Analisa Matriks (Balok Menerus) Revisi2 01-07-15

Hal.1 dari 115 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 17:19:17


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 17:19:17


- 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 17:19:17


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 17:19:17


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 17:19:17








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 17:19:17





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 17:19:17


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 17:19:17


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 17:19:17


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 17:19:17


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 17:19:17


-14062.5 0 0

0AM4 = 0

00

04/23/2016 17:19:17


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 17:19:17


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 17:19:17


-13925.926 kg/cm

2088888.9 kg

h

E

L4 = 2.5 m

M2=50000 Kg.cm

04/23/2016 17:19:18


-2088889 kg/cm

208888889 kg

13925.926 kg/cm

-2088889 kg

-2088889 kg/cm

417777778 kg

04/23/2016 17:19:18


-5875 kg/cm

1175000 kg

-1175000 kg/cm

156666667 kg

04/23/2016 17:19:18


5875 kg/cm

-1175000 kg

-1175000 kg/cm

313333333 kg

-13925.926 kg/cm

2088888.9 kg

04/23/2016 17:19:18


-2088889 kg/cm

208888889 kg

13925.926 kg/cm

-2088889 kg

-2088889 kg/cm

417777778 kg

04/23/2016 17:19:18


-24064 kg/cm

3008000 kg

-3008000 kg/cm

250666667 kg

04/23/2016 17:19:18


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 17:19:18


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 17:19:18


3

E

L4 = 2.5 m

M2

E

L4 = 2.5 m

AML4

04/23/2016 17:19:18


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 17:19:18


04/23/2016 17:19:18


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

E

L4 = 2.5 m

M2=50000 Kg.cm

AR6

Data beban

P1 = 20 kNq1 = 10 kN/m

Properties penampangBahan / material :

No. Element E G b (m) h (m)1 27000 0.3 0.62 27000 0.3 0.63 27000 0.3 0.6

PENYELESAIAN :

STEP 1 : Discretize and Global Degrees of Freedom (DOF)

DOF sesuai sumbu global struktur dan tentukan Beban atau Momen pada Titik Joint Struktur dalam Arah Sumbu Global/Sumbu Struktur

No.titik (Joint) Dis.Global Axis DOF

11 1 0 Trans. Y2 2 0 Rot. Z

23 3 0 Trans. Y4 d1 0 Rot. Z

35 5 0 Trans. Y6 d2 0 Rot. Z

47 d3 -20 Trans. Y8 d4 0 Rot. Z

Nomor joint untuk setiap element & Displacement titik sumbu global struktur pada tiap element

Element (i)Joint

Sudut (deg)Joint

Ujung j Ujung k j1 1 2 0 12 2 3 0 33 3 4 0 5

Rekap Elemen dalam Matriks Kekakuan BatangBatang 4.EI / L 2.EI / L 6.EI / L^2 12.EI / L^3

1 72.9 36.45 13.66875 3.41718752 291.6 145.8 218.7 218.73 291.6 145.8 218.7 218.7

Menghitung Gaya dan Momen Ujung (Akibat beban luar di sepanjang bentang dengan ujung dikekang/dijepit) (DALAM ARAH SUMBU LOKAL) :ELEMENT #1

Karena tidak ada gaya disepanjang bentang freebody maka rekasi ujung terkekang semuanya = 0

Gaya dan Momen Ujung pada Element #1 :Unit : kN-m

Va Ma Vb Mb0 0 0 0

ELEMENT #2

Beban / Momen

Arah Beban/Momen

A B

q1

L1

AML4

AML1

AML2

AML3


Va Ma Vb Mb10 3.3333333333 10 -3.3333333333

ELEMENT #3



Va Ma Vb Mb0 0 0 0

Rekap Gaya Ujung Batang (Sumbu Lokal)

No. BtgUjung kiri (j) Ujung kanan (k)

Beban FY(j) Beban MZ(j) Beban FY(k) Beban MZ(k)1 0 0 0 02 10 3.3333333333 10 -3.33333333333 0 0 0 0

Local Axis 1 2 3Element ke 1 1 2 3Element ke 2 3 4 5Element ke 3 5 6 d3

Element #1Local axis 1 2 3 4

Global axis 1 2 3 4DOF 0 0 0 d1


Global axis 1 2 3 4DOF 0 0 0 d2


Global axis 1 2 3 4DOF 0 0 d3 d4

===> DALAM SUMBU LOKAL

Batang = 1Elemen dalam matriks kekakuan batang :

4.EI / L 2.EI / L 6.EI / L^2 12.EI / L^372.9 36.45 13.66875 3.4171875

Kesesuaian SB LOKAL element dengan SB GLOBAL dan DOF arah global

Hitung matriks kekakuan batang 1 [SM1]

Cx = cos theta = 1Cy = sin theta = 0

Local axis 1 2 31 3.4171875 13.66875 -3.4171875

2 13.66875 72.9 -13.66875

3 -3.4171875 -13.66875 3.41718754 13.66875 36.45 -13.66875

1 2 3Global axis 1 2 3

1 3.4171875 13.66875 -3.4171875

2 13.66875 72.9 -13.668753 -3.4171875 -13.66875 3.41718754 13.66875 36.45 -13.66875



4.EI / L 2.EI / L 6.EI / L^2 12.EI / L^3291.6 145.8 218.7 218.7


Local axis 1 2 31 218.7 218.7 -218.7

2 218.7 291.6 -218.73 -218.7 -218.7 218.74 218.7 145.8 -218.7

3 d1 5Global axis 3 4 5

3 218.7 218.7 -218.7

4 218.7 291.6 -218.75 -218.7 -218.7 218.76 218.7 145.8 -218.7



4.EI / L 2.EI / L 6.EI / L^2 12.EI / L^3291.6 145.8 218.7 218.7


[SM1] =

[SMS1

] =


[SM2] =

[SMS2

] =


Local axis 1 2 31 218.7 218.7 -218.7

2 218.7 291.6 -218.73 -218.7 -218.7 218.74 218.7 145.8 -218.7

5 d2 d3Global axis 5 6 7

5 218.7 218.7 -218.7

6 218.7 291.6 -218.77 -218.7 -218.7 218.78 218.7 145.8 -218.7

IFERROR(VLOOKUP($D68,$D$49:$J$54,MATCH(E$67,$D$48:$J$48,0),FALSE),0)+IFERROR(VLOOKUP($D68,$D$58:$J$63,MATCH(E$67,$D$57:$J$57,0),FALSE),0)Urutkan sesuai sumbu global/struktur dengan DOF yang didahulukan

DOF : 1 2 3Dis.Global axis : 1 2 3

Ditata ulang manual ==> d1 d2 d34 6 7

4 364.5 145.8 06 145.8 583.2 -218.77 0 -218.7 218.78 0 145.8 -218.71 13.66875 0 02 36.45 0 03 205.03125 218.7 05 -218.7 0 -218.7

Hitung Matriks Beban Joint/ NodalNodal forces:

a. - Akibat beban di sepanjang bentang (kedua ujung dikekang)

Element: 1

0

= 000

Karena pada balok menerus sumbu lokal batang = sumbu global struktur maka Transformasi sb lokal ke global TIDAK DIPERLUKAN

Global axis1

[SM3] =

[SMS3

] =

MATRIKS KEKAKUAN STRUKTUR [SMS]

[SMS] =

[AML]1

= 234

Element: 2

10

= 3.333333333310

-3.3333333333


Global axis3

= 456

Element: 3

0

= 000


Global axis5

= 678

Susun matriks beban ekivalen

Joint Sb Global Element 11 1 01 2 02 3 02 4 0

-Ae = 3 5 0 +3 6 04 7 04 8 0

Ae = 0 1 10 2 2

-10 3 3

[AMS]1

[AML]2

[AMS]2

[AML]3

[AMS]3

-3.3333333333 4 d1-10 5 5

3.3333333333 6 d20 7 d30 8 d4

b. - Akibat Beban di joint (kedua ujung dikekang)

Joint Sb Global1 1 01 2 02 32 4 0

Aj = 3 5 03 6 04 7 -204 8 0

c. - Beban kombinasi atau gabungan (Ae + Aj)

Joint Sb Global DOF1 1 11 2 22 3 32 4 d1

Ac = 3 5 53 6 d24 7 d34 8 d4

Tata ulang Ac : Displacement / Deformasi Struktur

d1 -3.3333333333d2 3.3333333333

d3 -20d4 0

Ac = 1 02 03 -105 -10

Reactions:

GAYA-GAYA DALAM UJUNG ELEMEN

Batang: 1Deformasi struktur dalam arah SUMBU GLOBAL di batang 1, sbb :

Tentukan no. Sb global dahulu

1 1 0.0000002 2 0.000000

3 3 0.0000004 d1 0.051440

Deformasi struktur dalam arah SUMBU LOKAL di batang 1, sbb :

Karena pada balok menerus sumbu lokal batang = sumbu global struktur maka Transformasi GLOBAL ke LOKAL adalah TIDAK DIPERLUKAN

sb lokal DOF0.000000 1 10.000000 2 2

DM 1 = 0.000000 3 30.051440 4 d1

Gaya ujung batang

0 3.41718750 13.66875

0 + -3.41718750 13.66875

0 0.7031250 1.875

0 + -0.7031250 3.75


Tentukan no. Sb global dahulu3 3 0.0000004 d1 0.051440

5 5 0.0000006 d2 -0.151463



sb lokal DOF0.000000 1 30.051440 2 4

DM 2 = 0.000000 3 5-0.151463 4 d2

Gaya ujung batang

[DMS]1 =

AM 1 = AML 1 + SM 1 . DM 1

[AM1] =

[AM1] =

[DMS]2 =

10 218.73.3333333333 218.7

10 + -218.7-3.3333333333 218.7

10 -21.8753.3333333333 -7.0833333333

10 + 21.875-3.3333333333 -36.666666667


Tentukan no. Sb global dahulu5 5 0.0000006 d2 -0.151463

7 d3 -0.6687248 d4 -0.425812



sb lokal DOF0.000000 1 5-0.151463 2 6

DM 3 = -0.668724 3 d3-0.425812 4 d4

Gaya ujung batang

0 218.70 218.7

0 + -218.70 218.7

0 200 40

0 + -200 -1.42109E-014

SIM 3 v = 0.2 shear 2 = 0TABLE: Element Forces - Frames

AM 2 = AML 2 + SM 2 . DM 2

[AM2] =

[AM2] =

[DMS]3 =

AM 3 = AML 3 + SM 3 . DM 3

[AM3] =

[AM3] =

Frame Station OutputCase CaseTypeText m Text Text

1 0 DEAD LinStatic1 8 DEAD LinStatic2 0 DEAD LinStatic2 2 DEAD LinStatic3 0 DEAD LinStatic3 2 DEAD LinStatic

SIM 4 v = 0.25 shear 2 = 0TABLE: Element Forces - Frames

Frame Station OutputCase CaseTypeText m Text Text

1 0 DEAD LinStatic1 8 DEAD LinStatic2 0 DEAD LinStatic2 2 DEAD LinStatic3 0 DEAD LinStatic3 2 DEAD LinStatic

ELEMENT #1 MATRIKS SAP20000.703 -0.7031.875 -1.875

-0.703 -0.7033.750 3.750

ELEMENT #2 MATRIKS SAP2000-11.875 11.875-3.750 3.75031.875 31.875

-40.000 -40.000

ELEMENT #3 MATRIKS SAP200020.000 -20.00040.000 -40.000

-20.000 -20.0000.000 0.000

JIKA SHEAR LOCKING DIPERHITUNGAKAN MAKA :

Properties penampangBahan / material :

No. Element E G b (m) h (m)

1 27000 10800 0.3 0.62 27000 10800 0.3 0.63 27000 10800 0.3 0.6

12 EI / k G A

k = 5 = 0.83333333336

v = 0.25

G = E = 27000 =2(1+v) 2.5

12 4848 325.12

0.1369065505 -12 -4848 58.88

1 21 2

1 1.6428786058 6.57151442312 6.5715144231 44.51105769233 -1.642878606 -6.57151442314 6.5715144231 8.0610576923

=

=

IFERROR(VLOOKUP($D68,$D$49:$J$54,MATCH(E$67,$D$48:$J$48,0),FALSE),0)+IFERROR(VLOOKUP($D68,$D$58:$J$63,MATCH(E$67,$D$57:$J$57,0),FALSE),0)

d1 d2 d34 6 7

4 222.555288462 32.244230769 06 32.2442307692 356.08846154 -105.144230777 0 -105.1442308 105.144230778 0 32.244230769 -105.144230771 6.5715144231 0 02 8.0610576923 0 03 98.5727163462 105.14423077 05 -105.14423077 0 -105.14423077

d1 -3.3333333333d2 3.3333333333

d3 -20d4 01 02 03 -105 -10

DF = S

FF-1.A

FC

DF =

A (m2) L (m) I (m4) Keterangan1.800000E-01 8 5.400000E-03 A1 (30x60)1.800000E-01 2 5.400000E-03 A1 (30x60)1.800000E-01 2 5.400000E-03 A1 (30x60)

Defined DOF

after boundary conditions

DOF sesuai sumbu global struktur dan tentukan Beban atau Momen pada Titik Joint Struktur dalam Arah Sumbu Global/Sumbu Struktur

Nomor joint untuk setiap element & Displacement titik sumbu global struktur pada tiap elementJoint

j k2 3 4 ==> Global axis4 5 6 ==> Global axis6 7 8 ==> Global axis

Sudut (deg) Cx Cy0 1 00 1 00 1 0

Menghitung Gaya dan Momen Ujung (Akibat beban luar di sepanjang bentang dengan ujung dikekang/dijepit) (DALAM ARAH SUMBU LOKAL) :


==> perjanjian tanda(+)(-) berdasarkan sumbu lokal


4d1 <=== DOF sesuai sb global d2 <=== DOF sesuai sb global d4 <=== DOF sesuai sb global


SM11 =

SM21 =

413.66875

36.45 SM31 =

-13.6687572.9

SM41 =

d14

13.66875

36.45 SM1-13.66875

72.9


4218.7

145.8-218.7291.6

d26

218.7

145.8-218.7291.6


4218.7

145.8-218.7291.6

d48

218.7

145.8-218.7291.6


d1 5 d2 d3 d44 5 6 7 8

d4 1 2 3 58 1 2 3 5

0 13.66875 36.45 205.03125 -218.7145.8 0 0 218.7 0-218.7 0 0 0 -218.7291.6 0 0 0 218.7

0 3.4171875 13.66875 -3.4171875 00 13.66875 72.9 -13.66875 00 -3.4171875 -13.66875 222.1171875 -218.7

218.7 0 0 -218.7 437.4

Local axis1

234


0

000

Local axis1

234


10

3.333333333310

-3.3333333333

Local axis1

234


0

000

Element 2 Element 30 0 00 0 010 0 10

3.3333333333 0 3.333333333310 + 0 = 10

-3.3333333333 0 -3.33333333330 0 00 0 0

DOF1 12 13 2d1 25 3d2 3d3 4d4 4

0 0 00 0 0

-10 0 -10-3.3333333333 0 -3.3333333333

-10 + 0 = -103.3333333333 0 3.3333333333

0 -20 -200 0 0

Displacement / Deformasi Struktur

d1 = 0.051440d2 = -0.151463

d3 = -0.668724d4 = -0.425812

ADA MASALAHHipotesis :Nilai v berpengaruh pada displacement jika shear2 dan 3 =1Nilai v tidak berpengaruh pada displacement jika shear2 dan 3 =0

DF = S

FF-1.A

FC

DF =


13.66875 -3.4171875 13.66875 0.00000072.9 -13.66875 36.45 0.000000

-13.66875 3.4171875 -13.66875 x 0.00000036.45 -13.66875 72.9 0.051440

sb lokal DOF / sb global0.703 1 11.875 2 2

= -0.703 3 33.750 4 d1


218.7 -218.7 218.7 0.000000291.6 -218.7 145.8 0.051440

-218.7 218.7 -218.7 x 0.000000145.8 -218.7 291.6 -0.151463

sb lokal DOF / sb global-11.875 1 3-3.750 2 4

= 31.875 3 5-40.000 4 d2


218.7 -218.7 218.7 0.000000291.6 -218.7 145.8 -0.151463

-218.7 218.7 -218.7 x -0.668724145.8 -218.7 291.6 -0.425812

sb lokal DOF / sb global20.000 1 540.000 2 6

= -20.000 3 d30.000 4 d4

shear 3 = 0

P V2 V3 T M2KN KN KN KN-m KN-m

0 -0.703 0 0 00 -0.703 0 0 00 11.875 0 0 00 31.875 0 0 00 -20 0 0 00 -20 0 0 0

shear 3 = 0

P V2 V3 T M2KN KN KN KN-m KN-m

0 -0.703 0 0 00 -0.703 0 0 00 11.875 0 0 00 31.875 0 0 00 -20 0 0 00 -20 0 0 0

SELISIH0.000 0.0001250.000 00.000 -0.0001250.000 0

0.000 00.000 00.000 00.000 0

0.000 00.000 00.000 00.000 -1.065785E-014

A (m2) L (m) I (m4) Keterangan k1.800000E-01 8 5.400000E-03 A1 (30x60) 0.83333333331.800000E-01 2 5.400000E-03 A1 (30x60) 0.83333333331.800000E-01 2 5.400000E-03 A1 (30x60) 0.8333333333

10800

-12 48 12-48 58.88 1212 -48 8.7620192308 -12

-48 325.12 12

3 d1 33 4 3

-1.6428786058 6.5715144231 3 105.1442307692-6.5715144231 8.0610576923 4 105.14423076921.6428786058 -6.5715144231 5 -105.1442307692

-6.5715144231 44.5110576923 6 105.1442307692


d4 1 2 3 58 1 2 3 5

0 6.5715144231 8.0610576923 98.5727163462 -105.144230769232.2442307692 0 0 105.1442307692 0

-105.1442307692 0 0 0 -105.1442307692178.0442307692 0 0 0 105.1442307692

0 1.6428786058 6.5715144231 -1.6428786058 00 6.5715144231 44.5110576923 -6.5715144231 00 -1.6428786058 -6.5715144231 106.787109375 -105.1442307692

105.1442307692 0 0 -105.144230769 210.2884615385

MMULT(MINVERSE(D224:G227),D331:D334)

d1 = 0.0152600174d2 = -0.2087049607

d3 = -0.8819732502d4 = -0.4830533832

SALAH

12.Ec.Ix SM12 = 6.Ec.Ix


L1

- 12.Ec.Ix SM32 = - 6.Ec.Ix


L1

SM11 SM12 SM13 SM14

= SM21 SM22 SM23 SM24SM31 SM32 SM33 SM34SM41 SM42 SM43 SM44

L13 L12

L12

L13 L12

L12


11

4 d1 3.41718756 d2 13.668757 d3 -3.41718758 d4 13.668751 12 23 35 5

1 1 3.41718752 2 13.668753 3 -3.4171875

d1 4 13.668755 5 0

d2 6 0d3 7 0d4 8 0

11

364.5145.8

00

13.66875

36.45205.03125

-218.7

SIM3 SIM 1 v = 0.252 -0.051440 TABLE: Joint Displacements3 0.151463 Joint OutputCase

4 -0.668724 Text Text4 0.425812 1 DEAD

2 DEAD3 DEAD4 DEAD

Nilai v berpengaruh pada displacement jika shear2 dan 3 =1Nilai v tidak berpengaruh pada displacement jika shear2 dan 3 =0 SIM 2 v = 0.2

TABLE: Joint DisplacementsJoint OutputCaseText Text

1 DEAD2 DEAD3 DEAD4 DEAD

SIM 3 v = 0.2

TABLE: Joint DisplacementsJoint OutputCaseText Text

1 DEAD2 DEAD3 DEAD4 DEAD

SIM 4 v = 0.25TABLE: Joint Displacements

Joint OutputCaseText Text

1 DEAD2 DEAD

3 DEAD4 DEAD

M3 FrameElem ElemStationKN-m Text m

-1.875 1-1 03.75 1-1 83.75 2-1 0-40 2-1 2-40 3-1 0

-3.553E-015 3-1 2

M3 FrameElem ElemStationKN-m Text m

-1.875 1-1 03.75 1-1 83.75 2-1 0-40 2-1 2-40 3-1 0

-3.553E-015 3-1 2

1.080000E+001.080000E+001.080000E+00

12 -12 12 1220.32 -12 3.68 12

-12 12 -12 8.7620192308 -123.68 -12 20.32 12

d1 5 d2 54 5 6 5

105.1442307692 -105.1442308 105.144231 5 105.144231178.0442307692 -105.1442308 32.2442308 6 105.144231

-105.1442307692 105.1442308 -105.14423 7 -105.1442332.2442307692 -105.1442308 178.044231 8 105.144231

SM13 = - 12.Ec.Ix SM14 = 6.Ec.Ix

SM23 = -6.Ec.Ix SM24 = 2.Ec.Ix

L1

SM33 = 12.Ec.Ix SM34 = - 6.Ec.Ix

SM43 = -6.Ec.Ix SM44 = 4.Ec.Ix

L1

L13 L12

L12

L13 L12

L12

2 3 d1 5 d2 d3 d42 3 4 5 6 7 8

13.66875 -3.4171875 13.6687572.9 -13.66875 36.45

-13.66875 3.4171875 -13.6687536.45 -13.66875 72.9

218.7 218.7 -218.7 218.7218.7 291.6 -218.7 145.8

-218.7 -218.7 218.7 -218.7218.7 145.8 -218.7 291.6

218.7 218.7 -218.7 218.7218.7 291.6 -218.7 145.8

-218.7 -218.7 218.7 -218.7218.7 145.8 -218.7 291.6

13.66875 -3.4171875 13.66875 0 0 0 072.9 -13.66875 36.45 0 0 0 0

-13.66875 222.117188 205.03125 -218.7 218.7 0 0

36.45 205.03125 364.5 -218.7 145.8 0 00 -218.7 -218.7 437.4 0 -218.7 218.70 218.7 145.8 0 583.2 -218.7 145.80 0 0 -218.7 -218.7 218.7 -218.70 0 0 218.7 145.8 -218.7 291.6

2 3 d1 5 d2 d3 d42 3 4 5 6 7 8

145.8 0 0583.2 -218.7 145.8

-218.7 218.7 -218.7145.8 -218.7 291.6

0 0 0

0 0 0218.7 0 0

0 -218.7 218.7

shear 2 = 1 shear 3 = 1TABLE: Joint Displacements

CaseType U1 U2 U3 R1 R2 R3

Text m m m Radians Radians RadiansLinStatic 0 0 0 0 0 0LinStatic 0 0 0 0 -0.041617 0LinStatic 0 0 0 0 0.166457 0LinStatic 0 0 -0.723404 0 0.440806 0


CaseType U1 U2 U3 R1 R2 R3Text m m m Radians Radians Radians

LinStatic 0 0 0 0 0 0LinStatic 0 0 0 0 -0.042005 0LinStatic 0 0 0 0 0.165868 0LinStatic 0 0 -0.721237 0 0.440216 0

shear 2 = 0 shear 3 = 0

TABLE: Joint DisplacementsCaseType U1 U2 U3 R1 R2 R3

Text m m m Radians Radians RadiansLinStatic 0 0 0 0 0 0LinStatic 0 0 0 0 -0.05144 0LinStatic 0 0 0 0 0.151463 0LinStatic 0 0 -0.668724 0 0.425812 0


CaseType U1 U2 U3 R1 R2 R3Text m m m Radians Radians Radians

LinStatic 0 0 0 0 0 0LinStatic 0 0 0 0 -0.05144 0

LinStatic 0 0 0 0 0.151463 0LinStatic 0 0 -0.668724 0 0.425812 0

- -

- -

12 -12 1220.32 -12 3.68

-12 12 -123.68 -12 20.32

d2 d3 d46 7 8

105.144231 -105.14423 105.144231178.044231 -105.14423 32.2442308-105.14423 105.144231 -105.1442332.2442308 -105.14423 178.044231

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

Run on 4/1/2016 10:09

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Analisa Matriks (Balok Menerus) Revisi2 01-07-15

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