2357,5 TH s xt n nh hng ca un dc:= 11trNN = 1281, 5612357, 5 = 1,136e0 = 1,136 21,27 = 24,16 cme

= e0 + 0,5 h a = 24,16 + 0,5 80 5 = 59,16 cm.Gi thit l nn lch tm ln : 2a = 10 cm x R h0= 0,563 75 = 42,225 cmXc nh s b chiu cao vng chu nn x1:x1 = btNR b = 281560145 50 = 38,84 cm tho mn.Tnh AS = AS theo cng thc :SVTH : dng Ch linhlp 48k1_ksxd 47 n b tng ct thp s ii k.cong ngh - i hc vinh AS = AS= 0'0. . . ( 0, 5 )( )bSCN e Rb x h xR h a = 00( 0, 5 )( ')SCNe x hR h a+ = 281560(59,16 0, 5.38,84 75)3650(75 5)+ = 3,945 cm2Kim tra = = AS/b ho = 3,945/(50 75) = 0,0011 = 0,11%vi mnh h = 10,22 c min = 0,05%, m bo min

b. Tnh ct thp i xng vi cp 2 : Vi cp 2 c e0/h = 19,67/80 = 0,246H s xt n nh hng ca ti trng tc dng di hn :

l =1+(0, 5 ) 0, 914 281, 56(0, 5 0,8 0, 05)1(0, 5 ) 58, 94 353, 6806(0,5 0,8 0, 05)dh dhM N h aM N h a+ + ++ + = 1,5443H s xt n nh hng ca lch tm e0 :S = 0,110,10,1 0, 246 ++ = 0,418Lc dc ti hn:Ntr = 206, 4( lSl Eb Jb + Ea Ja) = 26, 4 0, 418(817, 5 1, 5443 300.103 2133333 + 200.104 45937,5 ) = 2538,7 TH s xt n nh hng ca un dc:= 11trNN = 1353, 680612538, 7 = 1,162e0 = 1,162 19,67 = 22,86 cme

= e0 + 0,5 h a = 22,86 + 0,5 80 5 = 57,86 cm.Gi thit l nn lch tm ln : 2a = 10 cm x R h0= 0,563 75 = 42,225 cmXc nh s b chiu cao vng chu nn x1:x1 = btNR b = 353680, 6145 50 = 48,78 cm >Rh0 gi thit khng ng. Xy ra l nn lch tm b.Tnh li x theo phng php ng dn:Vi x= x1 , tnh AS v t l AS* :SVTH : dng Ch linhlp 48k1_ksxd 48 n b tng ct thp s ii k.cong ngh - i hc vinhAS* = 00( 0, 5 )( ')SCNe x hR h a+ = 353680, 6(57, 86 0, 5.48, 78 75)3650(75 5)+ = 11,28 cm2x = 001[ 2 . *( 1)].12 . *. .1S SRS SbRN R A hR ARb h+ += 1[353680, 6 2.3650.11, 28( 1)].751 0, 5632.3650.11, 28145.50.751 0, 563+ += 47,096 cm.Tho mn iu kin :Rh0 = 42,225 cm < x 270,57 TVy ct kh nng chu lc theo phng ngoi mt phng un.- Phn ct di vi Nmax = 406,5178 T , chiu di tnh ton: l0 = 1,2 Hd = 1,2 545 = 654 cm. mnh b = l0/b = 654/50 = 13,08 H s un dc tra bng (ni suy) c= 0,9425Tnh ton kim tra cu kin theo cu kin chu nn ng tm Ab = 50 80 = 4000 cm2Tng din tch ct thp: ASt= 2 15,71= 31,42 cm2, tnh ton kim tra theo cng thc:N (Rb Ab + RSC ASt)SVTH : dng Ch linhlp 48k1_ksxd 51 n b tng ct thp s ii k.cong ngh - i hc vinh ( Rb Ab + RSC ASt) = 0,9425 (145 4000 + 3650 31,42) = 654739 kG = 654,739 T > 406,5178 TVy ct kh nng chu lc theo phng ngoi mt phng unVII . TNH TON CT TRC B THEO CC IU KIN KHC :a) Kim tra theo kh nng chu ct : phn ct di, lc ct ln nht xc nh t bng t hp Qmax = -12,3344 T K1 Rbt b h0 = 0,6 10,5 50 76,5 = 24097,5 kG =24,0975 T > Qmax nn b tng kh nng chu ct, t ct ai theo cu to. max1 1.20 54 45d mmdmm ' ; chn ai 6 min15 15 2, 0 305050d cma b cmcm ' chn a =30 cmc. Kim tra v nn cc b :nh ct chu lc nn do 2 mi truyn xung gc ln nh ct l : N1 = Gm1+Pm1 = 116,94 + 12,285= 129,225 TN2 = Gm2+Pm2 = 125,68 + 12,285 = 137,965 TV N2 > N1 nn ly N2 kim tra nn cc bBrngdnmiklnct24cm, bdi tnhtoncaonk26cm. Dintch trc tip chu nn cc b Fcb = 24 26 = 624 cm2, din tch tnh ton ca tit din ly i xng qua Fcb tnh c Ft = 50 30 = 1500 cm2H s tng cng c xc nh:mcb= 3cbtFF= 36241500= 1,34 < 2 vi cb= 0,75 th kh nng chu p cc b vai l:cb mcb Rb Fcb= 0,75 1,34 145 62 = 90932,4 kG = 90,9324 T Ft = 1500 cm2T s ct thp ca li tnh theo cng thc1 = (n1 f1 l1 + n2 f2 l2)/F1 S1 = (9 0,283 48 + 11 0,283 58)/2576 12 = 0,0098c = (1 RSl)/Rb = (0,0098 2250)/145 = 0,152k1= (5 + c)/(1 + 4,5 c) = (5 + 0,152)/(1 + 4,5 0,152) = 3,0594Tnh 1 vi F1 khng qu Ft 1= 4,5 3,5 (Fcb/Ft) = 4,5 3,5 (624/1500) = 3,044Kim tra kh nng chu lc theo cng thc:N (mcb Rb + k1 1 RSl 1) Fcb= (1,34 145 + 3,0594 0,0098 2250 3,044) 624 = 249380 kG = 249,38 Tvi N= 137,965 T < 249,38 T nn m bo kh nng chu lc cc b. 20 260 40 260 20130240130600500Fcb Ft 600560 20 205004602020150 15060020360 (gia c li thp u ct gia)c . Tnh ton vai ct :Kch thc v s tnh thp trong vai th hin hnh di SVTH : dng Ch linhlp 48k1_ksxd 53 n b tng ct thp s ii k.cong ngh - i hc vinhChiu cao lm vic h0=116 cm, b di vai Lv= 60cm c Lv < 0,9 h0=104,4cm nn vai ct thuc kiu cng son ngn.Lc tc dng ln vai:P = Dmax + Gd= 71,42 + 14,41 = 85,83 TKim tra kch thc vai ct theo cc iu kin: P 2,5 Rk b h0(1) P 1,2 Kv Rk b h20/av(2) 600 400400 750750 85,83 85,83 600600 350350 1200 45 600 2 20 2 20 8a150 2 25 (S tnh vai ct gia)P = 85,83 < 2,5Rbt bh0 = 2,5 10,5 50 116 = 152250 kG = 152,25 Ttho mn (1)CutrccchlmvicnngKv=0,75, khongcchtlcPnmpct diav = 75 - 40 = 35 cmP = 85,83 < 1,2 Kv Rbt b h20/av = 1,2 0,75 10,5 x 50 1162/35SVTH : dng Ch linhlp 48k1_ksxd 54 n b tng ct thp s ii k.cong ngh - i hc vinh = 181656 kG = 181,656 T tho mn (2)* Tnh ct dc:M men un ti tit din mp ct 1-1:M1 = P av = 85,83 0,35 = 30,0405 (T. m)Tnh ct thp vi m men tng 25%Mtt = 1,25 M1 = 1,25 30,0405 = 37,55 (T. m) m =20. .bMRb h = 23755000145 50 116 = 0,0385, tra bng c = 0,98 AS = 0. .SMRh = 37550000, 98 3650 116 = 9,05 cm2 chn 2 25, AS = 9,82 cm2* Tnh ct ai v ct xin:V P = 85,83 (T) > Rbt b h0 = 10,5 50 116 = 60900 kG = 60,9 T v h = 120cm > 2,5 av = 2,5 35 = 87,5 cm nn trong vai ct dng ct ai nm ngang v ct xin. Ct ai chn 8, khong cch 15cm, tho mn khng qu h/4 = 30cm v 15cmDin tch ct xin ct qua na trn on Lx khng b hn : 0,002 b h0 = 0,002 50 116 = 11,6 cm, chn 4 20 t thnh 2 lp. ng knh ct xin tho mn b hn 25 mm v Lx/ 15 = 83 mm (Lx =2 2120 35 + = 125 cm)*Kim tra p mt ln vai :Dm cu trc lp ghp, lc nn ln nht t 1 dm truyn vo vai l: N = 0,5 Gd + Dmax1Gitr Dmax1doPmaxgyranhngch tnhcho1bndm. Davongnh hng v phn xc nh hot ti ng dm cu trc ta tnh c :Dmax1 = n Pmax(y1+ y3+ y4) = 1,1 22 (1 + 0,842 + 0,475) = 56,07 TN = 0,5 14,41 + 56,07 = 63,275 (T)SVTH : dng Ch linhlp 48k1_ksxd 55 n b tng ct thp s ii k.cong ngh - i hc vinhB rng dm cu trc trong on gi c m rng ra 34cm, on dm gi ln vai 23 cm.Fcb=34 23 =782 cm2. Din tch tnh ton khi nn cc b l Ft ly theo hnh sau: 100340100 600600600 230 500 F t F cb Ft = 54 23 = 1242 cm2h s tng cng : mcb = 3cbtFF =31242782 = 1,17 < 2 vi cb = 0,75 th kh nng chu p cc b vai l:cb mcb Rb Fcb = 0,75 1,17 145 782 = 99500 kG = 99, 5 TvN = 63,275 T < 99, 5 T nn tho mn iu kin v kh nng chu p cc bnn khng phi gia c thm cc li thp mt trn vai ct, ch phi gia c vai ct theo cu to. y ta dng tm thp bn m c chiu dy 1 cm d. Kim tra ct khi chuyn ch, cu lp :Lcnyct b un, ti trnglybngtrnglngbnthnct nhnvi hs ng lc 1,5.on di: g1 = 1,5 0,5 0,8 2,5 = 1,5 (T/ m)SVTH : dng Ch linhlp 48k1_ksxd 56 n b tng ct thp s ii k.cong ngh - i hc vinhon trn: g2 = 1,5 0,5 0,6 2,5 = 1,125 (T/ m)Xt cc trng hp bc xp, treo buc chn ra 2 s tnh hnh trang bn.* Khi chuyn ch v bc xp. Ct c t nm theo phng ngang, cc im k hoc treo buc cch mt di mt on a1 = 3 m, cch mt trn 1 on a2=3,9m. M men m ti gi: M1 = 0,5 1,125 3,92 = 8,56 (T . m) M3 = 0,5 1,5 32 = 6,75 (T . m) M men dng ln nht on gia phn ct di tm c ti tit din cch gi 1,4 m ti M2 = 5,4 ( T. m)Qua so snh m men v tit din, ta thy ch cn kim tra vi M1 v M3 Kim tra kh nng chu lc vi tit din nm ngang, h = 50 cm, h0 = 45 cm+ khi tnh vi M1 = 8,56 (T. m)c 1 16 + 2 28 Kim tra theo cng thc: Mtd = RS AS (h0 - a') = 3650 (2,011 + 12,32) (45 - 5) = 2092326 kG.cm = 20,92 (T. m)V Mtd = 20,92 (T. m) > M1 = 8,56(T. m) nn ct kh nng chu lc.+ khi tnh vi M3 = 6,75 (T. m)c 1 16 +2 20 Kim tra theo cng thc Mtd = RS AS (h0 - a') = 3650 (2,011 + 6,28) (45 - 5) = 1210486 kG.cm = 12,1 (T. m) V Mtd = 12,1 (T. m) > M3 = 6,75 (T. m) nn ct kh nng chu lc.* Khi cu lp, lt ct theo phng nm nghing ri mi cu. im cu t ti vaict, cch mt trn 4,3 m. Chn ct t ln t. M men ln nht phn ct trn, ch tip gip vi vai ct:M4 = 0,5 1,125 4,12 = 9,46 (T. m)Vi tit din ct AS = 26,07 cm2 (3 28+2 22), h = 60 cm, h0 = 55 cm Mtd = RS AS (h0 - a') = 3650 26,07 (55 - 5) = 4757775 kG.cm = 47,58 (T. m), Vy ct kh nng chu lc.SVTH : dng Ch linhlp 48k1_ksxd 57 n b tng ct thp s ii k.cong ngh - i hc vinh+ phn di m men ln nht tm c cch chn ct 1 on 1,9 m. M6 = 2,7 (T. m). Tit din c h = 80 cm, h0 = 75 cm ct thp 5 20, AS = 15,71 cm2, tnh c:Mtd = RS AS (h0 - a') = 3650 15,71 (75 5) = 4013905 kG.cm = 40,14 (T. m) > M6 = 2,7 (T. m), nh vy ct kh nng chu lc.* chn ct khi lm vic chu lc tp trung kh ln cn c li thp gia c chn ct. Chn dng 4 li thp ng knh 6 .khong cch gia cc li l 12 cm+ theo phng cnh ngnb = 50 cm dng 9 6chiu di mi thanh l = 48 cm + theo phng cnh dih = 80 cm dng 14 6chiu di mi thanhl = 78 cm 3x120l u i t hp gia c chn ctgia8002080050020202020SVTH : dng Ch linhlp 48k1_ksxd 58 n b tng ct thp s ii k.cong ngh - i hc vinh1166050 430041001900M6 =2,7 T.mM4 =9.46 T.mg2=1,125 T/mM5 =10,4 T.mg1=1,5 T/m1165008003000300030003282228001400M2 =5,4 T.m6003250 41003450 3900M1 =8,56 T.mg2=1,125 T/m500228g1=1,5 T/m500520M3 =6,75 T.m220600500SVTH : dng Ch linhlp 48k1_ksxd 59 n b tng ct thp s ii k.cong ngh - i hc vinhSVTH : dng Ch linhlp 48k1_ksxd 60