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Computation of Stresses
If the state of deformation is known, the stress distribution follows from the material law
and kinematic relations. From Eq. (6.25) for element
i
, the stress anywhere, i.e., for any
value of
ξ
and
η
, is given by
v
i
σ
=
E
=
EB
(ξ
,
η)
(9)
where the vector
v
i
contains the nodal displacements, including boundary conditions, of
element
i
. Care must be taken here because, as explained previously, the entries in the vector
v
i
have been reordered for part of this example problem. Thus, the entries in
B
should be
rearranged accordingly.
As an example, compute the stresses at the center of the element where
ξ
=
η
=
.
0
5. The
relationship for the strains in terms of locally numbered displacements is
v
i
=
B
(ξ
=
0
.
5
,
η
=
0
.
5
)
u
x
1
u
y
1
u
x
2
u
y
2
u
x
3
u
y
3
u
x
4
u
y
4
−
0
.
50
0
.
50
0
.
50
−
0
.
50
x
=
0
−
0
.
50
−
0
.
50
0
.
50
0
.
5
(10)
y
γ
−
0
.
5
−
0
.
5
−
0
.
50
.
50
.
50
.
50
.
5
−
0
.
5
xy
The displacements of (8) for a particular element can be inserted into (10) to find the strains
in the center of that element. From (9) we can compute the stresses in the middle of an
element.
Equation (6.27), with
E
30 GN/m
2
=
and
ν
=
0, becomes
σ
x
σ
y
τ
xy
100
010
000
x
y
γ
xy
=
10
10
3
.
0
(
)
(11)
.
5
σ
=
E
The principal stresses for the center of each element, which are illustrated in Fig. 6.19,
are computed to be
Element
σ
-Max (kPa)
σ
-Min (kPa)
Angle
10
3
10
3
1
0
.
2522
×
−
0
.
5753
×
−
20
.
28
10
3
10
2
2
0
.
3820
×
−
0
.
3106
×
−
7
.
47
(12)
−
.
×
10
3
−
.
×
10
3
−
.
3
0
7267
0
3992
37
54
4
−
0
.
1427
×
10
3
−
0
.
4786
×
10
3
−
56
.
27
The stress can be expected to experience jumps in value at the boundaries of the elements.
The magnitudes of these discontinuities provide one indication of the errors involved in
this approximate solution.
Another indication of the accuracy of the solution can be obtained by comparing the
results here with the response of this structure as found by other methods. For example,
if this structure is treated as the simply supported beam of Fig. 6.20, simple statics gives a
bending moment at the center of the beam to be 133 kNm. A theory of elasticity solution
provides a center deflection along the bottom of the structure of 1
.
×
10
−
4
86
m.
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