SKRIPTA RIJEÅ ENIH ZADATAKA IZ OTPORNOSTI MATERIJALA
SKRIPTA RIJEÅ ENIH ZADATAKA IZ OTPORNOSTI MATERIJALA
SKRIPTA RIJEÅ ENIH ZADATAKA IZ OTPORNOSTI MATERIJALA
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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