CHAP 2 BAR & TRUSS FINITE ELEMENT
Direct Stiffness Method
FINITE ELEMENT ANALYSIS AND DESIGN
Nam-Ho Kim
1
INTRODUCTION TO FINITE ELEMENT METHOD
• What is the finite element method (FEM)?
– A technique for obtaining approximate solutions of differential
equations.
– Partition of the domain into a set of simple shapes (element)
– Approximate the solution using piecewise polynomials within the
element
F
Structure
  xx  xy
 x  y  bx  0


  xy   yy  b  0
y
 x
y
u
Piecewise-Linear Approximation
Element
x
2
INTRODUCTION TO FEM cont.
• How to discretize the domain?
– Using simple shapes (element)
– All elements are connected using “nodes”.
5
6
1
1
7
2
2
8
3
3
Nodes
Elements
4
– Solution at Element 1 is described using the values at Nodes 1, 2, 6,
and 5 (Interpolation).
– Elements 1 and 2 share the solution at Nodes 2 and 6.
3
INTRODUCTION TO FEM cont.
• Methods
– Direct method: Easy to understand, limited to 1D problems
– Variational method
– Weighted residual method
• Objectives
– Determine displacements, forces, and supporting reactions
– Will consider only static problem
4
1-D SYSTEM OF SPRINGS
F2
F4
2
1
2
6
3
4
1
5
5
3
4
F3
• Bodies move only in horizontal direction
• External forces, F2, F3, and F4, are applied
• No need to discretize the system (it is already discretized!)
• Rigid body (including walls)
• Spring
NODE
ELEMENT
5
SPRING ELEMENT
• Element e
– Consist of Nodes i and j
e
u i , fi(e )
– Spring constant k(e)
i
– Force applied to the nodes: fi e ,fj e
u j , f j(e )
j
– Displacement ui and uj
– Elongation: (e)  u j  ui

– Force in the spring: P e  k  e  e  k  e u j  ui

– Relation b/w spring force and nodal forces: fj e   P e 
– Equilibrium: fi e  fj e  0
or
fi   f j
e
e
6
SPRING ELEMENT cont.
• Spring Element e
– Relation between nodal forces and displacements
u  u 
f    k    u  u 
fi   k 
e
e
j
e
i
 k e
 e
 k
j
e
i
– Matrix notation:
j
e
k    ui  fi(e) 


 e  u  f (e) 
k   j   j 
(e)
u




f

i
i
(e)
k      (e) 
u j  f j 
   
[k (e) ] q(e)  f (e)
k q  f
– k: stiffness matrix
– q: vector of DOFs
– f: vector of element forces
7
SPRING ELEMENT cont.
• Stiffness matrix
– It is square as it relates to the same number of forces as the
displacements.
– It is symmetric.
– It is singular, i.e., determinant is equal to zero and it cannot be
inverted.
– It is positive semi-definite
• Observation
– For given nodal displacements, nodal forces can be calculated by
   
[k (e) ] q(e)  f (e)
– For given nodal forces, nodal displacements cannot be determined
uniquely
8
SYSTEM OF SPRINGS
cont.
F
2
• Element equation
and assembly
F4
2
1
2
6
3
4
1
5
5
3
F3
4
 k1 k1   u1  f1(1) 
 k
 u    (1) 
k
1  2
 1
f2 
 k2
 k
 2
k 2  u2  f
 
k 2  u4  f
(2)
2
(2)
4



 k1 k1
 k
k1
 1
 0
0

0
 0
 0
0
k1
 k1
 k k  k
2
 1 1
 0
0

k 2
 0
 0
0
0 0 0   u1  f1(1) 
 
0 0 0  u2  f2(1) 
   
0 0 0  u3    0 

0 0 0  u4   0 
   
0 0 0  u5   0 
0
0
0 k 2
0
0
0
k2
0
0
0   u1   f1(1) 


0  u2  f2(1)  f2(2) 
  

0  u3    0 

0  u4   f4(2) 
  

0  u5   0 
9
SYSTEM OF SPRINGS cont.
 k3
 k
 3
k 3  u2  f
 
k 3  u3  f
(3)
2
(3)
3
 k4
 k
 4
k 4   u1  f
 

k 4  u3  f
 k5
 k
 5
k 5  u3  f
 

k 5  u4  f
(4)
1
(4)
3









k1  k 4
 k
1

 k 4

 0
 0
(5)
3
(5)
4
k1
 k1
 k k  k  k
2
3
 1 1
 0
k 3

k 2
 0
 0
0
k1  k 4
 k
1

 k 4

 0
 0
0
0
k 3
k 2
k3
0
0
k2
0
0
k1
k 4
0
k1  k 2  k 3
k 3
k 2
k 3
k3  k 4
0
k 2
0
k2
0
0
0
k1
k 4
0
k1  k 2  k 3
k 3
k 2
k 3
k3  k 4  k5
k 5
k 2
k 5
k2  k5
0
0
0
0   u1  

f1(1)


0  u2  f2(1)  f2(2)  f2(3) 
  

0  u3   
f3(3)



0  u4  
f4(2)
  

0  u5  
0

0   u1   f1(1)  f1(4) 


0  u2  f2(1)  f2(2)  f2(3) 
  

0  u3    f3(3)  f3(4) 


0  u4  
f4(2)
  


u
0   5  
0

0   u1   f1(1)  f1( 4) 


0  u2   f2(1)  f2(2 )  f2(3) 
  

0  u3   f3( 3 )  f3(4)  f3(5) 

0  u4   f4(2)  f4(5) 
  


u
0   5  
0

10
SYSTEM OF SPRINGS cont.
 k6
 k
 6
k 6  u4  f4(6) 
    (6) 

k 6  u5  f5 
k1  k 4
 k
1

 k 4

 0
 0
k1
k 4
0
k1  k 2  k 3
k 3
k 2
k 3
k3  k5  k 4
k 5
k 2
k 5
k2  k5  k6
0
0
k 6
F2
F4
2
1
2
0   u1   f1(1)  f1(4 ) 


0  u2   f2(1)  f2(2)  f2(3) 
   

0  u3   f3(3)  f3(4 )  f3(5) 

k 6  u4   f4( 2)  f4(5)  f4(6) 
  

( 6)

k 6  u5  
f5

6
3
4
1
5
5
3
4
F3
11
SYSTEM OF SPRINGS cont.
• Relation b/w element
forces and external force
• Force equilibrium
F2
F4
2
1
2
6
3
4
1
5
5
3
ie
Fi   fi
e
0
4
F3
e 1
ie
Fi   fi  , i  1,...ND
e
f3(5)
f3(3)
e 1
3
f3(4)
• At each node, the summation of
element forces is equal to
the applied, external force
F3
 f1(1)  f1(4)  R1 
 (1) (2) (3)   
 f2  f2  f2   F2 
(3)
(4)
(5)
f3  f3  f3    F3 
f (2)  f (5)  f (6)   F 
4
4
4
4



(6)
f5

 R5 
12
SYSTEM OF SPRINGS cont.
• Assembled System of Matrix Equation:
k1  k 4
 k
1

 k 4

 0
 0
k1
k 4
0
k1  k 2  k 3
k 3
k 2
k 3
k3  k5  k 4
k 5
k 2
k 5
k2  k5  k6
0
0
k 6
0   u1  R1 
0  u2   F2 
    
0  u3    F3 

k 6  u4   F4 
   
k 6  u5  R5 
[Ks ]{Qs }  {Fs }
• [Ks] is square, symmetric, singular and positive semi-definite.
• When displacement is known, force is unknown
u1  u5  0
R1 and R5 are unknown reaction forces
13
SYSTEM OF SPRINGS cont.
• Imposing Boundary Conditions
– Ignore the equations for which the RHS forces are unknown and strike
out the corresponding rows in [Ks].
– Eliminate the columns in [Ks] that multiply into zero values of
displacements of the boundary nodes.
k1  k 4
 k
1

 k 4

 0
 0
k1
k 4
0
k1  k 2  k 3
k 3
k 2
k 3
k3  k5  k 4
k 5
k 2
k 5
k2  k5  k6
0
0
k 6
0   u1  R1 
0  u2   F2 
    
0  u3    F3 

k 6  u4   F4 
   

k 6  u5  R5 
14
SYSTEM OF SPRINGS cont.
• Global Matrix Equation
k1  k 2  k 3

k 3


k 2
k 3
k 2
 u2  F2 
 u   F 
k 5
 3  3
k 2  k 5  k 6  u4  F4 
k3  k 4  k5
k 5
[K]{Q}  {F}
• Global Stiffness Matrix [K]
– square, symmetric and positive definite and hence non-singular
• Solution
{Q }  [K]1 {F}
• Once nodal displacements are obtained, spring forces can be
calculated from
P   k      k 
e
e
e
e
u  u 
j
i
15
UNIAXIAL BAR
• For general uniaxial bar, we need to divide the bar into a set
of elements and nodes
• Elements are connected by sharing a node
• Forces are applied at the nodes (distributed load must be
converted to the equivalent nodal forces)
• Assemble all elements in the same way with the system of
springs
p(x)
F
• Solve the matrix equation
x
for nodal displacements
Statically indeterminate
• Calculate stress and strain
p(x)
using nodal displacements
F
Statically determinate
16
1D BAR ELEMENT
L
• Two-force member
f1
• Only constant
cross-section
x
• Element force is
proportional to
(e ) Node i
f
K=EA/L
i
relative displ
• First node: i
ui
second code: j
• Force-displacement relation
A
f2
Node j
f j(e )
uj
(e)
fi(e)
 AE 

(ui  u j )

 L 
Similar to the spring element
(e)
 AE 
fj(e)  fi(e)  
(u j  ui )

 L 
17
1D BAR ELEMENT cont.
fi(e ) Node i
– Matrix notation
f

f
(e)
i
(e)
j
  AE 


L

 
(e)
K=EA/L
ui
 1 1 ui 
 1 1  u 

 j
Node j
f j(e )
uj
{f (e) }  [k (e) ]{q(e) }
– Either force or displacement (not both) must be given at each node.
– Example: ui = 0 and fj = 100 N.
– What happens when fi and fj are given?
• Nodal equilibrium
– Equilibrium of forces acting on Node I
Fi  fi(e)  fi(e1)  0
fi(e)  fi(e1)  Fi
– In general
ie
Fi   f
e 1
(e)
i
Fi
Element e
fi(e)
fi(e+1)
Node i
Element e+1
18
1D BAR ELEMENT cont.
• Assembly
– Similar process as spring elements
– Replace all internal nodal forces with External Applied Nodal Force
– Obtain system of equations
[Ks]: Structural stiffness matrix
[Ks ]{Qs }  {Fs }
{Qs} Vector of nodal DOFs
{Fs}: Vector of applied forces
• Property of [Ks]
– Square, symmetric, positive semi-definite, singular, non-negative
diagonal terms
• Applying boundary conditions
– Remove rigid-body motion be fixing DOFs
– Striking-the-nodes and striking-the-columns (Refer to sprint elements)
[K]{Q}  {F}
[K]: Global stiffness matrix
{Q} Vector of unknown nodal DOFs
{F}: Vector of known applied forces
19
1D BAR ELEMENT cont.
• Applying boundary conditions cont.
– [K] is square, symmetric, positive definite, non-singular, invertible, and
positive diagonal terms
– Can obtain unique {Q}
• Element forces
– After solving nodal displacements, the element force can be calculated
P
(e)
 AE 


 L 
(e)
u
– Element stress
• Reaction Forces
j
 ui   f
P(e)
  (e)
A
(e)
j
(e)
(e)

 Pi 
  AE   1 1 ui 
 (e)   
 
 u 

P
L

1
1


 j 


 j

Note Pi = Pj
– Use [Ks]{Qs} = {Fs}: the rows that have been deleted (strike-the-rows)
– Or, use
ie
Fi   fi(e)
e 1
20
EXAMPLE
• 3 elements and 4 nodes
• At node 2:
F2  f2(1)  f2(2)  f2(3)
K2
K1
Element 2
Element 1
K3
Element 3
F1
x
• Equation for each element:
f1(1)   K1 K1   u1 
 (1)   
 u 

K
K
1  2
f2   1
f

f
(2)
2
(2)
3
f

f
(3)
2
(3)
4
  K2

  K 2
  K3

  K 3
K 2  u2 
 
K 2  u3 
K 3  u2 
 
K 3  u4 
K2
Element 2 N3
F1
N1
K1
Element 1
u1
N2
F3
u3
u2
N4
Element 3
K3
F4
u4
21
EXAMPLE cont.
• How can we combine different element equations? (Assembly)
– First, prepare global matrix equation:
0  0
0  0
  
 
0  0
0  0
0 0 0  0 
0 0 0  0 
 
0 0 0  0 

0 0 0  0 
Displacement vector
Stiffness matrix
Applied force vector
– Write the equation of element 1 in the corresponding location
f1(1)   K1 K1
 (1)  
f2   K1 K1
 
0
0  0
 0   0
0
0 0   u1 
0 0  u2 
 
0 0  u3 

0 0  u4 
K2
Element 2 N3
F1
N1
K1
Element 1
u1
N2
F3
u3
u2
N4
Element 3
K3
F4
u4
22
EXAMPLE cont.
– Write the equation of element 2:
 0  0 0
f (2)  0 K
2  
2
 (2)  
f3  0 K 2
 0  0 0
0
K 2
K2
0
K2
Element 2 N3
0   u1 
0  u2 
 
0  u3 

0  u4 
F1
N1
K1
Element 1
u1
N2
F3
u3
u2
N4
Element 3
K3
F4
u4
– Combine two equations of elements 1 and 2
 f1(1)   K1
K1
 (1) (2)  
f2  f2   K1 K1  K 2
 (2)  
K 2
 f3
  0
 0   0
0
0
K 2
K2
0
0   u1 
0  u2 
 
0  u3 

0  u4 
23
EXAMPLE cont.
– Write the equation of element 3
 0  0 0
f (3)  0 K
2  
3
 
 0  0 0
f4(3)  0 K 3
0   u1 
0 K 3  u2 
 
0 0  u3 

0 K 3  u4 
0
K2
Element 2 N3
F1
N1
K1
Element 1
u1
N2
F3
u3
u2
N4
Element 3
K3
F4
u4
– Combine with other two elements
K1
0
  K1
F1  
f1(1)
F   (1) (2) (3)   K (K  K  K ) K
 2  f2  f2  f2   1
1
2
3
2

 

(2)
K 2
K2
F
f
3
3
  
  0
F4  
  0
K 3
0
f4(3)
0   u1 
K 3  u2 
 
0  u3 

K 3  u4 
Structural Stiffness Matrix
24
EXAMPLE cont.
• Substitute boundary conditions and solve for the unknown
displacements.
– Let K1 = 50 N/cm, K2 = 30 N/cm, K3 = 70 N/cm and f1 = 40 N.
50
0
0   u1 
F1   50
F   50 (50  30  70) 30 70  u 
 2 
  2 
 
30
30
0  u3 
F3   0

F4   0
70
0
70  u4 
– Knowns: F1, F2, u3, and u4
– Unknowns: F3, F4, u1, and u2
50
0
0   u1 
40   50
 0   50 (50  30  70) 30 70  u 
  
  2 
 
30
30
0 0 
 F3   0

 F4   0
70
0
70   0 
25
EXAMPLE cont.
– Remove zero-displacement columns: u3 and u4.
40   50
 0   50
  
 
 F3   0
 F4   0
50 
150   u1 
 
30  u2 

70 
– Remove unknown force rows: F3 and F4.
40   50 50   u1 
 
 u 
0

50
150
  
 2
– Now, the matrix should not be singular.
Solve for u1 and u2.
– Using u1 and u2, Solve for F3 and F4.
u1  1.2 cm
u2  0.4 cm
F3  0u1  30u2  12 N
F4  0u1  70u2  28 N
26
EXAMPLE cont.
• Recover element data
f1(1)   K1 K1  u1   50 50  1.2   40 
 (1)   
 
 




K
K
u

50
50
0.4

40
f

  

1  2
2   1
f2(2)   K 2
 (2)   
f3   K 2
K 2  u2   30 30  0.4   12 
 
 

K 2  u3   30 30  0.0  12
f2(3)   K 3
 (3)   
f4   K 3
K 3  u2   70 70  0.4   28 
 
 



K 3  u4   70 70  0.0  28 
K2
F1 = 40 N
K1
Element
force
-12 N
0.4 cm
-28 N
1.2 cm
K3
27
EXAMPLE
•
•
•
•
•
Statically indeterminate bars
E = 100 GPa
RL
F = 10,000 N
A1 = 10−4 m2, A2 = 2×10−4 m2
Element stiffness matrices:
1011  10 4
[k ] 
0.25
(1)
• Assembly
B
C
F
0.25 m
RR
0.4 m
4  u1
 1 1
7  4
 1 1   10  4 4  u



 2
1011  2  10 4
[k ] 
0.4
(2)
A
5  u2
 1 1
7  5

10
 1 1 
 5 5  u



 3
 4 4 0   u1   F1 
  

107  4 9 5  u2   10,000 


 0 5 5  u3   F3 
28
EXAMPLE cont.
• Applying BC
107 9u2   10,000  u2  1.11 10 4 m
• Element forces or Element stresses
P
AE
u j  ui 

L


P(1)  4  107 u 2  u1  4,444N
P(2)  5  107  u3  u2   5,556N
• Reaction forces
RL  P(1)  4,444N
RR  P(2)  5,556N
29
PLANE TRUSS ELEMENT
• What is the difference between 1D and 2D finite elements?
– 2D element can move x- and y-direction (2 DOFs per node).
– However, the stiffness can be applied only axial direction.
• Local Coordinate System
– 1D FE formulation can be used
if a body-fixed local coordinate system
is constructed along the length of
the element
Y
– The global coordinate system
(X and Y axes) is chosen to
X
represent the entire structure
– The local coordinate system (x and y axes)
is selected to align the x-axis along the length
of the element
50 N
8 cm
12 cm
 f1x  EA  1 1  u1 
 
 1 1  u 
f
L

 2
 2x 
30
PLANE TRUSS ELEMENT cont.
• Element Equation (Local Coordinate System)
– Axial direction is the local x-axis.
– 2D element equation
 f1x 
1
f 

 1y  EA  0
 
f2x  L  1

f2y 
0
 
y
Local coordinates
0 1 0   u1 
0 0 0   v1 
 
0 1 0   u2 

0 0 0   v 2 
y
v1
u1
1
f1x
x
f
v 2 u2
f2x
2
Global coordinates
{ f }  [k ]{q}
x
– [k ] is square, symmetric, positive semi-definite, and non-negative
diagonal components.
• How to connect to the neighboring elements?
– Cannot connect to other elements because LCS is different
– Use coordinate transformation
31
COORDINATE TRANSFORMATION
• Transform to the global coord. and assemble
u1   cos f sin f  u1 
 
 v 
v

sin
f
cos
f
  1
 1 
u2   cos f sin f  u2 
 
 v 
v

sin
f
cos
f
 2
 2 
v
v
u
f
f
u
• Transformation matrix
u1   cos f sin f
0
0  u1 
  
 

sin
f
cos
f
0
0
v
 1 
 v1 


 
 


0
0
cos
f
sin
f
u2 
u2 

 v 2   0
0
 sin f cos f  v 2 
local
global
{q}  [T]{q}
Transformation matrix
32
COORDINATE TRANSFORMATION cont.
• The same transformation for force vector
 f1x   cos f sin f
0
0   f1x 
f  
 f 

sin
f
cos
f
0
0
 1y  
 1y 


 
 


f
0
0
cos
f
sin
f
 2x 
f2x 
 
f2y   0
0

sin
f
cos
f
 f2y 
 
local
{ f }  [T ]{f }
global
• Property of transformation matrix
[T ]1  [T ]T
{ f }  [T ]{f }
{ f }  [ T ]T { f }
33
ELEMENT STIFFNESS IN GLOBAL COORD.
• Element 1
y
 f1x 
1
f 

 1y  EA  0
 
L  1
f2x 

f2y 
0
 
0 1 0 
0 0 0

0 1 0

0 0 0
element stiffness matrix
v2
u1 
 
 v1 
 
u2 
 v 2 
v1
u1
K
f
u2
f2x
N2
N1
f1x
x
{ f }  [k ]{q}
• Transform to the global coordinates
[T ]{f }  [k ][T]{q}
{f }  [T ]1[k ][T ] {q}
global
[k ]  [T ]1[k ][T ]
global
{f }  [k]{q}
34
ELEMENT STIFFNESS IN GLOBAL COORD. cont.
• Element stiffness matrix in global coordinates
[k ]   T  [k ]  T 
T
 cos2 f
cos f sin f
 cos2 f
 cos f sin f 


sin2 f
 cos f sin f
 sin2 f 
EA  cos f sin f
[ k] 
L   cos2 f
 cos f sin f
cos2 f
cos f sin f 


2
2

cos
f
sin
f

sin
f
cos
f
sin
f
sin
f


– Depends on Young’s modulus (E), cross-sectional area (A), length (L),
and angle of rotation (f)
– Axial rigidity = EA
– Square, symmetric, positive semi-definite, singular, and non-negative
diagonal terms
35
N2
EXAMPLE
50 N
• Two-bar truss
Element 1
– Diameter = 0.25 cm
– E = 30106 N/cm2
8 cm
Element 2
• Element 1
– In local coordinate
N1
N3
12 cm
{ f (1) }  [k (1) ]{q(1) }
 f1x 
1
f 

 1y  EA  0
 
f2x  L  1

f2y 
0
 
0 1 0  u1 
 
0 0 0   v1 
 
0 1 0  u2 

0 0 0   v 2 
y
v2
v1
f1x
u1 K
N1
u2
f2x
N2
f1 f = 33.7o
1
E = 30 x 106 N/cm2
A = pr2 = 0.049 cm2
L = 14.4 cm x
36
EXAMPLE cont.
• Element 1 cont.
– Element equation in the global coordinates
f1x(1) 
 0.692 0.462 0.692 0.462  u1 
 (1) 
 0.462 0.308 0.462 0.308   
f1y 
 v1  {f (1) }  [k (1) ]{q(1) }



102150
 (1) 
 



0.692

0.462
0.692
0.462
f
 2x 
u 2 

 
(1) 
f2y

0.462

0.308
0.462
0.308

 v 2 
 
f2x
y
• Element 2
(2)
f2x

0 0
 (2) 
0 1
f2y 
 (2)   184125 
0 0
f3x 

(2) 
f3y
0 1
 
N2
0  u 2 
 
0 1 v 2 
 
0 0  u3 

0 1 
v 3 

0
f2 = –90o
E = 30 x 106 N/cm2
A = pr2 = 0.049 cm2 K
L = 8 cm
f2
v2
u2
x
N3
v3
f3x
u3
37
EXAMPLE cont.
• Assembly
– After transforming to the global coordinates
F1x   70687
47193 70687 47193
F  
 1y   47193 31462 47193 31462
F2x   70687 47193 70687
47193
 
F2y   47193 31462 47193 215587
F   0
0
0
0
3x
  
0
0
184125
F3y   0
Element 1
0
0
0
0
0
0
 u1 
 v 
0
 1
 u2 
0
 
184125  v 2 
 u3 
0
 
184125  v 3 
0
Element 2
• Boundary Conditions
– Nodes 1 and 3 are fixed.
– Node 2 has known applied forces: F2x = 50 N, F2y = 0 N
38
EXAMPLE cont.
• Boundary conditions (striking-the-columns)
F1x   70687
47193
F  
 1y   47193 31462
 50   70687 47193
 
 0   47193 31462
F3x   0
0
  
0
F3y   0
70687
47193 0
47193
31462 0
70687
47193 0
47193
215587 0
0
0
0
0
184125 0
 0 
 
0
 0 
 u 2 
0
 
184125  v 2 
 0 
0
 
184125  0 
0
– Striking-the-rows
50  70687 47193  u2 
 
 v 
0
47193
215587
  
 2
• Solve the global matrix equation
u2  8.28  104 cm
v 2  1.81 104 cm
39
EXAMPLE cont.
• Support reactions
F1x   70687 47193 
 50 
F  



4
 1y   47193 31462   8.28  10  33.39 

 

N
4 
 1.81 10   0 
0
F3x   0

F3y   0
 33.39 

184125

 
– The reaction force is parallel to the element length (two-force member)
• Element force and stress (Element 1)
– Need to transform to the element local coordinates
0
0  0  
0
 u1   .832 .555

 v   .555 .832

0
0   0  
0
 1 

   
 
4 
u
u


0
0
.832
.555
5.89

10
 2
 2 



4
 v 2   0
0
.555 .832  v 2  6.11 10 
40
EXAMPLE cont.
• Element force and stress (Element 1) cont.
– Element force can only be calculated using local element equation
 f1x 
1
f 

 1y  EA  0
 
f2x  L  1

f2y 
0
 
–
–
–
–
0 1 0  
0
 60.2
  0 
0 0 0  
0






N
4 
0 1 0   5.89  10   60.2 

0 0 0  6.11 10 4   0 
There is no force components in the local y-direction
In x-direction, two forces are equal and opposite
The force in the second node is equal to the element force
Normal stress = 60.2 / 0.049 = 1228 N/cm2.
–60.2 N
60.2 N
1
2
41
OTHER WAY OF ELEMENT FORCE CALCULATION
• Element force for plane truss
P(e)
 AE 


 L 
(e)
 AE 


 L 
(e)
(e)
u  u 
j
i
– Write in terms of global displacements
P(e)
 AE 


 L 
(e)
 AE 


L


(e)
lu  mv   lu mv 
j

j
i
i
l  u j  ui   m  v j  v i 

l  cos f
m  sin f
42
EXAMPLE
2
• Directly assembling global matrix equation
(applying BC in the element level)
• Element property & direction cosine table
Elem AE/L i -> j
1 206×105 1 -> 3
2 206×105 1 -> 2
3 206×105 1 -> 4
f
-30
90
210
l = cosf
0.866
0
0.866
m = sinf
0.5
1
0.5
F
2
45
3
1
1
3
4
• Since u3 and v3 will be deleted after assembly, it is not
necessary to keep them
u1
v1
 l2
lm
(1) 
lm m2
EA


(1)

 k   

2
lm
 L   l

2
 lm m
u3
v3
lm  u1

lm m2  v1
l2
lm  u3

lm m2  v 3
l2
u1
v1
2
lm  u1
 EA   l
(1)
 k   
 
2
 L  lm m  v1
(1)
43
SPACE TRUSS ELEMENT
• A similar extension using coordinate transformation
v2 u
2
Y
• 3DOF per node
w2
– u, v, and w
– fx, fy, and fz
v1 u
1
f1x
K
fX
N2
fZ
w1
• Element stiffness
matrix is 6x6
fY
f2x
N1
X
Z
• FE equation in the local coord.
 fix  AE  1 1  ui 
 
 1 1  u 
f
L

 j
 jx 
{ f }  [k ]{q}
44
SPACE TRUSS ELEMENT cont.
• Relation between local and global displacements
– Each node has 3 DOFs (ui, vi, wi)
 ui 
v 
 i
 ui   l m n 0 0 0   w i 
 
  uj 
u
0
0
0
l
m
n
 
 j 
 vj 
 
 w j 
– Direction cosines
l  cos fx 
L
x
x j  xi
L
y j  yi
 x i    y j  y i    z j  zi 
2
j
, m  cos f y 
2
L
{q}
 [T ]  {q}
(2  1) (2  6) (6  1)
, n  cos fz 
z j  zi
L
2
45
SPACE TRUSS ELEMENT cont.
• Relation between local and global force vectors
 fix   l 0 
f  

 iy  m 0 
 fiz   n 0   f 
 
ix
f   
 f 
 jx   0 l   jx 
f jy   0 m 
  

0
n


 f jz 
 
{ f }  [ T ]T { f }
• Stiffness matrix
{ f }  [k ]{q}
l2


EA 
[k ] 

L 



[T]T { f }  [T]T [k ][T]{q}
u
 ln  i

m2 mn lm m2 mn  v i
n2  ln mn n2  w i

l2
lm
ln  u j
sym
m2
mn  v j

n2  w
lm
ln
l2
{f }  [k ]{q}
lm
j
[k ]  [T ]T [k ][T ]
46
THERMAL STRESSES
• Temperature change causes thermal strain
L
L
L
No stress, no strain
No stress, thermal strain
(a) at T = Tref
Thermal stress, no strain
(b) at T = Tref + T
• Constraints cause thermal stresses
• Thermo-elastic stress-strain relationship
 = E    T 
=

+ T
E
Thermal expansion coefficient
47
THERMAL STRESSES cont.
• Force-displacement relation
L
 L

P = AE 
 T   AE
 AET
L
 L

• Finite element equation
u
Thermal force vector 1 i
 0 v
(e)
(e)
(e)
(e)
  i
(e)
{ f }  [k ]{q }  { fT }
{ fT }  AET  
u
1 j

0
vj
• For plane truss, transform to the global coord.
{f }  [k]{q}  {fT }
[k]{q}  {f }  {fT }
 l  ui
 m  v
  i
{fT }  AET  
u
 l  j
m v j
[Ks ]{Qs }  {Fs }  {FTs }
48
EXAMPLE
3
z
• Space truss
Node
1
2
3
4
Elem
x
0
0
0
1
EA/L
y
0
–1
1
0
i -> j
z
0
1
1
1
E2
1/ 2
0
1/ 2
0
0
35 2  105 2 -> 4
1/ 2
1/ 2
3
35 2  105 3 -> 4
1/ 2
1/ 2
4
1
n
2
y
E1
m
35 2  105 1 -> 4
l2


EA 
[ k] 

L 



2
l
1
E3
10,000N
x
u
ln  i

m2 mn lm m2 mn  v i
n2 ln mn n2  w i

l2
lm
ln  u j
sym
m2
mn  v j

n2  w
lm
ln
l2
lm
j
49
Project 1
• Fully-stressed design of a ten-bar truss
– Most efficient usage of material = all members are at yield stress
– Modify cross-sectional areas (individually) to minimize weight
e
A new 
old
e
 
A old
e

allowable
e
b
Parameters
Dimension, b
Safety factor, SF
Load, P1
Load, P2
Density, 
Young’s modulus, E
Allowable stress,
Initial area
Min. CS area
Values
360 inches
1.5
66.67 kips
66.67 kips
0.1 lb/in3
104 ksi
25 ksi*
1.0 in2
0.1 in2
*for Element 9, allowable stress is 75 ksi
5
b
3
1
8
1
0
9
7
3
6
6
5
y
x
1
2
b
4
4
2
P1
P2
50
PROJECT 1 cont.
• Schedule: Project due: 2/23 (Wed)
– Submission: Submit report (Word or PDF file, Max 10 pages), CAE file
and/or programs by 1:55 PM in Sakai (penalty for late submission)
• Report
– Formal report, including title, summary, introduction, approach, results,
discussion, appendix (input file, programs), and/or references
• What to include in the report
– Use the same definition of nodes and elements in the problem
definition. (You need to interpret your Abaqus results accordingly)
– Plot of overlapped undeformed and deformed geometry with stress
contour
– FE analysis results (tables) for initial design (nodal displacements and
element stresses, S11)
– Your methodology for design updates (algorithms or flowchart, etc)
– Fully-stressed design (cross-sectional areas and stress of each
element, use tables)
51
– Discussions (what you learned through the project)
PROJECT 1
• Analysis and Design of Plane Truss
– E = 100 GPA, A = 1.0 cm2, L = 0.3 m or 0.3 2 m
2
y
7
4
13
8
14
20
6
8
9
15
21
10
16
22
10
11
17
12
12
18
23
14
19
24
25
x
1
1
3
2
5
3
7
4
9
5
11
6
13
– Apply axial loading, transverse shear, and bending loads
– Calculate equivalent axial rigidity, flexural rigidity, and shear rigidity
– Use average deflection of Nodes 13 and 14
FxL
utip 
EA eq
v tip 
FyL3
3 EIeq

FyL
 GA eq
v tip
CL2

2 EIeq
– Verify the equivalent properties by adding two more bays
52