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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32 28. Zadatak Odrediti potrebne duljine l1 i l2 zavara spajanjem kutnih metalnih profila s plo�om. ( )( ) ( ) ( ) cm l cm l cm l cm l l cm l l l l h h l l h l h l h a l h a l h F h F M cm t F l l t l l F al F A F MPa cm h cm h cm t kN F dop dop A dop dop dop 5 , 5 5 , 13 19 5 , 13 4 , 1 19 19 4 , 1 19 4 , 0 19 4 , 0 4 , 0 6 , 3 4 , 1 2 2 0 19 10 90 10 5 , 0 4 , 1 10 120 7 , 0 2 45 cos 2 90 4 , 1 6 , 3 5 , 0 120 1 2 2 2 2 2 1 2 1 1 2 2 1 2 2 1 1 2 2 1 1 2 2 1 1 6 2 3 2 1 2 1 2 1 = − = → = = → = → = + → = + = → ≅ = = ⇒ ⋅ = ⋅ ⇒ ⇒ = → ⋅ = ⋅ → = ∑ = ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ≥ + ⇒ ≤ ° + = ∑ = = = = = = = − τ τ τ τ τ τ l 1 l 2 h 1 h2 t a F

t 29. Zadatak Odrediti potreban broj zakovica promjera 20 mm, ako je F F A F/2 F/2 F = 300kN 1 = 8mm t = 12mm τ σ dop dop τ = = 100MPa = 280MPa s F A F nmA F ndt F F = ≤ τ 2 d π nm 4 F/2 F/2 d n – broj zakovica m – reznost zakovica n ≥ 5 → n = 6 → 2n = 12 → 2reda σ = p = = 1 dop 3 300 ⋅10 = −3 6 ⋅ 20 ⋅10 ⋅12 ⋅10 F → n ≥ 2 d π m τ 4 −3 dop = 208MPa ≤ σ = dop t 1 t t 2 2 −3 ( 20 ⋅10 ) 4 F 300 ⋅10 2 = 280MPa 3 F π 100 ⋅10 6 33

t<br />

29. Zadatak<br />

Odrediti potreban broj zakovica promjera 20 mm, ako je<br />

F<br />

F<br />

A<br />

F/2<br />

F/2<br />

F = 300kN<br />

1<br />

= 8mm<br />

t = 12mm<br />

τ<br />

σ<br />

dop<br />

dop<br />

τ =<br />

= 100MPa<br />

= 280MPa<br />

s<br />

F<br />

A<br />

F<br />

nmA<br />

F<br />

ndt<br />

F<br />

F<br />

= ≤ τ 2<br />

d π<br />

nm<br />

4<br />

F/2<br />

F/2<br />

d<br />

n – broj zakovica<br />

m – reznost zakovica<br />

n ≥ 5 → n = 6 → 2n<br />

= 12 → 2reda<br />

σ =<br />

p<br />

=<br />

=<br />

1<br />

dop<br />

3<br />

300 ⋅10<br />

=<br />

−3<br />

6 ⋅ 20 ⋅10<br />

⋅12<br />

⋅10<br />

F<br />

→ n ≥ 2<br />

d π<br />

m τ<br />

4<br />

−3<br />

dop<br />

= 208MPa<br />

≤ σ<br />

=<br />

dop<br />

t 1<br />

t<br />

t 2<br />

2<br />

−3<br />

( 20 ⋅10<br />

)<br />

4<br />

F<br />

300 ⋅10<br />

2<br />

= 280MPa<br />

3<br />

F<br />

π<br />

100 ⋅10<br />

6<br />

33

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