48
2 Single Degree of Freedom (SDOF) Systems
and
d KE =
m
l
0
ϕ(x)
2 dx
˙
ξ(t) ¨
ξ(t) dt = m eq ˙
ξ(t) ¨
ξ(t) dt,
(2.82)
where m eq has the meaning of an equivalent mass of the beam.
– Strain energy: It relates to deformation and has the expression [2]
SE =
E I
2
l
0
w
(x, t)
2 dx =
1
2
E I
l
0
ϕ
(x)
2 dx
ξ(t)
2
so that
d SE =
E I
l
0
ϕ
(x)
2 dx
˙
ξ (t) ξ(t) dt = k eq ˙
ξ (t) ξ(t) dt,
(2.83)
where E is the modulus of elasticity, I denotes the moment of inertia of the beam,
and k eq denotes the beam equivalent stiffness. Note that the equivalent mass and
stiffness depend on the beam deformation.
– External work: The external work in a small time interval (t, t + dt) is
f
∂w(x, t)
∂x
(v 0 dt) +
∂w(x, t)
∂t
dt
= f
ξ(t) ϕ
(x) v 0 + ˙
ξ(t) ϕ(x)
dt
f ˙
ξ(t) ϕ(x) dt
(2.84)
for x = v 0 t, where the latter approximation holds under the assumption of small
deformation such that ϕ (x) ϕ(x).
These results and the work-energy equation d
KE + SE
= d W give
m eq ˙
ξ(t) ¨
ξ(t) dt + k eq ˙
ξ (t) ξ(t) dt = f ˙
ξ(t) ϕ(v 0 t) dt
so that ξ(t) is the solution of
m eq ¨
ξ(t) + k eq ξ(t) = f ϕ(v 0 t), t ≥ 0,
(2.85)
which is the equation of motion for a SDOF with mass m eq , stiffness k eq and
no damping which is subjected to the force f ϕ(v 0 t). Once ξ(t) is calculated, the
displacement of the beam can be obtained from Eq. 2.81.
Example 2.18 The function ϕ(x) = sin
π x/l
is valid since it satisfies all
boundary conditions. The displacement function w(x, t) is zero at supports, i.e.,
w(0, t) = w(l, t) = 0, at all times since ϕ(0) = ϕ(l) = 0 and so are the bending
moments since ϕ (0) = ϕ (l) = 0. The equivalent mass and stiffness are