Civil Engineering Reference
In-Depth Information
elementary strength of materials that the deflection
of an elastic bar of length
L
and uniform cross-sectional area
A
when subjected to axial load
P
is given by
PL
AE
=
(2.27)
where
E
is the modulus of elasticity of the material. Using Equation 2.27, we
obtain the equivalent spring constant of an elastic bar as
P
AE
L
k
=
=
(2.28)
and could, by analogy with the linear elastic spring, immediately write the stiff-
ness matrix as Equation 2.6. While the result is exactly correct, we take a more
general approach to illustrate the procedures to be used with more complicated
element formulations.
Ultimately, we wish to compute the nodal displacements given some loading
condition on the element. To obtain the necessary equilibrium equations relating
the displacements to applied forces, we proceed from displacement to strain,
strain to stress, and stress to loading, as follows. In uniaxial loading, as in the bar
element, we need consider only the normal strain component, defined as
d
u
d
x
ε
x
=
(2.29)
which, when applied to Equation 2.25, gives
u
2
−
u
1
ε
x
=
(2.30)
L
which shows that the spar element is a constant strain element. This is in accord
with strength of materials theory: The element has constant cross-sectional area
and is subjected to constant forces at the end points, so the strain does not vary
along the length. The axial stress, by Hooke's law, is then
E
u
2
−
u
1
x
=
E
ε
x
=
(2.31)
L
and the associated axial force is
AE
L
P
=
x
A
=
(
u
2
−
u
1
)
(2.32)
Taking care to observe the correct algebraic sign convention, Equation 2.32 is
now used to relate the applied nodal forces
f
1
and
f
2
to the nodal displacements
u
1
and
u
2
. Observing that, if Equation 2.32 has a positive sign, the element is in
tension and nodal force
f
2
must be in the positive coordinate direction while
nodal force
f
1
must be equal and opposite for equilibrium; therefore,
AE
L
f
1
=−
(
u
2
−
u
1
)
(2.33)
AE
L
f
2
=
(
u
2
−
u
1
)
(2.34)
