1..prvi_dio
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JU UNIVERZITET U TUZLI
Rudarsko-geološko-građ evinski fakultet
Teoretske osnove za izradu zadataka u narednom poglavlju nalaze se u knjizi «Otpornostmaterijala sa teorijom elastič nosti», autora Dr sci. Sadudina Hodžića, redovnog profesora na
Rudarsko-geološko-građ evinskom fakultetu u Tuzli, na stanicama broj 20 do 24.
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1. Prostorno stanje napona
Zadatak 1.1
Za prizmu prikazanu na slici 1. odrediti sile koje djeluju u pravcu osa (x,y,z), ako su poznatedilatacije u pravcu ovih osa (x,y,z).ε y=0,718⋅ε x ε z =0,155⋅ε x ε x =0,000224a=10cm E=21000kN/cm2 µ=0,33ε x =0,000224ε y=0,718⋅ε x =0,718⋅0,000224=0,000161ε z =0,155⋅ε x =0,155⋅0,000224=0,0000348
Slika 1.Određivanje napona u prizmi:
εx⋅E=σx-µσy-µσz
εy⋅E=σy-µσx-µσz
εz⋅E=σz-µσx-µσy
σx-µσ
y-µσ
z=ε
x⋅E
σy-µσx-µσz=εy⋅Eσz-µσx-µσy=εz⋅E
z
2a
a
a
FxFx
Fy
Fy
Fz
Fz
x
y
σx-µσy-µσz=0,000224⋅21000
σy-µσx-µσz=0,000161⋅21000
σz-µσx-µσy=0,0000348⋅21000
σx-µσy-µσz=4,7
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σy-µσx-µσz=3,38
σz-µσx-µσy=0,73
σx-0,33σy-0,33σz=4,7
σy-0,33σx-0,33σz=3,38
σz-0,33σx-0,33σy=0,73
σx-0,33σy-0,33σz=4,7
-0,33σx +σy-0,33σz=3,38
-0,33σx-0,33σy+σz =0,73
6015,03267,00718,0109,0109,0109,00359,00359,0133,0
1
33,0
33,0
33,0
1
133,033,0
33,0133,0
33,033,01
133,033,0
33,0133,0
33,033,01
=−−−−−−−=−
−
−
−
−−
−−
−−
=
−−
−−
−−
=
D
D
985,524,0512,011,10368,00795,07,4
33,0
1
33,0
73,0
38,3
7,4
133,073,0
33,0138,3
33,033,07,4
133,073,0
33,0138,3
33,033,07,4
=+−+++=
−
−
−
−
−−
=
−
−
−−
=
x
x
D
D
2/95,96015,0
985,5cmkN
D
D x x ===σ
554,5368,024,055,124,0512,038,3
73,0
38,3
7,4
33,0
33,0
1
173,033,0
33,038,333,0
33,07,41
173,033,0
33,038,333,0
33,07,41
=−++++=
−
−
−
−−
−
=
−
−−
−
=
y
y
D
D
2
/2,96015,0
554,5
cmkN D
D y
y ===σ
194,455,1115,10795,0511,0368,073,0
33,0
1
33,0
33,0
33,0
1
73,033,033,0
38,3133,0
7,433,01
73,033,033,0
38,3133,0
7,433,01
=++−++=
−
−
−
−
−−
−
−
=
−−
−
−
=
z
z
D
D
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2/97,66015,0
194,4cmkN
D
D z z ===σ
kN a F
aa
F
kN a F
aa
F
kN a F
aa
F
z z
z z
y y
y
y
x x
x x
69710297,62
2
18401022,92
2
199010295,92
2
22
22
22
=⋅⋅=⋅=⋅
=
=⋅⋅=⋅=⋅
=
=⋅⋅=⋅=⋅
=
σ
σ
σ
σ
σ
σ
Zadatak 1.2
U nekoj tač ki napregnutog tijela imamo napone:σ x =2kN/cm2;
τ xy=0;τ xz =1kN/cm2;
τ yx =0;σ y=2kN/cm2; τ yz =1kN/cm2; τ zx =1kN/cm2; τ zy=1kN/cm2; σ z =1kN/cm2 .-Utvrditi o kakvom se stanju napona radi, tj. dokazati da li je naponsko stanje ravno ili je prostorno.-Odrediti pravce i veli č ine glavnih napona.-Odrediti kosinuse pravaca glavnih napona.-Odrediti maksimalni napon smicanja.
Formiramo tenzor napona:
=
111
120
102
S
Ako je veličina determinante matrice tenzora napona različita od nule, stanje napona je prostorno, a
ako je jednaka nuli, radi se o ravnom naponskom stanju.
Pored toga bar jedan od njenih minora mora biti različit od nule.
0
0220004
11
20
02
111
120
102
=
=−−−++==
D
D
40420
021 =−== D
Dakle, vrijednost determinante je jednaka nuli, a jedan od minora je različit od nule.
Prema tome u pitanju je ravno stanje napona.
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Veličine glavnih napona σ i (i=1,2,3), dobijamo rješavanjem sekularne jednačine karakteristične
jednačine (*).
.....(*)....................032
2
1
3 =−⋅+⋅− I I I iii σ σ σ
Ovdje su I 1 , I 2 i I 3 , invarijante stanja napona.
21 5122cm
kN I z y x =++=++= σ σ σ
4
2
2
222
2
6110211222cm
kN I
I zx yz xy x z z y y x
=−−−⋅+⋅+⋅=
−−−⋅+⋅+⋅= τ τ τ σ σ σ σ σ σ
0001212122
2
3
222
3
=+−⋅−⋅−⋅⋅=
⋅⋅⋅+⋅−⋅−⋅−⋅⋅=
I
I yz xz xy xy z zx y yz x z y x τ τ τ τ σ τ σ τ σ σ σ σ
Uvrštavanjem invarijanti u karakterističnu jednačinu, dobijamo:
22
21
2
2,1
2
23
22
4
32
6
2
1
2
56
2
5
2
5
065
1065
cm
kN
cm
kN
ii
i
iii
==
==
±=−
±=
=+⋅−
⋅=⋅+⋅−
σ
σ
σ
σ σ
σ σ σ σ
03 =σ jer je u pitanju ravno naponsko stanje.
Da bi našli kosinuse uglova koje glavni napon σ i zatvara sa koordinatnim osama, postavljamo uvjet
u vidu proporcije:
( )( ) ( )
( )
( )( )i z zy
yz i y
i
zy zx
i y yx
i z zx
yz yx
i z zy
yz i y
A
k
σ σ τ
τ σ σ
τ τ
σ σ τ
γ
σ σ τ
τ τ
β
σ σ τ
τ σ σ
α
−
−=
=−=−
=−
−111 coscoscos
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( )
( )
2221
iii
zy zx
i y yx
i
i z zx
yz yx
i
C B Ak
C
B
++±=
−=
−=
τ τ
σ σ τ
σ σ τ
τ τ
Pošto je i=1,2,3, najpogodnije je da se račun predstavi tabelarno (nakon izračunavanja) za
23
cm
kN =σ računamo Ai , Bi i C i.
( )( )
( )
( )1
11
320
1311
10
112311
132
1
1
1
−=−
=
−=−=
=−=−
−=
C
B
A
22 2cm
kN =σ
( )( )
( )
( )0
11
220
1211
10
1211
122
2
2
2
=−
=
−=−
=
−=−
−=
C
B
A
03 =σ
( )
( )
( )
( )2
11
020
1011
10
1
011
102
3
3
3
−=−
=
−=−
=
=
−
−=
C
B
A
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Tabela 1. Kosinusi pravaca glavnih napona
i Ai Bi Ci 2
i A 2
i B 2
iC 222iii C B A ++ cosαi cosβi cosγi
1 1 -1 -1 1 1 1 33
1±
3
1±
3
1±
2 -1 -1 0 1 1 0 22
1±
2
1± 0
3 1 -1 -2 1 1 4 66
1±
6
1±
6
2±
Maksimalni napon smicanja:
( ) ( )231max 5,103
2
1
2
1
cm
kN =−⋅=−⋅= σ σ τ
Stanje glavnih napona u posmatranoj tački, predstavljeno je Morovim krugovima.
Slika 2.
Uσ =1(kN/cm )=2cm2
τ (kN/cm2)
σ (kN/cm2)σ3
σ1
τ m a x
σ2
Zadatak 1.3
U nekoj tač ki napregnutog tijela, date su komponente deformacije:
555
555
108110011050
105010021001
−−−
−−−
⋅=⋅−=⋅=
⋅=⋅=⋅−=
,γ; ,γ; ,γ
; ,ε; ,ε; ,ε
zx yz xy
z y x
Traži se:-Tenzor deformacija.-Veli č ine i pravci glavnih dilatacija.-Stanje glavnih dilatacija prikazati grafi č ki preko Morovih krugova dilatacija.
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Tenzor deformacija:
−−
−
=
= −
5,05,09,05,0225,0
9,025,00,1
10
2
1
2
12
1
2
12
1
2
1
5
z zy zx
yz y yx
xz xy x
D
ε γ γ
γ ε γ
γ γ ε
Karakteristična jednačina glasi:
*).........(....................032
2
1
3 =−⋅+⋅− I I I iii ε ε ε
I 1 , I 2 i I 3 su prva, druga i treća invarijanta stanja deformacija.
( )
( )44
1
4
1
222
3
222
2
1
zx yz xy
xy z zx y zy x z y x
zx yz xy x z y z y x
z y x
I
I
I
γ γ γ γ ε γ ε γ ε ε ε ε
γ γ γ ε ε ε ε ε ε
ε ε ε
⋅⋅+⋅+⋅+⋅⋅−⋅⋅=
++⋅−⋅+⋅+⋅=
++=
Voditi računa da je po zakonu konjugovanosti:
xz zx zy yz yx xy γ γ γ γ γ γ === ;;
( ) 55
1 105,1105,021 −− ⋅=⋅++−= I
( ) ( ) ( )
⋅+⋅−+⋅⋅−⋅⋅⋅−⋅⋅⋅+⋅⋅⋅−= −−−−−−−−− 252525555555
2 108,1101105,04
1101105,0102105,0102101 I
10
2 106225,2 −⋅−= I
( ) ( ) ([ ]( )
)
4
108,1101105,0
105,0105,0108,11021011014
1105,0102101
555
255255255555
3
−−−
−−−−−−−−−
⋅⋅⋅−⋅⋅+
+⋅⋅⋅+⋅⋅⋅+⋅−⋅⋅−⋅−⋅⋅⋅⋅⋅−= I
15151515
3 10626,210225,0104012,1101 −−−− ⋅−=⋅−⋅−⋅−= I
Karakteristična jednačina glasi:
010626,2106225,2105,1 1510253 =⋅+⋅⋅−⋅⋅− −−−iii ε ε ε
Za rješavanje prethodne sekularne jednačine trećeg stepena, slijede matematske upute.
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Matematske osnove potrebne za rješavanje zadatka.
Tabela 2. Tabela smijena
pr =
p<0
032 ≤+ pq
3cos
r
q=ϕ
3cos23 ⋅⋅−= r y
−°⋅⋅=
360cos21
ϕ r y
+°⋅⋅=
360cos22
ϕ r y
Jednač ina trećeg stepena u obliku:
03 =+⋅+⋅+⋅ d xc xb xa
rješava se uvođ enjem nove varijable:
a
b x y
⋅+=
3
Zatim vršimo zamjenu:
32
2
2
23
3
3
33
327
22
pq D
a
bca p
a
d
a
cb
a
bq
+=⋅−⋅⋅
=⋅
+⋅⋅
−⋅⋅
=⋅
Za diskriminantu D 0, postoje 3 realna rješenja. Kako za probleme rješavanja «karakteristič ne
jednač ine (*)» imamo uvijek takav sluč aj, dalje rješavanje jednač ine će se odvijati uz ovu predpostavku (D<0) i (p<0).
≤
Kada izrač unamo y1; y2 i y3 , vratimo se na «staru» varijablu:a
b x y
⋅+=
3 , te za (y1 ), izrač unamo
(x1 ), za (y2 ), izrač unamo (x2 ) i za (y3 ), izrač unamo (x3 ), i to su konač na rješenja.
15
10
5
23
3
10626,2
106225,2
105,1
1
327
22
−
−
−
⋅=
⋅−=
⋅−=
=
+⋅⋅
−⋅⋅
=⋅
d
c
b
a
a
d
a
cb
a
bq
( ) ( ) ( )[ ]
( )15
1515
1510535
1053237,0
1006475,110626,231125,125,02
10626,23
106225,2105,1
27
105,122
−
−−
−−−−
⋅=
⋅=⋅+−−=⋅
⋅+⋅−⋅⋅−
−⋅−⋅
=⋅
q
q
q
Nova varijabla:
a
b y ii ⋅
+=3
ε
( ) ( )
1010
102510
2
2
10124167,13
103725,3
3
101175,10
3
105,1106225,23
3
33
−−
−−−
⋅−=⋅
−=
⋅−=
⋅−−⋅⋅−=
⋅−⋅⋅
=⋅
p
a
bca p
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Diskriminanta:
0101372,11042064,1102834,0 30303032 <⋅−=⋅−⋅=+= −−− pq D
Pošto je diskriminanta manja od nule, sekularna jednačina ima tri realna korjena:
510 10060267,110124167,1 −− ⋅=⋅== pr
°=
=⋅⋅
== −
−
47,63
44665,0101919176,1
1053237,0cos
15
15
3
ϕ
ϕ r
q
55
33
55
3
104776,1103
5,1
9776,13
109776,13
47,63cos10060267,12
3cos2
−−
−−
⋅−=⋅
+−=−=
⋅−=°
⋅⋅⋅−=⋅⋅−=
b
y
r y
ε
555
22
55
2
555
11
55
1
10826,0103
5,1
10326,03
10326,03
47,6360cos10060267,12
360cos2
101514,2103
5,11065145,1
3
1065145,13
47,6360cos10060267,12
360cos2
−−−
−−
−−−
−−
⋅=⋅+⋅=−=
⋅=
°+°⋅⋅⋅=
+°⋅⋅=
⋅=⋅+⋅=−=
⋅=
°−°⋅⋅⋅=
−°⋅⋅=
b
y
r y
b y
r y
ε
ϕ
ε
ϕ
Dakle, glavne dilatacije iznose:
ε 1=2,1514⋅ 10-5
ε 2=0,826 ⋅ 10-5
ε 3=-1,4776 ⋅ 10-5
Prethodna 3 rješenja su korijeni karakteristične (sekularne) jednačine.Određivanje kosinusa smijerova glavnih dilatacija:
( )
( )
( )
−
−
−
=
i z zy zx
yz i y yx
xz xyi x
D
ε ε γ γ
γ ε ε γ
γ γ ε ε
2
1
2
12
1
2
12
1
2
1
Minore Ai , Bi i C i, određivat ćemo za i=1,2,3; tj. ε 1 , ε 2 i ε 3.
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Za:
ε 1=2,1514⋅ 10-5
ε x=-1,0⋅ 10-5→ zadato
ε y=2,0⋅ 10-5→ zadato
ε z =0,5⋅ 10-5→ zadato
( )
( )
( ) 1025
1
1
1
1
10000202,0106514,15,0
5,01514,0
2
12
1
−− ⋅=⋅−−
−−=
−
−=
A
A
z zy
yz y
ε ε γ
γ ε ε
( )
( ) 1025
1
1
1
1003715,0106514,19,0
5,025,0
2
12
1
2
1
−− ⋅=⋅−
−=
−
=
B
B
z zx
yz yx
ε ε γ
γ γ
( )
( )1025
1
1
1
1001126,0105,09,0
1514,025,0
2
1
2
12
1
−−
⋅=⋅−
−
=
−=
C
C
zy zx
y yx
γ γ
ε ε γ
Pošto je:
1coscoscos
coscoscos
1
2
1
2
1
2
1
1
1
1
1
1
=++
===
γ β α
γ β α k
C B A
0052,0cos
03882,0
10000202,010cos
03882,0
10
03882,010
1
01126,003715,0000202,010
1
1
1
1010
11
10
1022210
2
1
2
1
2
1
=
⋅⋅=⋅=
=
⋅=
++⋅
=
++=
−
−−
α
α Ak
k
k
C B Ak
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957,0cos
03882,0
1003715,010cos
1
1010
11
=
⋅⋅=⋅=
−
β
β Bk
29,0cos
03882,0
1001126,010cos
1
1010
11
=
⋅⋅=⋅=
−
γ
γ C k
( )
( )
( ) 1025
2
2
2
2
106327,010326,05,0
5,0174,1
2
12
1
−− ⋅−=⋅−−
−=
−
−=
A
A
z zy
yz y
ε ε γ
γ ε ε
( )
( ) 1025
2
2
2
103685,010326,09,0
5,025,0
2
12
1
2
1
−− ⋅=⋅−
−=
−=
B
B
z zx
yz yx
ε ε γ
γ γ
( )
( ) 1025
2
2
2
101816,1105,09,0
174,125,0
2
1
2
121
−− ⋅−=⋅−
=
−=
C
C
zy zx
y yx
γ γ
ε ε γ
( ) ( )
( )
455,0cos
39,1
106327,010cos
39,1
10
39,1101
1816,13685,06327,0101
1
2
1010
22
10
1022210
2
2
2
2
2
2
−=
⋅−⋅=⋅=
=
⋅=−++−⋅=
++=
−
−−
α
α Ak
k
k
C B Ak
12
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265,0cos
39,1
103685,010cos
2
1010
22
=
⋅⋅=⋅=
−
β
β Bk
( )
85,0cos
03882,0
101816,110cos
2
1010
22
−=
⋅−⋅=⋅=
−
γ
γ C k
( )
( )
( ) 1025
3
3
3
3
1011273,7109776,15,0
5,04776,3
2
12
1
−− ⋅=⋅−
−=
−
−=
A
A
z zy
yz y
ε ε γ
γ ε ε
( )
( ) 1025
3
3
3
10944,0109776,19,0
5,025,0
2
12
1
2
1
−− ⋅=⋅−
=
−=
B
B
z zx
yz yx
ε ε γ
γ γ
( )
( ) 1025
3
3
3
102548,3105,09,0
4776,325,0
2
1
2
12
1
−− ⋅−=⋅−
=
−=
C
C
zy zx
y yx
γ γ
ε ε γ
( )
903,0cos
8778,7
101127,710cos
8778,7
10
8778,710
1
944,02548,31127,710
1
1
3
1010
33
10
1022210
2
3
2
3
2
3
=
⋅⋅=⋅=
=
⋅
=
+−+⋅
=
++=
−
−−
α
α Ak
k
k
C B Ak
13
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12,0cos
8778,7
10944,010cos
3
1010
33
=
⋅⋅=⋅=
−
β
β Bk
( )
413,0cos
8778,7
102548,310
cos
3
1010
33
−=
⋅−⋅
=⋅=
−
γ
γ C k
Dakle uglovi koje zaklapaju pravci glavnih dilatacija sa koordinatnim osama su:
α→sa osom (x); β→sa osom (y); γ→sa osom (z).
ε1→ (α1; β1; γ1 )
ε2→ (α2; β2; γ2 )
ε3→ (α3; β3; γ3 )
Tabela 3. Tabelarni prikaz kosinusa smijerovai
iα cos i β cos iγ cos ε1
1 0,0052 0,957 0,29 2,1514⋅10-5
2 -0,455 0,265 -0,85 0,826⋅10-5
3 0,903 0,120 -0,413 -1,4776⋅10-5
Morov krug dilatacija
14
Slika 3.
Uε=1⋅10 =2cm-5
ε1
ε2 ε⋅10-5
½ ⋅γ⋅10
½ ⋅ γ m a x
-5
ε3
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Zadatak 1.4
Stanje deformacije u nekoj tač ki kvadra, ima tenzor deformacije:
00005,000005,000009,0
00005,00002,0000025,0
00009,0000025,00001,0
−
−
−
= D
Elasi č ne konstante materijala su: E=21000kN/cm2 ,µ
=1/3. Napisati tenzor napona za istu tač ku, i odrediti prvu invarijantu tenzora napona.
Modul klizanja:
( ) ( )
( ) ( ) ( ) ( )[ ]
( ) ( ) ( ) ( )[ ]
( ) ( ) ( ) ( )[ ]
)6....(................................................................................
)5....(................................................................................
)4....(................................................................................
)3.......(....................1211
)2.......(....................1211
)1.......(....................1211
78772333,01
21000
21 2
zx zx
yz yz
xy xy
y x z z
x z y y
z y x x
G
G
G
E
E
E
cm
kN E G
γ τ
γ τ
γ τ
ε ε µ ε µ µ µ
σ
ε ε µ ε µ µ µ
σ
ε ε µ ε µ µ µ
σ
µ
⋅=
⋅=
⋅=
+⋅+⋅−⋅−⋅+
=
+⋅+⋅−⋅−⋅+
=
+⋅+⋅−⋅−⋅+
=
=⋅+
=⋅+
=
15
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( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( )
( ) 0000335,000005,067,01
000134,00002,067,01
000067,00001,067,01
000033,00002,00001,033,0
0000165,00001,000005,033,0
0000825,000005,00002,033,0
67,01
452,066,0133,01211
=⋅=⋅−
=⋅=⋅−
−=−⋅=⋅−
=+−⋅=+⋅
−=−⋅=+⋅
=+⋅=+⋅
=−
=−⋅+=⋅−⋅+
z
y
x
y x
x z
z y
ε µ
ε µ
ε µ
ε ε µ
ε ε µ
ε ε µ
µ
( ) ( )
[ ]
[ ]
[ ]2
2
2
09,3000033,00000335,018,46460
459,50000165,0000134,018,46460
72,00000825,0000067,018,46460
18,46460452,0
21000
211
cm
kN
cm
kN
cmkN
E
y
y
x
=−⋅=
=−⋅=
=+−⋅=
==−⋅+
σ
σ
σ
µ µ
( )
2
2
2
41,100009,027877
788,000005,027877
394,0000025,027877
cm
kN cm
kN
cm
kN
zx
yz
xy
=⋅⋅=
−=−⋅⋅=
=⋅⋅=
τ
τ
τ
Tenzor napona:
2
09,3788,041,1
788,0459,5394,0
41,1394,072,0
cm
kN S
S
z zy zx
yz y yx
xz xy x
−
−=
=
σ τ τ
τ σ τ
τ τ σ
Prva invarijanta tenzora napona:
21 269,909,3459,572,0cm
kN I z y xn =++=++= σ σ σ
16
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17
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Teoretske osnove za izradu zadataka u narednom poglavlju nalaze se u knjizi «Otpornostmaterijala sa teorijom elastič nosti», autora Dr sci. Sadudina Hodžića, redovnog profesora na
Rudarsko-geološko-građ evinskom fakultetu u Tuzli, na stanicama broj 9 do 20.
18
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2. Dvoosno naponsko stanje
Zadatak 2.1
Kvadar prikazan na slici 4. optereć en je silama: F x ; F y; F T .
Odrediti normalne i napone smicanja u presjeku koji stoji pod uglom: ϕ=43,5o .Odrediti pravce i intenzitete glavnih napona. Analiti č ki dobijene rezultate provjeriti preko Morovog naponskog kruga. a=10cm
a/2
FTFx
Fy
yFy=100kN
Fx=333kN=100kN
a
2 / 3 a
Slika 4.
Analitič ko rješenje zadatka
22
1
22
3
502
100
22
33,333
100
33
2
2
cmaa
a A
cmaaa
A
===⋅=
===⋅=
222
3
222
1
222
3
3100
300300
3
100
2100
200200
2
100
10
100
10001000
3
333
cm
kN
aa A
F
cm
kN
aa A
F
cm
kN
aa A
F
T xy
y
y
x x
=====
−=−
=−
=−
=−
=
=====
τ
σ
σ
19
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Za pozitivan smičući napon:
( ) ( )
2222222282,7
100
2,7282,72899,0
300052,0
2001000
2
12001000
2
1
2sin2cos2
1
2
1
cm
kN
aaaaaan
xy y x y xn
===⋅+
++
−=
+−++=
σ
ϕ τ ϕ σ σ σ σ σ
99,05,43cos2sin
052,05,43cos2cos
==
==o
o
ϕ
ϕ
( ) ϕ τ ϕ σ σ τ 2cos2sin2
1 xy y xnl +−−=
2222222 784,5100
4,5784,5786,15594052,030099,0200100021
cmkN
aaaaaanl −=−=−=+−=⋅+
+−=τ
o
o
y x
xy
arctg
tg
28,13
;56,262
5,02
2
=
=
=+
=
α
α
σ σ
τ α
22424
2
4
2
22,18,670400
2
6,134140030041200
2
1400
aaaaaaa±=±=⋅+±=σ
221 78,10100
8,10708,1070
cm
kN
a===σ
222 708,2100
8,2708,270
cm
kN
a−=−=−=σ
20
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21
Provjera preko Morovog kruga
Grafič ko rješenje τ
σ2=-2,7kN/cm2 σ1=10,7kN/cm2Uσ=1kN/cm2=1cm
-σ =2kN/cm
2
σx=10kN/cm
2
(2)
(1)
τxyα=13,28o
σ
-τxy
τnl=-5,8kN/cm2
(2)
σn=7,3kN/cm2
Slika 5.
Zadatak 2.2
Tanka ploč a (slika 6.) je zategnuta u pravcu ose x i u pravcu ose y. Glavna naponska osa (1),stoji pod uglom =26,5°, u odnosu na osu (x). U nekom kosom presjeku ove ploč e poznati sunaponi:
σn=13,9kN/cm2;τnl =-0,95kN/cm2 . Takođ e je poznat napon smicanja:
τ xy=4kN/cm2 .Odrediti:-Glavne napone
σ1 i σ2 .-Ugao ( ϕ ), pod kojim stoji kosi presjek.-Napone u pravcu osa x i y ( σ x i σ y ). Izvršiti provjeru alternativnom metodom.
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22
y
σy
τyx
τxy
ϕ σx σx
x
Slika 6.σy
Grafič ko rješenje
Polupriječnik Morovog kruga može se izraziti i preko relacije: τ xy /R=2sinα
R=τ xy /2sinα =4/0,798=5kN/cm2. AO =R
Pošto su zadati: τ nl i σ n; nanesemo σn i na kraju iz tačke A/ nanesemo -τ nl .
Pošto smo izračunali « R», uzmemo u šestar dužinu «R», i iz tačke A presječemo osu σ . To je tačka «O» centar Morovog kruga, koji sada možemo nacrtati, a sa njega dobiti tražene
veličine.
σ O
Uσ=1kN/1cm2=1cm
σ =13,9kN/cm2
σ1=14kN/cm2
σx=12kN/cm2
σy=6kN/cm2σ2=4kN/cm2
τ n l
A
Aβ
- τ x y
τ x y = 4 k N / c m 2
(2)
(1)
2αϕ=32o
α=26,5o
R
τ
Slika 7.
Odgovor: σx=12kN/cm2; σy=6kN/cm
2; σ1=14kN/cm
2; σ2=4kN/cm
2; ϕ=32
o
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Analitič ka provjera rješenja
24
327,125,26
22
cm
kN
tg
tg
xy
o
y x
xy
=
==
−=
τ
α α
σ σ
τ α
( )
).......(..........6327,1
8
42327,1
22
2a
cm
kN
tg
y x
y x
xy y x
==−
⋅=−⋅
=−⋅
σ σ
σ σ
τ σ σ α
Pošto je:2
95,0cmkN
nl −=τ
( )
)1....(..........2cos33,1317,02sin
2cos21,412sin15,3
2cos21,42sin15,31
95,0
12cos42sin395,0
)1(2cos42sin395,0
2cos42sin62
195,0
2cos2sin2
1
ϕ ϕ
ϕ ϕ
ϕ ϕ
ϕ ϕ
ϕ ϕ
ϕ ϕ
ϕ τ ϕ σ σ τ
⋅+=
⋅+=⋅
⋅−⋅=
⋅⋅−⋅=
−⋅⋅+⋅−=−
⋅+⋅⋅−=−
⋅+⋅−−= xy y xnl
β α =−
β τ
2sin= R
nl (vidi se na Morovom krugu)
°−=°−=
−=−
=
48,595,102arcsin
19,0
5
95,02sin
β β
β
°=+= 32α β
44,02cos
9,02sin
=
=
ϕ
9,0438,033,1317,02sin =⋅+= što zadovoljava jednačinu (1)
23
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Iz jednačine (a) : y x σ σ += 6
( ) ( )
( ) ( )
( ) y y y
y y y y
xy y x y xn
σ σ σ
σ σ σ σ
ϕ τ ϕ σ σ σ σ σ
+=+++=++⋅+⋅=
⋅+⋅−+⋅+++⋅=
⋅+⋅−⋅++⋅=
9,76,33,136,33,1262
19,13
9,0444,06216
219,13
2sin2cos2
1
2
1
269,79,13
cm
kN y =−=σ
Pošto je:2
12666cm
kN y x =+=+= σ σ
Dakle: σx=12kN/cm2; σy=6kN/cm
2; ϕ=32
o; što se u potpunosti slaže sa grafičkim rješenjem.
Provjera glavnih napona:
( ) ( )
( ) ( ) 59446122
1612
2
1
42
1
2
1
22
2,1
22
2,1
±=⋅+−±+=
+−±+=
σ
τ σ σ σ σ σ xy y x y x
;4;142221
cm
kN
cm
kN == σ σ
Dakle, i glavni naponi se u potpunosti slažu sa grafičkim rješenjem.
Zadatak 2.3
Zadate su veli č ine glavnih napona: σ1=12,9kN/cm2; σ2=-4,9kN/cm2 i napona smicanja koji leži ukosoj ravnini pod uglom ϕ=30o . τnl =-4,9 kN/cm2 -Odrediti najpogodnijom metodom napone:
σ x , σ y , τ xy i σn .-Odrediti ugao ( ) i odrediti ose (1) i (2)
-Provjeriti drugom metodom tač nost rezultata dobijenih po prvoj metodi.
Najpovoljnija metoda je grafička metoda, primjenom Morovog naponskog kruga.
Očitane tražene veličine: σx=12kN/cm2; σy=-4kN/cm
2; σn=11,4kN/cm
2; τxy=4kN/cm
2; α=13,35
o
24
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25
τ Uσ=2kN/cm2=1cm
(1)
τ x y = 4 k N / c m 2
α=13,35°
σ
τ n l = - 4 , 9 k N / c m 2
ϕ=30°
σn=11,4kN/cm2
σy=-4kN/cm2 σx=12kN/cm2
σ2=-4,9kN/cm2 σ1=12,9kN/cm2
(2)
Slika 8.
Analitič ka provjera
Unesene su očitane veličine: σ x=12kN/cm2; σ y=-4kN/cm2; τ xy=4kN/cm2
( ) ( )
( ) ( ) 9,84444122
1412
2
1
42
1
2
1
22
2,1
22
2,1
±=⋅++±−=
+−±+=
σ
τ σ σ σ σ σ xy y x y x
;9,49,84;9,129,842221
cm
kN
cm
kN −=−==+= σ σ
S obzirom da smo dobili zadate veličine za σ 1 i σ 2, dobijeni parametri grafičkom metodom su tačni.
( )
( )2
9,45,04866,04122
1
2cos2sin2
1
cm
kN nl
xy y xnl
−=⋅+⋅+−=
+−−=
τ
ϕ τ ϕ σ σ τ
Iz prethodnog zaključujemo, da je i po drugoj provjeri tačan grafički način određenih, traženih
veličina.
Provjera očitanog ugla “α”
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o
o
y x
xy
arctg
tg
3,13
6,262
5,0412
4222
=
=
=+⋅
=−
=
α
α
σ σ
τ α
što se takođe slaže sa očitanom veličinom.
( ) ( )
( ) ( )2
46,11866,045,04122
1412
2
1
2sin2cos2
1
2
1
cm
kN n
xy y x y xn
=⋅+⋅+⋅+−⋅=
⋅+⋅−⋅++⋅=
σ
ϕ τ α σ σ σ σ σ
Dakle, očitana veličina i izračunata veličina (σ n) na bazi očitanja, identične su, pa je zaključak, da
su sve tražene veličine, dobijene grafičkom metodom, tačne.
Zadatak 2.4
Mjernim trakama postavljenim po pravcima (a), (b) i (c) na č eli č noj ploč i (slika 9. ) ustanovljenesu dilatacije:εa=0,00097εb= 0,00103εc=0 ,0011 Prema pravcima sila, oč ekuju se normalna naprezanja na zatezanje, a naprezanje na smicanjeć e takođ e imati pozitivan predznak.Odrediti naprezanja u pravcu ose (x) σ x , u pravcu ose (y) σ y i naprezanje na smicanje τ xy . takođ eodrediti glavne napone, kao i njihov položaj. E=21000kN/cm2; G=7875kN/cm2
26
Slika 9.
θ1=17°
θ2=22,5°
θ3=45°
(c)
(b)
(a)
θ3
θ2
θ1
x
y
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( )( ) )3..(..........2sin2cos2
)2..(..........2sin2cos2
)1..(..........2sin2cos2
33
22
11
θ γ θ ε ε ε ε ε
θ γ θ ε ε ε ε ε
θ γ θ ε ε ε ε ε
⋅+⋅−++=⋅
⋅+⋅−++=⋅
⋅+⋅−++=⋅
y x y xc
y x y xb
y x y xa
;00194,02 =⋅ aε ;00206,02 =⋅ bε ;0022,02 =⋅ cε
;0,02cos
;707,02cos
;829,02cos
3
2
1
=
=
=
θ
θ
θ
;0,12sin
;707,02sin
;559,02sin
3
2
1
=
=
=
θ
θ
θ
γ ε ε
γ ε ε ε ε
γ ε ε ε ε
++=
⋅+⋅−⋅++=
⋅+⋅−⋅++=
y x
y x y x
y x y x
0022,0
707,0707,0707,000206,0
559,0829,0829,000194,0
)3.....(........................................0022,0
)2.(..........707,0293,0707,100206,0
)1..(..........559,0171,0829,100194,0
γ ε ε
γ ε ε
γ ε ε
++=
⋅+⋅+⋅=
⋅+⋅+⋅=
y x
y x
y x
Formiramo determinante i razvijemo ih po Sarusu, te određujemo dilatacije εx i εy, kao i ugao
smicanja (klizanja) (γ).
138,01638,0293,1292,0954,0121,05359,0
1
293,0
171,0
1
707,1
829,1
111
707,0293,0707,1
559,0171,0829,1
111
707,0293,0707,1
559,0171,0829,1
−=−−−++=
==
D
D
000098,0
1
293,0
171,0
0022,0
00206,0
00194,0
110022,0
707,0293,000206,0
559,0171,000194,0
110022,0
707,0293,000206,0
559,0171,000194,0
−=
==
x
x
D
D
00071,0138,0
000098,0=
−−
== D
D x
xε
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28
00006,0
0022,01
00206,0707,1
00194,0829,1
10022,01
707,000206,0707,1
559,000194,0829,1
10022,01
707,000206,0707,1
559,000194,0829,1
−=
==
y
y
D
D
00044,0138,0
00006,0=
−−
== D
D y
yε
000139,0
11
293,0707,1
171,0829,1
0022,011
00206,0293,0707,1
00194,0171,0829,1
0022,011
00206,0293,0707,1
00194,0171,0829,1
−=
==
γ
γ
D
D
001,0138,0
000139,0=
−−
== D
D yγ
Naponi:
( ) ( )
( ) ( )
2
222
222
87,7001,07875
89,1533,01
00071,033,000044,021000
1
15,2033,01
00044,033,000071,021000
1
cm
kN G
cm
kN E
cm
kN E
xy
x y
y
y x
x
=⋅=⋅=
=−
⋅+⋅=
−
⋅+=
=−
⋅+⋅=
−
⋅+=
γ τ
µ
ε µ ε σ
µ
ε ε σ
Položaj glavnih osa napona:
0
0
42,37
85,742
69,389,1515,20
87,7222
=
=
=−⋅
=−
=
α
α
σ σ
τ α
arctg
tg y x
xy
( ) ( )
( ) ( ) 15,802,1887,7489,1515,202
189,1515,20
2
1
421
21
22
2,1
22
2,1
±=⋅+−±+=
+−±+=
σ
τ σ σ σ σ σ xy y x y x
;87,915,802,18;17,2615,802,182221
cm
kN
cm
kN =−==+= σ σ
;87,9;17,262min22max1
cm
kN
cm
kN ==== σ σ σ σ
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Zadatak 2.5
Odrediti prvu, drugu i treć u invarijantu tenzora napona, napisati matricu (tenzor napona) zakvadar, optereć en kao na slici. a=10cm
FT
A3
A2A1
Fx
Fy
y
Fy=50kN
=50kN
Fx=100kN
3a
2 a
0,75a
Slika 10.
222
1
222
2
222
1
2251025,225,275,03
150105,15,1275,0
600106632
cmaaa A
cmaaa A
cmaaa A
=⋅==⋅=
=⋅==⋅=
=⋅==⋅=
Naponi:
222
2
222
3
222
2
333,0
100
3,333,33
5,1
50
222,0100
2,222,22
25,2
50
667,0100
7,667,66
5,1
100
cm
kN
aa A
F
cm
kN
aa A
F
cm
kN
aa A
F
T xy
y
y
x
x
=====
−=−
=−
=−
=−
=
=====
τ
σ
σ
−=
−=
000
0222,0333,0
0333,0667,0
000
02,223,33
03,337,66
22
22
aa
aa
S
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0;0;0;0 ==== zx yz xy z τ τ τ σ
22221
1
445,0100
5,445,442,227,66
cm
kN
aaa I
I y x z y x
===−=
+=++= σ σ σ σ σ
4
2
4
2
2222
2222
2
258964,010000
64,258964,25893,332,227,66
cm
kN
aaaa I
I xy y x zx yz xy x z z y y x
−=−
=−
=
−
−⋅=
−⋅=−−−⋅+⋅+⋅= τ σ σ τ τ τ σ σ σ σ σ σ
02222
3 =⋅⋅⋅+⋅−⋅−⋅−⋅⋅= yz xz xy xy z zx y yz x z y x I τ τ τ τ σ τ σ τ σ σ σ σ
30
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Teoretske osnove za izradu zadataka u narednom poglavlju nalaze se u knjizi «Otpornost
materijala sa teorijom elastič nosti», autora Dr sci. Sadudina Hodžića, redovnog profesora na Rudarsko-geološko-građ evinskom fakultetu u Tuzli, na stanicama broj 62 do 73.
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3. Jednoosno naponsko stanje
Zadatak 3.1
Za štapni sistem prikazan na slici 11. odrediti sile u štapovima (1), (2) i (3). Popreč ni presjeci
štapova su: A1=2,075cm2; A2=5cm2; A3=2cm2 . Takođ e odrediti pomijeranje tač ke (C), poddjelovanjem sile G=100kN. Štapovi su od č elika. Težinu štapova zanemariti. l 2=200cm.
321
BA
C
G
35o 30o
l 3
l 2
l 1
S3
S2
S1
y
321
BA
C
G=100kN
35o30o
x
Slika 11. Slika 12.
331
31
15,1
30sin
35sin
035sin30sin
S S S
S S S
o
o
oo
x
⋅=⋅=
=⋅−⋅=Σ
0819,0866,0
035cos30sin
0
321
321
=−⋅++⋅
=−⋅++⋅
=Σ
GS S S
GS S S
S
oo
y
GS S
GS S S
=+⋅
=⋅++⋅⋅
23
323
81,1
819,0866,015,1
C/
Slika 13.
∆ l 2
∆ l 3
S3
S2
S1
C
35o30o
x35o
23 552,025,55 S S ⋅−=
23
2
3
0
2
3
819,0
819,0
35cos
l l
l
l l
l
∆⋅=∆
=∆∆
=∆
∆
2
222
3
333 ;
A E
l S l
A E
l S l
⋅⋅
=∆⋅⋅
=∆
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oo
o l l
l l
l 35cos;
30cos;30cos 2
32
1
1
2 === l
( )
kN S
S
S S
S S
A E
l S
A E
l S l
o
37,67
25,5582,0
268,0552,025,55
5
819,0
819,02
552,025,55
819,0
35cos
552,025,55
2
2
22
22
2
22
3
223
=
=⋅
⋅=⋅−
⋅=⋅
⋅−
⋅⋅⋅
=⋅
⋅⋅−=
∆
kN S 05,1837,67552,025,553 =⋅−=
kN S S 76,2005,1815,115,1 31 =⋅=⋅=
omijeranje tačke ( C ):
-C/=∆l2
P
C
cml l
l
A E
l
1282,0200000641,0000641,0
521000
37,67
22
2
2
22
=⋅=⋅=∆
⋅⋅
=⋅
S l 2 ⋅=∆
adatak 3.2
a stati č ki sistem prikazan na slici 15., odrediti dilatacije u štapovima
Z Z C A i C B , ako su isti od
g na eza :
z
Slika 14.
č elika. Štapove dimenzionirati na osnovu sila u štapovima i dozvoljeno pr nja na istezanje
σ
de=12kN/cm2 . Štapovi su okruglog popriječ nog presjeka. Greda A-B, č iju težinu trebaanemariti, optereć ena je silama F 1=60kN i F 2=30kN.
lA
B
C
l/2
l/2
l
α
α=60o
1
2l=2m
α
F2 F1
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roz tačku (A) postavimo koordinatni sistem i postavimo uvjete ravnoteže:
prva dva uvjeta odredimo sile u štapovima (kao funkcije ugla ϕ), a iz trećeg odredimo ugao (ϕ).
Slika 15.
K
0;0;0 =Σ=Σ= B y x M F F Σ
IzUgao (ϕ) je ravnotežni ugao. Zatim izračunamo sile FA i FB, i izvršimo dimenzioniranje štapova.
Kada su poznati popriječni presjeci štapova, određuje se njihova deformacija (∆li) i dilatacija (εi).
C BC A60cos
=°
==
F2
l
B
C
F1
l/2
l/2
l60o
60o
2
ϕ
FB
x
F 1+F 2=G=90kN
ml
45,0
2=
1
y
FA
A
ϕ
( )
( )
)3.(....................0cos2
3sin2
)2(........................................0cossin
)1(........................................0sincos
21 =⋅⋅
⋅+−⋅⋅⋅=Σ
=⋅−⋅+=Σ
=⋅−⋅−=Σ
ϕ α
ϕ α
α
l F F l F M
G F F Y
G F F X
A B
B Ai
B Ai
jednačine (1):Iz
α cossin⋅=− G F B A F
jednačine (2):Iz
α sin
cos⋅=+
G F B A F
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α α
α ϕ α ϕ
α α
α ϕ α ϕ
α
ϕ
α
ϕ
α
ϕ α
ϕ
α
ϕ
α
ϕ
α
cossin2
)coscossin(sin
cossin
coscossinsin
sin
cos
cos
sin2
sin
cossin
cos
cos
sin
cos
sin
cos
sin
⋅⋅⋅+⋅⋅
=
⋅⋅⋅+⋅⋅
=⋅
+⋅
=⋅
−⋅
=
−⋅
=⋅
−
⋅−=
−⋅
=−
G F
GGGG F
F G
F
F GG
F
G F F
F G
F
A
A
A B
A A
A B
A B
Matematske smjene:
( ) ( )[ ]
( ) ( )[ ]
( ) ( ) ( ) ( )[ ]
( ) ( )ϕ α ϕ α α ϕ α ϕ
ϕ α ϕ α ϕ α ϕ α α ϕ α ϕ
ϕ α ϕ α ϕ α
α α α
ϕ α ϕ α α ϕ
ϕ α ϕ α α ϕ
−=−⋅=⋅+⋅
++−++−−⋅=⋅+⋅
⋅+⋅=−
=⋅⋅
++−⋅=⋅
+−−⋅=⋅
coscos
2
2coscossinsin
coscoscoscos2
1coscossinsin
sinsincoscos)cos(
2sincossin2
coscos2
1coscos
coscos21sinsin
( )α
α
2sin
cos G F A
⋅−=
Analogijom slijedi:
( )α
α
2sin
cos G F B
⋅+=
Uvrštavanjem (FA) u jednačini (3), slijedi:
( )0cos
2
3sin2
2sin
cos21 =⋅⋅
⋅+−⋅⋅⋅−⋅
ϕ α α
ϕ α l F F l
G
Zamjenom poznatih veličina slijedi:
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( )( )
°=
=
=⋅+
⋅=⋅+⋅
⋅+⋅=⋅+⋅=−
⋅=−
47,5
096,0
583,0cos
sin866,05,0
cos583,0sin866,0cos5,0
sin866,0cos5,0sinsincoscoscos
cos583,0cos
ϕ
ϕ
ϕ
ϕ
ϕ ϕ ϕ
ϕ ϕ ϕ α ϕ α ϕ α
ϕ ϕ α
tg
( )kN F A 3,60
866,0
2,52
120sin
9047,560cos==
°⋅°−°
=
( )kN F
B2,43
866,0
36,37
120sin
9047,560cos==
°
⋅°+°=
Dimenzioniranje štapa (1)
2
1
1
025,512
3,60cm
F A
A
F
d
A
A
d
===
=
σ
σ
cmd
d A
53,2
14,3
025,54
4
1
2
11
=⋅
=
= π
Usvaja se d1=26mm.
Dimenzioniranje štapa (2)
2
2
2
6,312
2,43cm
F A
A
F
d
B
B
d
===
=
σ
σ
cmd
d A
14,214,3
6,34
4
2
2
2
2
=⋅
=
= π
Usvaja se d2=22mm.
cmC BC Al l 40021 ====
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38
( )
( )
( )
( )
z ⋅=−
2,0100
2040
( )
( ))(126,0
14,304,
14,32,020
22
2
2
cm z z
z
z
⋅+
⋅⋅
⋅⋅+
z h
r R z x
z
x
h
r R
dz h
r R x
dz h
r Rdx
⋅=−
⋅=
=−
⋅−
=
⋅−
=
∫
Tabela 4. Tabelaizračunatog pritiska
za z→0 do 100z
(cm)
A
(cm2)σ p
(kN/cm2)
0 1256 7,96
10 1520 6,6
20 1808 5,5
30 2123 4,7
40 2462 4,0
50 2826 3,54
60 3215 3,1
70 3630 2,75
80 4069 2,45
90 4534 2,2
100 5024 2,0
( )
12,251256
08202
2
A
z A
xr A
x
x
x
⋅+=
+⋅+=
=⋅+= π
Zakonitosti promjene napona:
π ⋅
⋅
−+=
∫
2
0
dz
h
r Rr A
z
x
⋅
⋅
−+
==
∫22
0
cm
kN
dz h
r Rr
F
A
F
z x
x
π
σ Veličina (z) ima vrijednost od 0 do h.
Dijagram pritiska
Slika 18.
F
r
h
σmax
R
UL=10cm=1cm
Uσ=2kN/cm =1cm2
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Zadatak 3.4
Stub od betona u obliku zarubljene kupe prikazane na slici 19., optereć en je silom F=40kN. zapreminska masa stuba:
ρ
=2,5t/m3 . Stub se oslanja na č vrstu podlogu koja je od istogmaterijala kao i stub. Postaviti jednač inu za napon u proizvoljnom presjeku, a zatim nacrtati
dijagram napona za č itav stub, po visini.
y
F
R1=20cm;
R2=40cm z
R 2
h = 8 m
R 1 F=40kN
R 2
h = 8 m
R 1
∆y
Slika 19. Slika 20.
z y R R
z y
z y
z
y
h
R R
⋅+=∆+=
⋅=∆
⋅
−
=∆
∆=
−
025,02,0
025,0
8
2,04,0
1
12
( ) ( ))(00196,00314,0126,014,3000625,001,004,014,3025,02,0
22
222
m z z A z z z R A
⋅+⋅+= ⋅⋅+⋅+=⋅⋅+=⋅= π
Težina štapa dužine (z)
22
1
222
11
00098,00157,0126,02
00196,00314,0252,0
2
126,014,32,0
z z z z A A
A
m R A
sr ⋅+⋅+=⋅+⋅+
=+
=
=⋅=⋅= π
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Zapremina posmatranog dijela:
32 00098,00157,0126,0 z z z z AV sr ⋅+⋅+⋅=⋅=
Težina:
32 024,038,01,3
5,2481,95,2
z z z G
V V V G
⋅+⋅+⋅=
⋅=⋅⋅=⋅⋅= ρ
Opća formula napona za bilo koju veličinu (z)
( )
⋅+⋅+⋅+⋅+⋅+
−=+
−=22
32
00196,00314,0126,0
024,038,01,340
m
kN
z z
z z z
A
G F pσ
Za dijagram napona, uzete su veličine za (z). z=0; 2; 4; 6; 7 i 8m, i izračunati su naponi koji su prikazani tabelarno i nacrtani na dijagramu napona.
Tabela 5.
z (m) Ai (m2) (F+G) (kN) σi (kN/m
2)
0 0,126 40 -317,46
2 0,197 47,9 -243,6
4 0,283 60 -211,7
6 0,385 77,5 -201,3
7 0,441 88,5 -200,68
8 0,5024 101,4 -202,0
R 1 F=40kN
σmax=-317,46kN/cm2
h = 8 m
UL=100cm=1cm
Uσ=100kN/cm =1cm2
R 2 z Slika 21.
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Teoretske osnove za izradu zadataka u narednom poglavlju nalaze se u knjizi «Otpornostmaterijala sa teorijom elastič nosti», autora Dr sci. Sadudina Hodžića, redovnog profesora na
Rudarsko-geološko-građ evinskom fakultetu u Tuzli, na stanicama broj 81 do 95.
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4. Momenti inercije
Zadatak 4.1
Za lik prikazan na slici 22. odrediti glavne momente inercije, položaj glavnih osa inercije, glavne
polupriječ nike inercije i nacrtati elipsu inercije.
Slika 22. Slika 23.
10
1 5 c m
1 5 c m
6 0 c m
20cm
30cm
x
y
X3=33,3
X2=25
A3
A2
A1
X1=13,3
y 3 = 5 5
y 2 = 3 0
y 1 = 5
Određivanje položaja težišta lika:
Tabela 6.
Br. Površina
(cm2)
xi(cm) yi(cm) Sx
(cm3)
Sy
(cm3)
1 150 13,3 5 750 1995
2 600 25 30 18000 15000
3 75 33,3 55 4125 2497,5
Σ 825 - - 22875 19492,5
Koordinate težišta:
cm A
S x
cm A
S y
y
T
xT
6,23825
5,19492
7,27825
22875
==Σ
Σ=
==ΣΣ
=
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Pošto je određen položaj težišta, određuju se momenti inercije za težišne ose.
4
223333
3191755589777293187593770846112327
3,27757,2215036
1520
36
1510
3
7,2710
3
3,3210
cm I
I
x
x
=+++++=
⋅+⋅+⋅
+⋅
+⋅
+⋅
=
2 7 , 3
A2/V
A2/
A2
///
x
y
xT=23,6cm
A3
A2//
A1
3 2 , 3 c m
y T = 2
7 , 7 c m
2 2 , 7
6,43,6
Slika 24.
4
223333
3289670571591333344165243933
7,9753,1015036
2015
36
1015
3
4,660
3
6,360
cm I
I
y
y
=+++++=
⋅+⋅+⋅
+⋅
+⋅
+⋅
=
Napomena:
Kada se traži ( I x), onda su za sopstvene momente inercije trokutova (1) i (3) baze paralelne osi ( x);
tj. a1=20cm; a3=10cm. Za ( I y) a1=15cm; b3=15cm.36
3ha I xs
⋅=∆
Centrifugalni momenat inercije
Svaku segmentarnu površinu: A1; A2
/
; A2
//
; A2
///
; A2
/V
i A3 množimo sa koordinatama težišta te površine u odnosu na težište integralnog lika, vodeći računa o predznacima koodinata u pojedinim
kvadrantima.
A1=150cm2; A2
/ =116,3 cm
2; A2
// =206,7 cm
2;
A2 ///
=99,7 cm2; A2
/V =177,3 cm
2; A3=75 cm
2;
Sopstveni centrifugalni momenat inercije pravouglog trokuta:72
22 ha I xys
⋅−=
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( ) ( ) ( )
( ) ( ) ( )
4
22
22
2,5529878585,248512505,350713381106825,3127,19860
85,132,33,17772
152085,138,17,99
7,223,1015015,168,13,11615,162,37,20672
15103,277,975
cm I
I
xy
xy
=−+−+−+−=
−⋅⋅+⋅
−−⋅−⋅+
+−⋅−⋅+⋅−⋅+⋅⋅+⋅
−⋅⋅=
Položaj glavnih osa inercije:
o
o
y x
xy
I I
I tg
5,10
;12,212
386,032896319175
2,55298222
−=
−=
−=−
⋅−=
−
−=
α
α
α
Glavni momenti inercije:
( ) ( )
( ) ( )
153450176035
2,552984328963191752
132896319175
2
1
42
1
2
1
2,1
22
2,1
22
2,1
±=
⋅+−⋅±+⋅=
⋅+−⋅±+⋅=
I
I
I I I I I I xy y x y x
44
;22585;329485 4
2
4
1 cm I cm I ==
cm A
I i
cm A
I i
2,5825
22585
;20825
329485
22
11
==Σ=
==Σ
=
Elipsa inercije:
Slika 25
0
(2)
(1)
i2 D
C
B
A
i
x
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Zadatak 4.2
Za lik prikazan na slici 26. odrediti glavne momente inercije, položaj glavnih osa inercije, glavne polupriječ nike inercije i nacrtati elipsu inercije. R=10cm
Slika 26. Slika 27.
A3
y
A2
A4
3R
4 R
xA1 R
R
3R
4 R
Određujemo težište i zadati lik postavljamo u koordinatni sistem
Tabela 7.
R
y 1
y 1
Br. A(cm2) xi(cm) yi(cm) Sx
(cm3)
Sy
(cm3)1 157 10 5,8 910,6 1570
2 1200 10 40 48000 12000
3 600 35 60 36000 21000
4 157 54,2 60 9420 8509
Σ 2114 - - 94330 43079Slika 28.
44
11
11,0;8
58,0;42,0
R I R
I
R y R y
x y ⋅=⋅
=
⋅=⋅=′
π
Koodinate težišta:
R xilicm A
S x
R yilicm A
S y
T
y
T
T
x
T
⋅≅==Σ
Σ=
⋅≅==ΣΣ
=
2202114
43079
46,46,442114
94330
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Za težišne ose x i y određujemo momente inercije:
( ) ( )( )
( )
( ) ( )
444
44444444
2
24
22
32
24
33
832000103,823,82
7,34,01426,2311,06,279,10
54,128
54,16
12
2388,3
211,0
3
46,32
3
54,22
cm R I
R R R R R R R R I
R R R R R
R R R
R R
R R R R I
x
x
x
=⋅=⋅=
⋅+⋅+⋅+⋅+⋅+⋅+⋅+⋅=
⋅⋅⋅+⋅+⋅⋅⋅+
+⋅⋅⋅
+⋅⋅⋅
+⋅+⋅⋅⋅
+⋅⋅⋅
=
π π
π
A2
A3 A4
3R
3 , 8 8 R
x
y
2 , 5 4 R
1 , 5 4 R
3,42R
4 , 4 6 R
R
3 , 4 6 R
A1
Slika 29.
( ) ( )( )
444
444444
42
22
24
33
545000105,545,54
4,057,14,1811,01816
8242,3
211,0
3
32
3
26
cm R I
R R R R R R I
R R
R R
R R
R R R R I
y
y
y
=⋅=⋅=
⋅+⋅+⋅+⋅+⋅+⋅=
⋅+⋅
⋅+⋅⋅
⋅+⋅+
⋅⋅⋅+
⋅⋅⋅=
π π π
( ) ( ) ( )
444
44444
2
22
337000107,337,33
61245,63,886,13
88,32
73,1)(46,32
27,1)(254,242,354,125,154,16
cm R I
R R R R R I
R R R
R R R R
R R R R R R
R
R R R I
xy
xy
xy
=⋅=⋅=
⋅+⋅+⋅−⋅+⋅=
⋅−⋅−⋅⋅
+⋅−⋅⋅⋅⋅⋅+
+⋅⋅−⋅⋅⋅⋅+⋅⋅⋅⋅⋅
+⋅⋅⋅⋅⋅=π
π
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Položaj glavnih osa simetrije:
o
o
o
y x
xy
arctg
R R
R
I I
I tg
8,33
;58,672
58,672
42,25,543,82
7,33222
44
4
−=
−=
−=
−=⋅−⋅
⋅⋅−=
−
⋅−=
α
α
α
α
Glavni momenti inercije:
( ) ( )
( ) ( ) ( )
( )
;319000109,319,31
;1049000109,1049,104
5,364,680,732
5,543,822
7,3345,543,822
15,543,82
2
1
42
1
2
1
444
2
444
1
44
44
2,1
2424444
2,1
22
2,1
cm R I
cm R I
R R R R I
R R R R R I
I I I I I I xy y x y x
=⋅=⋅=
=⋅=⋅=
⋅±⋅=⋅±+⋅=
⋅⋅+⋅−⋅⋅±⋅+⋅⋅=
+−⋅±+⋅=
cm28,122114
= A
I i
cm A
I i
319000
;27,222114
1049000
22
11
=Σ
=
==Σ
=
Elipsa inercije:(2)
i 2
α=-33,8°
i1
(1)
Slika 30.
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Zadatak 4.3
Za lik prikazan na slici31. odrediti glavne momente inercije, položaj glavnih osa inercije, glavne polupriječ nike inercije i nacrtati elipsu inercije. =1cm
1 0 δ1
1 0 δ
5
5δ10δ 10δ
15δ 5δ
2
2 5 δ
1 0 δ 3
4
5δ 10δ
Slika 31.Određivanje težišta lika:
Tabela 8.Br. A(cm2) xi(cm) yi(cm) Sx (cm3) Sy (cm3)
1 250⋅δ2=250 12,5⋅δ=12,5 30⋅δ=30 7500⋅δ3=7500 3125⋅δ3=3125
2 75⋅δ2=75 12,5⋅δ=12,5 17,5⋅δ=17,5 1312,5⋅δ3=1312,5 937,5⋅δ3=937,5
3 100⋅δ2=100 10⋅δ=10 5⋅δ=5 500⋅δ3=500 1000⋅δ3=1000
4 25⋅δ2=25 3,33⋅δ=3,33 3,33⋅δ=3,33 83,25⋅δ3=83,25 83,25⋅δ3=83,25
5 25⋅δ2=25 26,67⋅δ=26,67 28,33⋅δ=28,33 708,25⋅δ3=708,25 666,75⋅δ3=666,75
Σ 475⋅δ2=475 - - 10104⋅δ3=10104 5812,5⋅δ3=5812,5
Koodinate težišta:
cm A
S x
cm AS y
y
T
x
T
3,2113,213,21475
10104
2,1212,122,12475
5,5812
2
3
2
3
=⋅=⋅=⋅⋅
=Σ
Σ=
=⋅=⋅=⋅⋅=
ΣΣ=
δ δ
δ
δ δ δ
Momenat inercije za osu (x). Lik je podijeljen na 6 segmenata.
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5δ
x
y
5
4
3
2
6
15δ
y T = 2 1 , 3
δ
1 0
δ
1 0
δ
1 0 δ
25δ
10δ
15δ 5δ
xT=12,2δ 2,8δ
Slika 32.
( )
( )( ) 422
3
)2(
4
3
)1(
210067,825012
1025
4,843
7,35
δ δ δ δ δ
δ δ δ
⋅=⋅⋅⋅+⋅⋅⋅
=
⋅=⋅⋅⋅
=
I
I
( )
( )( )
( )( )
( ) ( )
444
444444
)6()5()4()3()2()1(
422
3
)6(
422
3
)5(
422
3
)4(
4
3
)3(
6047316047360473
821213641370116106210064,84
821297,172536105
136472536
105
137013,165012
105
161063
3,215
cm I
I
I I I I I I I
I
I
I
I
x
x
x
=⋅=⋅=
⋅+⋅+⋅+⋅+⋅+⋅=
+++++=
⋅=⋅⋅⋅+⋅⋅⋅=
⋅=⋅⋅⋅+⋅⋅⋅
=
⋅=⋅⋅⋅+⋅⋅⋅
=
⋅=⋅⋅⋅
=
δ
δ δ δ δ δ δ
δ δ δ δ δ
δ δ δ δ δ
δ δ δ δ δ
δ δ δ
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Momenat inercije za osu (y). Lik je podijeljen na 7 segmenata.
1 5 δ
12,8δ
7
7,2δ
x
y
5
4
3
2
6
1
5δ
1 0 δ
1 0 δ
12,2δ
10δ
15δ 2δ
y T =
2 1 , 3
δ
1 0 δ
2,
2,8δxT=12,2δ
Slika 33.
( )
( ) 4
3
)2(
4
3
)1(
5,69903
8,1210
8,60523
2,1210
δ δ δ
δ δ δ
⋅=⋅⋅⋅
=
⋅=⋅⋅⋅
=
I
I
( )
( )
( )
( ) ( )
( )( )
444
4444444
)7()6()5()4()3()2()1(
422
3
)7(
422
3
)6(
4
3
)5(
4
3
)4(
4
3
)3(
8,2226818,222688,22268
7,526622288,15218,1552,535,69908,6052
7,526647,142536
510
222837,92536
510
8,15213
7,710
8,1553
8,23,21
2,533
2,215
cm I
I
I I I I I I I I
I
I
I
I
I
y
y
y
=⋅=⋅=
⋅+⋅+⋅+⋅+⋅+⋅+⋅=
++++++=
⋅=⋅⋅⋅+⋅⋅⋅
=
⋅=⋅⋅⋅+⋅⋅⋅=
⋅=⋅⋅⋅
=
⋅=⋅⋅⋅
=
⋅=⋅⋅⋅
=
δ
δ δ δ δ δ δ δ
δ δ δ δ δ
δ δ δ δ δ
δ δ δ
δ δ δ
δ δ δ
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Centrifugalni momenat inercije za težišne ose. Lik je podijeljen na 9 segmenata.Segmenti (9) i (8) imaju i sopstvene centrifugalne momente inercije.
( ) ( )
( ) ( )422
)9(
422
)8(
4
22
)8()9(
6,24977,3447,14725
3,39457,3497,1786,825
7,3472
105
δ δ δ δ δ
δ δ δ δ δ
δ δ δ
⋅=⋅−⋅⋅⋅⋅⋅=
⋅=⋅−⋅−⋅⋅−⋅⋅=
⋅−=⋅⋅⋅−
==
xy
xy
sopstveni xy sopstveni xy
I
I
I I
9
8
1 5 δ
7
7,2δ
x
5
4 3
2
6
1
1 0 δ
1 0 δ
12,2δ
10δ
15δ 2δ
2,8δ
y T = 2 1 , 3
δ
2,
1 0 δ
5δ
xT=12,2δ
12,8δ
Slika 34.
( )
( )
( )
( ) ( )
( ) ( ) 4
)7(
4
)6(
4
)5(
4
)4(
4
)3(
4
)2(
4
)1(
42253,166,3102,7
5,1541,165,52,23,11
2,8894,165,108,23,21
3,1785,118,17,32,2
8,2685,14,17,38,2
5,64741,67,8102,12
71274,67,8108,12
δ δ δ δ δ
δ δ δ δ δ
δ δ δ δ δ
δ δ δ δ δ
δ δ δ δ δ
δ δ δ δ δ
δ δ δ δ δ
⋅=⋅−⋅⋅−⋅⋅⋅⋅=
⋅=⋅−⋅⋅−⋅⋅⋅⋅=
⋅−=⋅⋅⋅−⋅⋅⋅⋅=
⋅−=⋅⋅⋅−⋅⋅⋅⋅=
⋅=⋅⋅⋅⋅⋅⋅⋅=
⋅−=⋅−⋅⋅⋅⋅⋅⋅=
⋅=⋅⋅⋅⋅⋅⋅⋅=
xy
xy
xy
xy
xy
xy
xy
I
I
I
I
I
I
I
)9()8()7()6()5()4()3()2()1( xy xy xy xy xy xy xy xy xy xy I I I I I I I I I I ++++++++=
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444
444444444
5,1056015,105605,10560
6,2497394542255,1542,8893,178,265,64747127
cm I
I
xy
xy
=⋅=⋅=
⋅+⋅+⋅+⋅+⋅−⋅−⋅+⋅−⋅=
δ
δ δ δ δ δ δ δ δ δ
Položaj glavnih osa simetrije:
o
o
o
y x
xy
arctg
I I
I tg
47,14
;93,282
93,282
5528,08,2226860473
5,10560222
44
4
−=
−=
−=
−=⋅−⋅
⋅⋅−=−
−=
α
α
α
δ δ
δ α
Glavni centralni momenti inercije:
( ) ( )( ) ( ) ( )
11954419544
;6319816319863198
2182741371
58,1056048,22268604732
18,2226860473
2
1
42
1
2
1
44
2
444
1
44
2,1
2424444
2,1
22
2,1
I
cm I
I
I
I I I I I I xy y x y x
⋅=⋅=
=⋅=⋅=
⋅±⋅=
⋅⋅+⋅−⋅⋅±⋅+⋅⋅=
⋅+−⋅±+⋅=
δ
δ
δ δ
δ δ δ δ δ
;19544 4cm=
cm;
cm A
I i
A
I i
4,6475
19544
5,11475
63198
22
11
==Σ=
==Σ
=
(2)
y
Elipsa inercije:
α=-14,47° x
i 1
(1)
i2
Slika 35.
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Zadatak 4.4
Za popriječ ni presjek nosač a prikazanog na slici 36. odrediti momenat inercije za težišnu osu (x). R1=4
; R2=2
; R3=8
; R4=6
; =1cm.
-x x
R 1R 2
R 4 R 3
2 0 δ
2δ
Slika 36.
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Određivanje težišta popriječnog presjeka
R1=4 ; R2=2 ; R3=8 ; R4=6 ;
=1cm
5
2
3
1
2δ
2 0 δ
R 2R 1 4
x
R 4
R 3
Slika 37.
Površina (1):222
1 4014040220 cm A =⋅==⋅⋅⋅= δ δ δ
Površina (2): 222222
32 5,10015,1005,100
2
14,38
2cm
R A =⋅=⋅=
⋅⋅=
⋅= δ
δ π
Površina (3):222
222
43 5,5615,565,56
2
14,36
2cm
R A −=⋅−=⋅−=
⋅⋅−=
⋅−=− δ
δ π
Površina (4): 222222
14 2512525
2
14,34
2cm
R A =⋅=⋅=
⋅⋅=
⋅= δ
δ π
Površina (5): 222222
25 3,613,63,6
2
14,32
2cm
R A −=⋅−=⋅−=
⋅⋅=
⋅−=− δ
δ π
Površina presjeka:
2222222
54321
7,1027,1023,6255,565,10040 cm A
A A A A A A
=⋅=⋅−⋅+⋅−⋅+⋅=Σ
−+−+=Σ
δ δ δ δ δ δ
Koordinate površina (1,2,3,4,5):
x1=7⋅δ=7⋅1=7cm; y1=14⋅δ=14⋅1=14cm;
x2=14⋅δ=14⋅1=14cm; y2=(20⋅δ+4⋅δ+0,42⋅R 3)= 27,4⋅δ=27,4⋅1=27,4cm;
x3=14⋅δ=14⋅1=14cm; y3=(20⋅δ+4⋅δ+0,42⋅R 4)= 26,5⋅δ=26, 5⋅1=26,5cm;
x4=4⋅δ=4⋅1=4cm; y4=0,57⋅R 1=2,3⋅δ=2,3⋅1=2,3cm;
x5=4⋅δ=4⋅1=4cm; y5=(2⋅δ+0,57⋅R 2)= 3,14⋅δ=3,14⋅1=3,14cm;
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Statički momenti površina:
3332
222
3332
111
7,275317,27537,27534,275,100
56015605601440
cm y AS
cm y AS
x
x
=⋅=⋅=⋅⋅⋅=⋅=
=⋅=⋅=⋅⋅⋅=⋅=
δ δ δ
δ δ δ
333
54321
3332
555
3332
444
3332
333
4,185414,18544,1854
8,1918,198,1914,33,6
5,5715,575,573,225
14971149714975,265,56
cmS S S S S S
cm y AS
cm y AS
cm y AS
x x x x x x
x
x
x
=⋅=⋅=−+−+=Σ
−=⋅−=⋅−=⋅⋅⋅=⋅=−
=⋅=⋅=⋅⋅⋅=⋅=
−=⋅−=⋅−=⋅⋅⋅=⋅=−
δ
δ δ δ
δ δ δ
δ δ δ
cm A
S y x
T 18118187,102
4,18542
3
=⋅=⋅=⋅⋅
=ΣΣ
= δ δ
δ
3332
222
3332
111
1407114071407145,100
2801280280740
cm x AS
cm x AS
y
y
=⋅=⋅=⋅⋅⋅=⋅=
=⋅=⋅=⋅⋅⋅=⋅=
δ δ δ
δ δ δ
333
54321
3332
555
3332
444
3332
333
8,97018,9708,970
2,2512,252,2543,6
1001100100425
7911791791145,56
cmS S S S S S
cm x AS
cm x AS
cm x AS
y y y y y y
y
y
y
=⋅=⋅=−+−+=Σ
−=⋅−=⋅−=⋅⋅⋅=⋅=−
=⋅=⋅=⋅⋅⋅=⋅=
−=⋅−=⋅−=⋅⋅⋅=⋅=−
δ
δ δ δ
δ δ δ
δ δ δ
cm A
S x
y
T 45,9145,945,97,102
8,9702
3
=⋅=⋅=⋅⋅
=Σ
Σ= δ
δ
δ
Momenti inercije:
Slika 38.xT=9,45δ
R1=4 R2=2 R3=8 R4=6
=1cm
T 6 δ
y
x
1 4 δ
T = 1 8 δ
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( ) ( )( )
( ) ( )
( )
444
444
4444444
22
34
2
22
14
1
22
44
4
22
34
3
33
1231511231512315
13828,18,6685
2840985,14288025,4501829144
242,014211,0
442,0142
11,0642,062
11,0
842,062
11,03
142
3
62
cm I
I
R
R
R R
R R
R R I
x
x
x
=⋅=⋅=
⋅−⋅−⋅+
+⋅+⋅−⋅−⋅+⋅+⋅+⋅=⋅⋅+⋅⋅
⋅
−⋅−
−⋅⋅+⋅⋅⋅
+⋅+⋅⋅+⋅⋅⋅
−⋅−
−⋅⋅+⋅⋅⋅
+⋅+⋅⋅⋅
+⋅⋅⋅
=
δ
δ δ δ
δ δ δ δ δ δ δ
δ δ
π
δ δ π
δ δ π
δ δ π δ δ δ δ
Zadatak 4.5
Nacrtati elipsu inercije za presjek prikazan na slici 39. =2cm
56
Slika 39.3δ
1 0 δ
7,5δ 3 δ
δ
δ
δ
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Određivanje koordinata težišta:
A4
A3
A2
A1
x1
x2
y 3 y 2
y 1
x3
x4
y 4
´ 1=3δ 2;
´ 2=6,5δ 2;
´ 5=2δ 2;
1=4,2δ ;
2=3,2δ ;
5=6,8δ
y 3
A/5
A/4
A/3
A/2
A/2
7 , 3
δ
y 2
y 1
4,45δ
y 5
3 , 7
δ
y 4
Slika 40. Slika 41.
A1=3⋅δ 2=3⋅ 2
2=12cm
2; x1=10⋅δ =10⋅ 2=20cm; y1=11,5⋅δ =11,5⋅ 2=23cm;
A2=7,5⋅δ 2=7,5⋅ 2
2=30cm
2; x2=5,75⋅δ =5,75⋅ 2=11,5cm; y2=10,5⋅δ =10,5⋅ 2=21cm;
A3=9⋅δ 2=9⋅ 2
2=36cm
2; x3=2,5⋅δ =2,5⋅ 2=5cm; y3=5,5⋅δ =5,5⋅ 2=11cm;
A4=3⋅δ 2=3⋅ 2
2=12cm
2; x4=1,5⋅δ =1,5⋅ 2=3cm; y4=0,5⋅δ =0,5⋅ 2=1cm;
A= A1+A2+A3+A3= 22,5⋅δ 2=22,5⋅ 2
2=90cm
2
cm y
A
y A y A y A y A y
cm x
A
x A x A x A x A x
T
T
T
T
6,1423,73,7
5,22
5,035,595,105,75,113
9,8245,445,4
5,22
5,135,2975,55,7103
2
222244332211
2
222244332211
=⋅=⋅=
⋅⋅⋅⋅+⋅⋅⋅+⋅⋅⋅+⋅⋅⋅
=⋅+⋅+⋅+⋅
=
=⋅=⋅=⋅
⋅⋅⋅+⋅⋅⋅+⋅⋅⋅+⋅⋅⋅=⋅+⋅+⋅+⋅=
δ
δ
δ δ δ δ δ δ δ δ
δ
δ
δ δ δ δ δ δ δ δ
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Određivanje momenata inerercije za težišne ose:
( ) ( ) ( )
( ) ( ) ( )
44
444444444
22
4444
2222
4
2
5
/
5
33332
2
/
2
2
1
/
1
3
599225,374
5,37448,9217,07,12985,1654,06,799,5225,2
8,62122
3389
365,50
125,65,35,62,43
1227
12
2
3
3,7
3
7,3
12
5,6
12
3
cm I
I
I
y A y A y A I
x
x
x
x
=⋅=
⋅=+++++++=
⋅+++++⋅+⋅+=
⋅+⋅
+⋅
+⋅
+⋅
+⋅+⋅+⋅
=
δ δ δ δ δ δ δ δ δ
δ δ δ δ δ δ δ δ δ δ δ
δ δ δ δ δ δ δ δ δ δ
( )( ) ( ) ( )
( ) ( )3
45,1
3
05,555,53
12
395,111
12
1145,3
12
233
223
223
2/
5
3δ δ δ δ
δ δ δ δ
δ δ δ δ
δ δ δ ⋅
+⋅
+⋅+⋅
+⋅+⋅
+⋅+⋅
= A I y
44
444444444
8,326028,203
8,20393,424,9225,08,4192,08,237,0
cm I
I
y
y
=⋅=
⋅=+++++++= δ δ δ δ δ δ δ δ δ
44
4444444
22
2222
8,308428,192
8,19292,4695,5134,1334,372,4093,69
)8,6()45,3(2)95,1()65,3(3,7
)95,1(85,17,3)72,0(2,345,152,22,305,555,52,43
cm I
I
I
xy
xy
xy
=⋅=
⋅=++−−+=
−⋅−⋅+−⋅−⋅+
+−⋅⋅+−⋅⋅+⋅⋅+⋅⋅=
δ δ δ δ δ δ δ
δ δ δ δ δ δ
δ δ δ δ δ δ δ δ δ δ δ δ
Glavni centralni momenti inercije:
( ) ( )
( ) ( ) ( )
;2,125122,782,78
;8,799628,4998,499
8,2100,289
8,19248,2035,3742
18,2035,374
2
1
4
2
1
2
1
444
2
444
1
44
2,1
2424444
2,1
22
2,1
cm I
cm I
I
I
I I I I I I xy y x y x
=⋅=⋅=
=⋅=⋅=
⋅±⋅=
⋅⋅+⋅−⋅±⋅+⋅=
+−±+=
δ
δ
δ δ
δ δ δ δ δ
Polupriječnici inercije:
;6,328,18,15,22
2,78
;4,927,47,45,228,499
2
4
22
2
4
11
cm A
I i
cm A I i
=⋅=⋅=⋅⋅
==
=⋅=⋅=⋅⋅==
δ δ
δ
δ δ δ
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Položaj glavnih osa simetrije:
°−=
°−=°−=
−=⋅−⋅
⋅⋅−=
−
−=
06,33
;12,66212,662
258,28,2035,374
8,192222
44
4
α
α α
δ δ
δ α
arctg
I I
I tg
y x
xy
Elipsa inercije:
i2 i 1
(2)
(1)
y
x
Slika 42.
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Zadatak 4.6
Za lik prokazan na slici 43. odrediti glavne momente inercije kao i položaj glavnih osa inercije.=2cm
x
2,5δ
2 0 , 5
δ
10δ
y
δ δ
δ
1 5 , 7 5 δ
6 , 7 5 δ
1 0 δ
6δ2δ
11,5δ
1
2
3
T
4
4,25δ
10δ δ δ
2δ
5 δ
2 0 δ
δ
Slika 43. Slika 44.
Određivanje težišta lika:
222
4
222
3
222
2
222
1
8022020
4021010
20255
20255
cm A
cm A
cm A
cm A
=⋅=⋅=
=⋅=⋅=
=⋅=⋅=
=⋅=⋅=
δ
δ
δ
δ
cm y
y
T
T
5,31275,1575,1540
2002055,1125,112
40
2010105,2055,2255,222
2222
=⋅=⋅=⋅+⋅+⋅+⋅
=
⋅
⋅⋅⋅+⋅⋅⋅+⋅⋅⋅+⋅⋅⋅=
δ δ δ δ δ
δ
δ δ δ δ δ δ δ δ
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cm x
x
T
T
5,8225,425,440
50605,575,2
40
205,210655,115,052
2222
=⋅=⋅=⋅+⋅+⋅+⋅
=
⋅⋅⋅⋅+⋅⋅⋅+⋅⋅⋅+⋅⋅⋅
=
δ δ δ δ δ
δ
δ δ δ δ δ δ δ δ
Slika 45.
y
x
2,5δ
4 , 7 5 δ
7,25δ 3,75δ
2δ
6 , 7 5 δ
1 5 , 7 5 δ
δ
1 5 , 7 5 δ
6 , 7 5 δ
7 , 8 7 5 δ
4,25δ
2,5δ
1,75δ
4
, 2 5 δ
2 , 1
2 5 δ
3,25δ 6,75δ
Momenat inercije za osu (x):
( ) ( ) ( )( ) ( )2
322
333
75,41012
1075,652
12
52
3
25,4
3
7,15δ δ
δ δ δ δ
δ δ δ δ δ δ ⋅⋅⋅+
⋅⋅+⋅⋅⋅⋅+
⋅⋅⋅+
⋅⋅+
⋅⋅= x I
244
444444
6,3248926,20306,2030
6,22583,06,455216,251302
cm I
I
x
x
=⋅=⋅=
⋅+⋅+⋅+⋅+⋅+⋅=
δ
δ δ δ δ δ δ
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Momenat inercije za osu (y):
( ) ( )( )
( ) ( )
244
44444444
223
22
322
333
81762511511
617,126342,03,7042,05,1024,11
75,12012
2025,75
12
575,35
12
5
3
75,6
3
25,3
cm I
I
I
y
y
y
=⋅=⋅=⋅+⋅+⋅+⋅+⋅+⋅+⋅+⋅=
⋅⋅⋅++⋅⋅
+⋅⋅⋅+
+⋅⋅
+⋅⋅⋅+⋅⋅
+⋅⋅
+⋅⋅
=
δ
δ δ δ δ δ δ δ δ
δ δ δ δ
δ δ
δ δ δ δ
δ δ δ δ δ δ
Centrifugalni momenat inercije:
( ) ( )
( ) ( )
444
444444
22
22
22
8,643623,4023,402
7,2441082178,1525126
25,775,6537,375,475,6
75,1)875,7(75,1575,1125,225,4
62,175,425,375,675,35
cm I
I
I
xy
xy
xy
=⋅=⋅=
⋅+⋅+⋅+⋅−⋅−⋅−=
⋅⋅⋅⋅⋅+⋅⋅⋅⋅⋅+
+⋅−⋅⋅−⋅⋅+⋅−⋅⋅⋅⋅+
+⋅−⋅⋅⋅⋅+⋅⋅⋅−⋅⋅=
δ
δ δ δ δ δ δ
δ δ δ δ δ δ
δ δ δ δ δ δ
δ δ δ δ δ δ
Glavni momenti inercije:
( ) ( )
( ) ( ) ( )
;6,657721,4111,411
;34080221302130
7,8598,1270
3,40245116,20302
15116,2030
2
1
42
1
2
1
4442
444
1
44
2,1
2424444
2,1
22
2,1
cm I
cm I
I
I
I I I I I I xy y x y x
=⋅=⋅=
=⋅=⋅=
⋅±⋅=
⋅⋅+⋅−⋅±⋅+⋅=
+−±+=
δ
δ
δ δ
δ δ δ δ δ
Polupriječnici inercije:
;4,622,32,340
1,411
;58,14229,729,740
2130
2
4
22
2
4
11
cm A
I i
cm A
I i
=⋅=⋅=⋅⋅
==
=⋅=⋅=⋅⋅
==
δ δ
δ
δ δ
δ
Položaj glavnih osa simetrije:
xy I tg 5290
3,402222
4
=⋅⋅−
=−
=δ
α