4. Zadatak
Za prikazanu dispoziciju krova dimenzionisati veza č usvajajući da su gornji pojas, donji pojas, dijagonale i vertikale, svaka grupa štapova posebno, konstantnog preseka. Veze u čvoru ostvariti zavarivanjem a čvor usvojiti i dokazati na najoptere ćenijem mestu. Popre čne preseke štapova usvojiti kako je zadato. B
q= a= h= B=
Vezač B Vezač B e
č
a j n ž o r
e
č
2,0 2,7 2,0 9,0
kN/m2 m m m
pojasevi:
e
č
a j a j n n ž ž o o r r
B
ispuna:
B
Osnovni materijal: Č0361 Slučaj opterećenja: h
8xa
Radionički crtež prva dva polja dati u razmeri: 1:10
Analiza opterećenja na rožnjaču
λ = 2,7m
q = 2,0kN/m2x2,7m = 5,4kN/m q
9,0m R V R V = 5,4kN/m x 9,0m/2 = 24,3kN
R V
I
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Određivanje merodavnih uticaja u veza ču
2xR R 2xR R /2 a αD
R v Dmax a
2xR R /2 b
Оmin
Vmin
R Umax
Dmin b
Umax
R v
2,0 m
R Omin
8x2, 7 = 21, 6 m
Reakcije veza ča: R V = 2x24,3x(7+2*1/2)/2 2x24,3x(7+2*1/2)/2 = 194,4kN Moment oko Rihter-ovog čvora za merodavni štap donjeg pojasa (polje do sredine raspona) MR,Umax = (194,4-24,3)x4x2,7-48,6x(2,7 (194,4-24,3)x4x2,7-48,6x(2,7+5,4+8,1)=1049, +5,4+8,1)=1049,8kN 8kN Moment oko Rihter-ovog čvora za merodavni štap gornjeg pojasa (polje do sredine raspona) MR,Omin = (194,4-24,3)x3x2,7-48,6x(2,7 (194,4-24,3)x3x2,7-48,6x(2,7+5,4)=984,1kN +5,4)=984,1kN Transferzalna sila u prvom polju do oslonca (za merodavnu silu zatezanja u dijagonalama) Ta-a = 194,4-24,3 = 170,1kN Transferzalna sila u drugom polju do oslonca (za merodavnu silu pritiska u dijagonalama) T b-b = 194,4-24,3-48,6 = 121,5kN Merodavna sila u donjem pojasu Omin = -MR,Omin/h = -984,15/2,0 = 492,0kN Merodavna sila u gornjem pojasu Umax = MR,Umax/h = 1049,8/2,0 = 524,9kN Merodavne sile u dijagonalama Dmax = Ta-a / sinα = 170,1/sin36,5 = 286,0kN Dmin = -T b-b / sinα = -524,9/sin36,5 = -204,3kN - 204,3kN Merodavna sila u vertikali Vmin = -2xR R R = = -48,8kN * (-) pritisnut štap Osnovni materijal: Č0361 σdop = 16,0 kN/cm 2
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Dimenzionisanje
≈ Umax / σdop = 524,9 / 16,0 = 32,8 cm 2 2 potr AO ≈ Omin / σi,dop = 492,0 / 14,5 = 33,9 cm ⎧⎪ Dmax / σ dop = 286,0 / 16,0 = 17,9cm 2 potr A D ≈ ⎨ 2 ⎪⎩ Dmin / σ i , dop = −204,3 / 12,0 = 17,0cm potr AU
potr AV
≈ Vmin / σi,dop = -48,6 / 8,0 = 6,1 cm 2
Dopušteni napon izvijanja - σi,dop u prethodnim izrazima mora se pretpostaviti. Ovde, u principu, nema nekih naro čitih pravila pošto problem nije jednozna čno određen pa se primenjuju iskustveni podaci (za neku prvu iteraciju može se uzeti da je σi,dop ≈ 0,75σdop što je u ovom primeru ura đeno u slučaju dijagonale, za pojas je pretpostavljeno σi,dop ≈ 0,90σdop a za vertikalu σi,dop ≈ 0,50σdop) pretpostavljaju se slede ći profili: gornji pojas donji pojas dijagonale vertikale
180x180x5 – 160x160x5,6– 120x80x5 100x40x2,5 -
Č0361 Č0361 Č0361 Č0361
*- Prilikom izbora profila rukovoditi se slede ćim pravilima:
• •
Širina gornjeg i donjeg pojasa treba da je ista ili približno ista Ako su kutije pravougaone onda je njihova “Prirodna” orijentacija horizontalna – kako za pojaseve tako i za ispunu (ovo proizilazi iz zahteva da krutost štapova u ravni rešetke bude što manja kao i iz činjenice da je dužina izvijanja u ravni rešetke ista ili manja od dužine izvijanja van ravni rešetke). • Prilikom izbora profila za ispunu voditi ra čuna da “horizontalna” dimenzija profila - b i (dimenzija upravna na ravan rešetke) bude u okvirina 0,4b max ≤ bi ≤ bmin gde je sa “b max” odnosno “b min” označena minimalna odnosno maksimalna širina profila pojasnih štapova (ovo proizilazi iz zahteva da je mogu će fizičko naleganje kao i iz uslova da ne do đe do lokalmog deformisanja zida kutije na mestima spajanja) POS:U (donji pojas)
y 160x160x5,6
x
x A=34,2 cm2 ix = iy = 6,3cm y
Kontrola napona:
U
524,9
15 4 kN/cm2
=16,0 kN/cm2
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POS: O (gornji pojas)
y 180x180x5
x
x A=34,7 cm2 ix = iy = 7,1cm y
λ x ( y ) =
l x ( y ) i x ( y )
=
270,0 = 38,0 7,1
λ x ( y ) =
λ x ( y ) λ v
=
(
49,9 = 0,409 92,9
)
β = 1 + α ⋅ λ x − 0,2 + λ x
kriva izvijanja C α=0,489
κ =
2 2 ⎞ ⎛ 2 ⎜ β + β − 4λ x ⎟ ⎝ ⎠
2
= 0,892
Kontrola napona: O 492,0 σ = min = = 13,2 kN/cm 2 <σ i ,dop= κ ⋅σ dop= 0,892 ⋅16,0 = 14,3 kN/cm 2 A 34,7
POS: D (dijagonale)
y 120x80x5
x
x A=18,3 cm2 ix = 3,2 cm iy = 4,4 cm y
li , x = 0,8l = 0,8
270,0 2 + 200,0 2 = 268,8cm
270,0 2 + 200,0 2 = 336,0cm ⎧λ x = li , x / i x = 268,8 / 3,2 = 84,0 λ = max ⎨ ⎩λ y = li , y / i y = 336,0 / 4,4 = 76,4 li , y = l =
λ = λ / λ v = 84,0 / 92,9 = 0,904
kriva izvijanja C α=0,489
(
κ =
Kontrola napona:
)
β = 1 + α ⋅ λ − 0,2 + λ
2 ⎛ β β 2 λ 2 ⎞ −4 ⎜ + ⎟ ⎝ ⎠
2
= 0,892
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POS: V ( vertikala )
y 100x40x2,5
x
x A=6,6 cm2 ix = 1,7 cm iy = 3,4 cm y
li , x = 0,8l = 0,8 x 200 = 160,0cm
li , y = l = 200,0cm
⎧λ x = li , x / i x = 160,0 / 1,7 = 94,1 λ = max ⎨ ⎩λ y = li , y / i y = 200,0 / 3,4 = 58,8 λ = λ / λ v = 94,1 / 92,9 = 1,010
kriva izvijanja C α=0,489
(
)
β = 1 + α ⋅ λ − 0,2 + λ κ =
2 2 ⎞ ⎛ 2 ⎜ β + β − 4λ ⎟ ⎝ ⎠
2
= 0,534
Kontrola napona: V 48,6 σ = min = = 7,4 kN/cm 2 <σ i ,dop= κσ dop= 0,534 ⋅16,0 = 8,5 kN/cm 2 A 6,6
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