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