n08_fea
1 of 10
Finite Element Analysis – One Dimensional Example
straight cantilever beam with constant thickness in pure tension
w = 1 in
L = 10 in
t = 0.5 in
aluminum
E = 10.4e6 psi
P = 1000 lbf


1000 lbf 10 in  
PL
PL
in 2
 = 0.001923 in = 1.923 mil


6

0.5 in 1 in   10.4x10 lbf 
AE twE
Pk

k
twE
L
1000 lbf 
P
P


= 2.0 ksi
A t w 0.5 in 1 in 
n08_fea
2 of 10
tapered cantilever beam with constant thickness in pure tension
L = 10 in
b = 2 in
x
t = 0.5 in
a = 1 in
aluminum
E = 10.4e6 psi
P = 1000 lbf
w b

b  a  x  b  m x
L
P
P

E
A tw
L
L
0
0
    dx  


m
b  a 
L
0xL
P
Etw
 P 
P

 dx 
Et
E t w
L

1000 lbf 10 in  
in 2
0.5 in 2 in  1 in   10.4x10 6 lbf
L
0
 1 
PL
P  1

b

 dx 
ln  
  ln b  m x   
Et m
 0 E t b  a   a 
bm x
  2 in
 ln 
  1 in

 = 1.333 mil

(NOT 1.923 / 1.5 = 1.282 mil)
n08_fea
3 of 10
five elements – six nodes
width 2.0
node
length
element
width
1
1.8
1.6
1.4
1.2
1.0
2
3
4
5
6
2.0
2.0
2.0
2.0
2.0
1
2
3
4
5
1.9
1.7
1.5
1.3
1.1
n08_fea
4 of 10
five elements – six nodes
u1
u2
u3
u4
u5
u6
F1
F2
F3
F4
F5
F6
element1
element2
element3
element4
element5
ui = displacement of node i
Fi = external force applied at node i
li = length of element i
Ai = cross sectional area of element i
ti = thickness of element i
wi = width of element i
Ei = modulus for element i
ki = stiffness of element i
fi = internal force on element i
ui
ui+1
NO LOAD
u
WITH LOAD
u 
f i li
fi li

Ai Ei t i w i Ei
f i  k i u i 1  u i 
fi
fi 
fi
t i wi Ei
u   k u   k i u i 1  u i 
li
ki 
t i wi Ei
li
n08_fea
5 of 10
nodei
Fi
fi-1
fi
elementi-1
elementi

F on node 1
F1  f1  FWALL  0
F on node 2
F2  f 2  f1  0
F on node 3
F3  f 3  f 2  0
F on node 4
F4  f 4  f 3  0
F on node 5
F5  f 5  f 4  0
F5  k 4 u 4  k 4  k 5 u 5  k 5 u 6
F on node 6
F6  f 5  0
F6  k 5 u 5  k 5 u 6
 k 1
 k
 1
 0

 0
 0

 0
F1  k1u1  k1u 2  FWALL
F2  k1u1  k1  k 2 u 2  k 2 u 3
F3  k 2 u 2  k 2  k 3 u 3  k 3 u 4
F4  k 3 u 3  k 3  k 4 u 4  k 4 u 5
 k1
0
0
0
k1  k 2
 k2
0
0
 k2
k2  k3
 k3
0
0
 k3
k3  k4
 k4
0
0
 k4
k4  k5
0
0
0
 k5
0   u 1  F1  FWALL 
 

0  
F2
u 2  

u 3 
 


0 
F3

   

0  u 4  
F4






 k5 u5
F5
  


 k 5  
F6
u 6 
 


impose restraints u1 = 0 and F1 = 0 in first row of equations
 1
 k
 1
 0

 0
 0

 0
0
0
0
0
k1  k 2
 k2
0
0
 k2
k2  k3
 k3
0
0
 k3
k3  k4
 k4
0
0
 k4
k4  k5
0
0
0
 k5
0 u1   0 
  
0  
u 2  F2 
0 
u 3 
 
F3 

    
0  u 4  F4 
 k 5  u 5  F5 
   
 k 5  
u 6 
 
F6 

[K] = system stiffness matrix
{u} = vector of nodal displacements
{F} = vector of external forces on nodes
displacement analysis - solve for u  K 
1
F
Ku  F
n08_fea
6 of 10
stress analysis
 i = strain in element i
 i = stress in element i
i 
u i 1 u i
i  Ei i 
li
  1  E 1 / l 1
   0
 2  
 3    0
   0
 4 
 5   0
0
0
0
E2 / l2
0
0
0
E3 / l3
0
0
0
E4 / l4
0
0
0
Ei
li
u i  1
ui

u 2  u 1 
0  u 3  u 2 


0  u 4  u 3 

0  u 5  u 4 


E 5 / l 5  u 6  u 5 
0
{} = vector of stresses on each element
[C] = elasticity matrix
{u} = vector of differences in nodal displacements
  Cu
n08_fea
7 of 10
SUMMARY FOR THREE-DIMENSIONAL ANALYSIS
1) Know material properties – E, G,
 (Poisson’s ratio)
2) Generate mesh – create elements and nodes
reduce mesh size by factor of 2, increases number of nodes by factor of 8
3) Compute stiffness matrix – order n = number of degrees of freedom (DOF)
n = 3 * number of nodes
4) Define restraints – modify stiffness matrix to force specific nodal displacements to zero
5) Define loading – form right-hand-side (RHS) vector
6) Invert stiffness matrix and pre-multiply loading RHS vector to solve for nodal displacements
computation time proportional to n2
7) Compute nodal deformations RHS vector (differences of nodal displacements)
8) Compute elasticity matrix
9) Pre-multiply elasticity matrix times deformation RHS vector to solve for stresses at nodes
find von Mises stress, maximum shear stress, normal stress, Dowling stress
10) Know strength of material – SY, SUT, SUC
11) Compute factor of safety at each node
n08_fea
8 of 10
% fea_1d.m - one dimensional finite element analysis
% HJSIII, 12.09.05
%
P
L
t
E
axial load,
= 1000;
= 10.0;
= 0.5;
= 10.4e6;
geometry, material
% axial load [lbf]
% length [inch]
% thickness [inch]
% Young's modulus for aluminum [psi]
% analytical solution - constant width
w = 1.0;
% width [inch]
delta_const_w_mil = P *L /w /t /E *1000;
% [mils]
% analytical solution - tapered
a = 1.0;
% width at tip [inch]
b = 2.0;
% width at root [inch]
delta_taper_mil = P *L /t /E /(b-a) *log(b/a) *1000;
% element stiffness - tapered
n = 5;
li = L/n * ones(n,1);
ti = t * ones(n,1);
Ei = E * ones(n,1);
wi = [ 1.9 1.7 1.5 1.3 1.1 ]';
%
%
%
%
%
% [mils]
number of finite elements
length [inch]
thickness [inch]
Young's modulus for aluminum [psi]
width [inch]
k = wi .*ti .*Ei ./li;
% global stiffness
K = [ +1
0
-k(1) +k(1)+k(2)
0
-k(2)
0
0
0
0
0
0
% external forces
F = [ 0 0 0 0 0
0
-k(2)
+k(2)+k(3)
-k(3)
0
0
0
0
-k(3)
+k(3)+k(4)
-k(4)
0
P ]';
% displacements
u = inv(K) * F;
delta_fea_mil = u(6) *1000;
delta_const_w_mil
delta_taper_mil
delta_fea_mil
% stress
C = diag( Ei ./li );
strain = diff( u );
stress = C * strain;
stress_ksi = stress /1000
% [mils]
0
0
0
-k(4)
+k(4)+k(5)
-k(5)
0
0
0
0
-k(5)
+k(5)
;
;
;
;
;
];
n08_fea
>> fea_1d
delta_const_w_mil =
1.9231
delta_taper_mil =
1.3330
delta_fea_mil =
1.3306
stress_ksi =
1.0526
1.1765
1.3333
1.5385
1.8182
9 of 10
n08_fea
FEA sensitivity study for
10 of 10
tee_01.sldprt
mesh control
size parameter
[inch]
max von Mises
stress
[ksi]
DOF
relative
computations
0.086
0.06
0.04
0.02
0.01
45.59
50.80
51.55
50.38
50.42
35496
61335
87162
205329
762018
1
2.99
6.03
33.46
460.86