2/21/2015 6:18 AM
CONTOH SOAL ANALISA MATRIKS METODE KEKAKUAN LANGSUNG
GAMBAR BALOK MENERUS
Data Properties Penampang
Tinggi balok, h = 40 cm
Lebar balok, b = 25 cm
Mutu beton, fc' = 250 kg/cm2
Modulus elastisitas beton, Ec =4700 x sqrt (fc'/10) x 10 Ec = 235000 kg/cm2
Momen inersia balok, Ix = 1/12 x bh3
Ix = 133333.3 cm4
Span (bentang) balok, L1 = 300 cm
Span (bentang) balok, L2 = 400 cm
Span (bentang) balok, L3 = 300 cm
Span (bentang) balok, L4 = 250 cm
Jarak beban, a3 = L3/2 a3 = 150 cm
Beban-beban yang bekerja
q1 = 7.5 kg/cm
q2 = 6 kg/cm
P = 1000 kg
M1 = 100000 kg.cm
M2 = 50000 kg.cm
I. HITUNG MATRIKS KEKAKUAN BATANG [SM]
1. Matriks Kekakuan untuk Batang 1
- Perpindahan/Displacement arah 1 --> D1 (Translasi arah sb-Y)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM11 = 12.Ec.Ix = 13925.9259 kg/cm SM31 = - 12.Ec.Ix = -13925.93 kg/cm
L13
L13
SM21 = 6.Ec.Ix = 2088888.89 kg SM41 = 6.Ec.Ix = 2088888.9 kg
L12
L12
h
b
j
D1
D2 k
i
D3
D4
A B D C E
P = 1000 Kg q1 = 7.5 Kg/cm q2 = 6 Kg/cm
L1 = 3 m L2/2 = 2 m L3/2 = 1.5 m L4 = 2.5 m L2/2 = 2 m L3/2 = 1.5 m
M1=-100000
M2=50000 Kg.cm
EI
L1
∆
SM11
SM21
SM31
SM41 A
B
Hal.1 dari 12 Sondra Raharja, ST
2/21/2015 6:18 AM
- Perpindahan/Displacement arah 2 --> D2 (Rotasi arah sb-Z)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM12 = 6.Ec.Ix = 2088888.89 kg/cm SM32 = - 6.Ec.Ix = -2088889 kg/cm
L12
L12
SM22 = 4.Ec.Ix = 417777778 kg SM42 = 2.Ec.Ix = 208888889 kg
L1 L1
- Perpindahan/Displacement arah 3 --> D3 (Translasi arah sb-Y)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM13 = - 12.Ec.Ix = -13925.9259 kg/cm SM33 = 12.Ec.Ix = 13925.926 kg/cm
L13
L13
SM23 = -6.Ec.Ix = -2088888.89 kg SM43 = -6.Ec.Ix = -2088889 kg
L12
L12
- Perpindahan/Displacement arah 4 --> D4 (Rotasi arah sb-Z)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM14 = 6.Ec.Ix = 2088888.89 kg/cm SM34 = - 6.Ec.Ix = -2088889 kg/cm
L12
L12
SM24 = 2.Ec.Ix = 208888889 kg SM44 = 4.Ec.Ix = 417777778 kg
L1 L1
θ = 1
θ = 1 EI
L1
A B
SM12
SM42
SM32
SM22
EI
L1
∆
SM13
A
B SM23
SM33
SM43
EI
L1
θ = 1
θ = 1
SM14
SM24 SM44
SM34
A B
Hal.2 dari 12 Sondra Raharja, ST
2/21/2015 6:18 AM
Susun matriks kekakuan batang 1
SM11 SM12 SM13 SM14
SM1 = SM21 SM22 SM23 SM24
SM31 SM32 SM33 SM34
SM41 SM42 SM43 SM44
13925.92593 2088888.889 -13925.92593 2088888.889
SM1 = 2088888.889 417777777.8 -2088888.889 208888888.9
-13925.92593 -2088888.889 13925.92593 -2088888.889
2088888.889 208888888.9 -2088888.889 417777777.8
2. Matriks Kekakuan untuk Batang 2
- Perpindahan/Displacement arah 1 --> D1 (Translasi arah sb-Y)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM11 = 12.Ec.Ix = 5875 kg/cm SM31 = - 12.Ec.Ix = -5875 kg/cm
L23
L23
SM21 = 6.Ec.Ix = 1175000 kg SM41 = 6.Ec.Ix = 1175000 kg
L22
L22
- Perpindahan/Displacement arah 2 --> D2 (Rotasi arah sb-Z)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM12 = 6.Ec.Ix = 1175000 kg/cm SM32 = - 6.Ec.Ix = -1175000 kg/cm
L22
L22
SM22 = 4.Ec.Ix = 313333333 kg SM42 = 2.Ec.Ix = 156666667 kg
L2 L2
- Perpindahan/Displacement arah 3 --> D3 (Translasi arah sb-Y)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
EI
L2
∆
SM1
SM2
SM3
SM4B
C
θ = 1
θ = 1 EI
L2
B C
SM12
SM42
SM32
SM22
EI
L2
∆
SM13
B
C SM23
SM33
SM43
Hal.3 dari 12 Sondra Raharja, ST
2/21/2015 6:18 AM
SM13 = - 12.Ec.Ix = -5875 kg/cm SM33 = 12.Ec.Ix = 5875 kg/cm
L23
L23
SM23 = -6.Ec.Ix = -1175000 kg SM43 = -6.Ec.Ix = -1175000 kg
L22
L22
- Perpindahan/Displacement arah 4 --> D4 (Rotasi arah sb-Z)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM14 = 6.Ec.Ix = 1175000 kg/cm SM34 = - 6.Ec.Ix = -1175000 kg/cm
L22
L22
SM24 = 2.Ec.Ix = 156666667 kg SM44 = 4.Ec.Ix = 313333333 kg
L2 L2
Susun matriks kekakuan batang 2
SM11 SM12 SM13 SM14
SM2 = SM21 SM22 SM23 SM24
SM31 SM32 SM33 SM34
SM41 SM42 SM43 SM44
5875 1175000 -5875 1175000
SM2 = 1175000 313333333.3 -1175000 156666666.7
-5875 -1175000 5875 -1175000
1175000 156666666.7 -1175000 313333333.3
3. Matriks Kekakuan untuk Batang 3
- Perpindahan/Displacement arah 1 --> D1 (Translasi arah sb-Y)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM11 = 12.Ec.Ix = 13925.9259 kg/cm SM31 = - 12.Ec.Ix = -13925.93 kg/cm
L33
L33
SM21 = 6.Ec.Ix = 2088888.89 kg SM41 = 6.Ec.Ix = 2088888.9 kg
L32
L32
EI
L2
θ = 1
θ = 1
SM14
SM24 SM44
SM34
B C
EI
L3
∆
SM1
SM2
SM3
SM4C
D
Hal.4 dari 12 Sondra Raharja, ST
2/21/2015 6:18 AM
- Perpindahan/Displacement arah 2 --> D2 (Rotasi arah sb-Z)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM12 = 6.Ec.Ix = 2088888.89 kg/cm SM32 = - 6.Ec.Ix = -2088889 kg/cm
L32
L32
SM22 = 4.Ec.Ix = 417777778 kg SM42 = 2.Ec.Ix = 208888889 kg
L3 L3
- Perpindahan/Displacement arah 3 --> D3 (Translasi arah sb-Y)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM13 = - 12.Ec.Ix = -13925.9259 kg/cm SM33 = 12.Ec.Ix = 13925.926 kg/cm
L33
L33
SM23 = -6.Ec.Ix = -2088888.89 kg SM43 = -6.Ec.Ix = -2088889 kg
L32
L32
- Perpindahan/Displacement arah 4 --> D4 (Rotasi arah sb-Z)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM14 = 6.Ec.Ix = 2088888.89 kg/cm SM34 = - 6.Ec.Ix = -2088889 kg/cm
L32
L32
SM24 = 2.Ec.Ix = 208888889 kg SM44 = 4.Ec.Ix = 417777778 kg
L3 L3
θ = 1
θ = 1 EI
L3
C D
SM12
SM42
SM32
SM22
EI
L3
θ = 1
θ = 1
SM14
SM24 SM44
SM34
C D
EI
L3
∆
SM13
C
D SM23
SM33
SM43
Hal.5 dari 12 Sondra Raharja, ST
2/21/2015 6:18 AM
Susun matriks kekakuan batang 3
SM11 SM12 SM13 SM14
SM3 = SM21 SM22 SM23 SM24
SM31 SM32 SM33 SM34
SM41 SM42 SM43 SM44
13925.92593 2088888.889 -13925.92593 2088888.889
SM3 = 2088888.889 417777777.8 -2088888.889 208888888.9
-13925.92593 -2088888.889 13925.92593 -2088888.889
2088888.889 208888888.9 -2088888.889 417777777.8
4. Matriks Kekakuan untuk Batang 4
- Perpindahan/Displacement arah 1 --> D1 (Translasi arah sb-Y)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM11 = 12.Ec.Ix = 24064 kg/cm SM31 = - 12.Ec.Ix = -24064 kg/cm
L43
L43
SM21 = 6.Ec.Ix = 3008000 kg SM41 = 6.Ec.Ix = 3008000 kg
L42
L42
- Perpindahan/Displacement arah 2 --> D2 (Rotasi arah sb-Z)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM12 = 6.Ec.Ix = 3008000 kg/cm SM32 = - 6.Ec.Ix = -3008000 kg/cm
L42
L42
SM22 = 4.Ec.Ix = 501333333 kg SM42 = 2.Ec.Ix = 250666667 kg
L4 L4
- Perpindahan/Displacement arah 3 --> D3 (Translasi arah sb-Y)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
EI
L4
∆
SM1
SM2
SM3
SM4D
E
θ = 1
θ = 1 EI
L4
D E
SM12
SM42
SM32
SM22
EI
L4
∆
SM13
D
E SM23
SM33
SM43
Hal.6 dari 12 Sondra Raharja, ST
2/21/2015 6:18 AM
SM13 = - 12.Ec.Ix = -24064 kg/cm SM33 = 12.Ec.Ix = 24064 kg/cm
L43
L43
SM23 = -6.Ec.Ix = -3008000 kg SM43 = -6.Ec.Ix = -3008000 kg
L42
L42
- Perpindahan/Displacement arah 4 --> D4 (Rotasi arah sb-Z)
GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG
BATANG
SM14 = 6.Ec.Ix = 3008000 kg/cm SM34 = - 6.Ec.Ix = -3008000 kg/cm
L42
L42
SM24 = 2.Ec.Ix = 250666667 kg SM44 = 4.Ec.Ix = 501333333 kg
L4 L4
Susun matriks kekakuan batang 4
SM11 SM12 SM13 SM14
SM4 = SM21 SM22 SM23 SM24
SM31 SM32 SM33 SM34
SM41 SM42 SM43 SM44
24064 3008000 -24064 3008000
SM4 = 3008000 501333333.3 -3008000 250666666.7
-24064 -3008000 24064 -3008000
3008000 250666666.7 -3008000 501333333.3
II. SUSUN MATRIKS KEKAKUAN TITIK KUMPUL [Sj]
Matriks Sj disusun dari matriks SM
13925.92593 2088888.889 -13925.92593 2088888.889
SM1 = 2088888.889 417777777.8 -2088888.889 208888888.9
-13925.92593 -2088888.889 13925.92593 -2088888.889
2088888.889 208888888.9 -2088888.889 417777777.8
5875 1175000 -5875 1175000
SM2 = 1175000 313333333.3 -1175000 156666666.7
-5875 -1175000 5875 -1175000
1175000 156666666.7 -1175000 313333333.3
13925.92593 2088888.889 -13925.92593 2088888.889
SM3 = 2088888.889 417777777.8 -2088888.889 208888888.9
-13925.92593 -2088888.889 13925.92593 -2088888.889
2088888.889 208888888.9 -2088888.889 417777777.8
24064 3008000 -24064 3008000
SM4 = 3008000 501333333.3 -3008000 250666666.7
-24064 -3008000 24064 -3008000
3008000 250666666.7 -3008000 501333333.3
EI
L4
θ = 1
θ = 1
SM14
SM24 SM44
SM34
D E
Hal.7 dari 12 Sondra Raharja, ST
2/21/2015 6:18 AM
GAMBARKAN POSISI DOF UTK TATAULANG SJ
1 2 3 4 5 6 7 8 9 10
1 13925.92593 2088888.889 -13925.92593 2088888.889 0 0 0 0 0 0 5
2 2088888.889 417777777.8 -2088888.889 208888888.9 0 0 0 0 0 0 6
3 -13925.92593 -2088888.889 19800.92593 -913888.8889 -5875 1175000 0 0 0 0 7
4 2088888.889 208888888.9 -913888.8889 731111111.1 -1175000 156666667 0 0 0 0 D1
Sj = 5 0 0 -5875 -1175000 19800.926 913888.89 -13925.92593 2088888.89 0 0 8
6 0 0 1175000 156666666.7 913888.89 731111111 -2088888.889 208888889 0 0 D2
7 0 0 0 0 -13925.93 -2088889 37989.92593 919111.111 -24064 3008000 9
8 0 0 0 0 2088888.9 208888889 919111.1111 919111111 -3008000 250666667 D3
9 0 0 0 0 0 0 -24064 -3008000 24064 -3008000 10
10 0 0 0 0 0 0 3008000 250666667 -3008000 501333333 D4
5 6 7 D1 8 D2 9 D3 10 D4
Bentuk matriks Sj yang ditataulang (re-arrangement ) ---> berdasarkan posisi DOF
1 2 3 4 5 6 7 8 9 10
1 731111111.1 156666666.7 0 0 2088888.9 208888889 -913888.8889 -1175000 0 0
2 156666666.7 731111111.1 208888888.9 0 0 0 1175000 913888.889 -2088889 0
3 0 208888888.9 919111111.1 250666666.7 0 0 0 2088888.89 919111.11 -3008000
4 0 0 250666666.7 501333333.3 0 0 0 0 3008000 -3008000
Sj = 5 2088888.889 0 0 0 13925.926 2088888.9 -13925.92593 0 0 0
6 208888888.9 0 0 0 2088888.9 417777778 -2088888.889 0 0 0
7 -913888.8889 1175000 0 0 -13925.93 -2088889 19800.92593 -5875 0 0
8 -1175000 913888.8889 2088888.889 0 0 0 -5875 19800.9259 -13925.93 0
9 0 -2088888.889 919111.1111 3008000 0 0 0 -13925.926 37989.926 -24064
10 0 0 -3008000 -3008000 0 0 0 0 -24064 24064
SFF
SFR
Sj = SRF
SRR
Didapatkan matriks SFF
1
2
3
4
3
4
5
6
Pertemuan joint dijumlahkan
5
6
7
8
7
8
9
10
D1
1
2
A B D C E
3
4
5
6
7
8
9
10 D2 D1 D3 D4
A B D C E
D1
L1 = 3 m L2 = 4 m L4 = 2.5 m L3 = 3 m
D2 D3 D4
Hal.8 dari 12 Sondra Raharja, ST
2/21/2015 6:18 AM
731111111.1 156666666.7 0 0
156666666.7 731111111.1 208888888.9 0
SFF
= 0 208888888.9 919111111.1 250666666.7
0 0 250666666.7 501333333.3
Hitung invers matriks SFF
1.43924E-09 -3.33483E-10 8.77586E-11 -4.38793E-11
-3.33483E-10 1.55625E-09 -4.0954E-10 2.0477E-10
SFF
(-1)
= 8.77586E-11 -4.0954E-10 1.36757E-09 -6.83786E-10
-4.38793E-11 2.0477E-10 -6.83786E-10 2.33657E-09
III. SUSUN MATRIKS VEKTOR AKSI (GAYA) KOMBINASI [Ac]
Ac = Aj + AE
Aj ---> Beban aksi di joint
GAMBARKAN POSISI BEBAN LUAR PADA DOF
0 0
0 0
0 0
0 0
Aj = 0 = 0
0 0
0 0
- M1 -100000
0 0
M2 50000
Hitung reaksi di ujung batang freebody (AML) ---> Beban dimasukkan kecuali beban aksi di joint
GAMBARKAN BALOK SEMULA DG BEBAN, KECUALI BEBAN DIJOINT
Freebody A - B :
GAMBARKAN FREEBODY, BEBAN DAN REAKSINYA
AML1 = q1.L1/2 = 1125 kg
AML2 = 1/12 x q1.L12
= 56250 kg.cm
AML3 = q1.L1/2 = 1125 kg
AML4 = -1/12 x q1.L12
= -56250 kg.cm
1
2
3
4 Urutan Penomoran
A B D C E
L1 = 3 m L2 = 4 m L4 = 2.5 m L3 = 3 m
M M2
A B D C E
P = 1000 Kg q1 = 7.5 Kg/cm q2 = 6 Kg/cm
L1 = 3 m L2/2 = 2 m L3/2 = 1.5 m L4 = 2.5 m L2/2 = 2 m L3/2 = 1.5 m
A B
q1 = 7.5 Kg/cm
L1 = 3 m
AML4
AML1
AML2
AML3
DI TITIK A
DI TITIK B
DI TITIK C
DI TITIK D
DI TITIK E
Hal.9 dari 12 Sondra Raharja, ST
2/21/2015 6:18 AM
Freebody B - C :
GAMBARKAN FREEBODY, BEBAN DAN REAKSINYA
AML1 = P/2 = 500 kg
AML2 = P.L2 / 8 = 50000 kg.cm
AML3 = P/2 = 500 kg
AML4 = - P.L2 / 8 = -50000 kg.cm
Freebody C - D :
GAMBARKAN FREEBODY, BEBAN DAN REAKSINYA
AML1 = 13/32. q2.L3 = 731.25 kg
AML2 = 11/192.q2.L32
= 30937.5 kg.cm
AML3 = 3/32. q2.L3 = 168.75 kg
AML4 = - 5/192. q2.L32
= -14062.5 kg.cm
Freebody D - E :
GAMBARKAN FREEBODY, BEBAN DAN REAKSINYA
AML1 = 0 = 0 kg
AML2 = 0 = 0 kg.cm
AML3 = 0 = 0 kg
AML4 = 0 = 0 kg.cm
Susun matriks AE dari matriks AML
1125
AM1 = 56250
1125
-56250
1125 -1125
500 56250 -56250
AM2 = 50000 1625 -1625
500 -6250 6250
-50000 1231.25 -1231.25
AE = - -19062.5 AE = 19062.5
731.25 168.75 -168.75
AM3 = 30937.5 -14062.5 14062.5
168.75 0 0
-14062.5 0 0
0
AM4 = 0
0
0
B C
P = 1000 Kg
L2/2 = 2 m L2/2 = 2 m
AML2 AML4
AML1 AML3
D C
q2 = 6 Kg/cm
L3/2 = 1.5 m L3/2 = 1.5 m
AML2 AML4
AML1 AML3
D E
L4 = 2.5 m
AML2 AML4
AML1 AML3
Hal.10 dari 12 Sondra Raharja, ST
2/21/2015 6:18 AM
Susun matriks Ac
1 0 -1125 -1125 5
2 0 -56250 -56250 6
3 0 -1625 -1625 7
4 0 6250 6250 D1
Ac = 5 0 + -1231.25 = -1231.25 8
6 0 19062.5 19062.5 D2
7 0 -168.75 -168.75 9
8 -100000 14062.5 -85937.5 D3
9 0 0 0 10
10 50000 0 50000 D4
Tata ulang (re-arrangement ) matriks Ac
1 6250
2 19062.5 ---> AFC
3 -85937.5
4 50000 Ac = AFC
Ac = 5 -1125 ARC
6 -56250
7 -1625 ---> ARC
8 -1231.25
9 -168.75
10 0
Didapat matriks AFC dan ARC
6250 -1125
AFC = 19062.5 -56250
-85937.5 ARC = -1625
50000 -1231.25
-168.75
0
IV. HITUNG PERPINDAHAN (DISPLACEMENT) [ DF ]
DF = S
FF
(-1)
. AFC
1.43924E-09 -3.33483E-10 8.77586E-11 -4.38793E-11 6250
DF = -3.33483E-10 1.55625E-09 -4.0954E-10 2.0477E-10 x 19062.5
8.77586E-11 -4.0954E-10 1.36757E-09 -6.83786E-10 -85937.5
-4.38793E-11 2.0477E-10 -6.83786E-10 2.33657E-09 50000
-7.09748E-06
DF = 7.30152E-05
-0.000158973
0.000179221
V. HITUNG REAKSI PERLETAKAN [AR]
AR = -ARC + SRF
.DF
1125 2088888.889 0 0 0 -7.09748E-06
56250 208888888.9 0 0 0 7.30152E-05
AR = 1625 + -913888.8889 1175000 0 0 x -0.000158973
1231.25 -1175000 913888.8889 2088888.889 0 0.000179221
168.75 0 -2088888.889 919111.1111 3008000
0 0 0 -3008000 -3008000
Hal.11 dari 12 Sondra Raharja, ST
2/21/2015 6:18 AM
1125 -14.8258478 1110.17415 1110.17415 --> AR1
56250 -1482.58478 54767.4152 54767.4152 --> AR2
1625 + 92.279159 = 1717.27916 1717.27916 --> AR3
AR = 1231.25 -257.010484 974.239516 974.239516 --> AR4
168.75 240.461159 409.211159 409.211159 --> AR5
0 -60.9039872 -60.9039872 -60.9039872 --> AR6
A B D C E
P = 1000 Kg q1 = 7.5 Kg/cm q2 = 6 Kg/cm
L1 = 3 m L2/2 = 2 m L3/2 = 1.5 m L4 = 2.5 m L2/2 = 2 m L3/2 = 1.5 m
M1=-100000
M2=50000 Kg.cm
AR1
AR2
AR3 AR4 AR5 AR6
Hal.12 dari 12 Sondra Raharja, ST
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