ABSTRACT
FIGURE 6.24: Deformation of the beam due to the rotation of the beam cross section.
To obtain the second boundary condition at x = L, we will now apply Newton’s second rotational law (the sum of all moments equals to the product of the moment of inertia and the angular acceleration). Note that since it is a free end at the tip mass, it is free to rotate. Hence we have ∑
MI = Jα(t). (6.16)
This expression is the counterpart to Newton’s second law for rectilinear motion, which states that the sum of all applied forces equals to the product of mass and acceleration. In equation (6.16), J is the moment of inertia about the axis of rotation (which describes the spread of distribution of a region about an axis) and is given by
J = ∫ ∫
∫ ρ¯r2dV,
where ρ¯ is the tip mass density. In addition, the angular acceleration, α(t), can be computed from the following expressions
α(t) = d
dt ( dθ
dt ),
≈ d dt
( d
dt
∂y
∂x (t, L)),
= ∂3
∂t2∂x y(t, L). (6.17)
Finally, from Figure 6.25, we obtain that ∑ MI = −M(t, L)+S(t, L)δ+
h(t), where h is the external applied moment about the center of mass of the tip body. Substituting this expression for the moment of inertia
as well as the formula for the angular acceleration (6.17) into Newton’s second rotational law (6.16), we obtain
J ∂3
∂t2∂x y(t, L) = −M(t, L) + S(t, L)δ + h(t)
or
J ∂3
∂t2∂x y(t, L) + δ
∂
∂x M(t, L) +M(t, L) = h(t),
which gives us the required second boundary condition at x = L.
