Contoh Soal Analisa Matriks Kekakuan Langsung (Balok Menerus)

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




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)


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



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



BATANG


L12

L12


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


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




BATANG

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

L23

L23


L22

L22



BATANG


L22

L22


L2 L2



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


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



BATANG


L22

L22


L2 L2


SM11 SM12 SM13 SM14


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




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


2/21/2015 6:18 AM



BATANG


L32

L32


L3 L3



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



BATANG


L32

L32


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


2/21/2015 6:18 AM


SM11 SM12 SM13 SM14


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




BATANG


L43

L43


L42

L42



BATANG


L42

L42


L4 L4



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


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



BATANG


L42

L42


L4 L4


SM11 SM12 SM13 SM14


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


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


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


2/21/2015 6:18 AM

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 = 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 :


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


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


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


Contoh Soal Analisa Matriks Kekakuan Langsung (Balok Menerus)

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