Page 69 - Dasar Perencanaan Struktur beton bertulang OK 1_Neat

Lihat diagram strain
c ε
b = cu = 0.003
d ε cu +ε y 0.003+f / E s
y
Kalikan dengan Es = 2 x 10 Mpa
5
600
c = 600+f .d
b
y
a =  .c  a =  .c
1 b 1 b

c  Garis netral kondisi balance
b

Kesetimbangan Gaya
 H = 0

Tsb = Cc
As .Fy = 0,85.fc'.a .b
b b

0,85.fc'.a .b 0,85.fc'. .c .b
As = b = 1 b
b Fy Fy
As
ρ = b
b b.d


0,85.f'c. .c .b 0,85.f'c. .c
ρ = 1 b = 1 b ;
b f .b.d f .d
y
y
600
c = .d
b 600+f y
 .0,85.f'c 600
ρ = 1 .
b f y 600+f y

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