Modul 6 PERENCANAAN KUDA-KUDA (2) BEBAN ANGIN AKIBAT ANGIN KIRI Faktor koefisien angin :

1. Modul 6 PERENCANAAN KUDA-KUDA (2) BEBAN ANGIN AKIBAT ANGIN KIRI Faktor koefisien angin : 1 = 30,9637o , maka faktor koefisien angin = (+ 0,021 - 0,4) 2 = 54,4623o , maka faktor koefisien angin = (+ 0,022 - 0,4) Untuk angin hisap, faktor koefisien angin = (- 0,4) q angin = 40 kg/m2 x 3,5 m x Faktor koefisien = 140 kg/m x Faktor koefisien Pa1 = ½ x 1,72 m x 140 kg/m x (0,02 x 30,9637o – 0,4) = 26,4 kg Pa2 = ½ x 1,72 m x 140 kg/m x (0,02 x 30,9637o – 0,4) = 26,4 kg Pa3 = ½ x 2,332 m x 140 kg/m x (0,02 x 54,4623o – 0,4) = 112,51 kg 1 http://www.mercubuana.ac.id

2. VB = 16,06 kg ( ) GAYA BATANG AKIBAT ANGIN KIRI TITIK BUHUL A      Fy = 0, VA + S1 sin1 – Pa1 cos1 = 0 39,64 + (S1) (0,8137) – (26,4) (0,5812) = 0 0,8137 S1 + 24,29 = 0 S1 = - 29,86 kg (tekan)  Fx = 0, -HA + S1 cos1 + Pa1 sin1 + S8 = 0 -304,28 + (-29,86) (0,5812) + (26,4) (0,8137) + S8 = 0 -304,28 – 17,35 + 21,48 + S8 = 0 S8 = 300,15 kg (tarik) TITIK BUHUL H  http://www.mercubuana.ac.id 3

3.              Fy = 0, S2 sin2 – S10 sin3 – S1 sin1 – Pa2 cos1 – Pa3 cos2 – S9 = 0 0,5145 S2 – 0,5734 S10 – (-29,86) (0,8137) – (26,4) (0,5812) – (112,51) (0,8575) – 0 =0 0,5145 S2 – 0,5734 S10 + 24,29 – 15,34 – 96,48 = 0 0,5145 S2 – 0,5734 S10 = 87,53 S2 = 1,1145 S10 + 170,13 ........................................................................(i)  Fx = 0, Pa2 sin1 + S2 cos2 + Pa3 sin2 + S10 cos3 – S1 cos1 = 0 (26,4) (0,8137) + 0,8575 S2 + (112,51) (0,5145) + 0,8193 S10 – (-29,86) (0,5812) = 0 21,48 + 0,8575 S2 + 57,89 + 0,8193 S10 + 17,35 = 0 0,8575 S2 + 0,8193 S10 + 96,72 = 0 S10 = - 1,0466 S2 – 118,05 .......................................................................(ii) (i) dan (ii) S2 = (1,1145) (-1,0466 S2 – 118,05) + 170,13 S2 = - 1,1664 S2 – 131,57 + 170,13 2,1664 S2 = 38,56 5 http://www.mercubuana.ac.id