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SOAL: SISTEM PORTAL 2 DIMENSI QUIZ
Penyelesaian :
A. ELEMEN AB(1)
Diketahui :A : 0.1 cm²E : 200000 kg/cm² = 200 t/cm²
I : 500 cmα : 45 °
L1 : (x+y) = 10 mL2 : (z) = 8 mL3 : (x) = 1 mq : (y+z) = 17 t/mP : (x+z) = 9 ton
Tentukan Kekakuan elemen Metode Matrix !
AEL 0 0
−AEL 0 0
0
12EI
L36 EI
L2 0
−12EIL3
6 EI
L2
0
6 EI
L24 EIL 0
−6 EIL2
EIL
[K L1 ] =
−AEL 0 0
AEL 0 0
0
−12EIL3
−6 EIL2 0
12EI
L3−6 EIL2
0
6 EI
L22EIL 0
−6 EIL2
4 EIL
0,1 .2001000 0 0
−0,1 .2001000 0 0
0
12.200 .500
100036 .200 .500
10002 0
−12.200 .50010003
6 .200 .500
10002
0
6 .200 .500
100024 .200 .5001000 0
−6 .200 .50010002
200 .5001000
[K L1 ]=
−0,1 .2001000 0 0
0,1 .2001000 0 0
0
−12.200 .50010003
−6 .200 .50010002 0
12.200 .500
10003−6 .200 .50010002
0
6 .2000 .500
100022.200 .5001000 0
−6 .200 .50010002
4 .2000 .5001000
i = B
j = A
α = 270
Cos270 Sin 270 0 0 0 0
-Sin270 Cos270 0 0 0 0
0 0 1 0 0 0
[ T ] = 0 0 0 Cos 270 Sin 270 0
0 0 0 - Sin270 Cos 270 0
0 0 0 0 0 1
0 1 0 0 0 0-1 0 0 0 0 00 0 1 0 0 0
[ T ] T= 0 0 0 0 1 00 0 0 -1 0 00 0 0 0 0 1
[K g1 ] = [ T ] T . [K L
1 ] . [ T ]
[K g1 ] =
0 -1 0 0 0 0 20 0 0 -20 0 01 0 0 0 0 0 0 0.001 3 0 -0.001 3
0 0 1 0 0 0 x 0 3 400 0 -3 1000 0 0 0 -1 0 -20 0 0 20 0 00 0 0 1 0 0 0 -0.001 -3 0 0.001 -30 0 0 0 0 1 0 3 200 0 -3 400
0 1 0 0 0 0-1 0 0 0 0 0
x 0 0 1 0 0 00 0 0 0 1 00 0 0 -1 0 00 0 0 0 0 1
UB VB B UA VA A0 0 -1 0 0 -1 UB0 20 0 0 -20 0 VB-1 0 400 1 0 100 B0 0 1 0 0 1 UA0 -20 0 0 20 0 VA-1 0 200 1 0 400 A
[K g1 ] =
B. ELEMEN BC (2)
AEL 0 0
−AEL 0 0
0
12EI
L36 EI
L2 0
−12EIL3
6 EI
L2
0
6 EI
L24 EIL 0
−6 EIL2
EIL
[K L2 ] =
−AEL 0 0
AEL 0 0
0
−12EIL3
−6 EIL2 0
12EI
L3−6 EIL2
0
6 EI
L22EIL 0
−6 EIL2
4 EIL
0,1 .200800 0 0
−0,1 .200800 0 0
0
12.200 .500
80036 .200 .500
8002 0
−12.200 .5008003
6 .200 .500
8002
0
6 .2000 .500
80024 .200 .500800 0
−6 .200 .5008002
200 .500800
[K L2 ] =
−0,1 .200800 0 0
0,1 .200800 0 0
0
−12.200 .5008003
−6 .200 .5008002 0
12.200 .500
8003−6 .200 .500
8002
0
6 .200 .500
80022.200 .500800 0
−6 .200 .5008002
4 .200 .500800
Cos 0 Sin 0 0 0 0 0
i = B
j = c
α = 0
-Sin 0 Cos 0 0 0 0 0
0 0 1 0 0 0
[ T ] = 0 0 0 Cos 0 Sin 0 0
0 0 0 - Sin 0 Cos 0 0
0 0 0 0 0 1
1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 0
[ T ] T= 0 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1
[K g2 ] = [ T ] T . [K L
2 ] . [ T ]
[K g2 ] =
X
[K g2 ]=
1 0 0 0 0 0 25 0 0 -25 0 00 1 0 0 0 0 0 0.002 1 0 -0.002 1
0 0 1 0 0 0 x 0 1 500 0 -1 1250 0 0 1 0 0 -25 0 0 25 0 00 0 0 0 1 0 0 -0.002 -1 0 0.002 -10 0 0 0 0 1 0 1 250 0 -1 500
1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1
UB VB B UC VC C25 0 0 -25 0 0 UB0 0 1 0 0 1 VB0 1 500 0 -1 125 B-25 0 0 25 0 0 UC0 0 -1 0 0 -1 VC0 1 250 0 -1 500 C
C. ELEMEN CD (3)
sinα=COCD
sin 45= 9CD
CD= 9sin 45
=12,728m
AEL 0 0
−AEL 0 0
0
12EI
L36 EI
L2 0
−12EIL3
6 EI
L2
0
6 EI
L24 EIL 0
−6 EIL2
EIL
[K L3 ] =
−AEL 0 0
AEL 0 0
0
−12EIL3
−6 EIL2 0
12EI
L3−6 EIL2
i = C
j = D
α = 315
0
6 EI
L22EIL 0
−6 EIL2
4 EIL
0,1 .2001272,8 0 0
−0,1 .2001272,8 0 0
0
12.200 .500
1272,836 .200 .500
1272,82 0
−12.200 .5001272,83
6 .200 .500
1272,82
0
6 .2000 .500
1272,824 .200 .5001272,8 0
−6 .200 .5001272,82
200 .5001272,8
[K L2 ] =
−0,1 .2001272,8 0 0
0,1 .2001272,8 0 0
0
−12.200 .5001272,83
−6 .200 .5001272,82 0
12.200 .500
1272,83−6 .200 .5001272,82
0
6 .200 .500
1272,822.200 .5001272,8 0
−6 .2000.5001272,82
4 .200 .5001272,8
[ T ] T=
[K g3 ] = [ T ] T . [K L
3 ] . [ T ]
Cos 315 Sin 315 0 0 0 0-Sin 315 Cos 315 0 0 0 0
0 0 1 0 0 0[ T ] = 0 0 0 Cos 315 Sin 315 0
0 0 0 - Sin 315 Cos 315 00 0 0 0 0 1
0.707107 0.707107 0 0 0 0-0.707107 0.707107 0 0 0 0
0 0 1 0 0 00 0 0 0.707107 0.707107 00 0 0 -0.707107 0.707107 00 0 0 0 0 1
0.707 -0.707 0 0 0 00.707 0.707 0 0 0 0
0 0 1 0 0 00 0 0 0.707 -0.707 00 0 0 0.707 0.707 00 0 0 0 0 1
x
[K g3 ]=
16 0 0 -16 0 0 0.7071 0.7071 0 0 0 0
0 0.001 0 0 -0.001 0-
0.7071 0.7071 0 0 0 0
0 0 314 0 0 79 x 0 0 1 0 0 0-16 0 0 16 0 0 0 0 0 0.7071 0.7071 00 -0.001 0 0 0.001 0 0 0 0 -0.7071 0.7071 00 0 157 0 0 314 0 0 0 0 0 1
[K g3 ]=
D. GLOBAL
UA VA ѲA UB VB ѲB UC VC ѲC UD VD ѲD0 0 1 0 0 1 0 0 0 0 0 0 UA0 20 0 0 -20 0 0 0 0 0 0 0 VA1 0 400 -1 0 200 0 0 0 0 0 0 ѲA0 0 -1 25 0 -1 -25 0 0 0 0 0 UB0 -20 0 0 20 1 0 0 1 0 0 0 VB1 0 100 -1 1 900 0 -1 125 0 0 0 ѲB0 0 0 -25 0 0 33 8 0 -8 -8 0 UC0 0 0 0 0 -1 8 8 -1 -8 -8 0 VC0 0 0 0 1 250 0 -1 814 0 0 79 ѲC
UC VC C UD VD D8 8 0 -8 -8 0 UC8 8 0 -8 -8 0 VC0 0 314 0 0 79 c-8 -8 0 8 8 0 UD-8 -8 0 8 8 0 VD0 0 157 0 0 314 D
0 0 0 0 0 0 -8 -8 0 8 8 0 UD0 0 0 0 0 0 -8 -8 0 8 8 0 VD0 0 0 0 0 0 0 0 157 0 0 314 ѲD
KONDISI BATAS
[F ] = [ K ]. [U ]
{F AGA
M A
FBGB
M B
FCGCMC
FDGDMD
} {U A
V A
θ AU B
V BθBUC
V CθCUD
V D
θD
}RE-ARRANGEMENT
UB VB ѲB UC VC ѲC UA VA ѲA UD VD ѲD0.9 25 0 -1 -25 0 0 0 0 -1 0 0 0 UB-68 0 20 1 0 0 1 0 -20 0 0 0 0 VB
-89.8567 -1 1 900 0 -1 125 1 0 100 0 0 0 ѲB0 -25 0 0 33 8 0 0 0 0 -8 -8 0 UC
-68 0 0 -1 8 8 -1 0 0 0 -8 -8 0 VC90.6667 0 1 250 0 -1 814 0 0 0 0 0 79 ѲCRHA+8.1 0 0 1 0 0 0 0 0 1 0 0 0 0
RVA 0 -20 0 0 0 0 0 20 0 0 0 0 0MA-7.29 -1 0 200 0 0 0 1 0 400 0 0 0 0
RHD 0 0 0 -8 -8 0 0 0 0 8 8 0 0RVD 0 0 0 -8 -8 0 0 0 0 8 8 0 0MD 0 0 0 0 0 157 0 0 0 0 0 314 0
REDUKSI
0.9 25 0 -1 -25 0 0 UB-68 0 20 1 0 0 1 VB
-89.8567 -1 1 900 0 -1 125 ѲB0 -25 0 0 33 8 0 UC
-68 0 0 -1 8 8 -1 VC90.6667 0 1 250 0 -1 814 ѲC
INVERS
UB 226 0 0 226 -226 0 0.9VB 0 0 0 0 0 0 -68ѲB 0 0 0 0 0 0 -89.857UC 226 0 0 226 -226 0 0VC -226 0 0 -226 226 0 -68ѲC 0 0 0 0 0 0 90.667
RHA+8.1 0 0 1 0 0 0RVA 0 -20 0 0 0 0
MA-7.29 -1 0 200 0 0 0RHD 0 0 0 -8 -8 0RVD 0 0 0 -8 -8 0MD 0 0 0 0 0 157
=
RAH = -21,69 + 8,1 = -13,59
RAV = 93,85
-21.6993.85
-10340.1220.7942.15
-9122.90
MA = -10340,12- 7,29 = -10347,41
RHD = 20,79
RVD = 42.15
MD = -9122,90
CROSS CHECK
V = RAV + RDV – (q.l)Ʃ
= 93,85+ 42.15– (17 . 8)
= 0
H = RAH +RDH + PƩ
= -13,59+20,79+9
= 0
SOAL: SISTEM RANGKA BATANG QUIZ
Nama : Viorenza Everlyn
Npm : 123 110 198
P1 = 10 TON
P2 = 5 TON
P3 = 2 TON
L = 10 m
= 60α
Tentukan Kekakuan elemen Metode Matrix !
Penyelesain:
ELEMEN AB (1) = ELEMEN CD (5)
[KL1] = AE10 [ 1 0 −1 00 0 0 0
−1 0 1 00 0 0 0
] [T1] = [ cos60 sin 60 0 0
−sin 60 cos60 0 00 0 cos60 sin600 0 −sin 60 cos 60
]ELEMEN AB (1)
[KG1] = [T1]T . [KL1] . [T1]
0.5 -0.866 0 0 1 0 -1 0 0.5 0.866 0 00.866 0.5 0 0 AE 0 0 0 0 X -0.866 0.5 0 00 0 0.5 -0.866 10 -1 0 1 0 0 0 0.5 0.8660 0 0.866 0.5 0 0 0 0 0 0 -0.866 0.5
0.5 0.866025 0 0-0.866025 0.5 0 0
0 0 0.5 0.8660250 0 -0.86603 0.5
UA VA UB VB0.25 0.433 -0.25 -0.433 UA0.433 0.75 -0.433 -0.75 VA-0.25 -0.433 0.25 0.433 UB-0.433 -0.75 0.433 0.75 VB
ELEMEN CD (5)
[KG5] = [T5]T . [KL5] . [T5]
0.5 -0.866 0 0 1 0 -1 0 0.5 0.866 0 00.866 0.5 0 0 AE 0 0 0 0 X -0.866 0.5 0 00 0 0.5 -0.866 10 -1 0 1 0 0 0 0.5 0.8660 0 0.866 0.5 0 0 0 0 0 0 -0.866 0.5
UC VC UD VD0.25 0.433 -0.25 -0.433 UC0.433 0.75 -0.433 -0.75 VC-0.25 -0.433 0.25 0.433 UD-0.433 -0.75 0.433 0.75 VD
ELEMEN AC(2) = ELEMEN BD(4) = ELEMEN CE (6)
i = A , B , C
j = C , D , E
α = 0
[KL2] = AE10 [ 1 0 −1 00 0 0 0
−1 0 1 00 0 0 0
] [T2] = [ cos0 sin 0 0 0
−sin 0 cos0 0 00 0 cos0 sin 00 0 −sin 0 cos0
]=¿
Jadi [KL2] =
Jadi [KL4]=
Jadi [KL6] =
ELEMEN BC (3) = ELEMEN DE (7)
i = C , E
j = B , D
α = 300
1 0 0 00 1 0 00 0 1 00 0 0 1
UA VA UC VC1 0 -1 0 UA0 0 0 0 VA-1 0 1 0 UC0 0 0 0 VC
UB VB UD VD1 0 -1 0 UB0 0 0 0 VB-1 0 1 0 UD0 0 0 0 VD
UC VC UE VE1 0 -1 0 UC0 0 0 0 VC-1 0 1 0 UE0 0 0 0 VE
[KL2] = AE10 [ 1 0 −1 00 0 0 0
−1 0 1 00 0 0 0
] [T2] =
ELEMEN BC (3)
[KG3] = [T3]T . [KL3] . [T3]
0.5 0.866 0 0 1 0 -1 0 0.5 -0.866 0 0-0.866 0.5 0 0 AE 0 0 0 0 X 0.866 0.5 0 0
0 0 0.5 0.866 10 -1 0 1 0 0 0 0.5 -0.8660 0 -0.866 0.5 0 0 0 0 0 0 0.866 0.5
UC VC UB VB0.25 -0.433 -0.25 0.433 UC-0.433 0.75 0.433 -0.75 VC-0.25 0.433 0.25 -0.433 UB0.433 -0.75 -0.433 0.75 VB
ELEMEN DE (7)
[KG7] = [T7]T . [KL7] . [T7]
0.5 0.866 0 0 1 0 -1 0 0.5 -0.866 0 0-0.866 0.5 0 0 AE 0 0 0 0 X 0.866 0.5 0 0
0 0 0.5 0.866 10 -1 0 1 0 0 0 0.5 -0.8660 0 -0.866 0.5 0 0 0 0 0 0 0.866 0.5
UE VE UD VD0.25 -0.433 -0.25 0.433 UE-0.433 0.75 0.433 -0.75 VE-0.25 0.433 0.25 -0.433 UD0.433 -0.75 -0.433 0.75 VD
0.5 -0.866 0 00.866 0.5 0 00 0 0.5 -0.8660 0 0.866 0.5
GLOBAL
UA VA UB VB UC VC UD VD UE VE1.25 0.43 -0.25 -0.43 -1 0 0 0 0 0 UA0.43 0.75 -0.43 -0.75 0 0 0 0 0 0 VA-0.25 -0.43 1.50 0.00 -0.25 0.43 -1 0 0 0 UB-0.43 -0.75 0.00 1.50 0.43 -0.75 0 0 0 0 VB-1 0 -0.25 0.43 2.50 0.00 -0.25 -0.43 -1 0 UC0 0 0.43 -0.75 0.00 1.50 -0.43 -0.75 0 0 VC0 0 -1 0 -0.25 -0.43 1.50 0.00 -0.25 0.43 UD0 0 0 0 -0.43 -0.75 0.00 1.50 0.43 -0.75 VD0 0 0 0 -1 0 -0.25 0.43 0.25 -0.43 UE0 0 0 0 0 0 0.43 -0.75 -0.43 0.75 VE
KONDISI BATAS
[F ] = [ K ]. [U ]
{F AGA
FBGB
FCGCFDGDF EGE
} {RHARVA
P=10 ton00
P=−2 ton0
P=5 ton0RVE
} {U A
V A
U B
V BUC
V CUD
V D
U E
FE
}RE-ARRANGEMENT
UB VB UC VC UD VD UE UA VA VE10 1.50 0 -0.25 0.43 -1 0 0 -0.25 -0.43 0 UB0 0 1.50 0.43 -0.75 0 0 0 -0.43 -0.75 0 VB0 -0.25 0.43 2.50 0 -0.25 -0.43 -1 -1 0 0 UC-2 0.43 -0.75 0 1.50 -0.43 -0.75 0 0 0 0 VC0 AE -1.00 0 -0.25 -0.43 1.50 0 -0.3 0 0 0.43 UD5 10 0 0 -0.43 -0.75 0 1.50 0.4 0 0 -0.75 VD0 0 0 -1 0 -0.25 0.43 0.25 0 0 -0.43 UE
RHA -0.25 -0.43 -1 0 0 0 0 1.25 0.43 0 UARVA -0.43 -0.75 0 0 0 0 0 0.43 0.75 0 VARVE 0 0 0 0 0 -1 0 0 0 0.75 VE
REDUKSI MATRIX
10 1.50 0 -0.25 0.43 -1 0 0 UB0 0 1.50 0.43 -0.75 0 0 0 VB0 AE -0.25 0.43 2.50 0 -0.25 -0.43 -1 UC-2 10 0.43 -0.75 0 1.50 -0.43 -0.75 0 VC0 -1 0 -0.25 -0.43 1.50 0 -0.25 UD5 0 0 -0.43 -0.75 0 1.50 0.43 VD0 0 0 -1 0 -0.25 0.43 0.25 UE
INVERS
UB 1.875 -0.505 0.750 -0.577 1.375 -0.361 1 10VB -0.505 1.292 -0.433 1 -0.217 0.542 -0.577 0UC 10 0.750 -0.433 1 -0.289 0.750 -0.144 1 0VC AE -0.577 1 -0.289 1.833 0 1 -0.577 -2UD 1.375 -0.217 0.750 0 1.875 -0.072 1 0VD -0.361 0.542 -0.144 1 -0.072 1.292 -0.577 5UE 1 -0.577 1 -0.577 1 -0.577 2 0
UA -0.25 -0.43 -1 0 0 00
VA AE -0.43 -0.75 0 0 0 0 0VE 10 0 0 0 0 0.43 -0.75 -0.43
-10
=-
2.080199.08018
9
CROSS CHECK
ΣV= 0 H= 0Σ
RVA + RVE –P2 – P3 = 0 RHA + P1 = 0
-2.08019 + 9.080189 – 5 – 2 = 0 -10 + 10 = 0
0 = 0 OK! 0 = 0 OK!
GAYA BATANG
ELEMEN 1 i = A j = B
{FAGAFBGB
} = [T1] [KG1] {UAVAUBVB
}{FAGAFBGB
} = [ 0,5 0,87 0 0−0,87 0,5 0 00 0 0,5 0,870 0 −0,87 0,5
] AE5 [ 0,25 0,435 −0,25 −0,4350,435 0,757 −0,435 −0,757−0,25 −0,435 0,25 0,435−0,435 −0,757 0,435 0,757
] 5AE { 0
021,709−9,760
}¿
ELEMEN 2i = Aj = C
-2.4002.400
{FAGAFCGC
} = [T2] [KG2] {UAVAUCVC
}{FAGAFCGC
} = [1 0 0 00 1 0 00 0 1 00 0 0 1
] AE5 [ 1 0 −1 00 0 0 0
−1 0 1 00 0 0 0
] 5AE { 00
8,799−14,440
}=
ELEMEN 3i = Cj = B
{FCGCFBGB
} = [T3] [KG3] {UCVCUBVB
}{FBGBFCGC
} = [ 0,5 −0,87 0 00,87 0,5 0 00 0 0,5 −0,870 0 0,87 0,5
] AE5
[ 0,25 −0,435 −0,25 0,435−0,435 0,757 0,435 −0,757−0,25 0,435 0,25 −0,4350,435 −0,757 −0,435 0,757
] 5AE { 8,799−14,44021.709−9.760
}= 2.400-2.400
ELEMEN 4 i = B j = D
{FBGBFDGD
} = [T4] [KG4] {UBVBUDVD
}{FBGBFDGD
} = [1 0 0 00 1 0 00 0 1 00 0 0 1
] AE5 [ 1 0 −1 00 0 0 0
−1 0 1 00 0 0 0
] 5AE { 21.709−9.76014.111
−12.067} =
ELEMEN 5i = Cj = D
{FCGCFDGD
} = [T5] [KG5] {UCVCUDVD
}{FCGCFDGD
} = [ 0,5 0,87 0 0−0,87 0,5 0 00 0 0,5 0,870 0 −0,87 0,5
] AE5 [ 0,25 0,435 −0,25 −0,4350,435 0,757 −0,435 −0,757−0,25 −0,435 0,25 0,435−0,435 −0,757 0,435 0,757
] 5AE= -4.7104.710
ELEMEN 6i = C j = E
{FCGCFE¿
} = [T6] [KG6] {UCVCUEVE
}{FCGCFE
¿} = [1 0 0 0
0 1 0 00 0 1 00 0 0 1
] AE5 [ 1 0 −1 00 0 0 0
−1 0 1 00 0 0 0
] 5AE =
ELEMEN 7i = Dj = E
{FDGDFE¿
} = [T7] [KG7] {UDVDUEVE
}{FDGDFE
¿} = [ 0,5 −0,87 0 0
0,87 0,5 0 00 0 0,5 −0,870 0 0,87 0,5
] AE5 [ 0,25 −0,435 −0,25 0,435−0,435 0,757 0,435 −0,757−0,25 0,435 0,25 −0,4350,435 −0,757 −0,435 0,757
] 5AE =
-5.2405.240
18.350-18.350
