3872 ewahyuni 301 asmm english

09/02/2014

1

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


09/02/2014

2

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 Manual4Copyright: Dr Endah Wahyuni

09/02/2014

3

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...aaa...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


09/02/2014

4

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)]


09/02/2014

5

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

aa

A 21

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


09/02/2014

6

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

bb00

wkwhwgvfvevducubua

khgfedcba

wv

u

000000

kwhvgufwevducwbvau

wv

u

khgfedcba

000000


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

aaaaaa

A

aaa

aaaaaa

A

.........

...

...

.........

...

...

21

22212

12111

21

22221

11211

A symmetric matrix We can verify that AT = A


09/02/2014

7

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

121212321

B

21A


09/02/2014

8

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.


09/02/2014

9

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


09/02/2014

10

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



09/02/2014

11


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

=


09/02/2014

12

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)24Copyright: Dr Endah Wahyuni

09/02/2014

13

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



09/02/2014

14


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 LEI 6

LEI 12-

LEI 6

LEI 12

LEI 2

LEI 6-

LEI 4

LEI 6

22 K =

2323 LEI 6 -

LEI 12

LEI 6

LEI 12 -

Matriks Kekakuan LEI 4

LEI 6-

LEI 2

LEI 6

22

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

09/02/2014

15

Jika pada batang bekerja gaya aksial :

L, EA

U1,P1 U2,P2

P1+A=0 P2-A=0 =E(u2 u1)/LP1+EA (u2-u1)/L=0

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

K11 = LEA K21 = L

EA

U1= 1

K12 = - LEA U2= 1

K22 = L

EA

U2, P2 U5, P5

U1, P1 U4, P4

U3, P3 U6, P6

1 1 2


1111

////

2

1

2

1

LEAK

PP

uu

LEALEALEALEA

For beam with 6 dofFor beam with 6 dof

1 2

2

1

3

45

6


09/02/2014

16

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

00EA-00EA

6 x 6

K =

2323 LEI 6

LEI 12- 0

LEI 6

LEI 12 0

LEI 2

LEI 6- 0

LEI 4

LEI 6 0 22

0 0 L

0 0 L

0 0 L

EA- 0 0 L

EA

2323 LEI 6 -

LEI 12 0

LEI 6

LEI 12 0 -

LEI 4

LEI 6- 0

LEI 2

LEI 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}


09/02/2014

17

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 LEI 6

LEI 12-

LEI 6

LEI 12 0

LEI 2

LEI 6-

LEI 4

LEI 6

22 0 LLLL

2323 LEI 6 -

LEI 12

LEI 6

LEI 12 - 0

LEI 4

LEI 6-

LEI 2

LEI 6

22 1

K1 =

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

[ K1 ] =

0 LEI 4

2 x 2 0 0


09/02/2014

18

Elemen 2

0 1 0 2

2323 LEI 6

LEI 12-

LEI 6

LEI 12 0

LEI 2

LEI 6-

LEI 4

LEI 6

22 1 K2 =

2323 LEI 6 -

LEI 12

LEI 6

LEI 12 - 0

LEI 4

LEI 6-

LEI 2

LEI 6

22 2


K2

2 x 2

[ K2 ] =

LEI 4

LEI 2

LEI 2

LEI 4


= + =

Matriks Kekakuan Global Struktur

[ Ks ] = [ K1 ] + [ K2 ]

[ Ks ]LEI 2

LEI 40

LEI 4

LEI 2

LEI 8

0 0

[ Ks ] 2 x 2

LEI 4

LEI 2

LEI 4

LEI 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)


09/02/2014

19

q

0 0

Untuk contoh di atas, maka :

Ps =

2L q 121 2L q

121

2L q 121

2Lq1

Menghitung invers matrik kekakuan global [ Ks ]-1

[ Ks ] =

[ K ] 12- 4 L1 2- 4 L

LEI 4

LEI 2

LEI 2

LEI 8

L q 12

[ Ks ]-1 = 8 2-

EIL

2 . 2 - 4 . 81

=

8 2-

EI 28L

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

Us = 8 2-2- 4

EI 28

L

2L q 121

2L q 121


Us = EI 28

L

Us =

22 L q 61 - L q

31

22 L q 64 L q

61

EIL q

1683 3 Rotasi di joint 2

U11U12 U13 U14

0 0 0

Us

Deformasi untuk masing-masing elemen

Elemen 1 : U1 = =

EIL q

1685 3 Rotasi di joint 3

EIL q

1683 3

U21 U22 U23 U24

0 0

Elemen 2 : U2 = =

EIL q

1683 3

EIL q

1685 3


09/02/2014

20

q

0 0

Reaksi akibat beban luar :

2L q

1212L q

121

0 0 0

PR2 = PR1 =

1212

2L q

2L q

2L q 121

2L q

2L q

0

2L q 1212


0 0

0

0

Gaya akhir elemen :

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

2323 LEI 6

LEI 12-

LEI 6

LEI 12

EI2EI6EI4EI6 0 0

0 0 0

P1 = +

LEI2

LEI6-

LEI 4

LEI 6

22

2323 LEI 6 -

LEI 12

LEI 6

LEI 12 -

LEI 4

LEI 6-

LEI 2

LEI 6

22 EIL q

1683 3

L 6 3

P1 = =

2L q

564

2L q 562

L q 56

L q 56

6

2L q 282

2L q 281

L q 283

L q 28

3


09/02/2014

21

0

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

P2 = +

2323 LEI 6

LEI 12-

LEI 6

LEI 12

LEI 2

LEI 6-

LEI 4

LEI 6

22

EI6EI12EI6EI12

EIL q

1683 3 2L q

121

2L q

Lq0 P2 = +

2323 LEI6 -

LEI12

LEI 6

LEI 12 -

LEI 4

LEI 6-

LEI 2

LEI 6

22 EIL q

1685 3

2L q 564

L q 56

32

2L q 282

L q 28

16

2L q 121

2Lq

0 0

P2 = =

L q 56

2428

L q 28

12


q 0

Free Body Diagram :

2L q 2822L q

281

L q 283

2L q 282

L q 2816 L q

283 L q

2812

- -+

Menggambar gaya-gaya dalam :

Bidang D :

L q

283 L q

283

L q 2816

L q 2812

-

+ +

Bidang M :

2L q 282

2L q 281


09/02/2014

22

Example 2Example 2



09/02/2014

23


q 0

Free Body Diagram :

2L q 2822L q

281

L q 283

2L q 282

L q 2816 L q

283 L q

2812

- -+

Menggambar gaya-gaya dalam :

Bidang D :

L q

283 L q

283

L q 2816

L q 2812

-

+ +

Bidang M :

2L q 282

2L q 281


09/02/2014

24

Contoh 2Contoh 2



09/02/2014

25



09/02/2014

26

Example 4Example 4



09/02/2014

27



09/02/2014

28


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


09/02/2014

29

Elemen 1

0 1

0

2 x 2 1

Matriks Tujuan { T1 } = { 0 1 }T

K1 = LEI 2

LEI 4

LEI 4

LEI 2

[ K1 ] = 0 0

0

2 x 2

Elemen 2

1 2

1

2

LEI 4

K2 = LEI 2

LEI 4

EI 4EI 2 2 x 2 2


2 x 2

[ K2 ] =

LEI 4

LEI 2

LEI 2

LEI 4

L

L


= +

0

=


[ Ks ] = [ K1 ] + [ K2 ]

[ Ks ]LEI 2

LEI 4

LEI 2

LEI 8

LEI 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 :

LEI 4

LEI 2

LEI 4

LEI 2

dimana :

Us = deformasi ujung-ujung aktif

Ks = kekakuan struktur

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


09/02/2014

30

P

Untuk contoh di atas, maka :

0 LP1 LP1

0

LP8

LP8

Ps =

L P 81

L P 81



[ 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

LP8

1

LP81


09/02/2014

31

U11 U12

0


Elemen 1 : U1 = =

L P32

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

Us = 8 2-2- 4

EI 28

L

Us = EI 28

L

L P 81

L P 81

22 L q 61 - L q

31

22 L q 64 L q

61

LP3 2

1

U21 U22

Elemen 2 : U2 = =

EI

112

EIL P

1123 2

EIL P

1125 2

Us =

EIL P

1123

EIL P

1125 2

Rotasi di joint B

Rotasi di joint C


P Reaksi akibat beban luar :

0

L P 81L P

81

0

0

0

0

P1 = +

Gaya akhir elemen :

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

EIL P

1123 2

LEI 2

LEI 4

LEI 4

LEI 2

P1 = Hasil perhitungan hanya momen saja

L P

566

L P 563


09/02/2014

32

P2 = +

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

2 1

L P 81

EIL P

1123 2

LEI 2

LEI 4

P2

Hasil perhitungan

EILP

1125 2 L P

81

LEI4

LEI 2

2L q 566 2L q

283

P2 = =

0 0

Hasil perhitungan hanya momen saja

56 28


P 0

Dihitung lagi

Free Body Diagram :

P 569

L P 566

P 2817 P

2811

P 569

P 569

P 2817

L P 566

Dihitung lagi

P 569

P 2817

L P 563

Bidang M :

-

L P 566

Bidang D :

Bidang N :

-

+ P 2817

-

P 569

P 2811

P

- +

L P 563

L P 5611

+

P 2817

-

P 569-


09/02/2014

33

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 LEI 6

LEI 12-

LEI 6

LEI 12 0

LEI 2

LEI 6-

LEI 4

LEI 6

22 0 LLLL

2323 LEI 6 -

LEI 12

LEI 6

LEI 12 - 1

LEI 4

LEI 6-

LEI 2

LEI 6

22 2


K1 =

[ K1 ] =

2 x 2

23 LEI 6-

LEI 12

LEI 4

LEI 6- 2


09/02/2014

34

Elemen 2

2 3

2

2 x 2 3 K2 =

LEI 2

LEI 4

LEI 4

LEI 2


[ K2 ] = EI 4EI 2LEI 2

LEI 4

LL

2 x 2


[ Ks ] = [ K1 ] + [ K2 ]

L

L


= + =

1 2 2 3

[ Ks ] 3 x 3

LEI 4

LEI 2

LEI 2

LEI 4

LEI 2

LEI 8

LEI 62

0 LEI 6-

LEI 12

2323 LEI 6-

LEI 12

LEI 4

LEI 6- 2

LEI 4

LEI 2 0

0

Ps = L P 81

L P 81


09/02/2014

35


[ Ks ]-1 =

EIL 12 -

EIL 24

EIL 28 223

EIL 24 -

EIL 48

EIL 24 2

EI60L

EIL 24-

EIL 12 2

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

Us =

L P 81

L P 81

EIEIEI

EIL 12 -

EIL 24

EIL 28 223

EIL 24 -

EIL 48

EIL 24 2

60LL 24L 12 2

0

Us =

8

EIL P

1283 3

EIL P

12810 2 Rotasi di joint B

Rotasi di joint C

EI

EI-

EI

EIL P

1287 2

Dilatasi di joint B


U11

2

0


U12 U13 U14

0

Elemen 1 : U1 =

EIL P

1283 3

EIL P

1286 2

U21 U22

Elemen 2 : U2 = =

EIL P

1286 2

EIL P

1287 2


09/02/2014

36

0 0

0 0

Gaya akhir elemen :

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

2323 LEI 6

LEI 12-

LEI 6

LEI 12

EI2EI6EI4EI6 0

+

0 0 0

0

P1 = LEI2

LEI6-

LEI 4

LEI 6

22

2323 LEI 6 -

LEI 12

LEI 6

LEI 12 -

LEI 4

LEI 6-

LEI 2

LEI 6

22

EIL P

1283 3

EIL P

1286 2

0

0

P1 =

L P 643

L P 643


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

L P 81

EIL P

1286 2

LEI 2

LEI 4

P2 = +

P2 = = Hasil perhitungan hanya momen saja

EIL P

1287 2 L P

81

LEI 4

LEI 2

L P 128

6 2L q 643

0 0


09/02/2014

37

P 0

Dihitung lagi

Free Body Diagram :

L P 643

P35 P29

P 6435

L P 643

Dihitung lagi

P64

P64

L P 643

-+

Bidang M :

L P 643

L P 12829

P

6435

-

128

L P 643


-

+ Bidang D :

P 6435

P 6429

P

Bidang N :

-

P 6435


09/02/2014

38

Deformasi aksial diabaikan



09/02/2014

39



09/02/2014

40

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



09/02/2014

41


Transformation of CoordinatesTransformation of Coordinates2

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

3U4 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 LEI 6

LEI 12- 0

LEI 6

LEI 12 0

0 0 L

EA- 0 0 L

EA

6 x 6

k = LEI 2

LEI 6- 0

LEI 4

LEI 6 0 22

2323 LEI 6 -

LEI 12 0

LEI 6

LEI 12 0 -

4626

0 0 L

EA- 0 0 L

EA

LEI 4

LEI6- 0

LEI2

LEI6 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 ]

LEI 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


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 = 331 12) . 10(

50 . 30.000 LEI = 0,87

1 = 5012) . (10 . 5

ILA 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


09/02/2014

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 = 331 12) . 10(

50 . 30.000 LEI = 0,87

2 = 5012) . (10 . 5

ILA 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 =


09/02/2014

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 =


09/02/2014

49

u11

u12 u13

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

u23

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

09/02/2014

51

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

= 45C 0 707 Cos = 0.707

Sin = 0.707

1 = C S 0-S C 00 0 1

0.707 0.707 0

-0.707 0.707 0

0 0 1

=

09/02/2014

52

MATRIX TRANSFORMASI MATRIX TRANSFORMASI BATANGBATANG

= 0C 1 Cos = 1

Sin = 0

2 = C S 0-S C 00 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 3g1 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

= EI/L3 = 5,773 = AL2/I = 4286,44

09/02/2014

53


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

g1 g2 -g4 0

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

[k1] = g2 g3 g5-g4 -g5 g6 0

0 0 0 0

g6 g7 = 2 L2


Batang 2 1 2 3 0 0 4g1 g2 g4 -g1 -g2 g4 1g1 g2 g4 g1 g2 g4g2 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 g4g2 g3 g5 g5g4 g5 g6 g7g4 g5 g7 g6

= EI/L3 = 3,528 = AL2/I = 5952,381

09/02/2014

54


g1 = ( C2 + 12 S2) [k ] =

g1 g2 g4 g4g2 g3 g5 g5

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

[k2] = g2 g3 g5 g5g4 g5 g6 g7g4 g5 g7 g6

g7 = 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 0,0118U21 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

09/02/2014

56

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

{U1} =1 0

{U1}0 1

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 00,00073 -

0 00076U1

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} =2 0

{U2}0 2

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 0U26 0 0 1 0,0211 0,0211

09/02/2014

57

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

4286,44 0 0 -4286,44 0 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 =

-4286,44 0 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

-5952,38

10 0

0,000730 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

} =3,52

8 + =0 30 100 0 -30 50 5,208 4,58

-5952,38

10 0 5952,381 0 0

00 -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

58

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

59



09/02/2014

60



09/02/2014

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



09/02/2014

62



09/02/2014

63



09/02/2014

64



09/02/2014

65



09/02/2014

66



09/02/2014

67



09/02/2014

68



09/02/2014

69



09/02/2014

70



09/02/2014

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 = 2ij2ij 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 LAE

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 4Hanya 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


09/02/2014

74

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 :

LAE

-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


09/02/2014

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 ]

LAE

-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 0.1280.000 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


09/02/2014

76

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+001.28E-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+001.94E+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+020 1.07E+04


09/02/2014

77

Matriks {D} Global

0.00E+000.00E+00-1.43E+04

Matriks Gaya Dalam

F1H -0.02F1V 2500.00F2H 0.001.43E 04

2.13E+04[D]10X1 = -1.46E+04

0.00E+00-3.15E+043.64E+046.23E+021 07E+04

F2V -5000.00F3H 0.00F3V 2500.00

[F]12X1 = F4H = 0.00F4V 0.00F5H 0.00F5V 0.00

1.07E+04


3872 ewahyuni 301 asmm english

Documents