Chapter 6. Plane Stress /
Plane Strain Problems
Element types:
Line elements (spring, truss, beam, frame) –
chapters 2-5
2-D solid elements – chapters 6-10
3-D solid elements – chapter 11
Plate / shell elements – chapter 12
1
2-D Elements
Triangular elements – plane stress/plane strain:
CST – “constant strain triangle” – chap. 6
LST – “linear strain triangle” – chap. 8
Axisymmetric elements – chap. 10
Isoparametric elements – chap. 11
4-node quadrilateral element (linear interpolation)
8-node quadrilateral element (quadratic interpolation)
2
Plane stress
x  0
y 0
 xy  0
 z   xz   yz  0
3
Plane Strain
 z   xz   yz  0
x  0
y 0
 xy  0
z  0
 xz   yz  0
4
2-D Stress States
Matrix form:
 x 
 
 x    y 
 
 xy 
5
Principal Stresses
1,  2 
 x  y
2
  x  y
 
2

 2
1

xy
 p  tan 1 
2
  x  y

2

  2
xy







6
Displacements and Strains
Displacement field
u ( x, y )
u  

v
(
x
,
y
)


Strains
x 
u
v
u v
,y 
,  xy 

x
y
y x
x 
     y 
 
 xy 
7
Stress-Strain Relations
   D 
Recall:
E – Young’s modulus
 - Poisson’s Ratio
G – Shear modulus
8
Stress-Strain Relations (cont.)
Plane stress
 x 
E
 
 y  
2
  1 
 xy 

 
1

0

  x 
 1
0   y 

1    
0
0

  xy
2  

Plane strain


 x 
1 
0   x 

E
 
 
 1
0   y 
 y  
  1  1  2  0 0 1  2   

 xy
 xy 
2  

Note, in both cases
D33 
E
G
21  
9
Derivation of “Constant Strain Triangle”
(CST) Element Equations
Step 1 – Select element type
 ui 
v 
 i
 u j 
d    
v j 
u m 
 
vm 
Note – x-y are global coordinates (will not need to transform
from local to global
10
Displacement Interpolation
Assume “bi-linear” interpolation – guarantees that
edges remain straight => inter-element compatibility
u ( x, y )  a1  a2 x  a3 y 



v( x, y )  a4  a5 x  a6 y 
11
Displacement Interpolation (cont.)
As before, rewrite displacement interpolation in terms of nodal
displacements (see text for details)
u ( x , y )  N i ( x , y ) u i  N j ( x , y ) u j  N m ( x, y ) u m
v( x, y )  N i ( x, y ) vi  N j ( x, y ) v j  N m ( x, y ) vm
where
1
 i   i x   i y 
2A
1
Nj 
 j   jx  j y
2A
1
 m   m x   m y 
Nm 
2A
Ni 


12
Displacement Interpolation (cont.)
and
 i  x j ym  y j ym
 i  y j  ym
 i  xm  x j
1
xi
1
A  1 xj
2
1 xm
yi
yj
ym
13
Displacement Interpolation (cont.)
u   N i
      
v   0
0
Nj
0
Nm
Ni
0
Nj
0
 ui 
v 
 i
0   u j 
   N d 

N m  v j 
u m 
 
vm 
14
Displacement Interpolation (cont.)
1
 i   i x   i y 
2A
1
Nj 
 j   jx  j y
2A
1
 m   m x   m y 
Nm 
2A
Ni 


Graphically:
15
Step 3 – Strain-Displacement and StressStrain Relations
x 
u
v
u v
,y 
,  xy 

x
y
y x
x 
     y 
 
 xy 
From which it can be shown
 i
   1  0
2A 
 i
0
j
0
m
i
i
0
j
j
0
j
m
 ui 
v 
0  i 
  u j 
 m     B d 
vj 


m 
u m 
 
vm 
16
Strain-Displacement Relations (cont.)
• Note – the strain within each element is
constant (does not vary with x & y)
• Hence, the 3-node triangle is called a
“Constant Strain Triangle” (CST) element
17
Stress-Strain Relations
   D 
   DBd 
3x1
3x3
3x6 6x1
18
Step 4 – Derive Element Equations
 p  U   p  S   B
which will be used to derive
k   tA B DB
T
6x6
6x3
3x3
3x6
19
Derive Element Equations (cont.)
Strain energy:
20
Derive Element Equations (cont.)
Potential energy of applied loads:
21
Derive Element Equations (cont.)
Potential energy:
22
Derive Element Equations (cont.)
Substitute
to yield
23
Derive Element Equations (cont.)
Apply principle of minimum potential energy
To obtain
24
Derive Element Equations (cont.)
Element stiffness matrix
25
Steps 5-7
5. Assemble global equations
6. Solve for nodal displacements
7. Compute element stresses (constant
within each element)
26
Example – CST element stiffness matrix
27
CST Element Stiffness Matrix
k   tA B DB
T
where
[B] – depends on nodal coordinates
[D] – depends on E, 
See text for details
28
Body and Surface Forces
Replace distributed body forces and surface
tractions with work equivalent
concentrated forces.
{ fb }
{ fs }
29
Work Equivalent Concentrated Forces –
Body Forces
For a uniformly distributed body forces Xb and Yb:
 f ix   X b 
f   
 iy   Yb 
 f jx   X b  At 
 f b        
 f jy   Yb  3 
 f mx   X b 

  
 f my   Yb 
30
Work Equivalent Concentrated Forces –
Surface Forces
For a uniform surface loading, p, acting on a vertical
edge of length, L, between nodes 1 and 3:
 f six   pLt / 2
f  

0
 siy  

 f sjx   0  At 
fb      
 
f
 sjy   0  3 
 f smx   pLt / 2

 

f

 smy   0 
31
Example 6.2
32
Example 6.2 - Solution
 d 3 x  609 .6
d  

 3 y   4.2 
6


  10 in
d 4 x  663 .7
d 4 y  104 .1


Element 1
Element 2
 x  1005 
  



301
 y 

   2.4 

 xy  
 x   995 
  




1
.
2
 y 

   2.4

 xy  
33
In-class Abaqus Demonstrations
•
•
Example 6.2
Finite width plate with circular hole
(ref. “Abaqus Plane Stress Tutorial”)
34
Chapter 7 - Practical Considerations in
Modeling; Interpreting Results; and
Examples of Plane Stress/Strain Analysis
Discussion of Example 6.2:
35
Example 6.2 - discussion
36