MUSLIM ENGINEER: CIVIL ENGINEERING

Thursday, November 21, 2013

ANALISA STRUKTUR IV DENGAN MATLAB (RANGKA BATANG)

SOAL

PEMODELAN

PERSIAPAN DATA

· Data Propertis Batang

6 cm²= 6e-4 m²

210 Gpa = 210e9 N/m²

α = 45⁰

Gaya arah Y

Sin 45 = 0,7071

350 x 0,7071 = 247.458 N

Gaya arah X

Cos 45 = 0,7071

350 x 0,7071 = 247.458 N

· Data Koordinat

No Titik	X	Y
1	0	0
2	3.5	0
3	7	0
4	0	3.5
5	3.5	3.5
6	7	3.5
7	10.5	3.5

· Data Batang

No. Batang	No. titik kiri (bawah)	No. titik kanan (atas)	No. Propertis
1	1	2	1
2	2	3	1
3	2	4	1
4	2	5	1
5	2	6	1
6	3	6	1
7	3	7	1
8	4	5	1
9	5	6	1
10	6	7	1

· Data Tumpuan

No. Titilk	Restrain RX	Restrain RY	Restrain R
1	1	1	1
4	1	1	1

· Data Beban Titik

No. Titik	Beban Fx	Beban Fy	Beban Mz
5	0	-300	0
6	0	-200	0
7	247.458	-247.458	0

1. INPUT DATA

%****RANGKA BATANG 2D---TEKNIK SIPIL UNISSULA SEMARANG----

%****nama file : soal5Reqsi_s----oleh :ZAITUN ANDARWAN (022113198)----

%****struktur : SOAL2 R. BATANG

clear all

clc

%****General Data--------------

type='t2d'

nfile='SOAL2 R. BATANG'

%****Property---i,E,A--------------

prop = [ 1 210e9 6e-4 ];

%****Coordinates------------

%****--- Joint X Y ---

coord=[ 1 0 0

2 3.5 0

3 7 0

4 0 3.5

5 3.5 3.5

6 7 3.5

7 10.5 3.5 ];

%****Element data------------------

%****--- Element J1 J2 No.prop---

element=[ 1 1 2 1

2 2 3 1

3 2 4 1

4 2 5 1

5 2 6 1

6 3 6 1

7 3 7 1

8 4 5 1

9 5 6 1

10 6 7 1 ];

%****Nodal Restraint----------

%****---- Joint JR1 JR2 --

Support=[ 1 1 1 1

4 1 1 1 ];

%****Joint Load-------------

%****---Joint FX FY --

JL=[ 5 0 -300

6 0 -200

7 247.458 -247.458 ];

%****Element Load Data----------------

AML=[ 1 0 0 0 0];

%****Call function for analysis trs2d 2d--------------------

[dof,index,coord,element]=TRUSS2D_sdata(prop,element,coord,type);

[S,Sm,SmS,Cx,Cy,RT,L,A,Joint,Xj,Xk,Yj,Yk]=...

TRUSS2D_stiff(prop,element,coord,index,nfile);

[IR,IF,Support]=TRUSS2D_ldata(Support,dof);

[DF,AR,AM]=TRUSS2D_analysis_result...

(element,dof,index,IF,IR,S,Sm,JL,AML,RT,Support,type,nfile);

[joint_disp,support_reaction,beam_endforces]=TRUSS2D_print_result...

(JL,AML,dof,Support,element,IF,IR,DF,AR,AM,type,nfile);

%****end data------------------------------------------------

TRUSS2D_xplot;

%-------------------

2. OUTPUT DAN CEK

STIFFNESS MATRIX METHOD : PROGRAM TRUSS2D

File Name : SOAL2 R. BATANG_2.txt

Joint Loads :

Joint Fx Fy

5 0.000 -300.000

6 0.000 -200.000

7 247.458 -247.458

Member Loads :

Member Fx1 Fy1 Fx2 Fy2

1 0.000 0.000 0.000 0.000

Support Restraints :

Joint Rx Ry

1 1 1

1 4 1

1 1

Joint Displacements :

Joint Dx Dy

1 0.0000e+000 0.0000e+000

2 -4.0066e-005 -9.8792e-005

3 -4.6940e-005 -2.3324e-004

4 0.0000e+000 0.0000e+000

5 2.6177e-005 -1.0713e-004

6 5.2354e-005 -2.2637e-004

7 6.6102e-005 -3.6572e-004

Support Reactions :

Joint Rx Ry

1 1.4424e+003 0.0000e+000

4 -1.6898e+003 7.4746e+002

Member End Forces (axial/normal):

Member axial force : + tension - compression

1 -1442.374

2 -247.458

3 1057.065

4 -300.000

5 -632.801

6 247.458

7 -349.958

8 942.374

9 942.374

10 494.916

<<< end of file >>>

CEK KESEIMBANGAN

Support Reactions :

Joint Rx Ry

1 1.4424e+003 0.0000e+000

4 -1.6898e+003 7.4746e+002

ϵFX = 0
(1442.4 – 1689.8) + 247.4 = 0 (OKE)

ϵFY = 0

747.4 - ( 300 + 200 + 247.4 ) = 0 (OKE)

FREE BODY DIAGRAM

GAMBAR BIDANG NORMAL

Sunday, November 10, 2013

ANALISA STRUKTUR IV DENGAN SOFTWARE MATLAB (PORTAL)

1. Soal

2. Penomoran koordinat, titik simpul dan batang

3. Persiapan Data

· Data Propertis Batang

E=2.0e8 (ton/m2) A= 1.00*0.8 (m2) I=1/12*1.00*0.8^3 (m4)

· Data Koordinat

No. Titik	X	Y
1	0	0
2	0	4
3	0	8
4	5	8
5	10	8
6	10	4
7	10	0
8	5	0

· Data Batang

No. Batang	No. titik kiri (bawah)	No. titik kanan (atas)	No. Propertis
1	1	2	1
2	2	3	1
3	3	4	1
4	4	5	1
5	6	5	1
6	7	6	1
7	8	7	1
8	1	8	1

· Data Tumpuan

No. Titilk	Restrain RX	Restrain RY	Restrain R
1	1	1	0
7	1	1	0

· Data Beban Titik

No. Titik	Beban Fx	Beban Fy	Beban Mz
2	2	0	0
4	0	-6	0
6	-2	0	0
8	0	-5	0

· Data Beban Batang

No. Batang	Beban FX1	Beban FY1	Beban MZ1	Beban FX2	Beban FY2	Beban MZ2
1	0	8	0	5.33	8	-5.33
2	0	8	0	5.33	8	-5.33
3	0	20	0	16.67	20	-16.67
4	0	20	0	16.67	20	-16.67
5	0	-8	0	-5.33	-8	5.33
6	0	-8	0	-5.33	-8	5.33
7	0	5	0	4.167	5	-4.167
8	0	5	0	4.167	5	-4.167

4. Input Data

%****PORTAL 2D---TEKNIK SIPIL UNISSULA SEMARANG----

%****nama file : PORTALZ----oleh :zaitun andarwan 022113198 ----

%****struktur :

clear all

clc

%****General Data---

type='FRAME2D'

nfile='PORTALZ,'

%****Material Properties---i,E,A,I---

prop = [ 1 2e8 1.00*0.8 1/12*1.00*0.8^3];

%****Coordinates---

%****--- Joint X Y ---

coord=[ 1 0 0

2 0 4

3 0 8

4 5 8

5 10 8

6 10 4

7 10 0

8 5 0];

%****Element data---

%****--- Element J1 J2 prop---

element=[ 1 1 2 1

2 2 3 1

3 3 4 1

4 4 5 1

5 6 5 1

6 7 6 1

7 8 7 1

8 1 8 1];

%****Nodal Restraint---

%****---- Joint JR1 JR2 JR3---

Support=[ 1 1 1 0

7 1 1 0 ];

%****Joint Load---

%****---Joint FX FY MZ---

JL=[ 2 2 0 0

4 0 -6 0

6 -2 0 0

8 0 -5 0];

%****Load Data---

%****

AML=[ 1 0 8 5.33 0 8 -5.33

2 0 8 5.33 0 8 -5.33

3 0 20 16.67 0 20 -16.67

4 0 20 16.67 0 20 -16.67

5 0 -8 -5.33 0 -8 5.33

6 0 -8 -5.33 0 -8 5.33

7 0 5 4.167 0 5 -4.167

8 0 5 4.167 0 5 -4.167 ];

%****Call function for analysis portal 2d--------------------

[dof,index,coord,element]=FRAME2D_sdata(prop,element,coord,type);

[S,Sm,SmS,Cx,Cy,RT,L,A,Joint,Xj,Xk,Yj,Yk]=FRAME2D_stiff...

(prop,element,coord,index,nfile);

[IR,IF,Support]=FRAME2D_ldata(Support,dof);

[DF,AR,AM]=FRAME2D_analysis_result...

(element,dof,index,IF,IR,S,Sm,JL,AML,RT,Support,type,nfile);

[joint_disp,support_reaction,beam_endforces]=FRAME2D_print_result...

(JL,AML,dof,Support,element,IF,IR,DF,AR,AM,type,nfile);

%****end data------------------------------------------------

FRAME2D_xplot

5. Output Data dan cek

STIFFNESS MATRIX METHOD : PROGRAM FRAME2D

File Name : PORTALZ,_2.txt

Joint Loads :

Joint Fx Fy Mz

2 2.000 0.000 0.000

4 0.000 -6.000 0.000

6 -2.000 0.000 0.000

8 0.000 -5.000 0.000

Member Loads :

Member Fx1 Fy1 Mz1 Fx2 Fy2 Mz2

1 0.000 8.000 5.330 0.000 8.000 -5.330

2 0.000 8.000 5.330 0.000 8.000 -5.330

3 0.000 20.000 16.670 0.000 20.000 -16.670

4 0.000 20.000 16.670 0.000 20.000 -16.670

5 0.000 -8.000 -5.330 0.000 -8.000 5.330

6 0.000 -8.000 -5.330 0.000 -8.000 5.330

7 0.000 5.000 4.167 0.000 5.000 -4.167

8 0.000 5.000 4.167 0.000 5.000 -4.167

Support Restraints :

Joint Dx Dy Rz

1 1 1 0

7 1 1 0

Joint Displacements :

Joint Dx Dy Rz

1 0.0000e+000 0.0000e+000 -5.4719e-006

2 4.8437e-006 -1.0750e-006 2.8794e-006

3 8.2956e-007 -2.1500e-006 -6.6680e-006

4 -5.1689e-022 -4.6896e-005 -2.1794e-022

5 -8.2956e-007 -2.1500e-006 6.6680e-006

6 -4.8437e-006 -1.0750e-006 -2.8794e-006

7 0.0000e+000 0.0000e+000 5.4719e-006

8 0.0000e+000 -2.2835e-005 -3.3553e-022

Support Reactions :

Joint Rx Ry Mz

1 -7.4540e+000 5.5500e+001 0.0000e+000

7 7.4540e+000 5.5500e+001 0.0000e+000

Member End Forces :

Member Fx1 Fy1 Mz1 Fx2 Fy2 Mz2

1 43.000 7.454 -13.578 -43.000 8.546 11.394

2 43.000 -10.546 -11.394 -43.000 26.546 -62.790

3 26.546 43.000 62.790 -26.546 -3.000 52.210

4 26.546 -3.000 -52.210 -26.546 43.000 -62.790

5 43.000 10.546 11.394 -43.000 -26.546 62.790

6 43.000 -7.454 13.578 -43.000 -8.546 -11.394

7 0.000 -2.500 -23.922 0.000 12.500 -13.578

8 0.000 12.500 13.578 0.000 -2.500 23.922

<<< end of file >>>

Deformade Geometri

6. Cek Keseimbangan

Support Reactions :

Joint          Rx             Ry             Mz

    1   -7.4540e+000    5.5500e+001    0.0000e+000

    7    7.4540e+000    5.5500e+001    0.0000e+000

ϵFY = 0

(55.5 + 55.5) – (80+20+5+6) = 0

111 – 111 = 0 (oke)

ϵFX = 0

(-7.454+7.454) – (32 +2 -32 -2) = 0 (oke)

7. Free Body Diagram

8.      Gambar Bidang Momen, Lintang dan Normal (M, D & N)

·        BIDANG MOMEN

·       BIDANG LINTANG

·        GAYA NORMAL

Pages

Thursday, November 21, 2013

ANALISA STRUKTUR IV DENGAN MATLAB (RANGKA BATANG)

Sunday, November 10, 2013

ANALISA STRUKTUR IV DENGAN SOFTWARE MATLAB (PORTAL)