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170 cm

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.

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170 cm

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  2. 170 cm 170 cm 5 170 cm

  3. 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

  4. 2.1.3Beban Hidup(H) +Mati (M) • Total Beban Terbagi Rata berfaktor Momen Yg Terjadi 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) P’=16154kg/ lebar balok T q= 3873 kg/m /lebar balok T diapragma 7.5m ¼ x (16154) x 7.5 3

  5. 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 bm=150cm t=20 a Ac Tinggi bidang tekan (a) bw=30 As 1.70M 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 4 5

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  8. 2.1.1 ALTERNATIF 2 f p Ditentukan : Luas Besi Tarik (As): 6 22=6x(1/4 x22^2)=2281mm2 Dicari : tinggi h’ Perkiraan h’ : 800 mm Dari referensi ALTERNATIF 1 f Referensi :Untuk Luas besi : 8 22 , h’= 650mm f 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 Ternyata Momen yg bisa ditahan (Mt ) > mm < t (200 mm ) a = 943786.56 / 35164.5=26.8 Momen yg terjadi (Mn ) (memenuhi ) a < t ( perhitungan blk T, bisa dianggap balok Persegi ) ( 7410 x10^5Nmm >7190.1x10^5 Nmm) (h’-1/2 a ) b . Momen yg bisa ditahan= Mt =T x - Perkiraan h’ =800mm,dgn besi 6  22 (memenuhi) Mt As x fy x (h’ -a/2) 2281 X 413.76 x (800 - 13.4) = = [ Kalau MT < Mn , maka h’ harus ditambah ) =741018889.8 Nmm =7410 x10^5Nmm Keseimbangan Momen yg terjadi c. h = 800mm+50mm (beton deking )= 850 mm [Apakah Mt > MnYG TERJADI] Momen yg terjadi =Mn =7 1901 kgm=7190.1x10^5 Nmm ( lihat Hal sebelum) 8

  9. Gaya geser rata rata yg terjadi ,1m dari tepi (Vu ) 1m P’=16154 kg/lebar balok T q=3873 kg/m /lebar balok T 7.5 m 3.75 m 1m 6.5 m 16154 kg 6.5 / 7.5 x 16154 =14000kg 2.75 m ½ x 7.5 x 3873 = 14524 kg 3.75 m 2.75 / 3,75 x 14524 kg = 10651 kg 7500 kg rata = (14524+10651)/2=12588 kg Bidang ' D ' D tot rata rata 1m dr tepi =Vu= 14000 + 12588 =26588 kg =265880 Newton.

  10. Kuat geser ( Vn ),hanya beton dan sengkang , tanpa tul miring 28 150mm(sengkang) ( s = 150 mm ). 650 mm ( d ) fy= 414 N / mm² fc’ = 27,5 N / mm² 300 mm ( bw ) Tahanan geser max yg bisa ditahan beto (  t c )= fc’ / 6N / 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= 180261,1Newton • Vn =  ( Vc + Vs ) = 0,8 .( 156000 +180261.1 ) = 269008,9Newton Vu <  VnOke (265880 Newton.< 269008,9Newton)

  11. dipilih :Alt 1 1m 1m 1m f 2 8 - 20mm f f 8 22 mm 2 8 - 150mm f f 2 8 - 150mm 4 22 mm Tul praktis 200mm 650 mm f 22 2 700mm f 2 8 - 150mm f 2 22 f 8 22 300mm

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