4.3 Longitudinal Vibrations of Rods

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It can be shown that the governing equation of the longitudinal vibration of line structures such as rods

and bars is identical to that of the transverse vibration of cables and strings. Hence, it is not necessary to

repeat the complete analysis here. We will first develop the equation of motion, then consider BCs, next

identify the similarity with the cable vibration problem, and will conclude with an illustrative example.

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Box 4.1

TRANSVERSE VIBRATION OF STRINGS AND

CABLES

Equation of motion:

mðxÞ

›2vðx; tÞ

›t2 ¼ T

›2vðx; tÞ

›x2 þ f ðx; tÞ

Separable (modal) solution for free vibration:

vðx; tÞ ¼

X

YiðxÞqiðtÞ

with

d2YiðxÞ

dx2 þ l2i

YiðxÞ ¼ 0 ðneeds two boundary conditionsÞ

and

d2qiðtÞ

dt2 þ v2i

qiðtÞ ¼ 0 ðneeds two initial conditionsÞ

Natural frequency: vi ¼ lic

Wave speed: c ¼

ffiffiffiffi

T

m

r

Traveling-wave solution (long cable, independent of end conditions):

vðx; tÞ ¼ v1ðx 2 ctÞ þ v2ðx þ ctÞ

Orthogonality:

ðl

0

YiðxÞYjðxÞdx ¼

0 for i – j

l

2

for i ¼ j

8><

>:

Initial conditions:

(for initial displacement dðxÞ and speed sðxÞ)

qið0Þ ¼

2

l

ðl

0

dðxÞYiðxÞdx

q_ið0Þ ¼

2

l

ðl

0

sðxÞYiðxÞdx

Variable-tension problem:

m

›2v

›t2 ¼ T

›2v

›x2 þ

›T

›x

›v

›x

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4.3.1 Equation of Motion

Consider a rod that is mounted horizontally (so that the gravitational effects can be neglected) as shown

in Figure 4.4(a). A small element of length dx (the limiting case of dx) at position x is shown in

Figure 4.4(b). The longitudinal strain at x is given by

1 ¼

›u

›x ð4:53Þ

where uðx; tÞ ¼ longitudinal displacement of the rod at x from a fixed reference.

Note that the fixed reference may be chosen arbitrarily but, if the assumption of small u is needed, the

relaxed (unstrained) position of the element must be chosen as the reference. The longitudinal stress at

the cross section at x is s ¼ E1 and, hence, the longitudinal force is

P ¼ EA

›u

›x ð4:54Þ

where

E ¼ Young’s modulus of the rod

A ¼ area of cross section

It is not necessary at this point to assume a uniform rod. Hence, A may depend on x:

The equation of motion for the small element shown in Figure 4.4(b) is

rA dx

›2uðx; tÞ

›t2 ¼ P þ dP 2 P þ f ðx; tÞdx

or

rA

›2u

›t2 dx ¼ dP þ f ðx; tÞdx ð4:55Þ

Now, from Equation 4.54, we have

dP ¼

›

›x

EAðxÞ

›u

›x

dx ð4:56Þ

(a)

(b)

x

0

x x+dx

l

x

x

x+dx

Mass density = r

Force per unit length = f (x,t)

0

u(x,t)

P P+dP

Area of cross section = A(x)

FIGURE 4.4 (a) A rod with distributed loading and in longitudinal vibration; (b) a small element of the rod.

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which when substituted into Equation 4.55 gives

rA

›2uðx; tÞ

›t2 ¼

›

›x

EAðxÞ

›uðx; tÞ

›x þ f ðx; tÞ ð4:57Þ

For the case of a uniform rod (constant A) in free vibration ðf ðx; tÞ ¼ 0Þ; we have

›2u

›t2 ¼ c2 ›2uðx; tÞ

›x2 ð4:58Þ

which is identical to the cable vibration equation 4.3, but with the wave speed parameter given by

c ¼

ffiffiffi

E

r

s

ð4:59Þ

which should be compared with Equation 4.4. The analysis of the present problem may be carried out

exactly as for the cable vibration. In particular, the traveling wave solution will hold. Mode shape

orthogonality will hold also. Even the BCs are similar to those of the cable vibration problem.

4.3.2 Boundary Conditions

As for the cable vibration problem, two BCs will be needed along with two initial conditions in order to

obtain the complete solution to the longitudinal vibration of a rod. Both free and forced vibration may be

analyzed as before. For a fixed end at x ¼ x0; we will have no deflection. Hence,

uðx0; tÞ ¼ 0 ð4:60Þ

with the corresponding modal end condition

Xiðx0Þ ¼0 for i ¼ 1; 2; 3; … ð4:61Þ

For a free end at x ¼ x0; there will not be an end force. Hence, in view of Equation 4.54, the applicable BC

will be

›uðx0; tÞ

›x ¼ 0 ð4:62Þ

with the corresponding modal boundary condition

dXiðx0Þ

dx ¼0 for i ¼ 1; 2; 3; … ð4:63Þ

The mode shapes XiðxÞ will satisfy the orthogonality property

ðl

0

XiðxÞXjðxÞdx ¼

0 for i – j

lj for i ¼ j

(

ð4:64Þ

as before. It can be easily verified, for example, that, for a rod with both ends fixed

XiðxÞ ¼ sin

ipx

l ð4:65Þ

Example 4.3

A uniform structural column of length l; mass M; and area of cross section A hangs from a rigid platform

and is supported on a flexible base of stiffness k: A model is shown in Figure 4.5. Initially, the system

remains stationary, in static equilibrium. Suddenly an axial (vertical) speed of u0 is imparted uniformly

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on the entire column due to a seismic jolt.

Determine the subsequent vibration motion of

the column from its initial equilibrium

configuration.

Solution

The gravitational force corresponds to a force per

unit length

f ðx; tÞ ¼

Mg

l

and Equation 4.57 becomes

›2uðx; tÞ

›t2 ¼ c2 ›2uðx; tÞ

›x2 þ

Mg

rAl

Since M ¼ rAl, we have

›2uðx; tÞ

›t2 ¼ c2 ›2uðx; tÞ

›x2 þ g ð4:66Þ

Boundary conditions are

uð0; tÞ ¼ 0 ð4:67Þ

EA

›uðl; tÞ

›x þ kuðl; tÞ ¼ 0 ð4:68Þ

Initial conditions are

uðx; 0Þ ¼ 0 ð4:69Þ

›uðx; 0Þ

›t ¼ u0 ð4:70Þ

We seek the modal summation solution

uðx; tÞ ¼

X

XiðxÞqiðtÞ ð4:71Þ

where the mode shapes XiðxÞ satisfy

d2XiðxÞ

dx2 þ l2i

XiðxÞ ¼ 0 ð4:72Þ

whose solution is

XiðxÞ ¼ C1 sin lix þ C2 cos lix ð4:73Þ

According to Equation 4.67 and Equation 4.68, the modal BCs are

Xið0Þ ¼ 0 ð4:74Þ

EA

dXiðlÞ

dx þ kXiðlÞ ð4:75Þ

Substitute Equation 4.74 into Equation 4.73. We have C2 ¼ 0: Next, use Equation 4.75. We obtain

EAliC1 cos lil þ kC1 sin lil ¼ 0

Since, C1 – 0 for a nontrivial solution, the required condition is

EAli cos lil þ k sin lil ¼ 0

0

x

l

k

FIGURE 4.5 A column suspended from a fixed platform

and supported on a flexible base.

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which may be expressed as

tan lil þ

EA

k

li ¼ 0 ð4:76Þ

This transcendental equation has an infinite number of solutions li; which correspond to the modes of

vibration. The solution may be made computationally and the corresponding natural frequencies are

obtained using

vi ¼ lic ¼ li

ffiffiffi

E

r

s

¼ li

ffiffiffiffiffiffi

EAl

M

s

ð4:77Þ

Substitute Equation 4.71 into Equation 4.66 and use Equation 4.72 to obtain

X

XiðxÞq€iðtÞ ¼ 2c2

X

l2i

XiðxÞqiðtÞ þ g ð4:78Þ

Multiply Equation 4.78 by XjðxÞ and integrate from x ¼ 0 to l; using the orthogonality property

(Equation 4.64), to obtain

ljq€jðtÞ þ c2l2j

ljqjðtÞ ¼ g

ðl

0

XjðxÞdx ð4:79Þ

We normalize the mode shapes as

XiðxÞ ¼ sin lix ð4:80Þ

where the constant multiplier ðC1Þ has been absorbed into qiðtÞ in Equation 4.71. Then,

lj ¼

ðl

0

sin2ljx dx ¼

ðl

0

1

2 ½1 2 cos 2ljx􀀉dx ¼

1

2

x 2

1

2lj

sin 2ljx

" #l

0¼

1

2

l 2

1

2lj

sin 2ljl

" #

ð4:81Þ

and

ðl

0

sin ljx dx ¼

1

lj ½1 2 cos ljl􀀉

Accordingly, Equation 4.79 becomes

q€jðtÞ þv2j

qjðtÞ ¼

g

ljlj ½1 2 cos ljl􀀉 ð4:82Þ

where the right-hand side is a constant and is completely known from Equation 4.81 and Equation 4.76,

and vj is given by Equation 4.77. Now Equation 4.82, which corresponds to a simple oscillator with

a constant force input, may be solved using any convenient approach. For example, the particular

solution is

qjp ¼

g

v2j

ljlj ½1 2 cos ljl􀀉 ð4:83Þ

and the overall solution is

qjðtÞ ¼ Aj sin vjt þ Bj cos vjt þ qjp ð4:84Þ

The constants Aj and Bj are determined using the initial conditions qjð0Þ and q_jð0Þ: These are obtained by

substituting Equation 4.71 into Equation 4.67 and Equation 4.68, multiplying by XjðxÞ; and integrating

from x ¼ 0 to l; making use of the orthogonality property (Equation 4.64). Specifically, we obtain

qjð0Þ ¼ 0 ð4:85Þ

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and

q_jð0Þ ¼

u0

lj

ðl

0

sin ljx dx ¼

u0

ljlj ½1 2 cos ljl􀀉 ð4:86Þ

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