# TORSIONAL OSCILLATIONS OF A CANTILEVER BEAM

The equation of static equilibrium of a straight, cantilever beam in torsion is (Fig. 1.21), according to Eq. 5a of § 1.1,

Ms** |

where в is the angle of twist, T the torque, GJ the torsional rigidity, and у the axial coordinate. If the beam is in free oscillation, ЗГ/Эу arises

dy-

Fig. 1.21. Moments acting on an element of a beam in

free torsional oscillation.

from the inertia, and is /а(Э20/Эг2), Ia being the mass moment of inertia about the shear center of the cross section. Hence, the equation of motion is

For a cantilever beam, the boundary conditions are (1) no rotation at the root, and (2) no twisting moment at the tip:

0 = 0 at у = 0

= 0 at y = l dy

A Beam of Constant Cross Section. Consider the simplest case when GJ and Ia are constants. Assuming a solution of the form

%, 0 = Ф(у)еш (3)

we obtain, from Eq. 1

The general solution of Eq. 4 is

ф(у) = A sin ку + В cos ку (6)

where A and В are arbitrary constants. From the boundary conditions (Eqs. 2), we see that a nontrivial solution (ф(г/) not identically vanishing) exists only if

В = 0, cos кі = 0, A = arbitrary constant

This requires that КІ = n — |
(« = |
1, 3, 5, • – •) |
(7) |

І. Є., |
|||

Ig) 1 /« |
(n = 1, 3, 5, • • •) |
(8) |

The corresponding function ф(у) is then

. П7Т

Фп(У) = Asm—у

A general solution of Eq. 1 can be obtained by combining Eqs. 9 and 3. Writing the positive root of Eq. 8 as con and sum over n, we obtain

sin ^ (Апеіш-С + Впе~ІШп1) (10)

П

where An and Bn are arbitrary constants. The real part of Eq. 10 gives the general solution as a real-valued function in the form

6(2/, 0 ==2sin ^ (Cn cos a>nt + Dn sin qjJ) (11)

П

The particular values con are called the eigenvalues (or the characteristic values) of the problem, and the corresponding фп(у) are the eigenfunctions (or the characteristic functions). As long as the coefficient A in Eq. 9 is arbitrary, the eigenfunctions have unspecified magnitudes. When some special rule is given, so that the amplitudes can be definitely defined, the eigenfunctions are said to be “normalized.” For example, we may take A to be 1. Or we may define A so that фп2(у) integrated over the span / is equal to 1; or in such a way that фп(у) becomes 1 at у = /; etc.

It should be observed that all points along the span reach their maximum or minimum amplitude at the same time; i. e., they are “inphase.” The period of the motion is Tn — 2тт/соп, and the number of oscillations per second is n = aijln.

Torsional Oscillations of an Arbitrary Beam with Damping. The general case of torsional oscillation of an arbitrary beam with damping force proportional to the angular velocity can be solved approximately by a number of methods. Here we shall not discuss the methods of solution, but will investigate a general feature of the oscillation of such a beam. The equation of motion is

where GJ, /?, and Ia are functions of y. Equation 12 can be solved by assuming, as in Eq. 3,

%, t) = Ф(у)еш (13)

and determining the eigenvalue со from the solution ф(у) and the boundary conditions. Depending on the sign of (}, the motion is either damped or amplified as time increases. The eigenvalue со is, in general, a complex number. If we substitute Eq. 13 into Eq. 12, and cancel the factor elat, the result may be written as

where yfy) and щіу) are real functions. Since this is an equation with complex coefficients, the solution <f>(y) will be complex and can be written as

ф(.У) = Фі(у) + і Ф4-У) (15)

where фііу) and ф2(у) are real functions. Therefore a solution of (12) that satisfies the boundary conditions is

Oi = ШУ) + іфМУШІ (16)

But Eq. 12 is an equation with real coefficients; hence, there must exist another solution, conjugate to 6V which also satisfies the boundary conditions because the real and imaginary parts of вг satisfy these conditions separately. The general solution is the sum of these two particular solutions multiplied by two arbitrary constants. The solution can be written in the real form %> 0 = {Ifi Фі(У) ~ c2 ФзШ cos qt – [q ф2(у) + c2 фг(у)] sin qt}ept (17) where

ico = p + iq Let

ф(У) = ^ФЛУ) + ФЛУ)> ‘і%) = arc tan

Фі(У)

then Eq. 17 can be written as

%, t) = 2Ф(y)ept{c1 cos [qt + T(y)] – c2 sin [qt + T(y)]} (18)

The meaning of the function T(y) can be clarified by considering the time at which в(у, t) becomes zero; i. e., when the cross section at у passes through the equilibrium position. Setting 6(y, t) equal to zero and solving for t, we obtain

1 c

t = – arc tan — — ‘F(y) q l c2

where n is any integer. The right-hand side of Eq. 19 depends on y, which means that different cross sections of the beam pass through the equilibrium position at different instants of time; the motion of the beam is out of phase.

If Y(y) is a constant, Eq. 19 will be independent of y; the beam then oscillates in phase. T(y) is a constant if the ratio yfy)lv{y) is a constant. It can be shown that the necessary and sufficient condition for an oscillation of the beam to be in phase is that the ratio of the damping coefficient fl(y) to the sectional moment of inertia Ify) be constant along the entire span.6-13

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