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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GRA�EVINSKI FAKULTET<br />
SVEU�ILIŠTA U RIJECI<br />
<strong>SKRIPTA</strong> RIJEŠENIH <strong>ZADATAKA</strong> <strong>IZ</strong><br />
<strong>OTPORNOSTI</strong> <strong>MATERIJALA</strong><br />
NEIRA TORI�,dipl.ing.gra�.
1. Zadatak<br />
Izra�unati naprezanja i nacrtati dijagram naprezanja na konzolnom nosa�u ako je zadano :<br />
F<br />
F<br />
A = 10cm<br />
a = 2m<br />
b = 1m<br />
N<br />
1Pa<br />
= 1 2<br />
m<br />
N<br />
1MPa<br />
= 1<br />
mm<br />
2<br />
Vrijednosti uzdužnih sila na pojedinim segmentima :<br />
N<br />
N<br />
N<br />
1<br />
3<br />
III<br />
II<br />
I<br />
= F<br />
2<br />
= 4 ⋅10<br />
E = 2,<br />
2<br />
= −F<br />
= F<br />
3<br />
= 2 ⋅10<br />
⋅10<br />
1<br />
1<br />
2<br />
4<br />
N<br />
11<br />
Pa<br />
= 10<br />
= −2<br />
⋅10<br />
= −F<br />
+ F = 0<br />
2<br />
= 4 ⋅10<br />
4<br />
4<br />
N<br />
−3<br />
N<br />
4<br />
N<br />
Vrijednosti naprezanja na pojedinim segmentima :<br />
N III σ III =<br />
A<br />
6<br />
= −20<br />
⋅10<br />
Pa = −20MPa<br />
N II σ I =<br />
A<br />
= 0<br />
N I σ I =<br />
A<br />
6<br />
= 40 ⋅10<br />
Pa = 40MPa<br />
m<br />
2<br />
2
2. Zadatak<br />
Odrediti naprezanja u štapu pri promjeni temperature za +�T :<br />
Rješenje :<br />
Ukoliko nema vanjskog otpora izduženju štapa, nema niti naprezanja u štapu. To zna�i da se<br />
štap AC može slobodno izdužiti pod utjecajem temperature za veli�inu �. Sve dok je ∆l t ≤ δ u<br />
štapu su naprezanja jednaka nuli.<br />
Ukoliko je tendencija štapa da se izduži za ∆l t > δ , aktivira se u štapu sila koja opet<br />
“ vra���“ štap na realnu dužinu AB, kao što se vidi na slici. Može se zaklju�iti i da �e sila u štapu<br />
biti tla�na , odnosno s predznakom – (minus).<br />
∆l<br />
t<br />
∆l<br />
t<br />
∆l<br />
t<br />
δ = ∆l<br />
∆l<br />
t<br />
∆l<br />
B<br />
= α l ∆T<br />
+ α l ∆T<br />
1 1<br />
≤ δ → ε > 0,<br />
σ = 0<br />
> δ → ε > 0,<br />
σ ≠ 0<br />
t<br />
= ( α l<br />
=<br />
F<br />
E<br />
− ∆l<br />
1 1<br />
B 1<br />
1<br />
l<br />
A<br />
1<br />
B<br />
K<br />
2<br />
( 1)<br />
2<br />
2<br />
2<br />
+ α l ) ∆T<br />
K<br />
2<br />
FBl<br />
2<br />
+<br />
E A<br />
2<br />
K<br />
( 2)<br />
( 3)<br />
��– realno izdužrnje<br />
��B – nerealno izduženje<br />
( 2)<br />
; ( 3)<br />
→ ( 1)<br />
σ<br />
σ<br />
x1<br />
x2<br />
F<br />
= −<br />
A<br />
B<br />
1<br />
F<br />
= −<br />
A<br />
B<br />
2<br />
⇒ F<br />
B<br />
=<br />
[ ( α<br />
l + α l ) ∆T<br />
− δ ]<br />
1 1<br />
2 2<br />
⎛ l<br />
l1<br />
⎜<br />
⎜1+<br />
⎝ l<br />
2<br />
1<br />
E1A<br />
E A<br />
2<br />
1<br />
2<br />
E1A<br />
⎞<br />
⎟<br />
⎠<br />
1<br />
3
3. Zadatak<br />
Odrediti pomak to�ke C ukoliko se temperatura štapa 1 pove�a za �T :<br />
Rješenje :<br />
ε<br />
∆l<br />
sin 2β<br />
= 1 ; ∆l<br />
δ<br />
δ<br />
t<br />
C<br />
∆lt<br />
=<br />
l<br />
u = δ<br />
v = δ<br />
⇒ ∆l<br />
∆l<br />
α∆<br />
= 1 Tl<br />
=<br />
sin 2β<br />
sin 2β<br />
C<br />
C<br />
C<br />
cos β<br />
sinβ<br />
t<br />
= ε l = α∆Tl<br />
2<br />
t<br />
= 0<br />
4
4. Zadatak<br />
Odrediti naprezanja u štapovima ukoliko se temperatura štapa 2 pove�a za �T :<br />
Rješenje<br />
Iz uvjeta ravnoteže sila :<br />
ΣY = 0 → 2F1<br />
cosα<br />
− F2<br />
= 0K<br />
Iz plana pomaka :<br />
l<br />
∆l<br />
2 1<br />
cosα = = →∆l<br />
1 = ∆l2<br />
l1<br />
∆l2<br />
Iz Hookovog zakona :<br />
F1l1<br />
∆l<br />
1 = K(<br />
3)<br />
E1A1<br />
F2l<br />
2 ∆l2 = + α 2 ⋅ ∆T<br />
⋅l<br />
2 K<br />
E A<br />
2<br />
( 1)<br />
; ( 3);<br />
( 4)<br />
2<br />
→<br />
( 2)<br />
( 1)<br />
cosαK<br />
( 4)<br />
( 2)<br />
2<br />
α 2∆TE<br />
1 A1<br />
cos α<br />
⇒ F1<br />
=<br />
E1<br />
A1<br />
3<br />
1 − 2 cos α<br />
E A<br />
3<br />
2α<br />
2∆TE1<br />
A1<br />
cos α<br />
( 1)<br />
→ F2<br />
= 2F1<br />
cosα<br />
⇒ F2<br />
=<br />
E1<br />
A1<br />
3<br />
1 − 2 cos α<br />
E2<br />
A2<br />
Zadatak F 5<br />
1 F2<br />
σ<br />
1 = ; σ 2 = −<br />
A A<br />
1<br />
2<br />
2<br />
2<br />
Iz plana pomaka i optere�enja (ovdje promjena<br />
temperature) možemo zaklju�iti da se štap 2 nastoji<br />
izdužiti, no njegovo potpuno izduženje sprje�avaju druga<br />
dva štapa spojena u �voru D, koji nisu direktno optere�eni i<br />
nemaju tendenciju izduženja. Štap 2 se izduži za �lt pri<br />
����������������������������������������������������������<br />
���������������������������������������2����������������������<br />
D' u D''.<br />
Budu�i da su sva tri štapa me�usobno spojena u �voru D,<br />
izduženjem štapa 2 prisilno se izdužuju i štapovi s brojem<br />
1. Kao posljedica izduženja u njima se javlja vla�na sila<br />
F1(vla�na sila uzrokuje izduženje ako nema nikakvog drugog<br />
optere�enja).<br />
5
5. Zadatak<br />
Odrediti naprezanja u štapovima ako je srednji štap kra�i za � od predvi�ene duljine l.<br />
Rješenje<br />
Iz uvjeta ravnoteže sila :<br />
ΣY<br />
= 0 → F2<br />
− 2F1<br />
cos α = 0 → F2<br />
= 2F1<br />
cos αK<br />
Iz plana pomaka :<br />
l<br />
cos α =<br />
l<br />
δ = ∆l<br />
2<br />
( 1)<br />
→ ( 2)<br />
F<br />
1<br />
( 1)<br />
=<br />
l<br />
2<br />
⇒ F<br />
2<br />
1<br />
∆l1<br />
=<br />
δ − ∆l<br />
2<br />
∆l1<br />
F2l2<br />
+ =<br />
cos α E A<br />
2<br />
2<br />
F1l1<br />
+<br />
E A<br />
1 1 2 2<br />
3<br />
( 2E<br />
A cos α + E A )<br />
2<br />
E<br />
F l<br />
⇒ δ = 2<br />
E A<br />
A<br />
1<br />
=<br />
l<br />
2<br />
E<br />
1<br />
A<br />
2E<br />
1<br />
1<br />
1<br />
cos<br />
1 1 2 2<br />
3<br />
( 2E<br />
A cos α + E A )<br />
1<br />
A<br />
1<br />
1 2<br />
2<br />
E<br />
2<br />
δ cos<br />
A<br />
2<br />
F l<br />
cos α +<br />
E A<br />
α<br />
2<br />
2<br />
δ cos<br />
3<br />
2<br />
α<br />
1 1<br />
1<br />
2<br />
K<br />
α<br />
1<br />
1<br />
cos<br />
( 2)<br />
( 1)<br />
⎛ l2<br />
l1<br />
= F1<br />
⎜<br />
⎜2<br />
cos α +<br />
α ⎝ E2A<br />
2 E1A<br />
F1<br />
σ 1 = −<br />
A<br />
1<br />
∆l<br />
1<br />
∆l<br />
2<br />
F2<br />
σ 2 = −<br />
A<br />
F1l1<br />
=<br />
E A<br />
2<br />
1<br />
2<br />
1<br />
F2l2<br />
=<br />
E A<br />
2<br />
1<br />
1<br />
cos<br />
⎞<br />
⎟<br />
α ⎠<br />
6
6. Zadatak<br />
Odrediti naprezanja u štapovima 1 i 2 te vertikalni pomak to�ke D.<br />
Rješenje<br />
Sustav je jedanput stati�ki neodre�en (ima jednu prekomjernu veli�inu), odnosno nepoznate su<br />
4 veli�ine ( RH , RV , S1 , S2 ).<br />
Uz tri uvjeta ravnoteže postavljamo dodatnu jednadžbu na deformiranom sustavu na principu<br />
sli�nosti trokuta.<br />
Iz uvjeta ravnoteže sila :<br />
ΣM<br />
A<br />
= 0 → Fl − S a<br />
ΣX<br />
= 0 → R<br />
ΣY<br />
= 0 → S<br />
2<br />
H<br />
Iz plana pomaka :<br />
∆l1<br />
δ C ≡ ∆l2;<br />
δ B =<br />
sinα<br />
Iz sli�nosti trokuta :<br />
δ C δ B = K(<br />
4)<br />
a a<br />
2<br />
1<br />
1<br />
2<br />
2<br />
− S cosα<br />
= 0 → R<br />
− F + S sinα<br />
+ R<br />
Iz Hookovog zakona :<br />
S1l1<br />
S2l<br />
2<br />
l 1 = ; ∆l2<br />
=<br />
E A E A<br />
1<br />
1<br />
2<br />
1<br />
− S a sinα<br />
= 0K<br />
∆ → ( 4)<br />
2<br />
1<br />
1<br />
V<br />
H<br />
1<br />
= 0K<br />
( 1)<br />
= S cosα<br />
K<br />
( 3)<br />
S<br />
S<br />
1<br />
1<br />
∆l<br />
a<br />
( 2)<br />
= S<br />
2<br />
→<br />
2<br />
2<br />
( 1)<br />
S1<br />
σ1<br />
=<br />
A<br />
1<br />
S2<br />
σ 2 =<br />
A<br />
2<br />
δ<br />
=<br />
l<br />
E1A1<br />
l2<br />
a1<br />
sinα<br />
E A l a<br />
2<br />
2<br />
⇒ S<br />
2<br />
1<br />
D ⇒ δ D = ∆ 2 =<br />
a2<br />
2<br />
Fl<br />
=<br />
⎡ E1A1<br />
l2<br />
a<br />
a2<br />
⎢1+<br />
⎣ E2<br />
A2<br />
l1<br />
a<br />
l<br />
l<br />
2<br />
2<br />
2<br />
1<br />
2<br />
2<br />
S2l2l<br />
E A a<br />
2<br />
sin<br />
2<br />
⎤<br />
α ⎥<br />
⎦<br />
7
8<br />
7. Zadatak<br />
Dimenzionirati štapove AB i DG kružnog popre�nog presjeka napravljene od �elika, te odrediti<br />
njihova produljenja.<br />
m<br />
c<br />
m<br />
b<br />
m<br />
a<br />
kN<br />
F<br />
Mpa<br />
E<br />
Mpa<br />
dop<br />
1<br />
2<br />
3<br />
100<br />
10<br />
2<br />
140<br />
5<br />
=<br />
=<br />
=<br />
=<br />
⋅<br />
=<br />
=<br />
σ<br />
)<br />
15<br />
,<br />
6<br />
(<br />
28<br />
61<br />
,<br />
2<br />
4<br />
4<br />
36<br />
,<br />
5<br />
10<br />
536<br />
,<br />
0<br />
10<br />
140<br />
10<br />
75<br />
)<br />
85<br />
,<br />
13<br />
(<br />
42<br />
01<br />
,<br />
4<br />
4<br />
4<br />
6<br />
,<br />
12<br />
10<br />
26<br />
,<br />
1<br />
10<br />
140<br />
10<br />
8<br />
,<br />
176<br />
8<br />
,<br />
176<br />
45<br />
sin<br />
3<br />
5<br />
75<br />
45<br />
sin<br />
3<br />
5<br />
0<br />
3<br />
45<br />
sin<br />
5<br />
0<br />
75<br />
100<br />
4<br />
3<br />
4<br />
3<br />
0<br />
3<br />
4<br />
0<br />
2<br />
2<br />
1<br />
2<br />
2<br />
2<br />
2<br />
2<br />
2<br />
2<br />
3<br />
6<br />
3<br />
2<br />
2<br />
2<br />
1<br />
1<br />
1<br />
1<br />
2<br />
1<br />
1<br />
2<br />
2<br />
3<br />
6<br />
3<br />
1<br />
1<br />
2<br />
1<br />
1<br />
2<br />
2<br />
2<br />
cm<br />
A<br />
mm<br />
d<br />
usvojeno<br />
cm<br />
A<br />
d<br />
d<br />
A<br />
cm<br />
m<br />
S<br />
A<br />
cm<br />
A<br />
mm<br />
d<br />
usvojeno<br />
cm<br />
A<br />
d<br />
d<br />
A<br />
cm<br />
m<br />
S<br />
A<br />
S<br />
A<br />
A<br />
S<br />
ranje<br />
Dimenzioni<br />
kN<br />
S<br />
S<br />
S<br />
S<br />
M<br />
kN<br />
F<br />
S<br />
F<br />
S<br />
M<br />
dop<br />
pot<br />
dop<br />
pot<br />
dop<br />
pot<br />
dop<br />
x<br />
C<br />
H<br />
=<br />
→<br />
=<br />
→<br />
→<br />
=<br />
=<br />
⇒<br />
=<br />
=<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
≥<br />
=<br />
→<br />
=<br />
→<br />
→<br />
=<br />
=<br />
⇒<br />
=<br />
=<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
≥<br />
≥<br />
⇒<br />
≤<br />
=<br />
=<br />
°<br />
⋅<br />
=<br />
°<br />
⋅<br />
=<br />
→<br />
=<br />
⋅<br />
°<br />
⋅<br />
−<br />
⋅<br />
⇒<br />
=<br />
∑<br />
=<br />
=<br />
=<br />
→<br />
=<br />
⋅<br />
−<br />
⋅<br />
⇒<br />
=<br />
∑<br />
−<br />
−<br />
π<br />
π<br />
σ<br />
π<br />
π<br />
σ<br />
σ<br />
σ<br />
σ
9<br />
cm<br />
m<br />
A<br />
E<br />
l<br />
S<br />
l<br />
cm<br />
m<br />
A<br />
E<br />
l<br />
S<br />
l<br />
MPa<br />
MPa<br />
A<br />
S<br />
MPa<br />
MPa<br />
A<br />
S<br />
Kontrola<br />
dop<br />
x<br />
dop<br />
x<br />
12<br />
,<br />
0<br />
10<br />
195<br />
,<br />
12<br />
10<br />
15<br />
,<br />
6<br />
10<br />
2<br />
2<br />
10<br />
75<br />
18<br />
,<br />
0<br />
10<br />
056<br />
,<br />
18<br />
707<br />
,<br />
0<br />
10<br />
85<br />
,<br />
13<br />
10<br />
2<br />
2<br />
10<br />
8<br />
,<br />
176<br />
140<br />
9<br />
,<br />
121<br />
10<br />
15<br />
,<br />
6<br />
10<br />
75<br />
140<br />
6<br />
,<br />
127<br />
10<br />
85<br />
,<br />
13<br />
10<br />
8<br />
,<br />
176<br />
4<br />
4<br />
11<br />
3<br />
2<br />
2<br />
2<br />
2<br />
2<br />
4<br />
4<br />
11<br />
3<br />
1<br />
1<br />
1<br />
1<br />
1<br />
4<br />
3<br />
2<br />
2<br />
2<br />
4<br />
3<br />
1<br />
1<br />
1<br />
=<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
=<br />
∆<br />
=<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
=<br />
∆<br />
=<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
=<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
−<br />
−<br />
−<br />
−<br />
−<br />
−<br />
σ<br />
σ<br />
σ<br />
σ<br />
p<br />
p
8. Zadatak<br />
Odrediti pomak to�aka B i D na zadanom sustavu.<br />
Σ<br />
M A<br />
= 0<br />
⇒ F<br />
( a + b)<br />
( a + b)<br />
S ⋅ h F ⋅ h ⋅<br />
∆h<br />
= =<br />
E ⋅ A a ⋅ E ⋅ A<br />
( a + b)<br />
∆h<br />
∆B<br />
F ⋅ h ⋅<br />
= ⇒ ∆B<br />
= 2<br />
a a + b a ⋅ E ⋅ A<br />
a + b<br />
− Sa = 0 → S = F<br />
a<br />
2<br />
10
11<br />
9. Zadatak<br />
( )<br />
( )<br />
( ) ( )<br />
MPa<br />
m<br />
N<br />
A<br />
S<br />
MPa<br />
m<br />
N<br />
A<br />
S<br />
N<br />
S<br />
N<br />
S<br />
A<br />
l<br />
A<br />
l<br />
F<br />
S<br />
a<br />
S<br />
A<br />
l<br />
a<br />
A<br />
l<br />
S<br />
Fa<br />
A<br />
l<br />
A<br />
l<br />
S<br />
S<br />
A<br />
E<br />
l<br />
S<br />
A<br />
E<br />
l<br />
S<br />
l<br />
l<br />
a<br />
l<br />
a<br />
l<br />
a<br />
S<br />
a<br />
S<br />
Fa<br />
M<br />
N<br />
F<br />
E<br />
E<br />
m<br />
cm<br />
A<br />
m<br />
cm<br />
A<br />
m<br />
l<br />
m<br />
l<br />
A<br />
7<br />
,<br />
8<br />
10<br />
5<br />
,<br />
1<br />
35<br />
,<br />
1304<br />
43<br />
,<br />
10<br />
10<br />
1<br />
5<br />
,<br />
1043<br />
35<br />
,<br />
1304<br />
5<br />
,<br />
1043<br />
3<br />
3<br />
4<br />
0<br />
3<br />
3<br />
4<br />
1<br />
2<br />
2<br />
3<br />
2<br />
2<br />
3<br />
2<br />
3<br />
3<br />
2<br />
1<br />
0<br />
3<br />
2<br />
0<br />
6000<br />
10<br />
5<br />
,<br />
1<br />
5<br />
,<br />
1<br />
10<br />
1<br />
1<br />
8<br />
,<br />
1<br />
1<br />
2<br />
4<br />
2<br />
2<br />
2<br />
2<br />
4<br />
1<br />
1<br />
1<br />
2<br />
1<br />
2<br />
1<br />
1<br />
2<br />
2<br />
2<br />
2<br />
1<br />
1<br />
2<br />
2<br />
2<br />
1<br />
1<br />
2<br />
2<br />
1<br />
2<br />
2<br />
2<br />
2<br />
1<br />
1<br />
1<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
4<br />
2<br />
2<br />
2<br />
4<br />
2<br />
1<br />
2<br />
1<br />
=<br />
⋅<br />
=<br />
=<br />
=<br />
⋅<br />
=<br />
=<br />
=<br />
=<br />
+<br />
=<br />
⇒<br />
=<br />
−<br />
−<br />
⇒<br />
→<br />
=<br />
⇒<br />
=<br />
∆<br />
=<br />
∆<br />
⇒<br />
∆<br />
=<br />
∆<br />
=<br />
−<br />
−<br />
⇒<br />
=<br />
Σ<br />
=<br />
=<br />
⋅<br />
=<br />
=<br />
⋅<br />
=<br />
=<br />
=<br />
=<br />
−<br />
−<br />
−<br />
−<br />
σ<br />
σ<br />
K<br />
K
10. Zadatak<br />
Kruta greda AB obješena je o tri �eli�na štapa istog popre�nog presjeka površine 10 cm 2 .<br />
Dužina štapova je h = 1 m, s tim da je srednji štap (2) napravljen kra�i od projektirane dužine za<br />
� = 0,6 mm. Odrediti sile u štapovima i izduženje srednjeg štapa ako je izvršena prinudna<br />
montaža sustava.<br />
E<br />
l<br />
1<br />
1<br />
A<br />
1<br />
= l<br />
= E<br />
2<br />
2<br />
= l<br />
3<br />
A<br />
2<br />
= h<br />
= E<br />
3<br />
A<br />
3<br />
= EA<br />
S1h<br />
∆l1<br />
= ; ∆l2<br />
EA<br />
S 2h<br />
= ; ∆l3<br />
EA<br />
S3h<br />
=<br />
EA<br />
∆l3<br />
∆l1<br />
= ⇒ ∆l3<br />
3a<br />
a<br />
S3h<br />
S1h<br />
= 3∆l1<br />
⇒ = 3 ⇒ S<br />
EA EA<br />
∆l<br />
a<br />
ΣM<br />
S<br />
S<br />
∆l<br />
∆ − ∆l<br />
=<br />
2a<br />
A<br />
= 90kN<br />
= 54kN<br />
⇒ 2∆l<br />
= ∆ − ∆l<br />
= 3S<br />
K<br />
( 1)<br />
S 2h<br />
S1h<br />
⇒ = ∆ − 2 ⇒ S<br />
EA EA<br />
= 0 → S a − S 2a<br />
+ S 3a<br />
= 0 ⇒ S = 2S<br />
− 3S<br />
K<br />
3<br />
S 2h<br />
90 ⋅10<br />
⋅1,<br />
00<br />
= = 11<br />
EA 2,<br />
1⋅10<br />
⋅1⋅10<br />
( 3)<br />
−4<br />
11 −3<br />
∆EA<br />
6 ⋅10<br />
⋅ 2,<br />
1⋅10<br />
⋅1⋅10<br />
3<br />
= c =<br />
= 126 ⋅10<br />
N = 126kN<br />
h<br />
1,<br />
00<br />
1 126<br />
( 1)<br />
, ( 2)<br />
→ ( 3)<br />
⇒ S1<br />
= 2c<br />
− 4S1<br />
− 9S1<br />
→ S1<br />
= c = = 18kN<br />
7 7<br />
3 3<br />
S1<br />
→ ( 1)<br />
⇒ S3<br />
= c = 126 = 54kN<br />
7 7<br />
2 5<br />
S1<br />
→ ( 2)<br />
⇒ S 2 = c − 2S1<br />
= c − c = 126 = 90kN<br />
7 7<br />
S = 18kN<br />
1<br />
2<br />
3<br />
1<br />
2<br />
2<br />
1<br />
2<br />
1<br />
3<br />
−3<br />
2<br />
1<br />
−4<br />
= 4,<br />
3⋅10<br />
m = 0,<br />
4mm<br />
3<br />
2<br />
1<br />
3<br />
2<br />
∆EA<br />
= − 2S1<br />
K<br />
h<br />
( 2)<br />
12
11. Zadatak<br />
Cilindri�ni stup promjera 4 cm, dužine 120 cm optere�en je u presjecima (1), (2) i (3) na<br />
udaljenostima zi = 0; 40 i 80 cm (i=1,2,3) od slobodnog kraja aksijalnim silamaFi(i=1,2,3) = 15 kN,<br />
10 kN i 5 kN.<br />
Izra�umati naprezanje u pojedinim dijelovima stupa i pomak slobodnog kraja.<br />
E = 2 ⋅10<br />
d = 4cm<br />
L<br />
L = 120cm;<br />
l1<br />
= l2<br />
= l3<br />
= l = = 40cm<br />
3<br />
2 2 −4<br />
d π 4 ⋅10<br />
π<br />
−4<br />
2<br />
A = = = 12,<br />
56 ⋅10<br />
m<br />
4 4<br />
K0<br />
≤ z < 40cm<br />
( 1)<br />
N<br />
σ<br />
( 1)<br />
( 1)<br />
( 2)<br />
N<br />
σ<br />
( 2)<br />
( 2)<br />
( 3)<br />
N<br />
σ<br />
( 3)<br />
( 3)<br />
∆l<br />
= ∆l<br />
k =<br />
∆l<br />
∆l<br />
∆l<br />
N<br />
l<br />
AE<br />
N<br />
N<br />
∆l<br />
= ∆l<br />
= −F<br />
= −15kN<br />
( 1)<br />
A<br />
K40<br />
≤ z<br />
= −F<br />
− F = −25kN<br />
( 2)<br />
A<br />
K80<br />
≤ z<br />
1<br />
2<br />
3<br />
=<br />
=<br />
= −F<br />
− F − F = −30kN<br />
=<br />
1<br />
( 3)<br />
A<br />
= k ⋅ N<br />
+ ∆l<br />
= k ⋅ N<br />
= k ⋅ N<br />
1<br />
5<br />
1<br />
1<br />
1<br />
1<br />
MPa<br />
−15⋅10<br />
=<br />
12,<br />
56 ⋅10<br />
2<br />
−8<br />
3<br />
( 1)<br />
= 0,<br />
159 ⋅10<br />
⋅ ( −15⋅10<br />
)<br />
−8<br />
3<br />
( 2)<br />
= 0,<br />
159 ⋅10<br />
⋅ ( − 25⋅10<br />
)<br />
−8<br />
3<br />
( 3)<br />
= 0,<br />
159 ⋅10<br />
⋅ ( − 30 ⋅10<br />
)<br />
+ ∆l<br />
< 80cm<br />
− 25⋅10<br />
=<br />
12,<br />
56 ⋅10<br />
3<br />
< 120cm<br />
− 30 ⋅10<br />
=<br />
12,<br />
56 ⋅10<br />
2<br />
2<br />
2<br />
2<br />
+ ∆l<br />
+ ∆l<br />
3<br />
3<br />
3<br />
=<br />
3<br />
−4<br />
3<br />
−4<br />
3<br />
−4<br />
l<br />
AE<br />
N<br />
m<br />
2<br />
N<br />
m<br />
2<br />
N<br />
m<br />
2<br />
3<br />
( N ( 1)<br />
+ N ( 2)<br />
+ N ( 3)<br />
) = ∑ k ⋅ N ( i )<br />
i=<br />
1<br />
= −0,<br />
112mm<br />
= −12<br />
⋅10<br />
6<br />
= −24<br />
⋅10<br />
Pa = −12MPa<br />
6<br />
= −20<br />
⋅10<br />
Pa = −20MPa<br />
0,<br />
4<br />
=<br />
= 0,<br />
159 ⋅10<br />
−4<br />
11<br />
12,<br />
56 ⋅10<br />
⋅ 2 ⋅10<br />
6<br />
=<br />
=<br />
=<br />
Pa = −24MPa<br />
−8<br />
m<br />
N<br />
−2,<br />
415⋅10<br />
−4,<br />
00 ⋅10<br />
−4,<br />
80 ⋅10<br />
−5<br />
−5<br />
−5<br />
m = −0,<br />
024mm<br />
m = −0,<br />
04mm<br />
m = −0,<br />
048mm<br />
13
12. Zadatak<br />
Dimenzionirati �eli�nu zategu kružnog pore�nog presjeka (A1) i drveni kosnik pravokutnog<br />
popre�nog presjeka (h=2b) površine A2 = 10A1 ako je F=155 kN a dopušteni naponi za �elik 12 ·10 7<br />
N/m 2 i drvo 6·10 6 N/m 2 .<br />
7 N<br />
σ 1dop<br />
= 12 ⋅10<br />
m<br />
6 N<br />
σ 2dop<br />
= 6 ⋅10<br />
2<br />
m<br />
11 N<br />
E1<br />
= 2 ⋅10<br />
2<br />
m<br />
10 N<br />
E2<br />
= 1⋅10<br />
2<br />
m<br />
E = 20E<br />
A<br />
1<br />
2<br />
= 10 A<br />
h = 2b<br />
2<br />
1<br />
Plan pomaka :<br />
2<br />
l<br />
/ 2 3<br />
sinα<br />
= =<br />
5l<br />
/ 6 5<br />
2l<br />
/ 3 4<br />
cosα<br />
= =<br />
5l<br />
/ 6 5<br />
l<br />
l<br />
2<br />
1<br />
2<br />
= l<br />
3<br />
5<br />
= l<br />
6<br />
sin β = cos β =<br />
14<br />
2<br />
2
15<br />
Sile u štapovima :<br />
( )<br />
( )<br />
( )<br />
( )<br />
kN<br />
S<br />
kN<br />
S<br />
F<br />
F<br />
F<br />
S<br />
S<br />
S<br />
S<br />
l<br />
A<br />
E<br />
l<br />
A<br />
E<br />
S<br />
S<br />
l<br />
A<br />
E<br />
l<br />
A<br />
E<br />
S<br />
A<br />
E<br />
l<br />
S<br />
A<br />
E<br />
l<br />
S<br />
A<br />
E<br />
l<br />
S<br />
l<br />
A<br />
E<br />
l<br />
S<br />
l<br />
l<br />
l<br />
l<br />
S<br />
l<br />
S<br />
Fl<br />
M<br />
D<br />
C<br />
C<br />
D<br />
D<br />
c<br />
A<br />
4<br />
,<br />
353<br />
;<br />
9<br />
,<br />
58<br />
38<br />
,<br />
0<br />
2<br />
6<br />
1<br />
5<br />
12<br />
sin<br />
4<br />
sin<br />
3<br />
1<br />
1<br />
6<br />
2<br />
5<br />
2<br />
6<br />
10<br />
2<br />
3<br />
3<br />
5<br />
20<br />
2<br />
sin<br />
sin<br />
2<br />
sin<br />
2<br />
sin<br />
2<br />
sin<br />
sin<br />
;<br />
sin<br />
sin<br />
2<br />
2<br />
3<br />
2<br />
3<br />
1<br />
1<br />
0<br />
3<br />
1<br />
sin<br />
3<br />
2<br />
sin<br />
0<br />
1<br />
2<br />
2<br />
2<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
2<br />
2<br />
1<br />
1<br />
1<br />
2<br />
2<br />
2<br />
2<br />
1<br />
1<br />
1<br />
1<br />
1<br />
1<br />
1<br />
1<br />
1<br />
2<br />
2<br />
2<br />
2<br />
2<br />
2<br />
1<br />
=<br />
=<br />
→<br />
=<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
=<br />
+<br />
=<br />
⇒<br />
⇒<br />
→<br />
=<br />
⇒<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
=<br />
→<br />
=<br />
⇒<br />
=<br />
∆<br />
=<br />
=<br />
∆<br />
=<br />
=<br />
→<br />
=<br />
=<br />
−<br />
−<br />
→<br />
=<br />
∑<br />
α<br />
β<br />
β<br />
α<br />
β<br />
α<br />
α<br />
α<br />
δ<br />
β<br />
β<br />
δ<br />
δ<br />
δ<br />
δ<br />
δ<br />
β<br />
α<br />
K<br />
K<br />
Dimenzioniranje :<br />
cm<br />
h<br />
cm<br />
b<br />
cm<br />
A<br />
b<br />
cn<br />
b<br />
h<br />
b<br />
A<br />
cm<br />
A<br />
A<br />
cm<br />
d<br />
cm<br />
m<br />
d<br />
S<br />
d<br />
S<br />
d<br />
A<br />
A<br />
S<br />
dop<br />
dop<br />
dop<br />
74<br />
,<br />
27<br />
87<br />
,<br />
13<br />
87<br />
,<br />
13<br />
2<br />
8<br />
,<br />
384<br />
2<br />
8<br />
,<br />
384<br />
2<br />
8<br />
,<br />
384<br />
48<br />
,<br />
38<br />
10<br />
10<br />
7<br />
6<br />
06<br />
,<br />
0<br />
10<br />
12<br />
10<br />
4<br />
,<br />
353<br />
4<br />
4<br />
4<br />
2<br />
2<br />
2<br />
2<br />
2<br />
1<br />
2<br />
7<br />
3<br />
1<br />
1<br />
1<br />
1<br />
2<br />
1<br />
1<br />
1<br />
1<br />
1<br />
=<br />
=<br />
=<br />
=<br />
=<br />
⇒<br />
=<br />
=<br />
⋅<br />
=<br />
=<br />
⋅<br />
=<br />
=<br />
=<br />
→<br />
=<br />
≥<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
≥<br />
→<br />
≥<br />
=<br />
→<br />
≤<br />
=<br />
π<br />
πσ<br />
σ<br />
π<br />
σ<br />
σ<br />
Kontrola napona :<br />
MPa<br />
MPa<br />
A<br />
S<br />
MPa<br />
MPa<br />
A<br />
S<br />
dop<br />
dop<br />
6<br />
53<br />
,<br />
1<br />
10<br />
87<br />
,<br />
13<br />
2<br />
10<br />
9<br />
,<br />
58<br />
120<br />
84<br />
,<br />
91<br />
4<br />
10<br />
7<br />
10<br />
4<br />
,<br />
353<br />
2<br />
4<br />
2<br />
3<br />
2<br />
2<br />
2<br />
1<br />
4<br />
2<br />
3<br />
1<br />
1<br />
1<br />
=<br />
≤<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
=<br />
=<br />
≤<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
−<br />
−<br />
σ<br />
σ<br />
σ<br />
π<br />
σ
13. Zadatak<br />
Odrediti sile u �eli�nim kružnim zategama CF, DF i EG uslijed djelovanja nazna�enog<br />
optere�enja.<br />
Fl<br />
M =<br />
4<br />
Plan pomaka<br />
l<br />
l<br />
1<br />
2<br />
=<br />
=<br />
2 2<br />
l l<br />
+ =<br />
4 16<br />
2<br />
l<br />
2<br />
4<br />
=<br />
l3<br />
=<br />
9 2<br />
2 l =<br />
16<br />
l<br />
sinα<br />
=<br />
2<br />
=<br />
5<br />
l<br />
4<br />
l<br />
sin β =<br />
2<br />
=<br />
2<br />
l<br />
2<br />
2<br />
2<br />
l<br />
3<br />
4<br />
2<br />
5<br />
1<br />
5<br />
16<br />
2<br />
l<br />
2l<br />
2<br />
=<br />
5<br />
4<br />
l<br />
16
17<br />
Iz uvjeta ravnoteže sila :<br />
( )<br />
[ ] ( ) ( )<br />
1<br />
0<br />
sin<br />
3<br />
2<br />
sin<br />
2<br />
0<br />
sin<br />
3<br />
2<br />
sin<br />
4<br />
4<br />
0 3<br />
2<br />
1<br />
3<br />
2<br />
1<br />
K<br />
=<br />
+<br />
−<br />
−<br />
→<br />
=<br />
+<br />
+<br />
−<br />
+<br />
→<br />
=<br />
∑ β<br />
α<br />
β<br />
α S<br />
S<br />
S<br />
F<br />
S<br />
S<br />
S<br />
l<br />
M<br />
Fl<br />
M A<br />
Iz plana pomaka :<br />
( )<br />
;<br />
sin<br />
;<br />
sin<br />
;<br />
sin<br />
2<br />
4<br />
3<br />
2<br />
4<br />
3<br />
2<br />
1<br />
β<br />
δ<br />
β<br />
δ<br />
α<br />
δ<br />
δ<br />
δ<br />
δ<br />
l<br />
l<br />
l<br />
l<br />
l<br />
l<br />
E<br />
D<br />
C<br />
E<br />
D<br />
C<br />
∆<br />
=<br />
∆<br />
=<br />
∆<br />
=<br />
=<br />
= K<br />
( )<br />
( ) ( )<br />
F<br />
F<br />
S<br />
S<br />
F<br />
S<br />
S<br />
F<br />
S<br />
S<br />
S<br />
S<br />
F<br />
S<br />
S<br />
l<br />
l<br />
l<br />
l<br />
l<br />
l<br />
S<br />
S<br />
l<br />
S<br />
l<br />
S<br />
l<br />
l<br />
l<br />
l<br />
E<br />
D<br />
E<br />
D<br />
D<br />
C<br />
D<br />
C<br />
38<br />
,<br />
0<br />
47<br />
,<br />
0<br />
5<br />
4<br />
5<br />
4<br />
47<br />
,<br />
0<br />
sin<br />
5<br />
sin<br />
5<br />
4<br />
2<br />
0<br />
sin<br />
3<br />
2<br />
sin<br />
5<br />
4<br />
2<br />
1<br />
2<br />
3<br />
sin<br />
2<br />
sin<br />
3<br />
2<br />
3<br />
4<br />
3<br />
2<br />
5<br />
4<br />
sin<br />
sin<br />
2<br />
sin<br />
sin<br />
2<br />
2<br />
2<br />
4<br />
2<br />
2<br />
1<br />
3<br />
2<br />
2<br />
2<br />
2<br />
2<br />
3<br />
2<br />
3<br />
2<br />
3<br />
2<br />
2<br />
1<br />
2<br />
2<br />
1<br />
1<br />
2<br />
1<br />
=<br />
=<br />
=<br />
=<br />
=<br />
→<br />
+<br />
=<br />
→<br />
=<br />
+<br />
−<br />
−<br />
→<br />
=<br />
→<br />
∆<br />
=<br />
∆<br />
→<br />
∆<br />
=<br />
∆<br />
→<br />
=<br />
→<br />
=<br />
=<br />
→<br />
=<br />
→<br />
∆<br />
=<br />
∆<br />
→<br />
=<br />
→<br />
=<br />
→<br />
β<br />
α<br />
β<br />
α<br />
β<br />
β<br />
δ<br />
δ<br />
δ<br />
δ<br />
β<br />
α<br />
β<br />
α<br />
δ<br />
δ<br />
δ<br />
δ
ρ<br />
14. Zadatak<br />
Zadani tenzor naprezanja prikazati kao zbroj sfernog tenzora naprezanja i tenzora devijatorskog<br />
naprezanja. Odrediti normalno, posmi�no i puno naprezanje u ravnini s normalom koja ima<br />
kosinuse smjera :<br />
cos(x,n)=2/3 ; cos(y,n)=1/3 ; cos(z,n)=2/3.<br />
[ σ ]<br />
⎡σ<br />
x<br />
⎢<br />
⎢τ<br />
yx<br />
⎢<br />
⎣τ<br />
zx<br />
[ σ ]<br />
n<br />
ij<br />
ij<br />
⎡30<br />
=<br />
⎢<br />
⎢<br />
10<br />
⎢⎣<br />
15<br />
τ<br />
σ<br />
τ<br />
xy<br />
y<br />
zy<br />
⎡56,<br />
7<br />
=<br />
⎢<br />
⎢<br />
10<br />
⎢⎣<br />
15<br />
10<br />
− 50<br />
− 20<br />
τ xz ⎤ ⎡σ<br />
x −σ<br />
s<br />
⎥ ⎢<br />
τ yz ⎥ = ⎢ τ yx<br />
σ ⎥ ⎢<br />
z ⎦ ⎣ τ zx<br />
10<br />
− 23,<br />
3<br />
− 20<br />
15 ⎤<br />
− 20<br />
⎥<br />
⎥<br />
MPa<br />
− 60⎥⎦<br />
τ<br />
zy<br />
15 ⎤ ⎡−<br />
26,<br />
7<br />
− 20<br />
⎥<br />
+<br />
⎢<br />
⎥ ⎢<br />
0<br />
− 33,<br />
3⎥⎦<br />
⎢⎣<br />
0<br />
y<br />
τ<br />
xy<br />
σ −σ<br />
s<br />
τ xz ⎤ ⎡σ<br />
s<br />
⎥<br />
τ +<br />
⎢<br />
yz ⎥ ⎢<br />
0<br />
σ − ⎥<br />
⎦<br />
⎢<br />
z σ s ⎣ 0<br />
srednje naprezanje devijatorski dio sferni dio<br />
σ s<br />
σ x<br />
=<br />
+ σ y + σ z<br />
3<br />
30 − 50 − 60<br />
=<br />
= −26,<br />
7MPa<br />
3<br />
Komponente devijatorskog tenzora naprezanja :<br />
s<br />
s<br />
s<br />
=<br />
xx<br />
yy<br />
zz<br />
[ σ ]<br />
ij<br />
⋅ n<br />
⎡ρ<br />
nx ⎤ ⎡30<br />
⎢ ⎥<br />
=<br />
⎢<br />
⎢<br />
ρ ny ⎥ ⎢<br />
10<br />
⎢⎣<br />
ρ ⎥⎦<br />
⎢ nz ⎣15<br />
= σ<br />
−σ<br />
= 30 −<br />
x<br />
= σ −σ<br />
= −50<br />
z<br />
y<br />
= σ −σ<br />
= −60<br />
s<br />
s<br />
s<br />
10<br />
− 50<br />
− 20<br />
( − 26,<br />
7)<br />
= 56,<br />
7MPa<br />
− ( − 26,<br />
7)<br />
= −23,<br />
3<br />
− ( − 26,<br />
7)<br />
= −33,<br />
3Pa<br />
⎡ ⎤<br />
⎢ ⎥<br />
⎤ ⎡ 33,<br />
3<br />
⎢ ⎥<br />
⎤<br />
⎥ 1<br />
− ⎢ ⎥ =<br />
⎢ ⎥<br />
⎥<br />
⋅<br />
⎢<br />
− 23,<br />
3<br />
⎢3<br />
⎥<br />
⎥<br />
− ⎥⎦<br />
⎢2<br />
⎥<br />
⎢⎣<br />
− 36,<br />
7⎥⎦<br />
⎢<br />
⎣3<br />
⎥<br />
⎦<br />
3<br />
2<br />
15<br />
20<br />
60<br />
MPa<br />
0<br />
− 26,<br />
7<br />
0<br />
ρ<br />
σ<br />
τ<br />
n<br />
n<br />
n<br />
=<br />
0 ⎤<br />
0<br />
⎥<br />
⎥<br />
− 26,<br />
7⎥⎦<br />
=<br />
ρ<br />
n<br />
2<br />
nx<br />
2<br />
n<br />
0<br />
σ<br />
0<br />
s<br />
+ ρ<br />
− σ<br />
2<br />
n<br />
0 ⎤<br />
0<br />
⎥<br />
⎥<br />
σ ⎥ s ⎦<br />
2<br />
ny<br />
= ρ ⋅ n = 33,<br />
3 ⋅<br />
ρ<br />
=<br />
+ ρ<br />
2<br />
3<br />
2<br />
nz<br />
54,<br />
8<br />
2<br />
=<br />
−10<br />
33,<br />
3<br />
1<br />
− 23,<br />
3 ⋅ − 36,<br />
7 ⋅<br />
3<br />
2<br />
2<br />
=<br />
+<br />
23,<br />
3<br />
2<br />
3<br />
53,<br />
9<br />
2<br />
+<br />
= −10MPa<br />
MPa<br />
36,<br />
7<br />
2<br />
= 54,<br />
8MPa<br />
18
15. Zadatak<br />
Za zadano stanje naprezanja �x = 100M Pa , �y = -80M Pa , txy = 50 Mpa odrediti analiti�ki<br />
a) glavna naprezanja<br />
b) maksimalno posmi�no naprezanje s odgovaraju�im normalnim naprezanjem<br />
c) naprezanja �n i tn u presjeku s normalom koja s osi x zatvara kut � = -25°.<br />
Glavna naprezanja i smjerovi naprezanja :<br />
σ 1,<br />
2<br />
σ x<br />
=<br />
+ σ y 1<br />
±<br />
2 2<br />
σ = 113MPa<br />
1<br />
σ = −93MPa<br />
2<br />
tgϕ<br />
tgϕ<br />
01<br />
02<br />
τ xy<br />
=<br />
σ −σ<br />
1<br />
=<br />
2 2 100 − 80 1<br />
( σ −σ<br />
) + 4τ<br />
= ± ( 100 + 80)<br />
50<br />
= 0,<br />
2591 ⇒ ϕ01<br />
113 + 80<br />
= 14,<br />
5°<br />
τ xy 50<br />
= = = −3,<br />
8462 ⇒ ϕ02<br />
= −75,<br />
5°<br />
σ −σ<br />
− 93 + 80<br />
2<br />
y<br />
y<br />
Kontrola preko invarijante naprezanja :<br />
+ ϕ<br />
= 14,<br />
5°<br />
+ 75,<br />
5°<br />
= 90°<br />
x<br />
+ σ = σ + σ →113−<br />
93 = 10 −80<br />
→ 20 = 20<br />
σ 1 2 x y<br />
ϕ<br />
01<br />
02<br />
y<br />
xy<br />
2<br />
2<br />
2<br />
+ 4 ⋅50<br />
2<br />
= 10 ± 103<br />
Maksimalno posmi�no naprezanje, smjer normale ravnine i normalno naprezanje :<br />
ϕ = ϕ<br />
1<br />
01<br />
+ 45°<br />
= 14,<br />
5°<br />
+ 45°<br />
= 59,<br />
5°<br />
σ 1 −σ<br />
2 113 + 93<br />
τ max = = = 103MPa<br />
2 2<br />
σ 1 + σ 2 113 − 93<br />
σ s = = = 10MPa<br />
2 2<br />
Naprezanja u ravnini pod kutem :<br />
σ<br />
n<br />
σ x<br />
=<br />
+ σ y σ x<br />
+<br />
2<br />
−σ<br />
y<br />
2<br />
cos 2ϕ<br />
+ τ xy sin 2ϕ<br />
= σ x cos ϕ + σ y sin<br />
2<br />
100 − 80 100 + 80<br />
σ n = + cos(<br />
− 50°<br />
) + 50sin(<br />
− 50°<br />
) = 29,<br />
6MPa<br />
2 2<br />
τ n<br />
σ y<br />
=<br />
−σ<br />
x<br />
− 80 −100<br />
sin 2ϕ<br />
+ τ xy cos 2ϕ<br />
= sin<br />
2<br />
2<br />
2<br />
ϕ + τ<br />
( − 50°<br />
) + 50cos(<br />
− 50°<br />
) = 101,<br />
1MPa<br />
xy<br />
sin 2ϕ<br />
19
20<br />
Prikaz Mohrove kružnice naprezanja za zadano stanje naprezanja :<br />
)<br />
1<br />
,<br />
101<br />
;<br />
6<br />
,<br />
29<br />
(<br />
)<br />
,<br />
(<br />
;<br />
1<br />
,<br />
101<br />
;<br />
6<br />
,<br />
29<br />
)<br />
103<br />
,<br />
10<br />
(<br />
)<br />
,<br />
(<br />
;<br />
5<br />
,<br />
59<br />
;<br />
103<br />
;<br />
10<br />
)<br />
0<br />
,<br />
93<br />
(<br />
)<br />
0<br />
,<br />
(<br />
;<br />
5<br />
,<br />
75<br />
;<br />
93<br />
)<br />
0<br />
,<br />
113<br />
(<br />
)<br />
0<br />
,<br />
(<br />
;<br />
5<br />
,<br />
14<br />
;<br />
113<br />
)<br />
50<br />
,<br />
80<br />
(<br />
)<br />
,<br />
(<br />
)<br />
50<br />
,<br />
100<br />
(<br />
)<br />
,<br />
(<br />
1<br />
max<br />
3<br />
1<br />
max<br />
1<br />
2<br />
2<br />
02<br />
2<br />
1<br />
1<br />
1<br />
01<br />
1<br />
−<br />
→<br />
=<br />
=<br />
→<br />
°<br />
=<br />
=<br />
=<br />
−<br />
→<br />
°<br />
−<br />
=<br />
−<br />
=<br />
→<br />
°<br />
=<br />
=<br />
−<br />
→<br />
−<br />
→<br />
K<br />
K<br />
MPa<br />
MPa<br />
N<br />
N<br />
MPa<br />
MPa<br />
N<br />
N<br />
MPa<br />
N<br />
N<br />
MPa<br />
N<br />
N<br />
N<br />
N<br />
n<br />
n<br />
n<br />
n<br />
s<br />
s<br />
y<br />
xy<br />
y<br />
y<br />
x<br />
xy<br />
x<br />
x<br />
τ<br />
σ<br />
τ<br />
σ<br />
τ<br />
σ<br />
ϕ<br />
τ<br />
σ<br />
σ<br />
ϕ<br />
σ<br />
σ<br />
ϕ<br />
σ<br />
τ<br />
σ<br />
τ<br />
σ
16. Zadatak<br />
Za zadano stanje naprezanja odrediti glavna normalna naprezanja, glavna posmi�na naprezanja i<br />
oktaedarska naprezanja.<br />
[ σ ] =<br />
⎢<br />
50 120 0<br />
⎥[<br />
MPa]<br />
ij<br />
⎡− 80<br />
⎢<br />
⎢⎣<br />
σ<br />
−σ<br />
0<br />
50<br />
0<br />
0 ⎤<br />
⎥<br />
60⎥⎦<br />
2<br />
[ −τ<br />
]<br />
( σ −σ<br />
) ( σ −σ<br />
)( σ −σ<br />
)<br />
z<br />
x<br />
( σ −σ<br />
)( σ −σ<br />
)<br />
σ<br />
σ<br />
2<br />
m<br />
m<br />
z<br />
x<br />
yx<br />
zx<br />
m<br />
m<br />
m<br />
m<br />
( σ + σ )<br />
x<br />
y<br />
y<br />
x<br />
y<br />
xy<br />
zy<br />
m<br />
m<br />
m<br />
m<br />
m<br />
1<br />
2<br />
y<br />
2<br />
xy<br />
( σ −σ<br />
)<br />
x<br />
z<br />
z<br />
x<br />
xz<br />
yz<br />
y<br />
m<br />
y<br />
m<br />
= 0<br />
= 0 ⇒<br />
2<br />
xy<br />
2<br />
xy<br />
′ − 80 + 120 1<br />
σ 1,<br />
2 = ±<br />
2 2<br />
2<br />
2<br />
[ ( − 80 −120)<br />
] + 4 ⋅50<br />
= 20 ± 111,<br />
8<br />
′<br />
′<br />
′<br />
σ = 131,<br />
8MPa;<br />
σ = −91,<br />
8MPa;<br />
σ = 60MPa<br />
1<br />
1<br />
2<br />
3<br />
τ<br />
τ<br />
−<br />
σ x + σ y<br />
=<br />
2<br />
σ = 60MPa<br />
τ<br />
σ −σ<br />
τ<br />
±<br />
σ = 131,<br />
8MPa<br />
σ<br />
σ = −91,<br />
8MPa<br />
2<br />
−τ<br />
+ σ σ −τ<br />
2<br />
xy<br />
+ 4τ<br />
x<br />
= 0<br />
′<br />
σ −σ<br />
= 0 ⇒ σ = σ = 60MPa<br />
= σ<br />
τ<br />
τ<br />
σ −σ<br />
= 0 → σ<br />
3<br />
σ −σ<br />
3<br />
τ<br />
yx<br />
0<br />
m<br />
m<br />
y<br />
τ<br />
xy<br />
σ −σ<br />
0<br />
m<br />
σ x + σ y ±<br />
=<br />
z<br />
0<br />
0<br />
σ −σ<br />
2<br />
2<br />
[ − ( σ + σ ) ] − 4(<br />
−τ<br />
+ σ σ )<br />
x<br />
m<br />
= 0<br />
y<br />
2<br />
xy<br />
x<br />
y<br />
21
Kontrola naprezanja :<br />
σ + σ + σ = σ + σ + σ<br />
1<br />
2<br />
− 80 + 120 + 60 = 131,<br />
8 + 60 − 91,<br />
8<br />
100MPa<br />
= 100MPa<br />
3<br />
x<br />
y<br />
z<br />
Pravac glavnog naprezanja �2 podudara se s pravcem osi z. Pravci glavnih naprezanja �1 i �3<br />
okomiti su na pravac naprezanja �2, a njihov se položaj u ravnini okomitoj na pravac �2 odre�uje<br />
prema izrazima :<br />
tgϕ<br />
tgϕ<br />
ϕ<br />
01<br />
01<br />
03<br />
τ xy<br />
=<br />
σ −σ<br />
τ xy<br />
=<br />
σ −σ<br />
+ ϕ<br />
03<br />
1<br />
3<br />
y<br />
y<br />
50<br />
= = 4,<br />
237<br />
131,<br />
8 −120<br />
50<br />
=<br />
= −0,<br />
236 ⇒ ϕ03<br />
= −13,<br />
3°<br />
− 91,<br />
8 −120<br />
= 76,<br />
7°<br />
+ 13,<br />
3°<br />
= 90°<br />
Maksimalna posmi�na naprezanja :<br />
τ<br />
max<br />
= τ<br />
2<br />
= 111,<br />
8MPa<br />
⇒ ϕ<br />
01<br />
= 76,<br />
7°<br />
σ 2 −σ<br />
3 60 + 91,<br />
8<br />
τ 1 = ± = ± = ± 75,<br />
9MPa<br />
2 2<br />
σ 3 −σ<br />
1 − 91,<br />
8 −131,<br />
8<br />
τ 2 = ± = ±<br />
= ± 111,<br />
8MPa<br />
2<br />
2<br />
σ 1 −σ<br />
2 131,<br />
8 − 60<br />
τ 3 = ± = ± = ± 35,<br />
9MPa<br />
2 2<br />
Oktaedarska naprezanja :<br />
Budu�i da normala oktaedarske ravnine zatvara s koordinatnim osima kuteve :<br />
2<br />
2<br />
2<br />
2<br />
1<br />
cos α<br />
+ cos β + cos γ = 1 → 3cos<br />
α = 1 → cosα<br />
= ± → α = 54,<br />
7°<br />
3<br />
1<br />
1<br />
1<br />
σ okt = ( σ 1 + σ 2 + σ 3 ) = ( σ x + σ y + σ z ) = ( 131,<br />
8 + 60 − 91,<br />
8)<br />
3<br />
3<br />
3<br />
σ = 33,<br />
3MPa<br />
τ<br />
τ<br />
ρ<br />
okt<br />
okt<br />
okt<br />
okt<br />
1<br />
= 1<br />
3<br />
= 93,<br />
2MPa<br />
=<br />
σ<br />
2<br />
2<br />
2<br />
2<br />
2<br />
( σ −σ<br />
) + ( σ −σ<br />
) + ( σ −σ<br />
) = ( 131,<br />
8 − 60)<br />
+ ( 60 + 91,<br />
8)<br />
+ ( − 91,<br />
8 −131,<br />
)<br />
2<br />
okt<br />
+ τ<br />
2<br />
2<br />
okt<br />
=<br />
2<br />
33,<br />
3<br />
2<br />
3<br />
+<br />
93,<br />
2<br />
2<br />
3<br />
1<br />
1<br />
3<br />
= 99,<br />
0MPa<br />
2<br />
22
17. Zadatak<br />
Plo�a je izložena djelovanju vla�nog naprezanja u oba pravca i tangencijalnog naprezanja u<br />
bo�nim ravninama.Odrediti komponente naprezanja u presjeku �ija normala zatvara kut od 30° s<br />
osi x i smjer i veli�inu glavnih naprezanja.<br />
σ = 8MPa<br />
σ<br />
τ<br />
xy<br />
n<br />
x<br />
y<br />
n<br />
n<br />
n<br />
n<br />
= 4MPa<br />
= −2MPa<br />
ϕ = 30°<br />
2<br />
σ = σ cos ϕ + σ sin<br />
σ<br />
ρ<br />
=<br />
=<br />
x<br />
5,<br />
27<br />
σ<br />
2<br />
n<br />
MPa<br />
+ τ<br />
ρ = 5,<br />
93MPa<br />
2<br />
n<br />
=<br />
y<br />
2<br />
5,<br />
27<br />
ϕ + τ<br />
2<br />
+<br />
xy<br />
sin 2ϕ<br />
=<br />
( − 2,<br />
73)<br />
2<br />
8cos<br />
2<br />
30°<br />
+<br />
4sin<br />
τ n<br />
σ y<br />
=<br />
−σ<br />
x<br />
4 − 8<br />
sin 2ϕ<br />
+ τ xy cos 2ϕ<br />
= sin 60°<br />
− 2cos60°<br />
2<br />
2<br />
τ = −2,<br />
73MPa<br />
Glavna naprezanja :<br />
σ 1,<br />
2<br />
σ x<br />
=<br />
+ σ y 1<br />
±<br />
2 2<br />
σ = 8,<br />
83MPa<br />
σ<br />
1<br />
2<br />
tgϕ<br />
tgϕ<br />
=<br />
01<br />
02<br />
3,<br />
17<br />
MPa<br />
τ xy<br />
=<br />
σ −σ<br />
1<br />
τ xy<br />
=<br />
σ −σ<br />
2<br />
y<br />
y<br />
=<br />
2<br />
30°<br />
− 2sin<br />
60°<br />
2 2 8 + 4 1<br />
2<br />
( σ −σ<br />
) + 4τ<br />
= ± ( 8 − 4)<br />
+ 4 ⋅ ( − 2)<br />
x<br />
y<br />
− 2<br />
= = −0,<br />
414 ⇒ ϕ01<br />
= −22,<br />
5°<br />
8,<br />
83 − 4<br />
xy<br />
− 2<br />
= 2,<br />
41 ⇒ ϕ02<br />
3,<br />
17 − 4<br />
Kontrola preko invarijante naprezanja :<br />
01<br />
+ ϕ<br />
= 22,<br />
5°<br />
+ 67,<br />
5°<br />
= 90°<br />
2<br />
= 67,<br />
5°<br />
+ σ = σ + σ → 8,<br />
8 + 3,<br />
2 = 8 + 4 → 12 = 12<br />
σ 1 2 x y<br />
ϕ<br />
02<br />
2<br />
2<br />
= 6 ± 2,<br />
83<br />
Maksimalno posmi�no naprezanje, smjer normale ravnine i pripadaju�e normalno naprezanje :<br />
ϕ<br />
= ϕ<br />
τ<br />
σ<br />
1<br />
max<br />
s<br />
01<br />
+ 45°<br />
= −22,<br />
6°<br />
+ 45°<br />
= 22,<br />
4°<br />
σ 1 −σ<br />
2 8,<br />
8 − 3,<br />
2<br />
= = = 2,<br />
8MPa<br />
2 2<br />
σ 1 + σ 2 8,<br />
8 + 3,<br />
2<br />
= = = 6MPa<br />
2 2<br />
23
18. Zadatak<br />
Kratki betonski stup, kvadratnog popre�nog presjeka 30x30 cm pritisnut je silom F.Odrediti<br />
veli�inu ove sile ako je normalno naprezanje u kosom presjeku pod kutem od 30°prema normali<br />
1,5 Mpa.Koliko je tangencijalno naprezanje u tom presjeku?<br />
σ = 1,<br />
5MPa<br />
n<br />
ϕ = 30°<br />
a = 30cm<br />
= 0,<br />
3m<br />
A = a<br />
n<br />
2<br />
= 0,<br />
09m<br />
2<br />
σ = σ cos ϕ + σ sin<br />
F = −180kN<br />
x<br />
2<br />
y<br />
2<br />
ϕ + τ<br />
xy<br />
sin 2ϕ<br />
6<br />
2 F 2<br />
σ n A 1,<br />
5⋅10<br />
⋅ 0,<br />
09<br />
σ n = σ x cos ϕ = − cos ϕ → F = − = − 2<br />
2<br />
A<br />
cos ϕ cos 30°<br />
3<br />
F 180 ⋅10<br />
σ x = = − = −2MPa<br />
A 0,<br />
09<br />
σ y −σ<br />
x<br />
τ n = sin 2ϕ<br />
+ τ xy cos 2ϕ<br />
2<br />
6<br />
−σ<br />
x − 2 ⋅10<br />
τ n = sin 2ϕ<br />
= − sin 60°<br />
= 0,<br />
86MPa<br />
2<br />
2<br />
19. Zadatak<br />
Stup kružnog popre�nog presjeka optere�en je vla�nom silom intenziteta 20 kN. Tangencijalno<br />
naprezanje na bilo kojem presjeku (ravnini) ne smije prije�i vrijednost od 7Mpa. Odrediti polumjer<br />
stupa.<br />
F = 20kN<br />
τ<br />
= 7MPa<br />
n<br />
σ x F 2F<br />
2F<br />
τ n ≤ sin 2ϕ<br />
= sin 2ϕ<br />
= sin 2ϕ<br />
→ d ≥ sin 2ϕ<br />
=<br />
2<br />
2 2A<br />
d π<br />
τ π<br />
(sin 2ϕ)<br />
d ≥<br />
max<br />
= 1<br />
4<br />
2 ⋅ 2 ⋅10<br />
→ d ≥ 4,<br />
3cm<br />
→ d = 4,<br />
5cm<br />
6<br />
7 ⋅10<br />
π<br />
n<br />
4<br />
2 ⋅ 2 ⋅10<br />
sin 2ϕ<br />
6<br />
7 ⋅10<br />
π<br />
24
20. Zadatak<br />
U to�ki A napregnutog elementa izmjerene su dužinske deformacije u smjerovima x, y i n �xx = -<br />
8·10 -4 , �yy = 4·10 -4 , �nn = 6·10 -4 . Odrediti promjene pravog kuta izme�u osi x i y, te veli�inu i<br />
smjerove glavnih deformacija.<br />
ε<br />
nn<br />
⇒ ε<br />
ε<br />
γ<br />
xy<br />
xy<br />
= ε<br />
xy<br />
xx<br />
= 2ε<br />
xy<br />
cos<br />
2<br />
1<br />
=<br />
sin 2ϕ<br />
= 12,<br />
7 ⋅10<br />
ϕ + ε<br />
−4<br />
2<br />
2 1<br />
−4<br />
−4<br />
−4<br />
( ε − ε cos ϕ − ε sin ϕ)<br />
= ( 6 ⋅10<br />
+ 8⋅10<br />
cos30°<br />
− 4 ⋅10<br />
sin 30°<br />
)<br />
nn<br />
yy<br />
= 25,<br />
4 ⋅10<br />
Glavne deformacije :<br />
1,<br />
2<br />
1,<br />
2<br />
1<br />
2<br />
ε + ε 1<br />
2 2<br />
⎛ − 8 + 4 1<br />
= ⎜ ±<br />
⎝ 2 2<br />
−4<br />
−4<br />
−4<br />
sin<br />
−4<br />
xx<br />
2<br />
rad<br />
ϕ + ε<br />
( ε − ε )<br />
xy<br />
sin 2ϕ<br />
⇒<br />
xx yy<br />
2 2<br />
ε1, 2 = ± xx yy + 4ε<br />
xy<br />
ε<br />
ε<br />
= −2<br />
⋅10<br />
ε = 12 ⋅10<br />
ε = −16<br />
⋅10<br />
Kontrola :<br />
+ ε = ε + ε<br />
ε xx yy<br />
− 8⋅10<br />
− 4 ⋅10<br />
−4<br />
−4<br />
1<br />
+ 4 ⋅10<br />
± 14 ⋅10<br />
2<br />
−4<br />
= − 4 ⋅10<br />
−4<br />
yy<br />
2 ( − 8 − 4)<br />
+ 4(<br />
12,<br />
7)<br />
−4<br />
= 12 ⋅10<br />
−4<br />
Smjerovi glavnih deformacija :<br />
tgϕ<br />
tgϕ<br />
ϕ<br />
01<br />
01<br />
02<br />
ε xy<br />
=<br />
ε − ε<br />
=<br />
+ ϕ<br />
02<br />
1<br />
2<br />
ε<br />
xy<br />
yy<br />
ε − ε<br />
yy<br />
−16<br />
⋅10<br />
−4<br />
12,<br />
7 ⋅10<br />
= −4<br />
12 ⋅10<br />
− 4 ⋅10<br />
−4<br />
−4<br />
12,<br />
7 ⋅10<br />
=<br />
−4<br />
−16<br />
⋅10<br />
− 4 ⋅10<br />
−4<br />
= 57,<br />
8°<br />
+ 32,<br />
4°<br />
= 90,<br />
2°<br />
≈ 90°<br />
2<br />
⎞<br />
⎟ ⋅10<br />
⎠<br />
−4<br />
sin 60°<br />
= 1,<br />
588 ⇒ ϕ = 57,<br />
8°<br />
−4<br />
01<br />
= −0,<br />
635 ⇒ ϕ<br />
02<br />
= −32,<br />
4°<br />
25
26<br />
21. Zadatak<br />
Odrediti deformaciju dijagonale �eli�nog elementa.<br />
5<br />
5<br />
5<br />
5<br />
5<br />
5<br />
5<br />
5<br />
10<br />
5<br />
,<br />
19<br />
2<br />
10<br />
75<br />
,<br />
9<br />
15<br />
10<br />
2<br />
3<br />
,<br />
0<br />
1<br />
1<br />
2<br />
10<br />
9<br />
10<br />
2<br />
60<br />
3<br />
,<br />
0<br />
10<br />
30<br />
10<br />
2<br />
60<br />
15<br />
60<br />
3<br />
,<br />
0<br />
10<br />
2<br />
−<br />
−<br />
−<br />
−<br />
⋅<br />
=<br />
=<br />
⋅<br />
=<br />
⋅<br />
+<br />
=<br />
+<br />
=<br />
=<br />
⋅<br />
−<br />
=<br />
⋅<br />
−<br />
=<br />
−<br />
=<br />
⋅<br />
=<br />
⋅<br />
=<br />
=<br />
=<br />
=<br />
=<br />
⋅<br />
=<br />
xy<br />
xy<br />
xy<br />
xy<br />
xy<br />
x<br />
yy<br />
x<br />
xx<br />
xy<br />
x<br />
E<br />
G<br />
E<br />
E<br />
MPa<br />
MPa<br />
MPa<br />
E<br />
ε<br />
γ<br />
τ<br />
ν<br />
τ<br />
ε<br />
σ<br />
ν<br />
ε<br />
σ<br />
ε<br />
τ<br />
σ<br />
ν<br />
Relativna deformacija dijagonale :<br />
cm<br />
d<br />
d<br />
d<br />
xy<br />
yy<br />
xx<br />
d<br />
10<br />
30<br />
sin<br />
5<br />
5<br />
30<br />
sin<br />
10<br />
69<br />
,<br />
28<br />
60<br />
sin<br />
10<br />
75<br />
,<br />
9<br />
30<br />
sin<br />
10<br />
9<br />
30<br />
cos<br />
10<br />
30<br />
2<br />
sin<br />
sin<br />
cos<br />
5<br />
5<br />
2<br />
5<br />
2<br />
5<br />
2<br />
2<br />
=<br />
°<br />
=<br />
→<br />
=<br />
°<br />
⋅<br />
=<br />
°<br />
⋅<br />
+<br />
°<br />
⋅<br />
−<br />
°<br />
⋅<br />
=<br />
+<br />
+<br />
=<br />
−<br />
−<br />
−<br />
−<br />
ε<br />
ϕ<br />
ε<br />
ϕ<br />
ε<br />
ϕ<br />
ε<br />
ε<br />
Apsolutna deformacija :<br />
mm<br />
d<br />
d d<br />
3<br />
5<br />
10<br />
69<br />
,<br />
28<br />
100<br />
10<br />
69<br />
,<br />
28<br />
−<br />
−<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
=<br />
∆ ε
22. Zadatak<br />
Pravokutna plo�ica dimenzija 120x90 mm optere�ena je u dva pravca tako da su<br />
5<br />
E = 2⋅10<br />
MPa<br />
ν =<br />
0,<br />
3<br />
N<br />
σ x = 10000 2<br />
cm<br />
= 100MPa<br />
N<br />
σ y = −4000<br />
2<br />
cm<br />
= −40MPa<br />
Odrediti kolika su naprezanja u kosoj ravnini koja se poklapa s dijagonalom koja s osi x zatvara<br />
oštri kut i za koliko �e se promijeniti dužina dijagonale.<br />
9<br />
tgα<br />
= = 0,<br />
75 → α = 37°<br />
⇒ ϕ = α − 90°<br />
= −53°<br />
12<br />
σ n<br />
σ x<br />
=<br />
+ σ y σ x<br />
+<br />
2<br />
−σ<br />
y<br />
100 − 40 100 + 40<br />
cos 2ϕ<br />
+ τ xy sin 2ϕ<br />
= + cos<br />
2<br />
2 2<br />
τ n<br />
σ y<br />
=<br />
−σ<br />
x<br />
− 40 −100<br />
sin 2ϕ<br />
+ τ xy cos 2ϕ<br />
= sin(<br />
−106°<br />
) = 67,<br />
2MPa<br />
2<br />
2<br />
ρ<br />
ε<br />
ε<br />
n<br />
xx<br />
yy<br />
=<br />
σ<br />
2<br />
n<br />
+ τ<br />
2<br />
n<br />
=<br />
10,<br />
7<br />
2<br />
67,<br />
2<br />
2<br />
( − 40)<br />
= 68,<br />
0MPa<br />
σ σ x y<br />
= −ν<br />
E E<br />
100<br />
= 5<br />
2 ⋅10<br />
− 0,<br />
3 5<br />
2 ⋅10<br />
−5<br />
= 56 ⋅10<br />
σ y σ x − 40<br />
= −ν<br />
= 5<br />
E E 2 ⋅10<br />
100<br />
− 0,<br />
3 5<br />
2 ⋅10<br />
= −35⋅10<br />
+<br />
−5<br />
( −106°<br />
)<br />
= 10,<br />
7MPa<br />
Budu�i da tražimo izduženje dijagonale, tražimo deformaciju u smjeru normale ravnine okomite<br />
na dijagonalu, te je sada kut ������������<br />
ε<br />
= ε<br />
d<br />
d =<br />
xx<br />
90<br />
= 150mm<br />
∆d<br />
= ε ⋅ d = 23,<br />
04 ⋅10<br />
d<br />
cos<br />
2<br />
2<br />
ϕ + ε<br />
+ 120<br />
2<br />
yy<br />
sin<br />
2<br />
−5<br />
ϕ + ε<br />
xy<br />
sin 2ϕ<br />
= 56 ⋅10<br />
⋅150<br />
= 3,<br />
46 ⋅10<br />
−2<br />
mm<br />
−5<br />
2<br />
cos ( 37°<br />
) − 35⋅10<br />
−5<br />
sin<br />
2<br />
( 37°<br />
) = 23,<br />
04 ⋅10<br />
−5<br />
27
23. Zadatak<br />
�eli�na kocka brida a = 10 cm postavljena je bez zazora izme�u dviju krutih stijenki na podlogu i<br />
na gornjoj plohi optere�ena s q = 60 MPa.<br />
Odrediti : a) naprezanja i deformacije u tri okomita smjera i deformaciju dijagonale kocke<br />
b) promjenu volumena kocke<br />
c) normalno i posmi�no naprezanje u presjeku pod kutem od 45° prema stjenki.<br />
E = 2 ⋅10<br />
ν =<br />
σ<br />
ε<br />
ε<br />
ε<br />
xx<br />
yy<br />
zz<br />
1<br />
=<br />
E<br />
1<br />
=<br />
E<br />
1<br />
=<br />
E<br />
d = a<br />
ε<br />
ε<br />
d<br />
d<br />
y<br />
z<br />
0,<br />
30<br />
MPa<br />
= −q<br />
= −60MPa<br />
σ = 0<br />
= ε<br />
=<br />
xx<br />
[ σ −ν<br />
( σ + σ ) ] = 0 → σ = νσ = 0,<br />
3⋅<br />
( − 60)<br />
1<br />
[ σ −ν<br />
( σ + σ ) ] = [ − 60 − 0,<br />
3(<br />
−18)<br />
]<br />
1<br />
[ σ −ν<br />
( σ + σ ) ] = − 0,<br />
3(<br />
−18<br />
− 60)<br />
3<br />
∆d<br />
= ε ⋅ d = −5,<br />
2 ⋅10<br />
d<br />
5<br />
x<br />
y<br />
z<br />
cos<br />
2<br />
−5,<br />
2 ⋅10<br />
−5<br />
z<br />
x<br />
z<br />
α + ε<br />
yy<br />
y<br />
y<br />
x<br />
cos<br />
−5<br />
2<br />
⋅10<br />
2 ⋅10<br />
2 ⋅10<br />
β + ε<br />
2<br />
⋅<br />
zz<br />
5<br />
5<br />
x<br />
y<br />
[ ]<br />
2<br />
cos γ = 0 − 27,<br />
3⋅10<br />
3 = −9<br />
⋅10<br />
Relativna i apsolutna promjena volumena :<br />
∆V<br />
ε V = = ε xx<br />
V<br />
+ ε yy + ε<br />
∆V<br />
= ε<br />
−<br />
⋅V<br />
= −15,<br />
6 ⋅10<br />
V<br />
⎡−18<br />
ρ<br />
n = [ σ ij ] ⋅ n =<br />
⎢<br />
⎢<br />
0<br />
⎢⎣<br />
0<br />
ρ = −12,<br />
72MPa<br />
ρ<br />
ρ<br />
n<br />
nx<br />
ny<br />
nz<br />
n<br />
n<br />
= −42,<br />
42MPa<br />
= 0<br />
=<br />
=<br />
n<br />
2<br />
nx<br />
2<br />
n<br />
+ ρ<br />
−σ<br />
2<br />
ny<br />
2<br />
n<br />
+ ρ<br />
zz<br />
0<br />
0<br />
2<br />
nz<br />
5<br />
− 60<br />
= 21MPa<br />
= 0 −<br />
⋅10<br />
=<br />
3<br />
27,<br />
3⋅10<br />
−5<br />
= −0,<br />
15cm<br />
0⎤⎡cos<br />
45°<br />
⎤<br />
0<br />
⎥⎢<br />
cos 45<br />
⎥<br />
⎥⎢<br />
°<br />
⎥<br />
0⎥⎦<br />
⎢⎣<br />
cos90°<br />
⎥⎦<br />
12,<br />
72<br />
2<br />
+<br />
−3<br />
42,<br />
42<br />
mm<br />
=<br />
+ 11,<br />
7 ⋅10<br />
2<br />
+ 0<br />
2<br />
−5<br />
= −18MPa<br />
−27,<br />
3⋅10<br />
= 11,<br />
7 ⋅10<br />
σ = ρ ⋅ n = −12,<br />
72 ⋅ cos 45°<br />
− 42,<br />
42 ⋅ cos 45°<br />
= −39MPa<br />
ρ<br />
τ<br />
ρ<br />
ρ<br />
−5<br />
=<br />
−5<br />
−5<br />
1<br />
⋅ + 11,<br />
7 ⋅10<br />
3<br />
−15,<br />
6 ⋅10<br />
= 44,<br />
28MPa<br />
−5<br />
−5<br />
1<br />
⋅<br />
3<br />
28
24. Zadatak<br />
Odrediti glavna naprezanja na stranicama kvadratnog elementa ako su tenzometri (ure�aji za<br />
mjerenje dužinske deformacije) A i B pokazali prirast �nA = 9,9 mm i<br />
�nB = 3,1 mm. Tenzometar A postavljen je pod kutem � = 30° prema pravcu glavnog naprezanja<br />
�1, a tenzometar B okomito na tenzometar A. Baza tenzometra l0 = 20 mm, a uve�anje<br />
tenzometra k = 1000. E = 0,8·10 5 Mpa, �=0,35<br />
Relativne deformacije :<br />
ε<br />
ε<br />
nnA<br />
nnB<br />
∆l<br />
∆n<br />
A 9,<br />
9<br />
= = = = 4,<br />
95⋅10<br />
3<br />
l k ⋅ l 10 ⋅ 20<br />
0<br />
∆n<br />
=<br />
k ⋅l<br />
B<br />
0<br />
0<br />
3,<br />
1<br />
= = 1,<br />
55⋅10<br />
3<br />
10 ⋅ 20<br />
−4<br />
Za ravninsko stanje deformacija :<br />
σ<br />
σ<br />
σ<br />
σ<br />
Kontrola :<br />
σ<br />
nA<br />
nB<br />
n<br />
nA<br />
nA<br />
E<br />
=<br />
1 −ν<br />
E<br />
=<br />
1 −ν<br />
+ σ<br />
nB<br />
−4<br />
5<br />
0,<br />
8 ⋅10<br />
−4<br />
−4<br />
( ε + νε ) = ( 4,<br />
95 ⋅10<br />
+ 0,<br />
35 ⋅1,<br />
55 ⋅10<br />
)<br />
5<br />
0,<br />
8 ⋅10<br />
−4<br />
−4<br />
( ε + νε ) = ( 1,<br />
55 ⋅10<br />
+ 0,<br />
35 ⋅ 4,<br />
95 ⋅10<br />
)<br />
2<br />
( cos(<br />
ϕ + 90°<br />
) ) = ( − sin ϕ )<br />
2<br />
( sin(<br />
ϕ + 90°<br />
) ) 2<br />
= ( cosϕ<br />
)<br />
σ<br />
nB<br />
x<br />
cos<br />
= σ cos<br />
= σ cos<br />
= 20MPa<br />
nnA<br />
nnB<br />
y<br />
sin<br />
nnB<br />
nnA<br />
sin<br />
( ϕ + 90°<br />
) + σ<br />
0,<br />
75σ<br />
+ 0,<br />
25σ<br />
= 50K<br />
0,<br />
25σ<br />
+ 0,<br />
75σ<br />
= 30K<br />
1<br />
2<br />
= σ<br />
1<br />
1<br />
1<br />
1<br />
2<br />
2<br />
2<br />
σ = 60MPa<br />
σ<br />
ϕ + σ<br />
2<br />
2<br />
ϕ + σ<br />
2<br />
2<br />
2<br />
2<br />
ϕ + τ<br />
( 1)<br />
( 2)<br />
1 − 0,<br />
35<br />
1 − 0,<br />
35<br />
= 50 + 30 = 60 + 20 = 80MPa<br />
2<br />
xy<br />
ϕ = σ cos<br />
2<br />
= sin<br />
= cos<br />
2<br />
sin<br />
⇒<br />
1<br />
2<br />
2<br />
sin 2ϕ<br />
2<br />
ϕ<br />
ϕ<br />
2<br />
2<br />
2<br />
30°<br />
+ σ<br />
sin<br />
( ϕ + 90°<br />
) = σ sin<br />
2<br />
1<br />
2<br />
30°<br />
K<br />
2<br />
( 1)<br />
30°<br />
+ σ<br />
2<br />
cos<br />
2<br />
= 50MPa<br />
= 30MPa<br />
30°<br />
K<br />
( 2)<br />
29
25. Zadatak<br />
Kocka dužine stranice “a“, izložena je djelovanju jednolikog tlaka intenziteta p = - 700 N/cm 2 .<br />
Odrediti Poissonov koeficijent ako je relativna promjena volumena -9·10 -5 , a modul elasti�nosti<br />
materijala 0,7·10 4 kN/cm 2 .<br />
N<br />
p = −700<br />
2<br />
cm<br />
6 N<br />
= −7<br />
⋅10<br />
2<br />
m<br />
4 kN<br />
E = 0,<br />
7 ⋅10<br />
2<br />
cm<br />
11 N<br />
= 0,<br />
7 ⋅10<br />
m<br />
ε<br />
−5<br />
= −9<br />
⋅10<br />
V<br />
1−<br />
2ν<br />
ε V = ε1<br />
+ ε 2 + ε 3 =<br />
E<br />
σ = σ = σ = σ = p<br />
1<br />
2<br />
3<br />
( σ + σ + σ )<br />
1−<br />
2ν<br />
1 ⎛ ε V E ⎞ 1 ⎛<br />
ε 3 ⎜1<br />
⎟ = 1<br />
2 3 2 ⎜<br />
V = p →ν<br />
= −<br />
−<br />
E<br />
⎝ p ⎠ ⎝<br />
ν = 0,<br />
35<br />
26. Zadatak<br />
1<br />
2<br />
2<br />
3<br />
−5<br />
( − 9 ⋅10<br />
) ⋅ 0,<br />
7 ⋅<br />
6<br />
3(<br />
− 7 ⋅10<br />
)<br />
U gredi �eli�nog mosta, pri prolazu vlaka, izmjerena su pomo�u tenzometra relativna izduženja u<br />
horizontalnom pravcu (u smjeru osi x ili paralelno s osi grede) 0,0004 i vertikalnom pravcu<br />
(u smjeru osi y ili okomito na os grede) -0,00012. Odrediti normalna naprezanja u pravcu osi<br />
grede i okomito na nju.<br />
ε<br />
ε<br />
x<br />
y<br />
E = 2,<br />
1⋅10<br />
ν =<br />
( 2)<br />
→ ( 1)<br />
y<br />
=<br />
=<br />
0,<br />
0004<br />
−0,<br />
00012<br />
0,<br />
3<br />
1<br />
ε x =<br />
E<br />
1<br />
ε y =<br />
E<br />
MPa<br />
( σ −νσ<br />
) → σ = Eε<br />
+ νσ K(<br />
1)<br />
( σ y −νσ<br />
x ) → σ y = Eε<br />
y + νσ x K(<br />
2)<br />
E(<br />
ε x + νε y ) 11<br />
2,<br />
1⋅10<br />
( 0,<br />
0004 − 0,<br />
3⋅<br />
0,<br />
00012)<br />
y<br />
x<br />
5<br />
⇒ σ x =<br />
2<br />
1−ν<br />
6 N<br />
σ x = 84 ⋅10<br />
2<br />
m<br />
= 84MPa<br />
x<br />
y<br />
x<br />
x<br />
=<br />
y<br />
1−<br />
0,<br />
3<br />
11<br />
6<br />
σ = Eε<br />
+ νσ = −0,<br />
00012 ⋅ 2,<br />
1⋅10<br />
+ 0,<br />
3⋅<br />
84 ⋅10<br />
= 0MPa<br />
2<br />
10<br />
11<br />
⎞<br />
⎟<br />
⎠<br />
30
27. Zadatak<br />
Za optere�enje na tlak betonske kocke ( stranice 7 cm) postavljene su na njene �etiri strane<br />
papu�ice od �elika spojene me�usobno zglobnim mehanizmom. Dvije sile veli�ine F djeluju u<br />
�vorovima A i D. Odrediti za koliko se promijeni volumen kocke ako je E =4 ·10 3 KN/cm 2 , � = 0,3 i F<br />
= 50 kN.<br />
�����<br />
Σy<br />
= 0:<br />
F + S<br />
Σx<br />
= 0:<br />
F<br />
�����<br />
F<br />
Σx<br />
= 0 → S B = − = S<br />
cosα<br />
Σy<br />
= 0 → F = F<br />
α = 45°<br />
∆V<br />
= εV<br />
⋅V<br />
= 1 2 3<br />
1−<br />
2ν<br />
E<br />
− S − 2F<br />
3<br />
σ1<br />
= σ 2 = = ; σ<br />
2<br />
2 3 = 0;<br />
V = a<br />
a 2a<br />
− 4<br />
∆V<br />
=<br />
∆V<br />
=<br />
AB<br />
+ S<br />
BC<br />
A<br />
( ε + ε + ε ) V = ( σ + σ + σ )<br />
( 1−<br />
2ν<br />
) aF − 4(<br />
1−<br />
2 ⋅ 0,<br />
3)<br />
2E<br />
−9,<br />
9 ⋅10<br />
sinα<br />
= 0 → S<br />
A<br />
−8<br />
⋅ cosα<br />
→ F<br />
m<br />
3<br />
=<br />
A<br />
A<br />
AB<br />
F<br />
= −<br />
sinα<br />
= F<br />
2 ⋅ 4 ⋅10<br />
1<br />
10<br />
2<br />
3<br />
⋅ 0,<br />
07⋅<br />
50⋅10<br />
3<br />
1−<br />
2ν<br />
V = 2σV<br />
E<br />
31
32<br />
28. Zadatak<br />
Odrediti potrebne duljine l1 i l2 zavara spajanjem kutnih metalnih profila s plo�om.<br />
( )( )<br />
( ) ( )<br />
cm<br />
l<br />
cm<br />
l<br />
cm<br />
l<br />
cm<br />
l<br />
l<br />
cm<br />
l<br />
l<br />
l<br />
l<br />
h<br />
h<br />
l<br />
l<br />
h<br />
l<br />
h<br />
l<br />
h<br />
a<br />
l<br />
h<br />
a<br />
l<br />
h<br />
F<br />
h<br />
F<br />
M<br />
cm<br />
t<br />
F<br />
l<br />
l<br />
t<br />
l<br />
l<br />
F<br />
al<br />
F<br />
A<br />
F<br />
MPa<br />
cm<br />
h<br />
cm<br />
h<br />
cm<br />
t<br />
kN<br />
F<br />
dop<br />
dop<br />
A<br />
dop<br />
dop<br />
dop<br />
5<br />
,<br />
5<br />
5<br />
,<br />
13<br />
19<br />
5<br />
,<br />
13<br />
4<br />
,<br />
1<br />
19<br />
19<br />
4<br />
,<br />
1<br />
19<br />
4<br />
,<br />
0<br />
19<br />
4<br />
,<br />
0<br />
4<br />
,<br />
0<br />
6<br />
,<br />
3<br />
4<br />
,<br />
1<br />
2<br />
2<br />
0<br />
19<br />
10<br />
90<br />
10<br />
5<br />
,<br />
0<br />
4<br />
,<br />
1<br />
10<br />
120<br />
7<br />
,<br />
0<br />
2<br />
45<br />
cos<br />
2<br />
90<br />
4<br />
,<br />
1<br />
6<br />
,<br />
3<br />
5<br />
,<br />
0<br />
120<br />
1<br />
2<br />
2<br />
2<br />
2<br />
2<br />
1<br />
2<br />
1<br />
1<br />
2<br />
2<br />
1<br />
2<br />
2<br />
1<br />
1<br />
2<br />
2<br />
1<br />
1<br />
2<br />
2<br />
1<br />
1<br />
6<br />
2<br />
3<br />
2<br />
1<br />
2<br />
1<br />
2<br />
1<br />
=<br />
−<br />
=<br />
→<br />
=<br />
=<br />
→<br />
=<br />
→<br />
=<br />
+<br />
→<br />
=<br />
+<br />
=<br />
→<br />
≅<br />
=<br />
=<br />
⇒<br />
⋅<br />
=<br />
⋅<br />
⇒<br />
⇒<br />
=<br />
→<br />
⋅<br />
=<br />
⋅<br />
→<br />
=<br />
∑<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
≥<br />
+<br />
⇒<br />
≤<br />
°<br />
+<br />
=<br />
∑<br />
=<br />
=<br />
=<br />
=<br />
=<br />
=<br />
=<br />
−<br />
τ<br />
τ<br />
τ<br />
τ<br />
τ<br />
τ<br />
l 1<br />
l 2<br />
h 1<br />
h2<br />
t<br />
a<br />
F
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
30. Zadatak<br />
Na vijak promjera d djeluje tla�na sila F koja izaziva naprezanje u vijku � i površinski pritisak p<br />
izme�u prstena promjera D i lima. Odrediti promjer D, te posmi�no naprezanje u prstenu ako je<br />
prsten debljine<br />
d = 10cm<br />
σ<br />
= 100MPa<br />
p = 40MPa<br />
t = 5cm<br />
p =<br />
σ =<br />
D =<br />
τ =<br />
F<br />
A<br />
F<br />
A<br />
F<br />
A<br />
s<br />
p<br />
d<br />
=<br />
4F<br />
2 2 ( D − d )<br />
2<br />
d π<br />
→ F = σA<br />
= σ<br />
4<br />
2<br />
2<br />
4σd<br />
π<br />
+ = d<br />
4πp<br />
2<br />
σd<br />
π σd<br />
= = =<br />
4dπt<br />
4t<br />
→ D =<br />
π<br />
d<br />
2<br />
σ<br />
1+<br />
= 10<br />
p<br />
6<br />
100 ⋅10<br />
⋅<br />
4 ⋅ 0,<br />
05<br />
0<br />
4F<br />
+<br />
πp<br />
, 1<br />
100<br />
1+<br />
= 18,<br />
7cm<br />
40<br />
= 5⋅10<br />
7<br />
= 50MPa<br />
34
31. Zadatak<br />
Metalna vilica i plo�ica 1 i 2 spojene su vijkom i optere�ene prema slici. Odrediti promjer vijka d,<br />
ako je<br />
F = 150 kN<br />
τ<br />
p<br />
a = 15mm<br />
b = 35mm<br />
p =<br />
p<br />
p<br />
p<br />
sdop<br />
dop<br />
τ =<br />
1<br />
2<br />
1<br />
=<br />
=<br />
><br />
= 70MPa<br />
= 150 MPa<br />
F<br />
A<br />
s<br />
F<br />
A<br />
F<br />
A<br />
F<br />
A<br />
p<br />
p<br />
p1<br />
p 2<br />
2<br />
=<br />
≤<br />
⇒<br />
F<br />
mA<br />
p<br />
=<br />
dop<br />
F<br />
=<br />
2ad<br />
d<br />
F<br />
= ≤ τ 2<br />
d π<br />
m<br />
4<br />
F<br />
bd<br />
≥<br />
2<br />
F<br />
ap<br />
dop<br />
sdop<br />
→ d ≥<br />
0,<br />
0369 m > 0,<br />
033m<br />
⇒ d = 37mm<br />
4F<br />
mπτ<br />
3<br />
150 ⋅10<br />
=<br />
2 ⋅ 0,<br />
015 ⋅150<br />
⋅10<br />
6<br />
sdop<br />
=<br />
=<br />
0,<br />
033<br />
3<br />
4 ⋅150<br />
⋅10<br />
2 ⋅π<br />
⋅ 70 ⋅10<br />
m<br />
6<br />
=<br />
0,<br />
0369<br />
m<br />
35
Zavareni spojevi<br />
Lom zavarenih spojeva optere�enih na smicanje nastaje po najslabijem (smi�nom) presjeku A-B,<br />
tj. po presjeku za koji je smi�na površima najmanja.<br />
Naprezanja u zavaru iznose :<br />
- normalno naprezanje<br />
σ =<br />
F<br />
al<br />
F<br />
- posmi�no naprezanje τ =<br />
al<br />
F<br />
ili op�enito σ =<br />
Aσ<br />
F F<br />
= , τ =<br />
∑(<br />
a ⋅l<br />
) σ Aτ<br />
F<br />
=<br />
∑(<br />
a ⋅l<br />
) τ<br />
gdje je A�, A� … veli�ina površine zavara koja<br />
se odnosi na normalno, odnosno posmi�no naprezanje.<br />
Za spoj na slici posmi�no naprezanje je<br />
τ =<br />
F<br />
A<br />
τ<br />
=<br />
∑<br />
F<br />
a<br />
=<br />
F<br />
( ⋅l<br />
) a(<br />
2l<br />
+ l)<br />
1<br />
Normalno naprezanje jednako je posmi�nom.<br />
Kod zavarenih spojeva uobi�ajeno se provjeravaju<br />
normalna naprezanja ako sila djeluje okomito na površinu<br />
zavarenog spoja, odnosno na posmi�na naprezanja dok sila<br />
djeluje u samoj površini zavarenog spoja.<br />
36
37<br />
33. Zadatak<br />
Odrediti potrebnu duljinu zavara prema slici.<br />
5<br />
40<br />
40<br />
84<br />
,<br />
0<br />
7<br />
,<br />
0<br />
2<br />
,<br />
1<br />
135<br />
50<br />
2<br />
1<br />
2<br />
1<br />
2<br />
×<br />
×<br />
=<br />
=<br />
≥<br />
=<br />
=<br />
B<br />
t<br />
a<br />
t<br />
a<br />
t<br />
t<br />
mm<br />
N<br />
kN<br />
F<br />
dop<br />
τ<br />
Iz priru�nika za profile :<br />
2<br />
2<br />
379<br />
6<br />
,<br />
11<br />
40<br />
mm<br />
A<br />
mm<br />
e<br />
mm<br />
b<br />
=<br />
=<br />
=<br />
( )<br />
( )<br />
( )<br />
mm<br />
l<br />
mm<br />
l<br />
usvojeno<br />
mm<br />
a<br />
A<br />
l<br />
mm<br />
a<br />
A<br />
l<br />
mm<br />
a<br />
mm<br />
a<br />
usvojeno<br />
mm<br />
t<br />
a<br />
mm<br />
t<br />
a<br />
mm<br />
A<br />
A<br />
A<br />
mm<br />
b<br />
Ae<br />
A<br />
b<br />
Fe<br />
F<br />
mm<br />
F<br />
A<br />
A<br />
F<br />
Fe<br />
b<br />
F<br />
M<br />
F<br />
F<br />
F<br />
F<br />
dop<br />
dop<br />
B<br />
z<br />
70<br />
,<br />
40<br />
62<br />
,<br />
62<br />
,<br />
56<br />
,<br />
30<br />
4<br />
,<br />
3<br />
2<br />
,<br />
4<br />
5<br />
84<br />
,<br />
0<br />
84<br />
,<br />
0<br />
5<br />
,<br />
3<br />
5<br />
7<br />
,<br />
0<br />
7<br />
,<br />
0<br />
263<br />
107<br />
40<br />
6<br />
,<br />
11<br />
370<br />
2<br />
370<br />
135<br />
10<br />
50<br />
2<br />
0<br />
1<br />
0<br />
2<br />
1<br />
2<br />
2<br />
2<br />
1<br />
1<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
2<br />
2<br />
1<br />
2<br />
1<br />
2<br />
3<br />
2<br />
1<br />
2<br />
1<br />
=<br />
=<br />
→<br />
=<br />
=<br />
=<br />
=<br />
=<br />
=<br />
→<br />
=<br />
⋅<br />
=<br />
=<br />
=<br />
⋅<br />
=<br />
=<br />
=<br />
−<br />
=<br />
=<br />
⋅<br />
=<br />
=<br />
→<br />
=<br />
→<br />
=<br />
⋅<br />
=<br />
≥<br />
→<br />
≤<br />
=<br />
=<br />
→<br />
=<br />
∑<br />
=<br />
+<br />
→<br />
=<br />
∑<br />
τ<br />
τ<br />
τ<br />
K<br />
K
34. Zadatak<br />
Odrediti najve�a naprezanja zakovica.<br />
Q = 20kN<br />
d = 1,<br />
4cm<br />
a = 4cm<br />
b = 8cm<br />
H = 12kN<br />
Sila na jednu zakovicu od djelovanja sile H<br />
H<br />
FH =<br />
h<br />
Moment u težištu sustava zakovica<br />
M = Q a + b<br />
( )<br />
Sila na jednu zakovicu od djelovanja sile Q<br />
Q<br />
FQ =<br />
h<br />
Sila u i-toj zakovici od momenta M<br />
F = k ⋅ ρ<br />
Mi<br />
a<br />
a<br />
i<br />
4<br />
3<br />
a<br />
5<br />
a<br />
1<br />
2<br />
b<br />
Q<br />
H<br />
F M<br />
4<br />
F Q<br />
3<br />
F M<br />
F Q<br />
F H<br />
FH 5<br />
F Q<br />
F M<br />
1<br />
F H<br />
2<br />
F Q<br />
F Q<br />
45°<br />
38
39<br />
Moment od sile FMi s obzirom na težište sustava<br />
( ) ( ) ( )<br />
kN<br />
a<br />
b<br />
a<br />
Q<br />
a<br />
a<br />
b<br />
a<br />
Q<br />
F<br />
F<br />
F<br />
F<br />
F<br />
F<br />
a<br />
M<br />
F<br />
M<br />
k<br />
k<br />
M<br />
M<br />
k<br />
F<br />
M<br />
M<br />
M<br />
M<br />
M<br />
M<br />
M<br />
i<br />
i<br />
i<br />
i<br />
Mi<br />
i<br />
i<br />
i<br />
i<br />
i<br />
i<br />
i<br />
Mi<br />
i<br />
6<br />
,<br />
10<br />
04<br />
,<br />
0<br />
8<br />
08<br />
,<br />
0<br />
04<br />
,<br />
0<br />
20<br />
2<br />
8<br />
2<br />
2<br />
2<br />
4<br />
0<br />
,<br />
2<br />
2<br />
4<br />
3<br />
2<br />
1<br />
5<br />
4<br />
3<br />
2<br />
5<br />
1<br />
2<br />
5<br />
1<br />
2<br />
2<br />
5<br />
1<br />
2<br />
=<br />
⋅<br />
+<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
+<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
+<br />
⋅<br />
=<br />
=<br />
=<br />
=<br />
=<br />
=<br />
=<br />
=<br />
=<br />
=<br />
⋅<br />
=<br />
=<br />
⇒<br />
⋅<br />
=<br />
∑<br />
=<br />
⋅<br />
=<br />
⋅<br />
=<br />
∑<br />
∑<br />
∑<br />
=<br />
=<br />
=<br />
ρ<br />
ρ<br />
ρ<br />
ρ<br />
ρ<br />
ρ<br />
ρ<br />
ρ<br />
ρ<br />
ρ<br />
ρ<br />
U srednjoj zakovici ne djeluje sila od momenta!<br />
0<br />
5 =<br />
M<br />
F<br />
kN<br />
F<br />
F<br />
R<br />
kN<br />
F<br />
F<br />
F<br />
F<br />
R<br />
kN<br />
F<br />
F<br />
F<br />
F<br />
R<br />
kN<br />
F<br />
F<br />
F<br />
F<br />
R<br />
kN<br />
F<br />
F<br />
F<br />
F<br />
R<br />
kN<br />
F<br />
kN<br />
F<br />
Q<br />
H<br />
M<br />
Q<br />
M<br />
H<br />
M<br />
Q<br />
M<br />
H<br />
M<br />
Q<br />
M<br />
H<br />
M<br />
Q<br />
M<br />
H<br />
Q<br />
H<br />
66<br />
,<br />
4<br />
49<br />
,<br />
10<br />
2<br />
2<br />
2<br />
2<br />
17<br />
,<br />
6<br />
2<br />
2<br />
2<br />
2<br />
57<br />
,<br />
12<br />
2<br />
2<br />
2<br />
2<br />
15<br />
,<br />
15<br />
2<br />
2<br />
2<br />
2<br />
4<br />
5<br />
20<br />
4<br />
,<br />
2<br />
5<br />
12<br />
2<br />
2<br />
5<br />
2<br />
2<br />
4<br />
2<br />
2<br />
3<br />
2<br />
2<br />
2<br />
2<br />
2<br />
1<br />
=<br />
+<br />
=<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
⋅<br />
−<br />
+<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
⋅<br />
+<br />
=<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
⋅<br />
−<br />
+<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
⋅<br />
−<br />
=<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
⋅<br />
+<br />
+<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
⋅<br />
−<br />
=<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
⋅<br />
+<br />
+<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
⋅<br />
+<br />
=<br />
=<br />
=<br />
=<br />
=<br />
Najve�e optere�enje ima zakovica 1 : R1 = 15,15 kN<br />
MPa<br />
d<br />
R<br />
mA<br />
F<br />
s<br />
20<br />
,<br />
49<br />
4<br />
014<br />
,<br />
0<br />
2<br />
15150<br />
4<br />
2<br />
2<br />
2<br />
1<br />
=<br />
=<br />
=<br />
=<br />
π<br />
π<br />
τ
35. Zadatak<br />
Unutarnji promjer vanjskog �eli�nog prstena je za D manji od vanjskog promjera unutarnjeg<br />
aluminijskog valjka, pa je �eli�ni prsten u zagrijanom stanju nataknut na aluminijski valjak.<br />
Odrediti pritisak prstena na valjak i naprezanje u prstenu nakon hla�enja prstena.<br />
Kolika sila djeluje u prstenu?<br />
D = 1,<br />
1mm<br />
E<br />
E<br />
ν<br />
ν<br />
1<br />
2<br />
Al<br />
�<br />
5 N<br />
= E Al = 0,<br />
7 ⋅10<br />
mm<br />
5 N<br />
= E = 2 ⋅10<br />
�<br />
2<br />
mm<br />
= 0,<br />
34 = ν<br />
=<br />
0,<br />
3<br />
t = 5mm<br />
= ν<br />
Uvjet deformacija :<br />
∆D = ∆D<br />
+ ∆D<br />
K 1<br />
2<br />
1<br />
( )<br />
∆D<br />
= D − D = 1,<br />
1mm<br />
1<br />
1<br />
Uvjet ravnoteže :<br />
2 ⋅ S − p ⋅ D = 0K<br />
2<br />
2<br />
p ⋅ D<br />
S 2 =<br />
2<br />
R = p ⋅ D<br />
2<br />
2<br />
2<br />
2<br />
⋅t<br />
2<br />
( )<br />
Izraz za deformacije :<br />
( 1)<br />
( 1)<br />
1 σ 2<br />
−ν<br />
1<br />
1 E1<br />
2<br />
σ<br />
p p p ∆D1<br />
ε1<br />
= = −ν<br />
1 = ( 1−ν<br />
1 ) =<br />
E<br />
E E E D<br />
Za valjak :<br />
F = D<br />
2<br />
2<br />
⋅ h<br />
( 2)<br />
∆D2<br />
2 = ,<br />
D E<br />
σ<br />
ε<br />
Za prsten :<br />
2<br />
2<br />
D p D2<br />
p ∆D<br />
ε<br />
2 = = =<br />
2tE D 2tE<br />
D<br />
2<br />
2<br />
1<br />
1<br />
1<br />
S 2 ∆D2<br />
S 2 ( 2)<br />
D2<br />
p<br />
= ⇒ = ⎯⎯→<br />
∆D2<br />
=<br />
F ⋅ E D E ⋅ F<br />
2tE<br />
2<br />
1<br />
2<br />
2<br />
2<br />
1<br />
1<br />
2<br />
1<br />
40
∆D<br />
( 1)<br />
1 ( 1−ν<br />
)<br />
1<br />
1<br />
∆D<br />
2<br />
= ε D<br />
2<br />
1<br />
2<br />
1<br />
1<br />
= ε D<br />
pD<br />
⎯⎯→<br />
E<br />
D<br />
≅ D<br />
p =<br />
1600<br />
0,<br />
7 ⋅10<br />
2<br />
5<br />
=<br />
pD<br />
E<br />
1<br />
1<br />
1<br />
( 1−ν<br />
)<br />
2<br />
2<br />
pD<br />
=<br />
2tE<br />
2<br />
≅ 1600mm<br />
1,<br />
1<br />
( 1−<br />
0,<br />
34)<br />
1<br />
2<br />
2<br />
pD<br />
+<br />
2tE<br />
2<br />
= 1,<br />
1 → p =<br />
D<br />
E<br />
2<br />
1600<br />
+<br />
2 ⋅5<br />
⋅ 2 ⋅10<br />
5<br />
=<br />
1<br />
1<br />
0,<br />
85<br />
1,<br />
1<br />
( 1−ν<br />
)<br />
N<br />
mm<br />
2<br />
1<br />
2<br />
2<br />
D<br />
+<br />
2tE<br />
= σ<br />
Naprezanje u prstenu (homogeno stanje naprezanja) :<br />
( 2)<br />
pD2<br />
0,<br />
85⋅1600<br />
N<br />
σ 1 = = = 136,<br />
0 2<br />
2t 2 ⋅5<br />
mm<br />
Sila u prstenu :<br />
( 2)<br />
1<br />
S = t ⋅ h ⋅σ<br />
= 5⋅<br />
200 ⋅136,<br />
0 = 136kN<br />
Kontrola :<br />
∆D<br />
2<br />
∆D<br />
1<br />
( 2)<br />
1<br />
σ<br />
= D2<br />
E<br />
pD1<br />
=<br />
E<br />
1<br />
1<br />
2<br />
136,<br />
0<br />
= 1600 = 1,<br />
087mm<br />
5<br />
2 ⋅10<br />
0,<br />
85⋅1600<br />
5<br />
0,<br />
7 ⋅10<br />
( 1−ν<br />
) = ( 1−<br />
0,<br />
34)<br />
∆D<br />
= ∆D<br />
+ ∆D<br />
= 1,<br />
087 +<br />
2<br />
1<br />
0,<br />
013<br />
= 1,<br />
1mm<br />
= 0,<br />
013mm<br />
2<br />
( 1)<br />
( 1)<br />
1 = σ 2<br />
41
36. Zadatak<br />
U parnom kotlu promjera D djeluje pritisak p. Kotlovski lim debljine t spojen je u uzdužnom<br />
smjeru dvorednim zakovicama promjera d na me�usobnom razmaku e. Odrediti :<br />
a) naprezanja u sastavu<br />
b) koliki maksimalni tlak pare može kotao podnijeti ako su zadana dopuštena naprezanja.<br />
D = 1600mm<br />
N<br />
p = 1,<br />
0<br />
mm<br />
t = 10mm<br />
d = 20mm<br />
e = 100mm<br />
τ<br />
σ<br />
σ<br />
dop<br />
dop<br />
O dop<br />
2<br />
N<br />
= 70 2<br />
mm<br />
N<br />
= 100 2<br />
mm<br />
N<br />
= 160<br />
mm<br />
2<br />
a) Naprezanje stjenke kotla (u smjeru tangente na stjenku)<br />
pD 1,<br />
0 ⋅1600<br />
N<br />
σ = = = 80 2<br />
2δ 2 ⋅10<br />
mm<br />
Sila na razmaku e (sila koja djeluje na jedan red zakovica)<br />
S = σ ⋅δ<br />
⋅ e = 80 ⋅10<br />
⋅100<br />
= 80kN<br />
Naprezanje zakovice<br />
S 80000 N<br />
posmik τ = = = 127,<br />
32<br />
2<br />
2<br />
2<br />
d π 20 π mm<br />
m 2<br />
4 4<br />
S 80000 N<br />
bo�ni pritisak σ O = = = 200 2<br />
2dδ 2 ⋅ 20 ⋅10<br />
mm<br />
Vlak na oslabljenom presjeku<br />
σ<br />
=<br />
S<br />
=<br />
e − d δ<br />
80000<br />
= 100<br />
100 − 20 10<br />
N<br />
mm<br />
( ) ( ) 2<br />
42
43<br />
b) Nosivost zakovice<br />
( ) ( )<br />
2<br />
max<br />
2<br />
max<br />
2<br />
2<br />
200<br />
200<br />
160<br />
1600<br />
100<br />
10<br />
2<br />
2<br />
2<br />
80<br />
80<br />
10<br />
20<br />
100<br />
100<br />
32<br />
20<br />
10<br />
160<br />
22<br />
991<br />
,<br />
21<br />
4<br />
20<br />
70<br />
4<br />
mm<br />
N<br />
p<br />
mm<br />
N<br />
D<br />
p<br />
pD<br />
kN<br />
N<br />
N<br />
kN<br />
d<br />
e<br />
N<br />
kN<br />
d<br />
N<br />
kN<br />
kN<br />
d<br />
N<br />
V<br />
dop<br />
V<br />
Odop<br />
O<br />
dop<br />
S<br />
=<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
⇒<br />
=<br />
=<br />
=<br />
=<br />
−<br />
=<br />
−<br />
=<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
≅<br />
=<br />
=<br />
=<br />
δσ<br />
δ<br />
σ<br />
δ<br />
σ<br />
δ<br />
σ<br />
π<br />
π<br />
τ
44<br />
37. Zadatak<br />
Za satavljeni štap na slici potrebno je odrediti najve�a posmi�na naprezanja i potencijalnu<br />
energiju deformacija ako su u to�ki A zadana glavna naprezanja.<br />
30<br />
,<br />
0<br />
10<br />
2<br />
8<br />
8<br />
2<br />
5<br />
2<br />
2<br />
2<br />
1<br />
=<br />
⋅<br />
=<br />
−<br />
=<br />
=<br />
ν<br />
σ<br />
σ<br />
mm<br />
N<br />
E<br />
mm<br />
N<br />
mm<br />
N<br />
kNm<br />
Nmm<br />
I<br />
M<br />
I<br />
M<br />
m<br />
mm<br />
D<br />
I<br />
m<br />
mm<br />
D<br />
I<br />
A<br />
p<br />
A<br />
C<br />
A<br />
p<br />
C<br />
A<br />
p<br />
p<br />
4<br />
,<br />
5<br />
10<br />
4<br />
,<br />
5<br />
30<br />
10<br />
3<br />
,<br />
20<br />
8<br />
10<br />
3<br />
,<br />
20<br />
10<br />
3<br />
,<br />
20<br />
32<br />
120<br />
32<br />
10<br />
02<br />
,<br />
4<br />
10<br />
02<br />
,<br />
4<br />
32<br />
80<br />
32<br />
6<br />
6<br />
2<br />
2<br />
1<br />
4<br />
6<br />
4<br />
6<br />
4<br />
4<br />
2<br />
2<br />
4<br />
6<br />
4<br />
6<br />
4<br />
4<br />
1<br />
1<br />
=<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
=<br />
→<br />
=<br />
=<br />
⋅<br />
=<br />
⋅<br />
=<br />
=<br />
=<br />
⋅<br />
=<br />
⋅<br />
=<br />
=<br />
=<br />
−<br />
−<br />
ρ<br />
τ<br />
ρ<br />
σ<br />
τ<br />
π<br />
π<br />
π<br />
π<br />
Uvjet ravnoteže<br />
T<br />
C<br />
B<br />
M<br />
M<br />
M =<br />
+<br />
Uvjet deformacije<br />
0<br />
0<br />
,<br />
2<br />
4<br />
,<br />
1<br />
8<br />
,<br />
0<br />
0<br />
1<br />
2<br />
1<br />
=<br />
⋅<br />
⋅<br />
−<br />
⋅<br />
⋅<br />
−<br />
⋅<br />
⋅<br />
→<br />
=<br />
p<br />
C<br />
p<br />
C<br />
p<br />
T<br />
C<br />
I<br />
G<br />
M<br />
I<br />
G<br />
M<br />
I<br />
G<br />
M<br />
ϕ<br />
kNm<br />
M<br />
M<br />
M<br />
kNm<br />
M<br />
I<br />
I<br />
M<br />
I<br />
I<br />
M<br />
I<br />
M<br />
C<br />
T<br />
B<br />
T<br />
p<br />
p<br />
C<br />
p<br />
p<br />
C<br />
p<br />
T<br />
10<br />
4<br />
,<br />
15<br />
0<br />
,<br />
2<br />
3<br />
,<br />
20<br />
02<br />
,<br />
4<br />
4<br />
,<br />
1<br />
8<br />
,<br />
0<br />
413<br />
,<br />
5<br />
0<br />
,<br />
2<br />
4<br />
,<br />
1<br />
8<br />
,<br />
0<br />
0<br />
,<br />
2<br />
4<br />
,<br />
1<br />
8<br />
,<br />
0 2<br />
1<br />
1<br />
2<br />
1<br />
=<br />
−<br />
=<br />
=<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎣<br />
⎡<br />
+<br />
=<br />
⎥<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎢<br />
⎣<br />
⎡<br />
+<br />
=<br />
⎥<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎢<br />
⎣<br />
⎡<br />
+<br />
=
45<br />
( ) ( )<br />
Nmm<br />
kNm<br />
U<br />
U<br />
I<br />
G<br />
M<br />
I<br />
G<br />
M<br />
I<br />
G<br />
M<br />
U<br />
m<br />
kN<br />
mm<br />
N<br />
E<br />
G<br />
mm<br />
N<br />
D<br />
I<br />
M<br />
p<br />
C<br />
p<br />
C<br />
p<br />
B<br />
p<br />
B<br />
198000<br />
198<br />
,<br />
0<br />
013<br />
,<br />
0<br />
056<br />
,<br />
0<br />
129<br />
,<br />
0<br />
10<br />
3<br />
,<br />
20<br />
10<br />
77<br />
,<br />
0<br />
2<br />
4<br />
,<br />
1<br />
4<br />
,<br />
5<br />
10<br />
02<br />
,<br />
4<br />
10<br />
77<br />
,<br />
0<br />
2<br />
2<br />
,<br />
1<br />
4<br />
,<br />
5<br />
10<br />
02<br />
,<br />
4<br />
10<br />
77<br />
,<br />
0<br />
2<br />
8<br />
,<br />
0<br />
10<br />
2<br />
4<br />
,<br />
1<br />
2<br />
2<br />
,<br />
1<br />
2<br />
8<br />
,<br />
0<br />
10<br />
77<br />
,<br />
0<br />
10<br />
77<br />
,<br />
0<br />
3<br />
,<br />
0<br />
1<br />
2<br />
10<br />
2<br />
1<br />
2<br />
5<br />
,<br />
99<br />
2<br />
80<br />
10<br />
02<br />
,<br />
4<br />
10<br />
10<br />
2<br />
6<br />
8<br />
2<br />
6<br />
8<br />
2<br />
6<br />
8<br />
2<br />
2<br />
2<br />
1<br />
2<br />
1<br />
2<br />
2<br />
8<br />
2<br />
5<br />
5<br />
2<br />
6<br />
6<br />
1<br />
1<br />
max<br />
=<br />
=<br />
+<br />
+<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
+<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
+<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
+<br />
⋅<br />
⋅<br />
⋅<br />
+<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
=<br />
⋅<br />
=<br />
+<br />
⋅<br />
=<br />
+<br />
=<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
−<br />
−<br />
−<br />
ν<br />
τ
38. Zadatak<br />
Osovina kružnog popre���������������������ena je momentom torzije Mt. Temperatura osovine<br />
se promijeni za �t. Odrediti smjer i veli�inu glavnih naprezanja u to�ki C.<br />
E = 2,<br />
0<br />
ν<br />
=<br />
α<br />
t<br />
∆t<br />
= + 30K<br />
D = 120mm<br />
M<br />
A<br />
G =<br />
2<br />
E<br />
=<br />
Mt<br />
0,5 m<br />
2 ⋅10<br />
( 1+<br />
ν ) 2(<br />
1+<br />
0,<br />
3)<br />
5<br />
5<br />
= 0,<br />
77 ⋅10<br />
N<br />
mm<br />
1,7 m<br />
4<br />
4<br />
D π 120 π<br />
6 4<br />
I p = = = 20,<br />
4 ⋅10<br />
mm = 20,<br />
36 ⋅10<br />
32 32<br />
2<br />
2<br />
D π 120 π<br />
4 2<br />
−2<br />
2<br />
A = = = 1,<br />
1⋅10<br />
mm = 1,<br />
1⋅10<br />
m<br />
4 4<br />
Uvjet ravnoteže :<br />
M + M = M<br />
A<br />
B<br />
t<br />
K(<br />
1)<br />
Uvjet deformacije :<br />
M A ⋅l<br />
M t ⋅b<br />
ϕ A = − = 0K<br />
G ⋅ I G ⋅ I<br />
M<br />
M<br />
t<br />
A<br />
B<br />
0,<br />
30<br />
=<br />
1,<br />
25<br />
⋅10<br />
= 16kNm<br />
( 2)<br />
b 1,<br />
2<br />
= M t = 16 = 11,<br />
29kNm<br />
l 1,<br />
7<br />
= M − M = 4,<br />
71kNm<br />
t<br />
⋅10<br />
p<br />
5<br />
N<br />
mm<br />
−5<br />
A<br />
K<br />
2<br />
−1<br />
p<br />
C<br />
2<br />
1,2 m<br />
8 kN<br />
= 0,<br />
77 ⋅10<br />
2<br />
m<br />
−6<br />
m<br />
4<br />
t<br />
0,8 m<br />
D<br />
B<br />
C<br />
46
47<br />
( )<br />
( )<br />
°<br />
−<br />
=<br />
→<br />
°<br />
−<br />
=<br />
→<br />
−<br />
=<br />
−<br />
⋅<br />
=<br />
−<br />
=<br />
+<br />
=<br />
±<br />
−<br />
=<br />
⋅<br />
+<br />
−<br />
±<br />
−<br />
=<br />
+<br />
±<br />
=<br />
−<br />
=<br />
⋅<br />
⋅<br />
−<br />
=<br />
−<br />
=<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
∆<br />
⋅<br />
=<br />
⇒<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
∆<br />
⋅<br />
=<br />
∆<br />
−<br />
215<br />
,<br />
10<br />
43<br />
,<br />
20<br />
2<br />
3725<br />
,<br />
0<br />
52<br />
,<br />
74<br />
88<br />
,<br />
13<br />
2<br />
2<br />
02<br />
,<br />
77<br />
50<br />
,<br />
2<br />
76<br />
,<br />
39<br />
26<br />
,<br />
37<br />
88<br />
,<br />
13<br />
4<br />
52<br />
,<br />
74<br />
2<br />
1<br />
2<br />
52<br />
,<br />
74<br />
4<br />
2<br />
1<br />
2<br />
52<br />
,<br />
74<br />
10<br />
1<br />
,<br />
1<br />
10<br />
23<br />
,<br />
848<br />
88<br />
,<br />
13<br />
2<br />
120<br />
10<br />
36<br />
,<br />
20<br />
10<br />
71<br />
,<br />
4<br />
2<br />
23<br />
,<br />
848<br />
10<br />
1<br />
,<br />
1<br />
10<br />
2<br />
30<br />
10<br />
25<br />
,<br />
1<br />
2<br />
2<br />
2<br />
1<br />
2<br />
2<br />
2<br />
2<br />
2<br />
,<br />
1<br />
2<br />
4<br />
3<br />
2<br />
6<br />
6<br />
4<br />
5<br />
5<br />
ϕ<br />
ϕ<br />
ϕ<br />
σ<br />
σ<br />
τ<br />
σ<br />
σ<br />
σ<br />
σ<br />
τ<br />
α<br />
α<br />
tg<br />
mm<br />
N<br />
mm<br />
N<br />
tlak<br />
mm<br />
N<br />
A<br />
F<br />
mm<br />
N<br />
D<br />
I<br />
M<br />
kN<br />
F<br />
A<br />
E<br />
t<br />
F<br />
A<br />
E<br />
l<br />
F<br />
l<br />
t<br />
l<br />
C<br />
C<br />
C<br />
C<br />
C<br />
C<br />
C<br />
p<br />
B<br />
C<br />
t<br />
t
39. Zadatak<br />
Za spoj dva vratila odrediti potreban broj vijaka krute spojke. Zanemariti utjecaj spojke na<br />
uvijanje.<br />
d = 12mm<br />
τ<br />
V<br />
M<br />
M<br />
D<br />
D<br />
dop<br />
t1<br />
t 2<br />
1<br />
2<br />
= 50MPa<br />
= 10kNm<br />
= 15kNm<br />
= 100mm<br />
= 140mm<br />
2R<br />
= 0,<br />
2m<br />
I<br />
I<br />
pI<br />
pIII<br />
= I<br />
pII<br />
= I<br />
pIV<br />
4<br />
1<br />
4<br />
D π 0,<br />
1 π<br />
−5<br />
= = = 0,<br />
9817 ⋅10<br />
m<br />
32 32<br />
4<br />
2<br />
Uvjet ravnoteže :<br />
4<br />
D π 0,<br />
14 π<br />
−5<br />
= = = 3,<br />
771⋅10<br />
m<br />
32 32<br />
∑ M z = 0 → M A + M B + M t1<br />
− M t 2<br />
Uvjet deformacije :<br />
ϕ<br />
4<br />
= ∑<br />
1<br />
= i<br />
AB<br />
i=<br />
ϕ = 0 →ϕ<br />
+ ϕ<br />
i<br />
I<br />
II<br />
+ ϕ<br />
III<br />
+ ϕ<br />
= 0K<br />
IV<br />
4<br />
( 1)<br />
= 0K<br />
4<br />
( 2)<br />
48
( 2)<br />
⎛ M A − ⎜<br />
⎝ GI pI<br />
M A + M t1<br />
M A + M t1<br />
M A + M t1<br />
− M<br />
+ + +<br />
GI pII GI pIII GI pIV<br />
I = I , I = I<br />
pI<br />
M<br />
M<br />
A<br />
A<br />
→<br />
pII<br />
M<br />
=<br />
t 2<br />
I<br />
pIII<br />
pI<br />
2<br />
M t1(<br />
2I<br />
pI + I pIII )<br />
( I + I )<br />
−<br />
pI<br />
15 ⋅0,<br />
9817<br />
=<br />
2<br />
pIV<br />
pIII<br />
−10(<br />
2 ⋅0,<br />
9817 + 3,<br />
771)<br />
( 0,<br />
9817 + 3,<br />
771)<br />
Torzijski moment u presjeku C :<br />
M C = M A + M t<br />
1<br />
= 4,<br />
483 −10<br />
= −5,<br />
517kNm<br />
t 2<br />
= −4,<br />
483kNm<br />
⎞<br />
⎟ ⋅ a = 0<br />
⎟<br />
⎠<br />
Zbog djelovanja momenta MC vijci spojke su optere�eni na smicanje :<br />
M C = n ⋅ F ⋅ R<br />
n – broj vijaka<br />
F – posmi�na sila koja djeluje na jedan vijak<br />
R – krak djelovanja sile F<br />
Potreban broj vijaka u spoju :<br />
τ<br />
=<br />
s<br />
n ≥<br />
d<br />
F<br />
A<br />
v<br />
2<br />
v<br />
4F<br />
= ≤ τ 2<br />
d π<br />
4M<br />
πτ<br />
C<br />
v<br />
sdop<br />
sdop<br />
=<br />
R 0,<br />
012<br />
d v π<br />
→ F ≤ τ<br />
4<br />
2<br />
2<br />
4 ⋅5517<br />
⋅π<br />
⋅50<br />
⋅10<br />
6<br />
sdop<br />
=<br />
⋅ 0,<br />
1<br />
9,<br />
76<br />
→ n = 10vijaka<br />
49
50<br />
40. Zadatak<br />
Sastavljena osovina kružnog popre�nog presjeka optere�ena je momentom torzije Mt . Odrediti<br />
najve�a naprezanja u pojedinim dijelovima osovine i potencijalnu energiju deformacije.<br />
mm<br />
d<br />
mm<br />
D<br />
mm<br />
D<br />
mm<br />
N<br />
E<br />
kNm<br />
M t<br />
64<br />
120<br />
80<br />
3<br />
,<br />
0<br />
10<br />
2<br />
15<br />
2<br />
1<br />
5<br />
=<br />
=<br />
=<br />
=<br />
⋅<br />
=<br />
=<br />
ν<br />
( ) ( )<br />
4<br />
6<br />
4<br />
4<br />
2<br />
2<br />
4<br />
6<br />
4<br />
4<br />
4<br />
4<br />
1<br />
1<br />
10<br />
3<br />
,<br />
20<br />
32<br />
120<br />
32<br />
10<br />
4<br />
,<br />
2<br />
32<br />
64<br />
80<br />
32<br />
mm<br />
D<br />
I<br />
mm<br />
d<br />
D<br />
I<br />
p<br />
p<br />
⋅<br />
=<br />
=<br />
=<br />
⋅<br />
=<br />
−<br />
=<br />
−<br />
=<br />
π<br />
π<br />
π<br />
π<br />
Uvjet ravnoteže<br />
( )<br />
1<br />
0 K<br />
t<br />
B<br />
A<br />
z<br />
M<br />
M<br />
M<br />
M =<br />
+<br />
→<br />
=<br />
∑<br />
Uvjet deformacije<br />
( )<br />
kNm<br />
M<br />
M<br />
M<br />
kNm<br />
I<br />
I<br />
I<br />
M<br />
M<br />
GI<br />
M<br />
GI<br />
M<br />
GI<br />
M<br />
B<br />
t<br />
A<br />
p<br />
p<br />
p<br />
t<br />
B<br />
p<br />
B<br />
p<br />
B<br />
p<br />
t<br />
A<br />
458<br />
,<br />
9<br />
54<br />
,<br />
5<br />
0<br />
,<br />
2<br />
4<br />
,<br />
1<br />
1<br />
8<br />
,<br />
0<br />
2<br />
0<br />
0<br />
,<br />
2<br />
4<br />
,<br />
1<br />
8<br />
,<br />
0<br />
0<br />
1<br />
2<br />
1<br />
1<br />
2<br />
1<br />
=<br />
−<br />
=<br />
=<br />
+<br />
=<br />
=<br />
⋅<br />
−<br />
⋅<br />
−<br />
⋅<br />
→<br />
= K<br />
ϕ<br />
( ) ( )<br />
kNm<br />
GI<br />
M<br />
GI<br />
M<br />
GI<br />
M<br />
U<br />
E<br />
m<br />
kN<br />
mm<br />
N<br />
E<br />
G<br />
mm<br />
N<br />
D<br />
I<br />
M<br />
mm<br />
N<br />
D<br />
I<br />
M<br />
p<br />
B<br />
p<br />
B<br />
p<br />
A<br />
p<br />
p<br />
A<br />
p<br />
B<br />
307<br />
,<br />
0<br />
0137<br />
,<br />
0<br />
0996<br />
,<br />
0<br />
1936<br />
,<br />
0<br />
2<br />
4<br />
,<br />
1<br />
2<br />
2<br />
,<br />
1<br />
2<br />
8<br />
,<br />
0<br />
10<br />
77<br />
,<br />
0<br />
10<br />
77<br />
,<br />
0<br />
3<br />
,<br />
0<br />
1<br />
2<br />
10<br />
2<br />
1<br />
2<br />
63<br />
,<br />
157<br />
2<br />
80<br />
10<br />
4<br />
,<br />
2<br />
10<br />
458<br />
,<br />
9<br />
2<br />
37<br />
,<br />
16<br />
2<br />
120<br />
10<br />
3<br />
,<br />
20<br />
10<br />
54<br />
,<br />
5<br />
2<br />
2<br />
2<br />
1<br />
2<br />
1<br />
2<br />
2<br />
8<br />
2<br />
5<br />
5<br />
2<br />
6<br />
6<br />
1<br />
1<br />
1<br />
max<br />
2<br />
6<br />
6<br />
2<br />
2<br />
2<br />
max<br />
=<br />
+<br />
+<br />
=<br />
⋅<br />
+<br />
⋅<br />
+<br />
⋅<br />
=<br />
=<br />
⋅<br />
=<br />
⋅<br />
=<br />
+<br />
⋅<br />
=<br />
+<br />
=<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
ν<br />
τ<br />
τ
41. Zadatak<br />
Odrediti potrebne dimenzije pojedinih odsje�aka te kut uvijanja slobodnog kraja.<br />
a = 400mm<br />
G = 80GPa<br />
M<br />
M<br />
M<br />
τ<br />
t1<br />
t 2<br />
t3<br />
dop<br />
= 4kNm<br />
= 7kNm<br />
= 3kNm<br />
= 60MPa<br />
Uvjet ravnoteže :<br />
∑ M = 0 → M − M − M + M = 0K<br />
( 1)<br />
M<br />
→ M<br />
A<br />
z<br />
A<br />
= M<br />
= 6kNm<br />
t1<br />
t3<br />
+ M<br />
t 2<br />
t 2<br />
− M<br />
t3<br />
t1<br />
A<br />
( 1)<br />
= 3 + 7 − 4 = 6kNm<br />
Vrijednost momenata torzije na pojedinim odsje�cima :<br />
M<br />
M<br />
M<br />
I<br />
II<br />
III<br />
= M<br />
= M<br />
t1<br />
= M<br />
t1<br />
t1<br />
= 4kNm<br />
− M<br />
− M<br />
t 2<br />
t 2<br />
Kriterij �vrsto�e :<br />
3<br />
i<br />
= 4 − 7 = −3kNm<br />
− M<br />
t3<br />
= 4 − 7 − 3 = −6kNm<br />
M i 16M<br />
i<br />
τ i =<br />
= ≤ τ dop , i = I,<br />
II,<br />
III,<br />
IV → Di<br />
≥<br />
W d π<br />
pi<br />
3<br />
16M<br />
πτ<br />
i<br />
dop<br />
51
52<br />
Dimenzioniranje :<br />
mm<br />
D<br />
m<br />
M<br />
D<br />
mm<br />
D<br />
m<br />
M<br />
D<br />
mm<br />
D<br />
m<br />
M<br />
D<br />
III<br />
dop<br />
III<br />
III<br />
II<br />
dop<br />
II<br />
II<br />
I<br />
dop<br />
I<br />
I<br />
80<br />
0798<br />
,<br />
0<br />
10<br />
60<br />
10<br />
6<br />
16<br />
16<br />
64<br />
0634<br />
,<br />
0<br />
10<br />
60<br />
10<br />
3<br />
16<br />
16<br />
70<br />
0698<br />
,<br />
0<br />
10<br />
60<br />
10<br />
4<br />
16<br />
16<br />
3<br />
6<br />
3<br />
3<br />
3<br />
6<br />
3<br />
3<br />
3<br />
6<br />
3<br />
3<br />
=<br />
→<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
≥<br />
=<br />
→<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
≥<br />
=<br />
→<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
≥<br />
π<br />
πτ<br />
π<br />
πτ<br />
π<br />
πτ<br />
Kutevi uvijanja :<br />
( )<br />
( )<br />
rad<br />
D<br />
G<br />
a<br />
M<br />
I<br />
G<br />
a<br />
M<br />
rad<br />
D<br />
G<br />
a<br />
M<br />
I<br />
G<br />
a<br />
M<br />
rad<br />
D<br />
G<br />
a<br />
M<br />
I<br />
G<br />
a<br />
M<br />
III<br />
III<br />
III<br />
p<br />
III<br />
AB<br />
III<br />
II<br />
II<br />
II<br />
p<br />
II<br />
BC<br />
II<br />
I<br />
I<br />
I<br />
p<br />
I<br />
CD<br />
I<br />
3<br />
4<br />
9<br />
3<br />
4<br />
3<br />
4<br />
9<br />
3<br />
4<br />
3<br />
4<br />
9<br />
3<br />
4<br />
10<br />
46<br />
,<br />
7<br />
08<br />
,<br />
0<br />
10<br />
80<br />
4<br />
,<br />
0<br />
10<br />
6<br />
32<br />
32<br />
10<br />
07<br />
,<br />
9<br />
064<br />
,<br />
0<br />
10<br />
80<br />
4<br />
,<br />
0<br />
10<br />
3<br />
32<br />
32<br />
10<br />
99<br />
,<br />
16<br />
07<br />
,<br />
0<br />
10<br />
80<br />
4<br />
,<br />
0<br />
10<br />
4<br />
64<br />
64<br />
2<br />
−<br />
−<br />
−<br />
⋅<br />
−<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
−<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
⋅<br />
−<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
−<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
π<br />
π<br />
ϕ<br />
ϕ<br />
π<br />
π<br />
ϕ<br />
ϕ<br />
π<br />
π<br />
ϕ<br />
ϕ<br />
Kut uvijanja slobodnog kraja nosa����<br />
( ) rad<br />
III<br />
II<br />
I<br />
AD<br />
uk<br />
3<br />
3<br />
10<br />
46<br />
,<br />
0<br />
10<br />
46<br />
,<br />
7<br />
07<br />
,<br />
9<br />
99<br />
,<br />
16<br />
−<br />
−<br />
⋅<br />
=<br />
⋅<br />
−<br />
−<br />
=<br />
+<br />
+<br />
=<br />
= ϕ<br />
ϕ<br />
ϕ<br />
ϕ<br />
ϕ
42. Zadatak<br />
Za nosa� oblika I-profila odrediti dimenzije popre�nog presjeka prema maksimalnim normalnim i<br />
tangencijalnim naprezanjima.<br />
F = 4kN<br />
kN<br />
q = 2<br />
m<br />
h = 20t<br />
b = 8t<br />
σ<br />
τ<br />
dop<br />
dop<br />
= 120MPa<br />
= 60MPa<br />
Reakcije oslonaca :<br />
∑<br />
F<br />
∑ M<br />
F<br />
F<br />
B<br />
Ay<br />
y<br />
= 0 → −F<br />
A<br />
1<br />
=<br />
5<br />
= F<br />
= 0 → F<br />
( 2F<br />
+ 4,<br />
5q)<br />
A<br />
Ay<br />
B<br />
− F<br />
⋅5<br />
− F ⋅ 2 − q ⋅3<br />
⋅1,<br />
5 = 0<br />
= 3,<br />
4kN<br />
= F + 3q<br />
= 6,<br />
6kN<br />
B<br />
+ F + q ⋅3<br />
= 0<br />
53
Momenti savijanja :<br />
M<br />
M<br />
x1<br />
x2<br />
= M<br />
= F<br />
l<br />
x1<br />
B<br />
= F<br />
Ay<br />
⋅ 2 = 6,<br />
8kNm<br />
⋅ 2 − q ⋅ 2 ⋅1<br />
= 9,<br />
2kNm<br />
Ekstremni momenti savijanja na udaljenosti z od oslonca A :<br />
z = 0K2m<br />
M<br />
dM<br />
M<br />
x<br />
dz<br />
dM<br />
dz<br />
( z)<br />
= F<br />
= F<br />
z = 2K3m<br />
x<br />
x<br />
2<br />
( z)<br />
= F z − qz − F(<br />
z − 2)<br />
x<br />
= F<br />
Ay<br />
Ay<br />
z 1<br />
z − qz = FAy<br />
z −<br />
2 2<br />
FAy<br />
− qz = 0 → z =<br />
q<br />
Ay<br />
Ay<br />
1<br />
2<br />
− qz − F = 0 → z =<br />
qz<br />
= 3,<br />
3m<br />
F<br />
2<br />
Ay<br />
− F<br />
= 1,<br />
3m<br />
q<br />
Iz dijagrama D(M) i D(Q) je vidljivo da se maksimalni moment savijanja javlja u presjeku 1, dok se<br />
maksimalna popre����sila javlja u presjeku A<br />
M<br />
Q<br />
x max<br />
y max<br />
= M<br />
= Q<br />
yA<br />
x1<br />
= 9,<br />
2kNm<br />
= 6,<br />
6kN<br />
Aksijalni moment površine drugog reda i moment otpora<br />
3 ( 22t)<br />
7t<br />
⋅ ( 20t)<br />
8t<br />
⋅<br />
I x =<br />
12<br />
−<br />
12<br />
= 2432t<br />
I x<br />
Wx<br />
=<br />
ymax<br />
4<br />
2432t<br />
3<br />
= = 221,<br />
09t<br />
⎛ h ⎞<br />
⎜ + t ⎟<br />
⎝ 2 ⎠<br />
3<br />
4<br />
54
55<br />
Iz kriterija �vrsto�e za maksimalna normalna i posmi�na naprezanja :<br />
m<br />
Q<br />
t<br />
t<br />
t<br />
Q<br />
t<br />
t<br />
h<br />
t<br />
h<br />
t<br />
h<br />
bt<br />
b<br />
S<br />
b<br />
S<br />
I<br />
Q<br />
m<br />
M<br />
t<br />
t<br />
M<br />
W<br />
M<br />
dop<br />
y<br />
dop<br />
y<br />
zy<br />
x<br />
dop<br />
x<br />
x<br />
y<br />
zy<br />
dop<br />
x<br />
dop<br />
x<br />
x<br />
x<br />
z<br />
3<br />
max<br />
2<br />
4<br />
max<br />
max<br />
2<br />
max<br />
max<br />
max<br />
max<br />
3<br />
3<br />
max<br />
3<br />
max<br />
max<br />
max<br />
10<br />
46<br />
,<br />
2<br />
2432<br />
134<br />
134<br />
2432<br />
134<br />
4<br />
2<br />
2<br />
10<br />
03<br />
,<br />
7<br />
09<br />
,<br />
221<br />
09<br />
,<br />
221<br />
−<br />
−<br />
⋅<br />
=<br />
≥<br />
≤<br />
=<br />
=<br />
+<br />
+<br />
=<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
≤<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
=<br />
⋅<br />
=<br />
⋅<br />
≥<br />
≤<br />
=<br />
=<br />
τ<br />
τ<br />
τ<br />
τ<br />
τ<br />
σ<br />
σ<br />
σ<br />
Zaklju�uje se da je za dimenzioniranje mjerodavan kriterij maksimalnih normalnih naprezanja i<br />
usvaja se :<br />
4<br />
6<br />
10<br />
18<br />
,<br />
6<br />
,<br />
8<br />
,<br />
56<br />
8<br />
,<br />
142<br />
20<br />
,<br />
1<br />
,<br />
7 mm<br />
I<br />
mm<br />
t<br />
b<br />
mm<br />
t<br />
h<br />
mm<br />
t x<br />
−<br />
⋅<br />
=<br />
=<br />
=<br />
=<br />
=<br />
=<br />
Provjera naprezanja u presjecima 1 i 2 :<br />
MPa<br />
MPa<br />
b<br />
S<br />
I<br />
Q<br />
MPa<br />
MPa<br />
y<br />
I<br />
M<br />
kN<br />
Q<br />
kNm<br />
M<br />
MPa<br />
MPa<br />
b<br />
S<br />
I<br />
Q<br />
MPa<br />
MPa<br />
y<br />
I<br />
M<br />
kN<br />
Q<br />
kNm<br />
M<br />
dop<br />
x<br />
x<br />
mj<br />
y<br />
zymj<br />
dop<br />
mj<br />
x<br />
mj<br />
x<br />
mj<br />
z<br />
mj<br />
y<br />
mj<br />
x<br />
dop<br />
x<br />
x<br />
mj<br />
y<br />
zymj<br />
dop<br />
mj<br />
x<br />
mj<br />
x<br />
mj<br />
z<br />
mj<br />
y<br />
mj<br />
x<br />
60<br />
026<br />
,<br />
0<br />
0071<br />
,<br />
0<br />
84<br />
10<br />
18<br />
,<br />
6<br />
3400<br />
120<br />
94<br />
,<br />
85<br />
0781<br />
,<br />
0<br />
10<br />
18<br />
,<br />
6<br />
6800<br />
4<br />
,<br />
3<br />
8<br />
,<br />
6<br />
60<br />
051<br />
,<br />
0<br />
0071<br />
,<br />
0<br />
134<br />
10<br />
18<br />
,<br />
6<br />
6600<br />
120<br />
26<br />
,<br />
116<br />
0781<br />
,<br />
0<br />
10<br />
18<br />
,<br />
6<br />
9200<br />
6<br />
,<br />
2<br />
2<br />
,<br />
9<br />
3<br />
6<br />
max<br />
2<br />
6<br />
2<br />
2<br />
1<br />
2<br />
3<br />
6<br />
max<br />
1<br />
6<br />
1<br />
1<br />
1<br />
1<br />
=<br />
≤<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
=<br />
=<br />
≤<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
=<br />
=<br />
=<br />
≤<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
=<br />
=<br />
≤<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
=<br />
=<br />
−<br />
−<br />
−<br />
−<br />
τ<br />
τ<br />
σ<br />
σ<br />
τ<br />
τ<br />
σ<br />
σ
Dijagrami naprezanja :<br />
56
43. Zadatak<br />
Za nosa� zadan i optere�en prema slici potrebno je:<br />
a) odrediti reakcije u osloncima<br />
b) skicirati dijagrame popre�nih sila i momenta savijanja<br />
c) dimenzionirati nosa� ( t zaokružiti na prvi ve�i cijeli broj u milimetrima )<br />
d) u presjeku x skicirati raspodjelu normalnog i tangencijalnog naprezanja po visini<br />
presjeka<br />
a = 2.2 m<br />
b = 1.8 m<br />
c = 1.2 m<br />
x = a / 2<br />
F = 8 kN<br />
q = 7 kN/m<br />
σDOP = 180 Mpa<br />
57
a) reakcije u osloncima<br />
b) dijagram popre�nih sila i momenta savijanja<br />
I. podru�je: 0 m
II. podru�je: 2.2 m < x < 4 m<br />
III. podru�je: 4 m < x < 5.2 m<br />
59
Mmax = 5.488 kNm<br />
Mmin = -9.6 kNm<br />
Najve�i moment po apsolutnoj vrijednosti je u to�ki x = 4 m i iznosi My = - 9.6 kNm .<br />
Stoga treba dimenzionirati nosa� za iznos momenta |My| = 9.6 kNm !<br />
60
c) dimenzioniranje nosa�a<br />
-������������������������������������<br />
- moment tromosti presjeka<br />
61
d) raspodjela normalnog i tangencijalnog naprezanja u presjeku za x = 1.1 m<br />
- normalno naprezanje<br />
62
- tangencijalno naprezanje<br />
63
44. Zadatak<br />
Za nosa� sa slike odrediti dimenzije popre�nog presjeka ako je on:<br />
a) kružnog oblika<br />
b) pravokutnog oblika<br />
c) oblika I-profila<br />
prema uvjetu maksimalnih normalnih naprezanja.<br />
Za I-profil napraviti kontrolu dobivenih dimenzija<br />
prema kriteriju dopuštenih tangencijalnih naprezanja.<br />
Reakcije oslonaca :<br />
∑ F<br />
∑ M<br />
F<br />
F<br />
Ay<br />
B<br />
y<br />
= 0 → −F<br />
A<br />
= F<br />
A<br />
= 112,<br />
5kN<br />
Ay<br />
B<br />
− F<br />
= −12,<br />
5kN<br />
B<br />
+ F + q ⋅ 4 = 0<br />
= 0 → F ⋅ 4 − F ⋅1<br />
− q ⋅ 4 ⋅ 2 + M = 0<br />
Momenti savijanja :<br />
M<br />
M<br />
x1<br />
x2<br />
= M<br />
= F<br />
l<br />
x1<br />
B<br />
= F<br />
Ay<br />
⋅ 2 = 6,<br />
8kNm<br />
⋅ 2 + q ⋅ 2 ⋅1<br />
= 9,<br />
2kNm<br />
F = 4kN<br />
kN<br />
q = 2<br />
m<br />
h = 20t<br />
b = 8t<br />
σ<br />
τ<br />
dop<br />
dop<br />
= 120MPa<br />
= 60MPa<br />
64
Maksimalni moment savijanja na udaljenosti z od oslonca B :<br />
M<br />
dM<br />
M<br />
x<br />
dz<br />
( z)<br />
x<br />
max<br />
1 2<br />
= FB<br />
z − qz<br />
2<br />
FB<br />
= FB<br />
− qz = 0 → z = = 2,<br />
25m<br />
q<br />
=<br />
112,<br />
5<br />
⋅<br />
2,<br />
25<br />
Moment savijanja u to�ki A :<br />
M xA<br />
= F ⋅1<br />
= 100kNm<br />
2,<br />
25<br />
− 50 ⋅ 2,<br />
25⋅<br />
= 126,<br />
56kNm<br />
2<br />
Za dimenzioniranje prema najve�im normalnim naprezanjima koristi se maksimalni moment<br />
savijanja :<br />
M = M x = 126,<br />
56kNm<br />
max<br />
1<br />
Za kontrolu dimenzija prema dopuštenim tangencijalnim naprezanjima koristi se vrijednost<br />
popre�ne sile :<br />
Qy max = QyB<br />
= FB<br />
= 112,<br />
5kN<br />
a) kružni popre�ni presjek<br />
W<br />
x min<br />
M x<br />
≥<br />
σ<br />
max<br />
dop<br />
D ≥ 234,<br />
47mm<br />
3<br />
D π<br />
= → D ≥<br />
32<br />
USVOJENO → D = 235mm<br />
b) pravokutni popre�ni presjek<br />
W<br />
x<br />
2 3<br />
h b h M x<br />
= = ≥<br />
6 12 σ<br />
h ≥ 247,<br />
6mm<br />
max<br />
dop<br />
→ h ≥<br />
3<br />
32M<br />
πσ<br />
12M<br />
σ<br />
USVOJENO → h = 248mm,<br />
b = 124mm<br />
3<br />
x max<br />
dop<br />
x max<br />
dop<br />
=<br />
=<br />
3<br />
3<br />
32 ⋅126,<br />
56 ⋅10<br />
π100<br />
12 ⋅126,<br />
56 ⋅10<br />
100<br />
6<br />
6<br />
65
66<br />
c) popre�ni presjek oblika I-profila<br />
( ) ( )<br />
mm<br />
b<br />
mm<br />
t<br />
h<br />
mm<br />
t<br />
USVOJENO<br />
mm<br />
t<br />
M<br />
t<br />
M<br />
t<br />
t<br />
t<br />
y<br />
I<br />
W<br />
t<br />
t<br />
t<br />
t<br />
t<br />
I<br />
dop<br />
x<br />
dop<br />
x<br />
x<br />
x<br />
x<br />
160<br />
,<br />
400<br />
25<br />
,<br />
16<br />
39<br />
,<br />
15<br />
100<br />
9<br />
,<br />
346<br />
10<br />
56<br />
,<br />
126<br />
9<br />
,<br />
346<br />
9<br />
,<br />
346<br />
5<br />
,<br />
13<br />
8<br />
,<br />
4683<br />
8<br />
,<br />
4683<br />
12<br />
25<br />
9<br />
12<br />
27<br />
10<br />
3<br />
6<br />
3<br />
max<br />
max<br />
3<br />
4<br />
max<br />
4<br />
3<br />
3<br />
=<br />
=<br />
=<br />
=<br />
→<br />
≥<br />
⋅<br />
⋅<br />
=<br />
≥<br />
→<br />
≥<br />
=<br />
=<br />
=<br />
=<br />
−<br />
=<br />
σ<br />
σ<br />
Provjera dimenzija prema dopuštenom tangencijalnom naprezanju :<br />
2<br />
2<br />
max<br />
2<br />
4<br />
4<br />
3<br />
max<br />
2<br />
max<br />
max<br />
max<br />
max<br />
60<br />
53<br />
,<br />
19<br />
53280<br />
16<br />
3<br />
,<br />
4683<br />
10<br />
5<br />
,<br />
112<br />
53280<br />
4<br />
2<br />
2<br />
mm<br />
N<br />
mm<br />
N<br />
mm<br />
mm<br />
N<br />
mm<br />
t<br />
h<br />
t<br />
h<br />
t<br />
h<br />
bt<br />
b<br />
S<br />
b<br />
S<br />
I<br />
Q<br />
dop<br />
zy<br />
zy<br />
x<br />
dop<br />
x<br />
x<br />
y<br />
zy<br />
=<br />
<<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
=<br />
+<br />
+<br />
=<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
≤<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
=<br />
τ<br />
τ<br />
τ<br />
τ<br />
τ
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
68<br />
max<br />
1<br />
180<br />
2<br />
3<br />
3<br />
40<br />
2<br />
140<br />
2<br />
70<br />
M<br />
kNm<br />
b<br />
b<br />
q<br />
M<br />
kNm<br />
a<br />
F<br />
M<br />
B<br />
A<br />
=<br />
−<br />
=<br />
⋅<br />
⋅<br />
−<br />
=<br />
⋅<br />
⋅<br />
−<br />
=<br />
=<br />
⋅<br />
=<br />
⋅<br />
=<br />
kN<br />
T<br />
kNm<br />
M<br />
130<br />
180<br />
max<br />
max<br />
−<br />
=<br />
−<br />
=<br />
Maksimalno normalno naprezanje :<br />
( )<br />
( ) 2<br />
8<br />
6<br />
2<br />
max<br />
1<br />
2<br />
8<br />
6<br />
1<br />
max<br />
1<br />
55<br />
,<br />
8<br />
10<br />
5<br />
,<br />
52<br />
10<br />
5<br />
,<br />
110<br />
10<br />
180<br />
48<br />
,<br />
4<br />
10<br />
5<br />
,<br />
52<br />
80<br />
10<br />
5<br />
,<br />
110<br />
10<br />
180<br />
mm<br />
N<br />
y<br />
I<br />
M<br />
mm<br />
N<br />
y<br />
I<br />
M<br />
z<br />
d<br />
z<br />
g<br />
−<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
−<br />
=<br />
=<br />
+<br />
=<br />
⋅<br />
−<br />
⋅<br />
⋅<br />
+<br />
=<br />
=<br />
σ<br />
σ
69<br />
Posmi������������������������ :<br />
2<br />
8<br />
6<br />
3<br />
max<br />
3<br />
6<br />
3<br />
933<br />
,<br />
0<br />
275<br />
10<br />
5<br />
,<br />
110<br />
10<br />
82<br />
,<br />
21<br />
10<br />
130<br />
275<br />
5<br />
,<br />
27<br />
10<br />
75<br />
,<br />
8<br />
2<br />
10<br />
2<br />
75<br />
,<br />
8<br />
60<br />
10<br />
5<br />
,<br />
52<br />
60<br />
10<br />
5<br />
,<br />
52<br />
8210<br />
,<br />
21<br />
21820<br />
3<br />
5<br />
,<br />
52<br />
2<br />
5<br />
,<br />
52<br />
75<br />
,<br />
8<br />
2<br />
2<br />
5<br />
,<br />
52<br />
5<br />
,<br />
52<br />
10<br />
mm<br />
N<br />
b<br />
I<br />
S<br />
T<br />
mm<br />
cm<br />
x<br />
b<br />
cm<br />
x<br />
x<br />
mm<br />
cm<br />
S<br />
z<br />
z<br />
z<br />
z<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
=<br />
+<br />
⋅<br />
=<br />
+<br />
=<br />
=<br />
⋅<br />
=<br />
→<br />
=<br />
=<br />
=<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛ ⋅<br />
+<br />
⋅<br />
⋅<br />
=<br />
τ<br />
Glavna naprezanja i njihov smjer u to�ki 3 :<br />
( )<br />
( )<br />
( )<br />
( )<br />
( )<br />
°<br />
=<br />
→<br />
°<br />
=<br />
→<br />
=<br />
⋅<br />
=<br />
−<br />
=<br />
+<br />
=<br />
±<br />
=<br />
⋅<br />
+<br />
±<br />
=<br />
+<br />
±<br />
=<br />
+<br />
=<br />
⋅<br />
−<br />
⋅<br />
⋅<br />
+<br />
=<br />
=<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
=<br />
=<br />
−<br />
⋅<br />
⋅<br />
=<br />
67<br />
,<br />
26<br />
35<br />
,<br />
53<br />
2<br />
34<br />
,<br />
1<br />
22<br />
,<br />
1<br />
82<br />
,<br />
0<br />
2<br />
2<br />
412<br />
,<br />
0<br />
632<br />
,<br />
1<br />
022<br />
,<br />
1<br />
61<br />
,<br />
0<br />
82<br />
,<br />
0<br />
4<br />
22<br />
,<br />
1<br />
2<br />
1<br />
2<br />
22<br />
,<br />
1<br />
4<br />
2<br />
1<br />
2<br />
22<br />
,<br />
1<br />
10<br />
5<br />
,<br />
52<br />
60<br />
10<br />
5<br />
,<br />
110<br />
10<br />
180<br />
82<br />
,<br />
0<br />
300<br />
10<br />
5<br />
,<br />
110<br />
10<br />
21<br />
10<br />
130<br />
10<br />
21<br />
21000<br />
5<br />
,<br />
52<br />
70<br />
60<br />
20<br />
2<br />
3<br />
2<br />
2<br />
3<br />
1<br />
2<br />
2<br />
2<br />
2<br />
3<br />
2<br />
,<br />
1<br />
2<br />
8<br />
6<br />
3<br />
max<br />
3<br />
2<br />
8<br />
6<br />
3<br />
max<br />
3<br />
3<br />
6<br />
3<br />
3<br />
ϕ<br />
ϕ<br />
ϕ<br />
σ<br />
σ<br />
τ<br />
σ<br />
σ<br />
σ<br />
σ<br />
τ<br />
tg<br />
mm<br />
N<br />
mm<br />
N<br />
mm<br />
N<br />
y<br />
I<br />
M<br />
mm<br />
N<br />
b<br />
I<br />
S<br />
T<br />
mm<br />
cm<br />
S<br />
z<br />
z<br />
z<br />
z
46. Zadatak<br />
Koriste�i analiti�ki postupak odre�ivanja elasti�ne linije nosa�a odrediti kut zaokreta u to�kama<br />
D i F te progib u to�ki E.<br />
a = 5m<br />
kN<br />
q =<br />
10<br />
m<br />
b = 2m<br />
P = 50kN<br />
c = d = 1m<br />
EI<br />
5 2<br />
= 8⋅10<br />
Nm<br />
Ekvivalentni sustav<br />
Reakcije u osloncima<br />
∑ M<br />
∑ M<br />
∑ F<br />
z<br />
d<br />
D<br />
A<br />
= 0 → F<br />
= 0 → q ⋅ 7 ⋅3,<br />
5 + P ⋅8<br />
− F ⋅5<br />
− F ⋅9<br />
= 0 → F<br />
= 0 → F<br />
A<br />
F<br />
y<br />
⋅ 2 − P ⋅1<br />
= 0 → F<br />
− q ⋅ 7 − P + F<br />
F<br />
C<br />
+ F<br />
C<br />
F<br />
=<br />
P 50<br />
= = 25kN<br />
2 2<br />
F<br />
= 0 → F<br />
A<br />
C<br />
= 11kN<br />
=<br />
420<br />
= 84kN<br />
5<br />
70
Momenti savijanja na pojedinim segmentima :<br />
( I ) 0 ≤ x ≤ a → M ( x)<br />
= F<br />
A<br />
2<br />
x<br />
⋅ x − q ⋅<br />
2<br />
( II ) a ≤ x ≤ ( a + b)<br />
→ M ( x)<br />
= F ⋅ x − q ⋅ + F ⋅ ( x − 5)<br />
A<br />
( III )( a + b)<br />
≤ x ≤ ( a + b + c)<br />
→ M ( x)<br />
= F ⋅ x − q ⋅ + F ⋅ ( x − 5)<br />
( IV )( a + b + c)<br />
≤ x ≤ ( a + b + c + d ) → M ( x)<br />
= F ⋅ x − q ⋅ + F ⋅ ( x − 5)<br />
2<br />
x<br />
2<br />
A<br />
C<br />
A<br />
2<br />
x<br />
2<br />
C<br />
2<br />
x<br />
2<br />
C<br />
+ q ⋅<br />
( x − 7)<br />
2<br />
2<br />
+ q ⋅<br />
Iz diferencijalne jednadžbe elasti�ne linije nosa�a slijedi za moment savijanja :<br />
2<br />
d w<br />
EI = −M<br />
2<br />
dx<br />
2<br />
d w<br />
EI = −F<br />
2<br />
dx<br />
2<br />
d w<br />
2<br />
dx<br />
( I )<br />
( x)<br />
2<br />
x<br />
⋅ x + q ⋅<br />
2<br />
2<br />
x<br />
2<br />
( II ) EI = −F<br />
⋅ x + q ⋅ − F ⋅ ( x − 5)<br />
2<br />
d w<br />
2<br />
dx<br />
2<br />
x<br />
2<br />
( III ) EI = −F<br />
⋅ x + q ⋅ − F ⋅ ( x − 5)<br />
2<br />
d w<br />
2<br />
dx<br />
A<br />
A<br />
A<br />
2<br />
x<br />
2<br />
( IV ) EI = −F<br />
⋅ x + q ⋅ − F ⋅ ( x − 5)<br />
Za kut zaokreta :<br />
2 3<br />
dw x x<br />
( I ) EI = −FA<br />
⋅ + q ⋅ + C<br />
dx 2 6<br />
( II )<br />
( III )<br />
( IV )<br />
( II )<br />
( III )<br />
( IV )<br />
A<br />
2 3<br />
dw x x<br />
EI = −FA<br />
⋅ + q ⋅ − FC<br />
⋅<br />
dx 2 6<br />
2 3<br />
dw x x<br />
EI = −FA<br />
⋅ + q ⋅ − FC<br />
⋅<br />
dx 2 6<br />
2 3<br />
dw x x<br />
EI = −FA<br />
⋅ + q ⋅ − FC<br />
⋅<br />
dx 2 6<br />
Za progib nosa�� :<br />
3 4<br />
x x<br />
( I ) EIw =<br />
−FA<br />
⋅ + q ⋅ + C1<br />
⋅ x + D<br />
6 24<br />
3 4<br />
x x<br />
EIw = −FA<br />
⋅ + q ⋅ − FC<br />
⋅<br />
6 24<br />
3 4<br />
x x<br />
EIw = −FA<br />
⋅ + q ⋅ − FC<br />
⋅<br />
6 24<br />
3 4<br />
x x<br />
EIw = −FA<br />
⋅ + q ⋅ − FC<br />
⋅<br />
6 24<br />
1<br />
C<br />
C<br />
C<br />
( x − 5)<br />
2<br />
− q ⋅<br />
− q ⋅<br />
2<br />
( x − 7)<br />
2<br />
( x − 7)<br />
2<br />
2 ( x − 5)<br />
( x − 7)<br />
2<br />
2<br />
2<br />
2<br />
+ P ⋅<br />
( x − 8)<br />
2<br />
3<br />
( x − 5)<br />
( x − 7)<br />
( x − 8)<br />
1<br />
2<br />
( x − 5)<br />
6<br />
3<br />
+ C<br />
− q ⋅<br />
− q ⋅<br />
6<br />
6<br />
3 ( x − 5)<br />
( x − 7)<br />
6<br />
2<br />
3<br />
+ C<br />
3<br />
+ P ⋅<br />
3<br />
4<br />
( x − 5)<br />
( x − 7)<br />
( x − 8)<br />
6<br />
+ C<br />
− q ⋅<br />
− q ⋅<br />
⋅ x + D<br />
24<br />
24<br />
2<br />
4<br />
+ C<br />
3<br />
+ P ⋅<br />
6<br />
2<br />
⋅ x + D<br />
3<br />
3<br />
2<br />
+ C<br />
+ C<br />
4<br />
4<br />
⋅ x + D<br />
( x − 7)<br />
4<br />
2<br />
2<br />
− P ⋅<br />
( x − 8)<br />
71
Uvjeti kompatibilnosti deformacija :<br />
ϕ<br />
ϕ<br />
w<br />
w<br />
w<br />
( I ) ( II ) ( a)<br />
= ϕ ( a)<br />
→ C<br />
( III ) ( IV ) ( a + b + c)<br />
= ϕ ( a + b + c)<br />
( I ) ( II ) ( a)<br />
= w ( a)<br />
→ D = D<br />
1<br />
1<br />
= C<br />
2<br />
2<br />
( II ) ( III ) ( a + b)<br />
= w ( a + b)<br />
→ C ⋅ 7 + D = C ⋅ 7 + D K(<br />
1)<br />
2<br />
→ C<br />
( III ) ( IV ) ( a + b + c)<br />
= w ( a + b + c)<br />
→ C3<br />
⋅8<br />
+ D3<br />
= C4<br />
⋅8<br />
+ D4<br />
→ D3<br />
= D4<br />
Rubni uvjeti :<br />
w<br />
w<br />
w<br />
( 0)<br />
( I ) ( a)<br />
= w ( x = 5)<br />
( IV ) ( l)<br />
= w ( x = 9)<br />
3<br />
2<br />
= C<br />
+ 9 ⋅C<br />
4 + D4<br />
= 0K(<br />
2)<br />
( 1)<br />
⇒ −6,<br />
25⋅<br />
7 + 0 = C4<br />
⋅ 7 + D4<br />
→ D4<br />
= −43,<br />
75 − 7 ⋅C<br />
4 K(<br />
3)<br />
( 3)<br />
→ ( 2)<br />
⇒ C = −229,<br />
58 = C<br />
502,<br />
917<br />
D<br />
4<br />
= 0 → D<br />
= 1563,<br />
31 = D<br />
Progib u to�ki E :<br />
w<br />
w<br />
w<br />
E<br />
E<br />
E<br />
= w<br />
1<br />
4<br />
= 0 → D<br />
3<br />
2<br />
= D<br />
1<br />
= 0<br />
3<br />
4<br />
9 9<br />
= 0 → −11⋅<br />
+ 10 ⋅ − 84 ⋅<br />
6 24<br />
3<br />
4<br />
3<br />
3 4<br />
5 5<br />
= 0 → −11<br />
+ 10 + C1<br />
⋅5<br />
= 0 → C1<br />
= −6,<br />
25 = C<br />
6 24<br />
( III ) ( IV ) ( x = 8)<br />
= w ( x = 8)<br />
3<br />
10<br />
=<br />
8⋅10<br />
6<br />
3<br />
4<br />
⎡ 8 8<br />
⎢−11⋅<br />
+ 10 ⋅ − 84 ⋅<br />
⎣ 6 24<br />
−3<br />
= 14,<br />
53⋅10<br />
m = 14,<br />
53mm<br />
Kut zaokreta u to�ki F :<br />
ϕ<br />
= ϕ<br />
ϕ<br />
ϕ<br />
F<br />
F<br />
F<br />
( IV ) ( x = 9)<br />
3<br />
10<br />
=<br />
8 ⋅10<br />
=<br />
6<br />
−15,<br />
05 ⋅10<br />
1<br />
=<br />
EI<br />
y<br />
⎡<br />
⎢−<br />
F<br />
⎣<br />
2<br />
3<br />
⎡ 9 9<br />
⎢−11⋅<br />
+ 10 ⋅ − 84 ⋅<br />
⎣ 2 6<br />
−3<br />
rad = −0,<br />
01505rad<br />
A<br />
3<br />
3<br />
4<br />
( 9 − 5)<br />
( 9 − 7)<br />
( 9 − 8)<br />
6<br />
3 ( 8 − 5)<br />
( 8 − 7)<br />
6<br />
−10<br />
⋅<br />
24<br />
−10<br />
⋅<br />
3 4<br />
1 ⎡ x x<br />
= ⎢−<br />
FA<br />
⋅ + q ⋅ − FC<br />
⋅<br />
EI y ⎣ 6 24<br />
2<br />
3<br />
x x<br />
⋅ + q ⋅ − FC<br />
⋅<br />
2 6<br />
4<br />
−<br />
24<br />
2<br />
+ 50 ⋅<br />
6<br />
3 ( x − 5)<br />
( x − 7)<br />
6<br />
− q ⋅<br />
24<br />
⎤<br />
229,<br />
58⋅<br />
8 + 1563,<br />
31⎥<br />
⎦<br />
2<br />
3<br />
( x − 5)<br />
( x − 7)<br />
( x − 8)<br />
− q ⋅<br />
2<br />
3<br />
( 9 − 5)<br />
( 9 − 7)<br />
( 9 − 8)<br />
2<br />
−10<br />
⋅<br />
2<br />
6<br />
+ 50 ⋅<br />
6<br />
2<br />
+ P ⋅<br />
2<br />
2<br />
3<br />
4<br />
2<br />
⎤<br />
− 229,<br />
58⎥<br />
⎦<br />
+ C<br />
4<br />
⋅9<br />
+ D<br />
⎤<br />
+ C3<br />
⋅ x + D3<br />
⎥<br />
⎦<br />
⎤<br />
+ C4<br />
⎥<br />
⎦<br />
4<br />
= 0<br />
72
73<br />
Kut zaokreta u to�ki D :<br />
( ) ( ) ( ) ( )<br />
( )<br />
( ) ( )<br />
( )<br />
( ) ( )<br />
rad<br />
rad<br />
rad<br />
rad<br />
C<br />
x<br />
q<br />
x<br />
F<br />
x<br />
q<br />
x<br />
F<br />
EI<br />
C<br />
x<br />
F<br />
x<br />
q<br />
x<br />
F<br />
EI<br />
x<br />
x<br />
D<br />
d<br />
D<br />
d<br />
D<br />
l<br />
D<br />
l<br />
D<br />
C<br />
A<br />
y<br />
d<br />
D<br />
C<br />
A<br />
y<br />
l<br />
D<br />
III<br />
II<br />
d<br />
D<br />
l<br />
D<br />
D<br />
00406<br />
,<br />
0<br />
01193<br />
,<br />
0<br />
01599<br />
,<br />
0<br />
01193<br />
,<br />
0<br />
10<br />
93<br />
,<br />
11<br />
58<br />
,<br />
229<br />
6<br />
7<br />
7<br />
10<br />
2<br />
5<br />
7<br />
84<br />
6<br />
7<br />
10<br />
2<br />
7<br />
11<br />
10<br />
8<br />
10<br />
01599<br />
,<br />
0<br />
10<br />
99<br />
,<br />
15<br />
25<br />
,<br />
6<br />
2<br />
5<br />
7<br />
84<br />
6<br />
7<br />
10<br />
2<br />
7<br />
11<br />
10<br />
8<br />
10<br />
6<br />
7<br />
2<br />
5<br />
6<br />
2<br />
1<br />
2<br />
5<br />
6<br />
2<br />
1<br />
7<br />
7<br />
3<br />
3<br />
2<br />
3<br />
2<br />
6<br />
3<br />
3<br />
2<br />
3<br />
2<br />
6<br />
3<br />
3<br />
3<br />
2<br />
3<br />
2<br />
2<br />
2<br />
3<br />
2<br />
=<br />
−<br />
=<br />
−<br />
=<br />
⋅<br />
−<br />
=<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎣<br />
⎡<br />
−<br />
−<br />
⋅<br />
−<br />
−<br />
⋅<br />
−<br />
⋅<br />
+<br />
⋅<br />
−<br />
⋅<br />
=<br />
=<br />
⋅<br />
=<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎣<br />
⎡<br />
−<br />
−<br />
⋅<br />
−<br />
⋅<br />
+<br />
⋅<br />
−<br />
⋅<br />
=<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎣<br />
⎡<br />
+<br />
−<br />
⋅<br />
−<br />
−<br />
⋅<br />
−<br />
⋅<br />
+<br />
⋅<br />
−<br />
=<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎣<br />
⎡<br />
+<br />
−<br />
⋅<br />
−<br />
⋅<br />
+<br />
⋅<br />
−<br />
=<br />
=<br />
+<br />
=<br />
=<br />
+<br />
=<br />
−<br />
−<br />
ϕ<br />
ϕ<br />
ϕ<br />
ϕ<br />
ϕ<br />
ϕ<br />
ϕ<br />
ϕ<br />
ϕ<br />
ϕ<br />
ϕ<br />
ϕ
74<br />
47. Zadatak<br />
Odrediti to�����������������������������������������������������������������������������<br />
m<br />
cm<br />
a<br />
m<br />
l<br />
kN<br />
F<br />
2<br />
,<br />
0<br />
20<br />
5<br />
10<br />
=<br />
=<br />
=<br />
=<br />
( )<br />
4<br />
2<br />
2<br />
'<br />
4<br />
2<br />
2<br />
3<br />
2<br />
2<br />
3<br />
'<br />
4<br />
2<br />
2<br />
3<br />
2<br />
2<br />
3<br />
'<br />
2<br />
2<br />
2<br />
2<br />
2<br />
2<br />
2<br />
2<br />
3<br />
1<br />
6<br />
5<br />
6<br />
7<br />
2<br />
3<br />
2<br />
0<br />
6<br />
5<br />
2<br />
6<br />
7<br />
2<br />
0<br />
12<br />
11<br />
6<br />
5<br />
2<br />
12<br />
2<br />
6<br />
5<br />
2<br />
12<br />
12<br />
11<br />
6<br />
7<br />
2<br />
3<br />
2<br />
12<br />
2<br />
6<br />
7<br />
2<br />
12<br />
6<br />
7<br />
3<br />
2<br />
3<br />
2<br />
2<br />
6<br />
5<br />
3<br />
2<br />
2<br />
3<br />
2<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
I<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
I<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
I<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
z<br />
a<br />
a<br />
a<br />
a<br />
a<br />
a<br />
y<br />
a<br />
a<br />
a<br />
a<br />
A<br />
yz<br />
z<br />
y<br />
T<br />
T<br />
=<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎣<br />
⎡<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛ −<br />
⋅<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
−<br />
⋅<br />
+<br />
+<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎣<br />
⎡<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛ −<br />
⋅<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛ −<br />
⋅<br />
+<br />
=<br />
=<br />
⎥<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎢<br />
⎣<br />
⎡<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛ −<br />
⋅<br />
+<br />
⋅<br />
+<br />
⎥<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎢<br />
⎣<br />
⎡<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛ −<br />
⋅<br />
+<br />
⋅<br />
=<br />
=<br />
⎥<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎢<br />
⎣<br />
⎡<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
−<br />
⋅<br />
+<br />
⋅<br />
+<br />
⎥<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎢<br />
⎣<br />
⎡<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛ −<br />
⋅<br />
+<br />
⋅<br />
=<br />
=<br />
⋅<br />
+<br />
⋅<br />
=<br />
=<br />
⋅<br />
+<br />
⋅<br />
=<br />
=<br />
+<br />
⋅<br />
=<br />
Glavne osi momenata površine drugog reda :<br />
( ) ( ) ( )<br />
0<br />
12<br />
7<br />
3<br />
1<br />
12<br />
11<br />
4<br />
5<br />
3<br />
1<br />
12<br />
11<br />
3<br />
1<br />
12<br />
11<br />
3<br />
1<br />
2<br />
2<br />
1<br />
12<br />
11<br />
2<br />
2<br />
1<br />
4<br />
2<br />
1<br />
2<br />
1<br />
45<br />
90<br />
2<br />
0<br />
3<br />
1<br />
2<br />
12<br />
11<br />
12<br />
11<br />
3<br />
1<br />
2<br />
2<br />
2<br />
4<br />
4<br />
2<br />
4<br />
4<br />
1<br />
4<br />
4<br />
4<br />
2<br />
'<br />
2<br />
'<br />
'<br />
'<br />
'<br />
2<br />
,<br />
1<br />
4<br />
4<br />
4<br />
4<br />
'<br />
'<br />
'<br />
=<br />
=<br />
=<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
−<br />
=<br />
=<br />
=<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
=<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
±<br />
=<br />
⋅<br />
±<br />
⋅<br />
⋅<br />
=<br />
+<br />
−<br />
±<br />
+<br />
=<br />
°<br />
=<br />
→<br />
°<br />
=<br />
→<br />
∞<br />
=<br />
⋅<br />
−<br />
=<br />
−<br />
⋅<br />
−<br />
=<br />
−<br />
−<br />
=<br />
yz<br />
z<br />
y<br />
yz<br />
z<br />
y<br />
z<br />
y<br />
z<br />
y<br />
yz<br />
I<br />
I<br />
a<br />
a<br />
I<br />
I<br />
a<br />
a<br />
I<br />
a<br />
a<br />
a<br />
I<br />
I<br />
I<br />
I<br />
I<br />
I<br />
a<br />
a<br />
a<br />
a<br />
I<br />
I<br />
I<br />
tg α<br />
α<br />
α
75<br />
Položaj neutralne osi :<br />
°<br />
−<br />
=<br />
→<br />
−<br />
=<br />
−<br />
=<br />
°<br />
−<br />
=<br />
−<br />
= 65<br />
14286<br />
,<br />
2<br />
7<br />
15<br />
45<br />
12<br />
7<br />
4<br />
5<br />
4<br />
4<br />
ϕ<br />
α<br />
ϕ tg<br />
a<br />
a<br />
tg<br />
I<br />
I<br />
tg<br />
z<br />
y<br />
Ekstremna naprezanja su u to�kama B i D :<br />
MPa<br />
a<br />
Fl<br />
a<br />
Fl<br />
a<br />
a<br />
a<br />
a<br />
Fl<br />
y<br />
I<br />
z<br />
I<br />
M<br />
MPa<br />
a<br />
Fl<br />
a<br />
Fl<br />
a<br />
a<br />
a<br />
a<br />
Fl<br />
y<br />
I<br />
z<br />
I<br />
M<br />
D<br />
D<br />
z<br />
D<br />
y<br />
D<br />
B<br />
B<br />
z<br />
B<br />
y<br />
B<br />
78<br />
,<br />
6<br />
2<br />
,<br />
0<br />
5<br />
10<br />
10<br />
35<br />
38<br />
35<br />
38<br />
7<br />
2<br />
5<br />
4<br />
12<br />
7<br />
6<br />
2<br />
4<br />
5<br />
2<br />
2<br />
2<br />
45<br />
sin<br />
45<br />
cos<br />
21<br />
,<br />
3<br />
2<br />
,<br />
0<br />
5<br />
10<br />
10<br />
35<br />
18<br />
35<br />
18<br />
7<br />
2<br />
5<br />
4<br />
12<br />
7<br />
6<br />
2<br />
4<br />
5<br />
2<br />
2<br />
2<br />
45<br />
sin<br />
45<br />
cos<br />
3<br />
3<br />
3<br />
3<br />
4<br />
4<br />
3<br />
3<br />
3<br />
3<br />
4<br />
4<br />
=<br />
⋅<br />
⋅<br />
=<br />
=<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛ +<br />
=<br />
⎟<br />
⎟<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛ °<br />
+<br />
°<br />
=<br />
−<br />
=<br />
⋅<br />
⋅<br />
−<br />
=<br />
−<br />
=<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
−<br />
=<br />
⎟<br />
⎟<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
−<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛ °<br />
+<br />
°<br />
=<br />
σ<br />
σ<br />
σ<br />
σ<br />
a<br />
a<br />
a<br />
a<br />
a<br />
y<br />
a<br />
a<br />
a<br />
a<br />
a<br />
z<br />
a<br />
a<br />
a<br />
a<br />
a<br />
y<br />
a<br />
a<br />
a<br />
a<br />
a<br />
z<br />
D<br />
D<br />
B<br />
B<br />
6<br />
2<br />
2<br />
2<br />
6<br />
5<br />
2<br />
2<br />
6<br />
7<br />
45<br />
sin<br />
6<br />
5<br />
45<br />
cos<br />
6<br />
7<br />
2<br />
2<br />
2<br />
6<br />
7<br />
2<br />
2<br />
6<br />
5<br />
45<br />
sin<br />
6<br />
7<br />
45<br />
cos<br />
6<br />
5<br />
6<br />
2<br />
2<br />
2<br />
6<br />
7<br />
2<br />
2<br />
6<br />
5<br />
45<br />
sin<br />
6<br />
7<br />
45<br />
cos<br />
6<br />
5<br />
2<br />
2<br />
2<br />
6<br />
5<br />
2<br />
2<br />
6<br />
7<br />
45<br />
sin<br />
6<br />
5<br />
45<br />
cos<br />
6<br />
7<br />
=<br />
−<br />
=<br />
°<br />
−<br />
°<br />
=<br />
=<br />
+<br />
=<br />
°<br />
+<br />
°<br />
=<br />
=<br />
+<br />
−<br />
=<br />
°<br />
−<br />
−<br />
°<br />
−<br />
=<br />
−<br />
=<br />
−<br />
+<br />
−<br />
=<br />
°<br />
−<br />
+<br />
°<br />
−<br />
=
48. Zadatak<br />
Za zadani tankostijeni presjek prikazan na slici potrebno je odrediti središte posmika u odnosu<br />
na to�ku A.<br />
x =<br />
35 ⋅sin<br />
25°<br />
= 14,<br />
8mm<br />
76
77<br />
( ) ( )<br />
( ) ( )<br />
( )<br />
( )<br />
T<br />
T<br />
y<br />
y<br />
I<br />
T<br />
dy<br />
t<br />
y<br />
t<br />
I<br />
T<br />
T<br />
mm<br />
S<br />
t<br />
y<br />
t<br />
I<br />
T<br />
S<br />
S<br />
t<br />
I<br />
T<br />
b<br />
I<br />
S<br />
T<br />
T<br />
T<br />
z<br />
z<br />
I<br />
t<br />
T<br />
dz<br />
z<br />
z<br />
I<br />
t<br />
T<br />
dz<br />
t<br />
z<br />
z<br />
I<br />
T<br />
T<br />
z<br />
z<br />
I<br />
T<br />
z<br />
t<br />
z<br />
t<br />
I<br />
T<br />
b<br />
I<br />
S<br />
T<br />
mm<br />
I<br />
I<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
xy<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
xz<br />
y<br />
y<br />
⋅<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
=<br />
⋅<br />
⋅<br />
+<br />
⋅<br />
=<br />
=<br />
+<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
+<br />
⋅<br />
=<br />
+<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
⋅<br />
=<br />
⋅<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
+<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
=<br />
⎥<br />
⎥<br />
⎦<br />
⎤<br />
⎢<br />
⎢<br />
⎣<br />
⎡<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
⋅<br />
⋅<br />
+<br />
⋅<br />
+<br />
⋅<br />
⋅<br />
+<br />
⋅<br />
+<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
−<br />
⋅<br />
⋅<br />
+<br />
⋅<br />
=<br />
∫<br />
∫<br />
∫<br />
184<br />
,<br />
0<br />
2<br />
149<br />
5<br />
,<br />
1672<br />
149<br />
5<br />
,<br />
1672<br />
5<br />
,<br />
1672<br />
2<br />
/<br />
15<br />
8<br />
,<br />
14<br />
15<br />
5<br />
8<br />
,<br />
29<br />
5<br />
,<br />
1672<br />
0324<br />
,<br />
0<br />
6<br />
2<br />
8<br />
,<br />
14<br />
2<br />
8<br />
,<br />
14<br />
2<br />
8<br />
,<br />
14<br />
2<br />
8<br />
,<br />
14<br />
2<br />
8<br />
,<br />
14<br />
10<br />
4364<br />
,<br />
3<br />
2<br />
8<br />
,<br />
14<br />
8<br />
,<br />
14<br />
5<br />
12<br />
8<br />
,<br />
14<br />
5<br />
8<br />
,<br />
29<br />
5<br />
20<br />
12<br />
5<br />
20<br />
5<br />
,<br />
7<br />
8<br />
,<br />
29<br />
15<br />
5<br />
12<br />
15<br />
5<br />
2<br />
2<br />
2<br />
20<br />
0<br />
2<br />
2<br />
2<br />
2<br />
20<br />
0<br />
2<br />
2<br />
3<br />
1<br />
2<br />
2<br />
1<br />
2<br />
2<br />
1<br />
15<br />
0<br />
3<br />
1<br />
2<br />
1<br />
15<br />
0<br />
1<br />
2<br />
1<br />
1<br />
1<br />
2<br />
1<br />
1<br />
15<br />
0<br />
1<br />
2<br />
1<br />
1<br />
1<br />
1<br />
1<br />
1<br />
4<br />
5<br />
2<br />
3<br />
2<br />
3<br />
2<br />
3<br />
τ<br />
τ<br />
( )<br />
( ) ( )<br />
mm<br />
e<br />
T<br />
T<br />
T<br />
e<br />
T<br />
T<br />
T<br />
T<br />
e<br />
T<br />
T<br />
T<br />
z<br />
z<br />
z<br />
I<br />
T<br />
dz<br />
t<br />
z<br />
z<br />
t<br />
I<br />
T<br />
T<br />
mm<br />
S<br />
z<br />
z<br />
t<br />
I<br />
T<br />
z<br />
t<br />
z<br />
t<br />
I<br />
T<br />
S<br />
S<br />
S<br />
t<br />
I<br />
T<br />
b<br />
I<br />
S<br />
T<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
y<br />
xz<br />
16<br />
,<br />
1<br />
244<br />
,<br />
0<br />
2<br />
72<br />
,<br />
31<br />
184<br />
,<br />
0<br />
6<br />
,<br />
59<br />
0324<br />
,<br />
0<br />
44<br />
,<br />
103<br />
25<br />
cos<br />
35<br />
2<br />
8<br />
,<br />
29<br />
2<br />
25<br />
cos<br />
35<br />
20<br />
2<br />
244<br />
,<br />
0<br />
6<br />
5<br />
2<br />
149<br />
5<br />
,<br />
4652<br />
2<br />
5<br />
149<br />
5<br />
,<br />
4652<br />
2980<br />
8<br />
,<br />
29<br />
20<br />
5<br />
2<br />
5<br />
149<br />
5<br />
,<br />
4652<br />
2<br />
8<br />
,<br />
29<br />
2980<br />
5<br />
,<br />
1672<br />
3<br />
2<br />
1<br />
3<br />
15<br />
0<br />
3<br />
3<br />
2<br />
3<br />
3<br />
3<br />
2<br />
3<br />
3<br />
15<br />
0<br />
3<br />
3<br />
2<br />
2<br />
3<br />
3<br />
3<br />
3<br />
3<br />
2<br />
1<br />
3<br />
3<br />
−<br />
=<br />
⋅<br />
⋅<br />
−<br />
⋅<br />
+<br />
⋅<br />
=<br />
⋅<br />
°<br />
⋅<br />
⋅<br />
⋅<br />
−<br />
⋅<br />
⋅<br />
+<br />
°<br />
⋅<br />
+<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
−<br />
+<br />
=<br />
⋅<br />
⋅<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
−<br />
+<br />
⋅<br />
=<br />
=<br />
⋅<br />
⋅<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
−<br />
+<br />
⋅<br />
=<br />
=<br />
⎟<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎜<br />
⎝<br />
⎛<br />
⎟<br />
⎠<br />
⎞<br />
⎜<br />
⎝<br />
⎛<br />
−<br />
⋅<br />
⋅<br />
+<br />
+<br />
⋅<br />
=<br />
+<br />
+<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
∫<br />
τ
49. Zadatak<br />
Za zadani nosa� i optere�enje odrediti mjerodavan koeficijent sigurnosti prema teoriji najve�ih<br />
normalnih naprezanja u to�kama 1 i 2 presjeka I-I ako je zadano kriti�no naprezanje materijala<br />
�K.<br />
P = 30kN<br />
H = 25kN<br />
q = 15kN<br />
/ m<br />
σ<br />
ΣF<br />
R<br />
ΣM<br />
1 1,<br />
5<br />
− q ⋅1⋅<br />
+ q ⋅<br />
2 2<br />
ΣF<br />
= 0 :<br />
R<br />
x<br />
Ax<br />
Ay<br />
y<br />
− H = 0 ⇒ R<br />
A<br />
=<br />
0 :<br />
=<br />
+ R<br />
0 :<br />
B<br />
Presjek I – I :<br />
N = −25kN<br />
T = −18,<br />
125kN<br />
2<br />
Ax<br />
M = 27,<br />
1875kNm<br />
I<br />
I<br />
= H = 25kN<br />
+ P ⋅1,<br />
5 − R ⋅3<br />
= 0 ⇒ R<br />
− q ⋅ 2,<br />
5 − P = 0 ⇒ R<br />
3<br />
70 ⋅10<br />
= + 70 ⋅10<br />
⋅<br />
12<br />
6 4<br />
= 5,<br />
83⋅10<br />
mm<br />
B<br />
Ay<br />
B<br />
= 49,<br />
375kN<br />
10 ⋅100<br />
12<br />
3<br />
= 18,<br />
125kN<br />
10 ⋅ 70 ⋅5<br />
+ 10 ⋅100<br />
⋅ 60 + 20 ⋅ 40 ⋅120<br />
159500<br />
yT<br />
=<br />
= ⇒ y<br />
10 ⋅ 70 + 10 ⋅100<br />
+ 20 ⋅ 40 2500<br />
2<br />
A = 2500mm<br />
S<br />
y<br />
y<br />
W<br />
K<br />
y max<br />
y max<br />
= 320MPa<br />
=<br />
z<br />
I<br />
y<br />
max<br />
= 40 ⋅ 20 ⋅<br />
5,<br />
83⋅10<br />
=<br />
66,<br />
2<br />
T<br />
= 63,<br />
8mm<br />
40 ⋅ 20<br />
12<br />
2<br />
2<br />
( 63,<br />
8 − 5)<br />
+ + 10 ⋅100<br />
⋅ ( 63,<br />
8 − 60)<br />
+ + 40 ⋅ 20 ⋅ ( 66,<br />
2 −10)<br />
6<br />
⇒ W<br />
y max<br />
3<br />
= 88,<br />
1⋅10<br />
mm<br />
3 3<br />
( 66,<br />
2 −10)<br />
+ ( 66,<br />
2 − 20)<br />
⋅10<br />
⋅ ⇒ S = 55,<br />
6 ⋅10<br />
mm<br />
3<br />
66,<br />
2 − 20<br />
2<br />
y<br />
y max<br />
T<br />
40<br />
2<br />
1<br />
30 10 30<br />
70 mm<br />
z<br />
3<br />
20<br />
100<br />
10<br />
130<br />
2<br />
78
79<br />
To�ka 1 :<br />
MPa<br />
mm<br />
N<br />
b<br />
I<br />
S<br />
T<br />
MPa<br />
mm<br />
N<br />
A<br />
N<br />
xz<br />
y<br />
y<br />
xz<br />
x<br />
x<br />
29<br />
,<br />
17<br />
29<br />
,<br />
17<br />
10<br />
10<br />
83<br />
,<br />
5<br />
10<br />
6<br />
,<br />
55<br />
10<br />
125<br />
,<br />
18<br />
10<br />
10<br />
2500<br />
10<br />
25<br />
1<br />
2<br />
6<br />
3<br />
3<br />
max<br />
1<br />
1<br />
2<br />
3<br />
1<br />
=<br />
⇒<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
−<br />
=<br />
⇒<br />
−<br />
=<br />
⋅<br />
−<br />
=<br />
=<br />
τ<br />
τ<br />
σ<br />
σ<br />
To�ka 2 :<br />
0<br />
0<br />
0<br />
40<br />
10<br />
83<br />
,<br />
5<br />
0<br />
10<br />
125<br />
,<br />
18<br />
6<br />
,<br />
318<br />
6<br />
,<br />
318<br />
10<br />
6<br />
,<br />
308<br />
2500<br />
10<br />
25<br />
10<br />
1<br />
,<br />
88<br />
10<br />
1875<br />
,<br />
27<br />
2<br />
2<br />
6<br />
3<br />
2<br />
2<br />
2<br />
2<br />
3<br />
3<br />
6<br />
2<br />
=<br />
=<br />
⇒<br />
=<br />
⋅<br />
⋅<br />
⋅<br />
⋅<br />
=<br />
⋅<br />
⋅<br />
=<br />
−<br />
=<br />
⇒<br />
−<br />
=<br />
−<br />
−<br />
=<br />
⋅<br />
−<br />
+<br />
⋅<br />
⋅<br />
−<br />
=<br />
+<br />
=<br />
y<br />
xz<br />
y<br />
y<br />
xz<br />
x<br />
y<br />
x<br />
S<br />
b<br />
I<br />
S<br />
T<br />
MPa<br />
mm<br />
N<br />
A<br />
N<br />
W<br />
M<br />
τ<br />
τ<br />
σ<br />
σ<br />
Provjera po teoriji najve�ih normalnih naprezanja (1. teorija �vrsto�e) :<br />
To�ka 1 :<br />
( ) ( )<br />
( ) ( ) ( )<br />
6<br />
,<br />
24<br />
13<br />
320<br />
00<br />
,<br />
13<br />
29<br />
,<br />
17<br />
4<br />
10<br />
5<br />
,<br />
0<br />
10<br />
5<br />
,<br />
0<br />
4<br />
5<br />
,<br />
0<br />
5<br />
,<br />
0<br />
1<br />
1<br />
1<br />
1<br />
2<br />
2<br />
1<br />
2<br />
1<br />
2<br />
1<br />
1<br />
1<br />
=<br />
⇒<br />
=<br />
=<br />
=<br />
⇒<br />
⋅<br />
+<br />
−<br />
⋅<br />
+<br />
−<br />
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=<br />
⋅<br />
+<br />
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=<br />
k<br />
k<br />
MPa<br />
ek<br />
K<br />
ek<br />
ek<br />
xz<br />
x<br />
x<br />
ek<br />
σ<br />
σ<br />
σ<br />
σ<br />
τ<br />
σ<br />
σ<br />
σ<br />
To�ka 2 :<br />
( ) ( )<br />
( ) ( ) ( )<br />
0<br />
,<br />
1<br />
6<br />
,<br />
318<br />
320<br />
6<br />
,<br />
318<br />
0<br />
4<br />
6<br />
,<br />
318<br />
5<br />
,<br />
0<br />
6<br />
,<br />
318<br />
5<br />
,<br />
0<br />
4<br />
5<br />
,<br />
0<br />
5<br />
,<br />
0<br />
2<br />
2<br />
2<br />
2<br />
1<br />
2<br />
2<br />
1<br />
2<br />
2<br />
2<br />
2<br />
2<br />
2<br />
=<br />
⇒<br />
=<br />
=<br />
−<br />
=<br />
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⋅<br />
+<br />
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+<br />
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+<br />
⋅<br />
=<br />
k<br />
k<br />
MPa<br />
ek<br />
K<br />
x<br />
ek<br />
ek<br />
xz<br />
x<br />
x<br />
ek<br />
σ<br />
σ<br />
σ<br />
σ<br />
σ<br />
τ<br />
σ<br />
σ<br />
σ
z<br />
q<br />
C A<br />
D<br />
1,0 m 1,5 m<br />
1,5 m<br />
-25<br />
34,375<br />
-<br />
-15<br />
-<br />
-7,5<br />
+<br />
4,0 m<br />
-18,125<br />
P<br />
I<br />
I<br />
11,875<br />
+ +<br />
27,1875<br />
B<br />
- D(N)<br />
-<br />
D(T)<br />
D(M)<br />
H<br />
x<br />
80