ECA - Ex1

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Ex.1. Para as lajes abaixo, dimensionar e detalhar as armaduras,
quanto a flexão simples.
Dados:
fck = 30 Mpa
ϒc = 25,0 kN/m³
Aço = CA 50
c = 2,0 cm
d‘ = 3,0 cm
Rev. = 1,5 kN/m²
SC = 3,0 kN/m²
20
380
20
280
Resolução:
L1
h=12
L2
h=12
L3
h=12
20
20
2020
380
280
lx
ly
lx
ly
lx
ly
L1
Tipo 2B
L2
Tipo 3 L3
Tipo 3
Laje
lx
(cm)
ly
(cm)
λ = 
ly / lx
L1
L2
L3
400 600 1,5
300 400 1,333
300 400 1,333
μx μ'x μy μ'y
4,24 (1,35) 9,65 (1,35) 2,50 (1,30) 7,88 (1,35)
5,24 11,09 2,12 -
4,24 (1,35) 9,65 (1,35) 2,50 (1,30) 7,88 (1,35)
p = ϒc*h+Ver+SC
p = 25*0,12+1,5+3,0
p = 7,5kN/m²
1. Determinação das Cargas:
2. Determinação das Momentos:
M = μ*plx²/100 => [kN.m/m]
Mx M'x My M'y
6,288 -13,308 2,544 -
2,862 -6,514 1,687 -5,319
2,862 -6,514 1,687 -5,319
M = MK
M = [kN.m/m]
MK = Momento Característico
MK = [kN.cm/m]
Portanto = M*100 
Md = Momento de Cálculo
ϒf = Coeficientes de ponderação: 1,4
Md = ϒf.Mk
K = Coeficiente Tabelado
Ks = Tabela
3. Dimensionamento das armaduras:
Laje 1.
MKx = 628,8 => Kc = 9,201 => Ks = 0,024 => Asx = 2,347 
MK'x = -1330,8 => Kc = 4,347 => Ks = 0,025 => As’x = 5,175
MKy = 254,4 => Kc = 22,743 => Ks = 0,025 => Asy = 0,910 => ***Asy = 1,200
Kc = bw.d²/ ϒf.Mk
Asx = Ks.ϒf.Mk/d
Laje 2 = Laje 3.
MKx = 286,2 => Kc = 20,216 => Ks = 0,023 => Asx = 1,024 => ***Asx = 1,200
MK'x = -651,4 => Kc = 8,882 => Ks = 0,024 => A’sx = 2,432
MKy = 168,7 => Kc = 34,296 => Ks = 0,023 => Asy = 0,603 => ***Asy = 1,200
MK'y = -531,9 => Kc = 10,877 => Ks = 0,024 => A’sy = 1,986
*** SEGUNDO A ABNT NBR 6118:2014. 
As (AREA DE AÇO (CA-50; ϒf=1,4; ϒs=1,15))
ARMADURA MÍNIMA
POSITIVA E NEGATIVA: 10%*h
4. Detalhamento das armaduras:
Laje 1.
Asx = 2,42 ; φ = 6,3 ; e = 13
As'x = 5,25 ; φ = 8,0 ; e = 9,5
Asy = 1,58 ; φ = 6,3 ; e = 20
Laje 2 = Laje 3.
Asx = 1,58 ; φ = 6,3 ; e = 20
As'x = 2,50 ; φ = 8 ; e = 20
Asy = 1,58 ; φ = 6,3 ; e = 20
As‘y = 2,10 ; φ = 6,3 ; e = 15
20
380
20
280
L1
h=12
L2
h=12
L3
h=12
20
20
2020
380
280
Positiva Negativa
lx
L1
h=12
L2
h=12
L3
h=12
ly
lx
lx
ly
1. φ 6,3 c/ 13 
2
. φ
 6
,3
 c
/ 
2
0
 
3
. φ
 6
,3
 c
/ 
2
0
 
4. φ 6,3 c/ 20 
4. φ 6,3 c/ 20 
5. φ 6,3 c/ 20 
5. φ 6,3 c/ 20 
5
. φ
 6
,3
 c
/ 
2
0
 
6. φ 6,3 c/ 20 
6
. φ
 6
,3
 c
/ 
2
0
 
6. φ 6,3 c/ 20 
7. φ 8 c/ 20 
8
. φ
 8
,0
 c
/ 
9
,5
 
Bordas => φ = 6,3 ; e = 20
Bordas => φ = 6,3 ; e = 20
4. 30 φ 6,3 c/ 20 c = 325
Q = (280/20+1)*2 => 30
c = (12-2-2-1)+(20-2)+380+20 => 425
7418
c
cobrimento = 2,0 cm
c h
380,0cm
20,0
20,0
12,0
1. 46 φ 6,3 c/ 13 c = 426
Q = 580/13+1 => 45,615 = 46~
c = (12-2-2)+(20-2)+380+20 => 426
8 418
2. 20 φ 6,3 c/ 20 c = 630
Q = 380/20+1 => 20
c = 2*(12-2-2-1)+2(20-2)+580 => 630
7 616 7
3. 20 φ 6,3 c/ 20 c = 632
Q = 380/20+1 => 20
c = 2*(12-2-2)+2(20-2)+580 => 632
8 616 8
c
cobrimento = 2,0 cm
c h
580,0cm
20,0
20,0
1. φ6,3
c
h
c
cobrimento = 2,0 cm
c
280,0cm
20,0
20,020,0
280,0cm
c
c
cobrimento = 2,0 cm
c
380,0cm
20,0
20,0
12,0 c 3. φ6,3
h
Armadura positiva:
Armadura negativa:
5. 70 φ 6,3 c/ 20 c = 94
Q = 580/20+1+2*380/20+2 => 73
c = 2*(12-2-2)+(20-2)+0,15*400 => 94
8 78 8
h
cobrimento = 2,0 cm
c
20,0
0,15 lx
6. 70 φ 6,3 c/ 20 c = 94
Q = 580/20+1+2*380/20+2 => 70
c = 2*(12-2-2)+(20-2)+0,15*300 => 79
8 63 8
h
cobrimento = 2,0 cm
c
20,0
0,15 lx
c
20,0
12,0
7. 20 φ 8,0 c/ 20 c = 157
Q = 380/20+1 => 20
c = 1,5(0,25*300+19)+2*(12-2-2)=> 157
8 141 8
0,25*lx
c
c
lb
c = 1,5(0,25*lx+lb)+lganchos
c
20,0
12,0
8. 62 φ 8,0 c/ 9,5 c = 157
Q = 580/20+1 => 62,052 = 62
c = 1,5(0,25*400+19)+2*(12-2-2)=> 194,5 = 195
8 179 8
0,25*lx
c
c
lb
c = 1,5(0,25*lx+lb)+lganchos
~
lx = Laje 1 => 400
lx = Laje 2;3 => 300
CONSIDERAR O MAIOR
~