Civil Engineering Reference
In-Depth Information
stress distribution on a cross section located at axial position
x
is given by
yE
d
2
[
N
]
d
x
2
x
(
x
,
y
)
=−
{}
(4.30)
Since the normal stress varies linearly on a cross section, the maximum and min-
imum values on any cross section occur at the outer surfaces of the element,
where distance
y
from the neutral surface is largest. As is customary, we take the
maximum stress to be the largest tensile (positive) value and the minimum to be
the largest compressive (negative) value. Hence, we rewrite Equation 4.30 as
y
max
E
d
2
[
N
]
d
x
2
x
(
x
)
=
{}
(4.31)
and it is to be understood that Equation 4.31 represents the maximum and mini-
mum normal stress values at any cross section defined by axial coordinate
x
. Also
y
max
represents the largest distances (one positive, one negative) from the neutral
surface to the outside surfaces of the element. Substituting for the interpolation
functions and carrying out the differentiations indicated, we obtain
y
max
E
12
x
L
3
v
1
+
6
x
L
2
6
L
2
v
2
6
L
2
4
L
12
x
L
3
x
(
x
)
=
−
−
1
+
−
6
x
L
2
2
2
L
+
−
(4.32)
Observing that Equation 4.32 indicates a linear variation of normal stress along
the length of the element and since, once the displacement solution is obtained,
the nodal values are known constants, we need calculate only the stress values
at the cross sections corresponding to the nodes; that is, at
x
=
0
and
x
=
L
. The
stress values at the nodal sections are given by
y
max
E
6
1
+
2
)
2
L
(2
x
(
x
=
0)
=
L
2
(
v
2
−
v
1
)
−
(4.33)
y
max
E
6
2
+
1
)
2
L
(2
x
(
x
=
L
)
=
L
2
(
v
1
−
v
2
)
+
(4.34)
The stress computations are illustrated in following examples.
4.4 FLEXURE ELEMENT STIFFNESS MATRIX
We may now utilize the discretized approximation of the flexure element dis-
placement to examine stress, strain, and strain energy exhibited by the element
under load. The total strain energy is expressed as
1
2
(4.35)
U
e
=
x
ε
x
d
V
V
