9 Resetkasti Nosaci N N
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Transcript of 9 Resetkasti Nosaci N N
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Rešetkasti nosa�i
9. dio
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• Rešetkasti nosa�i su važne inženjerske konstrukcije.
• Prema obliku razlikujemo:a) ravninske rešetkaste konstrukcijeb) prostorne rešetkaste konstrukcije
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• Rešetkasti nosa�i sastavljeni su od štapova koji tvore geometrijski nepromjenjiv oblike: – trokut - ravninska rešetka– tetraedar - prostorna rešetka
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• Stati�ki: - odre�ene rešetke-���������ene rešetke
• Stati�ki odre�ena:
• Labilna
• Stati�ki neodre�ena:
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Stati�ka odre�ena rešetka
n – broj �vorovas – broj štapova 32 −⋅= ns
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Veza izme�u broja �vorova i štapova
• n – broj �vorova n = 10• s – broj štapova s = 17 32 −⋅= ns
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Stati�ki odre�ene rešetke iznutra i izvana
Nestabilna rešetka - mehanizam
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Mehanizam - labilna “rešetka”
32 −⋅< ns
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Stati�ki neodre�ena rešetka(iznutra)
32 −⋅> ns
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• Pretpostavke kod prora�una rešetkastih nosa�a:
1. Štapovi rešetke su u �vorovima zglobno spojeni.
2. Aktivne sile (optereenje) djeluju u samo �vorovima rešetke.
vor se ozna�ava kružiem °
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• Posljedica ovih pretpostavki prora�una je da se prijenos aktivne sile na oslonce ostvaruje vla�nim (+) odnosno tla�nim (-) silama.
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Odre�ivanje sila u štapovima rešetke:
a) analiti�ki:– a1) za sve štapove rešetke: Metoda �vorova– a2) presjek rešetke kroz tri štapa: Ritterova metoda
b) grafi�ki:– b1) za sve štapove rešetke: Maxwell-Creamonin plan
(Poligoni sila)– b2) presjek rešetke kroz tri štapa: Culmannova metoda
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Primjer:
F1= 5 kN a=2m b=3m h=2mF2= 10 kNF3= 20 kN
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Reakcije – analiti�kiNekanonske jednadžbe ravnoteža:
+ dopunski uvjet:Os projekcija ne smijebiti okomita na spojnicu AB
0 3.
0 .2
0 .1
=
=
=
�
�
�
y
B
A
F
M
M
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kN ,003F 0FF F 0 3.
kN ,006F 03F5F 2F 2F 0 .2
kN 55,0 F 05F3F 2F 0 .1
Ay23Ay
Ax321Ax
B23B
=�=−−=
=�=⋅+⋅+⋅+⋅−=
=�=⋅+⋅+⋅−=
�
�
�
y
B
A
F
M
M
kNFFF AyAxA 1,673060 2222 =+=+=
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a1) Metoda �vorova
0 .2
0 .1
=
=
�
�
y
x
F
Fvorne sile: Si
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1. vor E:
045 scoSSF- ad.2.
045 sinS-F- ad.1.
F .
F .
211
22
x
y
=°⋅++=°⋅
=�
=�
02
01
kN , - S
kN , S
1414
0015
2
1
==
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2. vor C:
0SF- 2. ad.
0SS- 1. ad.
F .
F .1
33
41
y
x
=−=+
=�
=�
02
0
kN , S
kN ,S
0015
0020
4
3
=−=
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3. vor D:
kN ,S
kN ,S
0055
0854
6
5
−==
073345
073345
652
532
=+°⋅+°⋅−=°⋅++°⋅
S,cosSsinS
,sinSScosS
0 .2
0 .1
=
=
�
�
x
y
F
F
kN , S
kN ,S
0055
0854
6
5
−==
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21
4. vor A:
0
0733
02
01
7
75
=
=−°⋅−=�
=�
S
S,sinSF ad.1.
F .
F .
Ay
x
y
Kontrola:
00
F,cosS-S-
F 2. ad.
Ax54
x
==+°⋅
=�
0733
0
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22
5. vor B:
Kontrola:
0 .2
0 .1
=
=
�
�
y
x
F
F
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a2) Ritterova metoda• Odre�ivanje sila u presje�enim štapovima:
S2, S3 i S4
• Presjek samo kroz tri štapa
0 3.
0 .2
0 .1
3
2
1
=
=
=
�
�
�
R
R
R
M
M
M
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S2=? S3=? S4=?
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S
(tlak) kN , S
F S
F S
M .
3
3
3
E
0202
2202
2
022
01
3
3
−=
⋅−=⋅−=
=⋅+⋅� =
Sila S3
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Sila S4
(vlak) kN , S
FF S
FFS-
M .
4
4
4
D
0152
210252
22
0222
02
21
21
=
⋅+⋅=⋅+⋅=
=⋅+⋅+⋅� =
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Sila S2
(tlak) kN , S
rF
S
Fr S
M .
2
2
2
C
11422102
02
03
2
2
22
−=
⋅−=⋅−=
=⋅+⋅� =
22
2222 === d
r
r2r2
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28
Grafi�ko odre�ivanje sila u štapovima
b1) Poligoni sila – Maxwell-Creamonon plansila
1. vor E:
?S
?S
2
1
=
=
0SSF F
: silaRavnoteža
2121 =+++ �vor)(u tlak S
�vora)(iz vlak S
2
1
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2. vor C:
0SSS F
: silaRavnoteža
4313 =+++
?S
?S
4
3
=
=
�vora)(iz vlak S
�vor)(u tlak S
4
3
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3. vor D:
?S
?S
6
5
=
=
0SSS S
: silaRavnoteža
6523 =+++
�vor)(u tlak S
�vora)(iz vlak S
6
5
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5. vor B: Kontrola
4. vor A:
0SFS S
: silaRavnoteža
7A45 =+++ 0SS F
: silaRavnoteža
76B =+++
07 =S
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b2) Culmannova metoda
• Odre�ivanje sila u presje�enim štapovima: S4, S5 i S6 = ???
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3365
4R
SS L :pravac Culmannov
0LS F
: silaRavnoteža
+=
=++
R321 FFFF
: silavanjskih tatanzulRe
=++
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65
4 0
SS L
:pravac Culmannov
LS F
: silaRavnoteža
R
+=
=++
R321 FFFF
: silavanjskih tatanzulRe
=++
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Program
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Redoslijed sila u verižnom poligonu:
• Tipovi 1 i 3
• Tipovi 2 i 4
321 VVVH +++
HVVV 321 +++
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Zadatak:
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Reakcije
Mjerilo:
1cm :: 10kNGrafi�ko rješenje
Verižni poligon (v.p.)
Zaklju�na linija
verižnog poligona (s)
�Prvom zrakom (1) verižnog poligona kroz to�ku A
�Sjecište zadnje zrake (2) verižnog poligona s pravcem reakcije RB
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39kN5799RkN0099R
kN3695R
kN6410R
AAH
B
AV
, ,
,
,
===
=
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• Metoda �vorova:Rješenje:vor 10 N4 = 99,00 kN (vlak)
N13 = 0,00 kN
vor 9 N17 = 68,96 kN (vlak)N8 = - 56,11 kN (tlak)
vor 7 N7 = - 56,11 kN (tlak)N12 = - 95,36 kN (tlak)
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42 �vor)(u tlak N
�vor)(u tlak N �vora)(iz vlak N
�vor)(u tlak N �vora)(iz vlak N
8
717
124
10
97
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zrakesi3sjecištuu
kN11510511F
FRHV
D
DB3
)( )(
,
: sila Rezultanta
=⋅==++
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167 NNK +=
: pravac Culmannov
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: sila Ravnoteža
167
3D
NNK
0NKF
+=
=++
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• Ritterova metoda Presjek D-DRješenja:
�vora)(iz vlak kN ,N
�vor)(u tlak kN ,N
�vora)(iz vlak kN ,N
2895
0056
5077
16
7
3
=−=
=