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ANALISA DAN DESAIN BALOK UJI
Gambar Balok dan Penampang
Data-data teknis
Mutu baja
tulangan utama 10mm ; fy = 400 MPa.
sengkang 6mm ; fy = 400 MPa
Mutu beton ; fc = 30 MPa
Berat balok uj ; !bs = 0.40 " 0.13 " #400 = 1#$ kg%m
!asio pen"langan
b = 13 cm
& = 40 cm
' = selmut beton ( 6 ( 10 ( ) *3+
= 4 cm ( 0,6 cm ( 1,0 cm ( 1,$ cm
= -,1 cm
' = ngg /fektf Balok
' = 40-,1
= 3#, 3# cm
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2aso enulangan lentur
= 5s%b.' =3#0130
10.4%14 #
x
xx = 0.00-$$ atau 0.-$$
7ontrol ter&a'a syarat raso enulangan
mn =fy
4.1=
400
4.1= 0.003$
ma" = 0.-$ balance
= 0.-$
+ fy
xfy
xfcx
600
600819$.0 1
= 0.-$
+400600600
400
309$.09$.0x
xx
ma"= 0.0#439
kontrol mn : 0.00--$ : ma" ..............ok
2aso enulangan bag
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?c = dbMu
dVucf w ..
-
1..1#08
+
= ( ) 3
10
1.3#0.130.
-
1.100--$.01#030 xx+
= 39,0-- k> * 3,909 ton+
Balok 'engan sengkang @ertkal
?s =S
dfyAv ..
tonkNmm
mmxMPaxmmx
#3$,-3$,-#10
1
100
3#0+400*+6*4
1#
3
#
==
=
Beban #aksim"m $Pmaks% berdasarkan geser "ltimit
Aua ttk beban 'engan rato a%' = 1,0
?u = ) P ( ) !bs.
Maka
3,909 = ) P ( ) *0,1#$+ *1,0+ 'erole& Pcrack = -,41 ton
*3,909 ( -,#3$+ = ) P ( ) *0,1#$+ *1,0+ 'erole& Pult = &'()*'ton *sgk @ertkal+
Beban #aksim"m $Pmaks% berdasarkan momen "ltimit
Aua ttk beban 'engan rato a%' = 1,0
Mu = ) P.a ( 1%9 !bs #
'mana a = ' = 0,3# m
Mu = Mn
3,-9 = ) P. 0,3# ( 1%9 *0,1#$+ *1.0#+
Pult = &+(,& ton
Aua ttk beban 'engan rato a%' = 0,-$
'mana a = 0,#4 m
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Mu = ) P.a ( 1%9 !bs #
3,-9 = ) P. 0,#4 ( 1%9 *0,1#$+ *1.0#+
Pult = +'(+. ton
Aua ttk beban 'engan rato a%' = 0,$
'mana a = 0,16 mMu = ) P.a ( 1%9 !bs #
3,-9 = ) P. 0,16 ( 1%9 *0,1#$+ *1.0#+
Pult = /(.,, ton