SKRIPTA RIJEÅ ENIH ZADATAKA IZ OTPORNOSTI MATERIJALA
SKRIPTA RIJEÅ ENIH ZADATAKA IZ OTPORNOSTI MATERIJALA SKRIPTA RIJEÅ ENIH ZADATAKA IZ OTPORNOSTI MATERIJALA
66 c) popre�ni presjek oblika I-profila ( ) ( ) mm b mm t h mm t USVOJENO mm t M t M t t t y I W t t t t t I dop x dop x x x x 160 , 400 25 , 16 39 , 15 100 9 , 346 10 56 , 126 9 , 346 9 , 346 5 , 13 8 , 4683 8 , 4683 12 25 9 12 27 10 3 6 3 max max 3 4 max 4 3 3 = = = = → ≥ ⋅ ⋅ = ≥ → ≥ = = = = − = σ σ Provjera dimenzija prema dopuštenom tangencijalnom naprezanju : 2 2 max 2 4 4 3 max 2 max max max max 60 53 , 19 53280 16 3 , 4683 10 5 , 112 53280 4 2 2 mm N mm N mm mm N mm t h t h t h bt b S b S I Q dop zy zy x dop x x y zy = < = ⋅ ⋅ ⋅ = = + + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≤ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = τ τ τ τ τ
45. Zadatak Za zadani nosa� i optere�enje odrediti maksimalna normalna i posmi�na naprezanja u visini osi “z“te smjer i veli�inu glavnih naprezanja u to�ki 3 popre�nog presjeka nad ležajem B. Položaj težišta popre�nog presjeka nosa�a : ⎛10 ⋅ 60 ⎞ 20 ⋅ 60 ⋅ 70 + 2⎜ ⎟ ⋅ 40 + 10 ⋅ 60 ⋅30 ∑ Ai yi 2 126000 yT = = ⎝ ⎠ = = 52, 5cm ∑ Ai ⎛10 ⋅ 60 ⎞ 2400 20 ⋅ 60 + 2⎜ ⎟ + 10 ⋅ 60 ⎝ 2 ⎠ T ( x , y ) = T T T ( 30; 52, 5) Moment površine drugog reda : I I I z z z = I z1 + A y 60 ⋅ 20 = 12 1 3 2 1 + 2 4 2 ( I + A y ) z2 + 60 ⋅ 20 ⋅ 4 = 110, 5⋅10 cm Reakcije i unutarnje sile : ∑ F F = 0 → F + F = 2F + qb − F 2 2 + I z3 2 ( 70 − 52, 5) + 2⎜ + ⋅ 60 − 52, 5 ⎟ + + 10 ⋅ 60 ⋅ ( 30 − 52, 5) 8 = 110, 5⋅10 mm − 2F − q ⋅b = 0 4 + A 3 y 2 3 ⎛10 ⋅ 60 ⎜ ⎝ 36 b ∑ M B = 0 → FA ⋅3a − F ⋅ 2a − F ⋅ a + q ⋅b ⋅ = 0 2 2 b 3Fa − q F 2 A = 3a 2 3 3⋅100 ⋅ 2 − 40 = 2 3⋅ 2 → FA = 70kN B y A A B = 2 ⋅100 + 40 ⋅3 − 70 → F B 3 10 ⋅ 60 2 = 250kN ⎛ 2 ⎜ ⎝ 3 ⎞ ⎟ ⎠ 2 ⎞ ⎟ ⎠ 10 ⋅ 60 12 3 67 2
- Page 15 and 16: 15 Sile u štapovima : ( ) ( ) ( )
- Page 17 and 18: 17 Iz uvjeta ravnoteže sila : ( )
- Page 19 and 20: 15. Zadatak Za zadano stanje naprez
- Page 21 and 22: 16. Zadatak Za zadano stanje naprez
- Page 23 and 24: 17. Zadatak Plo�a je izložena dj
- Page 25 and 26: 20. Zadatak U to�ki A napregnutog
- Page 27 and 28: 22. Zadatak Pravokutna plo�ica di
- Page 29 and 30: 24. Zadatak Odrediti glavna napreza
- Page 31 and 32: 27. Zadatak Za optere�enje na tla
- Page 33 and 34: t 29. Zadatak Odrediti potreban bro
- Page 35 and 36: 31. Zadatak Metalna vilica i plo�
- Page 37 and 38: 37 33. Zadatak Odrediti potrebnu du
- Page 39 and 40: 39 Moment od sile FMi s obzirom na
- Page 41 and 42: ∆D ( 1) 1 ( 1−ν ) 1 1 ∆D 2 =
- Page 43 and 44: 43 b) Nosivost zakovice ( ) ( ) 2 m
- Page 45 and 46: 45 ( ) ( ) Nmm kNm U U I G M I G M
- Page 47 and 48: 47 ( ) ( ) ° − = → ° − =
- Page 49 and 50: ( 2) ⎛ M A − ⎜ ⎝ GI pI M A
- Page 51 and 52: 41. Zadatak Odrediti potrebne dimen
- Page 53 and 54: 42. Zadatak Za nosa� oblika I-pro
- Page 55 and 56: 55 Iz kriterija �vrsto�e za mak
- Page 57 and 58: 43. Zadatak Za nosa� zadan i opte
- Page 59 and 60: II. podru�je: 2.2 m < x < 4 m III
- Page 61 and 62: c) dimenzioniranje nosa�a -��
- Page 63 and 64: - tangencijalno naprezanje 63
- Page 65: Maksimalni moment savijanja na udal
- Page 69 and 70: 69 Posmi���������
- Page 71 and 72: Momenti savijanja na pojedinim segm
- Page 73 and 74: 73 Kut zaokreta u to�ki D : ( ) (
- Page 75 and 76: 75 Položaj neutralne osi : ° −
- Page 77 and 78: 77 ( ) ( ) ( ) ( ) ( ) ( ) T T y y
- Page 79 and 80: 79 To�ka 1 : MPa mm N b I S T MPa
45. Zadatak<br />
Za zadani nosa� i optere�enje odrediti maksimalna normalna i posmi�na naprezanja u visini osi<br />
“z“te smjer i veli�inu glavnih naprezanja u to�ki 3 popre�nog presjeka nad ležajem B.<br />
Položaj težišta popre�nog presjeka nosa�a :<br />
⎛10 ⋅ 60 ⎞<br />
20 ⋅ 60 ⋅ 70 + 2⎜<br />
⎟ ⋅ 40 + 10 ⋅ 60 ⋅30<br />
∑ Ai<br />
yi<br />
2<br />
126000<br />
yT<br />
= =<br />
⎝ ⎠<br />
= = 52,<br />
5cm<br />
∑ Ai<br />
⎛10 ⋅ 60 ⎞<br />
2400<br />
20 ⋅ 60 + 2⎜<br />
⎟ + 10 ⋅ 60<br />
⎝ 2 ⎠<br />
T ( x , y ) = T<br />
T<br />
T<br />
( 30;<br />
52,<br />
5)<br />
Moment površine drugog reda :<br />
I<br />
I<br />
I<br />
z<br />
z<br />
z<br />
= I<br />
z1<br />
+ A y<br />
60 ⋅ 20<br />
=<br />
12<br />
1<br />
3<br />
2<br />
1<br />
+ 2<br />
4<br />
2 ( I + A y )<br />
z2<br />
+ 60 ⋅ 20 ⋅<br />
4<br />
= 110,<br />
5⋅10<br />
cm<br />
Reakcije i unutarnje sile :<br />
∑ F<br />
F<br />
= 0 → F<br />
+ F<br />
= 2F<br />
+ qb − F<br />
2<br />
2<br />
+ I<br />
z3<br />
2 ( 70 − 52,<br />
5)<br />
+ 2⎜<br />
+ ⋅ 60 − 52,<br />
5 ⎟ + + 10 ⋅ 60 ⋅ ( 30 − 52,<br />
5)<br />
8<br />
= 110,<br />
5⋅10<br />
mm<br />
− 2F<br />
− q ⋅b<br />
= 0<br />
4<br />
+ A<br />
3<br />
y<br />
2<br />
3<br />
⎛10<br />
⋅ 60<br />
⎜<br />
⎝<br />
36<br />
b<br />
∑<br />
M B = 0 → FA<br />
⋅3a<br />
− F ⋅ 2a<br />
− F ⋅ a + q ⋅b<br />
⋅ = 0<br />
2<br />
2<br />
b<br />
3Fa<br />
− q<br />
F<br />
2<br />
A =<br />
3a<br />
2<br />
3<br />
3⋅100<br />
⋅ 2 − 40<br />
=<br />
2<br />
3⋅<br />
2<br />
→ FA<br />
= 70kN<br />
B<br />
y<br />
A<br />
A<br />
B<br />
= 2 ⋅100<br />
+ 40 ⋅3<br />
− 70 → F<br />
B<br />
3<br />
10 ⋅ 60<br />
2<br />
= 250kN<br />
⎛ 2<br />
⎜<br />
⎝ 3<br />
⎞<br />
⎟<br />
⎠<br />
2<br />
⎞<br />
⎟<br />
⎠<br />
10 ⋅ 60<br />
12<br />
3<br />
67<br />
2
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