[ 指導教授 ] ：陳為仁
[ 製作學生 ] ：吳裕虎、杜建宗
[ 修正日期 ] ：2004年8月

§ Introduction to the displacement method

Displacement method

Element stiffness matrix of 1D spring

Assembly of system equation

Boundary conditions (BC)
 Summary
§ Principle of minimum potential energy
1 .Total potential energy ， π p
2. Stationary value of a function G(x)
3. Principle of minimum potential energy
4. Derivation of element equation of a spring
§ TRUSSES
 Definition
 Element stiffness matrix of a rod under axial load
 Transformation matrix and stiffness matrix in 2D

Stress for a bar (truss) element in 2D

§

Displacement method

Element stiffness matrix of ID spring

Assembly of system equation

Boundary conditions (BC)

Summary

§
A finite element approach method was proposed in the early 1950s for investigating the complex behavior of structural problems by considering the nodal displacements as the basic unknowns in the system of equations.
    
Stiffness matrix nodal displacements (unknowns)
Basic equations of the displacement method are derived from
(1) The equilibrium of nodal forces
(2) The compatibility of displacements
(3) The force-displacement relationship

§
(1) Element stiffness matrix
(2) Spring under axial load
(3) Procedures of derivation of stiffness matrix

(1)
  e
  e
by
  e
  e 
  e
  e

(2)
Consider a linear spring with spring constant K and length L under axial load.
u i f ip u j f jp
Positive x direction i,j ： nodal number u i
, u
： nodal displacements at nodal i and j f ip f jp
： nodal forces at nodal i and j

(3)
(A) Satisfy static equilibrium
(B) Relate the compatibility of displacements
(C) Relate force to displacement
(D) Derivation of element stiffness matrix

(A) Satisfy the static equilibrium
 f x
 f ip
 f jp

0 or f ip
  f jp
(1)
(B) Relate the compatibility of displacements
Assume that the spring has a total elongation
δ due to the tensile force.
  u
 j u i
(2)

(C) Relate force to displacement
For the linear spring , tensile force and elongation are related through Hooke ´ s law by
T
 k
  k ( u j
 u i
)
L
(3) u i u j

(D) Derivation of element stiffness matrix
By the sign convection for nodal force , we have f ip
 
T f ip

T
Using Esq..(2),(3),(4),we have
T
T
  f ip
 f jp

 k ( u j k ( u j
 u i
)
 u i
)


 f
In matrix form f ip jp







 k k
 k k


 

 u u i j


 where or f ip f jp
 k ( u i
 k ( u j
 u j
)
 u i
)
Two linear Eqns. with two unknowns or
 

   
 

  ij
=element stiffness matrix (local)
=vector of nodal forces
=vector of nodal displacement
(4)

§
 
：
Internal forces
 
：
External forces
Element 1


 f
11 f
31







 k
1 k
1 local node j i
 k
1 k
1


 

 u u
3
1


 global node
1
3
Element 2
f
32 f
22




 k
2 k
2 local node j i
 k k
2
2


 u
3 u
2 global node
3
2

§
Augmented stiffness equations
(Expand the element matrix to the order of the total stiffness matrix)
Element 1




 f f
11
21 f
31











 k
0 k
1
0
0
0

0 k
1 k
1

 u
1


 u u
2
3










 k
  aug u
  aug
Element 2


 f f
12
22 f
32






0



0
0
0 k
2
 k
2

0 k
2 k
2







 u u u
1
2
3



 k
    aug aug u

§
Nodal force equilibrium equation
( 每個 node 的力需平衡 )
  ext 


F
F
2
F
1
3



  int  e n 

1 f e f f


 f
11 f
31 f
22
 f
32


Compatibility (continuity)equation

    aug aug

§
Assembly
Imposed equilibrium and compatibility on global level

  aug
  aug

  aug
  aug

(
    aug aug
)
 





 k
1
0 k
1
0
0
 k
2
0
 k
2 k
0

 k
1 k
2
1
 k
2





 u
 u u
1
2
3




   
： global nodal force matrix
： global stiffness matrix
： global displacement matrix

§
 Direct stiffness method
A reliable method directly assembles individual element stiffness matrix to form the global stiffness matrix and global system equations.

Direct assembly of DSM (short cut)
According to the
D.O.F.

§ Boundary conditions (BC)
(1)Homogeneous BS ：
No movement at locations
(2)Nonhomogeneous BS ： specified nonzero displacement
Consider the displacement at node1 being zero (Fixed End)
∴ BC at node1 is homogeneous ； ∴ u
1

0
Therefore, the system equation becomes

§ Boundary conditions (BC)
Because u
1
= 0, so column 1 has no effects on {F} ext . By directly deleting the row and column of above system equation corresponding to the zero-displacement d.o.f., the equation reduces to


 k k
2
2 k
1

 k
2 k
2

 u
2 u
3

F
F
2
3
Two Eqns for two unknowns
Solve above Eq. for displacements u
2 u
2 u
3



 k
2 k
2 k
1

 k
2 k
2



1
F
2
F
3




 k
1
1 k
1
1 k
2 and u
3
.
1 k
1
1 k
1







F
F
3
2


Reaction force F
1
F
1
 k
1 u
1

0 u
2
 k
1 u
3
  k
1 u
3
  k
1


F
2 k
1

F
3 k
1


 

F
2

F
3


§ Boundary conditions (BC)
Element force on each spring element:


 f f ip jp






 k k
 k k




 u u i j


 f
11 f
31



 k
1 k
1
 k
1 k
1

 u
1 u
3



 k
1 k
1
 k
1 k
1




1 k
1
( F
2
0

F
3
)




 f
32 f
22






 k k
2
2
 k k
2
2


 u

 u
3
2




§ Boundary conditions (BC)
Now, consider the case of nonhomogeneous boundary conditions. Assume u
1
= δ, where δ is a known displacement.
Thus system equation becomes
Rewrite above Eq. in expanded form.
k
1
0






0 k
2
   u
2


 u
2 u
2 k
2


 k
1 k
2

 u
3 u
3 u
2

( k
1

F
1

 k
F
2
2
)
 u
3

F
3
Considering the 2 nd and 3 rd terms of above eq and transforming the known δ terms to the right side, we have k
2 k
2
 u
2
 u
2

 k
2
( k
1
 u
3
 k
2

)

F
2 u
3

F
3
 k
1
 

(1) Define analysis problem : ID linear spring
(2) Constitutive equation
F=ku
(3) Form element stiffness matrix

 f
1

 f
2



 k
1 k
1
 k
1 k
1



 u u
3
1


(4) Form system equation



F
F
2
F
3
1








 k
1
0 k
1
 k
1 k
1

 k
2 k
2
 k
2 k
0
2



 


 u u
2 u
3
1




(5)Apply load/boundary conditions u
1

0 F
2

F F
3

F




 k
1
0 k
1 k
1

 k
1 k
2
 k
2
 k
0 k
2
2





 u u u
1
2
3





F
F
2
F
1
3


(6) Solve the system equation
(a) Obtain nodal displacements u
2 u
3


 k
1

 k
2 k
2
 k
2 k
2



1
F
2
F
3
(b) Obtain element forces

 f
11 f
21

 


 k
1 k
1
 k k
1


1

 u u
1
2




 f f
22
32






 k k
2
2
 k k
2
2


 u

 u
3
2




§
1. Total potential energy ， π p
2. Stationary value of a function G(x)
3. Principle of minimum potential energy
4. Derivation of element equation of a spring

1.
，
p
 p

U
 
Where
U = Internal strain energy
V
= Potential energy of external forces
Strain energy
The capacity of internal forces (or stresses ) to do work through deformation (strain) in the structure.
Potential energy
The capacity of body forces, surface traction forces and applied nodal forces to do work through deformation of the structure.
Total strain energy of the spring
U
  dU
  fdx
  kxdx

1 kx
2
2
Potential energy of external force of the spring
  p

U
  
1
2 kx
2 
Fx

)
To find a value of x yielding a stationary value of G(x), we have dG ( x )

0 dx
An analogous process will be used by replacing G(x) with total potential energy π p
， and x with nodal displacements d i
.
min(
 p
)
   p

  p
 d
1
 d
1

  p
 d
2
 d
2
  
  p
 d n

0

3. Principle of minimum potential energy
The principle states that equilibrium exits when the nodal displacement d i defines a structure state such that for arbitrary admissible variations δd i
  p

0 from the equilibrium state.
  p

  p
 d
1
 d
1

  p
 d
2
 d
2
  
  p
 d n

0
δd i might be nonzero. Hence, to satisfy associated with δd i
 π p
=0, all coefficients must be zero independently. Thus
 
 d i p 
0 i=1.2. ‥n where n equations must be solved for n values of the d i static equilibrium state of the structure.
that define the

4. Derivation of element equation of a spring

P

U
  
1
2
K ( d

2 x
 d

1 x
)
2  f

1 x d

1 x
 f

2 x
 d
2 x

0
Min (Πp)=0

 
 d

1 x p
 
 d

2 p x


1
2
1
2 k ( 2 d

2 x k ( 2 d

2 x

2 d

1 x
)


2
 d
1 x
)
 f

1 x f

2 x

0

0
In matrix form


 k k
 k k




 d d


1 x
2 x






 f

2
 f
1 x x


 or k k (
(

 d

2 d x
2 x


 d d
1 x

1 x
)
)

 f

2 x
 f
1 x

§

Definition

Element stiffness matrix of a rod under axial load

Transformation matrix and stiffness matrix in 2D

Stress for a bar (truss) element in
2D

§
An engineering structure of members connected by frictionless hinge or pins. The member is a two-force member and only support axial force along its length.
The loads are applied to joints of member.
Assumptions:
(1) No flexural loading transmits between members.
(2) Cross-sectional dimension and elastic properties are constant.
(3) Displacement of material point along a members is a linear function of position .

§
Consider a rod (truss) element under axial load as shown.
Here, the rod element is assumed to have constant cress-sectional area A, elastic constant E, and initial length L.
u
1
, u
2
= axial displacements at nodes1 and 2.
f
1
, f
2
= axial nodal forces at nodes 1 and 2.

§
Steps for derivation of the stiffness matrix for the rod element:
A. Force equilibrium
 f x
 f
1
 f
2

0 or f
2
  f
1
 ( 1 )
B. Strain/Displacement relation

Assume strain is constant in element

Assume that elongation of the rod due to axial load is △ L
 x


L

L u
1
 u
2
L
 ( 2 )
C. Stress/Strain law

Assume that a homogenous, isotropic material
 Hooke’s law
(Axial stress = P/A)

X
  
X
 ( 3 )

§
D. Relate force to stress
 x
1
  f
1
A
&
 x
2
 f
2
A
 ( 4 )
E. Relate force to displacement
Substitute Eqs. (2) & (3) into Eq (4) to yield f
1 f
2
   x
A
  x
A

    x
A
  x
A
 
 

( u
2
( u
2

L
 u
1
)
A
L u
1
)
A
Matrix form f
1  e f
2


L



1
1

1
1


 u
1
 u
2
 e or
  e 
  e
  e
Element stiffness matrix
Note: AE/L for a rod element is analogous to the spring constant k for a spring element.

§ Transformation matrix and stiffness matrix in 2D
X,Y ： global coordinates
： local coordinates
Θ ： positive, C.C.W. from x to
Assumptions ：
(1) No shear or bending stiffness,
 f
1 y
 f

2 y

0
(2) Any effect of transverse displacement is ignored

( d
1 y
, d

2 y
)

§ Transformation matrix and stiffness matrix in 2D
Rod element equation in the local coordinate system


 f

2
 f
1 x x





L



1
1

1
1





 d
1 d

2 x x


 or
 

   
In the global coordinate system XY


 f
1 x f
2 x



 f
3 x f
4 x



 


 d d d
1
2
3 x x x
 d
4 x






§ Transformation matrix and stiffness matrix in 2D
General vector transformation law in 2D
 x
 y





A x
 x
A x
 y

 cos
 sin


A  y x
A  y y


 sin x y
 cos


 x y
： projection of unit i into

§ Transformation matrix and stiffness matrix in 2D
Hence,the relation between the global displacement ﹛ d ﹜ and the local displacement is



 d
 d
1 x
2 x





 cos
0
 sin

0
0 cos

0 sin






 d d d
1 x
1
2 d
2 y x y



 or
 

   
Similarly
 

   
Expand and to the order of the global coordinates.
or


 d
1 x




 1 d
 d
2
2 y x y







 cos
 sin
0
0


 

    sin
 cos

0
0
0
0
 cos
 sin

0
0 sin
 cos







 d
1 x d
1

 d d
2
2 y x y





§ Transformation matrix and stiffness matrix in 2D
Similarly
 

   



 f
 1 x




 1 f f
 2
2 y x y



AE
L





1
0

1
0
0
0
0
0

1
0
1
0
0

0


0


0 




 d
 1 x
 1 d d
2
2 y x y





   
 

   

     

   

   

     
Premultiply both sides by
 

1
       

 

1
     

§ Transformation matrix and stiffness matrix in 2D

 

1 
 
T

(For orthogonal matrix)

 
T
     

   
Hence,the global stiffness matrix for an 2D rod element is
   
T
   

AE
L





 
 cos cos
 cos cos

2
2
 sin
 sin


 cos
 sin sin cos

2
 sin


 sin
2


 cos
2
 cos
 sin cos
2
 cos
 sin


 cos
 sin
 cos sin

2
 sin

 sin
2







Total stiffness matrix and force matrix are
 

N   e e


1
 

N   e e


1
System equation is
 

   

§
Local force/local displacement relation is



 f f
 1 x
2 x





L



1
1

1
1





 d
1 d

2 x x



 

 
 
  
E
L


1 1
 
 cos
0


E
L

 cos
  sin
 sin

0
0 cos
 cos
 sin
0






 d d d d
1 x
1
2
2 y x y



 sin





 d d d
1 d
1
2
2 x y x y



 or
 
E
L
(
 d
2 x
 d
1 x
) cos
 
( d
2 y
 d
1 y
) sin



§
The global stiffness matrix for an 3D rod element is

AE
L









C x
2
C x
C y
C y
2
C x
C z
C y
C z
C z
2
Symm.

C x
2

C x
C y

C x
C z
C x
2

C x
C y


C y
2
C y
C z
C x
C y
C
2 y

C x
C z


C y
C z
C z
2
C
C x y
C
C
C z
2 z z









Where C x
= cos
 x
, C y
= cos
 y and C z
= cos
 z
.