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8/6/2019 Kt cu ton khi

1/25

I. Kt cu ton khi:

1. Mng M1:

- Din tch vn khun: 200*(1600 + 2000) = 0.72 m2

- Th tch b tng: (1.6*0.2 + 0.35*(1.6 +0.4)/2)*2 = 1.22 m3

2. Mng M2


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3. Mng M3


3/25

300

50

100400

250

- Vi bc ct 6 m, chn dm mng c chiu di bnh qun 4,85 m ( cc bcct u hi hoc cnh khe nhit , chiu di ca dm mng c ththp hn), trng lng 1,5 T, tit din nh hnh v.

- Phn dm mng nm trong nn t c din tch tit din:[(0,4 + 0,25)/2] x 0,05 + 0,3 x 0,25 = 0,08125 m2

- Dm mng c b tr xung quanh chu vi nh, do s dmmng l: 2 x 19 + 2 x 10 = 58 dm

- Th tch chim ch trong t ca cc dm mng:

58 x (0,08125 x 4,85) = 22,85 m3

.


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2. Ct thp:

TNCT

CHIU CAO(mm)

TIT DIN(mm)

DIN

TCHTITDIN(cm2)

THTCH1V(cm3)

TRNGLNG1V (T)

SLNG

TNGTRNGLNG

(T)

CS1 6000350

-650

x200

x 8 x12

8608 51,648 0.41 20 8.11

CS2 6200300

-300

x200

x 8 x 8 5472 33,926 0.27 40 10.65

CS3 6800 300 - 300 x 200 x 8 x 8 5472 37,209 0.29 40 11.68


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3. Cc cu kin khc:

a. Thng k dm:

TNDM

CHIUDI

(mm)

TIT DIN(mm)

DINTCHTITDIN(mm2)

THTCH1V

(mm3)

TRNGLNG

1V(T)

SLNG

TNGTRNGLNG

(T)

RB1_1

9000300

-300

x200

x 8 x 8 5472 49,248,000 0.39 80 30.93

RB1_ 6074 30 - 30 x 20 x 8 x 8 5472 33,236,928 0.26 80 20.87


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2 0 0 0

RB2 5700200

-200

x100

x 6 x 6 2328 13,269,600 0.10 200 20.83

b. Thng k x g mi:

TNDM

CHIU

DI(mm)

TIT DIN(mm)

DINTCHTITDIN(mm2)

THTCH1V

(mm3)

TRNG

LNG 1V

(T)

SLN

G

TNGTRN

GLNG (T)

Z-150 6000 150 x50 x

20 x 2 564 3384000 0.03 1092 29.01

c. Thng k ca mi:

RNG(mm)

CAO(mm)

TRNGLNG

1V(N/mm)

SLNG

TNGTRNGLNG

(T)5000 1500 1,219.36 60 7.32

d. Thng k tn mi:


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RNG(mm)

DI (mm)DY(mm)

TRNGLNG(kG/m)

SLNG

TNGTRNGLNG

(T)

30000 110000 0.56 0.13

2 29.01

C. THI CNG PHN NGM:

I. THI CNG O T:

1. Chn phng n o:

- Theo iu kin thi cng trn nn t st pha, tng i nng, nn chn phng n o vi mi dc tnhin, c h s mi dc l m = 1: 0,25.

- Chiu su h o (tnh c chiu dy lp b tng lt): H = 1,6 0,15 = 1,45 m.- B rng chn mi dc: B = 1,45 . 0,25 = 0,35 m.- Khong li thao tc ca cng nhn quanh mp mng l 0,5 m.


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2. Chn t hp my thi cng

- Chn my o gu nghch EO 2621A c cc thng s k thut sau:o Dung tch gu q = 0,25 m3

o Bn knh o ln nht Ro max= 5 mo Chiu su o ln nht Ho max= 3,3 mo Chiu cao ln nht H max= 2,2 mo Chu k k thut tck = 20 s


9/25

- Tnh nng sut ca my o:o H s y gu: kd = 1,1o H s ti ca t kt = 1,15

o H s quy v t nguyn th:d

1t

k 1

k 0,87k 1,15= = =

o H s s dng thi gian: ktg = 0,75

Khi o ti ch:

- Chu k o: tdck = tck = 20 s (gc quay 90o)- S chu k o trong 1 gi: nck= 3600/20 = 180


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- Nng sut ca ca my o:3

ca ck 1 tgW t.q.n .k .k 7.0,25.0,87.180.0,75 205 m /ca= = =

Khi o ln xe:

- Chu k o: tdck = tck.ktg = 20.1,1 = 22 s (gc quay 90o)- S chu k o trong 1 gi: nck= 3600/22 = 164- Nng sut ca my o:

3

ca ck 1 tgW t.q.n .k .k 7.0,25.0,87.164.0,75 187 m /ca= = =

- Thi gian o t bng my:o ng ti ch: t = (1645.48 361.93)/ 205 = 6.26 ca, chn 6,5

ca.o ln xe: tx = 361.93/187 = 2 ca.

- Cng trng rng v khng cn quan tm n c ly vn chuyn ca xe tnn ta xem nh thi gian ch tnh i vi my o.

Phnon Mng

Slng

o my(m3)

o thcng(m3)

1

M1 2 14.2 1.76

M2 4 49.8 6.16M3 16 98.24 10.4

2M1 3 21.3 2.64


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II. THIT K BIN PHP THI CNG TON KHI:

1. Xc nh c cu qu trnhMng nh cng nghip 1 tng c thit k l mng n, ti ch. Qu trnh

thi cng gm 4 phn on sau:- Gia cng lp t ct thp.- Gia cng lp dng vn khun.- b tng, bo dng.

- Tho d vn khun.

2. Tnh khi lng cng tc

- Hm lng ct thp tnh bng 100 kg/m3 b tng mng.- Cng tc lp dng vn khun nh cng tc tho vn khun.- Tng hp khi lng cng tc ca tng mng cho trong bng sau:

Loimng

Vn khun (m2) Btng (m3) Btng lt (m3) Ct thp (tn)

M2 0.72 1.22 0.40 0.72

M1 1.50 3.13 0.72 1.50

M3 0.64 1.07 0.32 0.64

3. Chia phn on thi cng

D i k h d l i i h i h


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Phnon

Mng

Slng

Cp pha(m2)

B tng(m3)

B tng lt(m3)

Ct thp(Kg)

Tho Cp pha (m2)

1M1 2 1.44 2.45 0.79 0.24 1.44M2 4 6.00 12.53 2.88 1.25 6.00M3 16 10.24 17.06 5.18 1.71 10.24

2M1 3 2.16 3.67 1.19 0.37 2.16M2 6 9.00 18.80 4.32 1.88 9.00

3M1 3 2.16 3.67 1.19 0.37 2.16M2 6 9.00 18.80 4.32 1.88 9.00

4M1 3 2.16 3.67 1.19 0.37 2.16M2 6 9.00 18.80 4.32 1.88 9.00

5M1 3 2.16 3.67 1.19 0.37 2.16M2 6 9.00 18.80 4.32 1.88 9.00M3 8 5.12 8.53 2.59 0.85 5.12

6M1 3 2.16 3.67 1.19 0.37 2.16M2 6 9.00 18.80 4.32 1.88 9.00M3 8 5.12 8.53 2.59 0.85 5.12

7M1 3 2.16 3.67 1.19 0.37 2.16M2 6 9.00 18.80 4.32 1.88 9.00M3 8 5.12 8.53 2.59 0.85 5.12


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4. Tnh nhp cng tc ca dy chuyn b phn

- Chn t hp chuyn nghip thi cng cc qu trnh thnh phn theo nh mc 726:

TT T th chuyn nghip St th S th(1 t) Phn loi theo bc th2 3 4 51 Gia cng, t ct thp 3 10 4 3 2 1

2Gia cng, dng vnkhun

2 4 1 1 2 -

3 b tng 5 9 4 3 1 14 Tho vn khun 2 4 1 1 2 -

- Chi ph lao ng cho cc cng vic theo nh mc 1242 (ai):o

b tng mng (MH: HA-1210): 1,64 cng/m3

.o Gia cng, lp t ct thp (MH: IA-1120): 8,34 cng/tn.o Gia cng, lp dng, tho d vn khun mng (MH: IA-1220): 29,7 cng/100m2.

- Nu chn t th chuyn nghip vi s lng v c cu theo nh mc 726, ta tnh c nhp cng tc ca ccdy chuyn b phn trn cc phn on theo cng thc:

ij i

ij

c i

P .ak

t.n .N= (ngy)

nc: h s ca lm vic trong ngy, chn nc = 1.

ai : nh mc chi ph lao ng cho cng vic i.t : thi gian lm vic trong 1 ca.

Qu trnhPhn on

Ct thpLp

vn khunB tng

Thovn khun

1-5 1 1 1 16 1 1 1 1


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7 1 1 1 1

Theo biu tin , tnh c thi gian ca dy chuyn k thut l T = 14 ngy.

Ghi ch:

1. Dy chuyn lp vn khun.

2. Dy chuyn an ct thp.

3. Dy chuyn b tng.

4. Dy chuyn tho vn khun.


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5. Chn t hp my thi cng

a) Hnh thc cung ng b tng:

- Vi quy m cng trnh nh trn, tit kim nhn cng v tng nng sut btng ta chn phng n dngbtng ti do nh my cung cp. Vi phng n ny, cn xc nh th tch btng ln nht cho mt ln , ck hoch cung ng btng hp l.

- Song song vi , cn tin hnh t chc giao thng phc v cho xe btng v chn loi xe c tay cn thchhp.

b) My m di: Chn my MIKASAPHW-40 c cc thng s k thut sau:

o ng knh x chiu di u di: 40 x 306 mmo

Bin rung: 3,1 mmo rung: 12.000 13.000 ln/pht.o Trng lng: 2,1 kg.

PHN 2: THI CNG LP GHP

A. THNG K CU KIN LP GHP

B.TNH TON THIT B CU LP1.Chn v tnh ton thit b treo buc

Lc cng cp c xc nh theo cng thc: ttk.P

Sm.n.cos

=

Trong :k: h s an ton k n lc qun tnh (k = 5-6).m: h s k n sc cng cc si cp khng u.


17/25

n: s si cp (s nhnh treo vt). : gc nghing ca cp so vi phng ng.

a) Ct:

Cc ct c trng lng chnh lch nhau khng ln nn ta ch cn tnh dy cu cho ct gia c trng lng lnnht, dy cu ny ng thi l dy cu chung cho cc ct cn li.

Ptt = 1,1. P = 1,1. 0,41 = 0451 TLc cng cp:

= = =

tt

o

k.P 6.0,451S 1.353T

m.n.cos 1.2.cos0

T bng tra chn c dy cp mm 6x37x1, ng knh D = 24 mm, cng chu ko = 150 kg/cm2.

b) Dm mng:S dng dng c treo buc dm c kho t ng.

Ptt = 1,1. P = 1,1 . 2,6 = 2,86 T.Lc cng cp:

tt

o

k.P 6.2,86S 12,14 T

m.n.cos 1.2.cos45= = =

T bng tra chn c dy cp mm 6x37x1, ng knh D = 19,5 mm, cng chu ko = 150 kg/cm2.

Tng t nh dm cu chy, ta cng chn dy cu cho dm mng l cp mm 6x37x1, ng knh D = 17,5mm, cng chu ko = 150 kg/cm2.

c) V ko v ca tri:

Tin hnh t hp v ko v ca tri sau cu lp ng thi. S dng n treo v dy treo t cn bng.


18/25

45

S

panelmai

Dn D2 v ca triPtt = 1,1. P = 1,1(15,2 + 2,5) = 19,47 T

Lc cng cp:tt

o

k.P 6.19,847S 40,3 Tm.n.cos 0,75.4.cos20

= = =

T bng tra chn dy cp mm 6x37x1, D = 32 mm, cng chu ko = 150kg/cm2.

Dm RB1Ptt = 1,1. P = 1,1. 0.65 = 0.715 T

Lc cng cp:

= = =

tt

o

k.P 6.0.715S 1,52T

m.n.cos 0,75.4.cos20

T bng tra chn dy cp mm 6x37x1, D = 22 mm, cng chu ko = 150kg/cm2.Dm RB2

Ptt = 1,1. P = 1,1. 0.1 = 0.11 TLc cng cp:

= = =

tt

o

k.P 6.0,11S 0,23T

m.n.cos 0,75.4.cos20

T bng tra chn dy cp mm 6x37x1, D = 22 mm, cng chu ko = 150kg/cm2.

d) Panel sn:

S dng chm dy cu c vng treo t cn bng:Ptt = 1,1. P = 1,1. 2,3 = 2,53 T

Lc cng cp:tt

o

k.P 6.2,53S 6,8 T

m.n.cos 0,75.4.cos45= = =


19/25

T bng tra ta chn c dy cp mm 6x37x1, ng knh D = 13 mm, cng chu ko = 150 kg/cm2.

2.Xc nh c cu qu trnh v chn s lp ghp kt cu

- Cn c vo c im kt cu ca cng trnh c th chia qu trnh lp ghp nh cng nghip 1 tng ra thnhcc qu trnh thnh phn sau:o Lp dm mng.o Lp ct.o Lp dm mi.o Lp panel sn, dm sn

- Phng php lp ghp l phng php hn hp. hai trc u hi nh c mt s ct sn tng, cc ct nyc th lp chung vi dn mi v tm mi.

- y chn s cu theo phng dc nh. Vi cng trnh ny c th chn 2 loi my cu lp ghp.o My cu c sc nng trung bnh lp cc loi cu kin nh nh dm mng, dm sn, dng s dc

bin nhp tn dng sc nng v gim chiu di tay cn.o My cu c sc nng ln hn lp ct (dng s dc bin nhp), dn v ko mi, dm mi dung

bin php u cu.3.Tnh ton cc thng s cu lp

Vic la chn s di chuyn ca cu trong qa trnh lp ghp l bcu rt quan trng, n nh hng n vic tnh ton cc thng s cu lp.Sau khi tnh cc thng s cu lp, chn cu ta s la chn s di chuynhp l nht m bo t thi gian lu thng khng cu. V d nh gcquay cu cng nh cng c li, cng 1 v tr lp nhiu cu kin cng li.

Tnh cho vic cu lp tng cu kin nh sau:

a) Lp dm mng:

Chiu cao nng mc cu:Hm = h1 + h2 + h3 = 0,5 + 0,45 + 1,2 = 2,15 m

h3

h2

h1

Lh4

1500

1000

972500

875

1200

450

500

1000


20/25

Chiu cao nh cn:H = Hm + h4 = 2,15 + 1,5 = 3,65 m

Chiu di tay cn ti thiu:c

min omax

H h 3,65 1,5L 2,23 m

sin sin75

= = =

Tm vi ti thiu:Rmin = r + Lmin.cos max

= 1,5 + 2,23.cos75o = 2,08 mKhi lp dm mng cha lp t khe mng, nn dm mng phi b tr cch mp mng t nht l 1 m.Khong cch t v tr xp n v tr thit k:

a = 1 + 0,35 + 1 = 2,35 mTm vi lm vic ca cu:

R = Rmin + a = 2,08 + 2,35 = 4,43 m, chn R = 5 mChiu di tay cn lm vic:2 2 2 2

cL (R r) (H h ) 3,5 2,15 4 m= + = + =

Sc nng yu cu:Q = Pck + Ptr= 1,5 + 0,5 = 2 T

b) Lp ct bin: h2 = 6 m.

Chiu cao nng mc cu:

Hm = h1 + h2 + h3 = 0,5 + 6 + 1,5 = 8 mChiu cao nh cn:H = Hm + h4 = 8 + 1,5 = 9,5 m

Chiu di tay cn ti thiu:

= = =

cmin o

max

H h 9,5 1,5L 8,28m

sin sin75

Tm vi ti thiu:Rmin = r + Lmin.cos max = 1,5 + 8,28.cos75o = 3,64 m


21/25

Sc nng yu cu:Q = Pck + Ptr= 0,41 + 0,5 = 0,91 T

c) Lp ct gia: h2 = 8.1 m.Chiu cao nng mc cu: Hm = h1 + h2 + h3 = 0,5 + 8.1+ 1,5 = 10.1 m

Chiu cao nh cn: H = Hm + h4 = 10.1 + 1,5 = 11.6 m

Chiu di tay cn ti thiu:

= = =

cmin o

max

H h 11,6 1,5L 10,5m

sin sin75

Tm vi ti thiu:Rmin = r + Lmin.cos max = 1,5 + 10,5.cos75o = 4,2 m

Sc nng yu cu:Q = Pck + Ptr= 1,85 + 0,5 = 2,35 T

C. THIT K BIN PHP THI CNG LP GHP:

1.Lp dm mng

a) Chun b:

- b tng cc khi m trn mng n cao trnh 0,5 m, vch tim trn cu kin v khi m b tng, vsinh cc bn thp ch trong mng v dm mng.

b) Cng tc dng lp:

- Treo buc cu kin ti 2 im cch u mt 1,25 m.


22/25

- My cu nng cu kin ln khi mt t cch 0,5 0,7 m th dng li 1 2 pht kim tra an ton treo buc,sau h tay cn xung a cu kin vo v tr thit k. Dng my kinh v hay dy di kim tra v tr ca cu kintheo cc vch tim. Nu sai lch th dng x beng iu chnh.

- C nh tm: hn im cc bn thp ch cu kin v gi .- C nh vnh vin: hn lin tc cc bn thp ch.

2.Lp ct

Dng phng php quay tin hnh lp ct.Phng php quay i hi vic xp ct sao cho tm cc mng,chn ct v im treo buc ct nm trn mt cung trn bn knh R.

Tnh ton v tr xp ct nh sau:

Ct ngn (hc = 6 m)

- V tr im treo buc nm cch chn ct mt khong ht = 3 m.- Chn gc 1= 45o, 2 = 15o.- Khong cch t nh ct n tm cc mng theo phng trc ngang nh:

a = (hc + 0,4).cos60o = (6 + 0,4).0,5 = 3.2 m(0,4 m l khong cch t chn ct n tm cc mng)

- Khong cch t v tr treo buc ct n tm cc mng theo phng trc ngang nh:a = (ht + 0,4).cos60o = (3 + 0,4).0,5 = 1,7 m

- Tm vi thao tc:+ +

= = =

t

o o

1 2

h 0, 4 3 0, 4

R 2.1m2.cos( ) 2.cos(45 15 )

- Khong cch t v tr my ng n tm cc mng theo phng trc ngang nh:B = R. cos 2= 2,1. cos15o = 2 m.

Ct gia (hc = 8.1 m)

- V tr im treo buc nm cch chn ct mt khong ht = 6 m.- Chn gc 1= 45o, 2 = 15o.


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- Khong cch t nh ct n tm cc mng theo phng trc ngang nh:a = (hc + 0,4).cos60o = (8.1 + 0,4).0,5 = 4,3 m(0,4 m l khong cch t chn ct n tm cc mng)

- Khong cch t v tr treo buc ct n tm cc mng theo phng trc ngang nh:

a = (ht + 0,4).cos60o = (6 + 0,4).0,5 = 3,2 m- Tm vi thao tc:

+ += = =

t

o o

1 2

h 0, 4 6 0, 4R 3,5m

2.cos( ) 2.cos(45 15 )

- Khong cch t v tr my ng n tm cc mng theo phng trc ngang nh:B = R. cos 2= 3,5. cos15o = 3,4 m.

3.LP DM MI:- S dng 2 cu trc (u cu) lp dm mi c khuch i t mt t


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