Projektni Zadatak Projektni Zadatak Projektni Zadatak: Projektovanje Dizalica Projektovanje Dizalica

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PR JEKTNI ZADATA GLA NI MA INSKI PROJEKA ELEKTRIČNE DVOGREDE MOSNE DIZALICE

Predmetni nastavnik prof. dr Nenad Zrnić Asistent doc. dr Vlada Gašić Saradnik u nastavi Goran Milojević ��

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UVOD

Dvogreda mosna dizalica Mosna Mosna diza dizalic licaa može može da im dva glavna nosa ča ili jedan glavni n sač (ako je u pitanju jednogreda mosna dizalica). Glavni nosači su vezani za čeone nos če (obično montažnim vezama). U čeonim nosačima su smešteni točkovi (pogonski i slobodni). Točkovi omogućavaju translatorno kre anje mosnog mosnog krana po kranski kranskim m šinama. Vertikalni točkovi su obično s obodima (ako nisu ugrađeni eni hori horizzont ontal i to čkovi). Horizontalni točkovi ovi pri prima maju ju hori horizzonta ontaln lnee sile, pa se tada ugrađuju vertikalni točkovi bez oboda. Po glavnim nosa čima mosnoe dizalice se kreću kolica (obično po g rnjem pojasu). Glavni nosači mogu biti limene i rešetkaste konstrukcije. Nosa či rešetkaste konstrukcije se upotrebljavaju za ve će raspone, jer su tada ekonomi čniji. Limeni nosa i su obično sandučasti (formirani obično zavarivanje ) ili valjani profili. Kolica se kreću translatorno po glavnim nosačima i na njima su s ešteni mehanizmi za dizanje i kretanje kolica. Čest postoji i glvno i pomo ćno dizanje. Duž glavnih nosa ča su posta ljene sta stazze, ko koje sl služe za za sme smešštan tanje pogona i trolnih vodova a takođe predstavljaju ukruće nje st strukture mos mosta u horizontal talnoj r vni. Struktura mosne dizalice može dobro da prihvata i horizontalna i vertikalna optere enja.Čeoni nosači su snab snabde deve veni ni sa sa odbo odbojn jnic icim imaa k ji u slučaju otkazivanja krajnjih isklju čivača stupaju u dejstvo. Ako je je upravljanje mosne di diz lice lice sa poda poda (brz (brzin inee tans tansla lato torn rnog og kret kretaa ja dizalice ne treba da su veće od 30) onda u sklopu krana nije ugrađena kabina. U suprotn m je izgrađena kabina u sklopu krana. Vertikalna pterećenja primaju točkovi kovi koli kolica ca.. S Silil od uticaja momenta ekscentriteta od težine kolica i korisnog tereta primaju horizontalni toč ovi ili takođe vertikalni točkovi. Zbog ekscentri čnosti olica sa teretom, glavni nosač je optere en na uvijanje.

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Konstr Konstrukc ukcija ija dvog dvogred redne ne mosn mosne dizalice se sastoji iz sledećih el elemenata: • • • • • • • • • •

Reduktor Doboš za dizanje Kočiona spojnica meh nizma za dizanje EM mehanizma za diz nje EM pogo pogona na za kret kretan anjj Spojnica Reduktor pogona kret nja Koturača Pogonski točak Slobodni točak

Noseća čelična konstrukcija mosnih kranova se sastoji od glavnih i čeonih nosača. Po glavnim nosačima se kreću k lica, a na čeonim nosačima je montiran ehanizam za kretanje dizalice. Glavni nosa či se izvode kao kutijaste konstrukcije. Mosne dizalice se dele u dve pšte klase : • Mosne dizalice opšte namene • Mosne osne diz dizalic alicee spec specijijal alne namene Uobičajen jena kla klasa sa mos mosnih dizalica opšte namene se izrađuju za nosivosti masa od Q=5; 6,3; 8; 10; 12,5; 16; 20; 25; 32; 40; 63; 80 i 100. Uobičajeno j da se mosne dizalice opšte namene projektuju za p gonsku klasu II.

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Projektni zadatak Prema važećim standardima i propis propisima ima uraditi uraditi glavni glavni mašins mašinski ki proj projek ek t električne dvogrede mosne dizalice slede ćih karakteristika : 1. Nosivost dizalice: Q = 6300 kg 2. Raspon: L = 23,5 m ffffffff 3. Brzina dizanja: V diz = 4 ff min

m ffffffff 4. Brzina kretanja di alice: V kd = 20 ff min 5. Visina dizanja: H = 10 m 6. Pogonska klasa: 1 Am (FEM) 7. Tež Težina ina koli kolica ca:: m k l = 395 kg 8. Rastojanje šina: l Spw = 2240 mm 9. Prečnik točkova dizalice: Dt = 200 mm 10. 10. irin irinaa šine šine kran kransk sk staze: bš ks = 40 mm (40 x 40) 11. Osno rastojanje točkova čeonog nosača: Lt = 4700 mm

Dizalica se postavlja na posto eću kransku stazu u objektu korisnika. lavni nosači mosne dizalice su limeni nosači kutij ste konstrukcije. Elektromotorno vitlo i č one nosače dizalice potrebno je usvojiti iz standar nog programa proizvo đača STAHL (ww (ww .stahlcranes.com .stahlcranes.com). ). Za nosivost nosivost dizalice 6300 kg svajam vitlo SH 4016 – 16 4/1 iz STAH kataloga. V diz

ffmffffffff fina , 4 ffmffffffffgl = 0,6 min

b c

min

b

c vna

Razmak između šina: L 23,5 l = ffff= ffffffff= 235 cm [ Sw = 1800 mm 10

10 Spw = 2240 mm L 23,5 = = = 4,7 m Lt = 5@7 5

ffffffffff ffffffff

Prikaz razmaka između šina, rastojanja između točkova i ra pon dizalice

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DIMENZIONISANJE GL GLAVN G NOSAČA Izbor geometrije glavnog nos ča dizalice (kutijasta konstrukcija)

Kutijasti nosač Određivanje orjentacionih vre nosti dimenzija nosa ča:

fffffLfffffffff = H =

14 @ 20 23,5 = = 1,38 m = 138 cm H = 17

ffffffff

H H 13 f f f= 55,2 cm D = ffffffffff[ D = ffffff[ D = ff f 2@3 D = 55 cm

2,5

2,5

L 23,5 D min = fffff= ffffffff 50 50 D min = 47 cm

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Preporučene debljine lima u zavisnosti od pogonske klase ( tab. 4.6 M talne konstrukcije 1)

Pr poručene debljine vertikalnih limova

P eporučene debljine pojasnih limova Debljina limova: Za H (1000mm ÷ 1600mm) ⇒ δ 1 6mm; δ 2 = 8mm

Gde je: •

δ 1 - debljina vertikalnih limova

•

δ 2 - debljina pojasnih li

ova

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Proračun opterećenja točkova Određivanje pritiska po točku, na osnovu formula iz kataloga. Ulazni podaci: • O2 = 800 mm • O10 = 316 mm • z1 = 1048 mm • e4 = 110 mm • mL = 6300 kg • mKa = 395 kg •

S PW − z1 O10 ⋅ ⋅ mL + 0,4 ⋅ mKa SPW O2

R11max =

2240 − 1048 316

R11max = R12max

⋅ 6300 + 0.4 ⋅ 395 = 14,82kN 800 S − z O 2 − O10 = PW 1 ⋅ ⋅ mL + 0,2 ⋅ mKa S PW O2 2240 − 1048 800 − 316

R12max = R21max = R21max = R22max

⋅

2240

⋅

2240

800

⋅ 6300 + 0.2 ⋅ 395 = 21kN

z1 + e4 O10 ⋅ ⋅ mL + 0,25 ⋅ mKa Spw O 2 1048 + 110 316

⋅ 6300 + 0, 25 ⋅ 395 = 13,85kN 800 z + e O 2 − O10 = 1 4⋅ ⋅ mL + 0,15 ⋅ mKa Spw O2

R22max =

2240

⋅

1048 + 110 800 − 316 2240

⋅

800

⋅ 300 + 0,15 ⋅ 395 = 20,3 kN

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Faktori φ Za srednju pogonsku klasu usvajamo φ1=1.05.

Vrednosti za srednju pogonsku klasu ϕ 1 = 1.05 ϕ 2 = ϕ 2 min + β 2 ⋅ vh

vh = 0,067

m sec

β 2 = 0.34 ϕ 2min = 1.1 ( HC 2) ϕ 2 = 1.1 + 0,34 ⋅ 0.067 = 1.12

Proračun sila koje optereću u nosač

Prikaz sila koje opterećuju nosač F 2 = ϕ 2 A 13,24 + ϕ 1 A 1,58 = 1,12 A 13,24 + 1,05 A 1,58 = 16,52 kN F 1 = ϕ 2 A 20,2 + ϕ 1 A 0,79 = 1,12 A 20,2 + 1,05 A 0,79 = 23,45 kN R = F 1 + F 2 R = 16,52 + 23,45 R = 39,97 kN

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Rastojanje od ose točka do mesta dejstva rezultujuće sile F 2fAff bff 16,52 ffffffffAff800 ffffff= 330,6 mm = 33,07 cm = e x = fff R 39,97 b = O2 = 800 mm

Određivanje potrebne visin nosača prema kriterijumu čvrstoće 40 R M f = 0,5 A ffffA L @ e x A 0,5 A L e x = 0,5 A fffffffffA 2350 @ 33 A 0,5 A 2350 @ 33 L 2350 M F = 22 844 kNcm

b c b c

`

a `

a

Određivanje potrebne visin nosača prema kriterijumu ugiba

v wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww u u u βfffffffffffffff 1 + α 1 ff6ffff u H = u F L fffffffffffffffffffff f f t 3

h lj

A

A 1

3

A

48 A

b E

b

2

A

c

c f

δ + 0,2 A δ 2

A 0,166 A 1

A

max

b = O2 = 800 mm L = 23500 mm kN ff E = 21000 ffffff cm 2 α =

F fff 16,52 f fffffffff= 0,7 = 2

F 1 23,45 b fff ff800 ffffffff= 0,034 β= = L 23500 L f max = ffffff= 4,7 cm 500

v wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww u u u 1 0,7 1 6 0,034 u f f f f f f f f f f f ff fffffffffffffffffffffffffffffffffffffff H = u 23,45 2350 t 3

h lj

A

3

A

A

48 A 2100

b

b

@

A

A 0,166 A 0,6

2

c

c

+ 0,2 A 0,8 A 4,7

H = 62,63 cm H =

3

3 ⋅ M f 2 ⋅ [σ ] ⋅ δ 1

=

3

3 ⋅ 22844 2 ⋅ 10 ⋅ 0.6

= 75cm

kN cm 2 H = 75cm

[σ ] = 10

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Usvaja se: H1 = 100cm D = 40cm D1 = 35cm

Momenti inercije preseka n sača Aksijalni momenti inercije preseka nosača 2  D ⋅ δ 23  H 1 + δ 2   + 2⋅  + D⋅ 2 ⋅ I X = 2 ⋅   12 12  2    2 3  40 ⋅ 0.83 100 ⋅ 0.6  100 + 0.8   4 + 2⋅  + 40 ⋅ 0.8 ⋅  I X = 2 ⋅   = 262563cm 12 2     12 2  H 1 ⋅ δ 13 D 3 ⋅ δ 2  D1 + δ 1   + 2⋅  + H 1 ⋅ δ 1 ⋅  IY = 2 ⋅   12 12  2    2 3 100 ⋅ 0.6 3 40 ⋅ 0.8  34 + 0.6   4 + 2⋅  + 100 ⋅ 0.6 ⋅  I Y = 2 ⋅   = 46557 cm 12 2     12

H13 ⋅ δ1

Otporni momenti inercije pr seka nosača I X 262563cm 4 = = 5169 m 3 W X = ymax 50,8cm IY 46557cm 4 = = 2328c WY = xmax 20cm

3

Presek glavnog nosa ča ��

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Statički moment odsečaka ko ture:

1-2

b S

c

0 @ 1,9

2

y1.2 = 50,4 cm δ = 0,8 cm 3-2

b S 2

S x12 = 76,61 cm 3

c

0 @ 17,8

y3.2 = 50,4 cm δ = 0,8 cm

S x32 = 717,69 cm 3

2

2-4

76, 1 + 717,69 = 794,3 cm³

b S 2

c

0 @ 50,4

y2.4 = 50,4 @ S δ = 0,6 cm 4-5

b S 2

S x24 = 1556,3 cm 3

c

0 @ 50,4

y4.5 = 0

S x45 = 794,3 m 3

δ = 0,6 cm

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Dija ram statičkih momenta odsečaka konture Sx

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DOKAZ NAPONA U PRESE U I - I

Prikaz delovanja opterećenja na nosa A = 2 ⋅ H1 ⋅ δ1 + 2 ⋅ D ⋅ δ 2 = 2 ⋅ 100 ⋅ 0.6 + 2 ⋅ 40 ⋅ 0.8 = 184 cm 2 kg ρ = 0,00785 3 cm q* = A ⋅ 7.85 × 10−5 q = q* ⋅ 1.2 = 184 ⋅ 1.2 ⋅ 7.85 × 10−5 = 0.0173

kN cm

Moment u vertikalnoj ravni 2 R L2 39,97 23 0 2 2 M V 1 = ⋅ ( L − ex ) + φ 1 ⋅ q ⋅ = ⋅ ( 2350 − 33,07 ) + 1.05 ⋅ 0.0173 ⋅ = 35365,65kNcm 4 L 8 4 ⋅ 2350 8

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Moment u horizontalnoj rav i k a = 0,1 F 1 H = F 1 A k a = 23,45? ,1 = 2,345 N F 2 H = F 2 A k a = 16,52 A 0,1 = 1,65 kN R H = F 1 H + F 2 H = 3,997 kN R H L2 3,997 23502 2 2 M H 1 = ⋅ ( L − ex ) + φ 1 ⋅ q ⋅ ⋅ K a = ⋅ ( 2350 − 33,07) + 1.05 ⋅ 0.0173 ⋅ ⋅ 0.1 = 3535, 42kNcm 4 ⋅ L 8 4 ⋅2350 8

Moment torzije D1ff δ ff ff 35 fffff +fff 0,6 fffffff= 17,8 c + 1= x1 = fff 2

2

2

H 1ff + 4 = 54 cm y1 = ffff 2

M t 1 = F 1 A x1 + F 1 H A y1 = 23,45 A 1 .8 + 2,345 A 54 = 544,04 kNcm M t 2 = F 2 A x1 + F 2 H A y1 = 16,52 A 1 .8 + 1,652 A 54 = 383,27 kNcm L @ e 2350 @ 33,07 M tA = M t 1 + M t 2 A ffffffff xfffA L = 544,04 + 383,27 A ffffffffffffffffffffffff= 457,13 k cm

b

M tB = M tA @

c

bM

2

t 1 +

c M t 2

c

2 A 2350 = 457,13 @ 544,04 + 383,27 = @ 470,18 kNcm

b

c

Reakcije oslonaca Σ M a = 0

hf fffffffg F j b 2350 2

i hf fff f fffg ek j c

23 0 + 0,5 A 2

i hf fffffffg e k F j O c b 2350 + 2

2 @ 0,5 A e x

@

F B A l

23,45 A 1175 @ 16,54 + 39,97 A 1175 + 16,54 + 16,52 A 1175 + 80 @ 16,54

@

F B A 2350 = 0

1

A

@ 0,5 A x

F B = 40,53 kN

+

x

+

2

A

Σ yi = 0

F A @ F 1 @ R @ F 2 + F B = 0 F A @ 23,45 @ 39,97 @ 16,52 + F B = 0 F A + F B = 79,94 F A = 79,94 @ 40,53 F A = 39,41 kN

M I = F A A 0,5 A

bl e c @ x

b

c

= 39,41 A ,5 A 2350 @ 33,07

M I = 45 666,7 kNcm

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M II = 1

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d e b c F l e b b d b c e A A

0,5 A

@ x

+

@

1

A

M II 1 = 39,41 A 0,5 A 2350 @ 33,07 + 80

@ 23,45 A 80

M II 1 = 46 932 kNcm

M II 2 = F B A 0,5 A l + e x

@

b

d b

M II = 40,53 A 0,5 A 2350 + 33,0 2

M II 2 = 45 050,5 kNcm

c e @ 80

Uticaj loklnog pritiska točka na vertikalni lim

Površina merodavna za proračun koraka ukrućen ja A = 10.2 ⋅ 0.8 + 4 ⋅ 4 = 24.16 cm 2 Y = X = I Y =

(10.2 ⋅ 0.8 ) ⋅ 0.4 + 4 ⋅ 4 ⋅ 2.8 24.16 (10.2 ⋅ 0.8) ⋅ 5.1 + 4 ⋅ 4 ⋅ 8 24.16 10.2 ⋅ 0.8 12

= .99cm

= 7. 2cm

3 2

+ 10.2 ⋅ 0.8 ⋅1.6 +

4⋅4 12

3

+ 4 ⋅ 4 ⋅ 0.82 = 52.9 cm 4

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Za tačku 1 M VI kN 35365.65 ⋅ y1 = ⋅ 50.8 = 6.84 2 I X 262563 cm M kN 3535.42 σ ZH 1 = HI ⋅ x1 = ⋅ 20 = 1.52 2 I Y cm 46557 kN σ U 1 = σ ZV + σ ZH = 6.84 + 1.52 = .36 cm 2 kN σ d = 16 2 cm kN σ U 1 = 8.36 2 ≤ σ d cm σ ZV 1 =

Dijagra napona u vertikalnoj i horizontalnoj rav i

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Za tačku 2 M VI H 1 35365, 65 100 kN ⋅ = ⋅ = 6,73 2 I X 2 cm 262563 2 M  D kN  3535, 2  35  σ ZH 2 = HI ⋅  1 + δ 2  = ⋅  + 0.6  = 1.37 2 I Y  2 cm  4655  2  F ⋅ S kN 20 ⋅ 806,4 τ 2 = tr X 2 = = 0.102 2 I X ⋅ δ 1 262563⋅ 0,6 cm D ⋅ δ 2 H 1 + δ 2 40 ⋅ 0.8 00 + 0.8 ⋅ = ⋅ = 806.4cm 3 S X 2 = σ ZV 2 =

2

2 2 Ftr ≅ 0.5R = 0.5 ⋅ 39,97 ≅ 20kN

2

M tA kN 457,13 = 0.106 cm 2 2 ⋅ A* ⋅ δ 1 2 ⋅ 3588, 48 ⋅ 0.6 A* = ( H − δ 2 ) ⋅ ( D1 − δ 1 ) = (101.6 − 0.8) ⋅ ( 35 − 0 .6 ) = 3588, 48cm 2 τ 2,, =

2

σ U 2 = (σ ZV 2 + σ ZH 2 ) + σ y2 − (σ

V 2 + σ ZH 2

2

) ⋅ σ y + 3 ⋅ (τ 2,, + τ 2 )

2

2

σ U 2 = ( 6.73 + 1.37 ) + 2.24 2 − ( 6.73 + 1.37 ) ⋅ 2.24 + 3 ⋅ (0.102 + 0.106 ) = 7.2

kN cm 2

Za tačku 3 kN cm2 kN τ 3,, = τ 2,, = 0.106 2 cm 100 100 H H ⋅ ⋅ 0.6 = 1556.4cm4 S x 3 = S x2 + 1 ⋅ 1 ⋅ δ 1 = 806.4 + σ ZH 3 = σ ZH 2 = 1.37

2

τ 3 =

4 20 ⋅1556.4

2

4

Ftr ⋅ S x 3 kN = = 0.2 2 I x ⋅ δ 1 262563⋅ 0.6 cm 2

2

2 ,, 2 σU 3 = σ ZH 3 ⋅ 3 ⋅ (τ 3 + τ 3 ) = 1.37 ⋅ 3 ⋅ ( 0.2 + 0.106) = 1.47

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DOKAZ NAPONA U PRESE U III

Prikaz sila koje opterećuju nosač er = 1m = 100cm ekol = er + e x = 100 + 33,07 = 133, 07cm F A ⋅ L = F1 ⋅ ( L − er ) + F2 ⋅ ( L − e r b ) L − er L − er − b + F 2 ⋅ F A = F1 ⋅ L 2 L − ekol 2350 − 133,07 F A = R ⋅ = 39,97 ⋅ = 37,7 kN L 23 0 1

1

2

2

FTIII = FA + φ 1 ⋅ ⋅ q ⋅ L = 37,7 + 1, 05 ⋅ ⋅ 0,0173 ⋅ 2350 = 59,05kN M TIII = ( M t1 + M t 2 ) ⋅ •

L − ekol 2350 − 133,07 = ( 544, 04 + 383, 27 ) ⋅ = 877 kNcm L 2350

Moment savijanja u preseku III

e3 = 128 − 25 = 103mm = 10.3cm M VIII = FA max ⋅ e3 = 59.05 ⋅10.3 = 08.22kNcm F A max = FTIII = 59.05kN

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L − ( er + ex ) 1 + φ 1 ⋅ ⋅ q ⋅ L 2 L 2350 − (100 + 33,07) 1 + 1.05 ⋅ ⋅ 0,0173 ⋅ 2350 = 25,11k N F HIII = ( 2,345 + 1,652) ⋅ F HIII = ( F1H + F 2H ) ⋅

M HIII

350 = FHIII ⋅ e3 = 25,11⋅ 10,3 = 258,63kNcm

2

Izgled profila u preseku III •

Moment inercije za pro il u preseku III

2 1 3  H 3 δ 2   I X 3 = 2 ⋅ H ⋅ δ1 + 2 ⋅  δ 2 ⋅ D + D ⋅ δ 2  +   12 2    2 12

1

3 3

2  40 ⋅ 0.8 3  50 0.8   4 + 2⋅  + 40 ⋅ 0.8 ⋅  + I X 3 = 2 ⋅   = 53793.64 cm 12 2    2  12 2 1 1 3  D1 δ 1   IY 3 = 2 ⋅ D ⋅ δ1 + 2 ⋅  H 3 ⋅ δ 1 + H 3 ⋅ δ 1  +   12  2 2   12

0.6 ⋅ 50

I Y 3 = 2 ⋅

3

3 40 ⋅ 0.6

12

2  50 ⋅ 0.63  35 0.6   4 + 2⋅  + 50 ⋅ 0.6 ⋅  +   = 25412.2 cm 2    2  12

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Otporni moment inercij za presek III

•

I X 3 53793,64 = = 2117, 6cm3 ymax 25.4 I 25412,2 = 1270,61cm 3 W Y III = Y 3 = xmax 20 W X III =

Za tačku 1

•

M VIII M HIII + ≤ σ dop W X W Y kN = 16 2 cm

σ U III 1 = σ dop

III σ U 1 =

608,22 2117.46

+

258,63 1270, 61

= 0,49

kN cm 2

σ U III 1 ≤ σ dop •

Za tačku 2

A* = ( D1 + δ1 ) ⋅ ( H 3 + δ 2 ) = ( 35 + 0.6 ) ⋅ ( 50 + 0.8 ) = 1808.48 cm 2 M tAIII 877 kN τ 3, = = 0.404 * cm 2 2 ⋅ A ⋅ δ 1 2 ⋅ 1808.48 ⋅ 0.6 FtIII ⋅ S III kN 59.05 ⋅ 593.85 X 3 τ 3 = = = 1.39 2 I X 3 ⋅ δ 1 53793.64 ⋅ 0.6 cm 2

2  M  III σ U =  HIII  + 3 ⋅ (τ 3 + τ 3, ) 3  W Y 

2

k 2  258.63    + 3 ⋅ (1.39 + 0.404 ) = 3.11 cm 2  1270.61 

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Statički moment odsečaka ko ture:

1-2

b

c

S 2 0 @ 1,9

y1.2 = 25,4 cm δ = 0,8 cm 3-2

b S 2

S x12 = 38,60 cm 3

c

0 @ 17,8

y3.2 = 25,4 cm δ = 0,8 cm

S x32 = 361,7 cm 3

2

2-4

38,60 + 361,7 = 400,3 cm³

b S 2

c

0 @ 25,4

y2.4 = 25,4 @ S δ = 0,6 cm 4-5

b S 2

S x24 = 593,85 cm 3

c

0 @ 25,4

y4.5 = 0

S x45 = 400,3 cm 3

δ = 0,6 cm

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Dijagram s atičkih momenta odsečaka konture Sx z presek III

DOKAZ DEFORMACIJE GL VNOG NOSAČA

ema opterećenja glavnog nosa ča

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F1* ⋅ L3  21 ⋅ 2350 3 2 ⋅ 1 + α ⋅ (1 − 6 ⋅ β )  = ⋅ 1 + 0.71 ⋅ (1 − 6 ⋅ 0.03 42 )  = 1.76cm f = 48 ⋅ E ⋅ I X 48 ⋅ 21000 ⋅262563 F1* = R12 = 14.82kN F1* = R11 = 21kN L = 23.5m = 2350cm I X = 262563cm 4 b = O 2 = 80cm F 2* 14.82 α = * < 1 ⇒ α = = 0.71 < 1 F 1 21 b 80 β = = = 0.034 L 2350 f dop = k ⋅ L = k =

1 500

⋅ 2350 = 4.7cm

1

500 f dop = 4.7cm

Dokaz dinamičke krutosti glavnih nosača dizalice

Prikaz amplitude oscilovanja

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m1 = 0.5 ( mQ + mkol ) + 0.486 ⋅ mnos = 0.5( 6300 + 395 ) + 0.486 ⋅ 3394.34 = 4997. 5kg mQ = Q = 6300kg mkol = 395kg mnos = ρ ⋅ A ⋅ L = 7850

kg ⋅ 0.018 ⋅ 23.5 = 3394.34 kg m3

A = 2 ⋅ H1 ⋅ δ1 + 2 ⋅ D ⋅ δ 2 = 2 ⋅ 100 ⋅ 0.6 + 2 ⋅ 40 ⋅ 0.8 = 184 cm 2 = 0.0184 m 2 τ γ

T = 3 ⋅ ≤ T d T = 3 ⋅

0.3109

= 7.77 s < T d = 12 15s 0.12 τ − period oscilovanja [s] γ = 0.12 − logaritamski dekrame t prigušenja (tab. 7.18; 251str. Dizalice) τ = 2 ⋅ π ⋅ δ11 ⋅ m1 = 2 ⋅ π ⋅ 4.9 ⋅ 10−7 ⋅ 4997.15 = 0.3109 δ 11 =

1.0 ⋅ L3 48 EI X

=

1.0 ⋅ 2350 3 48 ⋅ 21000 ⋅ 26256

= 4.9 ⋅10 −7

m N

Dokaz stabilnosti

Prikaz polja 1 i polja 2 glavnog nosa ča

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S V = R =

H 1 δ 1

=

Re R ei

100 0.6

=

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Tabela 5 = 166.667  → Kriterijum 3 ⇒ 160 R < SV < 265 R

23.5 23.5

=1

Položaj horizontalnih ukrućenja

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Vertikalno ukrućenje

Kratko vertikalno ukrućenje

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I R = 3 ⋅ H 1 ⋅ δ 13 = 3 ⋅ 100 ⋅ 0.6 3 = 64 .8cm 4 Usvojeno:

I R = 73cm 4 ⇒ L50x50x5 (tab. .16; 191str. MK1) b r = 5cm A1 = 4.8cm 2 e1 = 1.4cm I1 = 11cm 4 AU = A1 + (18 ⋅ δ1 ) ⋅ δ 1 = 4.8 + (18 ⋅ 0.6) ⋅ 0.6 = 11.28cm 2 e=

A1 ⋅ ( br − e1 + 0.5 ⋅ δ 1 ) 4.8 ⋅ ( − 1.4 + 0.5 ⋅ 0.6 ) = = 1.65cm AU 11.28

Površina merodavna za prora čun imenzija horizontalnog ukrućenja

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IU = 1.5 ⋅ δ14 + 18 ⋅ δ 12 ⋅ e 2 + I 1 + A1 ⋅ ( br − e1 + 0.5δ 1 − e )

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2

I U = 1.5 ⋅ 0.64 + 18 ⋅ 0.6 2 ⋅1.652 + 1 1 + 4.8 ⋅ (5 − 1.4 + 0.5 ⋅ 0.6 − 1.65 )

2

IU = 53.13cm 4 I U 53.13 = = 2.17 cm AU 11.28

iu = λ = −

λ =

a 200 = = 92.16 iu 2.17

i

λ 92.16 = = 0.99 λ v 92.9

riva B → α = 0.339

= a = 200cm

λ v = 92.9

a1 ≤

6 ⋅ W šin ⋅ [σ šin ]

F 1

σ šin =

Re 1.5

=

23.5 1.5

=

6 ⋅ 44 ⋅ 15.66 55.81

= 15.66

= 176.6cm

kN cm 2

−

−  β = 1 + α  λ − 0.2  + λ 2 = 1 + 0.3 9 ( 0.99 − 0.2 ) + 0.99 2 = 2.24   2

χ =

−

β β 2 − 4 λ 2

2

=

2.24 2.24

2

4 ⋅0.99

2

= 0.6

σ 2 ≤ σ KR

Iy ffff 52,9 fffffffff= 44 cm3 = W sin = ffffsin xmax 1,2 xmax = 2 A δ1 M V 1  1  35365 1 kN  ⋅  H 1  = ⋅  ⋅ 100  = 3.37 2 I X  4  262563 4 cm  M  D  3535  35 kN  σ ZH 2 = H 1 ⋅  1 δ 1  = ⋅  ⋅ 0.6  = 1.37 2 I Y  2  46557  2 cm  kN σ = γ (σ ZV 2 + σ ZH 2 ) = 1.5 ⋅ ( 3.37 1.37 ) = 7.11 2 cm k σ KR = χ ⋅ Re = 0.6 ⋅ 23.5 = 14.1 2 c σ ZV 2 =

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Rastojanja i među vertikalnih i kratkih vertikanih ukrućenja

Dijagram normalnih napona

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Provera vertikalnih limova na izbočavanje M X H 1 kN 35365 1 0 ⋅ =− ⋅ = − 6.73 2 I X 2 cm 262563 M  D kN 3535  35   σ ZH 2 = − Y ⋅  1 + δ 1  = − ⋅  + 0.6  = −1.37 2 I Y  2 cm 465 7  2   kN σ 1 = −γ (σ ZV 2 + σ ZH 2 ) = −1.5 ⋅ ( − .73 − 1.37 ) = 12.15 2 cm kN 0.25 H 1 0.25 ⋅100     σ 2 = −γ  σ ZV 2 ⋅ + σ ZH 2  −1.5  −6.73 ⋅ − 1.37  = 7.02 2 cm 0.5 H 0.5 ⋅101.6     σ ZV 2 = −

σ 3 = σ 2 σ 4 = γ ( σ ZV 2 − σ ZH 2 ) = 1.5 ( 6.73 − 1.37 ) = 8.04

kN cm 2

Dijagram normal ih napona merodavnih kod provere na i bočavanje

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Polje 1 α =

a1 85 = = 3.4 0.25 H 1 0.25 ⋅100

ψ =

σ 2 7.02 = = 0.57 σ 1 12.15 2

2

 δ 1  kN  0.6  σ E = 18980   = 18980   = 10.93 2 cm  0.25 ⋅100   0.25 H 1  Za α > 1 ; ψ < 1 ⇒ K σ =

K σ =

8.4 0.57 + 1.1

8.4

tab.1 245str.prilog, MKM1

ψ + 1.1

= 5.029

σ KR = K σ ⋅ σ E = 5.029 ⋅ 10.93 ≅ 5

kN cm 2

K σ ⋅ α 2 = 5.029 ⋅ 3.4 2 > 2 ⇒ 58.1 > 2 ⇒ f = 0 −

λ p = χ p =

Re

σ KR

=

23.5 55.76

0.6 − 2

λ p − 0.13

=

= 0.64 0.6 2

0.64 − 0.1

= 1.17

−

σ uz = χ p = 1.17

•

Granični napon:

C σ = 1.25 − 0.25ψ = 1.25 − 0.25 ⋅ .57 = 1.11 − kN σ uz = Cσ ⋅ σ uz ⋅ Re = 1.11⋅ 1.17⋅ 23. 5 = 30.51 2 cm •

Kontrola stabilnosti polja I: 2

2

 σ   10.93  σ 2 =  1  ≤ 1 ;   = 0.128 < 1 σ 30.51    uz  −

Stabilnost polja I je obezbeđe a.

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Polje II α =

a 170 = = 2.27 0.75 H 1 0.75 ⋅100

ψ =

σ 4 8.04 = = −1.14 σ 3 −7.02 2

2

 δ 1  kN  0.6  σ E = 18980   = 18980   = 1.21 2 cm  0.75 ⋅100   0.75 H 1  Za α > 1 ; ψ ≤ − 1 ⇒ K σ = 23.9 tabela 1 245str.prilog, MKM1

kN cm 2 K σ ⋅ α 2 = 23.9 ⋅ 2.27 2 = 135.15 > ⇒ f = 0

σ KR = K σ ⋅ σ E = 23.9 ⋅ 1.21 = 28.91

−

λ p = χ p =

Re

σ KR

=

23.5 28.91

0.6 − 2

λ p − 0.13

=

= 0.9 0.6 0.9 2 − 0.13

= 0.727

−

σ uz = χ p = 0.727 •

Granični napon

C σ = 1.25 − 0.25ψ = 1.25 − 0.25 ⋅ −1.14 ) = 1.54 − kN σ uz = Cσ ⋅ σ uz ⋅ Re = 1.54 ⋅ 0.727 ⋅ 2 .5 = 26.31 2 cm •

Kontrola stabilnosti polja II: 2

 σ  σ =  3  ≤ 1 ;  σ uz  −

2

2

 − 7.02    = 0 .071 < 1 26.31  

Stabilnost polja II je obezbeđena.

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Provera pojasnog lima na izbočavanje

Pojasni lim S p =

D1

S p =

23.5

≤ S p max = 65

δ 2 35 0.8

Re

= 65

= 43.75 < S p max

α =

a1 85 = = 2.43 D1 35

ψ =

σ 2 = 0.57 σ 1 2

2

 δ  kN  0.8  σ E = 18980  2  = 18980  = 9.91 2  cm  35   D1  Za α > 1 ; ψ < 1 ⇒ K σ =

K σ =

8.4 0.57 + 1.1

8.4

tabela 1 245str.prilog, MKM1

ψ + 1.1

= 5.02

σ KR = K σ ⋅ σ E = 5.02 ⋅ 9.91 = 49.58

kN cm 2

K σ ⋅ α 2 = 5.02 ⋅ 2.432 > 2 ⇒ 29.6 > 2 ⇒ f = 0 −

λ p = χ p =

Re

σ KR

=

23.5 49.58

0.6 − 2

λ p − 0.13

=

= 0.69 0.6 2

0.69 − 0.1

= 1.02

−

σ uz = χ p = 1.02

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Granični napon

C σ = 1.25 − 0.25ψ = 1.25 − 0.25 ⋅ .57 = 1.11 − kN σ uz = Cσ ⋅ σ uz ⋅ Re = 1.11⋅ 1.02 ⋅ 23. 5 = 26.6 2 cm

Kontrola stabilnosti polja II:

•

2

 σ  σ =  4  ≤ 1 ;  σ uz  −

2

2

 8.04    = 0. 9 < 1 26.6  

Stabilnost polja je zadovoljen .

Dimenzije polja uz oslonac H sr = 0.75 ⋅ H1 = 0.75 ⋅ 100 = 75c F TIII kN 59.05 τ sr = = = 0.66 2 cm 2 ⋅ H sr ⋅ δ 1 2 ⋅ 75 ⋅ 0.6 τ = D1 ⋅ (τ sr + τ 3, ) = 35 ⋅ ( 0.66 + 0. 7 ) = 32.55 τ 3, =

M tIII 877 = 2 ⋅ A* ⋅ δ 1 2 ⋅ 2698.48 ⋅ 0.6

0.27

b

0,8

c bD c b

A = H SR + δ2 C

A

+ δ1 = 75 1

kN cm 2

kN cm 2

cb A

c

35 + 0,6 = 2698,48 cm2

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α =

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H 1 100 = = 1.33 H sr 75

τ 3,, 0.27 Ψ= = = 0.44 τ SR 0.66

K σ = 23.9 K σ ⋅ δ 12 ≥ 2 23.9 ⋅ 0.6 2 ≥ 2 ⇒ 8.6 ≥ 2 ⇒ f = 2

 δ  kN  0.6  σ E = 18980  1  = 18980  = 1.21 2  cm  75   H sr  2

σ KR = K σ ⋅ σ E = 23.9 ⋅1.21 = 29.0

kN cm2

C σ = 1.25 − 0.25ψ = 1.25 − 0.25 ⋅ .44 = 1.14 − kN σ uz = Cσ ⋅ σ uz ⋅ Re = 1.14 ⋅ 0.73 ⋅ 23. 5 = 19.56 2 cm −

λ p = χ p =

Re

σ KR

23.5

=

29.03

0.6

=

− 2

λ p − 0.13 •

= 0.9 0.6 2

0.9 − 0.13

Kritični sičući napon kN cm 2

τ kr = k τ ⋅ σ E = 7.6 ⋅ 1.21 = 9.2

k τ = 5.34 + −

λ = χ p =

= 0.73

4.0

= 5.34 +

α 2

RE 3 ⋅ τ kr

3 ⋅ 9.2

=

−2

λ − 0.13 −

τ u = χ p = 0.51

2

1.33

23.5

=

0.6

4.0

= .6

= 1.21

0.6 2

1.21 − 0.13

= 0.51

kN cm 2

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Granični napon smican ja

C τ = 1.25 −

τ u = C τ ⋅ τ u ⋅

Re 3 2

= 1.25 ⋅ 0.51 ⋅

23.5 1.73

= 8.64

kN cm 2

2

 σ   τ  σ 2 =  E  ⋅   ≤ 1  σ uz   τ u   1.21  σ =    19.56  σ 2 ≤ 1 2

2

2

kN  32.55  ⋅  = 0.05 cm 2  8.64 

Provera stabilnosti glavnog nosača na bočno izvijanje

Polupresek glavnog nosa ča

ema opterećenja glavnog nosa ča

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AP = D ⋅δ 2 + 2 ⋅ D 3 ⋅ δ 2

I P =

12

H 1 2

⋅ δ 1 = 40 ⋅ 0.8 2 ⋅

100 2

⋅ 0.6 = 92cm 2

2

H1

�� . 3�

2

403 ⋅ 0.8 100  D1 + δ 1  35 + 0.6  4 + 2⋅ ⋅ δ 1  = + ⋅ ⋅ 2 0.6    = 2327 .4cm 2 12 2  2  2 

I P 23277.4 = = 15.9cm AP 92

i p =

1

LK = β ⋅ L =

⋅ L = li →

2.09 l 1625.5 λ = i = = 102.23 i p 15.9

λ =

��

�� . 1

1 2.09

⋅ 2350 = 1625.5cm

λ 102.23 = = 1.1 λ v 92.9

•

Ekvivalentna sila pritis a

M VI 35365 ⋅ S PX = ⋅ 3112. = 419.26kN I X 262563 H δ H H 100 0.8  100 100 + + ⋅ ⋅ = 3112.8cm 3 S PX = D ⋅ δ 2  1 + 2  + 1 ⋅ δ 1 ⋅ 1 = 40 ⋅ 0.8  0.6 

D* =

 2

•

2 

2

4

 2

2 

2

4

Radni napon usled ekvivalentne sile pritiska

D * 1 M R 24 kN σr = ⋅ + 0.9 ⋅ HI ≤ σ k = = = 16 2 AP χ WY v1 1.5 cm kN σ dop = 16 2 cm kriva.izvijanja → C → χ = 0.67 kN 419.26 1 3535.4 σ r = ⋅ + 0.9 ⋅ = 8.16 2 ≤ σ k cm 92 0.67 2328

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Provera veze pojasa sa vert kalnim limom

Sile koje deluju na šav kN ≤ σ Šdop ; cm 2 F 1 kN 21 σ y = V 1 = = = 2.42 2 cm 2 ⋅ z1 ⋅ a 2 ⋅ 14.46 ⋅ 0.3 σ u = V22 + V 12 = 2.952 + 2.42 2 = 3.82

σ Šdop = 12

kN cm 2

FT III S PIII kN 59.05 161 .8 = ⋅ = 2.95 2 < σ Šdop V 2 = III ⋅ I X 2 ⋅ a 53793.64 2 ⋅ 0 .3 cm a = δ 1 − 3 = 6 − 3 = 3mm = 0.3cm H 1 + δ 2 1 0 + 0.8 S III = 40 ⋅ 0.8 ⋅ = 1612.8 cm 3 P = D ⋅ δ 2 ⋅ 2

z1

2

v wwwwwwwwwwwwww u u I fffff = 3,25 t δ v wwwwwwwwwwwwwwwwwwwww u u 52,9 f fffffff = 14,46 = 3,25 A t 0,6 A

3

f g f g α 1

z1

3

I α = 52,9 cm 4 ,δ1 = 0,6 cm,F 1 =

max

= R12 = 21 kN

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Analiza opterećenja usled z košenja

ema opterećenja celokupnog krana

ema opt rećenja kada su kolica u krajnjem položaju

Gi =

1000 9.81 2

⋅

1000

= 4.9kN ; L = 2 50cm; ekol = 133.07cm; q = 0.0173

ΣFi = R11 + R12 = 39.97kN

L − ekol 1 1 2350 − 133.07 1 1 + q ⋅ L + ⋅ Gi = 39.97 ⋅ + 0.0173 ⋅ 2350 ⋅ 4.9 = 60.48kN L 2 2 2350 2 2 e 1 1 133.07 1 1 + 0.017 3 ⋅ 2350 + ⋅ 4.9 = 2 5.04kN F B = ( ΣFi ) ⋅ kol + q ⋅ L + ⋅ Gi 39.97 ⋅ L 2 2 2350 2 2 max R = 2 F A = 120.96kN min R = 2 F B = 50.08kN Σ R = 171.04kN F A = ( ΣFi ) ⋅

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Prikaz di enzija kada su kolica u krajnjem polož ju µ ' =

min R

=

50.08kN

= 0.29 Σ R 171.04kN µ = 1 − µ ' = 1 − 0.29 = 0.71 1 1 lS =  µ −  ⋅ L =  0.71 −  ⋅ 23 0 = 493.5cm 2   F y = v ⋅ f ⋅ ( ΣR )

v = 1−

Σd i

n

= 1−

2

Lt = 0.5 2 ⋅ Lt

DIN 4132 → f = 0.25 f = 0.3 1 − e( −250α )  

F y = 0.5 ⋅ 0.25 ⋅ 171.04 = 21.38 kN µ '  d 0. 9 ⋅ 0.25 ⋅ 171.04 = 6.2 kN F y11 = ⋅  1 − 1  ⋅ f ⋅ ( ΣR ) = n  n µ 0.71 ⋅ 0. 5 ⋅ 171.04 = 15.18kN F y12 = ⋅ (1) ⋅ f ⋅ ( ΣR ) =

n 2 SILE _ ZAKOSENJA : F L = µ ⋅ F y = 0.71⋅ 21.38 = 15.18 N F D = µ , ⋅ Fy = 0.29 ⋅ 21.38 = 6.2 k F L = F y12 F D = F y11 M HZ = F L ⋅

Lt − S PW 2

= 15.18 ⋅

470 − 224 2

= 1867.14kNcm

S PW = 2240mm = 224cm

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Proračun veze čeonog i gla nog nosača

Prikaz zavrtanjske veze

Prirubna ploča

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a2 = D1 + 2δ 1 + 6d = 35 + 2 ⋅ 0.6 + 6 ⋅ 2 = 48.2cm d = 20mm − prečnik zavrtnja a2 = 56cm; F y 2 AB = FTIII = 59.05kN F x 2 AB = 0.1 ⋅ F TIII = 0.1 ⋅ 59.05 = 5.905kN F 59.05 = 14.76kN F y1 = y 2 AB = 4

F x 2 AB

F x1 =

4 5.905

=

4

4

= 1.48kN

2

l  a 56 28 r 1 =  2  +  y  =   +   = 31.30 cm  2 2  2   2 2

Fr1 = M III ta ⋅

2

2

r 1 31.30 = ⋅ = 7kN 877 2 4 ⋅ r 1 4 ⋅ 31.30

F R = F r 21 + F y2 = 72 + 14.76 2 = 16.33kN M HZ 1 1867.14 1 kN ⋅ = ⋅ = 16. 7 2 a2 2 cm 56 2 FP = 154kN − SILA _ PRITEZAN A _ VIJKA − MK 1_ STR _ 63_ T .3.2. F Z =

` a

a1 = a2 + 2 A 3 A d = 56 + 2.3 A 2 = 68 cm

Fts =

µ ⋅ ( FP − F Z )

V 2 V2 = 1.4; m = 1 F τ = 2 R < τ dop d π

⋅m =

0.71⋅ (15 − 16.67 ) 1.4

⋅ 1 = 69.65kN

τ dop = 22.4

kN za zavrtan kvaliteta 8.8 cm 2

4

σ Z = τ=

F Z < σ dop AS

σ dop = 22

16.33 kN = < τ 5.19 22 π cm 2 dop 4 16.67

σ Z =

2.45

= 6.8

AS = 2.45cm 2 − ispitna površina zavrtnja

kN < σ dop cm 2

Nosivost vijka na smicanje –

dd fffe

F τ =

2

A

π 4

τ

kN za zavrtanj kvaliteta 8.8 cm 2

A doz >

F R ;2 2 A

slučaju da veza “proradi”

πff f 4

A 22.4

= 70,336 > 16,33

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PRORAČUN ČEONOG NOS ČA

Prikaz opterećenja i glavnih dimenzija

ematski prikaz opterećenja glavnih i čeonih nosača

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M ment torzije usled horizontalnih sila

Opterećenje čeonog nosača

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F3Q = 1.1 ⋅ F3G = 1.1 ⋅

Q 4

⋅ g = 1.1 ⋅

mkol

6300

�� . 45

⋅ 9.81 = 17kN

4 395

⋅ 9.8 = 1.07kN 4 Q 6300 ⋅ 9.81 = 14.68kN F4Q = 0.95 ⋅ ⋅ g = 0.95 ⋅ 4 4 m 39 ⋅ 9.81 = 0.92kN F 4 G = 0.95 ⋅ kol ⋅ 9.81 = 0.95 ⋅ 4 4 F3 = F3G ⋅ ϕ1 + F 3Q ⋅ ϕ1 ⋅ ϕ 2 = 1.07 ⋅ 1.05 + 17 ⋅1.05 ⋅1.12 = 21.12 kN 4

⋅ g = 1.1 ⋅

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F4 = F 4 G ⋅ ϕ1 + F 4Q ⋅ ϕ1 ⋅ ϕ 2 = 0.92 ⋅ 1.05 + 14.68 ⋅1.05 ⋅1.12 = 18.23kN F A max = ( F1 + F 2 ) ⋅

L − ekol 1 + ϕ 1 ⋅ ⋅ q ⋅ L L 2

F A max = ( 23.45 + 16.52 ) ⋅ FC max = ( F3 + F 4 ) ⋅

2350 − 133.07 2350 1

1

+ 1.05 ⋅ ⋅ 0.0173 ⋅ 2350 = 59.04 kN 2

L − ekol + ϕ 1 ⋅ ⋅ q ⋅ L L 2

F C max = ( 21.12 + 18.23) ⋅

2350 − 133.07 2350

1

+ 1.05 ⋅ ⋅ 0.0173 ⋅ 2350 = 58.45 kN 2

M 3 = M tIII = 877kNcm D δ 35 0.6  M t,1 = F3 ⋅  1 + 1  + F3H ⋅ y1 = 1.12 ⋅  +  + 1.81 ⋅ 54.8 = 475.12 kNcm 2   2 2  2 D δ 35 0.6  M t,2 = F4 ⋅  1 + 1  + F4 H ⋅ y1 = 8.23 ⋅  +  + 1.56 ⋅ 54.8 = 409.98 kNcm  2

2

 2

2 

F3 H = ( F3Q + F3G ) ⋅ K a = (17 + 1.0 ) ⋅ 0.1 = 1.81kN F4 H = ( F4Q + F4G ) ⋅ K a = (14.68 + 0.92) ⋅ 0.1 = 1.56kN L − ekol 2350 − 133.07 = ( 47 .12 + 409.98 ) ⋅ = 834.65kNcm L 2350 M t 3 = F3H ⋅ y1 = 1.81⋅ 54.8 = 99.18kNcm M t 4 = F4 H ⋅ y1 = 1.56 ⋅ 54.8 = 85.49kNcm H y1 = + H sin = 50.8 + 4 = 54.8c M 1 = ( M t,1 + M t ,2 ) ⋅

2

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Dokaz napona za presek II σ II =

f fM ffffffg y f fM fffffffg x IIV

I x

A

IIH

+

I y

A

Lt = 459,6 cm l = 224 cm l1 = 117,8 cm F Amax = 59,04 kN F Cmax = 58,45 kN F z = 16,67 kN @ Sila usled zako enja

d M M F bl l c F l e L b c ffffffffd b c b c f f F L l b c b c f f f F L l 1

F = fff D

@

t

1 F D = 459,6

1

+

3

@ 834,65

1 A t @ 2 1 M IIH = z A A t @ 2 I x3 = 53793,64 cm 4

M VII =

+

D A

Cmax A

1

+

Amax A 1

e

+ 877 + 58,45 A 224 + 117,8 + 59,04 A 117,8 = 58,7 k

= 58,7 A 0,

= 16,67 A

A 459,6 @ 224

= 6914,9 kNcm

1 A 459,6 @ 224 = 1963,7 kNcm 2

I 3y = 25412,2 cm 4 31,4 2 x = 13 cm 6914,9 σ II = 53793,64

y = fffffff= 15,7 cm

f fffffffffffffg

σ II = 3,02

A 15,7

+

f fffff f ffffffg 196 ,7 25412,2

A 13

fkN fffff cm 2

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Dijagram Sx čeonog nosača PRORAČUN POGONSKE G UPE Dt = 20cm D d = t = 5cm 4 µ = 0.01

f = 0.05cm η = 0.8 β ∈ ( 2 ÷ 4 )

vmosta = 20

Usvojeno: β = 3

m min

= 0.33

m s

F L = F A + FC + GČ N = 59.04 + 58. 45 + 4.9 = 122.39 kN  µ ⋅ d

Fw L = F L ⋅ 

 Dt

P L =

Fw L ⋅ vmosta η

=

+

2 ⋅ f 

 0.01 ⋅ 5 2 ⋅ 0.05  ⋅ β = ⋅ + 12 .39   ⋅ 3 = 2.75kN Dt  20   20

2.75 ⋅ 0.33 0.8

= 1.13kW

Potrebna snag elektromotora za kretanje dizalice: PW = 2 A P L u P W = 2 A 1.13 = 2. 6 kW

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Usvajaju se motori za kretanj dizalice: P = 2x1.5kW (STAHL katalog), Motor za pogon kolica: P=0.55kW (STAHL katalog). Snaga motora vitla: P = 4kW (S AHL katalog).

Pdiz

4 kW

Pkol

0.55 kW

2 ⋅ P Ldiz

3 kW

ΣPinst = 7.55kW

Instalisana snaga

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Rezultati dobijeni u progra u KRASTA

Rešetkasti model – mreža

Definisanje poprečnih preseka – izgled modela

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Prikaz poprečnog preseka glavnog nosača u programu

Prikaz karakteristika glavnog nosača ��

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Prikaz opterećenja sa faktorisanim silama

Prikaz opterećenja sa nefaktorisanim silama

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Uticaj sopstvene težine

Transverzalne sile u preseku I – I :

Prikaz transverzalne sile u preseku I – I sa tabelarnim vrednostima

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Analiza napona u preseku I – I :

Prikaz napona za karakteristične tačke preseka I – I sa tabelarnim vrednostima

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Moment savijanja:

Prikaz momenta savijanja u preseku I – I sa tabelarnim vrednostima Ugib:

Pri az ugiba sa tabelarnim vrednostima

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Rezultat # 1 2 3 4 5 6 7

Ix Iy Wx Wy M 1V 1 σ ZV

f

Proračun 262 563 46 557 5 196 2 328 35 365,65 kNcm 6,84kN/cm2 1,76 cm

KRASTA 264 989 46 858 52 7 23 2 32 801,56 kNc 6,24 kN/cm2 1,78 c

Razlika (%) -1% -1% -1% -1% +7% +9% -1%

Analitički dobijene vrednosti k je su kontrolisane razlikuju se za manje od 10% od vrednosti dobijenih pomoću programsk g paketa KRASTA, samim tim proračun je zadovoljio.

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LITERATURA 1. Ostrić D., Tošić S.: DIZALI E – Mašinski fakultet Beograd, Beograd, 2005. 2. Petković Z., Ostrić D.: MET LNE KONSTRUKCIJE U MAŠINOGRADNJI 1 - Mašinski fakultet Beograd, Beograd, 1996. 3. Petković Z.: METALNE KO STRUKCIJE U MA INOGRADNJI 2 - ašinski fakultet Beograd, Beograd, 2005. 4. Zrnić N.: IZVODI SA PRED VANJA - Mašinski fakultet Beograd, Beograd, 2012.

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Projektni Zadatak Projektni Zadatak Projektni Zadatak: Projektovanje Dizalica Projektovanje Dizalica

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