PRVI PROGRAM-1.pdf
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GRAĐEVNA STATIKA 2 ŠKARE LUKA 0178034980 1 ܧൌ 3 כ10 / ଶ / ൌ 50/50 NEPOZNANICE: u 23 , ϕ 2 l 12 = 3,30 m l 23 = 3,25 m l 24 = 3,30 m l ଷହ ൌ ඥ4,2 ଶ 1,65 ଶ = 4,512482687 m I ൌ ୠכ୦ య ଵଶ ൌ ,ହכ,ହ య ଵଶ ൌ ,ଶହ ଵଶ = 5,208333 כ10 ଷ m 4 EI = ܫܧൌ3 כ10 כ5,208333 כ10 ଷ = 156 250 kNm 2 k 12 = k 24 = EI / l 12 = 47 348,485 kNm k 23 = EI / l 23 = 48 076,923 kNm k 35 = EI / l 35 = 34 626,172 kNm¸
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Transcript of PRVI PROGRAM-1.pdf
GRAĐEVNA STATIKA 2 ŠKARE LUKA 0178034980
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3 10 / / 50/50
NEPOZNANICE: u23 , ϕ2 l12 = 3,30 m l23 = 3,25 m l24 = 3,30 m l 4,2 1,65 = 4,512482687 m
I , , , = 5,208333 10 m4 EI = 3 10 5,208333 10 = 156 250 kNm2 k12 = k24 = EI / l12 = 47 348,485 kNm k23 = EI / l23 = 48 076,923 kNm k35 = EI / l35 = 34 626,172 kNm¸
GRAĐEVNA STATIKA 2 ŠKARE LUKA 0178034980
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PLAN POMAKA
ΨIII = Ψ23 = ,
, , =
, = 0,12087912
ΨI = Ψ12 = ,
= 0,303030303
ΨII = Ψ24 = ,
= 0,303030303
ΨIV = Ψ35 = ,
= 0,238095238
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PRISILNI POMAK
Ψ24 = ∆
, = ,
, = 1,515151515 10
OSNOVNI SISTEM¸
GRAĐEVNA STATIKA 2 ŠKARE LUKA 0178034980
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I M12 = , 32,175
M21 = , 32,175
M12= 2 ϕ 6 Ψ + M12
M21= 4 ϕ 6 Ψ + M21
II Mc24 (Μ) = 33,5
Mc24 (∆h)= 3 Ψ
Mc
24 = Mc24 (Μ) + Mc
24 (∆h)
M24= 3 ϕ 3 Ψ + Mc24
M42 = 0
III M23 = 0
M23 = 3 ϕ 3 Ψ
M32 = 0
IV tg γ = ,,
γ ≈ 21,44773632°
F = 149 cos γ 138,6819725
Mc53 = – 3 = – 117,3375 kNm
M35= 0
M53= 3 Ψ + Mc53
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JEDNADŽBA RAVNOTEŽE ČVORA 2 ∑ 0 - M21 – M24 – M23 = 0 _______________________ M21 + M24 + M23 = 0 4 k12 ϕ2 - 6 k12 ψ12 + 32,175 + 3 k24 ϕ2 - 3 k24 ψ24 + (-33,5 - 3 k24 ψ24 + 3 k23 ϕ2 - 3 k23 ψ23 = 0 4 47 348,485 ϕ 6 47 348,485 0,303030303 32,175 3 47 348,485 ϕ
3 47 348,485 0,303030303 33,5 3 47 348,485 1,515151515 10 3 48 076,923 ϕ 3 48 076,923 0,12087912 0
189 393,94 ϕ 86 088,15454 32,175 142 045,455 ϕ 43 044,07727
33,5 215,2203863 144 230,769 ϕ 17 434,48843 0
475 670,164 ϕ 60 478,56614 216,5453863
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VIRTUALNI RAD δ W = 0
ψ 1,65 ψ 2,1 ψ ψ ψ
ψ ψ ψ 0 78 1,65
3,3149 2,1
4,221
3,323
8,27272727324
3,353
4,2
12, ,
0
, , , , , 15,1969697 3,3
0,39890198 0,785714285 50,15 4 k12 ϕ2 - 6 k12 ψ12 + 32,175 + 0,39890198 (3 k23 ϕ2 - 3 k23 ψ23) - 3 k24 ϕ2 + 3 k24 ψ24 - Mc
24 +
- 0,785714285 ( - 3 k35 ψ35 + Mc53 ) + 2 k12 ϕ2 - 6 k12 ψ12 - 32,175 = 50,15
4 47 348,485 ϕ2 6 47 348,485 23
3,332,175 57 533,93933 ϕ2
57 533,93933 0,12087912 3 47 348,485 ϕ 3 47 348,4853,3
33,5 3 47 348,4850,005
3,3 2,357142855 34 626,172 0,238095238
92,19374992 94 696,97 ϕ2 284 090,91 23
3,382,325
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189 393,94 ϕ2 86 088,15455 23 32,175 57 533,93933 ϕ2 6954,651956 23
142 045,455 ϕ 43 044,07727 33,5 215,2203864 19 433,05569
92,19374992 94 696,97 ϕ 86 088,15455 82,325
199 579,3943 ϕ 241 608,094 290,7641363 SUSTAV JEDNADŽBI : 475 670,164 ϕ 60 478,56614 216,5453863 (1) 199 579,3943 ϕ 241 608,094 290,7641363 (2)
iz (1) : ϕ , ,
,
ϕ uvrstimo u (2) :
199 579,3943216,5453863 60 478,56614
475 670,164241 608,094 290,7641363
290,7641363 199 579,3943 216,5453863475 670,164
199 579,3943 60 478,56614475 670,164 241 608,094
381,6212056216 232,7939
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KONAČNA RJEŠENJA NEPOZNANICA
0,001764862761
ϕ 0,0006796342928 ODSTUPANJE OD RJEŠENJA PO DiM-u : DiM 0,00176626 relativna pogreška = 0,079 % ϕ 0,000681179 relativna pogreška = 0,227 %
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MOMENTI U ČVOROVIMA : M12 = 2 ϕ 6 Ψ + M12 =
= 2 47 348,485 0,0006796342928 6 47 348,485 ,,
32,175
= -119,75 kNm
M21 = 4 ϕ 6 Ψ + M21 =
= 4 47 348,485 0,0006796342928 6 47 348,485 0,0017648627613,3 32,175
= 8,96 kNm M24 = 3 ϕ 3 Ψ + Mc
24 =
= 3 47 348,485 0,0006796342928 3 47 348,485 0,0017648627613,3
33,5 3 47 348,485 1,515151515 10 = = -76,21 kNm M42 = 0
M23 = 3 ϕ 3 Ψ =
= 3 48 076,923 0,0006796342928 3 48 076,923 0,0017648627618,272727273
= 67,26 kNm M32 = 0 M35 = 0 M53= 3 Ψ + Mc
53
3 34 626,172 ,,
117,3375
73,69
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PRILOG: PRORAČUN U DiM-u
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M - dijagram
T - dijagram
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N - dijagram
Pomaci
