EXERCíCIO SAPATA

Exercício

Dados: ap ≔ 30 cm ϕest ≔ 30

bp ≔ 20 cm
Pest ≔ 400 kN

cm

Aspilar ≔ 31.50

Concreto 30Mpa fck ≔ 3

fyk ≔ 50

fck fcd ≔ ――
= 2.14
1.4

cm²

N ≔ 717 kN

Aço CA-50

fyk
= 32.94 σyd ≔ ―――――
1.1 ⋅ 1.2 ⋅ 1.15

cob ≔ 4 cm

e ≔ 2.5 ⋅ ϕest

Geometria: a ≔ (2 ⋅ 15) + ϕest + ⎛⎝2.5 ⋅ ϕest⎞⎠

a = 135

b ≔ 15 + ϕest + 15

b = 60

cm cm Altura H:
45 ° < α < 55 ° αmín ≔ 45 °

αmáx ≔ 55 °

⎛ e ap ⎞ dmín ≔ ⎛⎝tan ⎛⎝αmín⎞⎠⎞⎠ ⋅ ⎜―− ―⎟
4⎠
⎝2

dmín = 30

⎛ e ap ⎞ dmáx ≔ ⎛⎝tan ⎛⎝αmáx⎞⎠⎞⎠ ⋅ ⎜―− ―⎟
4⎠
⎝2

dmáx = 42.84

Adota-se d=40 cm, pois esta no intervalo admitido d ≔ 40 h≔d+5 cm h = 45

d tanα ≔ ――― e ap
―− ―
2
4

ifi

cm tanα = 1.33

α ≔ 53.06 °

Verificação:

no pilar:

na estaca:

concreto:

1.4 ⋅ N σpilar ≔ ――――――
2
ap ⋅ bp ⋅ (sin (α))

σpilar = 2.62

1.4 ⋅ N σestaca ≔ ―――――――
⎛π ⋅ ϕ 2 ⎞
2
⎝ est ⎠
2 ⋅ ――――
⋅ sin (α)
4

kN/cm²

σestaca = 1.11 kN/cm²

Kr ≔ 0.95 σcd.lim ≔ 1.4 ⋅ Kr ⋅ fcd

σcd.lim = 2.85 kN/cm²

Armadura Principal:
1.4 ⋅ N ⎛⎝2 ⋅ e − ap⎞⎠
⋅ ――――
Rs ≔ ―――
8
d

Rs = 376.43

kN

Rs
As ≔ ―― σyd As = 11.43

cm²

larm ≔ ϕest

larm = 30

ϕprinc ≔ 20.0

ϕprinc
1
A1ϕ ≔ π ⋅ ―――
⋅ ――
= 3.14
100
4

cm
2

mm

As n ≔ ――
A1ϕ

n = 3.64

espaçamento:

smín ≔ 8 cm larm s ≔ ――― nadot − 1

nadot ≔ 4

barras

smáx ≔ 20

s = 10

cm²

cm

cm

Comprimento da armadura principal: cprincipal ≔ a − 2 ⋅ cob

cprincipal = 127 cm

ϕprinc ganchoprincipal ≔ 12 ⋅ ――
10

ganchoprincipal = 24

cm

Armadura Transversal ao Tirante:
1
Ast ≔ ―⋅ As
5

Ast = 2.29 cm²

ϕst ≔ 10.0 mm

Ast ntransversal ≔ ――――
= 2.91
2
⎛ ϕst ⎞
⎜――
⎟
⎝ 10 ⎠ π ⋅ ―――
4

nt ≔ 3

barras

ϕprinc st ≔ ――= 10 nt − 1

cm

ctransversal ≔ b − 2 ⋅ cob = 52cm

Armadura Superior do Bloco (estaca até 50tf) : ϕsuperior ≔ 10 mm

s ≔ 20

csuperior ≔ cprincipal = 127

cm

ϕsuperior ganchosuperior ≔ 12 ⋅ ―――
= 12 cm
10

cm

Estribos Verticais (estaca até 50tf) : ϕestribo ≔ 6.3

smáx.estribo ≔ 15

mm

cm

ldisp.estribo ≔ a − 2 ⋅ cob = 127 cm nestribos ≔ 9

129 sestribo ≔ ―――
= 14.33 cm nestribos estribos

Armadura de Pele (estaca até