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Q= |
12000 |
[N] |
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h= |
450 |
[mm] |
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Jako materiał śruby przyjmuję St 5 |
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l= |
450 |
[mm] |
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ls= |
318,2 |
[mm] |
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Obliczanie śruby z warunku na wyboczenie |
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Średnice śruby obliczamy z wzoru Tetmajera |
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R0,R1- współczynniki materiałowe; dla stali St5: |
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R0= |
330 |
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R1= |
0,61 |
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xw- |
współczynnik bezpieczeństwa |
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xw= |
3,5 |
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s- smukłość śruby |
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ls= |
318,2 |
[mm] |
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imin - promień bezwładności |
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imin=0,25d1 |
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d1= |
13,93168 |
[mm] |
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Z PN - 88/M - 02019 dobieram gwint trapezowy symetryczny: Tr 20x4 gdzie: |
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d= |
20 |
[mm] |
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D= |
20,5 |
[mm] |
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d2= |
18 |
[mm] |
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D2= |
18 |
[mm] |
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d1= |
15,5 |
[mm] |
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D1= |
16 |
[mm] |
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P= |
4 |
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Obliczam samohamowność gwintu: |
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g-kąt wzniosu lini zwoju |
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g= |
4,05 |
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ρ' - zastępczy kąt tarcia, |
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m - współczynnik tarcia pomiędzy współpracującymi powierzchniami; w naszym przypadku |
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stal - brąz |
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m= |
0,15 |
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a - dla gwintu trapezowego |
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a = |
15 |
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r = |
8,83 |
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Warunek został spełniony |
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Sprawdzam śrubę na naprężenia zastępcze: |
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Re - granica plastyczności |
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Re = |
295 |
[MPa] |
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xw= |
3,5 |
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kr= |
84,29 |
[MPa] |
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a - dla obciążeń statycznych : |
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a = |
3 |
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Q= |
12000 |
[N] |
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d= |
18 |
[mm] |
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sc= |
47,18 |
[MPa] |
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Ms= |
24685,48 |
[Nmm] |
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t = |
21,56822 |
[MPa] |
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sz= |
60,17978 |
[MPa] |
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sz<kr |
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Warunek został spełniony |
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Obliczam smukłość śruby: |
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imin= |
3,875 |
[mm] |
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ls= |
318,2 |
[mm] |
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s= |
82,12 |
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Obliczam współczynnik bezpieczeństwa |
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Rw= |
280 |
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S= |
254,34 |
[mm2] |
pole przekroju śruby |
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Q= |
12000 |
[N] |
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xw= |
5,93 |
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Obliczam nakrętkę: |
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Zakładam materiał nakrętki BK 331 dla którego według PN-79/H-87026 Rm=360 MPa. |
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Przyjmując współczynnik xm = 4 otrzymamy dopuszczalne naprężenia przy rozciąganiu: |
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Rm= |
360 |
[MPa] |
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xm= |
4 |
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kr= |
90 |
[MPa] |
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Nakrętka jest rozciągana: |
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Q= |
12000 |
[N] |
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d2= |
18 |
[mm] |
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D= |
20,5 |
[mm] |
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Dn= |
24,29 |
[mm] |
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Przyjmuję: |
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Dn= |
30 |
[mm] |
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Ponieważ wysokość kołnierza hk i średnica dk z obliczeń wychodzą bardzo małe dlatego: |
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Dk= |
40 |
[mm] |
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hk= |
5 |
[mm] |
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Wysokość nakrętki przy założeniu na gwincie nacisku: |
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pdop= |
12 |
[MPa] |
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P= |
4 |
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pdop= |
15 |
[MPa] |
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D2= |
20,5 |
[mm] |
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D1= |
16 |
[mm] |
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H= |
24,82 |
[mm] |
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Przyjmuje: |
H= |
35 |
[mm] |
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Obliczamy pokrętło: |
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Mc=Ms= |
24685,48 |
[Nmm] |
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Przyjmując siłę napędową N = 250 N przykładaną na końcu pokrętła |
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obliczymy jego wysięg: |
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N= |
250 |
[N] |
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Mc= |
24685,48 |
[Nmm] |
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R= |
98,74 |
[mm] |
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Przyjmuję długość R= |
200 |
[mm] |
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Jako materiał przyjmuję stal St3 dla której: |
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kr= |
94 |
[MPa] |
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kg= |
117,5 |
[MPa] |
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Obliczam średnice pokrętła: |
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N= |
250 |
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R= |
200 |
[mm] |
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kg= |
117,5 |
[MPa] |
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d≥ |
16,31 |
[mm] |
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Przyjmuję d= |
20 |
[mm] |
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Obliczam słupki: |
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Każdy słupek jest ściskany siłą Qs która: |
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Qs= |
6000 |
[N] |
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Siłę zginającą określimy z zależności: |
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Ms= |
24685 |
[Nmm] |
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L - minimalna odległość między słupkami |
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L= |
300 |
[mm] |
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Ps= |
82,28493 |
[N] |
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Przyjmując wysokość słupka Hs=550 mm obliczymy moment gnący działający na podstawę słupka: |
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Mg=Ps·Hs |
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Ps= |
82,285 |
[N] |
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Hs= |
550 |
[mm] |
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Mg= |
45256,71 |
[Nmm] |
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Przyjmujemy na słupki stal węglową St3 dla której: |
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Re = |
235 |
[MPa] |
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xw= |
2,5 |
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Rm = |
400 |
[MPa] |
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dopuszczalne naciski: |
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pdop= |
94 |
[MPa] |
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Obliczam średnice słupka ze wzoru: |
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Mg= |
45257 |
[Nmm] |
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pdop= |
94 |
[MPa] |
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Ds≥ |
22,0 |
[mm] |
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Z nierówności: |
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Przyjmujemy graniczne: |
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Qs= |
6000 |
[N] |
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Ds= |
22,0 |
[mm] |
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podstawiamy do warunku wytrzymałości na rozciąganie |
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Jeżeli założymy średnice otworu przelotowego w płycie do rowną średnicy gwintu d oraz średnicę |
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rdezenia gwintu d1=0,85d to po przekształceniu otrzymamy: |
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Qs= |
6000 |
[N] |
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Ds= |
22,0 |
[mm] |
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pdop= |
94 |
[MPa] |
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d≥ |
17,0 |
[mm] |
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Połączenie piasty pokrętła przyjmuję wpustowe: |
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ko= |
72 |
[MPa] |
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lo -toretyczna długość wpustu |
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T= |
24685 |
[Nmm] |
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d= |
18,0 |
[mm] |
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F= |
2742,831 |
[N] |
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Dobieram wstępnie wpust: |
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b= |
6 |
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h= |
6 |
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n - ilość wpustów |
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n= |
1 |
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lo= |
13 |
[mm] |
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l= |
19 |
[mm] |
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Przyjmuje wpust A 6x6x25 |
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