contoh soal analisa matriks (balok menerus) revisi2 01-07-15

Hal.1 dari 65 Sondra Raharja, ST

CONTOH SOAL ANALISA MATRIKS METODE KEKAKUAN BIASA

GAMBAR BALOK MENERUS

Data Properties Penampang

Tinggi balok, h = 40Lebar balok, b = 25Mutu beton, fc' = 250Modulus elastisitas beton, Ec =4700 x sqrt (fc'/10) x 10 Ec = 235000

Ix = 133333.3Span (bentang) balok, L1 = 300Span (bentang) balok, L2 = 400Span (bentang) balok, L3 = 300Span (bentang) balok, L4 = 250Jarak beban, a3 = L3/2 a3 = 150

Beban-beban yang bekerjaq1 = 7.5q2 = 6

P = 1000M1 = 100000M2 = 50000

Penyusunan matriks-matriks

GAMBAR 1 . GAMBAR BALOK MENERUS DIKEKANGBEBAN HILANGKAN, GBRKAN DOF

GAMBAR 2 . GAMBAR BALOK MENERUS DIKEKANGBEBAN HILANGKAN, KECUALI BEBAN SELARAS DOF

Momen inersia balok, Ix = 1/12 x bh3

A B C

P = 1000 Kgq1 = 7.5 Kg/cm q2 = 6 Kg/cm

L1 = 3 m L2/2 = 2 m L3/2 = 1.5 mL2/2 = 2 m L3/2 = 1.5 m

A B C

D1

L1 = 3 m L2 = 4 m L3 = 3 m

D2

A B C

L1 = 3 m L2 = 4 m L3 = 3 m


Susun matrik AD, ---> gaya luar yang selaras DOF

Dari perletakan didapat DOF = 4

0 0AD = 0 AD = 0

M1 -100000M2 50000

GAMBAR 3 . GAMBAR BALOK MENERUS DIKEKANGBEBAN DI MASUKAN, KECUALI YG SELARAS DOF, YG SELARAS DOF DIHILANGKAN, GAMBARKANREAKSI PERLETAKAN AKIBAT KEKANGAN (ARL DAN ADL)

Catatan : ARL adalah reaksi perletakan semula, akibat beban primerADL adalah rekasi perletakan akibat kekangan pada posisi DOF atau gaya akibat beban terjepit yg selaras dg DOF

Freebody A- B (Bentang 1) :

GAMBAR FREEBODY SESUAI BENTANGAN YG DIKEKANGBESERTA REAKSI AKIBAT KEKANGAN (ARL DAN ADL)

ARL1 = q1.L1/2 = 1125 kg

ARL2 = = 56250 kg.cm

ARL3 = q1.L1/2 = 1125 kg

ADL1 = = -56250 kg.cm

Freebody B- C (Bentang2 ) :

GAMBAR FREEBODY SESUAI BENTANGAN YG DIKEKANGBESERTA REAKSI AKIBAT KEKANGAN (ARL DAN ADL)

ARL3 = P/2 = 500 kg

ADL1 = P.L2 / 8 = 50000 kg.cm

ARL4 = P/2 = 500 kg

ADL2 = - P.L2 / 8 = -50000 kg.cm

Bh --> orde matriks d x 1 = AD

1/12 x q1.L12

-1/12 x q1.L12

A B C

P = 1000 Kgq1 = 7.5 Kg/cmq2 = 6 Kg/cm

L1 = 3 m L2/2 = 2 m L3/2 = 1.5 mL2/2 = 2 m

ADL1 ADL2

ARL1

ARL2

ARL3 ARL4

A

ARL1

ARL2

ARL3


Freebody C - D (Bentang 3) :

GAMBAR 2 . GAMBAR FREEBODY SESUAI BENTANGAN YG DIKEKANGBESERTA REAKSI AKIBAT KEKANGAN (ARL DAN ADL)

ARL4 = 13/32. q2.L3 = 731.25 kg

ADL2 = = 30937.5 kg.cm

ARL5 = 3/32. q2.L3 = 168.75 kg

ADL3 = = -14062.5 kg.cm

Freebody D - E (Bentang 4) :

GAMBAR 2 . GAMBAR FREEBODY SESUAI BENTANGAN YG DIKEKANGBESERTA REAKSI AKIBAT KEKANGAN (ARL DAN ADL)

Karena tidak ada gaya luar disepanjang bentang, maka reaksi perletakan tidak ada

ARL5 = 0 kgADL3 = 0 kg.cmARL6 = 0 kgADL4 = 0 kg.cm

Gabungkan seluruh reaksi ujung batang akibat gaya luar dan kekangan sehingga dapat disusun matriks ADL dan ARL

Pada joint B reaksi gaya ujung batang yg dijumlahkan adalah sbb :

ARL3 Btg1 + ARL3 Btg2 = 1625

ADL1 btg1 + ADL1 btg2 = -6250

Pada joint C reaksi gaya ujung batang yg dijumlahkan adalah sbb :

11/192.q2.L32

- 5/192. q2.L32

ADL2

ARL4

D

L4 = 2.5 m

ADL3

ARL5

A B

q1 = 7.5 Kg/cm

L1 = 3 m

ADL1

ARL1

ARL2

ARL3

B

L2/2 = 2 m

ADL1

ARL3

DC

q2 = 6 Kg/cm

L3/2 = 1.5 m L3/2 = 1.5 m

ADL2 ADL3

ARL4 ARL5


ARL4 btg2 + ARL4 btg3 = 1231.25

ADL2 btg2 + ADL2 btg3 = -19062.5

Pada joint D reaksi gaya ujung batang yg dijumlahkan adalah sbb :

ARL5 btg3 + ARL5 btg4 = 168.75

ADL3 btg3 + ADL3 btg4 = -14062.5

Susun matriks ADL dan ARL sbb :

ADL1 -6250ADL2 = -19062.5

ADL = ADL3 -14062.5ADL4 0

ARL1 1125ARL2 56250

ARL = ARL3 = 1625ARL4 1231.25ARL5 168.75ARL6 0

Susun matriks kekakuan [S] dan matriks reaksi perletakan [ARD]

Hitung matriks akibat displacement / perpindahan yaitu matriks kekakuan [S] dan matriksreaksi perletakan semula akibat displacement [ARD]

Perpindahan 1 ---> yaitu akibat D1

UNTUK MENDAPATKAN KEKAKUANGAMBAR PERPINDAHAN 1 SATUAN (PUT. SUDUT/ROTASI) PD TITIK B

A B C

L1 = 3 m L2 = 4 m L3 = 3 mARD11

ARD21

ARD31 ARD41

S11 S21


GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

ARD11 = 6.Ec.Ix = 2088888.889 kg

ARD21 = 2.Ec.Ix = 208888888.9 kg.cmL1

S11 = 4.Ec.Ix + 4.Ec.Ix = 731111111.1 kg.cmL1 L2

ARD31 = - 6.Ec.Ix + 6.Ec.Ix = -913888.8889 kg

ARD41 = - 6.Ec.Ix + 0 = -1175000 kg

S21 = 2.Ec.Ix + 0 = 156666666.7 kg.cmL2

ARD51 = 0 = 0 kg

S31 = 0 + 0 = 0 kg.cm

ARD61 = 0 = 0 kg

S41 = 0 = 0 kg.cm


L12

L12 L22

L22

EI

L1

θ = 1

θ = 1

ARD11

ARD21 S11

ARD31

A B

θ = 1

θ = 1B

ARD31

S11

EI

L3

ARD41

S21 S31

ARD51

C D

ARD51

S31

D

A B C

L1 = 3 m L2 = 4 m L3 = 3 mARD12=0

ARD22=0

ARD32 ARD42

S12 S22


UNTUK MENDAPATKAN KEKAKUANGAMBAR PERINDAHAN 1 SATUAN (PUT. SUDUT/ROTASI) PD TITIK C



ARD12 = 0 = 0 kg

ARD22 = 0 = 0 kg.cm

ARD32 = 0 + 6.Ec.Ix = 1175000 kg

S12 = 0 + 2.Ec.Ix = 156666666.7 kg.cmL2

ARD42 = - 6.Ec.Ix + 6.Ec.Ix = 913888.8889 kg


ARD52 = - 6.Ec.Ix + 0 = -2088888.889 kg

S32 = 2.Ec.Ix + 0 = 208888888.9 kg.cmL3

ARD62 = 0 = 0 kg

S42 = 0 = 0 kg.cm


L22

L22 L32

L32

EI

L1

ARD12

ARD22 S12

ARD32

A B

ARD32

S12

B

θ = 1

θ = 1EI

L3

C D

ARD42

S32

ARD52

S22

ARD52

S32

D

A B C

L1 = 3 m L2 = 4 m L3 = 3 mARD13=0

ARD23=0

ARD33=0 ARD43

S13=0S23


UNTUK MENDAPATKAN KEKAKUANGAMBAR PERINDAHAN 1 SATUAN (PUT. SUDUT/ROTASI) PD TITIK D



ARD13 = 0 = 0 kg

ARD23 = 0 = 0 kg.cm

ARD33 = 0 + 0 = 0 kg

S13 = 0 + 0 = 0 kg.cm

ARD43 = 0 + 6.Ec.Ix = 2088888.889 kg

S23 = 0 + 2.Ec.Ix = 208888888.9 kg.cmL3

ARD53 = - 6.Ec.Ix + 6.Ec.Ix = 919111.1111 kg


ARD63 = - 6.Ec.Ix = -3008000 kg

S43 = 2.Ec.Ix = 250666666.7 kg.cmL4


L32

L32 L42

L42

EI

L1

ARD13

ARD23 S13

ARD33

A B

ARD33

S13

B

EI

L3

θ = 1

θ = 1

ARD43

S23 S33

ARD53

C D

θ = 1D

ARD53

S33

A B C

L1 = 3 m L2 = 4 m L3 = 3 mARD14=0

ARD24=0

ARD34=0 ARD44=0

S14=0 S24=0


UNTUK MENDAPATKAN KEKAKUANGAMBAR PERINDAHAN 1 SATUAN (PUT. SUDUT/ROTASI) PD TITIK E



ARD14 = 0 = 0 kg

ARD24 = 0 = 0 kg.cm

ARD34 = 0 + 0 = 0 kg

S14 = 0 + 0 = 0 kg.cm

ARD44 = 0 + 0 = 0 kg

S24 = 0 + 0 = 0 kg.cm

ARD54 = 0 + 6.Ec.Ix = 3008000 kg

S34 = 0 + 2.Ec.Ix = 250666666.7 kg.cmL4

ARD64 = - 6.Ec.Ix = -3008000 kg

S44 = 4.Ec.Ix = 501333333.3 kg.cmL4

Matriks kekakuan sbb :

S11 S12 S13 S14S21 S22 S23 S24

= S31 S32 S33 S34S41 S42 S43 S44

731111111 156666667 0 0

156666667 731111111 208888888.89 0

L42

L42

S d x d

EI

L1

ARD14

ARD24 S14

ARD34

A B

ARD34

S14

B

EI

L2

ARD44

S24 S34

ARD54

C D

ARD54

S34

D


= 0 208888889 919111111.11 250666666.67

0 0 250666666.67 501333333.33

S d x d


Matriks reaksi perletakan (ARD) karena displacement (akibat beban rotasi 1 satuan) sbb :

ARD11 ARD12 ARD13 ARD14ARD21 ARD22 ARD23 ARD24

= ARD31 ARD32 ARD33 ARD34ARD41 ARD42 ARD43 ARD44ARD51 ARD52 ARD53 ARD54ARD61 ARD62 ARD63 ARD64

2088888.9 0 0 0208888889 0 0 0

= -913888.89 1175000 0 0-1175000 913888.89 2088888.889 0

0 -2088889 919111.1111 30080000 0 -3008000 -3008000

Mencari Displacement dari DOF

Invers matriks kekakuan

731111111 156666667 0 0

156666667 731111111 208888888.89 0

S = 0 208888889 919111111.11 250666666.67

0 0 250666666.67 501333333.33

1.4392E-009 -3.335E-010 8.775864E-011 -4.38793E-011

-3.335E-010 1.5563E-009 -4.09540E-010 2.047702E-010

= 8.7759E-011 -4.095E-010 1.367572E-009 -6.83786E-010

-4.388E-011 2.0477E-010 -6.83786E-010 2.336574E-009

1.4392E-009 -3.335E-010 8.775864E-011 -4.38793E-011

D = -3.335E-010 1.5563E-009 -4.09540E-010 2.047702E-010

8.7759E-011 -4.095E-010 1.367572E-009 -6.83786E-010

-4.388E-011 2.0477E-010 -6.83786E-010 2.336574E-009

1.4392E-009 -3.335E-010 8.775864E-011 -4.38793E-011

D = -3.335E-010 1.5563E-009 -4.09540E-010 2.047702E-010

8.7759E-011 -4.095E-010 1.367572E-009 -6.83786E-010

-4.388E-011 2.0477E-010 -6.83786E-010 2.336574E-009

-7.10E-006 radD = 7.302E-005 rad

-0.000159 rad0.0001792 rad

ARDr x d

ARDr x d

D = S-1 .(AD - ADL)

S-1


HITUNG REAKSI PERLETAKAN MATRIKS AR

AR = ARL + ARD.D

1125 2088888.8889 0 0

56250 208888888.89 0 0

AR = 1625 + -913888.88889 1175000 0

1231.25 -1175000 913888.88889 2088888.9

168.75 0 -2088888.8889 919111.11

0 0 0 -3008000

1125 -14.825847846 1110.1742

56250 -1482.5847846 54767.415

AR = 1625 + 92.279159028 = 1717.2792

1231.25 -257.0104835 974.23952

168.75 240.46115949 409.21116

0 -60.903987168 -60.903987

A B C

P = 1000 Kgq1 = 7.5 Kg/cm q2 = 6 Kg/cm

L1 = 3 m L2/2 = 2 m L3/2 = 1.5 mL2/2 = 2 m

AR1

AR2

AR3 AR4


cmcmkg/cm2kg/cm2

cm4cmcmcmcmcm

kg/cmkg/cmkgkg.cmkg.cm

h

b

D E

L4 = 2.5 mL3/2 = 1.5 m

M1=-100000 Kg.cmM2=50000 Kg.cm

D E

L4 = 2.5 mL3 = 3 m

D3 D4

D E

L4 = 2.5 mL3 = 3 m

M1 M2


Bh --> orde matriks d x 1 = AD4x1

D E

q2 = 6 Kg/cm

L4 = 2.5 mL3/2 = 1.5 m

ADL3 ADL4

ARL5 ARL6

A B

q1 = 7.5 Kg/cm

L1 = 3 m

ADL1

ARL1 ARL3

B C

P = 1000 Kg

L2/2 = 2 m L2/2 = 2 m

ADL1 ADL2

ARL4


Gabungkan seluruh reaksi ujung batang akibat gaya luar dan kekangan sehingga dapat disusun matriks ADL dan ARL

kg

kg.cm

DC

q2 = 6 Kg/cm

L3/2 = 1.5 m L3/2 = 1.5 m

ADL3

ARL4 ARL5

E

L4 = 2.5 m

ADL4

ARL6

C

P = 1000 Kg

L2/2 = 2 m

ADL2

ARL4


kg

kg.cm

kg

kg.cm

Hitung matriks akibat displacement / perpindahan yaitu matriks kekakuan [S] dan matriks

D E

L4 = 2.5 mARD51=0 ARD61=0

S31 =0 S41 =0


(kg/cm2 x cm4 )/ cm2

(kg/cm2 x cm4 )/ cm

(kg/cm2 x cm4 )/ cm

θ = 1

EI

L2

C

S21

ARD41

EI

L4

S41

ARD61

E

D E

L4 = 2.5 mARD52 ARD62=0

S32S42 =0


EI

L2

θ = 1

θ = 1S22

ARD42

C

EI

L4

ARD52

S42

ARD62

E

D E

L4 = 2.5 mL3 = 3 mARD53 ARD63

S33 S43


EI

L2

ARD33

S23

ARD43

C

θ = 1

EI

L4

E

ARD53

S43

ARD63

D E

L4 = 2.5 mL3 = 3 mARD54 ARD64

S34 S44


EI

L2

ARD34

S24

ARD44

C

EI

L3

θ = 1

θ = 1S44

ARD64

E


Matriks reaksi perletakan (ARD) karena displacement (akibat beban rotasi 1 satuan) sbb :

0 -6250

0 - -19062.5

-100000 -14062.5

50000 0

6250

19062.5

-85937.5

50000


0 -7.09748E-006

0 7.301519E-005

0 X -0.0001589734

0 0.0001792207

3008000

-3008000

kg -> AR1kg.cm -> AR2

kg -> AR3kg -> AR4kg -> AR5kg -> AR6

D E

L4 = 2.5 mL3/2 = 1.5 m

M1=-100000 Kg.cmM2=50000 Kg.cm

AR5 AR6

04/23/2016 17:25:45


CONTOH SOAL ANALISA MATRIKS METODE KEKAKUAN LANGSUNG

GAMBAR BALOK MENERUS

Data Properties Penampang

Tinggi balok, h = 40 cmLebar balok, b = 25 cmMutu beton, fc' = 250 kg/cm2Modulus elastisitas beton, Ec =4700 x sqrt (fc'/10) x 10 Ec = 235000 kg/cm2

Ix = 133333.3 cm4Span (bentang) balok, L1 = 300 cmSpan (bentang) balok, L2 = 400 cmSpan (bentang) balok, L3 = 300 cmSpan (bentang) balok, L4 = 250 cmJarak beban, a3 = L3/2 a3 = 150 cm

Beban-beban yang bekerjaq1 = 7.5 kg/cmq2 = 6 kg/cm

P = 1000 kgM1 = 100000 kg.cmM2 = 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 UJUNGBATANG

SM11 = 12.Ec.Ix = 13925.92593 kg/cm SM31 = - 12.Ec.Ix =

SM21 = 6.Ec.Ix = 2088888.889 kg SM41 = 6.Ec.Ix =

Momen inersia balok, Ix = 1/12 x bh3

L13 L13

L12 L12

b

j

D1

D2 k

i

D3

D4

A B DC

P = 1000 Kgq1 = 7.5 Kg/cm q2 = 6 Kg/cm

L1 = 3 m L2/2 = 2 m L3/2 = 1.5 m L4 = 2.5 mL2/2 = 2 m L3/2 = 1.5 m

M1=-100000 Kg.cm

EI

L1

∆

SM11

SM21

SM31

SM41A

B

04/23/2016 17:25:45


- Perpindahan/Displacement arah 2 --> D2 (Rotasi arah sb-Z)



SM22 = 4.Ec.Ix = 417777778 kg SM42 = 2.Ec.Ix =L1 L1

- Perpindahan/Displacement arah 3 --> D3 (Translasi arah sb-Y)


SM13 = - 12.Ec.Ix = -13925.9259 kg/cm SM33 = 12.Ec.Ix =

SM23 = -6.Ec.Ix = -2088888.89 kg SM43 = -6.Ec.Ix =





L12 L12

L13 L13

L12 L12

L12 L12

θ = 1

θ = 1EI

L1

A B

SM12

SM42

SM32

SM22

EI

L1

∆

SM13

A

BSM23

SM33

SM43

EI

L1

θ = 1

θ = 1

SM14

SM24 SM44

SM34

A B

04/23/2016 17:25:45


Susun matriks kekakuan batang 1

SM11 SM12 SM13 SM14SM1 = SM21 SM22 SM23 SM24

SM31 SM32 SM33 SM34SM41 SM42 SM43 SM44

13925.925926 2088888.8889 -13925.92593 2088888.8889

SM1 = 2088888.8889 417777777.78 -2088888.889 208888888.89

-13925.92593 -2088888.889 13925.925926 -2088888.889

2088888.8889 208888888.89 -2088888.889 417777777.78



SM11 = 12.Ec.Ix = 5875 kg/cm SM31 = - 12.Ec.Ix =

SM21 = 6.Ec.Ix = 1175000 kg SM41 = 6.Ec.Ix =






GAMBAR FREEBODY BALOK DG PERPINDAHAN DAN REAKSI UJUNG

L23 L23

L22 L22

L22 L22

EI

L2

∆

SM11

SM21

SM31

SM41B

C

θ = 1

θ = 1EI

L2

B C

SM12

SM42

SM32

SM22

EI

L2

∆

SM13

B

CSM23

SM33

SM43

04/23/2016 17:25:45


BATANG

SM13 = - 12.Ec.Ix = -5875 kg/cm SM33 = 12.Ec.Ix =

SM23 = -6.Ec.Ix = -1175000 kg SM43 = -6.Ec.Ix =








5875 1175000 -5875 1175000

SM2 = 1175000 313333333.33 -1175000 156666666.67

-5875 -1175000 5875 -1175000

1175000 156666666.67 -1175000 313333333.33




SM21 = 6.Ec.Ix = 2088888.889 kg SM41 = 6.Ec.Ix =

L23 L23

L22 L22

L22 L22

L33 L33

L32 L32

EI

L2

θ = 1

θ = 1

SM14

SM24 SM44

SM34

B C

EI

L3

∆

SM11

SM21

SM31

SM41C

D

04/23/2016 17:25:46








SM13 = - 12.Ec.Ix = -13925.9259 kg/cm SM33 = 12.Ec.Ix =

SM23 = -6.Ec.Ix = -2088888.89 kg SM43 = -6.Ec.Ix =





L32 L32

L33 L33

L32 L32

L32 L32

θ = 1

θ = 1EI

L3

C D

SM12

SM42

SM32

SM22

EI

L3

θ = 1

θ = 1

SM14

SM24 SM44

SM34

C D

EI

L3

∆

SM13

C

DSM23

SM33

SM43

04/23/2016 17:25:46





13925.925926 2088888.8889 -13925.92593 2088888.8889

SM3 = 2088888.8889 417777777.78 -2088888.889 208888888.89

-13925.92593 -2088888.889 13925.925926 -2088888.889

2088888.8889 208888888.89 -2088888.889 417777777.78




SM21 = 6.Ec.Ix = 3008000 kg SM41 = 6.Ec.Ix =







L43 L43

L42 L42

L42 L42

EI

L4

∆

SM11

SM21

SM31

SM41D

E

θ = 1

θ = 1EI

L4

D E

SM12

SM42

SM32

SM22

EI

L4

∆

SM13

D

ESM23

SM33

SM43

04/23/2016 17:25:46


SM13 = - 12.Ec.Ix = -24064 kg/cm SM33 = 12.Ec.Ix =

SM23 = -6.Ec.Ix = -3008000 kg SM43 = -6.Ec.Ix =








24064 3008000 -24064 3008000

SM4 = 3008000 501333333.33 -3008000 250666666.67

-24064 -3008000 24064 -3008000

3008000 250666666.67 -3008000 501333333.33

II. SUSUN MATRIKS KEKAKUAN TITIK KUMPUL [Sj]

Matriks Sj disusun dari matriks SM

13925.925926 2088888.8889 -13925.92593 2088888.8889

SM1 = 2088888.8889 417777777.78 -2088888.889 208888888.89

-13925.92593 -2088888.889 13925.925926 -2088888.889 PENJUMLAHAN GAYA DI JOINT DILAKUKAN2088888.8889 208888888.89 -2088888.889 417777777.78 HANYA PADA TITIK / JOINT YANG TERJADI PERPINDAHAN

(DISPLACEMENT)

5875 1175000 -5875 1175000

SM2 = 1175000 313333333.33 -1175000 156666666.67

-5875 -1175000 5875 -1175000

1175000 156666666.67 -1175000 313333333.33

13925.925926 2088888.8889 -13925.92593 2088888.8889

SM3 = 2088888.8889 417777777.78 -2088888.889 208888888.89

-13925.92593 -2088888.889 13925.925926 -2088888.889

2088888.8889 208888888.89 -2088888.889 417777777.78

24064 3008000 -24064 3008000

SM4 = 3008000 501333333.33 -3008000 250666666.67

-24064 -3008000 24064 -3008000

L43 L43

L42 L42

L42 L42

EI

L4

θ = 1

θ = 1

SM14

SM24 SM44

SM34

D E

04/23/2016 17:25:46


3008000 250666666.67 -3008000 501333333.33

GAMBARKAN POSISI DOF UTK TATAULANG SJ

1 2 3 4 5 6 7

1 13925.925926 2088888.8889 -13925.92593 2088888.8889 0 0 0

2 2088888.8889 417777777.78 -2088888.889 208888888.89 0 0 0

3 -13925.92593 -2088888.889 19800.925926 -913888.8889 -5875 1175000 0

4 2088888.8889 208888888.89 -913888.8889 731111111.11 -1175000 156666667 0

Sj = 5 0 0 -5875 -1175000 19800.926 913888.89 -13925.92593

6 0 0 1175000 156666666.67 913888.89 731111111 -2088888.889

7 0 0 0 0 -13925.926 -2088889 37989.925926

8 0 0 0 0 2088888.9 208888889 919111.11111

9 0 0 0 0 0 0 -24064

10 0 0 0 0 0 0 3008000

5 6 7 D1 8 D2 9

1 2 3 4 5 6 7

1 731111111.11 156666666.67 0 0 2088888.9 208888889 -913888.8889

2 156666666.67 731111111.11 208888888.89 0 0 0 1175000

3 0 208888888.89 919111111.11 250666666.67 0 0 0

4 0 0 250666666.67 501333333.33 0 0 0

Sj = 5 2088888.8889 0 0 0 13925.926 2088888.9 -13925.92593

6 208888888.89 0 0 0 2088888.9 417777778 -2088888.889

7 -913888.8889 1175000 0 0 -13925.926 -2088889 19800.925926

8 -1175000 913888.88889 2088888.8889 0 0 0 -5875

9 0 -2088888.889 919111.11111 3008000 0 0 0

10 0 0 -3008000 -3008000 0 0 0

Bentuk matriks Sj yang ditataulang (re-arrangement) ---> berdasarkan posisi DOF

SFF SFR

1

2

3

4

3

4

5

6

Pertemuan joint dijumlahkan (digabungkan)

5

6

7

8 8

D1

1

2

A B DC

3

4

5

6

7

8D2D1 D3 D4

A B DC

D1

L1 = 3 m L2 = 4 m L4 = 2.5 mL3 = 3 m

D2 D3

04/23/2016 17:25:46


Sj =

731111111.11 156666666.67 0 0

156666666.67 731111111.11 208888888.89 0

= 0 208888888.89 919111111.11 250666666.67

0 0 250666666.67 501333333.33

1.4392418E-009 -3.334828E-010 8.7758644E-011 -4.387932E-011

-3.334828E-010 1.5562533E-009 -4.095403E-010 2.0477017E-010

= 8.7758644E-011 -4.095403E-010 1.3675722E-009 -6.837861E-010

-4.387932E-011 2.0477017E-010 -6.837861E-010 2.3365739E-009

III. SUSUN MATRIKS VEKTOR AKSI (GAYA) KOMBINASI [Ac]

Ac = Aj + AE

Aj ---> Beban aksi di joint

GAMBARKAN POSISI BEBAN LUAR PADA DOF

0 00 00 00 0

Aj = 0 = 00 00 0

- M1 -1000000 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

SRF SRR

Didapatkan matriks SFF

SFF

Hitung invers matriks SFF

SFF(-1)

1

2 4Urutan Penomoran

A B DC

L1 = 3 m L2 = 4 m L4 = 2.5 mL3 = 3 m

M1

A B DC E

P = 1000 Kgq1 = 7.5 Kg/cmq2 = 6 Kg/cm

L1 = 3 m L2/2 = 2 m L3/2 = 1.5 m L4 = 2.5 mL2/2 = 2 m L3/2 = 1.5 m

q1 = 7.5 Kg/cmAML2

DI TITIK A

DI TITIK B

DI TITIK C

DI TITIK D

DI TITIK E

04/23/2016 17:25:46


AML2 = = 56250 kg.cm

AML3 = q1.L1/2 = 1125 kg

AML4 = = -56250 kg.cm

Freebody B - C :


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 :


AML1 = 13/32. q2.L3 = 731.25 kg

AML2 = = 30937.5 kg.cm

AML3 = 3/32. q2.L3 = 168.75 kg

AML4 = = -14062.5 kg.cm

Freebody D - E :


AML1 = 0 = 0 kg

AML2 = 0 = 0 kg.cm

AML3 = 0 = 0 kg

AML4 = 0 = 0 kg.cm

Susun matriks AE dari matriks AML

1125AM1 = 56250

1125-56250

1125 -1125500 56250 -56250

AM2 = 50000 1625 -1625500 -6250 6250

-50000 1231.25 -1231.25AE = - -19062.5 AE = 19062.5

731.25 168.75 -168.75AM3 = 30937.5 -14062.5 14062.5

168.75 0 0

1/12 x q1.L12

-1/12 x q1.L12

11/192.q2.L32

- 5/192. q2.L32

A

L1 = 3 mAML1

B

P = 1000 Kg

L2/2 = 2 m

AML2

AML1

C

q2 = 6 Kg/cm

L3/2 = 1.5 m

AML2

AML1

D

L4 = 2.5 m

AML2

AML1

04/23/2016 17:25:46


-14062.5 0 0

0AM4 = 0

00

04/23/2016 17:25:46


Susun matriks Ac

1 0 -1125 -1125 52 0 -56250 -56250 63 0 -1625 -1625 74 0 6250 6250 D1

Ac = 5 0 + -1231.25 = -1231.25 86 0 19062.5 19062.5 D27 0 -168.75 -168.75 98 -100000 14062.5 -85937.5 D39 0 0 0 1010 50000 0 50000 D4

1 62502 19062.5 ---> AFC3 -85937.54 50000 Ac = AFC

Ac = 5 -1125 ARC6 -562507 -1625 ---> ARC8 -1231.259 -168.7510 0

Didapat matriks AFC dan ARC

6250 -1125AFC = 19062.5 -56250

-85937.5 ARC = -162550000 -1231.25

-168.750

1.43924E-009 -3.33483E-010 8.77586E-011 -4.38793E-011 6250

-3.33483E-010 1.55625E-009 -4.09540E-010 2.04770E-010 x 19062.5

8.77586E-011 -4.09540E-010 1.36757E-009 -6.83786E-010 -85937.5

-4.38793E-011 2.04770E-010 -6.83786E-010 2.33657E-009 50000

-7.097480E-006

7.3015192E-005

-0.0001589734

0.0001792207

V. HITUNG REAKSI PERLETAKAN [AR]

1125 2088888.8889 0 0 0 -7.097480E-006

56250 208888888.89 0 0 0 7.3015192E-005

AR = 1625 + -913888.8889 1175000 0 0 x -0.0001589734

1231.25 -1175000 913888.88889 2088888.8889 0 0.0001792207

168.75 0 -2088888.889 919111.11111 3008000

0 0 0 -3008000 -3008000

Tata ulang (re-arrangement) matriks Ac

IV. HITUNG PERPINDAHAN (DISPLACEMENT) [ DF ]

DF = SFF (-1). AFC

DF =

DF =

AR = -ARC + SRF.DF

04/23/2016 17:25:46


1125 -14.8258478 1110.174152 1110.17415256250 -1482.58478 54767.41522 54767.415221625 + 92.27915903 = 1717.279159 1717.279159

AR = 1231.25 -257.010484 974.2395165 974.2395165168.75 240.4611595 409.2111595 409.2111595

0 -60.9039872 -60.9039872 -60.9039872

A B DC

P = 1000 Kgq1 = 7.5 Kg/cm q2 = 6 Kg/cm

L1 = 3 m L2/2 = 2 m L3/2 = 1.5 m L4 = 2.5 mL2/2 = 2 m L3/2 = 1.5 m

M1=-100000 Kg.cm

AR1

AR2

AR3 AR4 AR5

04/23/2016 17:25:46


-13925.926 kg/cm

2088888.9 kg

h

E

L4 = 2.5 m

M2=50000 Kg.cm

04/23/2016 17:25:46


-2088889 kg/cm

208888889 kg

13925.926 kg/cm

-2088889 kg

-2088889 kg/cm

417777778 kg

04/23/2016 17:25:46


-5875 kg/cm

1175000 kg

-1175000 kg/cm

156666667 kg

04/23/2016 17:25:46


5875 kg/cm

-1175000 kg

-1175000 kg/cm

313333333 kg

-13925.926 kg/cm

2088888.9 kg

04/23/2016 17:25:46


-2088889 kg/cm

208888889 kg

13925.926 kg/cm

-2088889 kg

-2088889 kg/cm

417777778 kg

04/23/2016 17:25:46


-24064 kg/cm

3008000 kg

-3008000 kg/cm

250666667 kg

04/23/2016 17:25:46


24064 kg/cm

-3008000 kg

-3008000 kg/cm

501333333 kg

PENJUMLAHAN GAYA DI JOINT DILAKUKANHANYA PADA TITIK / JOINT YANG TERJADI PERPINDAHAN(DISPLACEMENT)

04/23/2016 17:25:46


8 9 10

0 0 0 50 0 0 60 0 0 70 0 0 D1

2088888.89 0 0 8208888889 0 0 D2919111.111 -24064 3008000 9919111111 -3008000 250666667 D3-3008000 24064 -3008000 10

250666667 -3008000 501333333 D4

D3 10 D4

8 9 10

-1175000 0 0

913888.889 -2088889 0

2088888.89 919111.11 -3008000

0 3008000 -3008000

0 0 0

0 0 0

-5875 0 0

19800.9259 -13925.926 0

-13925.9259 37989.926 -24064

0 -24064 24064

7 9

10

E

9

10

E

L4 = 2.5 m

D4

04/23/2016 17:25:46


3

E

L4 = 2.5 m

M2

E

L4 = 2.5 m

AML4

04/23/2016 17:25:46


B

AML3

C

P = 1000 Kg

L2/2 = 2 m

AML4

AML3

D

q2 = 6 Kg/cm

L3/2 = 1.5 m

AML4

AML3

E

L4 = 2.5 m

AML4

AML3

04/23/2016 17:25:46


04/23/2016 17:25:47


04/23/2016 17:25:47


--> AR1 --> AR2 --> AR3 --> AR4 --> AR5 --> AR6

E

L4 = 2.5 m

M2=50000 Kg.cm

AR6

GAMBAR 1 . GAMBAR BALOK MENERUS DIKEKANGBEBAN HILANGKAN, GBRKAN DOF

GAMBAR 2 . GAMBAR BALOK MENERUS DIKEKANGBEBAN HILANGKAN, KECUALI BEBAN SELARAS DOF

GAMBAR 1 . GAMBAR BALOK MENERUS DIKEKANGBEBAN DI MASUKAN, KECUALI YG SELARAS DOF, YG SELARAS DOF DIHILANGKAN, GAMBARKANREAKSI PERLETAKAN AKIBAT KEKANGAN (ARL DAN ADL)

A B C

P = 1000 Kgq1 = 7.5 Kg/cm q2 = 6 Kg/cm

L1 = 3 m L2/2 = 2 m L3/2 = 1.5 mL2/2 = 2 m

EI

L

θ = 1

θ = 1

ARD11

ARD21 S11

ARD31

A B

EI

L3

ARD41

S21 S31

ARD51

C D

A B C

L1 = 3 m L2 = 4 mARD14=0

ARD24=0

ARD34=0 ARD44=0

S14=0 S24=0

EI

L

D E

q2 = 6 Kg/cm

L3/2 = 1.5 m L4 = 2.5 mL3/2 = 1.5 m

M1=-100000 Kg.cmM2=50000 Kg.cm

θ = 1

θ = 1EI

L

A B

ARD31

S21S11

EI

L4

ARD51

S31 S41

ARD61

D E

D E

L4 = 2.5 mL3 = 3 mARD54 ARD64

S34 S44

EI

B

S21

ARD41

contoh soal analisa matriks (balok menerus) revisi2 01-07-15

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