Civil Engineering Reference
In-Depth Information
Substituting for
f
2
in Equation 10.9 gives
k
1
u
1
u
2
−
1
f
1
=
(10.12)
−
11
mg
−
m
u
2
¨
d
t
2
. The dynamic effect of the inertia of the attached mass is
shown in the second of the two equations represented by Equation 10.12. Equa-
tion 10.12 can also be expressed as
00
0
d
2
u
2
/
where
u
2
=
¨
¨
k
1
u
1
u
2
f
1
mg
u
1
¨
−
1
+
=
(10.13)
m
u
2
−
11
where we have introduced the mass matrix
00
0
[
m
]
=
(10.14)
m
and the nodal acceleration matrix
¨
u
1
¨
{¨
u
}=
(10.15)
u
2
For the simple harmonic oscillator of Figure 10.1, we have the constraint (bound-
ary) condition
u
1
=
0
, so the first of Equation 10.13 becomes simply
−
ku
2
=
f
1
,
while the second equation is
mg
(10.16)
Note that Equation 10.16 is
not
the same as Equation 10.3. Do the two equations
represent the same physical phenomenon? To show that the answer is yes, we
solve Equation 10.16 and compare the results with the solution given in Equa-
tion 10.6.
Recalling that the solution of any differential equation is the sum of a homo-
geneous (complementary) solution and a particular solution, both solutions must
be obtained for Equation 10.16, since the equation is not homogeneous (i.e., the
right-hand side is nonzero). Setting the right-hand side to zero, the form of the
homogeneous equation is the same as that of Equation 10.3, so by analogy,
the homogeneous solution is
u
2
h
(
t
)
m
u
2
+
¨
ku
2
=
)
(10.17)
where
,
C
, and
are as previously defined. The particular solution must satisfy
Equation 10.16 exactly for all values of time. As the right-hand side is constant,
the particular solution must also be constant; hence,
=
C
sin(
t
+
mg
k
u
2
p
(
t
)
=
=
st
(10.18)
which represents the static equilibrium solution per Equation 10.1. The complete
solution is then
u
2
(
t
)
=
u
2
h
(
t
)
+
u
2
p
(
t
)
=
st
+
C
sin(
t
+
)
(10.19)
