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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