Ref : SNI - 03 - 2847 - 2002
HUBUNGAN BALOK KOLOM
H.Kartono Hd
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Ref : SNI - 03 - 2847 - 2002
Hubungan Balok Kolom
Mu
kolom
T2 = As. ( 1,25 . f y )
Vh T2
C 1= T1 Mpr balok kiri
Mpr
T1
balok kanan
C 2 = T2
n n g a u b h u g a n o n a a o m T u l k k o l o o l b a
Vh
Mu
kolom
HUBUNGAN BALOK KOLOM ( BEAM BEAM COLUMN JOINT )
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Ref : SNI - 03 - 2847 - 2002
Hubungan Hubungan Balok Kolom
Langkah desain tulangan hubungan balok balok - k olom 1. Ketahui Ketahui :
a. Panja Panjang ng kolom s erta dime nsi kolom b. Dime Dime nsi balok ser ta tulangannya tulangannya.. c. Mutu baja serta mutu beton berikut selimutnya
2 . Tentuk Tentuk an daerah pengekang kolom
o
berik ut penulanganny penulangannya a ( Ash )
3 . Tentukan luas efek tip hubungan hubungan balok - kolom ( A ) j 4. Hitung besar Mpr balok pr balok 5. Hitung M u kolom 6. Hitung besar V joint = ( V j ) 7. Hitung besar V n Dimens i hubungan hubungan balok - kolom me me nuhi syarat syarat untuk dipakai, dipakai, bila : Bes ar V n
>
V j
Dimana besar Vn ditentukan se bagai bagai berik ut ( pasal 23.5 23.5.3 .3 . 1 ) Untuk berat beton normal : ; terkekang pada keempat sisinya
Vn = 1,70 (
f' c ) . A j
Vn = 1,25 (
atau dua sisi yang ber lawanan f' c ) . A j ; terke kang pada ketiga atau
Vn = 1,0 (
f' c ) . A j
; terkekang pada hubungan lainnya
Untuk be rat be ton ringan : ( Pasal 25.5. 3 . 2 )
Vn =
3 .1,70 ( 4
f' c ) . A j
Vn =
3 .1,25 ( 4
f'
Vn =
3 .1,0 ( 4
c
; terkekang pada keempat sisinya
) . A j
f' c ) . A j
; terke kang pada ketiga atau dua sisi yang ber lawanan ; terkekang pada hubungan lainnya
8. Rencanakan tulangan hubungan balok - kolom se bes ar 0,50 0,50 Ash 9. Gamb Gamb ar tulangan hubungan balok - kolom
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Ref : SNI - 03 - 2847 - 2002
Hubungan Balok Kolom
Mu kolom
: dimensi kolom yang terbesar o
>
:
Vh
1 tinggi bers ih kolom 6
: 500 mm
Mpr balok kiri
Mpr balok kanan
a n u n g b u n h n g a l o a a o m l T u k k o o o b a l
Vh o
: dimensi kolom yang terbesar
>
:
1 tinggi bers ih kolom 6
: 500 mm
Mu kolom
o m k o l o g n e k a e n g p n a h o a a n g e r a T u l a a d a m d a l a Persyaratan tulangan tulangan transve rsal di daer ah o : ( Pasal 23.4.4.1.b ) 0,30 Ash > 0,09 Note :
s.hc.f'c
Ag
f yh
Ach
-1
s.hc.f'c f yh
Persyaratan tulangan tulangan longitudinal longitudinal pada pada joint joint kolom - balok untuk s truk tur tahan gem pa
2
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Ref : SNI - 03 - 2847 - 2002
Hubungan Balok Kolom
3,50 m 3,50 m 3,50 m 3,50 m 3,50 m 4,00 m
6m
6m
6m
6m
Portal II II 3,50 m 3,50 m 3,50 m 3,50 m 9
10 3,50 m
5
6
7
8 4,00 m
1
2
3
4
Portal B - B
A
B
C
D
E
IV
IV 6m
III
III 6m
II
II 6m
I
I A
B
C
D
E
3
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Ref : SNI SNI - 03 - 284 7 - 2002
Hubungan Hubungan Balok Kolom
Dime nsi Kolom 600 x 600 mm
Data :
4
2
Dimensi Balok 300 x 650 mm 2 Tulangan balok dimuka joint 5 , As = 8 D 19 , A' s = 6 D 19 Tulangan balok dimuka joint 6 , As = 8 D 19 , A' s = 4 D 19 fy = 400 Mpa ; f'c = 30 Mpa Selimut beton = 40 mm Design the confinem confinem ent reinforcem ent ( Perencananaan pengekangan kolom ) - Ash Untuk mencegah terjadinya buckling ( tekuk) , maka pada kolom harus dipasang tulangan pengekang sepanjang jarak o dari muka hubungn balok - kolom dimana besar : > dime dime nsi kolom (h penampang kolom) = 600 mm
o
> 1/6 tinggi bers ih kolom = 1/6 .L u = 1/6.3,6 1/6.3,675 75 = 0,6125 0,6125 m = 620 620 mm untuk k olom baw ah = 1/6. 2,85 = 0,475 m = 480 mm untuk untu k kolom kol om atas > 500 mm o
o
Dengan dem ikian panjang panjang
= 620 mm untuk kolom atas dan o = 500 500 mm untuk kolom bawah dengan jarak tulangan penge kangan = s , dimana s < 1/4 dimensi kolom ( h penampang kolom) = 1/4 . 600 = 150 mm s < 6 x diamete r tulangan longitudinal kolom = 6x 19 19 = 114 114 mm s > 100 mm .......D .......Dicoba icoba dipakai jarak tulangan se ngkang kolom D 12 dengan jarak = s = 100 mm Persyaratan tulangan transversal di daerah Ash > 0,30 Ash >
0,09
s.hc.f'c
Ag
f yh
Ach
o
: ( Pasal 23.4.4.1. 23.4.4.1. b )
-1
s.hc.f'c f yh
Dipakai penge kang : D 12 mm Ach = Luas penampang inti inti kolom dengan dimens i sisi ter luar luar s engkang = ( 600 - 2.40 ) = 520 mm hc = Dimensi inti kolom yang dihitung dari pusat ke pusat sengkang = 600 - ( 40 + 1/2.12) 1/2.12) 2 = 508 m m
Ash = 0,30
Ash > 0,09
100. 508.30 400
100. 508.30 400
( 600 . 600 )
-1
= 378 mm 2
sengkang D 12 - 100
pengekang kolom D 12 - 100
( 520 . 520 )
= 342mm 2 Dime nsi Kolom 600 x 600 mm 2
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Ref : SNI - 03 - 2847 - 2002
Hubungan Balok Kolom
Catatan : Menurut Note On A.C.I 318 - 99 , halaman 31-59 , Bab 31 - Tulangan pelat didalam didalam lebar effe ktip balok,tidak balok,tidak term asuk /tidak diperhitung- kan dalam me nghitung besar MPR - Tulangan pelat didalam didalam lebar effek tip balok, balok, tidak tidak term asuk/tidak diperhitung-k an dalam menghitung besar Mn ini kare na : 1. M PR (-) yang ter masuk tulangan pelat didalam didalam lebar e ffek tip balok , akan hasilnya yang lebih
bes ar dari pada M PR (-) yang tidak termasuk tulangan pelat, dimana hal ini akan berpengaruh pada pene ntuan kuat ges er yang diperluk an balok se rta pada pene ntuan tulangan Hubungan Balok - Kolom 2. Lazimnya , bila ke na gem pa rencana,tulangan pelat yang yang ada dalam dalam lebar effek tip balok, akan ter lepas dari baloknya,
Mene ntukan Luas efe ktip hubungan balok balok - kolom : ( Pasal : 23 23 . 5 . 3 .1 .1 ) Lebar efek tip hubungan hubungan balok balok - kolom : b balok + h kolom = 300 + 600 = 900 mm nilai ter ke cil dar i : b balok + 2 . x = b kolom + 2 . ( setengah selisih lebar lebar kolom terhadap lebar balok balok ) = 300 + 2 . 0,5 ( 600 - 300 ) = 600 mm Jadi ,lebar efe ktip hubungan balok balok - kolom = 600 600 mm Tinggi hubungan balok - kolom ,adalah tinggi total penampang kolom = 600 mm A j = Luas efek tip hubungan balok - kolom : 600 x 600 mm 2 = 360000 mm 2 Menghitung M PR balok (Mome nt Probable Probable ) , dimuk a joint 5 59,50 mm
8 D 19
118,50 mm
1500 mm 2 D12 650 mm 59,50 mm
Sengkang
10 mm
6 D 19 118,50 mm
120 mm
5
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Ref : SNI - 03 - 2847 - 2002
Hubungan Balok Kolom
Mu
= kolom
Mpr balok 2
=
551 2
6
= 275,50 kNm
Vh = 169 kN
T2 = AS . ( 1,25 . fy ) T
AS
A'S
C2 = T2
V horizontal =
Mu kolom=
Mpr balok 2
Mpr = 551 kNm
275,50 x 2 1 ( 2,85 + 3,675 ) 2
=
551 2
= 169 kN
= 275,50 kNm
HUBUNGAN HUBUNG AN BALOK - KOLOM EKSTERIOR ( LUAR ) T = As . 1,25 . f y = 8 . D 19 . 1,25 . 400 = 8 . 283 . 1,25 . 400 = 1132000 N = 1132 kN V joint= V j = T - Vh = 1132 - 169 = 963 k N
Kuat Geser Nominal hubungan balok - kolom yang terkekang pada dua sisinya = Vn V j Vn = 1,25 (
f'c ) . A j = 1,25 1, 25
30 . 600 . 600 = 2464751 N = 2465 k N
=
963 = 1284 kN 0,75
Didaerah hubungan balok balok - k olom , dipasang 1/2 ( se tengah) dari tulangan trans verval didaerah ( Pasal : 23.5.2.2 ) Tulangan transversal didaerah
= 4 D 12 - 100 mm
( Ok)
o
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Ref : SNI - 03 - 2847 - 2002
Hubungan Balok Kolom
Menghitung M PR balok (Mome nt Probable Probable ) , dimuk a joint 6 59,50 mm
8 D 19
118,50 mm
1500 mm 2 D12
Sengkang
650 mm 59,50 mm
10 mm
120 mm
4 D 19 300 mm
Penampang Balok di Potongan 3 - 3 Menghitung MPR (-) ds =
4.283.59,50 + 4.283.118,50 4.283 + 4.283
= 89 mm
dbaru= h - ds = 650 - 89 = 561 mm apr =
As. ( 1,25 f y) 0,85.f' c.b
=
( 8.283 ( 1,25 . 400 ) 0,85.30.300
MPR (- ) = As ( 1,25 fy ) . ( d -
= 147,97 mm
a) = ( 8.283) ( 1,25 .400 ) ( 561 2
= 551300980 551300980 Nm Nm m = 551 kNm Menghitung MPR ( + ) ds = 40 + 10 + 1/2 . 19 = 59,50 mm dbaru= h - ds = 650 - 59,50 59,50 = 590,50 590,50 mm a
=
As. ( 1,25 f y)
=
{ 4.283 ( 1,25 . 400 )}
= 14,80 mm
147,97 ) 2
7
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Ref : SNI - 03 - 2847 - 2002
Hubungan Balok Kolom
Mu
= kolom
Mpr balok 2
=
8
551 + 330 = 440,50 kNm 2
Vh = 270 kN
T2 = AS . ( 1,25 . fy )
C1 = T1
T
AS Mpr = 330 kNm A'S
C2 = T2
Mpr = 551 kNm
T1 = AS . ( 1,25 . fy ) V horizontal =
Mu kolom=
Mpr balok 2
=
440,50 x 2 1 ( 2,85 + 3,675 ) 2
= 270 kN
551 + 330 = 440,50 kNm 2
HUBUNGAN HUBUNGAN BALOK BALOK - KOLOM INTERIOR ( DALA DALAM M ) C1 = T1 = As . 1,25 . f y = 8 . D 19 . 1,25 . 400 = 8 . 283 . 1,25 . 400 = 1132000 N = 1132 kN C2 = T 2 = A' A' s . 1,25 . f y = 4 . D 19 . 1,25 . 400 = 4 . 183 . 1,25 . 400 = 566000 N = 566 kN V joint= V j = C1 + T2 - Vh = 1132 + 566 - 270 = 1428 1428 kN Kuat Geser Nominal hubungan balok - kolom yang terkekang pada empat sisinya = Vn V j Vn = 1,70 (
f'c ) . A j = 1,70 1, 70
30 . 600 . 600 = 3352062 N = 3352 k N
=
1428 = 1904 kN 0,75
Didaerah hubungan balok balok - k olom , dipasang 1/2 1/2 ( set engah) dari tulangan trans verval didaerah ( Pasal : 23.5.2.2 )
( Ok)
o
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Hubungan Hubungan Balok Kolom
S 2 e n D g 1 2 k a g - 1 n 0 0 m m
Ref : SNI - 03 - 2847 - 2002
2 S e D n 1 g k 2 a Sengkang - 1 n g berpengikat 0 0 600 mm 4 D 12 - 100 mm m m = 500 mm o
lom ng Ko lo Penampa
Joint 10
600 mm
Joint 9 Penampa
Sengkang berpengikat 4 D 12 - 100 mm
lom ng Ko lo
620 mm
= 620 mm
o
lom ng Ko lo Penampa
3500 mm
Sengkang 2 D 12 - 100 mm 575 mm
o m o l o g K n a a m p P e n
Sengkang berpengikat 4 D 12 - 100 mm 700 mm Sengkang 2 D 12 - 100 mm 575 mm
lom g Ko lo Penampan
Sengkang berpengikat 4 D 12 - 100 mm
lom g Ko lo Penampan
455 mm
700 mm
455 mm
620 mm
= 620 mm
o
5 D19
Joint 5 m g 0 m n a 1 0
lom ng Ko lo Penampa
5 D19
Sengkang berpengikat 4 D 12 - 100 mm
Joint 6 2
9
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