Note that the distributed force is equally divided into Nodes 3 and 4. The solution of the above equation
provides non-zero displacements. By combining with zero displacements, we have nodal displacements,
as
5
, 0, 0, 0.25, 0, 0.25, 0
{
}
10
{0, 0
} m
s
Q
The element strains can be calculated using Eq. (6.25), as
(1)
(1)
(1)
5
(2)
(2)
(2)
5
{
}
[
]{
}
10
{0, 0, 0.25}
{
}
[
]{
}
10
{0, 0, 0.25}
Bq
Thus, there are no normal strains and shear strains are same for both elements. The element stress can be
calculated using the stress-strain relation for plane stress, as
(1)
(1)
5
(2)
(2)
5
{
}
[ ]{
}
{0, 0, 10 } Pa
{
}
[ ]{
}
{0, 0, 10 } Pa
T
T
C
C
Note that only shear stress exists, which satisfy the pure shear condition. Since distributed force
f
= 10
kN/m
2
is applied at the top edge, the above shear stress is exact. Thus, the CST element can represent the
pure shear condition accurately. The figure below shows the deformed and undeformed shape of the
elements.
5. A structure shown in the figure is approximated with one triangular element. Plane strain assumption
is used.