Course
Year
: S0912 - Introduction to Finite Element Method
: 2010
1D OF FINITE ELEMENT METHOD
Session 4 – 6
COURSE 4
Content:
• 1D Element Types
• 1D Element Modelling
• 1D Solution
• Example/Case Study
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1D ELEMENT TYPES
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1D ELEMENT MODELLING
Forces and Moments on 1D Element
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APPLICATION TO FINITE ELEMENT
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1D SOLUTION
Global and Local Coordinate System
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1D SOLUTION
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1D SOLUTION
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1D ELEMENT EXAMPLE
u1
u2
Deformed shape
f1
x
Element
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f2
Node
(a hinge)
1D ELEMENT EXAMPLE
(i)
Conjecture a displacement function
u(x)
x
 a1 
u x   a1  a2 x  1 x    N a (1)
 
 a2 
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1D ELEMENT EXAMPLE
(ii)
Express u(x) in terms of nodal displacements by using boundary
conditions.
Deformed shape
u(0) = u1
 u1  1 0   a1 
u   1 L a  
 2 
 2 
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u(L) = u2
u  Aa (2)
1D ELEMENT EXAMPLE
Sub (2) into (1)
1
1 0 
1
 u
u x   NA u  1, x


1 L
 x
u x   1  ,
 L
x


u

C
u
L 
(3)
Displacement polynomial that satisfies boundary conditions
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Bar Element example
(iii)
Derive strain-displacement relationship by using mechanics
theory
du d
 1
 x  
  Cu   Bu  
dx dx
 L
Axial Strain
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1
u

L
(4)
1D ELEMENT EXAMPLE
(iv)
Derive stress-displacement relationship by using elasticity
theory
 x  E x  EB u
Elastic Modulus
Axial Stress
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(5)
1D ELEMENT EXAMPLE
(v)
Use principle of Virtual Work
Work = Stress x Strain x Volume
Internal work
Bar cross-sectional area A
WI     x . x dxdydz     x .  x dx  dydz
 A  x .  x dx
 EA B u.B u dx  EAu
External work
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WE  u1
 f1 
T


u2 
u f
 
 f2 
T
T


B
 Bdxu
1D ELEMENT EXAMPLE
Equate internal and external work
WE  WI
u f  EAu
T
f  k  u ,
Stiffness matrix
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T
T


B
 Bdxu
k   EA BT Bdx
(6)
1D ELEMENT EXAMPLE
Resultant stiffness matrix
k   EA

 EA
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L  1 
L
 1 
0 
 L 
L
1
2
L
 1
0  2
 L
 L1 L1 dx
1
L2 dx
1
L2 


EA  1  1

(7 )


L  1 1 
EXAMPLE
Axial deformation of a bar subjected to a uniform load
(1-D Poisson equation)
p x  = p0
x = 0, L
d 2u
EA 2 = p0
dx
u 0  = 0
du
EA
dx
=0
xL
u = axial displacement
E=Young’s modulus = 1
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A=Cross-sectional area = 1
EXAMPLE
Model the following shaft using two beam finite elements
neglecting axial deformation, given the following data:
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EXAMPLE
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EXAMPLE
Global and element coordinates are parallel. The Global
nodal coordinates are then defined as ui, i =1, 2, ......... , 6 .
Now assign a set of generalized coordinates qi along same
directions.
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EXAMPLE
Simple Form
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EXAMPLE
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EXAMPLE
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EXAMPLE
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EXAMPLE
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