Page 286 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

Sec. 9.3 Torsional Vibration of Rods 273

The two equations corresponding to these boundary conditions are, therefore,
= ^ —(C sin wr + D cos = 0
\dx),^o ^
[du\ oj ( ^ (x)l ^ ^ X ^

\ ^ \ = —\ A cos------ 5 sin — (C sin it>r + D cos M = 0

\dx 1,^1 c \ c c 1^ ^
Because these equations must be true for any time t, A must be equal to zero from
the first equation. Because B must be finite in order to have vibration, the second
equation is satisfied when
sin — = 0
c
or
¡ I p
----- = o)„l\ = 7T, Ztt, ITT,..., mr
c y E
The frequency of vibration is thus given by
niT / E
f - ^ I
I V P ~ 2lV J
where n represents the order of the mode. The solution of the free-free rod with zero
initial displacement can then be written as
niT . riTT / E

u = Wn cos —Î-JC sm — \ — t
i \ P
The amplitude of the longitudinal vibration along the rod is, therefore, a cosine wave
having n nodes.

9.3 TORSIONAL VIBRATION OF RODS
The equation of motion of a rod in torsional vibration is similar to that of
longitudinal vibration of rods discussed in the preceding section.
By letting jc be measured along the length of the rod, the angle of twist in
any length cbc of the rod due to torque T is
do = (9.3-1)

where IpG is the torsional stiffness given by the product of the polar moment of
inertia Ip of the cross-sectional area and the shear modulus of elasticity G. The
torque on the two faces of the element being T and T + (dT/dx) dx, as shown in
Fig. 9.3-1, the net torque from Eq. (9.3-1) becomes

ax = IpU — Tax (9.3-2)

By equating this torque to the product of the mass moment of inertia pIp dx of the
element and the angular acceleration d^O/dt^, where p is the density of the rod in