170 cm

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1.2Perencanaan G irder ( B alok T ) K reteria Design Beban H idup BerdasarBinam arga Beton :fc’=4000 lb /inch = 27,584 N /m m Baja :fy = 60 000 lb /inch = 413,76 N /m m Perhitungan Beton PBI 89 Lebar 7m Bentang 7.5 m 3 50cm 1.70M 20 30 7.5 m 1.2.1B eban 1 .2.1 .1B eb a n M ati PlatBeton 0.20x 1.7x 2400 =816 kg /m Beam 0.30x 0.50x2400 =360 kg/m A sphal 0.05 x 1.7 x2200 =187 kg/m A irhujan 0.03x 1.7x1000 =51 kg/m ---------------------------- 1414 kg/m Berat D iapragm a =0.15 x 0.25x 1.4 x 2400 =126 kg/1.7m Perkiraan D em ensi D iapragm a 15x25cm 1.4 m 20 cm 1

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1. 170 cm. 170 cm. 5. 170 cm. Momen & Geser Yg Terjadi. 1.2.1.1. Beban Mati ( M ). 1.7. 1.7. qM= 1414 kg / m P M = 126 KG. 1414 kg. 126 kg. Ly. Beban Mati & Hidup. P t. q t/m. Ly = 7.5m. Pt = Total Beban Garis (PM +PH ) qt = Total Beban Terbagi Rata (qM +q H). 2. - PowerPoint PPT Presentation

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Page 1: 170 cm

1.2Perencanaan Girder ( Balok T )Kreteria Design

Beban Hidup Berdasar BinamargaBeton : fc’=4000 lb / inch = 27,584 N / mmBaja : fy = 60 000 lb / inch = 413,76 N / mmPerhitungan Beton PBI 89Lebar 7mBentang 7.5 m

3

50cm

1.70M

20

30

7.5 m

1.2.1Beban1.2.1.1Beban Mati

Plat Beton 0.20x 1.7x 2400 =816 kg /mBeam 0.30x 0.50x2400 =360 kg/ mAsphal 0.05 x 1.7 x2200 =187 kg/mAir hujan 0.03x 1.7x1000 =51 kg/m

----------------------------1414 kg/m

Berat Diapragma =0.15 x 0.25x 1.4 x 2400 =126 kg/1.7m

Perkiraan Demensi

Diapragma 15x25cm

1.4 m

20 cm

1

Page 2: 170 cm

5

170 cm170 cm

170 cm

Page 3: 170 cm

1.71.7

qM= 1414 kg / m

P M = 126 KG

Ly

126 kg1414 kg

1.2.1.1

2

P t q t/m

Ly = 7.5m

Pt = Total Beban Garis (PM +PH )

qt = Total Beban Terbagi Rata (qM +q H)

Page 4: 170 cm

q= 3873 kg/m /lebar balok T

P’=16154 kg/ lebar balok T

7.5m

diapragma

1.2.1.3 Beban Hidup(H) +Mati (M)

Total Beban Terbagi Rata berfaktor

Total q = (1,6 )x1.360(qH) + (1,2)x1414(qM) =3873kg/m

Total Beban Garis (Terpusat) berfaktor =

Total P’ = (1,6)x(10002,16)(PH)+(1,2)x (126) (PM) =16154kg

1.2.2 Momen max yg terjadi 1.2.2.1 Momen Max yg Terjadi(Mn)

¼ x (16154) x 7.5

3

Page 5: 170 cm

Ac =Luas beton yg bisa menahan gaya tekan akibat Momen Lentur = bm x a bm= lebar manfaat sayap yg bisa menahan gaya tekan Momen Lentur

a =tinggi manfaat sayap yg bisa menahan gaya tekan Momen Lentur

T=As x fy C= 0.85 x bm x a x fc C = T

0.85 x bm x a x fc = As x fy

a= As x fy / ( 0.85 x bm x fc )

As = ? a= tahu

1.70M

t=20

bw=30

bm=150cm

a

1.2.5. Menentukan lebar manfaat sayap bm1 < =6 t+ bo =6 x 20 + 30 = 150cm bm2 <=bo + L /5 = 30 + 750 / 5 =180 cm bm3< =bo + L/10+ bt /2 = 30+750 /10 + 170/2 =190 cm bm4 <=bt =170 cm bm Dicari Yg Terkecil = 150cm

Ac

As Tinggi bidang tekan (a)

4

Page 6: 170 cm

Momen yg bisa ditahan(Mt)

Mt=T x (h’ - 1/2 a ) atau

=Cx ( h’ - 1/2 a )

MOMEN YG TERJADI MOMEN PENAHAN

Momen yg terjadi (Mn)

T=Kek tulangan seluas As( n x 1/4x x )

C=Kek beton seluas Ac( bm x a )

T =C C= 0.85xbmx a x fc T = As x fy

a = bisa dicari

Ly

Mu=Total Mo men Max Yg. Terjadi (Ditengah )= 1/8 x q x Ly 2 + 1/4 x Px Ly

Mn = Mu / 0.8

Mn harus < , = , Mt

Mn < = T x (h’ - 1/2 a )

Mn = sudah dicari (momen yg terjadi)

T = ditentukan

(h’ - 1/2 a)= bisa dicari

a=sudah dicari h’ =bisa diketahui

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bm

6

Page 7: 170 cm

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Page 8: 170 cm

2.1.1 ALTERNATIF 2 Ditentukan : Luas Besi Tarik (As): 6 22=6x(1/4 x22^2)=2281mm2

Dicari : tinggi h’

Perkiraan h’ : 800 mm

Dari referensi ALTERNATIF 1

Referensi :Untuk Luas besi : 8 22 , h’= 650mm

Perkiraan :Untuk 6 22 , h’ = 8 / 6x 650 =866.6 = 800 mm

a. Mencari tinggi a (Apakah a < t ,sarat balok TDianggap balok persegi)

C=0.85xfc’x bmx a =0.85x27.58x1500x a = 35164.5 a N T =As x fy = 2281x 413.76 = 943786.56 N

Persamaan C=T 35164.5 a = 943786.56

a = 943786.56 / 35164.5=26.8 mm < t (200 mm ) a < t ( perhitungan blk T, bisa dianggap balok Persegi )

b . Momen yg bisa ditahan= Mt =T x -(h’-1/2 a )

Mt = As x fy x (h’ -a/2) = 2281 X 413.76 x (800 -13.4)

=741018889.8 Nmm =7410 x10^5Nmm Keseimbangan Momen yg terjadi

[Apakah Mt > MnYG TERJADI]

Momen yg terjadi =Mn =71901 kgm=7190.1x10^5 Nmm ( lihat Hal sebelum)

Ternyata Momen yg bisa ditahan (Mt ) >

Momen yg terjadi (Mn ) (memenuhi )

( 7410 x10^5Nmm >7190.1x10^5 Nmm)

Perkiraan h’ =800mm,dgn besi 6 22 (memenuhi)

[ Kalau MT < Mn , maka h’ harus ditambah )

c. h = 800mm+50mm (beton deking )= 850 mm

8

Page 9: 170 cm

q=3873 kg/m /lebar balok TP’=16154 kg/lebar balok T

6.5 / 7.5 x 16154 =14000kg 16154 kg

½ x 7.5 x 3873 = 14524 kg

7500 kg2.75 / 3,75 x 14524 kg = 10651 kg

rata = (14524+10651)/2=12588 kg

7.5 m

3.75 m 1m 6.5 m

1m

D tot rata rata 1m dr tepi =Vu= 14000 + 12588 =26588 kg =265880 Newton.

Gaya geser rata rata yg terjadi ,1m dari tepi (Vu )

2.75 m

3.75 m

Page 10: 170 cm

65

0 m

m

( d

)

28 150mm(sengkang)

( s = 150 mm ).

fy = 414 N / mm²

fc’ = 27,5 N / mm²

Tahanan geser max yg bisa ditahan beto

( t c )= fc’ / 6 N / mm²= 27,5 / 6 =0.87N/ mm²(rumus)

Kuat geser max beton ( V c )= ( t c ) x bw x d

=0.87 x 6500 x 300 =156000N

Kuat geser max yg bisa ditahan 2 tul. baja ( sengkang )

=As x fy = 2 x ¼ 8² x 414 = 41598,72 Newton

Kuat geser max yg bisa ditahan tul. baja ( sengkang ) ,selebar bidang geser ( V s ) = As x fy x d /s =41598.72 x 650 /150= 180261,1 Newton

300 mm

( bw )

Vn = ( Vc + Vs ) = 0,8 .( 156000 +180261.1 ) =

269008,9 Newton Vu < Vn Oke

(265880 Newton.< 269008,9 Newton)

Kuat geser ( Vn ),hanya beton dan sengkang , tanpa tul miring

Page 11: 170 cm

2 8-150mm 2 8-150mm

8 22 mm

8 22

2 8-150mm

1m 1m

4 22 mm

2 22

2 8-20mm

1m

Tul praktis

650 mm

700mm

300mm

200mm

dipilih :Alt 1

2 22