Page 148 - Dasar Perencanaan Struktur beton bertulang OK 1_Neat

 H = 0

Tf = Cf
Asf.fy = 0,85.fc' b -b .hf
e w
0,85.fc'. b -b .hf
e w
Asf =
fy

0,85x20x 500-250 x125 2
Asf = = 1330mm
400
Sehingga ,
 hf   hf 


Mnf = Cf. d-  atau Tf. d- 
 2   2 
 hf   hf 
Mnf = 0,85.fc' b -b .hf. d-  atau Asf.fy. d- 


e w
 2   2 
 125 
Mnf = 1330x400x 610-  = 290KNm.

 2 
Balok badan
As = As total -As f
w
2
As = 3000-1330 = 1670mm
w
 H = 0
C = T w
w
0,85.fc'.b .a = As .fy
w
w
As .fy 1670x400
w
a = = =157mm
0,85.fc'.b w 0,85x20x250
Sehingga,

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