Contoh Soal Analisa Matriks Kekakuan Langsung (Balok Menerus)

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 C D E P = 1000 Kg q1 = 7.5 Kg/cm q2 = 6 Kg/cm L1 = 3 m L2/2 = 2 m L2/2 = 2 m L3/2 = 1.5 m L3/2 = 1.5 m L4 = 2.5 m M1=-100000 M2=50000 Kg.cm EI L1 ∆ SM11 SM21 SM31 SM41 A B

- 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

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 SM4 B C θ = 1 θ = 1 EI L2 B C SM12 SM42 SM32 SM22 EI L2 ∆ SM13 B C SM23 SM33 SM43

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 SM4 C D

- 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

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 SM4 D E θ = 1 θ = 1 EI L4 D E SM12 SM42 SM32 SM22 EI L4 ∆ SM13 D E SM23 SM33 SM43

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

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 C D E 3 4 5 6 7 8 9 10 D2 D1 D3 D4 A B C D E D1 L1 = 3 m L2 = 4 m L3 = 3 m L4 = 2.5 m D2 D3 D4

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 C D E L1 = 3 m L2 = 4 m L3 = 3 m L4 = 2.5 m M M2 A B C D E P = 1000 Kg q1 = 7.5 Kg/cm q2 = 6 Kg/cm L1 = 3 m L2/2 = 2 m L2/2 = 2 m L3/2 = 1.5 m L3/2 = 1.5 m L4 = 2.5 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

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

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 = SFF (-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

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 C D E P = 1000 Kg q1 = 7.5 Kg/cm q2 = 6 Kg/cm L1 = 3 m L2/2 = 2 m L2/2 = 2 m L3/2 = 1.5 m L3/2 = 1.5 m L4 = 2.5 m M1=-100000 M2=50000 Kg.cm AR1 AR2

AR3 AR4 AR5 AR6