Tugas Analisis Struktur V MEKTEK Sloop D
SOAL: SISTEM PORTAL 2 DIMENSI QUIZ
Diketahui :
A : 0.1 cm²
E : 200000 kg/cm² = 200 t/cm²
: 500
I cm
α : 45 °
L
1 : (x+y) = 10 m
L
2 : (z) = 8 m
L
3 : (x) = 1 m
q : (y+z) = 17 t/m
P : (x+z) = 9 ton
Tentukan Kekakuan elemen Metode Matrix
!
Penyelesaian :
A. ELEMEN AB(1)
AE
L
0
0
[
K L1
− AE
L
]=
0
0
0,1. 200
1000
0
0
[ KL
]=
1
−0,1 . 200
1000
0
0
0
12 EI
L3
6 EI
L2
0
−12 EI
3
L
6 EI
L2
0
− AE
L
0
6 EI
L2
4 EI
L
0
0
AE
L
0
−6 EI
2
L
2 EI
L
0
0
0
12 . 200 .500
10003
6 . 200 . 500
2
1000
6 . 200 . 500
10002
4 . 200 .500
1000
0
0
−12 .200 . 500
3
1000
6 . 2000 . 500
10002
−6 . 200 .500
2
1000
2 . 200 .500
1000
0
−12 EI
L3
−6 EI
L2
0
12 EI
3
L
−6 EI
L2
−0,1 . 200
1000
0
6 EI
L2
EI
L
0
−6 EI
2
L
4 EI
L
0
0
0
−12 .200 . 500
10003
−6 . 200 .500
2
1000
6 . 200 . 500
10002
200 . 500
1000
0,1. 200
1000
0
0
0
0
0
12 . 200 .500
3
1000
−6 . 200 .500
1000 2
−6 . 200 .500
2
1000
4 . 2000 .500
1000
Cos27
0
Sin27
0
Sin
270
0
0
0
0
Cos27
0
0
0
0
0
0
0
1
0
0
0
0
0
0
Cos 270
Sin 270
0
0
0
0
- Sin270
Cos 270
0
0
0
0
0
0
1
[T]=
[T]
T
=
0
-1
0
1
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
-1
0
1
0
0
0
0
1
[ K g1 ] = [ T ] T . [
KL
1
].[T]
0
1
0
0
0
0
20
0
0
1
0
0
0
0
0
0
0.001
0
0
0
0
1
0
0
0
0
-1
0
0
0
-20
0
0
0
1
0
0
0
3
0
0.001
3
40
0
0
0
0
0
0
0
1
0
3
0
0
0
0
-1
0
0
0
0
1
0
0
0
0
0
0
0
1
[Kg ] =
1
x
0
-1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
x
-3
20
0
20
0
0
0.001
0
20
-3
0
0
0.001
0
-3
0
3
10
0
0
-3
40
0
UB
0
0
-1
0
0
-1
[ K g1 ]
B.
AE
L
]=
− AE
L
0
0
0
0
0,1. 200
800
0
0
0
0
−12 EI
3
L
6 EI
L2
0
12 . 200 .500
800 3
6 . 2000 . 500
2
800
UA
0
0
1
0
0
1
− AE
L
6 EI
2
L
4 EI
L
[
2
0
B
-1
0
400
1
0
200
12 EI
3
L
6 EI
L2
0
KL
VB
0
20
0
0
-20
0
0
0
AE
L
−6 EI
2
L
2 EI
L
0
0
0
6 . 200 . 500
8002
4 . 200 .500
800
VA
0
-20
0
0
20
0
0
−12 EI
3
L
−6 EI
L2
0
12 EI
3
L
−6 EI
L2
−0,1 . 200
800
0
0
A
-1
0
100
1
0
400
UB
VB
B
UA
VA
A
=
ELEMEN
BC (2)
0
6 EI
2
L
EI
L
0
−6 EI
2
L
4 EI
L
0
−12 .200 . 500
8003
−6 . 200 .500
2
800
0
6 . 200 . 500
8002
200 . 500
800
[ K L2
]=
−0,1 . 200
800
0
0
−12 .200 . 500
8003
6 . 200 . 500
8002
0
0
−6 . 200 .500
8002
2 . 200 .500
800
0,1. 200
800
0
0
12 . 200 .500
800 3
−6 . 200 .500
8002
0
Cos 0
-Sin 0
0
Sin 0
Cos 0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Cos 0
- Sin 0
0
Sin 0
Cos 0
0
0
0
1
[T]
=
[T]
T
=
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
1
[ K g2 ] = [ T ] T . [
KL
2
−6 . 200 .500
8002
4 . 200 .500
800
].[T]
1
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
1
[ K g2 ] =
0
25
x
0
0
-25
0
0
0
0.00
2
1
0
0.00
2
1
0
-25
0
0
25
0
0.002
-1
0
1
500
0
-1
250
0
1
125
0
0
0
0.002
-1
-1
500
1
0
0
0
0
0
X
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
UB
25
0
0
-25
0
0
[Kg ] =
2
C. ELEMEN CD (3)
sin α =
CO
CD
sin 45=
9
CD
VB
0
0
1
0
0
1
0
0
0
0
0
1
B
0
1
500
0
-1
250
UC
-25
0
0
25
0
0
VC
0
0
-1
0
0
-1
C
0
1
125
0
-1
500
UB
VB
B
UC
VC
C
CD=
9
=12,728 m
sin 45
AE
L
0
0
[
K L3
− AE
L
]=
0
0
0,1. 200
1272,8
0
−0,1 . 200
1272,8
0
0
[T]=
12 EI
L3
6 EI
L2
0
−12 EI
L3
6 EI
2
L
0
12 . 200 .500
1272,83
6 . 2000 . 500
1272,82
0
[ K L2
]=
0
Cos
315
-Sin
315
0
0
0
0
0
−12 .200 . 500
1272,83
6 . 200 . 500
2
1272,8
Sin
315
Cos
315
0
0
0
0
− AE
L
0
6 EI
L2
4 EI
L
0
−12 EI
L3
−6 EI
L2
0
0
AE
L
0
−6 EI
L2
2 EI
L
0
6 EI
L2
EI
L
0
0
−6 EI
L2
4 EI
L
12 EI
L3
−6 EI
2
L
0
0
−0,1 . 200
1272,8
0
6 . 200 . 500
1272,82
4 . 200 .500
1272,8
0
0
0
−12 .200 . 500
1272,83
−6 . 200 .500
1272,8 2
0,1. 200
1272,8
0
−6 . 200 .500
1272,8 2
2 . 200 .500
1272,8
0
12 . 200 .500
1272,83
−6 . 2000 .500
2
1272,8
0
0
0
0
0
0
0
1
0
0
0
0
0
Cos 315
- Sin 315
0
0
0
Sin 315
Cos 315
0
0
0
0
0
1
0
6 . 200 . 500
1272,82
200 . 500
1272,8
0
−6 . 200 .500
1272,8 2
4 . 200 .500
1272,8
[T]
T
[T]
0.70710
7
0.70710
7
0
0.70710
7
0
0
0
0
0.70710
7
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0.707107
0.707107
0
0
0
0.70710
7
0.70710
7
0
0.707
x0.707
0
0
0
0
-0.707
i=B
0.707
j=A
0
0
0
-16
0
0
0.00
1
0
0
314
0
0.00
1
0
[ K g3 ]
0
0
1
0
0
α = 270
0
0
0
0
16
0
0
=
0
0
1
[ K g3 ] =
. [ K L3 ] . [ T ]
[ K g3 ] =
-16
T
0
0
0
i=A,B,C
0
0
0
i=C,Ei=B j=C,D,E
i=C
0
0
0
j
=
B
0.707
0 , Dj = c
j = D -0.707
α = 0
0.707
0.707
0
α = 300α = 0
0α = 315
0
1
0.707
1
0
0
0
0
0.707
1
0
0
1
0
0
0.707
1
0.707
1
0
0
0
0.707
1
0
0
0
0
0
0.00
1
0
0
79
0
16
0
0
0
0
0
0
157
0
0
0.00
1
0
0
314
0
0
0
0
0
0
C
0
0
314
0
0
157
UD
-8
-8
0
8
8
0
UC
8
8
0
-8
-8
0
VC
8
8
0
-8
-8
0
0
0.70
71
0.70
71
0
x
VD
-8
-8
0
8
8
0
D
0
0
79
0
0
314
UC
VC
c
UD
VD
D
=
0.707
1
0
0
0
1
D. GLOBAL
UA
VA
ѲA
UB
VB
ѲB
UC
VC
ѲC
UD
VD
ѲD
0
0
1
0
0
1
0
0
0
0
0
0
0
20
0
0
-20
0
0
0
0
0
0
0
1
0
400
-1
0
100
0
0
0
0
0
0
0
0
-1
25
0
-1
-25
0
0
0
0
0
0
-20
0
0
20
1
0
0
1
0
0
0
1
0
200
-1
1
900
0
-1
250
0
0
0
0
0
0
-25
0
0
33
8
0
-8
-8
0
0
0
0
0
0
-1
8
8
-1
-8
-8
0
0
0
0
0
1
125
0
-1
814
0
0
157
0
0
0
0
0
0
-8
-8
0
8
8
0
0
0
0
0
0
0
-8
-8
0
8
8
0
0
0
0
0
0
0
0
0
79
0
0
314
KONDISI BATAS
UA
VA
ѲA
UB
VB
ѲB
UC
VC
ѲC
UD
VD
ѲD
[F]
{}
= [ K ]. [ U
FA
GA
MA
FB
GB
MB
FC
GC
MC
FD
GD
MD
RE-ARRANGEMENT
]
{}
UA
VA
θA
UB
VB
θB
UC
VC
θC
UD
VD
θD
0.9
-68
89.856
7
0
-68
90.666
7
RHA+8
.1
RVA
MA7.29
RHD
RVD
MD
UB
25
0
VB
0
20
ѲB
-1
1
UC
-25
0
VC
0
0
ѲC
0
1
UA
0
0
VA
0
-20
ѲA
-1
0
UD
0
0
VD
0
0
ѲD
0
0
UB
VB
-1
1
900
0
-1
125
1
0
100
0
0
0
ѲB
-25
0
0
0
0
-1
33
8
8
8
0
-1
0
0
0
0
0
0
-8
-8
-8
-8
0
0
UC
VC
0
1
250
0
-1
814
0
0
0
0
0
79
ѲC
0
0
1
0
0
0
0
0
1
0
0
0
0
-20
0
0
0
0
0
20
0
0
0
0
0
200
0
0
0
1
0
400
0
0
0
0
0
0
0
0
0
-8
-8
0
-8
-8
0
0
0
157
0
0
0
0
0
0
0
0
0
8
8
0
8
8
0
0
0
314
-1
0
0
0
REDUKSI
0.9
-68
89.8567
0
-68
90.6667
25
0
0
20
-1
1
-25
0
0
0
0
1
UB
VB
-1
-25
0
0
1
0
0
1
900
0
-1
250
0
33
8
0
-1
8
8
-1
125
0
-1
814
ѲB
UC
VC
ѲC
INVERS
UB
VB
ѲB
UC
VC
ѲC
226
0
0
226
-226
0
0
0
0
0
0
0
0
0
0
0
0
0
226
0
0
226
-226
0
-226
0
0
-226
226
0
0
0
0
0
0
0
0.9
-68
-89.857
0
-68
90.667
0
0
0
0
0
0
RHA+8.
1
RVA
0
0
0
-20
MA-7.29
RHD
RVD
-1
0
0
MD
0
=
0
0
0
0
0
0
0
0
0
1
0
20
0
0
0
0
-8
-8
0
-8
-8
0
0
0
0
0
0
0
15
7
-21.69
93.85
10340.1
2
20.79
42.15
9122.90
RAH = -21,69 + 8,1 = -13,59
RAV = 93,85
MA
= -10340,12- 7,29 = -10347,41
RHD = 20,79
RVD = 42.15
MD = -9122,90
CROSS CHECK
ƩV = RAV + RDV – (q.l)
= 93,85+ 42.15– (17 . 8)
=0
ƩH = RAH +RDH + P
= -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
α = 60
Tentukan Kekakuan elemen Metode Matrix !
Penyelesain:
ELEMEN AB (1) = ELEMEN CD (5)
[
1
AE 0
10 −1
0
[KL1] =
[T
[
] =
1
0 −1 0
0 0 0
0 1 0
0 0 0
]
cos 60 sin 60
0
0
−sin 60 cos 60
0
0
0
0
cos 60 sin 60
0
0
−sin 60 cos 60
]
ELEMEN AB (1)
[KG1]
= [T1]T . [KL1] . [T1]
0.5
0.866
0
0
0.8
66
0.5
0
0
0
0
0
0.5
0.8
66
0
0.86
6
VA
UB
0.25
0.433
0.433
0.75
0.433
-0.75
-0.25
0.433
-0.25
-0.433
0.25
0.433
0
0.5
0
0
0
0
0
0.5
0.866
03
0
0.8660
25
0
0
0.5
0
0
0.5
0.866
-1
0
AE
0
0
0
0
10
-1
0
1
0
0
0
0
0
0
0
0
0
VB
0.433
UA
-0.75
VA
0.433
0.75
UB
VB
= [T5]T . [KL5] . [T5]
X
0.5
0.86
6
0.5
0.8
66
0
ELEMEN CD (5)
[KG5]
0
1
0.5
UA
0.8660
25
0.5
0.8660
25
0.86
6
0.5
0.5
0.866
0
0
0.8
66
0.5
0
0
0
0
0
0.5
0.8
66
0
0.86
6
0
-1
0
AE
0
0
0
0
10
-1
0
1
0
0
0
0
0
0.5
UC
VC
UD
0.25
0.433
0.433
0.75
0.433
-0.75
-0.25
0.433
-0.25
-0.433
1
0.25
0.433
VD
0.433
UC
-0.75
VC
0.433
0.75
UD
VD
X
ELEMEN AC(2) = ELEMEN BD(4) = ELEMEN CE (6)
[KL2] =
[
1
AE 0
10 −1
0
0 −1 0
0 0 0
0 1 0
0 0 0
]
0.8
66
0
0
0.5
0
0
0
0
0
0
0.5
0.866
0.5
0.86
6
0.86
6
0.5
[T2]
[
=
Jadi [KL ]
2
Jadi
Jadi [KL6]
]
1
0
0
0
cos 0 sin 0
0
0
−sin 0 cos 0
0
0
=¿
0
0
cos 0 sin 0
0
0
−sin 0 cos 0
UA
1
0
-1
0
VA
0
0
0
0
UC
-1
0
1
0
VC
0
0
0
0
UA
VA
UC
VC
UB
1
0
-1
0
VB
0
0
0
0
UD
-1
0
1
0
VD
0
0
0
0
UB
VB
UD
VD
UC
1
0
-1
0
VC
0
0
0
0
UE
-1
0
1
0
VE
0
0
0
0
UC
VC
UE
VE
ELEMEN BC (3) = ELEMEN DE (7)
0
1
0
0
=
[KL4]=
=
0
0
1
0
0
0
0
1
[KL2] =
[T2]
[
1
AE 0
10 −1
0
0 −1 0
0 0 0
0 1 0
0 0 0
]
0.5
0.866
0
0
=
0.866
0.5
0
0
0
0
0.5
0.866
0
0
-0.866
0.5
ELEMEN BC (3)
[KG3]
0.5
0.86
6
= [T3]T . [KL3] . [T3]
0.8
66
0
0
0.5
0
0
0.8
66
0
0
0
0
UC
0.5
0.86
6
0.25
-0.433
VC
0.433
0.75
-0.25
0.433
0.433
-0.75
0
0
0.8
66
0.5
0
1
0
-1
0
A
E
0
0
0
0
1
0
-1
0
1
0
0
0
0.5
0
0.86
6
0
0
0
0
0
0
0.8
66
0.5
0.5
0.8
0.86
6
0.5
0
0
0
0
0.5
UB
VB
-0.25
0.433
0.433
-0.75
0.433
UC
VC
0.75
VB
0.25
0.433
0.5
0.86
6
X
UB
ELEMEN DE (7)
[KG7]
0.5
-
= [T7]T . [KL7] . [T7]
0.8
66
0.5
0
0
0
0
A
1
0
0
0
-1
0
0
0
X
0.86
6
66
E
0
0
0
0
UE
0.5
0.86
6
0.25
-0.433
VE
0.433
0.75
-0.25
0.433
0.433
-0.75
0.8
66
1
0
-1
0
1
0
0
0
0.5
0.86
6
0
0
0
0
0
0
0.8
66
0.5
0.5
UD
VD
-0.25
0.433
0.433
-0.75
0.433
UD
0.75
VD
0.25
0.433
UE
VE
GLOBAL
UA
1.25
0.43
-0.25
-0.43
-1
0
0
0
0
0
VA
0.43
0.75
-0.43
-0.75
0
0
0
0
0
0
UB
-0.25
-0.43
1.50
0.00
-0.25
0.43
-1
0
0
0
VB
-0.43
-0.75
0.00
1.50
0.43
-0.75
0
0
0
0
UC
-1
0
-0.25
0.43
2.50
0.00
-0.25
-0.43
-1
0
VC
0
0
0.43
-0.75
0.00
1.50
-0.43
-0.75
0
0
UD
0
0
-1
0
-0.25
-0.43
1.50
0.00
-0.25
0.43
VD
0
0
0
0
-0.43
-0.75
0.00
1.50
0.43
-0.75
UE
0
0
0
0
-1
0
-0.25
0.43
0.25
-0.43
VE
0
0
0
0
0
0
0.43
-0.75
-0.43
0.75
UA
VA
UB
VB
UC
VC
UD
VD
UE
VE
KONDISI BATAS
[F]
= [ K ]. [ U
]
{} { } {}
FA
GA
FB
GB
FC
GC
FD
GD
FE
GE
RHA
RVA
P=10ton
0
0
P=−2ton
0
P=5ton
0
RVE
UA
VA
UB
VB
UC
VC
UD
VD
UE
FE
RE-ARRANGEMENT
UB
VB
10
1.50
0
UC
0.2
5
0
0
1.5
0
0.4
3
0.4
3
0.7
5
0
-0.25
2.5
0
0
0.43
0.4
3
0.7
5
0
0.2
5
0.4
3
1.5
0
0.4
3
0.7
5
-2
VC
UD
VD
UE
VA
0.4
3
0.7
5
VE
0
UA
0.2
5
0.4
3
-1
0
0
0
0.2
5
0.4
3
0
0.4
3
0.7
5
-1
-1
0
1.5
0
0
0
UB
0
VB
0
0
UC
0
0
0
VC
-0.3
0
0
1.5
0
0.4
0
0
0.4
3
0.2
5
0
0
0.4
3
0.7
5
0.4
3
0
UA
0
0.7
5
VA
0
AE
-1.00
0
5
10
0
0
0
0
-1
0
RHA
-0.25
-1
0
0
0
0
1.2
5
0.4
3
RVA
-0.43
0
0.4
3
0.7
5
0
0.2
5
0
0
0
0
0
0.4
3
0.7
5
RVE
0
0
0
0
0
-1
0
0
0
-1
0
0
UB
0
0
0
VB
0.25
0.43
0.43
0.75
-1
UC
VC
1.50
0
0
1.50
0
0.25
0.43
REDUKSI MATRIX
1.5
0
10
0
0
-2
0
5
A
E
1
0
0
0.25
0
0.2
5
0.4
3
1.50
0.43
0.43
0.75
0.43
0.75
2.50
0
-1
0
0
0
0
0.25
-
1.50
0.43
-
UD
VD
UD
VD
UE
VE
0
0
0.43
0.75
-1
0
0
0.25
0.43
0.25
UE
INVERS
UB
1.875
VB
-0.505
UC
1
0
0.750
VC
A
E
-0.577
UD
1.375
VD
-0.361
UE
1
UA
A
E
1
0
VE
=
-10
-
1.29
2
0.43
3
1
0.21
7
0.54
2
0.57
7
-0.43
0.4
3
0.7
5
0
0
-0.25
VA
0.50
5
0.75
0
0.43
3
1
0.28
9
0.75
0
0.14
4
1
0.57
7
1.37
5
0.21
7
1
0.28
9
0.75
0
1.83
3
0
1.87
5
0.07
2
0
1
0.57
7
1
0.36
1
0.54
2
0.14
4
1
0.07
2
1.29
2
0.57
7
-1
0
0
0
0
0
0
0
0
0
0.4
3
0
0.7
5
0
0.4
3
1
0.57
7
10
1
0.57
7
0
0
-2
1
0.57
7
0
2
0
5
2.080
19
9.080
189
CROSS CHECK
ΣV= 0
ΣH= 0
RVA + RVE –P2 – P3
=0
RHA + P1
=0
-2.08019 + 9.080189 – 5 – 2
=0
-10 + 10
=0
OK!
0
OK!
0
=0
=0
GAYA BATANG
ELEMEN 1
i=A
j=B
{}
FA
GA
FB
GB
= [T1] [KG1]
{}
UA
VA
UB
VB
{} [
FA
GA
FB
GB
[
=
0,5 0,87
0
0
−0,87 0,5
0
0
0
0
0,5 0,87
0
0
−0,87 0,5
0,25
0,435 −0,25 −0,435
0,435
0,757 −0,435 −0,757
−0,25 −0,435
0,25
0,435
−0,435 −0,757 0,435
0,757
-2.40
0
2.40
0
]
AE
5
] { }
0
5
0
AE 21,709
−9,760
¿
ELEMEN 2
i=A
j=C
{}
{}
{} [ ] [
FA
GA
FC
GC
= [T2] [KG2]
UA
VA
UC
VC
FA
GA
FC
GC
1
0
0
0
0
0 AE
0 5
1
=
0
1
0
0
0
0
1
0
1
0
−1
0
ELEMEN 3
i=C
j=B
{}
{}
{} [
]
FC
GC
FB
GB
= [T3] [KG3]
FB
GB
FC
GC
[
=
UC
VC
UB
VB
0,5 −0,87
0
0
AE
0,87
0,5
0
0
0
0
0,5 −0,87 5
0
0
0,87
0,5
]{ }
0,25
−0,435 −0,25
0,435
8,799
−0,435 0,757
0,435 −0,757 5 −14,440
−0,25
0,435
0,25
−0,435 AE 21.709
0,435 −0,757 −0,435 0,757
−9.760
=
2.40
0
-2.40
0
]{ }
0 −1 0
0
0 0 0 5
0
0 1 0 AE 8,799
0 0 0
−14,440
=
ELEMEN 4
i=B
j=D
{}
{}
{} [ ] [
FB
GB
FD
GD
= [T4] [KG4]
UB
VB
UD
VD
FB
GB
FD
GD
1
0
0
0
0
0
1
0
0
0 AE
0 5
1
= [T5] [KG5]
{}
=
0
1
0
0
ELEMEN 5
i=C
j=D
{}
FC
GC
FD
GD
UC
VC
UD
VD
1
0
−1
0
]{ }
0 −1 0
21.709
0 0 0 5 −9.760
0 1 0 AE 14.111
0 0 0
−12.067
=
{} [
FC
GC
FD
GD
[
=
0,5 0,87
0
0
−0,87 0,5
0
0
0
0
0,5 0,87
0
0
−0,87 0,5
0,25
0,435 −0,25 −0,435
0,435
0,757 −0,435 −0,757
−0,25 −0,435
0,25
0,435
−0,435 −0,757 0,435
0,757
=
]
-4.71
0
4.71
0
ELEMEN 6
i=C
j=E
{}
FC
GC
FE
¿
= [T6] [KG6]
{}
UC
VC
UE
VE
5
AE
]
AE
5
{}
FC
GC
FE
¿
[
1
0
−1
0
=
[ ]
1
0
0
0
0
1
0
0
0
0
1
0
-5.24
0
5.24
0
0
0 AE
0 5
1
]
0 −1 0
0 0 0 5
0 1 0 AE
0 0 0
=
ELEMEN 7
i=D
j=E
{}
{}
FD
GD
FE
¿
FD
GD
FE
¿
[
= [T7] [KG7]
=
[
{}
UD
VD
UE
VE
]
0,5 −0,87
0
0
AE
0,87
0,5
0
0
0
0
0,5 −0,87 5
0
0
0,87
0,5
]
0,25
−0,435 −0,25
0,435
−0,435 0,757
0,435 −0,757 5
−0,25
0,435
0,25
−0,435 AE
0,435 −0,757 −0,435 0,757
18.3
5
0
18.3
5
0
=
Diketahui :
A : 0.1 cm²
E : 200000 kg/cm² = 200 t/cm²
: 500
I cm
α : 45 °
L
1 : (x+y) = 10 m
L
2 : (z) = 8 m
L
3 : (x) = 1 m
q : (y+z) = 17 t/m
P : (x+z) = 9 ton
Tentukan Kekakuan elemen Metode Matrix
!
Penyelesaian :
A. ELEMEN AB(1)
AE
L
0
0
[
K L1
− AE
L
]=
0
0
0,1. 200
1000
0
0
[ KL
]=
1
−0,1 . 200
1000
0
0
0
12 EI
L3
6 EI
L2
0
−12 EI
3
L
6 EI
L2
0
− AE
L
0
6 EI
L2
4 EI
L
0
0
AE
L
0
−6 EI
2
L
2 EI
L
0
0
0
12 . 200 .500
10003
6 . 200 . 500
2
1000
6 . 200 . 500
10002
4 . 200 .500
1000
0
0
−12 .200 . 500
3
1000
6 . 2000 . 500
10002
−6 . 200 .500
2
1000
2 . 200 .500
1000
0
−12 EI
L3
−6 EI
L2
0
12 EI
3
L
−6 EI
L2
−0,1 . 200
1000
0
6 EI
L2
EI
L
0
−6 EI
2
L
4 EI
L
0
0
0
−12 .200 . 500
10003
−6 . 200 .500
2
1000
6 . 200 . 500
10002
200 . 500
1000
0,1. 200
1000
0
0
0
0
0
12 . 200 .500
3
1000
−6 . 200 .500
1000 2
−6 . 200 .500
2
1000
4 . 2000 .500
1000
Cos27
0
Sin27
0
Sin
270
0
0
0
0
Cos27
0
0
0
0
0
0
0
1
0
0
0
0
0
0
Cos 270
Sin 270
0
0
0
0
- Sin270
Cos 270
0
0
0
0
0
0
1
[T]=
[T]
T
=
0
-1
0
1
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
-1
0
1
0
0
0
0
1
[ K g1 ] = [ T ] T . [
KL
1
].[T]
0
1
0
0
0
0
20
0
0
1
0
0
0
0
0
0
0.001
0
0
0
0
1
0
0
0
0
-1
0
0
0
-20
0
0
0
1
0
0
0
3
0
0.001
3
40
0
0
0
0
0
0
0
1
0
3
0
0
0
0
-1
0
0
0
0
1
0
0
0
0
0
0
0
1
[Kg ] =
1
x
0
-1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
x
-3
20
0
20
0
0
0.001
0
20
-3
0
0
0.001
0
-3
0
3
10
0
0
-3
40
0
UB
0
0
-1
0
0
-1
[ K g1 ]
B.
AE
L
]=
− AE
L
0
0
0
0
0,1. 200
800
0
0
0
0
−12 EI
3
L
6 EI
L2
0
12 . 200 .500
800 3
6 . 2000 . 500
2
800
UA
0
0
1
0
0
1
− AE
L
6 EI
2
L
4 EI
L
[
2
0
B
-1
0
400
1
0
200
12 EI
3
L
6 EI
L2
0
KL
VB
0
20
0
0
-20
0
0
0
AE
L
−6 EI
2
L
2 EI
L
0
0
0
6 . 200 . 500
8002
4 . 200 .500
800
VA
0
-20
0
0
20
0
0
−12 EI
3
L
−6 EI
L2
0
12 EI
3
L
−6 EI
L2
−0,1 . 200
800
0
0
A
-1
0
100
1
0
400
UB
VB
B
UA
VA
A
=
ELEMEN
BC (2)
0
6 EI
2
L
EI
L
0
−6 EI
2
L
4 EI
L
0
−12 .200 . 500
8003
−6 . 200 .500
2
800
0
6 . 200 . 500
8002
200 . 500
800
[ K L2
]=
−0,1 . 200
800
0
0
−12 .200 . 500
8003
6 . 200 . 500
8002
0
0
−6 . 200 .500
8002
2 . 200 .500
800
0,1. 200
800
0
0
12 . 200 .500
800 3
−6 . 200 .500
8002
0
Cos 0
-Sin 0
0
Sin 0
Cos 0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Cos 0
- Sin 0
0
Sin 0
Cos 0
0
0
0
1
[T]
=
[T]
T
=
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
1
[ K g2 ] = [ T ] T . [
KL
2
−6 . 200 .500
8002
4 . 200 .500
800
].[T]
1
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
1
[ K g2 ] =
0
25
x
0
0
-25
0
0
0
0.00
2
1
0
0.00
2
1
0
-25
0
0
25
0
0.002
-1
0
1
500
0
-1
250
0
1
125
0
0
0
0.002
-1
-1
500
1
0
0
0
0
0
X
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
UB
25
0
0
-25
0
0
[Kg ] =
2
C. ELEMEN CD (3)
sin α =
CO
CD
sin 45=
9
CD
VB
0
0
1
0
0
1
0
0
0
0
0
1
B
0
1
500
0
-1
250
UC
-25
0
0
25
0
0
VC
0
0
-1
0
0
-1
C
0
1
125
0
-1
500
UB
VB
B
UC
VC
C
CD=
9
=12,728 m
sin 45
AE
L
0
0
[
K L3
− AE
L
]=
0
0
0,1. 200
1272,8
0
−0,1 . 200
1272,8
0
0
[T]=
12 EI
L3
6 EI
L2
0
−12 EI
L3
6 EI
2
L
0
12 . 200 .500
1272,83
6 . 2000 . 500
1272,82
0
[ K L2
]=
0
Cos
315
-Sin
315
0
0
0
0
0
−12 .200 . 500
1272,83
6 . 200 . 500
2
1272,8
Sin
315
Cos
315
0
0
0
0
− AE
L
0
6 EI
L2
4 EI
L
0
−12 EI
L3
−6 EI
L2
0
0
AE
L
0
−6 EI
L2
2 EI
L
0
6 EI
L2
EI
L
0
0
−6 EI
L2
4 EI
L
12 EI
L3
−6 EI
2
L
0
0
−0,1 . 200
1272,8
0
6 . 200 . 500
1272,82
4 . 200 .500
1272,8
0
0
0
−12 .200 . 500
1272,83
−6 . 200 .500
1272,8 2
0,1. 200
1272,8
0
−6 . 200 .500
1272,8 2
2 . 200 .500
1272,8
0
12 . 200 .500
1272,83
−6 . 2000 .500
2
1272,8
0
0
0
0
0
0
0
1
0
0
0
0
0
Cos 315
- Sin 315
0
0
0
Sin 315
Cos 315
0
0
0
0
0
1
0
6 . 200 . 500
1272,82
200 . 500
1272,8
0
−6 . 200 .500
1272,8 2
4 . 200 .500
1272,8
[T]
T
[T]
0.70710
7
0.70710
7
0
0.70710
7
0
0
0
0
0.70710
7
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0.707107
0.707107
0
0
0
0.70710
7
0.70710
7
0
0.707
x0.707
0
0
0
0
-0.707
i=B
0.707
j=A
0
0
0
-16
0
0
0.00
1
0
0
314
0
0.00
1
0
[ K g3 ]
0
0
1
0
0
α = 270
0
0
0
0
16
0
0
=
0
0
1
[ K g3 ] =
. [ K L3 ] . [ T ]
[ K g3 ] =
-16
T
0
0
0
i=A,B,C
0
0
0
i=C,Ei=B j=C,D,E
i=C
0
0
0
j
=
B
0.707
0 , Dj = c
j = D -0.707
α = 0
0.707
0.707
0
α = 300α = 0
0α = 315
0
1
0.707
1
0
0
0
0
0.707
1
0
0
1
0
0
0.707
1
0.707
1
0
0
0
0.707
1
0
0
0
0
0
0.00
1
0
0
79
0
16
0
0
0
0
0
0
157
0
0
0.00
1
0
0
314
0
0
0
0
0
0
C
0
0
314
0
0
157
UD
-8
-8
0
8
8
0
UC
8
8
0
-8
-8
0
VC
8
8
0
-8
-8
0
0
0.70
71
0.70
71
0
x
VD
-8
-8
0
8
8
0
D
0
0
79
0
0
314
UC
VC
c
UD
VD
D
=
0.707
1
0
0
0
1
D. GLOBAL
UA
VA
ѲA
UB
VB
ѲB
UC
VC
ѲC
UD
VD
ѲD
0
0
1
0
0
1
0
0
0
0
0
0
0
20
0
0
-20
0
0
0
0
0
0
0
1
0
400
-1
0
100
0
0
0
0
0
0
0
0
-1
25
0
-1
-25
0
0
0
0
0
0
-20
0
0
20
1
0
0
1
0
0
0
1
0
200
-1
1
900
0
-1
250
0
0
0
0
0
0
-25
0
0
33
8
0
-8
-8
0
0
0
0
0
0
-1
8
8
-1
-8
-8
0
0
0
0
0
1
125
0
-1
814
0
0
157
0
0
0
0
0
0
-8
-8
0
8
8
0
0
0
0
0
0
0
-8
-8
0
8
8
0
0
0
0
0
0
0
0
0
79
0
0
314
KONDISI BATAS
UA
VA
ѲA
UB
VB
ѲB
UC
VC
ѲC
UD
VD
ѲD
[F]
{}
= [ K ]. [ U
FA
GA
MA
FB
GB
MB
FC
GC
MC
FD
GD
MD
RE-ARRANGEMENT
]
{}
UA
VA
θA
UB
VB
θB
UC
VC
θC
UD
VD
θD
0.9
-68
89.856
7
0
-68
90.666
7
RHA+8
.1
RVA
MA7.29
RHD
RVD
MD
UB
25
0
VB
0
20
ѲB
-1
1
UC
-25
0
VC
0
0
ѲC
0
1
UA
0
0
VA
0
-20
ѲA
-1
0
UD
0
0
VD
0
0
ѲD
0
0
UB
VB
-1
1
900
0
-1
125
1
0
100
0
0
0
ѲB
-25
0
0
0
0
-1
33
8
8
8
0
-1
0
0
0
0
0
0
-8
-8
-8
-8
0
0
UC
VC
0
1
250
0
-1
814
0
0
0
0
0
79
ѲC
0
0
1
0
0
0
0
0
1
0
0
0
0
-20
0
0
0
0
0
20
0
0
0
0
0
200
0
0
0
1
0
400
0
0
0
0
0
0
0
0
0
-8
-8
0
-8
-8
0
0
0
157
0
0
0
0
0
0
0
0
0
8
8
0
8
8
0
0
0
314
-1
0
0
0
REDUKSI
0.9
-68
89.8567
0
-68
90.6667
25
0
0
20
-1
1
-25
0
0
0
0
1
UB
VB
-1
-25
0
0
1
0
0
1
900
0
-1
250
0
33
8
0
-1
8
8
-1
125
0
-1
814
ѲB
UC
VC
ѲC
INVERS
UB
VB
ѲB
UC
VC
ѲC
226
0
0
226
-226
0
0
0
0
0
0
0
0
0
0
0
0
0
226
0
0
226
-226
0
-226
0
0
-226
226
0
0
0
0
0
0
0
0.9
-68
-89.857
0
-68
90.667
0
0
0
0
0
0
RHA+8.
1
RVA
0
0
0
-20
MA-7.29
RHD
RVD
-1
0
0
MD
0
=
0
0
0
0
0
0
0
0
0
1
0
20
0
0
0
0
-8
-8
0
-8
-8
0
0
0
0
0
0
0
15
7
-21.69
93.85
10340.1
2
20.79
42.15
9122.90
RAH = -21,69 + 8,1 = -13,59
RAV = 93,85
MA
= -10340,12- 7,29 = -10347,41
RHD = 20,79
RVD = 42.15
MD = -9122,90
CROSS CHECK
ƩV = RAV + RDV – (q.l)
= 93,85+ 42.15– (17 . 8)
=0
ƩH = RAH +RDH + P
= -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
α = 60
Tentukan Kekakuan elemen Metode Matrix !
Penyelesain:
ELEMEN AB (1) = ELEMEN CD (5)
[
1
AE 0
10 −1
0
[KL1] =
[T
[
] =
1
0 −1 0
0 0 0
0 1 0
0 0 0
]
cos 60 sin 60
0
0
−sin 60 cos 60
0
0
0
0
cos 60 sin 60
0
0
−sin 60 cos 60
]
ELEMEN AB (1)
[KG1]
= [T1]T . [KL1] . [T1]
0.5
0.866
0
0
0.8
66
0.5
0
0
0
0
0
0.5
0.8
66
0
0.86
6
VA
UB
0.25
0.433
0.433
0.75
0.433
-0.75
-0.25
0.433
-0.25
-0.433
0.25
0.433
0
0.5
0
0
0
0
0
0.5
0.866
03
0
0.8660
25
0
0
0.5
0
0
0.5
0.866
-1
0
AE
0
0
0
0
10
-1
0
1
0
0
0
0
0
0
0
0
0
VB
0.433
UA
-0.75
VA
0.433
0.75
UB
VB
= [T5]T . [KL5] . [T5]
X
0.5
0.86
6
0.5
0.8
66
0
ELEMEN CD (5)
[KG5]
0
1
0.5
UA
0.8660
25
0.5
0.8660
25
0.86
6
0.5
0.5
0.866
0
0
0.8
66
0.5
0
0
0
0
0
0.5
0.8
66
0
0.86
6
0
-1
0
AE
0
0
0
0
10
-1
0
1
0
0
0
0
0
0.5
UC
VC
UD
0.25
0.433
0.433
0.75
0.433
-0.75
-0.25
0.433
-0.25
-0.433
1
0.25
0.433
VD
0.433
UC
-0.75
VC
0.433
0.75
UD
VD
X
ELEMEN AC(2) = ELEMEN BD(4) = ELEMEN CE (6)
[KL2] =
[
1
AE 0
10 −1
0
0 −1 0
0 0 0
0 1 0
0 0 0
]
0.8
66
0
0
0.5
0
0
0
0
0
0
0.5
0.866
0.5
0.86
6
0.86
6
0.5
[T2]
[
=
Jadi [KL ]
2
Jadi
Jadi [KL6]
]
1
0
0
0
cos 0 sin 0
0
0
−sin 0 cos 0
0
0
=¿
0
0
cos 0 sin 0
0
0
−sin 0 cos 0
UA
1
0
-1
0
VA
0
0
0
0
UC
-1
0
1
0
VC
0
0
0
0
UA
VA
UC
VC
UB
1
0
-1
0
VB
0
0
0
0
UD
-1
0
1
0
VD
0
0
0
0
UB
VB
UD
VD
UC
1
0
-1
0
VC
0
0
0
0
UE
-1
0
1
0
VE
0
0
0
0
UC
VC
UE
VE
ELEMEN BC (3) = ELEMEN DE (7)
0
1
0
0
=
[KL4]=
=
0
0
1
0
0
0
0
1
[KL2] =
[T2]
[
1
AE 0
10 −1
0
0 −1 0
0 0 0
0 1 0
0 0 0
]
0.5
0.866
0
0
=
0.866
0.5
0
0
0
0
0.5
0.866
0
0
-0.866
0.5
ELEMEN BC (3)
[KG3]
0.5
0.86
6
= [T3]T . [KL3] . [T3]
0.8
66
0
0
0.5
0
0
0.8
66
0
0
0
0
UC
0.5
0.86
6
0.25
-0.433
VC
0.433
0.75
-0.25
0.433
0.433
-0.75
0
0
0.8
66
0.5
0
1
0
-1
0
A
E
0
0
0
0
1
0
-1
0
1
0
0
0
0.5
0
0.86
6
0
0
0
0
0
0
0.8
66
0.5
0.5
0.8
0.86
6
0.5
0
0
0
0
0.5
UB
VB
-0.25
0.433
0.433
-0.75
0.433
UC
VC
0.75
VB
0.25
0.433
0.5
0.86
6
X
UB
ELEMEN DE (7)
[KG7]
0.5
-
= [T7]T . [KL7] . [T7]
0.8
66
0.5
0
0
0
0
A
1
0
0
0
-1
0
0
0
X
0.86
6
66
E
0
0
0
0
UE
0.5
0.86
6
0.25
-0.433
VE
0.433
0.75
-0.25
0.433
0.433
-0.75
0.8
66
1
0
-1
0
1
0
0
0
0.5
0.86
6
0
0
0
0
0
0
0.8
66
0.5
0.5
UD
VD
-0.25
0.433
0.433
-0.75
0.433
UD
0.75
VD
0.25
0.433
UE
VE
GLOBAL
UA
1.25
0.43
-0.25
-0.43
-1
0
0
0
0
0
VA
0.43
0.75
-0.43
-0.75
0
0
0
0
0
0
UB
-0.25
-0.43
1.50
0.00
-0.25
0.43
-1
0
0
0
VB
-0.43
-0.75
0.00
1.50
0.43
-0.75
0
0
0
0
UC
-1
0
-0.25
0.43
2.50
0.00
-0.25
-0.43
-1
0
VC
0
0
0.43
-0.75
0.00
1.50
-0.43
-0.75
0
0
UD
0
0
-1
0
-0.25
-0.43
1.50
0.00
-0.25
0.43
VD
0
0
0
0
-0.43
-0.75
0.00
1.50
0.43
-0.75
UE
0
0
0
0
-1
0
-0.25
0.43
0.25
-0.43
VE
0
0
0
0
0
0
0.43
-0.75
-0.43
0.75
UA
VA
UB
VB
UC
VC
UD
VD
UE
VE
KONDISI BATAS
[F]
= [ K ]. [ U
]
{} { } {}
FA
GA
FB
GB
FC
GC
FD
GD
FE
GE
RHA
RVA
P=10ton
0
0
P=−2ton
0
P=5ton
0
RVE
UA
VA
UB
VB
UC
VC
UD
VD
UE
FE
RE-ARRANGEMENT
UB
VB
10
1.50
0
UC
0.2
5
0
0
1.5
0
0.4
3
0.4
3
0.7
5
0
-0.25
2.5
0
0
0.43
0.4
3
0.7
5
0
0.2
5
0.4
3
1.5
0
0.4
3
0.7
5
-2
VC
UD
VD
UE
VA
0.4
3
0.7
5
VE
0
UA
0.2
5
0.4
3
-1
0
0
0
0.2
5
0.4
3
0
0.4
3
0.7
5
-1
-1
0
1.5
0
0
0
UB
0
VB
0
0
UC
0
0
0
VC
-0.3
0
0
1.5
0
0.4
0
0
0.4
3
0.2
5
0
0
0.4
3
0.7
5
0.4
3
0
UA
0
0.7
5
VA
0
AE
-1.00
0
5
10
0
0
0
0
-1
0
RHA
-0.25
-1
0
0
0
0
1.2
5
0.4
3
RVA
-0.43
0
0.4
3
0.7
5
0
0.2
5
0
0
0
0
0
0.4
3
0.7
5
RVE
0
0
0
0
0
-1
0
0
0
-1
0
0
UB
0
0
0
VB
0.25
0.43
0.43
0.75
-1
UC
VC
1.50
0
0
1.50
0
0.25
0.43
REDUKSI MATRIX
1.5
0
10
0
0
-2
0
5
A
E
1
0
0
0.25
0
0.2
5
0.4
3
1.50
0.43
0.43
0.75
0.43
0.75
2.50
0
-1
0
0
0
0
0.25
-
1.50
0.43
-
UD
VD
UD
VD
UE
VE
0
0
0.43
0.75
-1
0
0
0.25
0.43
0.25
UE
INVERS
UB
1.875
VB
-0.505
UC
1
0
0.750
VC
A
E
-0.577
UD
1.375
VD
-0.361
UE
1
UA
A
E
1
0
VE
=
-10
-
1.29
2
0.43
3
1
0.21
7
0.54
2
0.57
7
-0.43
0.4
3
0.7
5
0
0
-0.25
VA
0.50
5
0.75
0
0.43
3
1
0.28
9
0.75
0
0.14
4
1
0.57
7
1.37
5
0.21
7
1
0.28
9
0.75
0
1.83
3
0
1.87
5
0.07
2
0
1
0.57
7
1
0.36
1
0.54
2
0.14
4
1
0.07
2
1.29
2
0.57
7
-1
0
0
0
0
0
0
0
0
0
0.4
3
0
0.7
5
0
0.4
3
1
0.57
7
10
1
0.57
7
0
0
-2
1
0.57
7
0
2
0
5
2.080
19
9.080
189
CROSS CHECK
ΣV= 0
ΣH= 0
RVA + RVE –P2 – P3
=0
RHA + P1
=0
-2.08019 + 9.080189 – 5 – 2
=0
-10 + 10
=0
OK!
0
OK!
0
=0
=0
GAYA BATANG
ELEMEN 1
i=A
j=B
{}
FA
GA
FB
GB
= [T1] [KG1]
{}
UA
VA
UB
VB
{} [
FA
GA
FB
GB
[
=
0,5 0,87
0
0
−0,87 0,5
0
0
0
0
0,5 0,87
0
0
−0,87 0,5
0,25
0,435 −0,25 −0,435
0,435
0,757 −0,435 −0,757
−0,25 −0,435
0,25
0,435
−0,435 −0,757 0,435
0,757
-2.40
0
2.40
0
]
AE
5
] { }
0
5
0
AE 21,709
−9,760
¿
ELEMEN 2
i=A
j=C
{}
{}
{} [ ] [
FA
GA
FC
GC
= [T2] [KG2]
UA
VA
UC
VC
FA
GA
FC
GC
1
0
0
0
0
0 AE
0 5
1
=
0
1
0
0
0
0
1
0
1
0
−1
0
ELEMEN 3
i=C
j=B
{}
{}
{} [
]
FC
GC
FB
GB
= [T3] [KG3]
FB
GB
FC
GC
[
=
UC
VC
UB
VB
0,5 −0,87
0
0
AE
0,87
0,5
0
0
0
0
0,5 −0,87 5
0
0
0,87
0,5
]{ }
0,25
−0,435 −0,25
0,435
8,799
−0,435 0,757
0,435 −0,757 5 −14,440
−0,25
0,435
0,25
−0,435 AE 21.709
0,435 −0,757 −0,435 0,757
−9.760
=
2.40
0
-2.40
0
]{ }
0 −1 0
0
0 0 0 5
0
0 1 0 AE 8,799
0 0 0
−14,440
=
ELEMEN 4
i=B
j=D
{}
{}
{} [ ] [
FB
GB
FD
GD
= [T4] [KG4]
UB
VB
UD
VD
FB
GB
FD
GD
1
0
0
0
0
0
1
0
0
0 AE
0 5
1
= [T5] [KG5]
{}
=
0
1
0
0
ELEMEN 5
i=C
j=D
{}
FC
GC
FD
GD
UC
VC
UD
VD
1
0
−1
0
]{ }
0 −1 0
21.709
0 0 0 5 −9.760
0 1 0 AE 14.111
0 0 0
−12.067
=
{} [
FC
GC
FD
GD
[
=
0,5 0,87
0
0
−0,87 0,5
0
0
0
0
0,5 0,87
0
0
−0,87 0,5
0,25
0,435 −0,25 −0,435
0,435
0,757 −0,435 −0,757
−0,25 −0,435
0,25
0,435
−0,435 −0,757 0,435
0,757
=
]
-4.71
0
4.71
0
ELEMEN 6
i=C
j=E
{}
FC
GC
FE
¿
= [T6] [KG6]
{}
UC
VC
UE
VE
5
AE
]
AE
5
{}
FC
GC
FE
¿
[
1
0
−1
0
=
[ ]
1
0
0
0
0
1
0
0
0
0
1
0
-5.24
0
5.24
0
0
0 AE
0 5
1
]
0 −1 0
0 0 0 5
0 1 0 AE
0 0 0
=
ELEMEN 7
i=D
j=E
{}
{}
FD
GD
FE
¿
FD
GD
FE
¿
[
= [T7] [KG7]
=
[
{}
UD
VD
UE
VE
]
0,5 −0,87
0
0
AE
0,87
0,5
0
0
0
0
0,5 −0,87 5
0
0
0,87
0,5
]
0,25
−0,435 −0,25
0,435
−0,435 0,757
0,435 −0,757 5
−0,25
0,435
0,25
−0,435 AE
0,435 −0,757 −0,435 0,757
18.3
5
0
18.3
5
0
=