Matakuliah
Tahun
: S0725 – Analisa Struktur
: 2009
Computer Application on Structural Analysis
Session 23-26
Contents
•Structural Model
•Computer Application
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3
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa dapat membuat diagram / skema untuk
analisa struktur dengan bantuan program komputer
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4
Outline Materi
• Input
• Analize
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5
Input
Element stiffness are :
xyz415236EA,I129118710
Deflection in point can be presented on
3 rotation and 3 translation , in
general they have 6 degree of
freedom
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6
Input
y
4
1
3
6
11
5
2
8 7
EA,I
9
10
x
12
z
Because of there are 12 Vectors, so the
stiffness matrix become 12
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x 12
7
Input
[K ] 
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 EA
 L

12 EI ZZ
 0
L3


0
 0

 0
0


0
 0

6 EI ZZ
 0
L2

 EA
0
 L

12 EI ZZ

 0
L3

 0
0


0
 0

 0
0


6 EI ZZ
 0
L2

12 EI yy
L3
GJ
L
0


6 EI yy
0
L2
L
0
0
0
4 EI ZZ
L
0
0
0
0
0
0
0
6 EI ZZ
L2
12 EI yy
0
3
L
0

4 EI yy
6 EI yy
L2
0

GJ
L
0
0
6 EI yy
L
0
2 EI yy
L
0
EA
L
12 EI ZZ
L3
0
0
0
0
0
0
0
0
0
0
6 EI ZZ

L2
2 EI ZZ
L
12 EI yy
L3
0
6 EI yy
L2
0
GJ
L
0
0
4 EI yy
L
0



























4 EI ZZ 

L 
8
Input
dimana :
L = panjang balok
A = luas penampang balok
Iyy = momen ineria terhedap sumbu y
Izz = momen inersia terhadap sumbu z
J = momen inersia polar
E = modulus elastisitas bahan
G = modulus geser bahan
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Analize
10t
5t
5
3
C
C 2
C
d
6
d 4
Rangka batang dibebani 5t di C dan
akan 10t di D
D
3
1
A
B
4
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10
Analize
1
2
4
3
C
D
7
5
6
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8
11
Analize
AB:
1
5
7
6
8
1
0
( k1 )  
 1

0
0
1
0
0
0
1
0
0
0
 15

 0
0 AE

 15
0 4


0
 0
0
 15
0
0
0
15
0
0
0
0
 AE
0 60

0
 AB  4
 0
Cos   Cos 0  1
Sin   Sin 0  0
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Analize
AC:
1
2
k 2  
5
6
 AC  3
0
0

0

0
0
0
1
0
0
0
-1
0
0
0

- 1
 AE  0
0
0 3


- 1
0
0
0
20
0
0
0
 20
0
0 
 20
 AE
0  60

 20
  90
Cos 90  0
Sin 90  1
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Analize
BC
2
k3  
1
7
 0,64
- 0,48

- 0,64

- 0,48
- 0,48
0,64
0,36
0,48
0,48
0,64
- 0,36
- 0,48
- 0,48
- 0,36
 AE
- 0,48 5

0,36 
8
 BC  5
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tg   -
3
4
Cos  -
4
5
Sin  
3
5
14
Analize
AD
4
3
5
k3  
6
tg  
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3
4
Cos 
4
5
 0,64 0,48 - 0,64 - 0,48
 0,48 0,36 - 0,48 - 0,36

 AE
- 0,64 - 0,48 0,64 0,48  5


0,48
0,36
0,36
0,36


Sin  
3
5
15
Analize
4
3
7
8
  90
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k4  
1
0

- 1

0
0 - 1 0
0 0 0 AE
0 1 0 5

0 0 0
Cos90  0
Sin 90  1
16
Analize
k5  
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1
0

- 1

0
0 - 1 0
0 0 0 AE
0 1 0 4

0 0 0
 0
Cos0  1
Sin 0  0
17
Analize
1
2
3
4
5
6
1 
2 

3 

k11
k12
k 21
k 22
4
k Str   
5

6 
7 

8 
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7
8












18
Analize
 F1   5 
F   0 
 2  
  
F
 3  0 
 F4  10
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1
1  1
  2
 2
  3
 3 
 4  4






2
k11
3
4






1
5
0
 
 
0
10
19
Analize
5
 F5   H A 
F  V  6
 6  A  
 F7   H B  7
   
 F8   VB  8
5
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6



 k21


7
8
1

2

3

4
1
2



 k11


3
4






1
5
0
 
 
0
10
20
Analize
{F}={k}{}
Member forces
Member 1
 F5 
F 
 6 
 F7 
 
 F8 
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





k1






5 
 
 6



 7

8 

5
6
7
8
0
0
0
0
21
Analize
{F}={k}{}
Member forces
Member 2
 F5 
F 
 6 
 F1 
 
F2 
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





k2






5 
 
 6
 

 1
 2 
22