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Dr. A. S. Sayyad
Professor & Head
Department of Structural Engineering
Sanjivani College of Engineering, Kopargaon 423603.
(An Autonomous Institute, Affiliated to Savitribai Phule Pune University, Pune)
Finite Element Method In Civil Engineering
Natural coordinates of 1D bar element
(x-y coordinate system)
2 1
1 2
,
x x x x
L L
L L
 
 
Let us consider a two nodded bar/line element (1D). Node 1 and 2 have Cartesian
coordinates 1 x and 2 x respectively. Cartesian coordinate of any point ‘P’ is ‘ x ’.
Natural coordinates of any point ‘P’ are (L1, L2).
Natural coordinates of 1D bar element (x-y coordinate system)
From the definition of line element
In matrix form,
1
1
2 1 2
1 1 1
L
L x x x

     

   
 
 
   
1 2 2
2 1 1
2 1
1 ( )
1
1 1
1 ( )
L x x x
L x x x
x
x x L
 
     
 
 
     
 
 
  
     
1
2 1 2
1 1 1
L
L x x x
    

   
 
 
  
1 2
1 1 2 2
1
L L
L x L x x
 
 
Therefore natural coordinates at node 1 is (1, 0) and natural
coordinates at node 2 is (0, 1).
2
1
x x
L
L

  1
2
x x
L
L



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Lect15

  • 1. Dr. A. S. Sayyad Professor & Head Department of Structural Engineering Sanjivani College of Engineering, Kopargaon 423603. (An Autonomous Institute, Affiliated to Savitribai Phule Pune University, Pune) Finite Element Method In Civil Engineering Natural coordinates of 1D bar element (x-y coordinate system) 2 1 1 2 , x x x x L L L L    
  • 2. Let us consider a two nodded bar/line element (1D). Node 1 and 2 have Cartesian coordinates 1 x and 2 x respectively. Cartesian coordinate of any point ‘P’ is ‘ x ’. Natural coordinates of any point ‘P’ are (L1, L2). Natural coordinates of 1D bar element (x-y coordinate system)
  • 3. From the definition of line element In matrix form, 1 1 2 1 2 1 1 1 L L x x x                     1 2 2 2 1 1 2 1 1 ( ) 1 1 1 1 ( ) L x x x L x x x x x x L                                1 2 1 2 1 1 1 L L x x x                  1 2 1 1 2 2 1 L L L x L x x    
  • 4. Therefore natural coordinates at node 1 is (1, 0) and natural coordinates at node 2 is (0, 1). 2 1 x x L L    1 2 x x L L  