Bab Vii Stabilitas Plat Buhul [PDF]

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BAB VII Perhitungan Stabilitas Pelat Buhul



Gambar Buhul 1



Gambar Tinjauan Potongan 1 – 1



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Analisa Penampang Potongan 1 – 1 1. Abruto - IWF 428x407x20x35-283 (Diagonal) A1 = 3,5 x 49,36 A2 = 2 x 33,7 A3 =3,5 x 49,36 - IWF 428x407x20x35-283 (Horizontal) A4 =3,5 x 40,7 A5 = 2 x 33,7 A6 = 3,5 x 40,7 - A pelat buhul -A pelat buhul 1 = (119,04 x 2) -A pelat buhul 2 = (119,04 x 2) 2. Alubang baut ∅ 25,4 mm - A lub 1= 3 x ((2 + 3,5) x 2,64) - A lub 2= 3 x ((2 + 3,5) x 2,64)



= 172,76 cm2 = 67,4 cm2 = 172,76 cm2 = 142,45 cm2 = 67,4 cm2 = 142,45 cm2 = 238,08 cm2 = 238,08 cm2



=43,56 cm2 =43,56 cm2



3. Anetto



= Abruto - Alubang = 1241,38 – 87,12 =1154,26 cm2 Titik berat penampang Y = Yb = (172,76 x 40,95)+(67,4 𝑥 22,35)+(172,76 𝑥 3,75)+(142,45𝑥 40,95)+(67,4 𝑥 22,35 )+(142,45𝑥 3,75)+ (238,08 𝑥 43,7)+ (238,08 𝑥1) −(3 𝑥 43,56 𝑥 41,95)− (3 𝑥43,56𝑥 2,75) 1154,26



Ya



= 18,9762 = 44,7 - 18,9762= 25,7238 cm



Inersia Penampang Potongan 1 – 1 1. I bruto IWF 428x407x20x35 I1



1



= ((12 𝑥 𝑏 𝑥 ℎ3 )) + (A x y2)) 1 12



= ((



𝑥 49,36 𝑥 3,53 )) + (172,76 x 21,97382))



= 83593,160 cm4 I2



1



= ((12 𝑥 𝑏 𝑥 ℎ3 )) + (A x y2)) 1 12



= ((



𝑥 2 𝑥 33,73 )) + (67,4 x 3,37382))



= 7145,9744 cm4



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I3



1



= ((12 𝑥 𝑏 𝑥 ℎ3 )) + (A x y2)) 1



= ((12 𝑥 49,36 𝑥 3,53 )) + (172,76 x 15,22622)) = 40228,5480 cm4 I4



1



= ((12 𝑥 𝑏 𝑥 ℎ3 )) + (A x y2)) 1



= ((12 𝑥 40,7 𝑥 3,53 )) + (142,45 x 21,9738 2)) = 68927,099 cm4 I5



1



= ((12 𝑥 𝑏 𝑥 ℎ3 )) + (A x y2)) 1



= ((12 𝑥 2 𝑥 33,73 )) + (67,4 x3,37382)) = 7145,9744 cm4 I6



1



= ((12 𝑥 𝑏 𝑥 ℎ3 )) + (A x y2)) 1



= ((12 𝑥 40,7 𝑥 3,53 )) + (142,45 x 15,2262 2)) = 33170,6221 cm4 I pelat buhul 1



I pelat buhul 1 = ((12 𝑥 𝑏 𝑥 ℎ3 )) + (A x y2)) 1 12



= ((



𝑥 119,04 𝑥 23 )) + (238,08 x 24,72382))



= 145609,6375 cm4 1



I pelat buhul 2 = ((12 𝑥 𝑏 𝑥 ℎ3 )) + (A x y2)) 1



= ((12 𝑥 119,04 𝑥 23 )) + (238,08 x 17,97622)) = 77013,4279 cm4



2. Ilubang baut ∅ 25,4 mm I7



1 12



𝑥 𝑏 𝑥 ℎ3 )) + (A x y2))



1 12



𝑥 2,64𝑥 5,53 ) + (43,56 𝑥 22,72382 ))



= 3 x (( = 3 x ((



= 69082,1217 cm4 I8



1



= 3 x ((12 𝑥 𝑏 𝑥 ℎ3 )) + (A x y2)) 1



=3 x ((12 𝑥 2,64𝑥 5,53 ) + (43,56 𝑥 16,22622 )) = 34516,4880 cm4 3. Inetto = Ibruto – Ilubang = 462924,4425 – 103598,6097 = 359325,8328 cm4



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4. Modulus tampang bagian atas Watas



=



𝐼netto 𝑌𝑎



=



359325,8328 25,7238



= 13968,654 cm3



5. Modulus tampang bagian bawah Wbawah



=



𝐼netto 𝑌𝑏



=



359325,8328 18,9762



= 18935,6053 cm3



6. Gaya – gaya yang bekerja : N



= 78654,6750 + (-45678,8900 cos 30) = 46354,8221 kg



D



= -45678,8900 sin 30 = -22839,445



M



= [(−45678,8900 cos 30) 𝑥 (25,7238 − 22,35) + 78654,6750 𝑥 (18,9762 − 22,35] =-398829,5638 kg.cm



7. Tegangan yang terjadi : Akibat N 𝑁



𝜎n = 𝐴𝑛𝑒𝑡𝑡𝑜 =



46354,8221 1154,26



= 40,1598 kg/cm2



Akibat D 𝐷



𝜏 = 𝐴𝑛𝑒𝑡𝑡𝑜 =



22839,445 = 1154,26



-19,7871 kg/cm2



Akibat M 𝑀



𝜎atas = 𝑊𝑎𝑡𝑎𝑠 = 𝑀



−398829,5638 13968,654



𝜎bawah = 𝑊𝑏𝑎𝑤𝑎ℎ =



= -28,5517 kg/cm2



−398829,5638 18935,6053



= -21,0624 kg/cm2



8. Tegangan total : 𝜎atas = -28,5517 + 40,1598 = 11,6081 kg/cm2 𝜎bawah = -21,0624 + 40,1598 = 19,09874 kg/cm2 9. Tegangan Idiil : 𝜎idiil = √(19,0974 )2 + (1,56 𝑥 − 19,7871)= 18,2714 kg/cm2 10. Syarat keamanan : 𝜎idiil < 𝜎 18,2714 kg/cm2 < 1600 kg/cm2 . . . . OK



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