Analisa Struktur III

5/20/2018 Analisa Struktur III

1/106

Persamaan Linier Simultan Metode Iterasi Seidei

4 x + 3 y + 1 z = 13

1 x + 2 y + 3 z = 14

3 x + 2 y + 5 z = 22

Diubah menjadi

x = (13-3y-z) / 4

y = (14-x-3z) / 2

z = (22-3x-2y) / 5

x y z x y z

Nilai Awal 0 0 0 0 0 0

Iterasi 1 3.25 5.375 0.3 3.25 5.375 0.3

Iterasi 2 -0.856 6.978 2.123 -0.856 6.978 2.123

Iterasi 3 -2.514 5.073 3.879 -2.514 5.073 3.879

Iterasi 4 -1.525 1.944 4.537 -1.525 1.944 4.537

Iterasi 5 0.658 -0.135 4.059 0.658 -0.135 4.059

Iterasi 6 2.337 -0.257 3.101 2.337 -0.257 3.101

Iterasi 7 2.668 1.015 2.394 2.668 1.015 2.394Iterasi 8 1.891 2.464 2.280 1.891 2.464 2.280

Iterasi 9 0.832 3.164 2.635 0.832 3.164 2.635

Iterasi 10 0.218 2.938 3.094 0.218 2.938 3.094

Iterasi 11 0.273 2.223 3.347 0.273 2.223 3.347

Iterasi 12 0.746 1.606 3.310 0.746 1.606 3.310

Iterasi 13 1.218 1.426 3.099 1.218 1.426 3.099

Iterasi 14 1.406 1.649 2.897 1.406 1.649 2.897Iterasi 15 1.289 2.010 2.823 1.289 2.010 2.823

Iterasi 16 1.037 2.248 2.879 1.037 2.248 2.879

Iterasi 17 0.845 2.259 2.990 0.845 2.259 2.990


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Iterasi 36 1.005 1.995 2.999 1.005 1.995 2.999

Iterasi 37 1.004 2.000 2.998 1.004 2.000 2.998

Iterasi 38 1.000 2.003 2.998 1.000 2.003 2.998

Iterasi 39 0.998 2.003 3.000 0.998 2.003 3.000Iterasi 40 0.997 2.001 3.001 0.997 2.001 3.001

Iterasi 41 0.999 1.999 3.001 0.999 1.999 3.001

Iterasi 42 1.000 1.998 3.001 1.000 1.998 3.001

Iterasi 43 1.001 1.999 3.000 1.001 1.999 3.000

Iterasi 44 1.001 2.000 2.999 1.001 2.000 2.999

Iterasi 45 1.000 2.001 3.000 1.000 2.001 3.000

Iterasi 46 1.000 2.001 3.000 1.000 2.001 3.000

Iterasi 47 0.999 2.001 3.000 0.999 2.001 3.000

Iterasi 48 1.000 2.000 3.000 1.000 2.000 3.000

Iterasi 49 1.000 2.000 3.000 1.000 2.000 3.000

Iterasi 50 1.000 2.000 3.000 1.000 2.000 3.000


3/106

Mencari Invers Matriks metode Cholesky

A = 3 2 0 = a11 a12 a13

2 4 -1 a21 a22 a230 -1 1 a31 a32 a33

a11 a12 a13 l11 0 0 l11 l21 l31

a21 a22 a23 = l21 l22 0 0 l22 l32

a31 a32 a33 l31 l32 l33 . 0 0 l33

a11 = l11.l11 l11 = a11 = 1.732

a21 = l21.l11 l21 = a21 / l11 = 1.155

a31 = l31.l11 l31 = a31 / l11 = 0

a22 = l21.l21 + l22.l22 l22 = (a22-l212) = 1.633

a32 = l31.l21 + l32.l22 l32 = ( a32 - ) / l22 = -0.612

a33 = l31.l31 + l32.l32 + l33.l33 l33 = (a33-l312-l312-l322) = 0.791

A = L . L T

3 2 0 1.732 0 0 1.732 1.155 0

2 4 -1 = 1.155 1.633 0 0 1.633 -0.612

0 -1 1 0 -0.612 0.791 . 0 0 0.791

det L = 2.2361

1.633 0 1.155 0 1.155 1.633

-0.612 0.791 0 0.791 0 -0.612

M13M11 M12


4/106

C11 = 1.291 C21 = 0 C31 = 0

C12 = -0.913 C22 = 1.369 C32 = 0

C13 = -0.707 C23 = 1.061 C33 = 2.828

L + = 1.291 0 0

-0.913 1.369 0

-0.707 1.061 2.828

L

-1

= U = 0.577 0 0-0.408 0.612 0

-0.316 0.474 1.265

U T = 0.577 -0.408 -0.316

0 0.612 0.474

0 0 1.265

A -1 = U T . U

= 0.577 -0.408 -0.316 0.577 0 0

0 0.612 0.474 -0.408 0.612 0

0 0 1.265 -0.316 0.474 1.265

= 0.6 -0.4 -0.4

-0.4 0.6 0.6

-0.4 0.6 1.6


5/106

Balok Menerus

diketahui :

q = 600 kg/m'

EI1 = 1

EI2 = 1

L1 = 10 mL2 = 8 m

Momen Primer

MAB = - q . L12 = -5000 kgm

12

MBA = q . L12 = 5000 kgm

12

MBC = - q . L22 = -3200 kgm

12

L2

CBA

EI2

q

EI1

L1


6/106

S = EI 0.4 0.2 0 0

0.2 0.4 0 0

0 0 0.5 0.25

0 0 0.25 0.5

Matriks Kekakuan Struktur

K = B . S . A

= A T S . A

= 0.2 0.4 0.5 0.25 . 0 EI

1

1

0

= 0.9 EI

K -1 =

0.9 EI

Deformasi Struktur

D = K -1 . Q

D4 = . 1800

0.9 EI

= 2000

EI

1

1


7/106

MC = H4 - MCB = 500 - 3200 = -2700


8/106


9/106

.


10/106

Kgm


11/106

Portal Tanpa Goyangan

diketahui :

q = 300 kg/m'

p = 600 kg

EI1 = 1

EI2 = 2EI3 = 1

L = 5 m

a = 3 m

b = 2 m

Momen Primer

MAB = - P . ab2

= -288 kgm(a+b)

2

MBA = P . a2b = 432 kgm

a

b

L

C

D

B

A

EI2PP

q

EI3EI1


12/106

0 1 d4

0 1 d5

0 0 d6

D1 D2

A T = 0 1 1 0 0 0

0 0 0 1 1 0

Matriks Kekakuan Elemen

S = EI 0.8 0.4 0 0 0

0.4 0.8 0 0 0

0 0 1.6 0.8 0

0 0 0.8 1.6 0

0 0 0 0 0.8

0 0 0 0 0.4

= 2 EI 2 1 0 0 0

5 1 2 0 0 0

0 0 4 2 00 0 2 4 0

0 0 0 0 2

0 0 0 0 1


K = B . S . A

= A T S . A

= 2 EI 1 2 4 2 0


13/106

D1 = . 6 -2 .

D2 64 EI -2 6

= . -1544

64 EI 1544

= 1 . -120.625

EI 120.625

Gaya Dalam

H = S . A . D

H1 = 2 EI 1 0

H2 5 2 0

H3 4 2

H4 2 4

H5 0 2 1 .

H6 0 1 . EI

H1 = -48.25

H2 -96.5

H3 -96.5

H4 96.5

H5 96.5

H6 48.25

Momen Lentur

MA = H1 - MAB = -48.25 - -288 = 239.75

5

5


14/106


15/106

0

0

0

0

0.4

0.8

0

0

00

1

2

0 0 0


16/106

-193

193

-120.625

120.625

Kgm


17/106


18/106

Portal Sederhana Dengan Pergoyangan Mendatar

diketahui :

P1 = 1000 kg

P2 = 600 kg

P3 = 400 kg

EI1 = 1EI2 = 2

EI3 = 1

L1 = 2 m

L2 = 2 m

L3 = 4 m

Momen Primer

MAB = 0 kgm

MBA = 0 kgm

MBC = - P1. (L1 + L2) = -500 kgm

L3

L1

C

D

B

A

EI2

P3P2

P1

EI3EI1

L2


19/106

d6 0.25 0 0

A = 0.25 0 0

0.25 1 00 1 0

0 0 1

0.25 0 1

0.25 0 0

A T = 0.25 0.25 0 0 0.25 0.25

0 1 1 0 0 0

0 0 0 1 1 0


S = EI 1 0.5 0 0 0

0.5 1 0 0 0

0 0 2 1 0

0 0 1 2 0

0 0 0 0 10 0 0 0 0.5

= 1 EI 2 1 0 0 0

2 1 2 0 0 0

0 0 4 2 0

0 0 2 4 0

0 0 0 0 20 0 0 0 1



20/106

-48 -15

Deformasi Struktur

D = K-1

. Q

D1 = . 512 -48 -48 .

D2 156 EI -48 63 -15

D3 -48 -15 63

D1 = . 512000 = 1 .

D2 156 EI -87000 EI

D3 -9000

Gaya Dalam

H = S . A . D

H1 = 1 . 0.75 1 0 . 3282.051

H2 2 0.75 2 0 -557.692H3 0 4 2 -57.6923

H4 0 2 4

H5 0.75 0 2

H6 0.75 0 1

H1 = 951.9231

H2 673.0769H3 -1173.08

H4 -673.077

H5 1173.077

1

1


21/106


22/106


23/106

0

0

0

0

0.51

0

0

0

0

1

2


24/106

63

1000

-500

500

3282.051

-557.692

-57.6923


25/106

A = 11.20833 -6.25 -6

-6.25 30 10

-6 10 28

det A = 10165 - 3294.583

= 6870.417

30 10 -6.25 10 -6.25 30

10 28 -6 28 -6 10

-6.25 -6 11.20833333 -6 11.20833 -6.25

10 28 -6 28 -6 10

-6.25 -6 11.20833333 -6 11.20833 -6.25

30 10 -6.25 10 -6.25 30

M 1 1 = 740

M 1 2 = -115

M 1 3 = 117.5

M 2 1 = -115

M 2 2 = 277.8333

M 2 3 = 74.58333

M 3 1 = 117.5

M 3 2 = 74.58333M 3 3 = 297.1875

M11 M12 M13

M21 M22 M23

M31 M32 M33

NO PRINT !!!!!


26/106


diketahui :

P1 = 1000 kg

P2 = 400 kg

P3 = 500 kgP4 = 600 kg

P5 = 400 kg

PCB = 666.67 kg

PCD = 833.33 kg

EI1 = 1

EI2 = 2

EI3 = 1

L1 = 1 m

L2 = 2 m

L3 = 2 m

L5

L2

CE B

A

EI2

P5P4

P1

EI3EI1

L3

P2

L4L1

L6

D

L7

P3


27/106

Gaya Ekivalen

Q1 = P4+ P5 - PCB = 333.33 kgm

Q2 = P2- P3 = -100 kgm

Q2 = P3 = 500 kgm

Matriks Deformasi

d1 = 0.25 0 0 . D1

d2 0.25 1 0 D2

d3 - 1/3 1 0 D3

d4 - 1/3 0 1

d5 1/3 0 1

d6 1/3 0 0

A = 0.25 0 0

0.25 1 0

- 1/3 1 0

- 1/3 0 1

1/3 0 1

1/3 0 0

A T = 0.25 0.25 - 1/3 - 1/3 1/3 1/3

0 1 1 0 0 0

0 0 0 1 1 0


S = EI 1 0.5 0 0 0

0.5 1 0 0 0

0 0 2 1 0

0 0 1 2 0


28/106

0 0 10 20 8

= 1 EI 11.20833 -6.25 -6

10 -6.25 30 10

-6 10 28

K -1 = . 1 . 740 115

1 EI 6870.417 115 277.8333

117.5 -74.5833

Deformasi Struktur

D = K -1 . Q

D1 = . 740 115 117.5 .

D2 687.0417 EI 115 277.8333 -74.5833

D3 117.5 -74.5833 297.1875

D1 = . 293916.7 = 1 .

D2 687.0417 EI -26741.7 EI

D3 195218.8

Gaya DalamH = S . A . D

H1 = 1 3 75 5 0 427 8004

1

10

1


29/106

MBC = H3 - MBC = -221.502 - -500 = 278.4978

MCB = H4 - MCB = 101.5647 - 500 = -398.435

MCD = H5 - MCD = 398.4353 - 0 = 398.4353

MD = H6 - MDC = 284.7777 - 0 = 284.7777


30/106


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32/106

0

0

0

0


33/106

4 - 1/3 1 0

- 1/3 0 1

1/3 0 1

1/3 0 0

117.5

-74.5833

297.1875

333.3333

-100

500

427.8004

-38.9229

284.144


34/106

Kgm

Kgm

Kgm

Kgm


35/106

Rangka Batang : [K]-1

Metode Partisi

diketahui :

P1 = 1000 kg

P2 = 2000 kg

A1 = 1 E1 = 1

A2 = 1 E2 = 1A3 = 1 E3 = 1

A4 = 1 E4 = 1

A5 = 1 E5 = 1

L1 = 2 m

L2 = 2 m

L3 = 1.5 m

L4 = L1

2

+ L3

2

= 2.5 mL5 = L2

2+ L3

2 = 2.5 m

ik f i


36/106

0 0 0 4/5 - 4/5

Menurut Hukum Hooke

d . AE = H . L

H = (AE/L) . d

(AE/L) = Kekauan Axial Batang


S = (A1E1) / L1 0 0 0 0

0 (A2E2) / L2 0 0 0

0 0 (A3E3) / L3 0 0

0 0 0 (A4E4) / L4 00 0 0 0 (A5E5) / L5

d1 = 1 d2 = 1 d3 = 1 d4 = 1 d5 = 1

= 1/2 0 0 0 0

0 1/2 0 0 0

0 0 2/3 0 0

0 0 0 2/5 00 0 0 0 2/5


K = B . S . A

= A T S . A

= A . E 0 0 2/3 0 0

0 - 1/2 0 0 0

0 0 2/3 6/25 6/25


37/106

K21 = - 2/3 0

0 0

rumus metode partisi

E1 = K11-1

. K12 F22 =

E2 = K21 . K11-1 F21 =

E3 = E2 . K12 F12 =

E4 = K22 - E3 F11 =

operasi metode partisi

E1 = 1 . 1/2 0 . - 2/3

1/3 0 2/3 0

E2 = - 2/3 0 . 1 . 1/2

0 0 1/3 0

E3 = -1 0 . - 2/3 0

0 0 0 0

E4 = 358/375 0 - 2/3 0

0 64/125 0 0

F22 = 1 . 64/125 0

142/963 0 36/125

F21 = - 3.4722222 0 . -1 0

0 1.953125 0 0


38/106

Q = Q1 = -1000 Kg

Q2 0 Kg

Q3 0 Kg

Q4 2000 Kg

Deformasi Struktur (Lendutan akibat Gaya Luar)

D = K -1 . Q

D1 = 1 . 4.97222222 0 3.472222222 0

D2 A . E 0 2 0 0

D3 3.47222222 0 3.472222222 0

D4 0 0 0 1.953125

D1 = 1 . -4972.2222

D2 A . E 0

D3 -3472.2222

D4 3906.25

Gaya Dalam pada BatangH = S . A . D

H1 = 1 . 0 0 0 0

H2 A . E 0 -0.5 0 0

H3 0.66666667 0 -0.66666667 0

H4 0 0 -0.24 0.32

H5 0 0 -0.24 -0.32

H1 = 1 . 0

H2 A . E 0


39/106


40/106

H1

H2

H3

H4H5

. 0 0 0 0

0 -1 0 0

1 0 1 0


41/106

K22 = 358/375 0

0 64/125

E4-1

-F22 . E2

-E1 . F22

K11-1

- (F12 .E22)

0 = -1 0

0 0 0

0 = -1 0

2/3 0 0

= 2/3 0

0 0

= 36/125 0

0 64/125

= 3.4722222 0

0 1.953125

= 3.4722222 0

0 0


42/106

. -1000

0

0

2000

. -4972.222

0

-3472.222

3906.25


43/106


Metode Partisi

diketahui :

P1 = 1000 kg

P2 = 2000 kg

A1 = 1 E1 = 1

A2 = 1 E2 = 1A3 = 1 E3 = 1

A4 = 1 E4 = 1

A5 = 1 E5 = 1

L1 = 2 m

L2 = 2 m

L3 = 1.5 m

L4 = L1

2

+ L3

2

= 2.5 mL5 = L2

2+ L3

2 = 2.5 m

ik f i


44/106

0 0 0 4/5 - 4/5

Menurut Hukum Hooke

d . AE = H . L

H = (AE/L) . d



S = (A1E1) / L1 0 0 0 0

0 (A2E2) / L2 0 0 0

0 0 (A3E3) / L3 0 0

0 0 0 (A4E4) / L4 00 0 0 0 (A5E5) / L5

d1 = 1 d2 = 1 d3 = 1 d4 = 1 d5 = 1

= 1/2 0 0 0 0

0 1/2 0 0 0

0 0 2/3 0 0

0 0 0 2/5 00 0 0 0 2/5


K = B . S . A

= A T S . A

= A . E 0 0 2/3 0 0

0 - 1/2 0 0 0

0 0 2/3 6/25 6/25


45/106

Q2 0 Kg

Q3 0 Kg

Q4 2000 Kg


D = K -1 . Q

D1 = 1 . 4.97222222 0 3.472222222 0

D2 A . E 0 2 0 0

D3 3.47222222 0 3.472222222 0

D4 0 0 0 1.953125

D1 = 1 . -4972.2222

D2 A . E 0

D3 -3472.2222

D4 3906.25

Gaya Dalam pada Batang

H = S . A . D

H1 = 1 . 0 0 0 0

H2 A . E 0 -0.5 0 0

H3 0.66666667 0 -0.66666667 0

H4 0 0 -0.24 0.32

H5 0 0 -0.24 -0.32

H1 = 1 . 0

H2 A . E 0

H3 -1000


46/106


47/106

H1

H2

H3

H4H5

. 0 0 0 0

0 -1 0 0

1 0 1 0


48/106

. -1000

0

0

2000

. -4972.222

0

-3472.222

3906.25


49/106

Invers Matriks Ordo 4x4

Metode Gauss-Jordan

A = 2/3 0 - 2/3 0

0 1/2 0 0

- 2/3 0 358/375 0

0 0 0 64/125

2/3 0 - 2/3 0 1 0 0 0

0 1/2 0 0 0 1 0 0

- 2/3 0 358/375 0 0 0 1 0 H31 ( + 1 )

0 0 0 64/125 0 0 0 1

2/3 0 - 2/3 0 1 0 0 0 H1 ( x - 54/125 )

0 1/2 0 0 0 1 0 0

0 0 36/125 0 1 0 1 0

0 0 0 64/125 0 0 0 1

- 36/125 0 36/125 0 - 54/125 0 0 0 H13 ( - 1 )

0 1/2 0 0 0 1 0 0

0 0 36/125 0 1 0 1 0

0 0 0 64/125 0 0 0 1

- 36/125 0 0 0 -1 54/125 0 -1 0 H1 ( / - 36/125 )

0 1/2 0 0 0 1 0 0 H2 ( / 1/2 )0 0 36/125 0 1 0 1 0 H3 ( / 36/125 )

0 0 0 64/125 0 0 0 1 H4 ( / 64/125 )

1 0 0 0 4 35/36 0 3 17/36 0

0 1 0 0 0 2 0 0

0 0 1 0 3 17/36 0 3 17/36 0

0 0 0 1 0 0 0 1 61/64


50/106

A -1 = 4.9722222 0 3.47222222 0

0 2 0 0

3.4722222 0 3.47222222 0

0 0 0 1.953125


51/106


Metode Gauss-Jordan

diketahui :

P1 = 10 ton = 10000 kg

A1 = 1 E1 = 12.5

A2 = 1 E2 = 16.25

A3 = 1 E3 = 16.25A4 = 1 E4 = 12.5

A5 = 1 E5 = 9.1666667

L8 = 6 m

L7 = 6 m

L6 = 2.5 m

L5 = 5.5 m

L4 = L8

2

+ (L5+L6)= 10 m

L3 = L82+ L6

2 = 6.5 m

L2 = L72+ L6

2 = 6.5 m


52/106

D1 D2 D3 D4

A T = 0.8 0 0 0.8 1

0.6 0 0 0.6 00 0.3846154 0.3846154 0 -1

0 0.9230769 -0.923077 0 0

0 0 0.9230769 0.6 0

Menurut Hukum Hooke

d . AE = H . L

H = (AE/L) . d



S = (A1E1) / L1 0 0 0 0

0 (A2E2) / L2 0 0 0

0 0 (A3E3) / L3 0 0

0 0 0 (A4E4) / L4 0

0 0 0 0 (A5E5) / L5

d1 = 1 d2 = 1 d3 = 1 d4 = 1 d5 = 1

= 1.25 0 0 0 0

0 2.5 0 0 0

0 0 2.5 0 0

0 0 0 1.25 0

0 0 0 0 1.6666667



53/106

A . E 1.2 9/10 0 0

-1 2/3 0 2 206/507 0

0 0 0 4 44/169

K11 = 3 4/15 1.2

1.2 9/10

K21 = -1 2/3 0

0 0


E1 = K11-1. K12 F22 =

E2 = K21 . K11-1 F21 =

E3 = E2 . K12 F12 =

E4 = K22 - E3 F11 =


E1 = 1 . 9/10 -1.2 . -1 2/3

1 1/2 -1.2 3 4/15 0

E2 = -1 2/3 0 . 1 . 9/10

0 0 1 1/2 -1.2

E3 = -1 1.33333333 . -1 2/3 0

0 0 0 0

E4 = 2 206/507 0 - 1 2/3 0

0 4 44/169 0 0


54/106

maka K -1 = 1 . 1.952 -2.6026667

A . E -2.602667 4.5813333

1.352 -1.8026667

0 0

Gaya Ekivalen (Gaya Luar)

Q = Q1 = -10000 Kg

Q2 0 Kg

Q3 0 Kg

Q4 #REF! Kg


D = K -1 . Q

D1 = 1 . 1.952 -2.602667 1.352 0

D2 A . E -2.602667 4.5813333 -1.802667 0

D3 1.352 -1.802667 1.352 0

D4 0 0 0 0.2347222

D1 = 1 . #REF!

D2 A . E #REF!

D3 #REF!

D4 #REF!


H = S . A . D

H1 = 1 . 1 0.75 0 0

H2 A E 0 0 0 9615385 2 3076923


55/106


56/106

H1

H2

H3

H4

H5


57/106

K12 = -1 2/3 0

0 0

K22 = 2 206/507 0

0 4 44/169

E4-1

-F22 . E2

-E1 . F22

K11-1

- (F12 .E22)

0 = -1 0

0 1.3333333 0

-1.2 = -1 1.33333333

3 4/15 0 0

= 1 2/3 0

0 0

= 125/169 0

0 4 44/169

K21 K22


58/106

1.352 0

-1.8026667 0

1.352 0

0 0.2347222

. -10000

0

0

#REF!

. #REF!

#REF!


59/106

Balok Menerus

diketahui :

q = 11517.1875 kg/m'

EI1 = 1

EI2 = 1

L1 = 12 mL2 = 9 m

Momen Primer

MAB = - q . L12 = -138206 kgm

12

MBA = q . L12 = 138206.3 kgm

12

MBC = - q . L22 = -77741 kgm

12

L2

CBA

EI2

q

EI1

L1


60/106

S = EI 1/3 1/6 0 0

1/6 1/3 0 0

0 0 4/9 2/9

0 0 2/9 4/9


K = B . S . A

= A T S . A

= 1/6 1/3 4/9 2/9 . 0 EI

11

0

= 7/9 EI

K -1 =

7/9 EI

Deformasi Struktur

D = K -1 . Q

D4 = . 60465.23

7/9 EI

= 77741.0156

EI

1

1


61/106

MC = H4 - MCB = 17275.78 - 77741.02 = -60465.2


62/106


63/106

.


64/106

Kgm


65/106

Portal Tanpa Goyangan

diketahui :

q = 8100 kg/m'

p = 650 kg

EI1 = 1

EI2 = 2

EI3 = 1

L = 12 m

a = 6 m

b = 4 m

Momen Primer

MAB = - P . ab2 = -624 kgm

(a+b)2

MBA = P . a2b = 936 kgm

a

b

L

C

D

B

A

EI2PP

q

EI3EI1


66/106

0 1 d4

0 1 d5

0 0 d6

D1 D2

A T = 0 1 1 0 0 0

0 0 0 1 1 0


S = EI 2/5 1/5 0 0 0

1/5 2/5 0 0 0

0 0 2/3 1/3 00 0 1/3 2/3 0

0 0 0 0 2/5

0 0 0 0 1/5

= 1 EI 6 3 0 0 0

15 3 6 0 0 0

0 0 10 5 0

0 0 5 10 0

0 0 0 0 6

0 0 0 0 3


K = B . S . A

= A T S . A

= 1 EI 3 6 10 5 0


67/106

D1 = . 16 -5 .

D2 231 EI -5 16

= . -2021544

231 EI 2021544

= 1 . -131269

EI 131269.1

Gaya Dalam

H = S . A . D

H1 = 1 EI 3 0

H2 15 6 0

H3 10 5

H4 5 10

H5 0 6 1 .

H6 0 3 . EI

H1 = -26253.8

H2 -52507.6

H3 -43756.4

H4 43756.36

H5 52507.64

H6 26253.82

Momen Lentur

MA = H1 - MAB = -26253.8 - -624 = -25629.8

15

15


68/106


69/106

0

0

00

1/5

2/5

0

0

0

0

3

6

0 0 0


70/106

-96264

96264

-131269

131269.1

Kgm


71/106

K = 3 3 3

3 24 8

3 8 24

det K = 1872 - 624

= 1248

24 8 3 8 3 24

8 24 3 24 3 8

3 3 3 3 3 38 24 3 24 3 8

3 3 3 3 3 3

24 8 3 8 3 24

M 1 1 = 512

M 1 2 = 48M 1 3 = -48

M 2 1 = 48

M 2 2 = 63

M 2 3 = 15

M 3 1 = -48

M 3 2 = 15

M 3 3 = 63

M11 M12 M13

M21 M22 M23

M31 M32 M33


72/106


diketahui :

P1 = 2000 kg

P2 = 750 kg

P3 = 550 kg

EI1 = 1

EI2 = 2

EI3 = 1

L1 = 9 m

L2 = 3 m

L3 = 12 m

Momen PrimerMAB = 0 kgm

MBA = 0 kgm

L3

L1

C

D

B

A

EI2

P3P2P1

EI3EI1

L2


73/106

d5 1/4 0 1

d6 1/4 0 0

A = 1/4 0 0

1/4 1 0

0 1 0

0 0 1

1/4 0 1

1/4 0 0

A T = 1/4 1/4 0 0 1/4 1/4

0 1 1 0 0 00 0 0 1 1 0


S = EI 1/3 1/6 0 0 0

1/6 1/3 0 0 0

0 0 2/3 1/3 0

0 0 1/3 2/3 0

0 0 0 0 1/3

0 0 0 0 1/6

= 1 EI 2 1 0 0 0

6 1 2 0 0 0

0 0 4 2 0

0 0 2 4 0

0 0 0 0 20 0 0 0 1


74/106

1 EI 1248 -48 63

-48 -15

Deformasi Struktur

D = K -1 . Q

D1 = . 512 -48 -48 .

D2 52 EI -48 63 -15

D3 -48 -15 63

D1 = . 665600 = 1 .

D2 52 EI -296400 EID3 171600

Gaya Dalam

H = S . A . D

H1 = 1 . 0.75 1 0 . 12800

H2 6 0.75 2 0 -5700

H3 0 4 2 3300

H4 0 2 4

H5 0.75 0 2

H6 0.75 0 1

H1 = 650

H2 -300H3 -2700

H4 300

1

1


75/106


76/106


77/106

0

0

0

0

1/6

1/3

0

0

0

0

12


78/106

-15

63

1300

-3000

3000

12800

-57003300


79/106

K = 23.08333 -12.5 -11

-12.5 60 20

-11 20 58

det K = 85830 - 25555.83= 60274.17

60 20 -12.5 20 -12.5 60

20 58 -11 58 -11 20

-12.5 -11 23.08333333 -11 23.08333 -12.520 58 -11 58 -11 20

-12.5 -11 23.08333333 -11 23.08333 -12.5

60 20 -12.5 20 -12.5 60

M 1 1 = 3080

M 1 2 = -505M 1 3 = 410

M 2 1 = -505

M 2 2 = 1217.833

M 2 3 = 324.1667

M 3 1 = 410

M 3 2 = 324.1667

M 3 3 = 1228.75

M11 M12 M13

M21 M22 M23

M31 M32 M33


80/106

Portal Dengan Kaki Miring

diketahui :

P1 = 2000 kg

P2 = 1536 kg

P3 = 300 kg

P4 = 600 kg

P5 = 400 kg

PCB = (L4/L5) . P3 = 225 kg

PCD = (L7/L5) . P3 = 375 kg

EI1 = 1

EI2 = 2

EI3 = 1

L1 = 2 mL2 = 3 m

L3 = 6 m

L5

L2

CE B

A

EI2

P5P4P1

EI3EI1

L3

P2

L4L1

L6

D

L7

P3


81/106

Gaya Ekivalen

Q1 = P4+ P5 - PCB = 775 Kg

Q2 = P2- P3 = 1236 Kg

Q2 = P3 = 300 Kg

Matriks Deformasi

d1 = 1/4 0 0 . D1

d2 1/4 1 0 D2

d3 - 1/3 1 0 D3

d4 - 1/3 0 1

d5 1/3 0 1

d6 1/3 0 0

A = 1/4 0 0

1/4 1 0

- 1/3 1 0

- 1/3 0 1

1/3 0 1

1/3 0 0

A T = 1/4 1/4 - 1/3 - 1/3 1/3 1/3

0 1 1 0 0 0

0 0 0 1 1 0


S = EI 4/9 2/9 0 0 0

2/9 4/9 0 0 00 0 8/9 4/9 0

0 0 4/9 8/9 0


82/106

0 0 20 40 18

= 1 EI 23.08333 -12.5 -11

45 -12.5 60 20

-11 20 58

K -1 = . 1 . 3080 505

1 EI 60274.17 505 1217.833

410 -324.167

Deformasi Struktur

D = K -1 . Q

D1 = . 3080 505 410 .

D2 1339.426 EI 505 1217.833 -324.167

D3 410 -324.167 1228.75

D1 = . 3134180 = 1 .

D2 1339.426 EI 1799367 EI

D3 285705

Gaya Dalam

H = S . A . D

H1 = 1 7 5 10 0 2339 943

1

45

1


83/106

MBC = H3 - MBC = 248.9488 - -2250 = 2498.949

MCB = H4 - MCB = -253.31 - 2250 = -2503.31

MCD = H5 - MCD = 553.3102 - 0 = 553.3102

MD = H6 - MDC = 510.6494 - 0 = 510.6494


84/106


85/106


86/106

0

00

0


87/106

9 - 1/3 1 0

- 1/3 0 1

1/3 0 1

1/3 0 0

410

-324.167

1228.75

775

1236

300

2339.943

1343.387

213.3041


88/106

Kgm

Kgm

Kgm

Kgm


89/106


Metode Partisi

diketahui :

P1 = 1000 kg

P2 = 1000 kg

A1 = 1 E1 = 1

A2 = 1 E2 = 1

A3 = 1 E3 = 1

A4 = 1 E4 = 1

A5 = 1 E5 = 1

L1 = 2 m

L2 = 2 m

L3 = 1.5 m

L4 = L12+ L3

2 = 2.5 m

L5 = L22+ L32 = 2.5 m

ik f i


90/106

0 0 0 4/5 - 4/5

Menurut Hukum Hooke

d . AE = H . L

H = (AE/L) . d(AE/L) = Kekauan Axial Batang


S = (A1E1) / L1 0 0 0 0

0 (A2E2) / L2 0 0 0

0 0 (A3E3) / L3 0 0

0 0 0 (A4E4) / L4 00 0 0 0 (A5E5) / L5

d1 = 1 d2 = 1 d3 = 1 d4 = 1 d5 = 1

= 1/2 0 0 0 0

0 1/2 0 0 0

0 0 2/3 0 0

0 0 0 2/5 0

0 0 0 0 2/5


K = B . S . A

= A T S . A

= A . E 0 0 2/3 0 0

0 - 1/2 0 0 0

0 0 2/3 6/25 6/25


91/106

K21 = - 2/3 0

0 0


E1 = K11-1

. K12 F22 =

E2 = K21 . K11-1 F21 =

E3 = E2 . K12 F12 =

E4 = K22 - E3 F11 =


E1 = 1 . 1/2 0 . - 2/31/3 0 2/3 0

E2 = - 2/3 0 . 1 . 1/2

0 0 1/3 0

E3 = -1 0 . - 2/3 0

0 0 0 0

E4 = 358/375 0 - 2/3 0

0 64/125 0 0

F22 = 1 . 64/125 0

142/963 0 36/125

F21 = - 3.4722222 0 . -1 0

0 1.953125 0 0


92/106

Q = Q1 = -1000 Kg

Q2 0 Kg

Q3 0 Kg

Q4 1000 Kg


D = K -1 . Q

D1 = 1 . 4.97222222 0 3.47222222 0

D2 A . E 0 2 0 0

D3 3.47222222 0 3.47222222 0

D4 0 0 0 1.953125

D1 = 1 . -4972.2222

D2 A . E 0

D3 -3472.2222

D4 1953.125


H = S . A . D

H1 = 1 . 0 0 0 0

H2 A . E 0 -0.5 0 0

H3 0.66666667 0 -0.66666667 0

H4 0 0 -0.24 0.32

H5 0 0 -0.24 -0.32

H1 = 1 . 0

H2 A . E 0


93/106


94/106

H1

H2

H3

H4

H5

. 0 0 0 0

0 -1 0 0

1 0 1 0


95/106

K22 = 358/375 0

0 64/125

E4-1

-F22 . E2

-E1 . F22

K11-1

- (F12 .E22)

0 = -1 00 0 0

0 = -1 0

2/3 0 0

= 2/3 0

0 0

= 36/125 0

0 64/125

= 3.4722222 0

0 1.953125

= 3.4722222 0

0 0


96/106

. -1000

0

0

1000

. -4972.222

0

-3472.222

1953.125


97/106


Metode Partisi

diketahui :

P1 = 11342 kg

P2 = 9420 kg

A1 = 1 E1 = 1

A2 = 1 E2 = 1

A3 = 1 E3 = 1

A4 = 1 E4 = 1

A5 = 1 E5 = 1

L1 = 6 m

L2 = 6 m

L3 = 2.5 m

L4 = L12+ L3

2 = 6.5 m

L5 = L22+ L32 = 6.5 m

ik f i


98/106

0 0 0 12/13 - 12/13

Menurut Hukum Hooke

d . AE = H . L

H = (AE/L) . d(AE/L) = Kekauan Axial Batang


S = (A1E1) / L1 0 0 0 0

0 (A2E2) / L2 0 0 0

0 0 (A3E3) / L3 0 0

0 0 0 (A4E4) / L4 0

0 0 0 0 (A5E5) / L5

d1 = 1 d2 = 1 d3 = 1 d4 = 1 d5 = 1

= 1/6 0 0 0 0

0 1/6 0 0 0

0 0 2/5 0 0

0 0 0 1/6 0

0 0 0 0 1/6


K = B . S . A

= A T S . A

= A . E 0 0 2/5 0 0

0 - 1/6 0 0 0

0 0 2/5 10/169 10/169


99/106

K21 = - 2/5 0

0 0


E1 = K11-1

. K12 F22 =

E2 = K21 . K11-1 F21 =

E3 = E2 . K12 F12 =

E4 = K22 - E3 F11 =


E1 = 1 . 1/6 0 . - 2/51/15 0 2/5 0

E2 = - 2/5 0 . 1 . 1/6

0 0 1/15 0

E3 = -1 0 . - 2/5 0

0 0 0 0

E4 = 323/725 0 - 2/5 0

0 253/965 0 0

F22 = 1 . 253/965 0

5/419 0 33/725

F21 = - 21.97 0 . -1 0

0 3.8142361 0 0


100/106

Q = Q1 = -11342 Kg

Q2 0 Kg

Q3 0 Kg

Q4 9420 Kg


D = K -1 . Q

D1 = 1 . 24.47 0 21.97 0

D2 A . E 0 6 0 0

D3 21.97 0 21.97 0

D4 0 0 0 3.81423611

D1 = 1 . -277538.74

D2 A . E 0

D3 -249183.74

D4 35930.1042


H = S . A . D

H1 = 1 . 0 0 0 0

H2 A . E 0 -0.16666667 0 0

H3 0.4 0 -0.4 0

H4 0 0 -0.0591716 0.14201183

H5 0 0 -0.0591716 -0.1420118

H1 = 1 . 0

H2 A . E 0


101/106


102/106

H1

H2

H3

H4

H5

. 0 0 0 0

0 -1 0 0

1 0 1 0


103/106

K22 = 323/725 0

0 253/965

E4-1

-F22 . E2

-E1 . F22

K11-1

- (F12 .E22)

0 = -1 00 0 0

0 = -1 0

2/5 0 0

= 2/5 0

0 0

= 33/725 0

0 253/965

= 21.97 0

0 3.81423611

= 21.97 0

0 0


104/106

. -11342

0

0

9420

. -277538.7

0

-249183.7

35930.104

Invers Matriks Ordo 4x4

Metode Gauss-Jordan


105/106

Metode Gauss Jordan

A = 2/5 0 - 2/5 0

0 1/6 0 0

- 2/5 0 323/725 0

0 0 0 253/965

2/5 0 - 2/5 0 1 0 0 0

0 1/6 0 0 0 1 0 0

- 2/5 0 323/725 0 0 0 1 0 H31 ( + 1 )

0 0 0 253/965 0 0 0 1

2/5 0 - 2/5 0 1 0 0 0 H1 ( x - 33/290 )

0 1/6 0 0 0 1 0 0

0 0 33/725 0 1 0 1 00 0 0 253/965 0 0 0 1

- 33/725 0 33/725 0 - 33/290 0 0 0 H13 ( - 1 )

0 1/6 0 0 0 1 0 0

0 0 33/725 0 1 0 1 0

0 0 0 253/965 0 0 0 1

- 33/725 0 0 0 -1 33/290 0 -1 0 H1 ( / - 33/725 )

0 1/6 0 0 0 1 0 0 H2 ( / 1/6 )

0 0 33/725 0 1 0 1 0 H3 ( / 33/725 )

0 0 0 253/965 0 0 0 1 H4 ( / 253/965 )

1 0 0 0 24 47/100 0 21 97/100 0

0 1 0 0 0 6 0 0

0 0 1 0 21 97/100 0 21 97/100 0

0 0 0 1 0 0 0 3 469/576

A -1 = 24 47 0 21 97 0


106/106

A = 24.47 0 21.97 0

0 6 0 0

21.97 0 21.97 0

0 0 0 3.8142361

Documents

Analisa Struktur III