EI o o fV o

and so the boundary condition can now be written as

At the right end the sign is changed as the result of the bending moment sign convention to yield

We note here that when the attached mass is idealized as a particle, then Ic — 0; and the moment boundary condition reduces to be the same as indicated above for the translational mass, that is, the bending moment is zero.

Figure 2.31 Schematic of rotational inertia end conditions.

v\\\\\\W Figure 2.32 Schematic of pinned-pinned beam.

2.3.4 Example Solutions for Mode Shapes and Frequencies

In this section we consider several examples of the calculation of natural frequencies and mode shapes of vibrating beams in bending. One of the simplest cases is the pinned-pinned case, with which we begin. It is one of the few cases for beams in bending for which a numerical solution of the characteristic equation is not required. Next we treat the important clamped-free case, followed by the case of a hinged-free beam with a rotational restraint about the hinge. Finally, we consider the free-free case, illustrating the concept of the rigid-body mode.

Example 8: Solution for Pinned-Pinned Beam

Consider the pinned-pinned beam as shown in Fig. 2.32. The horizontal rollers at the right end indicate that the resultant axial force in the beam is zero. The boundary conditions reduce to conditions on X given by

X(0) = X"(0) = X(£) = X"(£) = 0. (2.233)

Substituting the first two of the boundary conditions into the general solution as found in Eqs. (2.220), one finds that

The constant a cannot be zero, because the form of the solution would change to a cubic polynomial, and the boundary conditions for this case do not yield a nontrivial solution of that form. Therefore, D2 — D4 = 0, and the solution for X


Using the last two boundary conditions, one obtains a set of homogeneous algebraic equations in D\ and D \:

A nontrivial solution can only exist if the determinant of the coefficients is equal to zero, and so

Since a / 0, we know that the only way this characteristic equation can be satisfied is for sm(a£) = 0, (2.238)

which has a denumerably infinite set of roots given by oti = — (¿ = 1,2,...). (2.239)

sin(af) sinh(or^)"

0 0

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