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Pengaruh Geser thd balok Beton
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2ton 2ton
2ton 2ton2ton
2ton 2ton
1
2
3
4
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2ton
2ton
Bid “ D “ ( “Vu “)
4ton
2ton 2ton
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30 ton
20 ton
20 ton
20 ton
10 ton
10 ton
10 ton
10 ton10 ton
20 ton20 ton
1/3L 2/3 L
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Ra Rb
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Ra Rb
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Ra Rb
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Ra Rb
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Ra Rb
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Ra Rb
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60 kg60kg
120 kg
120 kg
2,5 m 2,5 m
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2m3m
120 kg
48 kg72 kg
48 kg72 kg
120 kg
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30 kg
90 kg
120 kg
1m 4m
120 kg
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q= 30kg / m
L = 5 m
75 ton
75 ton
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2,5 m
0,5 m
90 kg 120 kg
75 kg
75 kg2 / 2,5 x 75 kg = 60 kg
2m 2,5 m
Gaya geser “Vu” max jarak 0m --- ½ m dr tepi
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9000 kg
12000 kg
7500kg
7500 kg
2 / 2,5 x 7500 kg = 6000 kg
2m
Bid “Vu” max jarak 0m --- ½ m dr tepi
3000 kg
Rata rata = (7500+6000)/2=6750 kg
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q=3873 kg/m /lebar balok TP’=16154 kg/lebar balok T
‘D’ maxyg terjadi dari tepi jarak ‘ h m’=0 m
7.5 / 7.5 x 16154 = 16154 kg
½ x 7.5 x 3873 = 14524 kg
14524kg
7.5 m
7.5 m
h m=0 m
D tot 0 m dr tepi = 16154 kg + 14524 kg = 30678 kg
0 kg
3.75 m
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q=3873 kg per 1 balok TP’=16154 kg per 1 balok T
‘D’ maxyg terjadi dari tepi jarak ‘ 1m’
6.5 / 7.5 x 16154 =14000kg
16154 kg
½ x 7.5 x 3873 = 14524 kg
2.75 / 3,75 x 14524 kg = 10651 kg
7.5 m
3.75 m
1m 6.5 m
1m
D tot 1m dr tepi = 14000+ 10651 = 24651kg
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q=3873 kg per 1 balok T
P’=16154 kg per 1 balok T
‘D’ maxyg terjadi dari tepi jarak 2m
5.5 / 7.5 x 16154 =11846 kg
16154 kg
14524 kg
14524 kg
10651 kg
7.5 m3.75 m
2m 5.5 m
2m
D tot 2m dr tepi = 11846 + 6778 = 18624 kg
1.75/3.75x14524=6778kg
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q=3873 kg per 1 balok T
P’=16154 kg per 1 balok T
‘D’ maxyg terjadi dari tepi jarak 3m
4.5 / 7.5 x 16154 = 9692 kg
16154 kg
14524 kg
14524 kg10651 kg
7.5 m3.75 m
3m 4.5 m
3m
D tot 3m dr tepi = 9692 + 2905 = 12597 kg
6778kg
0.75/3.75 x 14524 = 2905 kg
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Kuat geser Beton ( Vn )
fy = 414 N / mm²
fc’ = 27,5 N / mm²
Tahanan geser max yg bisa ditahan beton
c = fc’ / 6 N / mm²= 27,5 / 6 =0.87N/ mm²
. Kuat geser max yg bisa ditahan beton ( V c )= ( c ) x bw x d
Vc =0.87 x 600 x 300 =156600 N
Kuat geser max yg bisa ditahan 2 tul. baja 8 mm tegak( satu.begel)
Vb =As x fy = 2 x ¼ 8² x 414 = 41598,72 Newton
Jumlah begel selebar bid geser = nb =d/s batang
Kuat geser max yg bisa ditahan tul. baja ( begel/sengkang ) ,selebar
bidang geser ( V s ) = Vb x nb = As x fy x d/s
s = V s / (As x fy x d)
300 mm
( bw )
600 mm
( d )
As=luas pot tulangan begel
s=jarak begel /sengkang
d = tinggi balok
fy =kekuatan baja per mm²
28
60
0 m
m
( d
)
28
sd
d
300mm
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2.7
200mm
C = 200x 27,5=3500 N
(Kekuatan batas beton luas 200 mm )2
2
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animasi
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animasi
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bw
d
DD( Vu )
Gaya geser terfaktor yg terjadi pd pot tertentu (Vu) = D Newton
animasi
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animasi
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animasi
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Luas pot Geser = bw x dbw
d
D
animasi
D( Vu )
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animasi
Kuat geser 1 tulangan geser tegak = Asx fy As = ¼ x ²xfy
As =luas pot =1/4 x x ² =diameter tul (mm)
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Tul tegak
Tul tegak
Tul tegak
Kekuatan 1 baja tegak = As x fy ( newton)
Untuk d/s buah, bisa menaha geser= As x fy x d/s N
Sumbangan kuat geser tul tegak pada luas pot geser ,bila dipasang setiap jarak s mm .
Vs=. = As x fy x d / ( s ) Newton
Tul tegak
Jumlah tul tegak setinggi d = d/s buah
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Luas bid geser = bw x d mm²
Vc = fc’ / 6 x bw x d N
bw
dD
D( Vu )
bw
( c )= fc’ / 6 N / mm²
d
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Luas pot Geser = bw x d
bw
dD
Kekuatan 1 baja tegak = As x fy ( newton)
Untuk d/s buah, bisa menaha geser= As x fy x d/s N
Sumbangan kuat geser tul tegak pada luas pot geser ,bila dipasang setiap jarak s mm .
Vs=. = As x fy x d / ( s ) Newton
Jumlah tul tegak setinggi d = d/s buah
Kekuatan 1 baja tegak = As x fy (newton)
As = ¼ x ²
d
ss
D( Vu )
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Kekuatan 1 baja tegak = As x fy N
Untuk nb buah begel, bisa menahan geser= 2x nb x As x fy N =2xnb x (1/4x x D²)x fy N
Sumbangan kuat geser tul begel pada luas pot geser
Jumlah tul begel per potongan nb = d/s buah
nb =jml tul begel per pot geser
d= tinggi balok dlm mm
S=jarak pemasangan begel dlm mm
D= Diameter tulangan begel dlm mm
Fy=tegangan ijin baja dlm Newton / mm²
N =satuan newton
s
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Luas pot Geser = bw x dbw
d
D
Gaya geser terfaktor yg terjadi pd pot tertentu (Vu) = D Newton
animasi
D( Vu )
Total Penahan Geser:
Vn = ( Vc + Vs ) = 0,8 .
Vu harus < Vn Oke
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Gaya geser yg terjadi (Vu ) Kuat geser ( Vn )
1,5 m
8400 kg
12000 kg
7500kg
7500kg
1 / 2,5 x 750kg = 3000 kg
3600 kg
Gaya Geser max jarak 0.5m ---1½ m dr tepi
6000 kg
q= 3000kg / m
P = 12000 kg
L = 5 m
2,5 m
1,0 m
Bid ‘D’ pd muatan terbagi rata
Rata rata Gaya geser=6000+3000) / 2 = 4500 kg
Bid ‘D’pd muatan terpusat
Gaya geser = 8400 kg
Total Gaya geser =8400 kg + 4500 kg = 12900 kg = 129000 newton ( 1kg = 10 newton )
Vu = 129000 Newton
50
0 m
m
( d
)
300 mm
( bw )
28 150mm
( s = 150 mm ).
fy = 414 N / mm²
fc’ = 27,5 N / mm²
Tahanan geser max yg bisa ditahan beton
( c )= fc’ / 6 N / mm²= 27,5 / 6 =0.87N/ mm²
Kuat geser max beton ( V c )= ( c ) x bw x d
Vc =0.87 x 500 x 300 =130500 N
50
0 m
m
( d
)
Kuat geser max yg bisa ditahan 2 tul. baja ( sengkang )
=As x fy = 2 x ¼ 8² x 414 = 41598,72 Newton
Jumlah Sengkang selebar bid geser = d/s batang
Kuat geser max yg bisa ditahan tul. baja ( sengkang ) ,selebar bidang geser ( V s ) = As x fy x d /s = 41598.72 x 500 /150=138662,40 Newton
300 mm
( bw )
Vn = ( Vc + Vs ) = 0,8 .( 130500 +138662,40 ) = 215329,9 Newton Vu <
Vn Oke