# Mekanika Teknik Soal Dan ian Metode Clapeyron

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Tugas Mekanika Teknik By: Muh_Desmawan_EWD

Gambarkan moment dan lintangnya,.!! Penyelesaian: AKIBAT MUATAN LUAR: Dukungan C 1 + 2 = = 9 1 . 2 3 . 1 2 . + 384 24 16 Dukungan D 1 + 2 = = 7 1 . 2 3 2 . 3 3 . + 384 24 24

9 5. 43 10. 62 . + 384 24 16 7,5 360 = + 24 16 22,81 = Jepitan B 2 . 3 3 3. 43 2 = = = 24 24

7 5. 43 3. 43 . + 384 24 16 5,83 48 = + 24 16 2,24 =

AKIBAT MOMENT PERALIHAN: Dukungan C 1 +2 = = = Jepitan B = = = 3 3 + 3 () 6 () 4 4 + 3 () 6 () 1,3 0,67 + 1 2 2 + + 3 () 3 () 6 () 6 4 4 + + 3 () 3 () 6 () 3,3 0,67 + 3 () 6 () Dukungan D 1 +2 = = = 2 2 3 3 + + + 6 () 3 () 3 () 6 () 4 4 4 4 + + + 6 () 3 () 3 () 6 () 0,67 2,67 0,67 + +

Tugas Mekanika Teknik By: Muh_Desmawan_EWD Didapat persamaan belahan sbb: 1 + 2 1 + 2 Sehingga didapat: 22,81 2,24 2 = 3,3 + 0,67 = 0,67 + 2,67 + 0,67 = 1,3 + 0,67 Dengan cara eliminasi didapat: = , = , = , = 1 +2 = 1 +2 = 3,3 0,67 22,81 = + 3 () 6 () 2,24 0,67 2,67 0,67 = + + 2 1,3 0,67 = +

Bidang Moment Freebody AC RAV = RCV1 = 5 kN MMAX = RAV. l = 5.3m =15kNm (ditengah-tengah batang) Free body CD 1 . 2.3 5.2.3 30 2 = = = = 7,5 4 4 4 1 . 2.1 5.2.1 10 2 = = = = 2,5 4 4 4 2 7,5 = = = 1,5 ( ) 5 2 2 7,52 = = = 5,625 2 10 Free body DB 2 . 4.2 3.4.2 24 1 = = = = = 6 4 4 4 = . 2 = 6.2 = 12 ( Reaksi gaya lintang pada tiap-tiap sendi: 7,23 = = 3,795 1 6 = 1 + 2 + + + 1 2 2 7,23 7,23 1,56 = 5 + 7,5 + + = 15,9025 6 4 4 = 1 + 2 + + 2 2 3 3 7,23 (1,56) (1,56) 2,34 = 2,5 + 6 + + = 5,3275 4 4 4 4 (1,56) 2,34 = + =6 + = 6,975 3 3 4 4 =

Tugas Mekanika Teknik By: Muh_Desmawan_EWD Gaya Lintang (D) =0 = = 3,795 = = 3,795 = = 3,795 10 = 6,205 = = 6,205 = + = 6,205 + 15,9025 = 9,6975 = = 0,3025 = = 0,3025 = + = 0,3025 + 5,3275 = 5,025 = 2 . 4 = 5,025 3.4 = 6,975 = + = 6,975 + 6,975 = 0

= 1 . 2 = 9,6975 5.2 = 0,3025