01 -(eks)centricno zatezanje
TRANSCRIPT
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TEORIJA BETONSKIH KONSTRUKCIJA
ESPB: 6Semestar: V
Doc. dr Ivan Ignjatović
Miodrag Stojanović
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O PREDAVAČU
1. Diplomirao: 2004, Građevinski fakultet u Beogradu2. Magistrirao: 2009, Građevinski fakultet u Beogradu3. Doktorirao: 2013, Građevinski fakultet u Beogradu4. Uloga u nastavi: asistent na K-smeru, nastavnik na
predmetima Sanacije, rekonstrukcije i održavanje (K) i Betonske konstrukcije 1 (HVE, PIZ, MTI)
5. Istraživanje: “zeleni” betoni u AB konstrukcijama, trajnost, seizmika
6. Stručni rad: projektovanje, tehničke kontrole, stručna mišljenja
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JELENA DRAGAŠ
SARADNICI
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NIKOLA TОŠIĆ
SARADNICI
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VELJKO KOKOVIĆ
SARADNICI
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1. Odrediti potrebnu površinu armature i oblikovatipoprečni presek pravougaonog oblika centričnozategnutog elementa. Element se nalazi u uslovima umereno agresivne sredine.
Podaci za proračun:Zg = 305 kN (stalno opterećenje)Zp = 337 kN (povremeno opterećenje)
GA 240/360
Centrično zatezanje
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Centrično zatezanje
Zu = γu,g×Zg + γu,p ×Zp
γu,g=? γu,p=? (79,80)
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a a' dc e
g
f
h
ZATEZANJE
b
PRITISAK
37
d4
7 d
C
εb2 εa2
εa1 εb1
2 3.5
3.523 εv 0b
h - a
2a 1
d
Aa2
Aa1
γu,i
a 2
10‰
NAPONSKA STANJA PRESEKA (84)
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2
v
ua cm61.45
0.241094.6ZA ==
σ=
Centrično zatezanje
Zu = 1.6×Zg + 1.8×Zp
Zu = 1.6×305 + 1.8×337 = 1094.6 kNGA 240/360 ⇒ σv = 240 MPa = 24 kN/cm2
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2
v
ua cm61.45
0.241094.6ZA ==
σ=
usvojeno: 15Ø20 (47.12 cm2)
Centrično zatezanje
Zu = 1.6×Zg + 1.8×Zp
Zu = 1.6×305 + 1.8×337 = 1094.6 kNGA 240/360 ⇒ σv = 240 MPa = 24 kN/cm2
Oblikovanje preseka ?
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12ZAŠTITNI SLOJ BETONA DO ARMATURE (135,136)
Ø Zašitini sloj je minimalno rastojanje od bilo koje armatureu elementu do najbliže površine betona
Ø Osnovni parametri koji određuju debljinu zaštitnog sloja su:Øvrsta elementa (ploča-zid, greda-stub);Østepen agresivnosti sredine u kojoj se element nalazi;Ømarka betona;Øprečnik i vrsta armature;Ønačin ugrađivanja betona, odnosno izvođenja
konstrukcije
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Zaštitni sloj betona do armature
Ø Slabo agresivna sredina: unutrašnjost stambenih, administrativnih i drugih prostorija, u kojima AB elementi nisu izloženi vlazi, atmosferskim i korozivnim uticajima (elementi "unutra", veoma suve sredine, relativna vlažnost 40%)
Ø Srednje agresivna sredina: unutrašnjost prostorija sa većom vlažnošću i slabijim ili povremenim korozivnim isparenjima; ABelementi u slobodnom prostoru, izloženi uobičajenim atmosferskim uticajima, kao i elementi u direktnom kontaktu sa tekućom vodom ili običnim tlom (elementi "napolju", relativnavlažnost 70%).
Ø Jako agresivna sredina: prostori u kojima su AB elementi izloženi jačim tečnim ili gasovitim uticajima (slabo kisele tečnosti, slana voda, voda sa mnogo kiseonika, korozivna tla i gasovi, atmosfera zagađena korozivnim industrijskim gasovima i vazduh u blizini mora)
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MINIMALNE DEBLJINE ZAŠTITNOG SLOJA a0 [cm]
Ø REBRASTE I OLAKŠANE MEĐUSPRATNE KONSTRUKCIJE SE TRETIRAJU KAO PLOČE
Ø ČIST ZAŠTITNI SLOJ NE SME BITI MANJI OD PREČNIKA UPOTREBLJENE ARMATURE (a0 ≥ Ø)
3.03.5jaka
2.02.5srednja
1.52.0slaba
ploče, ljuske, zidovi
grede, stuboviagresivnost sredine
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22Ø Vrednosti iz tabele se, zavisno od posebnih uslova,
koriguju i to:Ø +0.5 cm ukoliko se izvode od betona MB < 25;Ø+0.5 cm ukoliko elementi nakon betoniranja nisu
dostupni kontroli;Ø+1.0 cm ako se površina betona naknadno obrađuje
postupcima koji izazivaju oštećenja zaštitnog sloja betona;Ø+1.0 cm za konstrukcije koje se izvode klizajućom
oplatom;Ø -0.5 cm ukoliko su elementi proizvedeni u fabričkim
uslovima.
Ø Korekcija minimalnog zaštitnog sloja betona vrši se simultano
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23Ø Ukoliko je potreban zaštitni sloj betona veći od 5 cm, mora se
armirati posebno tankom armaturnom mrežom.Ø Rastojanje te armature u zaštitnom sloju od spoljne površine
betona ne može biti manje od 2 cm.Ø Takva armatura zaštitnog sloja se ne uzima pri dokazu granične
nosivosti ili napona
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24OBLIKOVANJE ARMATURE (139 – 147)
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eH ≥ 5 cm
e V ≥
3 c
m m = 5
n = 3
b ≥ 2×2.5 + 2×0.8 + 5×2.0 + (5-1)×5.0 = 36.6 cm ⇒ b = 40 cmd ≥ 2×2.5 + 2×0.8 + 3×2.0 + (3-1)×3.0 = 18.6 cm ⇒ d = 20 cm
Centrično zatezanje
b ≥ 2a0 + 2Øu + m×Ø + (m-1)×eHd ≥ 2a0 + 2Øu + n×Ø + (n-1)×eV
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5.5
40
4.5
4.5
5.5
4.5
20
4.57.5 7.52x8=16
UØ8/305Ø20
5Ø20
5Ø20
aI = a0 + Øu + Ø/2 = 2.5 + 0.8 + 2.0/2 = 4.3 cm ⇒ usv. aI = 4.5 cm
Centrično zatezanje
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31Ekscentrično zatezanje - proračunski model
Zau1
MuZu
y b2
εa1 = 10‰
y b1
b
Aa1
εa2
Gb
"1"
"2"
Aa2Zau2
y a1
y a2
a 2a 1
e
d
2a1a
2a
v
u1a yy
eyZA++
×σ
=2a1a
1a
v
u2a yy
eyZA+−
×σ
=
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a a' dc e
g
f
h
ZATEZANJE
b
PRITISAK
37
d4
7 d
C
εb2 εa2
εa1 εb1
2 3.5
3.523 εv 0b
h - a
2a 1
d
Aa2
Aa1
γu,i
a 2
10‰
NAPONSKA STANJA PRESEKA (84)
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2. Dimenzionisati pravougaoni poprečni presek zadatihdimenzija, opterećen silom zatezanja i momentomsavijanja u fazi malog ekscentriciteta.
Podaci za proračun:Zg = 305 kN Mg = 6.6 kNm b = 30 cmZp = 337 kN d = 25 cmGA 240/360
GA 240/360 ⇒ σv = 240 MPa = 24 kN/cm2
Ekscentrično zatezanje
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Ekscentrično zatezanjeZu = γu,g×Zg + γu,p ×Zp
Mu= γu,g×Mg
γu,g=? γu,p=? (79,80)
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Ekscentrično zatezanje
cm96.01094.6
1010.56ZMe
2
u
u =×
==
pretp. a1 = a2 = 4.5 cmya1 = ya2 = d/2 – a1 = 25/2 – 4.5 = 8.0 cm
2
v
u2a1a cm61.45
24.01094.6ZAA ==
σ=+
Zu = 1.6×Zg + 1.8×Zp = 1.6×305 + 1.8×337 = 1094.6 kNMu = 1.6×Mg = 1.6×6.6 = 10.56 kNm
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≈ 6Ø25 ; a1 = 6.33 cmya1 = 25/2 – 6.33 = 6.17 cm
≈ 4Ø25 ; a2 = 4.5 cmya2 = 25/2 – 4.5 = 8.0 cm
2
2a1a
2a
v
u1a cm55.25
0.80.896.00.861.45
yyeyZA =
++
×=++
×σ
=
2
2a1a
1a
v
u2a cm05.20
0.80.896.00.861.54
yyeyZA =
+−
×=+−
×σ
=
Ekscentrično zatezanje
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2
2a1a
2a
v
u1a cm85.28
0.817.696.00.861.45
yyeyZA =
++
×=++
×σ
=
2
2a1a
1a
v
u2a cm75.16
0.817.696.017.661.54
yyeyZA =
+−
×=+−
×σ
=
usv. 6Ø25 (29.45 cm2) ; a1 = 6.33 cm
usv. 4Ø25 (19.63 cm2) ; a2 = 4.5 cm
Ekscentrično zatezanje
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30
3x7=21
25
4.5
5.5
4.5
4.5
4.5
10.5
UØ8/30
4Ø25
4Ø25
2Ø25
aI = a0 + Øu + Ø/2aI = 2.5 + 0.8 + 2.5/2 = 4.55 cmusv. aI = 4.5 cm
aII = aI + eV + 2×Ø/2aII = 4.5 + 3.0 + 2×2.5/2usv. aII = 10 cm
a1 = (4×4.5 + 2×10)/6a1 = 6.33 cm
Ekscentrično zatezanje
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39 RASPOREĐIVANJE ARMATURE U PRESECIMA ELEMENATA (137,138)
( )
×≥
MAX
EKVVH
D8.0ØØcm3
e,e
e V
a0 Øu Ø
aI
aII
eH
a 0Ø
uØ
Ø
Ø
aI
DMAX - PREČNIK MAKSIMALNOG ZRNA UPOTREBLJENOG AGREGATA
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42GRUPISANJE ARMATURE
Ø Ø=22
ØE
KV.=
44
ØEKV.=39.6
Ø=2
8Ø
ØEK
V.=
43.3
Ø=25
Ø Ø
ØE
KV.
=39.
6
Ø=28Ø
nØØØ
4Ø4Ø
i
2iEKV
i
2i
EKV
==
π××
π=
∑
∑
mm44ØEKV ≤
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