Thursday 22 March 2018

Metode Join




Diketahui : Sistem rangka batang kuda-kuda
                  Sistem perletakan sendi-roll,dengan jumlah reaksi, R = 3
                  Jumlah buhul/join (A-H) = 8 titik
                  Jumlah elemen/batang, M = 13 buah
                  Pembebanan berupa beban titik
                  Perbandingan jarak horizontal vertikal = 2(x), 1,5(x) dengan x = 1 digit                   terakhir NIM.

Pertanyaan:
1. selesaikan secara manual
2. buatlah flowchart pengerjaan
3. buatlah source code

Jawab:
Reaksi tumpuan
€mb=0
R1 . 16– 5292 . 16 – 10292 . 12 – 10292 . 8 – 10292 . 4 = 0
16 R1 = 84672 + 123504 + 82336 + 41168
     R1 = 331680  / 16
     R1 = 20730 N
€v=0
R1 + R3 – 5292 – 10292 – 10292 – 10292 – 5292 = 0
20730+  R2 – 41460 = 0
                 R3 = 20730 N

Metode join
£=36,87°
Titik A
€v=0
R1 + S1 . sin 36,87 – 5292 = 0
20730 + S1 . 0,60 - 5292 = 0
                 S1 . 0,60 = -15438
                 S1 = -15438/ 0,6
                 S1 = -25730 N (tekan)



€H=0
R2 + S2 + S1 cos 36,87 = 0
0 + S2 + (-20730) . 0,8 = 0
      S2 = 16584 N (tarik)

Titik C
€H=0
-S2 + S6 = 0
-16584 + S6 = 0
               S6 = 16584 N (tarik)
€v=0
S5 = 0

\Titik B
€H=0
-S1 . cos36,87 + S3 . cos36,87 + S4 . cos36,87 = 0
-(-25730) . 0,8 + S3 . 0,8 + S4 . 0,8 = 0
20584  = -S3 . 0,8 - S4 . 0,8...(1)
€v=0
-S5-10292 - S4.sin36,87 - S1.sin36,87 + S3sin36,87=0
- 0 - 10292 - S4 . 0,6 + 25730. 0,6 + S3 . 0,6 = 0
       5146 - S4 . 0,6 + S3 . 0,6 = 0
       5146 = S4 . 0,6 - S3 . 0,6...(2)


Eliminasi pernyataan (1) dan (2):
20584 = -S3 . 0,8 - S4 . 0,8 [x6] = 123504 = -S3 .4,8 - S4 . 4,8
 5146  = -S3 . 0,6 + S4 . 0,6[x8] =   41168 = -S3 . 4,8 + S4 . 4,8       +
                                                       164672 = -9,6 S3
                                                                  S3= -17115,33 N (tekan) 
Dari pernyataan(1)=
€H=0
20584 = -S4 . 0,8 - S3 . 0,8
20584 = -S4 . 0,8 - (-17115,33) . 0,8
20584 = -S4 . 0,8 + 13722,66
20584 – 13722,66 = -S4 . 0,8
                    S4 = -6861,34/ 0,8
                    S4 = -8576,6 N (tekan)
Titik D
€H=0
-S3 . cos36,87 + S8 . cos36,87 = 0
-(-17118,33) . 0,8 + S8 . 0,8 = 0
17118 . 0,8 + S8 . 0,8 = 0
                              S8 = -13694,66  / 0,8
                                   = -17115,33  N (tekan)
€v=0
-10292 - S7 - S3 . sin36,87 - S8 . sin36,87 = 0
-10292 - S7 - (-17115,33) . 0,6 - (-17115,33) . 0,6 = 0
-10292 - S7 + 10292 + 10292 = 0
                                            S7=10292 N (tarik)
Titik H
€v=0
-5292 + R3 + S12 . sin36,87 = 0
-5292 + 20667,5 + S12 . 0,6 = 0
15438+ S12 . 0,6 = 0
                 S12 . 0,6 = -15438
                         S12 = -15438/ 0,6
                               = -25730 N (tekan)

€v=0
-S13 - S12 . cos36,87 = 0
-S13 - (-25730) . 0,8 = 0
-S13 + 20584 = 0
                S13 = 20584 N (tarik)

Titik G
€H=0
-S11 + S13 = 0
-S11 + 20584 = 0
-S11 = -20584
  S11 = 20584 N (tarik)


€v=0
S10=0

Titik E
€H=0
-S6 + S11 - S4 . cos36,87 + S9 . cos36,87 = 0
-20584 + 20584 - (-8576,6) . 0,8 + S9 . 0,8 = 0
6847,34 + S9 . 0,8 = 0
                         S9 = -6847 / 0,8
                         S9 = -8576,6 N (tekan)


€v=0
S7 + S9 . sin36,87 + S4 .sin36,87 = 0

Titik F
€H=0
-S9 . cos36,87- S8 . 0,8 + S12. cos . 0,8 = 0
€v=0
-10292 - S10 - S9 . sin36,87 - S12 . sin36,87 + S8 . sin36,87 = 0



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