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Dane |
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obliczenia |
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udzwig 5kN |
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wysokość podnoszenia 0,3 m |
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1) obliczam średnicę rdzenia śruby |
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dobieram materiał |
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stal S235JR (St3S) |
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kcj=90 Mpa |
90000000 |
Pa |
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h |
0,3 |
m |
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udzwig |
5000 |
N |
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d3= |
0,008412574418751 |
m= |
8,41257441875115 |
mm |
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przyjmuje gwint |
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p= |
5 |
mm |
0,005 |
m |
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d2=D2= |
25,5 |
mm |
0,0255 |
m |
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d3= |
22,5 |
mm |
0,0225 |
m |
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D1= |
23 |
mm |
0,023 |
m |
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d= |
28 |
mm |
0,028 |
m |
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3. Obliczam kąt pochylenia linii śrubowej g |
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0,074063 |
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tgy= |
0,062445360309729 |
y= |
3,57 |
stopnia |
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dla połączenia stal-stal |
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cosar= |
0,965960168538399 |
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u'= |
0,310571812142039 |
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p'= |
17,25 |
stopnia |
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0,2 |
0,3 |
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r’=17,25o |
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5. Sprawdzanie samohamowności gwintu |
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g=3,57o |
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r’ >g |
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17,25o>3,57o |
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h=300 mm |
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6. Sprawdzanie śruby na wyboczenia |
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0,3 |
m |
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Obliczam długość wyboczenia śruby |
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d3=18,5 mm |
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Lwyb=1,2*h |
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0,36 |
m= |
360 |
mm |
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0,0185 |
m |
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Lwyb=360 mm |
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α= |
2 |
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Obliczam ramię bezwładności |
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ix=0,25*d3 |
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0,004625 |
m= |
4,625 |
mm |
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E= |
200000 |
MPa |
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obliczam smukłość śruby |
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155,675675675676 |
>100 |
stosujemy wzór Eulera |
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Dla λ>100 stosujemy wzór Eulera |
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Dla λ<100 stosujemy wzór Tetmajera σkr=a-b*λ |
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W naszym przypadku stosujemy wzór Eulera |
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81,3669005594136 |
Mpa |
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Obliczam rzeczywiste naprężenia w przekroju |
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12581583,7068491 |
Pa= |
12,5815837068491 |
Mpa |
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Sprawdzamy współczynnik bezpieczeństwa |
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6,46714296508789 |
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Warunek jest spełniony |
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tg(y+p') |
20,82 |
stopnia |
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& Obliczam moment skręcający |
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tg(20,82)= |
0,38005301341749 |
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ms=0,5*Q*tg(y*p')*d2 |
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Ms= |
24,228379605365 |
Nm |
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obliczam napręrzenia zastępcze |
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2,38777375E-06 |
m3 |
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10146848,965638 |
N/m2=Pa |
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12581583,7068491 |
N/m2=Pa |
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21614159,2565349 |
Pa= |
21,6141592565349 |
Mpa |
<90 |
Mpa |
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OBLICZENIA NAKRĘTKI |
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Pdop= |
13500000 |
Pa |
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13500000 |
Pa> |
3122268,01548645 |
Pa |
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n-ilość czynnych zwojów |
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H- wysokość nakrętki |
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H> |
0,00925116449033 |
m= |
9,25116449033022 |
mm |
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n= |
1,85023289806604 |
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n[6-10] |
przyjmujemy n= |
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8 |
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H [1,2-2,5]d |
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H>p*(n+2)+2*4 [mm] |
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H= |
58 |
mm |
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ostatecznie przyjmujemy H= |
60 |
mm |
0,06 |
m |
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obliczamy średnice zewnętrzną nakrętki z warunku na rozciąganie z uwzględnieniem skręcenia momentem Ms |
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Q'=Q*1,3 |
6500 |
N |
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Dn= |
0,029597344996745 |
m= |
29,5973449967448 |
mm |
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przyjmujemy średnice nakrętki Dn= |
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42 |
mm= |
0,042 |
m |
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F= |
0,0007693 |
m2 |
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8449239,56843884 |
Pa= |
8,44923956843884 |
MPa |
<90MPa |
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p= |
100000 |
<13500000 |
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Fkol= |
0,00037037037037 |
m2 |
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Dk= |
0,047284346130689 |
m= |
47,2843461306894 |
mm |
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przyjmujemy średnice kołnierza |
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50 |
mm= |
0,05 |
m |
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Fr= |
200 |
N |
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l= |
0,121141898026825 |
m= |
121,141898026825 |
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obliczamy średnice pokrętła |
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Przyjmuję materiał pokrętła: stalE 360 (St 7), dla której kgj=130MPa. |
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130 |
Mpa |
130000000 |
Pa |
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11247878,2220223 |
Pa= |
11,2478782220223 |
<130 |
Mpa |
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2,15404E-06 |
m3 |
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= |
0,012306286050574 |
m= |
12,3062860505741 |
mm |
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przyjmujemy średnicę pokrętła dpok= |
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14 |
mm |
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Mg=l*Fr |
24,228379605365 |
Nm |
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Zgrubienie śruby pod pokrętła |
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D=[1,2-1,4]d |
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0,0364 |
m= |
36,4 |
mm |
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Mo= |
24,228379605365 |
Nm |
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F= |
1101,28998206204 |
N= |
1,10128998206204 |
kN |
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Przyjmujemy dla połączenia wielowypustowego |
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z normy PN-63/M-85015 przyjmujemy |
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ho= |
0,002 |
m= |
2 |
mm |
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d= |
0,042 |
m |
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D= |
0,046 |
m |
|
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p= |
8343105,92471246 |
8,34310592471246 |
Mpa |
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n= |
8 |
z |
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b= |
8 |
mm |
|
ko= |
9000000 |
Pa= |
9 |
Mpa |
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a= |
0,4 |
mm |
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r= |
0,3 |
mm |
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p<ko |
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l= |
0,011 |
m |
|
|
8,34310592471246 |
< |
9 |
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Obliczam przekładnie |
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i= |
3 |
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de2=m*z= |
117 |
mm |
|
de1=m*z= |
39 |
mm |
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dae2=m*(z+2cosb)= |
|
118,900970526923 |
mm |
dae1=m*(z+2cosb1)= |
|
44,6924091992909 |
mm |
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|
dfe2=m*(z-2,5*cosb2)= |
|
114,623786841346 |
mm |
dfe1=m*(z-2,5*cosb1)= |
|
31,8844885008864 |
mm |
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δ2=63,43 |
71,565051177078 |
z= |
39 |
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dla koła 2 |
|
dla koła1 |
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δ1=26,57 |
18,434948822922 |
z= |
13 |
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m= |
3 |
|
|
0,048640147094755 |
|
0,04862661822384 |
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3,93 |
stopnia |
3,96 |
stopnia |
3,945 |
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2 |
75,510051177078 |
66,635051177078 |
|
0,060800183868444 |
|
0,0607832727798 |
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|
1 |
22,379948822922 |
13,504948822922 |
4,91 |
stopnia |
4,95 |
stopnia |
4,93 |
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ba2= |
75,495051177078 |
stopnia |
|
ba1= |
18,4835754411458 |
stopnia |
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bf2= |
66,655051177078 |
stopnia |
|
bf1= |
13,484948822922 |
stopnia |
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|
Re2=de2/2sinb2= |
|
61,6774450569763 |
mm |
Re1=de1/2sinb1= |
|
61,6946049217385 |
mm |
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b2=(0,25-0,33)Re |
|
20,3535568688022 |
mm |
b2=(0,25-0,33)Re |
|
20,3592196241737 |
mm |
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ha= |
3 |
mm |
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hf=1,25*m= |
3,75 |
mm |
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obliczam wysięgnik podnośnika |
|
|
zakładam stal E360 (st7) |
|
kgj= |
130 |
Mpa= |
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|
130000000 |
Pa |
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Q= |
5000 |
N |
|
l1= |
40 |
mm= |
0,04 |
m |
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l2 |
40 |
mm= |
0,04 |
m |
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l3= |
120 |
mm= |
0,12 |
m |
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|
obliczam moment gnący x-x |
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Mg=Q*l2 |
200 |
Nm |
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obliczam miń wskaznik przekroju |
|
|
zakładam gr scianki wspornika g1= |
|
|
3 |
mm= |
0,003 |
m |
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c-ilośś scianek |
|
a-wysokość wysięgnika |
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0,027735009811262 |
m= |
27,7350098112615 |
mm |
przyjmuje a= |
28 |
mm= |
0,028 |
m |
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Wx=2*c*g1*(1/6)*a*a<Mg/kgj |
|
|
1,53846153846154E-06 |
< |
1,53846153846154E-06 |
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