Page 285 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

272 Vibration of Continuous Systems Chap. 9

evident then that the element cbc in the new position has changed in length by an
amount (du/dx) cbc, and thus the unit strain is du/dx. Because, from Hooke’s law,
the ratio of unit stress to unit strain is equal to the modulus of elasticity E, we can
write

where A is the cross-sectional area of the rod. By differentiating with respect to x,
d^u dP
AE- (9.2-2)
dx^ dx
We now apply Newton’s law of motion for the element and equate the
unbalanced force to the product of the mass and acceleration of the element:
dP
dx = pAcbc- ^ (9.2-3)
dx ^ dt^
where p is the density of the rod, mass per unit volume. Eliminating dP/dx
between Eqs. (9.2-2) and (9.2-3), we obtain the partial differential equation
d^U 1E \ d^u
(9.2-4)
dt^ 1, P 1
or
d^u 1 d^U
(9.2-5)
dx^
which is similar to that of Eq. (9.1-2) for the string. The velocity of propagation of
the displacement or stress wave in the rod is then equal to
(9.2-6)

and a solution of the form
u{x,t) = U(x)G{t) (9.2-7)
will result in two ordinary differential equations similar to Eqs. (9.1-7) and (9.1-8),
with
A ^
U(x) A sm—X + B cos—X (9.2-8)
c c
G(t) =- C sin cot + D cos cot (9.2-9)
-
Example 9.2-1
Determine the natural frequencies and mode shapes of a free-free rod (a rod with
both ends free).
Solution: For such a bar, the stress at the ends must be zero. Because the stress is given by
the equation E du/dx, the unit strain at the ends must also be zero; that is,
du = 0 at X = 0 and x = /

dx