mekanika teknik terapan rc09 1341 - website...

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Mekanika Teknik TerapanMekanika Teknik TerapanRC09 1341RC09 1341

Endah Wahyuni S T (ITS) M Sc (UMIST) Ph D (UM)Endah Wahyuni, S.T. (ITS), M.Sc. (UMIST), Ph.D (UM)[email protected], [email protected], @end222

1Copyright: Dr Endah Wahyuni

SyllabusSyllabus Review of Matrices Structural Analysis of Matrix Methods y

(ASMM)◦ Balok Menerus◦ Portal◦ Transformasi Sumbu Koordinat◦ Rangka Batang◦ Rangka Batang

Introduction of Finite Element Method Application using SAP2000


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MaterialsMaterials

Books Web personal dosen: Web personal dosen:

http://personal.its.ac.id/material.php?userid=ewahyuni

Online courses

Copyright: Dr Endah Wahyuni 3

Books:Books: Structural Analysis: Using Classical and Matrix

Methods; Jack C McCormac; Wiley; 4 edition,London 2006

The image part with relationship ID rId2 was not found in the file.

Structural AnalysisRC Coates, MG Coutie, FK KongChapman & Hall, London, 1997

Computer Analysis of Structuresp yS.M. HolzerElsevier Science Publishing Co. Inc, London, 1985

SAP2000 Manual


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EvaluationEvaluation

HW = 10% ETS = 25% ETS = 25% EAS = 25% Project = 20% Quiz = 20%

Note: Percentage can be changed according to the results of student assessments


Review of MatricesReview of Matrices

Definition: A matrix is an array of ordered numbers. A general matrix consists of mn numbers arranged in m general matrix consists of mn numbers arranged in m rows and n columns, giving the following array:

mnmm

n

n

a...aa

......

a...aa

a...aa

A

21

22221

11211

It is usual to use a generic symbol for the element, the lower case letter corresponding to the capital used for the matrix


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Matrix AdditionMatrix Addition

Two matrices A and B can be added if they have the same number of rows (m) and have the same number of rows (m) and columns (n)

It is possible to be able to write:Ax + Bx = (A+B) x

If C = A+B, then crs= ars + brs

Subtraction is treated as negative addition


Multiplication by scalarMultiplication by scalar

To multiply a matrix by a scalar quantity, each element of the matrix is to be multiplied by element of the matrix is to be multiplied by the scalar, that is

k [ars] = [(k ars)]


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Special matricesSpecial matrices Row Matrix

If the matrix consists only of elements in a single row. For example, a 1 x n row matrix is written as:

A=[a1 a2 ... an][ 1 2 n] Column Matrix

A matrix with elements stacked in a single column. The m x 1 column matrix is

a

a

A 2

1

Square MatrixWhen the number of row equals the number of columns

9

ma

.

Copyright: Dr Endah Wahyuni

Unit (or identity ) matrixIs a square matrix, where all elements are zero except on the principal diagonal and there they are units.It is readily verified that

A I = A, and IA = AandI . I = I2 = IConsequently for n any positive integerConsequently, for n any positive integer

In = I


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Diagonal Matrix,of which unit matrices are particular cases, have important properties. For instance:

bb00

wkwhwg

vfvevd

ucubua

khg

fed

cba

w

v

u

00

00

00

kwhvgu

fwevdu

cwbvau

w

v

u

khg

fed

cba

00

00

00


A transpose matrix◦ A Matrix may be transposed by interchange its

rows and columns. The transpose of the matrix A is written as ATA is written as AT

nmmm

n

n

T

mnmm

n

n

aaa

aaa

aaa

A

aaa

aaa

aaa

A

...

......

...

...

...

......

...

...

21

22212

12111

21

22221

11211

A symmetric matrix◦ We can verify that AT = A


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Inverse Matrix◦ The inverse of matrix A is denoted by A-1.◦ Assume that the inverse exists, then the elements of

A-1 are such that A-1A=I and AA-1=I

Determinant of a matrix◦ A determinant is a square array of numbers

enclosed within vertical bars

naaa ... 11211

nmnn

n

aaa

aaaA

...

.

...

21

22221


Find the determinant then the inverse matrices for the matrices below:

53

A

121

212

321

B

21

A


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Structural Analysis of Matrix MethodsStructural Analysis of Matrix Methods

Matrix Method of Structural Analysis is a method to analyze the structure with the help of the matrix, which consists of: tiff t i di l t t i d f t i b stiffness matrix, displacement matrix, and force matrix, by

using the relationship:{ P } = [ K ] { U }

dimana :{ P } = force matrix[ K ] = stiffness matriks{ U } = displacement matriks{ U } displacement matriks

One of the ways used to solve the above equation, namely by using the stiffness method.


On the stiffness method, the unknown variable is the amount of: deformation node of structure (rotations and deflection) is certain / sure. Thus the number of variables in the stiffness method similar to the degrees of uncertainty of kinematics of structure.

The stiffness method is developed from the node point equilibrium equations written in: "Stiffness Coefficients" and "Displacement of nodes that are Coefficients and Displacement of nodes that are not known“.


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Types of ElementTypes of Element Spring elementsTruss elements (plane & 3D) Beam elements (2D &3D) Plane Frame Grid elements Plane Stress Plane StrainAxisymmetrics elementsAxisymmetrics elements Plate Shell


Degrees of Freedom (DOF)Degrees of Freedom (DOF)

Degrees of freedom which is owned by a structure.

Each type of element will have a certain amount and kind of freedom.

Commonly referred to as 'activity' of the joint


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To be able to form a structure stiffness matrix, it is To be able to form a structure stiffness matrix, it is necessary to determine the 'active' ends of the elements.necessary to determine the 'active' ends of the elements.

Note : 0 = not active; 1, 2, 3, …., =active

Example

3

21

2

1

0

DOF : 2

0

00

1

0

2



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Metode Kekakuan Langsung Metode Kekakuan Langsung (Direct Stiffness Method)(Direct Stiffness Method)

matriks kekakuan

U1, P1 U3, P3

{ P } [ K ] { U } { P } = [ K ] { U }

U2, P2 U4, P4 gaya perpindahan

P1 K11 K12 K13 K14 U1

P2 K21 K22 K23 K24 U2

1 1 2

=P3 K31 K32 K33 K34 U3

P4 K41 K42 K43 K44 U4

=


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P1 = K11 . U1 + K12 . U2 + K13 . U3 + K14 . U4Kesetimbangan gaya di arah U1

P2 = K21 . U1 + K22 . U2 + K23 . U3 + K24 . U4 Kesetimbangan gaya di arah U2




Jika U1 = 1 dan U2 = U3 = U4 = 0 , maka :P1 = K11 ; P2 = K21 ; P3 = K31 ; P4 = K41 Lihat Gambar (a)

Jika U2 = 1 dan U2 = U3 = U4 = 0 , maka :P1 = K12 ; P2 = K22 ; P3 = K32 ; P4 = K42

Lihat Gambar (b)


Lihat Gambar (c)


Lihat Gambar (d)


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U1’ = 1 P1’ = K11P2’ = K21P3’ = K31P4’ = K41

U 1 P1’ K12U2’ = 1 P1’ = K12P2’ = K22P3’ = K32P4’ = K42

U3’ = 1 P1’ = K13P2’ = K23P3’ = K33P3 K33P4’ = K43

U4’ = 1 P1’ = K14P2’ = K24P3’ = K34P4’ = K44



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Matrix Matrix kekakuankekakuan::

K11 K12 K13 K14

K21 K22 K23 K24

K31 K32 K33 K34

K K K K

K =

K41 K42 K43 K44

2323 L

EI 6

L

EI 12-

L

EI 6

L

EI 12

L

EI 2

L

EI 6-

L

EI 4

L

EI 622

K =

2323 L

EI 6 -

L

EI 12

L

EI 6

L

EI 12-

Matriks Kekakuan L

EI 4

L

EI 6-

L

EI 2

L

EI 622

Gambar (a) (b) (c) (d) 28Copyright: Dr Endah Wahyuni

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Jika pada batang bekerja gaya aksial :

L, EA

U1’,P1’ U2’,P2’

P1+A=0 P2-A=0 =E(u2 – u1)/L

P1+EA (u2-u1)/L=0

P2-EA (u2-u1)/L=0

K11 = L

EA K21 =

L

EA

U1’= 1

K12 = - L

EA

U2’= 1

K22 = L

EA

U2, P2 U5, P5

U1, P1 U4, P4

U3, P3 U6, P6

1 1 2


11

11

//

//

2

1

2

1

L

EAK

P

P

u

u

LEALEA

LEALEA

For beam with 6 dofFor beam with 6 dof

1 2

2

1

3

45

6


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Stiffness Matrix with Axial LoadsStiffness Matrix with Axial LoadsMatriks kekakuan elemen dengan melibatkan gaya aksial :

00EA

-00EA

6 x 6

K =

2323 L

EI 6

L

EI 12- 0

L

EI 6

L

EI 12 0

L

EI 2

L

EI 6- 0

L

EI 4

L

EI 6 0

22

0 0 L

0 0 L

0 0 L

EA- 0 0

L

EA

2323 L

EI 6 -

L

EI 12 0

L

EI 6

L

EI 12 0 -

L

EI 4

L

EI 6- 0

L

EI 2

L

EI 6 0

22

LL


Procedure to solve a problemProcedure to solve a problem

Calculate Structure DOF Create the stiffness matrix of element [ki] Create the objective matrix Create the stiffness matrix of structure [Ks]

Where [Ks]=[ki] Calculate the value of action received by the

structure {Ps} Calculate {Us}=[Ks]-1 {Ps} For each element, specify {ui} Forces of element: {pi}=[ki]{ui}+{freaksi}


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ExampleExample

q

Sebuah balok statis tak tentu seperti pada gambar

1 1 2 2 3

L, EI L, EI

Menentukan keaktifan ujung-ujung elemen

Menentukan matriks tujuan DOF : 2 2 rotasi

1 2 3

0

1 2

0

0

0

1 2

0 0 00

Matriks kekakuan struktur

[ Ks ] 2 x 2

Membuat matrik kekakuan elemen : [ Ks ] = [ K1 ] + [ K2 ]

1 2

3

0 0 0

1 2

0 1 1 2

0


Membuat matrik kekakuan elemen :

Elemen 1

0 0 0 1

2323 L

EI 6

L

EI 12-

L

EI 6

L

EI 12 0

L

EI 2

L

EI 6-

L

EI 4

L

EI 622

0 LLLL

2323 L

EI 6 -

L

EI 12

L

EI 6

L

EI 12- 0

L

EI 4

L

EI 6-

L

EI 2

L

EI 622

1

K1 =

Matriks Tujuan { T1 } = { 0 0 0 1 }T

[ K1 ] =

0 L

EI 4

2 x 2 0 0


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Elemen 2

0 1 0 2

2323 L

EI 6

L

EI 12-

L

EI 6

L

EI 12 0

L

EI 2

L

EI 6-

L

EI 4

L

EI 622

1

K2 =

2323 L

EI 6 -

L

EI 12

L

EI 6

L

EI 12- 0

L

EI 4

L

EI 6-

L

EI 2

L

EI 622

2


K2

2 x 2

[ K2 ] =

L

EI 4

L

EI 2

L

EI 2

L

EI 4


= + =

Matriks Kekakuan Global Struktur

[ Ks ] = [ K1 ] + [ K2 ]

[ Ks ]L

EI 2

L

EI 40 L

EI 4

L

EI 2

L

EI 8

0 0

[ Ks ]

2 x 2

L

EI 4

L

EI 2L

EI 4

L

EI 2

Untuk mendapatkan deformasi ujung-ujung aktif struktur, maka digunakan

hubungan :

{ Ps } = [ Ks ] { Us } { Us } = [ Ks ]-1 { Ps }

dimana :

Us = deformasi ujung-ujung aktif

Ks = kekakuan struktur

Ps = gaya-gaya pada ujung aktif elemen akibat beban luar (aksi)


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q

0 0

Untuk contoh di atas, maka :

Ps =

2L q 12

1 2L q

12

1

2L q 12

1

2Lq1

Menghitung invers matrik kekakuan global [ Ks ]-1

[ Ks ] =

[ K ] 1 2- 4 L1 2- 4 L

L

EI 4

L

EI 2

L

EI 2

L

EI 8

L q 12

[ Ks ]-1 = 8 2-

EI

L

2 . 2 - 4 . 8

1

=

8 2-

EI 28

L

Jadi : { Us } = [ Ks ]-1 { Ps }

Us = 8 2-

2- 4

EI 28

L

2L q 12

1

2L q 12

1


Us = EI 28

L

Us =

22 L q 6

1 - L q

3

1

22 L q 6

4 L q

6

1

EI

L q

168

3 3

Rotasi di joint 2

U11

U12

U13

U14

0 0 0

Us

Deformasi untuk masing-masing elemen

Elemen 1 : U1 = =

EI

L q

168

5 3

Rotasi di joint 3

EI

L q

168

3 3

U21

U22

U23

U24

0 0

Elemen 2 : U2 = =

EI

L q

168

3 3

EI

L q

168

5 3


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q

0 0

Reaksi akibat beban luar :

2L q

12

12L q

12

1

0 0 0

PR2 = PR1 =

1212

2

L q 2

L q

2L q 12

1

2

L q

2

L q

0

2L q 12

1

2


0 0

0

0

Gaya akhir elemen :

Elemen 1 : { P1 } = [ K1 ] {U1}+ { PR1 }

2323 L

EI 6

L

EI 12-

L

EI 6

L

EI 12

EI2EI6EI4EI6 0 0

0

0

0

P1 = +

L

EI2

L

EI6-

L

EI 4

L

EI 622

2323 L

EI 6 -

L

EI 12

L

EI 6

L

EI 12-

L

EI 4

L

EI 6-

L

EI 2

L

EI 622 EI

L q

168

3 3

L 6 3

P1 = =

2L q 56

4

2L q 56

2

L q 56

L q 56

6

2L q 28

2

2L q 28

1

L q 28

3

L q 28

3


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0

Elemen 2 : { P2 } = [ K2 ]{U2} + { PR2 }

P2 = +

2323 L

EI 6

L

EI 12-

L

EI 6

L

EI 12

L

EI 2

L

EI 6-

L

EI 4

L

EI 622

EI6EI12EI6EI12

EI

L q

168

3 3

2L q 12

1

2

L q

Lq0 P2 = +

2323 L

EI6 -

L

EI12

L

EI 6

L

EI 12-

L

EI 4

L

EI 6-

L

EI 2

L

EI 622 EI

L q

168

5 3

2L q 56

4

L q 56

32

2L q 28

2

L q 28

16

2L q 12

1

2

Lq

0 0

P2 = =

L q 56

2428

L q 28

12


q 0

Free Body Diagram :

2L q 28

22L q 28

1

L q 28

3

2L q 28

2

L q 28

16 L q

28

3 L q

28

12

- -+

Menggambar gaya-gaya dalam :

Bidang D :

L q

28

3 L q

28

3

L q 28

16

L q 28

12

-

+ +

Bidang M :

2L q 28

2

2L q 28

1


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Example 2Example 2



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q 0

Free Body Diagram :

2L q 28

22L q 28

1

L q 28

3

2L q 28

2

L q 28

16 L q

28

3 L q

28

12

- -+

Menggambar gaya-gaya dalam :

Bidang D :

L q

28

3 L q

28

3

L q 28

16

L q 28

12

-

+ +

Bidang M :

2L q 28

2

2L q 28

1


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Contoh 2Contoh 2



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Example 4Example 4



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Portal 2DPortal 2D

B C

P

EI

B C

2

Sebuah portal statis tak tentu seperti pada gambar

EI L

L/2 L/2

A A

1 DOF = 2

0

1 1 2

0

Matriks kekakuan struktur

[ Ks ] 2 x 2

[ Ks ] = [ K1 ] + [ K2 ]

Deformasi aksial diabaikan


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Elemen 1

0 1

0

2 x 2 1

Matriks Tujuan { T1 } = { 0 1 }T

K1 = L

EI 2

L

EI 4

L

EI 4

L

EI 2

[ K1 ] = 0 0

0

2 x 2

Elemen 2

1 2

1

2

L

EI 4

K2 = L

EI 2

L

EI 4

EI 4EI 2 2 x 2 2


2 x 2

[ K2 ] =

L

EI 4

L

EI 2L

EI 2

L

EI 4

L

L


= +

0

=


[ Ks ] = [ K1 ] + [ K2 ]

[ Ks ]L

EI 2

L

EI 4L

EI 2

L

EI 8 L

EI 4

0 0

[ Ks ]

2 x 2

Untuk mendapatkan deformasi ujung-ujung aktif struktur, maka digunakan

hubungan :

{ Ps } = [ Ks ] { Us } { Us } = [ Ks ]-1 { Ps }

dimana :

L

EI 4

L

EI 2L

EI 4

L

EI 2

dimana :

Us = deformasi ujung-ujung aktif

Ks = kekakuan struktur

Ps = gaya-gaya pada ujung aktif elemen akibat beban luar (aksi)


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P

Untuk contoh di atas, maka :

0 LP1

LP1

0

LP8

LP8

Ps =

L P 8

1

L P 8

1



[ Ks ] =

LEI4

LEI2

LEI2

LEI8

[ Ks ]-1 = 82-

2-4

EIL

2.2-4.81

=

82-

2-4

EI28L

Jadi : { Us } = [ Ks ]-1 { Ps }

LL

Us = 82-

2-4

EI28L

LP81

LP81


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U11

U1

2

0


Elemen 1 : U1 = =

L P3 2

Jadi : { Us } = [ Ks ]-1 { Ps }

Us = 8 2-

2- 4

EI 28

L

Us = EI 28

L

L P 8

1

L P 8

1

22 L q 6

1 - L q

3

1

22 L q 6

4 L q

6

1

LP3 2

1

U21

U2

2

Elemen 2 : U2 = =

EI

112

EI

L P

112

3 2

EI

L P

112

5 2

Us =

EI

L P

112

3

EI

L P

112

5 2

Rotasi di joint B

Rotasi di joint C


P Reaksi akibat beban luar :

0

L P 8

1L P

8

1

0

0

0

0

P1 = +

Gaya akhir elemen :

Elemen 1 : { P1 } = [ K1 ] + { PR1 }

EI

L P

112

3 2

L

EI 2

L

EI 4

L

EI 4

L

EI 2

P1 = Hasil perhitungan hanya momen saja

L P

56

6

L P 56

3


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P2 = +

Elemen 2 : { P2 } = [ K2 ] + { PR2 }

2 1

L P 8

1

EI

L P

112

3 2

L

EI 2

L

EI 4

P2

Hasil perhitungan

EI

LP

112

5 2

L P 8

1

L

EI4

L

EI 2

2L q 56

6 2L q 28

3

P2 = =

0 0

Hasil perhitungan hanya momen saja

56 28


P 0

Dihitung lagi

Free Body Diagram :

P 56

9

L P 56

6

P 28

17 P

28

11

P 56

9

P 56

9

P 28

17

L P 56

6

Dihitung lagi

P 56

9

P 28

17

L P 56

3

Bidang M :

-

L P 56

6

Bidang D :

Bidang N :

-

+ P 28

17

-

P 56

9

P 28

11

P

- +

L P 56

3

L P 56

11

+

P 28

17

-

P 56

9-


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Contoh 2Contoh 2

B C

P

EI

EI L

B C

1

2

DOF = 3

4 1 2

3

L/2 L/2

A A

1 DOF = 3

2

1

B C

2

2 2 3

1

A

1 DOF = 3

0

0

Matriks kekakuan struktur[ Ks ] 3 x 3[ Ks ] = [ K1 ] + [ K2 ]


Elemen 1

0 0 1 2

2323 L

EI 6

L

EI 12-

L

EI 6

L

EI 12 0

L

EI 2

L

EI 6-

L

EI 4

L

EI 622

0 LLLL

2323 L

EI 6 -

L

EI 12

L

EI 6

L

EI 12- 1

L

EI 4

L

EI 6-

L

EI 2

L

EI 622

2


K1 =

[ K1 ] =

2 x 2

23 L

EI 6-

L

EI 12

L

EI 4

L

EI 6-

2


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Elemen 2

2 3

2

2 x 2 3

K2 = L

EI 2

L

EI 4

L

EI 4

L

EI 2


[ K2 ] = EI 4EI 2

L

EI 2

L

EI 4

LL

2 x 2


[ Ks ] = [ K1 ] + [ K2 ]

L

L


= + =

1 2 2 3

[ Ks ]

3 x 3 L

EI 4

L

EI 2

L

EI 2

L

EI 4

L

EI 2

L

EI 8

L

EI 62

0 L

EI 6-

L

EI 122323 L

EI 6-

L

EI 12

L

EI 4

L

EI 6-

2

L

EI 4

L

EI 2 0

0

Ps = L P 8

1

L P 8

1


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[ Ks ]-1 =

EI

L 12 -

EI

L 24

EI

L 28 223

EI

L 24 -

EI

L 48

EI

L 24 2

EI

60L

EI

L 24-

EI

L 12 2

Jadi : { Us } = [ Ks ]-1 { Ps }

Us =

L P 8

1

L P 8

1

EIEIEI

EI

L 12 -

EI

L 24

EI

L 28 223

EI

L 24 -

EI

L 48

EI

L 24 2

60LL 24L 12 2

0

Us =

8

EI

L P

128

3 3

EI

L P

128

10 2

Rotasi di joint B

Rotasi di joint C

EI

EI-

EI

EI

L P

128

7 2

Dilatasi di joint B


U11

2

0


U12

U1

3 U1

4

0

Elemen 1 : U1 =

EI

L P

128

3 3

EI

L P

128

6 2

U21

U2

2

Elemen 2 : U2 = =

EI

L P

128

6 2

EI

L P

128

7 2


09/02/2014

36

0 0

0 0

Gaya akhir elemen :

Elemen 1 : { P1 } = [ K1 ] + { PR1 }

2323 L

EI 6

L

EI 12-

L

EI 6

L

EI 12

EI2EI6EI4EI6 0

+

0 0 0

0

P1 = L

EI2

L

EI6-

L

EI 4

L

EI 622

2323 L

EI 6 -

L

EI 12

L

EI 6

L

EI 12-

L

EI 4

L

EI 6-

L

EI 2

L

EI 622

EI

L P

128

3 3

EI

L P

128

6 2

0

0

P1 =

L P 64

3

L P 64

3


Elemen 2 : { P2 } = [ K2 ] + { PR2 }

L P

8

1

EI

L P

128

6 2

L

EI 2

L

EI 4

P2 = +

P2 = = Hasil perhitungan hanya momen saja

EI

L P

128

7 2

L P 8

1

L

EI 4

L

EI 2

L P 128

6 2L q 64

3

0 0


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37

P 0

Dihitung lagi

Free Body Diagram :

L P 64

3

P35

P29

P 64

35

L P 64

3

Dihitung lagi

P64

P64

L P 64

3

-+

Bidang M :

L P 64

3

L P 128

29

P

64

35-

128

L P 64

3


-

+ Bidang D :

P 64

35

P 64

29

P

Bidang N :

-

P 64

35


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Deformasi aksial diabaikan



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40

The image part with relationship ID rId2 was not found in the file.



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Transformation of CoordinatesTransformation of Coordinates

2’

2

1

1’

U3, P3

u3, p3 U1, P1

U2, P2

u1, p1 u2, p2

Koordinat Lokal dan Global

3 3’

u1

u2

u3

=

C S 0 -S C 0

0 0 1

U1

U2

U3

C = cos S = sin


09/02/2014

42

C S 0 -S C 0

C = cos S = sin

Atau dapat ditulis : u = U

Dimana :

=

0 0 1

u1

u2

u3

0 U1

U2

U3[ ] [ R ] [ U ]

Untuk transformasi sumbu sebuah titik dengan 6 dof dapat ditulis :

3

u4

u5

u6

= 0

3

U4

U5

U6

[ u ] = [ R ] [ U ] R = matriks rotasi


P1 p1

Transformasi sumbu juga berlaku untuk gaya :

p = P

P = -1 p -1 = T

P = T p

P2

P3

P4

P5

P6

=

0 0

p2

p3

p4

p5

p6

[ P ] = [ R ]T [ p ] R = matriks rotasi

p = k u ; u = R U

P = RT p P = K U

K

P = R p P = K U

= RT k u K = RT k R

= RT k R U


09/02/2014

43

Matriks kekakuan elemen untuk 6 dof :

2323 L

EI 6

L

EI 12- 0

L

EI 6

L

EI 12 0

0 0 L

EA- 0 0

L

EA

6 x 6

k = L

EI 2

L

EI 6- 0

L

EI 4

L

EI 6 0

22

2323 L

EI 6 -

L

EI 12 0

L

EI 6

L

EI 12 0 -

4626

0 0 L

EA- 0 0

L

EA

L

EI 4

L

EI6- 0

L

EI2

L

EI6 0

22


0 0 - 0 0 0 12 6L 0 -12 6L 0 6L 4L2 0 -6L 2L2

- 0 0 0 0 0 -12 -6L 0 12 -6L 0 6L 2L2 0 -6L 4L2

Dimana :

k =

Dimana :

= =

[ K ] = [ R ]T [ k ] [ R ]

L

EI

3 I

LA

2


09/02/2014

44

C -S 0 S C 0 0

0 0 - 0 0 0 12 6L 0 -12 6L

C S 0 S C 0 0

[K]=[R]T [k][R]

0 0 1

C -S 0

S C 0

0 0 1

0

0 6L 4L2 0 -6L 2L2 - 0 0 0 0 0 -12 -6L 0 12 -6L 0 6L 2L2 0 -6L 4L2

K =

-S C 0 0 0 1

C S 0

-S C 0

0 0 1

0

0


g1 g2 g4 -g1 -g2 g4

g3 g5 -g2 -g3 g5

g6 -g4 -g5 g7

g6 g4 g5 g7

g1 g2 -g4

g3 -g5

g6

Dimana :

K =

Dimana :

g1 = ( C2 + 12 S2 ) g5 = 6 L C

g2 = C S ( - 12 ) g6 = 4 L2

g3 = ( S2 + 12 C2 ) g7 = 2 L2

g4 = - 6 L S 88Copyright: Dr Endah Wahyuni

09/02/2014

45

Sebuah portal seperti gambar, dengan menggunakan transformasi sumbu Sebuah portal seperti gambar, dengan menggunakan transformasi sumbu hitunglah gayahitunglah gaya--gaya dalam yang bekerjagaya dalam yang bekerja

L = 10 ft

1

1

E = 30.000 ksi A = 5 in2 I = 50 in4 L = 10 ft

q = 1,68 k/ft

L = 10 ft

M = 14 kft = 168 kin

2 3 2

1 0

0

0 Sumbu Global

1 2

1 3 Sumbu Lokal

1

2

2 3

0

3

1 0

0

2

DOF [ Ks ] 3 x 3

1

2

2 3

4

5 4

5

6 6

1

3

2

2


1 x

= 270o

=

C S 0 -S C 0 =

0 -1 0 1 0 0

Matriks transformasi batang :

Batang 1 : = 270o cos 270o = 0

sin 270o = -1

2 x’

1 0 0 1 0 0 1

Batang 2 : = 0o cos 0o = 1

sin 0o = 0

2 3 x

x’

= 0o

=

C S 0 -S C 0 0 0 1

=

1 0 0 0 1 0 0 0 1


09/02/2014

46

C S 0 -S C 0 0 0 1

C S 0

0

0 -1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0

R1 = =

C S 0

-S C 0

0 0 1

0 0 0 0 1 0 0 0 0 0 0 0 1

C S 0 -S C 0 0

1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0

0 0 1

C S 0

-S C 0

0 0 1

0

0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1

R2 = =


Matriks kekakuan system struktur

Elemen 1 :

1 = 33

112) . 10(

50 . 30.000

L

EI = 0,87

1 = 50

12) . (10 . 5

I

LA 221 = 1.440

g1 g2 g4 -g1 -g2 g4

g3 g5 -g2 -g3 g5

g6 -g4 -g5 g7

0 0 0

C = 0 ; S = -1

{ T } = { 0 0 0 1 0 2 }T

0 0 0 1 0 2

K1 =

g1 g2 -g4

g3 -g5

-g4 g6

1 0 2

K1


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47

g1 -g4 0 -g4 g6 0 0 0 0

1 2 3

1 2 3

K1 =

10 44 626 4 0

g1 = ( C2 + 12 S2 ) = 0,87 [ 0 + 12 (-1)2 ] = 10,44

g4 = - 6 L S = -0,87 . 6 . 120 (-1) = 626,4

g6 = 4 L2 = 0,87 . 4 . 1202 = 50.112

Sehingga :

10,44 -626,4 0 -626,4 50.112 0 0 0 0

K1 =


Elemen 2 :

2 = 33

112) . 10(

50 . 30.000

L

EI = 0,87

2 = 50

12) . (10 . 5

I

LA 221 = 1.440

C = 1 ; S = 0

{ T } = { 1 0 2 0 0 3 }T

g1 g2 g4 -g1 -g2 g4

g3 g5 -g2 -g3 g5

g4 g6 -g4 -g5 g7

g1 g2 -g4

g3 -g5

1 0 2 0 0

1 0 2 0 0 3

K2 =

g4 g7 g6

3

g1 g4 g4

g4 g6 g7

g4 g7 g6

1 2 3

1 2 3

K2 =


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48

g1 = ( C2 + 12 S2 ) = 0,87 [ 1.440 . 12 + 12 (0)2 ] = 1.252,8

g4 = - 6 L S = -0,87 . 6 . 120 (0) = 0

g6 = 4 L2 = 0,87 . 4 . 1202 = 50.112

g7 = 2 L2 = 0,87 . 2 . 1202 = 25.056

1.252,8 0 0 0 50.112 25.056 0 25.056 50.112

Sehingga :

K2 =

1.263,24 -626,4 0 -626,4 100.224 25.056 0 25.056 50.112

KS =


q = 0,14 k/in

168 kin 168 kin 168 kin 0 0

0

Matriks beban :

8,4 8,4

0 168 0

1.263,24 -626,4 0 -626 4 100 224 25 056

- 1 0 168

PS =

{ Ps } = [ Ks ] { Us } { Us } = [ Ks ]-1 { Ps }

US = -626,4 100.224 25.056

0 25.056 50.112

168 0

0,00095 0,00192 -0,00096

Defleksi horizontal di 2 Rotasi di 2 Rotasi di 3

US

US =


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49

u11

u12

u1

3

u14

0

0 0

0,00095

=

0

0 0

0

Displacement masing-masing batang (koordinat lokal)

u1 = =

0 -1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0

u15

u16

0

0,00192

0,00095

0,00192

u21

u22

0,00095

0

0,00095

0

0 0 0 1 0 0 0 0 0 0 0 1

1 0 0 0 0 0 0 1 0 0 0 02

u2

3

u24

u25

u26

0,00192

0

0

-0,0096

=

0 0,00192

0

0

-0,0096

u2 = =

0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1


0

1,193 k 47,512 kin

0

1,193 k 3,959 kft

Gaya akhir batang :

Elemen 1 :

{ P1 } = [ k1 ] { u1 } + { 0 }

0

-1,193 k

95,620 kin

0

-1,193 k

7,968 kft

P1 = =

1,19 k 1,19 k

Elemen 2 :

{ P2 } = [ k2 ] { u2 } + { Faksi }

-7,8 k -95,84 kin

-1,19 k

-9 k

168 kin

-7,8 k -7,99 kft

-1,19 k

-9 k

14 kft

P2 = =


09/02/2014

50

1

1,193 k

0

3,959

7,968 kft

Free body diagram :

+

+

1,193

9

q = 1,68 k/ft 14 kft

7,8 k 9 k

2 1,19 k 1,19 k

7,99 kft 1,193 k

7,968 kft

3,959

+

-1,193

7,8

-

7,9914

+ +

-- 1,191,19


TRANSFORMASI SUMBU TRANSFORMASI SUMBU PADA BIDANG MIRINGPADA BIDANG MIRINGMEKANIKA TERAPAN

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CONTOH SOALCONTOH SOAL

A = 50 cm2

E = 2,1 . 106 kg/cm2, g

I = 2100 cm4

• DOF : 4

• [Ks] 4 x 4

MATRIX TRANSFORMASI MATRIX TRANSFORMASI BATANGBATANG

ϴ = 45ᵒC ϴ 0 707 Cos ϴ = 0.707

Sin ϴ = 0.707

λ1 = C S 0

-S C 0

0 0 1

0.707 0.707 0

-0.707 0.707 0

0 0 1

=

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52

MATRIX TRANSFORMASI MATRIX TRANSFORMASI BATANGBATANG

ϴ = 0ᵒC ϴ 1 Cos ϴ = 1

Sin ϴ = 0

λ2 = C S 0

-S C 0

0 0 1

1 0 0

0 1 0

0 0 1

=

Matrix kekakuan sistem strukturMatrix kekakuan sistem struktur

Batang 1 0 0 0 1 2 3

g1 g2 g4 -g1 -g2 g4 0g1 g2 g4 g1 g2 g4

g3 g5 -g2 -g3 g5 0

g6 -g4 -g5 g7 0

g1 g2 -g4 1

g2 g3 -g5 2

-g4 -g5 g6 3

[K1] =

[k1] =g1 g2 -g4 0

g2 g3 -g5 0

-g4 -g5 g6 0

0 0 0 0

α = EI/L3 = 5,773β = AL2/I = 4286,44

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53


g1 = α ( β C2 + 12 S2 ) C S ( β 12 ) [k ] =

g1 g2 -g4 0

g2 g3 -g5 0 g2 = α C S ( β - 12 ) g3 = α ( β S2 + 12 C2) g4 = -α 6 L S g5 = α 6 L C g6 = α 4 L2

[k1] = g2 g3 g5

-g4 -g5 g6 0

0 0 0 0

g6

g7 = α 2 L2


Batang 2 1 2 3 0 0 4

g1 g2 g4 -g1 -g2 g4 1g1 g2 g4 g1 g2 g4

g2 g3 g5 -g2 -g3 g5 2

g4 g5 g6 -g4 -g5 g7 3

g1 g2 -g4 0

g2 g3 -g5 0

g4 g5 g7 -g4 -g5 g6 4

[K2] =

[k2] =g1 g2 g4 g4

g2 g3 g5 g5

g4 g5 g6 g7

g4 g5 g7 g6

α = EI/L3 = 3,528β = AL2/I = 5952,381

09/02/2014

54


g1 = α ( β C2 + 12 S2

) [k ] =g1 g2 g4 g4

g2 g3 g5 g5

g2 = α C S ( β - 12 ) g3 = α ( β S2 + 12 C2

g4 = -α 6 L S g5 = α 6 L C g6 = α 4 L2

[k2] = g2 g3 g5 g5

g4 g5 g6 g7

g4 g5 g7 g6

g7 = α 2 L2

Persamaan kesetimbangan Persamaan kesetimbangan {Ps} = [Ks] {Us}{Ps} = [Ks] {Us}

[ks] = [k1] + [k2] =33403,7 12334,4 103,907 0

12334,445 12446,036 1,933 105,84

103 907 1 933 768 526 176 4103,907 1,933 768,526 176,4

0 105,84 176,4 352,8

[Ps] 0

-11,25=

-5,208

5,208

09/02/2014

55

Jika {Ps} = [Ks] {Us} ; maka {Us} Jika {Ps} = [Ks] {Us} ; maka {Us}

=33403,7 12334,4 103,907 0

12334,445 12446,036 1,933 105,84

103,907 1,933 768,526 176,4

0

-11,25

5 208

{Us} 103,907 1,933 768,526 176,4

0 105,84 176,4 352,8

{Us} =0.00073

-0.0018

-5,208

5,208

-0.0118

0.0211

Displacement tiap batang pada Displacement tiap batang pada sistem koordinat sistem koordinat GLOBALGLOBAL

U11 0 0

U12 0 01

{U1}

= U13 = 0 = 0

U14 Us1 0,00073

U15 Us2 -0,0018

U16 Us3 0,0118

U21 Us1 0,00073

U 2 U 2 0 0018U22 Us2 -0,0018

{U2}

= U23 = Us3 = 0,0118

U24 0 0

U25 0 0

U26 Us4 0,0211

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Displacement tiap batang pada Displacement tiap batang pada sistem koordinat sistem koordinat LOKALLOKAL

{U1} =λ1 0

{U1}0 λ1

U11

=

0.707 0.707 0

00

=

0

U12 -0.707 0.707 0 0 0

U13 0 0 1 0 0

U14 0.707 0.707 0

0,00073 -0 00076

U1

00.707 0.707 0

0.00076

U15 -0.707 0.707 0 -0,0018 -0.0018

U16 0 0 1 0,0118 -0.0118

Displacement tiap batang pada Displacement tiap batang pada sistem koordinat sistem koordinat LOKALLOKAL

{U2} =λ2 0

{U2}0 λ2

U21

=

1 0 0

00,00073

=

0,00073

U22 0 1 0 -0,0018 -0,0018

U23 0 0 1 0,0118 0,0118

U24

01 0 0 0 0

0U25 0 1 0 0 0

U26 0 0 1 0,0211 0,0211

09/02/2014

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Gaya BatangGaya Batang{p1} = [k1] {U1} + {f1}{p1} = [k1] {U1} + {f1}

4286,44 0 0-

4286,440 0 0 18.8

0 12 25,46 0 -12 25,46 0 -1.58

0 25,46 72,01 0 -25,46 36 0 -2.19

{p1} = 5,773 =-

4286,440 0 4286,44 0 0

-0.00076

-18.8

0 -12 -25,46 0 12 -25,46 -0.0018 1.58

0 -5,46 36 0 -25,46 72,01 -0.0118 -4.64

Gaya BatangGaya Batang{p2} = [k2] {U2} + {f2}{p2} = [k2] {U2} + {f2}

5952,381

0 0-

5952,381

0 00,00073

0 15,33

0 12 30 0 -12 30-0,0018

6.25 7,16

0 30 100 0 30 500,0118

5 208 4 58{p1

}=

3,528

+ =

0 30 100 0 -30 50 5,208 4,58

-5952,38

10 0

5952,381

0 00

0-

15,33

0 -12 30 0 12 300

6.25 5,34

0 -30 50 0 30 1000,0211

-5.208 0

09/02/2014

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Gambar Bidang M, N, DGambar Bidang M, N, D

RANGKA BATANGRANGKA BATANG In the Truss Construction (KRB), the calculation of

element stiffness matrix [K] is based on two dimensions f th t F th t k j t t i d of the truss case. Forces that works just tension and

compression axial only, is the moment and latitude force is not available.

Note the picture: a bar elements with area of A and modulus of elasticity of E is constant. The calculation of stiffness elements containing only element of A, E and f i t di t l i j i d jfour-point coordinates, namely: xi, xj, yi and yj.


09/02/2014

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60



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61

Example:Example: A truss structurewith arrangement as shown in the

figure as follows:3

4

10t

2t

12

600

With area of A = 20 cm2

◦ Modulus of elasticity : 2 x 106 kg/cm2 Question: Calculate the internal forces and deformation shape

4m



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71


y,v

L j

j

pj

c = cos

x,u

L

i

j

di

cui

ui qi pi

qjj

Elemen Rangka Batang, dengan sudut Elemen Rangka Batang setelah g g, g g g

pada bidang xy perpindahan titik ui > 0, titik lain tetap


09/02/2014

72

Pertama, harus menghitung :

L = 2ij2

ij y - y x- x

C = cos = L

x- x ij

S = sin = L

y - y ij

Perpendekan aksial cui menghasilkan gaya tekan aksial

F = icu L

AE

Dimana : x dan y merupakan komponen dari ;

pi = - pj = Fc

qi = - qj = Fs

Komponen ini menghasilkan kesetimbangan statis, sehingga diperoleh :

C2 CS -C2 -CS

ui =

pi

qi

pj

qj

L

AE


Hasil yang sama juga akan diperoleh dengan cara memberikan perpindahan pada vi,

uj, dan vj, dimana gaya bekerja sendiri-sendiri. Dan jika 4 dof dengan nilai tidak nol

bekerja bersama-sama, dan dengan superposisi masing-masing elemen matriks

kekakuan, dapat dihitung sebagai berikut :

K =

C2 CS -C2 -CS CS S2 -CS -S2 -C2 -CS C2 CS -CS -S2 CS S2

L

AE


09/02/2014

73

C2 CS -C2 -CS CS S2 -CS -S2 -C2 -CS C2 CS

ui

vi

uj

pi

qi

pj

Hubungan matriks kekakuan dengan gaya dapat ditulis sebagai berikut :

[ K ] { D } = { F }

=

L

AE

-CS -S2 CS S2 vj qj

Untuk kasus khusus :

1. Jika nilai = 0, sebagai batang horizontal, matriks kekakuan elemen [ K ] 4 x 4

Hanya berisi 4 komponen yang tidak bernilai nol, yaitu :

k11 = k33 = -k13 = -k31 =

L

AE

K =

1 0 -1 0 0 0 0 0 -1 0 1 0 0 0 0 0

L

AE


1. Jika nilai = 90, sebagai batang vertikal, matriks kekakuan elemen [ K ] 4 x 4

Hanya berisi 4 komponen yang tidak bernilai nol, yaitu :

k22 = k44 = -k24 = -k42 =

L

AE

K =

0 0 0 0 0 1 0 -1 0 0 0 0 0 -1 0 1

L

AE


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Sebuah Konstruksi Rangka Batang dengan luas A Sebuah Konstruksi Rangka Batang dengan luas A dan Modulus Elastisitas E yang sama, seperti dan Modulus Elastisitas E yang sama, seperti pada Gambarpada Gambar

L

L

L

L

1

4

3

7 6 5

2 3

4

2

5

1

v

u

L L

Hitunglah matriks kekakuaan masing-masing elemen


K =

C2 CS -C2 -CS CS S2 -CS -S2 -C2 -CS C2 CS

Perumusan untuk mencari nilai matriks kekakuan elemen dengan sudut :

L

AE

-CS -S2 CS S2

1 0 -1 0

Batang 1, 2 dan 3 merupakan batang horizontal, sehingga = 0o

Maka : [ K1 ] = [ K2 ] = [ K3 ]

K1 =

0 0 0 0 -1 0 1 0 0 0 0 0

L

AE


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75

K4 =

0,250 0,433 -0,250 -0,433 0,433 0,750 -0,433 -0,750 0 250 0 433 0 250 0 433

Batang 4 dan 6 merupakan batang diagonal dengan sudut = 60o

Dimana : C = cos 60o = 0,5

S = sin 60o = 0,866

Maka : [ K4 ] = [ K6 ]

L

AE

-0,250 0,433 0,250 -0,433 -0,433 -0,750 0,433 0,750

Batang 5 dan 7 merupakan batang diagonal dengan sudut = 300o

Dimana : C = cos 300o = 0,5

S = sin 300o = -0,866

Maka : [ K5 ] = [ K7 ]

L

K5 =

0,250 -0,433 -0,250 0,433 -0,433 0,750 0,433 -0,750 -0,250 0,433 0,250 -0,433 0,433 -0,750 -0,433 0,750

L

AE


0.335 0.128 -0.250 0.000 0.000 0.000 -0.085 -0.128 0.000 0.0000.128 0.192 0.000 0.000 0.000 0.000 -0.128 -0.192 0.000 0.000-0.250 0.000 0.671 0.000 -0.250 0.000 -0.085 0.128 -0.085 -0.1280.000 0.000 0.000 0.384 0.000 0.000 0.128 -0.192 -0.128 -0.192

[K]10X10 = A.E 0.000 0.000 -0.250 0.000 0.335 -0.128 0.000 0.000 -0.085 0.128

0.000 0.000 0.000 0.000 -0.128 0.192 0.000 0.000 0.128 -0.192-0.085 -0.128 -0.085 0.128 0.000 0.000 0.421 0.256 0.000 0.000-0.128 -0.192 0.128 -0.192 0.000 0.000 0.256 0.384 0.000 0.0000 000 0 000 -0 085 -0 128 -0 085 0 128 0 000 0 000 0 421 0 0000.000 0.000 0.085 0.128 0.085 0.128 0.000 0.000 0.421 0.0000.000 0.000 -0.128 -0.192 0.128 -0.192 0.000 0.000 0.000 0.384

3.35E-01 1.28E-01 -2.50E-01 0.00E+00 0.00E+00 0.00E+00 -8.53E-02 -1.28E-01 0.00E+00 0.00E+001.28E-01 1.92E-01 0.00E+00 0.00E+00 0.00E+00 0.00E+00 -1.28E-01 -1.92E-01 0.00E+00 0.00E+00-2.50E-01 0.00E+00 6.71E-01 0.00E+00 -2.50E-01 0.00E+00 -8.53E-02 1.28E-01 -8.53E-02 -1.28E-010.00E+00 0.00E+00 0.00E+00 3.84E-01 0.00E+00 0.00E+00 1.28E-01 -1.92E-01 -1.28E-01 -1.92E-01

[K]10X10 = 0.00E+00 0.00E+00 -2.50E-01 0.00E+00 3.35E-01 -1.28E-01 0.00E+00 0.00E+00 -8.53E-02 1.28E-010.00E+00 0.00E+00 0.00E+00 0.00E+00 -1.28E-01 1.92E-01 0.00E+00 0.00E+00 1.28E-01 -1.92E-01-8.53E-02 -1.28E-01 -8.53E-02 1.28E-01 0.00E+00 0.00E+00 4.21E-01 2.56E-01 0.00E+00 0.00E+00-1.28E-01 -1.92E-01 1.28E-01 -1.92E-01 0.00E+00 0.00E+00 2.56E-01 3.84E-01 0.00E+00 0.00E+000.00E+00 0.00E+00 -8.53E-02 -1.28E-01 -8.53E-02 1.28E-01 0.00E+00 0.00E+00 4.21E-01 0.00E+000.00E+00 0.00E+00 -1.28E-01 -1.92E-01 1.28E-01 -1.92E-01 0.00E+00 0.00E+00 0.00E+00 3.84E-01


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6.71E-01 0.00E+00 -2.50E-01 -8.53E-02 1.28E-01 -8.53E-02 -1.28E-010.00E+00 3.84E-01 0.00E+00 1.28E-01 -1.92E-01 -1.28E-01 -1.92E-01-2.50E-01 0.00E+00 3.35E-01 0.00E+00 0.00E+00 -8.53E-02 1.28E-01

[K]7X7 = -8.53E-02 1.28E-01 0.00E+00 4.21E-01 2.56E-01 0.00E+00 0.00E+00

1.28E-01 -1.92E-01 0.00E+00 2.56E-01 3.84E-01 0.00E+00 0.00E+00-8.53E-02 -1.28E-01 -8.53E-02 0.00E+00 0.00E+00 4.21E-01 0.00E+00-1.28E-01 -1.92E-01 1.28E-01 0.00E+00 0.00E+00 0.00E+00 3.84E-01

1.81E+00 2.85E+00 1.09E+00 -1.68E+00 1.94E+00 1.46E+00 1.67E+002.85E+00 -4.26E+00 2.92E+00 6.31E+00 -7.28E+00 -1.25E-01 -2.15E+001.09E+00 2.92E+00 4.12E+00 -2.24E+00 2.59E+00 1.94E+00 4.48E-01

[K]-1 = 1 68E+00 6 31E+00 2 24E+00 3 61E+00 6 12E+00 1 12E+00 3 34E+00[K] 7X7 = -1.68E+00 6.31E+00 -2.24E+00 -3.61E+00 6.12E+00 1.12E+00 3.34E+00

1.94E+00 -7.28E+00 2.59E+00 6.12E+00 -5.77E+00 -1.30E+00 -3.86E+001.46E+00 -1.25E-01 1.94E+00 1.12E+00 -1.30E+00 3.03E+00 -2.24E-011.67E+00 -2.15E+00 4.48E-01 3.34E+00 -3.86E+00 -2.24E-01 1.94E+00


Matriks Beban { R } {D} = [K]-1 {R}

0 -1.43E+04

-5000 2.13E+04

0 -1.46E+04{R}7X1 = 0 {D}7X1 = -3.15E+04

0 3.64E+04

0 6.23E+02

0 1.07E+04


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Matriks {D} Global

0.00E+00

0.00E+00

-1.43E+04

Matriks Gaya Dalam

F1H -0.02

F1V 2500.00

F2H 0.001.43E 042.13E+04

[D]10X1 = -1.46E+04

0.00E+00

-3.15E+04

3.64E+04

6.23E+02

1 07E+04

F2V -5000.00

F3H 0.00

F3V 2500.00

[F]12X1 = F4H = 0.00

F4V 0.00

F5H 0.00

F5V 0.001.07E+04


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