Page 234 - Dasar Perencanaan Struktur beton bertulang OK 1_Neat

D10 – 100 dengan luas tulangan terpasang,

1000
As terpasang = x As D 10
S terpasang
1000

As terpasang = x 1/4 x x 10 2 = 785 mm 2
100

 Tulangan daerah lapangan arah X


Mu = ρ.0,8.fy 1-0,588 fy 
bd x 2   f'c  

12,198x10 6  240 

2 = ρ.0,8.240 1-0,588 
1000x165  25 
2
0,45 = 192ρ-1083,8ρ
ρ = 0,0024
1.0 1.0
ρ = = = 0,0042  ρ < ρ
min fy 240 min

ρ = ρ = 0,0042
min
Luas tulangan daerah lapangan arah X,
2
As = ρ.b.d = 0,0042x1000x165 = 687,5 mm
Lx
x
digunakan D10 mm, jumlah tulangan =
As 687,5
n = tx = = 8,76

As diameter 10 1/ 4. .10 2
jarak tulangan perlu =

221