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MEKANIKA REKAYASA IVPRESENTASI KONSTRUKSI PORTAL TIDAK BERGOYANG
ANGGOTA1. AKBARRANI ARIPRATIWI ( 1441320091 )2. FIRDAUS AUNUROCHIM ( 1441320092 )3. RIDO AULIA RAKHMAN ( 1441320035 )
SUMBER REFERENSI : LITERATUR BAHASA INGGRIS
Langkah : Perhitungan konstruksi portal tidak bergoyang1. Mencari nilai kekakuan : Titik A KAB = μAB =
= =
= 0,67 ( + ) 1,34 = 0,5
KAD = μAD = = =
= 0,67 = 0,5 = 1
Langkah : Perhitungan konstruksi portal tidak bergoyang1. Mencari nilai kekakuan : Titik B KBA = μBA = = =
= 0,67 = 0,364
KBE = μBE = = ( + ) 1,84 =
= 0,5 = 0,364
KBC = μBC = = =
= 0,67 = 0,364
= 1
Tabel perataan momen :TITIK D A B C E
BATANG DA AD AB BA BE BC CB EB
μ 1 0.5 0.5 0.364 0.272 0.364 1 1
M - - 77.63 -54.11 - 36 -36 -
-19.41 -38.82 -38.82 -19.41
6.83 13.66 10.21 13.66 6.83 5.1
-1.71 -3.42 -3.42 -1.71
0.31 0.62 0.47 0.67 0.31 0.23
-0.08 -0.16 -0.16 -0.08
0.01 0.03 0.02 0.03 0.01 0.01
M ujung -21.2 -42.4 42.38 -61 10.7 50.31 28.85 5.34
Arah
MOMEN PRIMERMBA1 = MAB1 =
=
= -25,26 = -25,26
MBA2 = MAB2 =
= =
= - 28,85 = - 28,85
MBA = 77,63 MAB = 54,11
MOMEN PRIMERMBA1 =
= -36
MBA2 = - MBA
= - (-36)
= 36
-21,2 5,341,347,07
42,38
36,24 37,34
7,07 1,3442,40 10,70
36,24 37,3436,24 24,76
7,07
28,85
7,07
12,58 5,42
5,73
61 50,31
5,73
7,071.34
6
12,58
8,39
4,819
5,428,39
4,81
9
5 7
37,793,53
5,0836,24
23,213,53
5,0824,76
Q = 8 Q =
3 Q = 3
40 21
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