Stiffness and Mass Matrices of Bar-Finite Element Method-Assignment Solution, Exercises for Finite Element Method. Aligarh Muslim University

Finite Element Method

Description: This assignment solution was submitted to Amar Sharma for Finite Element Method course at Aligarh Muslim University. It includes: Tapered, Bar, Length, Cross-sectional, Area, Strain, Kinetic, Energy, Matrix, Axial, Displacement
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Assignment#02
Finite Element Methods
Statement:
1.9 For the tapered bar shown in Figure 1.20. The area of cross section changes along
the length as . where A0 is the cross-sectional area at x=0, and is
the length of the bar. By expressing the strain and kinetic energies of the bar in
matrix forms, identify the stiffness and mass matrices of a typical element. Assume
a linear model for the axial displacement of the bar element.
Figure:
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Solution:
Step 1: Idealization
As shown in the above figure the bar is distributed into three equal and same elements.
We take a typical element of length 𝑙 (e) with nodal unknowns
 
1
e
and
 
2
e
at nodes 1
and 2.we take linear displacement model of bar element.
 
12
.......... 1xx

 
Step 2: B.Cs
At x=0
 
 
1
e
x  
And at x=
 
 
2
e
x  
Putting these conditions in equation.1 and finding them values of constants
gives
 
11
e

and
Putting these values in equation 1 we get
 
     
 
21
1
ee
e
e
xx
 
 
Element strain:
 
   
 
 
21
ee
e
e
d
dx
 

Element stress:
     
 
21
ee
ee
e
E

 


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Step 3: Element stiffness Matrix
First we calculate the potential energy of the bar
 
12
p
w

 
Where
       
0
1
2
e e e e
A dx
 
Or
       
2
0
2
ee
ee
AE dx

         
 
2
21
0
2
ee
ee
e
e
AE dx

 



             
 
22
2 1 1 2
2
0
2
2
e e e e
ee
e
e
AE dx

   



     
         
 
22
2 1 1 2
2
2
ee
e e e e e
e
AE
   
       
1
2
e
e e T e
K

 

   
 
1
2
e
e
e






And
     
 
11
11
ee
e
e
AE
K



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Now
 
12
2
eAA
A
Also
 
xl
o
A x A e
Therefore
 
1
20.3678
0.3678 1.3678 0.684
22
o
o
eo o o o
AA
AA
A A A
AA
 
And
so
   
11
2.052 ....... 1
11
e
eo
AE
K Ans



Step 4: Mass Matrix
The element mass matrix can be found from kinetic energy equation of the beam
2
0
1..........2
2
T m dx
t




Now
     
 
21
1
ee
e
ex
t
 
  
Putting this value in equation 2
 
3
e
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