KTTC 2 Tạ Thọ

GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 -------------------------------------------------------------------------------------------------- n KTTC s 2: THI CNG LP GHP NH CNG NGHIP 1 TNGCHNG 1. C IM KIN TRC V KT CU CNG TRNH 1.1. c im kin trc cng trnh.

Tn cng trnh: Nh my ch to t 19/5 H Ni c im cng trnh: S n nguyn: 2 n nguyn. S khu : 3 khu . S bc gian: + n nguyn s 1: 15 bc gian. + n nguyn s 2: 10 bc gian. Kch thc: + Chiu di nhp: L=22m + Bc khung: a=6m + Chiu di cng trnh: n nguyn s 1 di 15 6 = 90m n nguyn s 2 di 10 6 = 60m

Hnh a.1. Mt ct ngang cng trnh

-------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG1

GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 ---------------------------------------------------------------------------------------------------

Hnh a.2. Mt bng cng trnh -------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG2

GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------1.2. c im cng trnh

Kt cu mng: Mng thuc dng mng cc, b tng ct thp lp ghp.

Hnh 1.3. Mng cc Kt cu phn thn: Dn mi.

Hnh 1.4. Dn ko thp Dm cu chy .



Hnh 1.5. Dm cu chy Ct BTCT lp ghp

Hnh 1.6. Ct BTCT lp ghp -------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG4


CHNG 2. C IM A HNH, A CHT, THY VN, NG VN

CHUYN VO CNG TRNH2.1. c im a hnh. 2.2. c im a cht, thy vn. 2.3. c im ng vn chuyn vo cng trnh. CHNG 3. C IM N V THI CNG CN TRNH CHNG 4. THIT K TREO BUC KT CU 4.1. Thit k treo buc mng BTCT lp ghp.

Trng lng mng: 11.625 T Dng c treo buc chn dy cu n 4 nhnh, c b phn t cn bng. 0 Gc treo dy cu chn = 45 Trn mng c cha sn 4 mc thp. Tnh ton treo buc kt cu: Trng lng mng: P=11.625 T Ti trng tnh ton: Ni lc trong mi nhnh ca dy cp:P tt = n P = 1.1 11.625 = 12.8T


45

GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------P tt 12.8 S= = 0 m n cos 45 0, 75 4 cos 450 S = 6.04T

Trong : n=4: S nhnh dy cp M=0.75: H s khng iu ha trong cc nhnh dy Lc thit k dy cp:R = K S = 6 6.04 = 36.24T

Trong : K=6: H s an ton dy cu c mc. Chn loi cp c cu trc 6 37 1 c cc ch s k thut nh sau: Hnh 4.1. Treo buc mng BTCT + ng knh cp: D= 28.5mm + Trng lng cp: 2.67kg/m + Lc lm t cp: 38.6T 2 + Cng chu ko ca cp: = 160kg / cm4.2. Thit k treo buc ct BTCT lp ghp.

thun tin cho vic c nh ct v gii phng mc cu sau khi lp ghp ta chn thit b treo buc l ai ma st. Nguyn tc lm vic ca ai ma st: Gm 1 n treo v 2 dy cp ni vo thanh ch U v tr cao hn trng tm ct. Khi cn trc ko cng dy cp th cc thanh ch U nn li, 2 ai ma st b p vo thn ct. Nh c ma st gia 2 mt b tng vi hai thanh ai m ct c treo thng ng mt v tr nht nh. Sau khi iu chnh xong tm ct, c nh tm ta h mc cu xung, ai ma st s t h xung chn ct. Tnh ton chn dy cu: thin v an ton, ta tnh chn dy cu cho ct gia ri ly s dng lun cho ct bin. Trng lng ct gia: Qgiua = 9.32T Ni lc trong mi nhnh dy cu:P tt 1,1 9.32 = m n cos 1 2 cos 0 = 5.126T S=

Lc thit k dy cp:

R = K S = 6 5.126 = 30.756T Chn loi cp cu trc 6 37 1 c cc ch s k thut nh sau:

+ ng knh cp: D= 26mm -------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG6

GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------+ Trng lng cp: 2.38kg/m + Lc lm t cp: 31.1T2 + Cng chu ko ca cp: = 150kg / cm

Hnh 4.2. Treo buc ct BTCT H thng ging gi, c nh tm Dng cc con chim c chiu di 30cm, vt gc theo thnh mng. V ct c chiu cao 12,15m > 8m v nng 7,28T nn phi ging gi ct bng dy vng c tng iu chnh, h thng dy vng ny c lp di h ct, sau khi lp dng chnh xc v tr ca ct ta phi neo chng vo di t hoc cc mng ln cn.

Hnh 4.3. C nh tm ct -------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG7

GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------4.3. Thit k treo buc dm cu chy BTCT lp ghp.

Dng c treo buc: V dm cu chy c chiu di a = 6m nn ta dng dy cu kp buc nng cu vt. tho d cc dng c treo buc dm m khng phi tro ln cao, ta dng thit b treo buc c kha bn t ng. Nguyn tc v cch lm: + Dy cu kp treo dm cu chy qua kha, mt vng quai u dy vng vo mc cu cn vng ai kia i vo kha, c cht ngang gi u li. + ngn nga dy cp c vo thnh mp ca dm b tng, phi b tr 4 ming thp gc m di ng c treo trn dy cp, p vo mp cnh dm. + Khi t dm vo v tr thit k v c nh xong, ngi cng nhn ng mt sn cng tc ko si dy rt cht ra, vng quai u dy s tut kha ra gii phng dng c treo buc khi dm. + Khong cch t dm ti dy cu : L = 0.1 a = 0.1 6 = 0.6m

Hnh 4.4. Treo buc dm cu chy Tnh ton chn dy cu Trng lng dm cu chy: P = 3.5T Ni lc trong mi nhnh dy cu: -------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG8

GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------P tt 1,1 3.5 S= = m n cos 0, 75 4 cos 45 = 1,82T

Lc thit k dy cp:

R = K S = 8 1,82 = 14.56T

Trong : K=8: H s an ton i vi dy cp b un cong. Chn cp cu trc 6 37 1 c cc ch s k thut nh sau: + ng knh cp: D=17.5mm + Trng lng cp: 1.06kg/m + Lc lm t cp: 14.6T 2 + Cng chu ko ca cp: = 160kg / cm

4.4. Thit k treo buc dn thp lp ghp

Chn dng c treo buc: Dn mi c khu 22m nn khi cu lp ta dng h dn treo buc nhm gim bt chiu di treo buc v lm gim ng sut pht sinh trong dn khng thay i qu nhiu so vi ng sut lm vic thc t ca dn. Chn phng php treo buc 2 im. Gia cng khi cu lp: Trong qu trnh cu lp, ng sut pht sinh c th vt qu s tnh ton cu lp. m bo cho cu kin khng b ph hoi ta phi gia cng chng bng cc thanh g c np cht hai bn dn, v c thanh cnh thng v thanh cnh h. Cc gng g gia cng ch c tho khi dn c nh vnh vin. Tnh ton n treo: n treo lm vic nh mt cu kin chu nn ng tm. Chn n treo lm bng thp trn rng ca VGPIPE c cc thng s hnh hc nh sau: + D = 75,6mm -------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG9

GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------+ d = 68,6mm + t = 3,5mm Cng : R = 3600kG/cm2. Cc c trng hnh hc ca tit din ngang n treo: 3,14 4 4 4 4 4J= 64 D d

(

)=

64

75, 6 68, 6

(

) = 516102mm

A= r=

3,14 ( D2 d 2 ) = ( 75, 62 68, 62 ) = 792,379mm 2 4 4J 516102 = = 25,52mm A 792,379 L 22 + 0, 2 = + 0, 2 = 5, 7 m 4 4

Chiu di ca n treo c th chn theo cch sau:Ld =

=

L 4 = 1 5, 7 1000 = 223.35 r 25,52

Tra bng quan h gia h s un dc ca cc thanh chu nn vi mnh , ni suy tnh ton ta c = 0,099. Lc chu nn tnh ton ca n treo:N= P tt 1.1 2.808 tg = tg 450 = 1,5444T 2 2

Khi c k n nh hng ng ca cu lp:N tt = 1.1 N = 1.11.5444 = 1.699T

Kim tra n treo:N tt 1.699 1000 R = 2165.82kG / cm 2 0,85 3600 = 3060kG / cm 2 A 0.099 7.92379

Vy vi tit din chn kh nng chu lc. Tnh ton chn dy cu: Trng lng dn: P = 2.808T Ni lc trong mi nhnh dy: -------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG1 0

GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------P tt 1,1 2,808 S= = m n cos 1 2 cos 450 S = 2.185T

Lc thit k dy cp:

R = K S = 6 2,185 = 13.11T Chn cp cu trc 6 37 1 c cc ch s k thut nh sau:

+ ng knh cp: D=17.5mm + Trng lng cp: 1.06kg/m + Lc lm t cp: 13.7T 2 + Cng chu ko ca cp: = 150kg / cm2l 100x8 2l 100x8 2l 100x8

tha n h g g ia c NG 2l 100x10 d c u n y

n tr eo n g th p

d c u k y p

Hnh 4.5.Treo buc dn thp lp ghp

-------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG1 1


CHNG 5. THIT K BIN PHP LP GHP KHI QUT CC KT

CU.5.1. Thng k kt cu:

Bng 1.1. Bng thng k k cu lp ghp TT 1 2 3 4 5 Tn kt cu Mng Ct bin Ct gia Dm cu chy Dn mi X g Hnh dng kch thc Mng cc Ct c vai ct Ct c vai ct Ch T Dn thp Thp hnh Trng lng (T) 11.625 3.079 9.32 3.5 2.808 S lng 108 54 54 150 81 325 Ghi ch

5.2. Thit k s lp ghp khi qut:

Cng trnh c s dng cc bin php kt cu nh mng, ct, dm cu chy l cc kt cu BTCT c sn, v dn thp v vy khi thi cng ta s dng mi ni t v mi ni kh tu thuc tng cu kin. Dng mi ni i khi lp ghp dm cu chy vo ct, ct vo mng sau khi thc hin lin kt ny phi c thi gian ch cho mi ni t 70% cng mi cho thi cng tip. -------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG1 2

GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------Do phng n lp ghp sau u c th l phng n ti u nht vi cc cu kin dng trong lp ghp. + Ln 1: Lp ghp mng. + Ln 2: Lp ghp ct. + Ln 3: Lp ghp dm cu chy. + Ln 4: Lp ghp dn, x g. Vi mt bng cng trnh di 3 khi tip gip nhau, nh c nhp L = 22m, bc ct a = 6m, dm cu chy khu 6m nn ta dng cn trc t hnh lp ghp. Trong cc phng n lp ghp hin nay, vi c im ca cng trnh, ta chn phng n lp ghp tun t l ph hp nht bi c a hnh tng i rng ri tp kt vt liu, nguyn liu tng i ln. u im ca phng n ny l: + Khng phi thay i dng c treo buc, ch lp ghp cc kt cu cng loi cho nn nng sut cao. + Lp ghp tng cng vic ring bit hon chnh nn thi cng n gin. Tuy nhin phng n ny cng c nhng nhc im sau: + ng i ca cn trc t hnh l tng i ln + C th phi dng nhiu loi cn trc khc nhau cho tng cu kin ring Tin hnh lp ghp tun t: Mng, ct, dm cu chy, dn mi, x g Ta chn phng n lp ghp dc nh theo tng khu , cho cn trc di chuyn gia lp ghp cc cu kin. C th s di chuyn ca cn trc nh sau: i gia nhp lp mng khu 1 sau lp khu 3. i gia nhp lp ct khu 1 sau lp khu 3. i gia nhp lp dm cu chy khu 1 sau lp khu 3. i gia nhp lp dn v x g khu 1, sau lp khu 2, di chuyn sang lp khu 3. S dng 2 cn trc lp ghp cc cu kin ln lt nh trn ti 2 n nguyn.



Hnh a.1. S di chuyn ca cn trc khi lp Mng, ct, dm cu chy -------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG1 4



GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------Hnh a.2. S di chuyn ca cn trc khi lp ghp dn mi v x g.CHNG 6. TNH TON LA CHN CN TRC LP GHP 6.1. Tnh ton la chn cn trc lp ghp mng.

Nhp L=22m, chn phng n cu i gia c th hin trn hnh 5.1 v 5.2, s tnh ca cu lp mng c th hin nh sau:

Hnh a.1. S tnh cu lp ghp mng. Tm vi yu cu ca cu:a L 6 22 Ryc = + = + =11.4m 2 2 2 2 2 2 2 2

Chiu cao t cao trnh my ng ti Puli cn trc l:

H = h1 + h2 + h3 + h4 = 0.5 + 1.35 + 0.775 + 1.5 = 4.125m

Trong : h1 =0.5m: on nng cao cu kin hn cao trnh my ng h2 =1.35m: Chiu cao mng h3 =0.775m: Chiu cao ca dy cp treo mng. h4 =1.5m: on puli, rng rc, mc cu u cn. Gc nng:


GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------tg = H 4.125 = = 0.42 ( Ryc r ) (11.4 1.5)

Chiu di tay cn:

=> = 22.80 Lyc = Lyc =

(R

yc

r ) + ( H hc )2 2

2

( 11.4 1.5)

+ ( 4.125 1.5 ) = 10.24m2

Trong : hc=1.5m: Chiu cao t cao trnh my ng ti im ni tay cn vi my. Sc nng: Qyc = Qck + qtb Trong : Qck = 12.8T : Trng lng mng qtb = 2.67 1.1 4 = 11.75kG = 0.01175T : Trng lng dy treo.=> Qyc = 12.8 + 0.01175 = 12.81175T

Vy, cc thng s chn cu lp ghp mng l: + Sc trc: Qyc = 12.81175T + Chiu di tay cn: Lyc = 10.24m + vi tay cn: Ryc = 11.4m + Chiu cao nng mc: H yc = 4.125m6.2. Tnh ton la chn cn trc lp ghp ct.

Vi cu lp ghp ct, do trng lng ct gia ln hn ct bin, ta tnh ton cu lp ghp cho ct gia ri p dng lun cho lp ghp ct bin.



Hnh 6.2. S lp cu lp ghp mng. Khi lp ghp ct, s di chuyn ca cu ging nh khi lp ghp mng vi tay cn:a L 6 22 Ryc = + = + =11.4m 2 2 2 2 2 2 2 2


H = h1 + h2 + h3 + h4 = 0.5 + 12.15 + 0.5 + 1.5 = 14.65m

Trong : h1 =0.5m: on nng cao cu kin hn cao trnh my ng h2 =12.15m: Chiu cao ct h3 =0.5m: Chiu cao ca dy cp treo ct. h4 =1.5m: on puli, rng rc, mc cu u cn. Gc nng:tg = H 14.65 = = 1.48 ( Ryc r ) (11.4 1.5)

Chiu di tay cn:

=> = 560 Lyc = Lyc =

(R

yc

r ) + ( H hc )2 2

2

( 11.4 1.5)

+ ( 14.65 1.5 ) = 16.46m2


GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------Trong : hc=1.5m: Chiu cao t cao trnh my ng ti im ni tay cn vi my. Sc nng: Qyc = Qck + qtb Trong : Qck = 1.1 9.32 = 10.252T : Trng lng ct qtb = 2.38 5 2 + 5 = 28.8kG = 0.0288T : Trng lng dy treo.=> Qyc = 10.252 + 0.0288 = 10.2808T

Vy, cc thng s chn cu lp ghp ct l: + Sc trc: Qyc = 10.2808T + Chiu di tay cn: Lyc = 16.46m + vi tay cn: Ryc = 11.4m + Chiu cao nng mc: H yc = 14.65m

6.3. Tnh ton la chn cn trc lp ghp dm cu chy.


GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------Hnh 6.3. S lp ghp dm cu chy. Khi lp ghp ct, s di chuyn ca cu ging nh khi lp ghp mng vi tay cn:Ryc = L 22 = 11m 2 2


H = H L + h1 + h2 + h3 + h4 = 8.65 + 0.5 + 0.75 + 3.25 + 1.5 = 14.65m

Trong : HL =8.65m: Chiu cao t cao trnh my ng ti im t cu kin. h1 =0.5m: on nng cu kin cao hn v tr lp t cu kin. h2 =0.75m: Chiu cao dm cu chy. h3 =3.25m: Chiu cao ca dy cp treo dm cu chy. h4 =1.5m: on puli, rng rc, mc cu u cn. Gc nng:tg = H 14.65 = = 1.54 ( Ryc r ) (11 1.5)

=> = 57 0

Chiu di tay cn:Lyc = Lyc =

(R

yc

r ) + ( H hc )2 2

2

( 11 1.5)

+ ( 14.65 1.5 ) = 16.2m2

Trong : hc=1.5m: Chiu cao t cao trnh my ng ti im ni tay cn vi my. Sc nng: Qyc = Qck + qtb Trong : Qck = 1.1 3.5 = 3.85T : Trng lng mng qtb = 1.06 5.32 4 + 2 4 = 30.6kG = 0.0306T : Trng lng dy treo.=> Qyc = 3.85 + 0.0306 = 3.8806T

Vy, cc thng s chn cu lp ghp dm cu chy l: + Sc trc: Qyc = 3.8806T + Chiu di tay cn: Lyc = 16.2m + vi tay cn: Ryc = 11m + Chiu cao nng mc: H yc = 14.65m -------------------------------------------------------------------------------------------------- n KTTC s2 THI CNG LP GHP NH CNG NGHIP 1 TNG2 0

GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------6.4. Tnh ton la chn cn trc lp ghp dn thp mi.

Hnh 6.4. S lp ghp dn thp mi Chiu cao t cao trnh my ng ti Puli cn trc l:

H = H L + h1 + h2 + h3 + h4 = 12.15 + 0.5 + 3.4 + 3.77 + 1.5 = 21.32m

Trong : HL =12.15m: Chiu cao t cao trnh my ng ti im t cu kin. h1 =0.5m: on nng cu kin cao hn v tr lp t cu kin. h2 =3.4m: Chiu cao dm dn h3 =3.77m: Chiu cao ca dy cp treo dm cu chy. h4 =1.5m: on puli, rng rc, mc cu u cn Gc nng: Chn gc nng l gc ln nht m cn trc t hnh c th thc hin c: = 750 Tm vi ca tay cn:Ryc = H 21.32 = = 5.72m tg tg 750


GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 -------------------------------------------------------------------------------------------------- Chiu di tay cn:Lyc = Lyc =

(R

yc

r ) + ( H hc )2 2

2

( 5.72 1.5)

+ ( 21.32 1.5) = 20.27 m2

Trong : hc=1.5m: Chiu cao t cao trnh my ng ti im ni tay cn vi my. Sc nng: Qyc = Qck + qtb Trong : Qck = 1.1 2.808 = 3.0888T : Trng lng mng qtb = 0.05T : Trng lng n treo, np g v dy treo=> Qyc = 3.0888 + 0.05 = 3.13888T

Vy, cc thng s chn cu lp ghp mng l: + Sc trc: Qyc = 3.13888T + Chiu di tay cn: Lyc = 20.27 m + vi tay cn: Ryc = 5.27 m + Chiu cao nng mc: H yc = 21.32m6.5. Bng so snh la chn cc phng n s dng cn trc lp ghp.

STT

Tn cu kin

Thng s yu cu Qyc = 12.81T Hyc = 4.125m Ryc = 11.4m Lyc = 10.24 m Qyc = 10.3T Hyc = 14.65m Ryc = 11.4m Lyc = 16.46m Qyc = 3.88T Hyc = 14.65m Ryc = 11m Lyc = 16.2m

Loi cu chn E-2508 (NGUYEN TIEN THU)

Thng s c chn Qmax= 15T H = 29m R= L = 30m Q= H= R= L= Q= H= R= L=

1

Mng

2

Ct

E-2508 (NGUYEN TIEN THU)

3

Dm cu chy

E-2508 (NGUYEN


GVHD: Th.s L Huy Sinh SVTH: T Vn Th- lp 08X6 --------------------------------------------------------------------------------------------------TIEN THU)4

Dn mi

Qyc = 3,14T Hyc = 21.32m Ryc = 5.27m Lyc = 20.27m

RDK25

Q = 18T H = 22M RMIN =5m L = 22,5m


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KTTC 2 Tạ Thọ