INTRODUCTION
From historical viewpoint, the problem of moving load back to the beginning
of the nineteenth century, the time of erection of the early railway bridges.
A lot of work has been done during the past years on the dynamic response
of railway bridges and later of highway bridges under the influence of
moving masses. A comprehensive review of work done, as a matter of fact,
can be found in Fryba (1972).
The problems of moving loads are being studied in technically advanced
countries world over especially Czechoslovakia, USA, Germany, Switzerland,
France and Japan to mention but a few. Both the theoretical and experimental
information on the effect of moving on structures were made available.
Relevant publications are; Stokes (1849), Krylov (1995), Jeffcolt (1929),
Inglis (1934), Kolousek (1961), Bolotin (1964), Steele (1971), Knowles
(1968), Oni (1991), Ghorashi and Esmailzadeh (1995), Krylov (1995), Lee
(1994), Lin (1996), Idowu (1998), Dada (2002), Adetunde (2003) and Akinpelu
(2003).
The present study is an extension of the earlier research of Adetunde
(2007), in which the axial force is taken into consideration (Axial force
beams simply means beams which do experience compression when no external
force is applied i.e., artificial creation of stresses in structure before
loading.
The purpose of this paper is therefore to:
• 
Find the effects of moving mass and moving load in connection
with the length of the beam. 
• 
Find the effect of the length of the load on the beam 
• 
Make a comparison between the deflections due to the moving mass
and that of the moving force. 
Under the assumption that the beam is prismatic while rotary and damping
are taken into consideration.
MATHEMATICAL FORMULATION
Consider the simply supported axial force Rayleigh beam, shown in Fig.
1above, of length L having a uniform crosssection with constant mass
per unit length m and flexural stiffness EI. The beam is traversed by
a constant load P having mass M moving at a constant velocity V, which
assumed to strike a finite axial force Rayleigh beam from the left end
of the beam at time t = 0 (where, t is measured from the time the load
enters the beam) and advancing uniformly along the beam. Before the instant,
the deflection throughout the length of the beam is assumed to be zero.
The governing differential equation of motion for axial force Rayleigh
beam when rotatory and damping are considered is given as:

Fig. 1: 
Mathematical model of the problem 
Where:
a_{1} 
= 
The stiffness proportionality facto(damping complex
or radius of gyration) 
W(x,t) 
= 
The transverse displacement response 
X 
= 
The spatial coordinate 
c(x) 
= 
The external damping force per unit length 
N(x) 
= 
The axial force 
P(x,t) 
= 
The transverse loading inertia 
The transverse load inertia takes the form described below.
And
Where:
g 
= 
The gravitational force 
H 
= 
The heavyside unit function of the beam 
ε 
= 
The fixed length of the load 
ξ 
= 
A particular distance along the length of the beam. We employ the
Dirac delta function 
The mass M of the load P is not negligible but of comparable magnitude
with the mass of the total beam mL. As a result of this, we consider the
effect of Coriolis force (Complementary acceleration) and of centripetal
force (Acceleration related to curvature R of the deflection curve) associated
with the mass M of the moving load P = Mg.
Substituting Eq. 4 into Eq. 2 we have
OPERATIONAL SIMPLIFICATION OF THE GOVERNING EQUATION
A series solution in terms of normal modes can be sought in the form.
Where
φ_{i} (x)’s 
= 
The shape function for the nth mode of the freely Vibrating
prismatic beam while 
Y_{i}(t) 
= 
The corresponding modal amplitude that has to be determined. Introducing
Eq. 7 into Eq. 1 and 6, we have 
Multiply Eq. 8 by φ_{n}(x) and integrate
along the length of the beam and applying the two orthogonality relationships,
to the Eq. 8
Note
Putting equations (9ag) into Eq. 9,
we have
Remarks: Clearly all terms in the series in the third term of
Eq. 9 go to zero except for i = n the modes are obviously
uncoupled as far as the stiffness proportional damping is concerned. Coupling
will be present, however, due to c(x), unless it takes on a form allowing
only the term with i = n to remain in the series. This is indeed the case
for massproportional damping, that is, if we let c(x) = a_{0}m(x)
= a_{0}m, in which the proportionality constant a_{0}
has a dimension of t^{1}_{,} then we have
Introducing the damping ratio for the n^{th} mode, we have
Hence Eq. 11 now becomes
SIMULATIONS
Simply supported beam: We shall now consider a simply supported
beam configuration whose normalized deflection curve is given as
We can write
By putting Eq. 14 into the r.h.s. of Eq.
13, after a lot of simplification had been done, we finally have
The above Eq. 15 is the exact governing equation of a simply
supported viscously damped Rayleigh beam. We now use the finite difference method
to solve the above Eq. 15 numerically. To obtain the results,
we make use of central difference formula, which finally resulted into a system
of equations which was in turn solved by a Visual BASIC program.
NUMERICAL RESULTS AND DISCUSSION
In this section, numerical results in both tabular and graphical forms
are presented. The numerical analysis is in two folds. The first one concerns
a viscously damped axial force Rayleigh beam moving mass whilst the second
concerns a viscously damped axial force Rayleigh beam moving force.
The mathematical model discussed herein is related to the work done by
Adetunde (2007) in which the following data were used.
m (Mass per unit length of beam) 
= 
70 kg m^{1} 
M (Mass of the load) 
= 
7.04, 8, 10 kg m^{1} 
g 
= 
9.81 m sec^{2} 
π 
= 
22/7 
L 
= 
10 m 
ξ 
= 
vt + ε/2 
ε 
= 
0.001 m, 0.01 m, 0.1 m 
v 
= 
3.33 m sec^{1} 
t 
= 
0.5 sec 
t 
= 
1.0 sec 
t 
= 
1.5 sec 
h 
= 
0.01 
I 
= 
1.04x10^{6}m^{4} 
E 
= 
2.07x10^{11}N m^{2} 
N 
= 
0.5 
Hence, we have the followings.
Table 14 and Fig.
25 show the variation of deflection of the beam
acted upon by a moving mass and moving force.
It is observed from Fig. 2above that as the mass of
the moving load increases the deflection along the length of the beam
increases.
It is observed from Fig. 3 above that as the length(Eps)
of the moving load increases the deflection along the length of the beam
increases.
It is observed from Fig. 4 above that as the mass of
the moving force increases the deflection along the length of the beam
increases.
Table 1: 
Effect of mass of load on deflection of the beam under
moving mass 

Table 2: 
Effect of load length on deflection of the beam under
moving mass 

Table 3: 
Effect of mass of load on deflection of beam under
moving force 

Table 4: 
Effect of load length on the deflection of beam under
moving force 


Fig. 2: 
Deflection of beam under moving mass for different masses
of load 

Fig. 3: 
Deflection of beam under moving mass for different load
lengths 

Fig. 4: 
Deflection of beam under moving force for different
masses of load 

Fig. 5: 
Deflection of beam under moving force for different
lengths of the load 
It is observed from Fig. 5 above that as the length
(Eps) of the moving load increases the deflection along the length of
the beam increases.
CONCLUSIONS
The dynamic response of loads on viscously damped axial force Rayleigh
beam was carried out.
The results obtained can be summarized as follows:
1. 
The deflection of a viscously damped axial force Rayleigh
beam under a moving mass or moving force increases with increasing
mass of load 
2. 
The deflection of a viscously damped axial force Rayleigh beam under
a moving mass or moving force increases with increasing span of load 
3. 
The deflection of beam due to moving mass is greater than the deflection
due to moving force. 
The dynamic response of loads on viscously damped axial force Rayleigh
beam is studied. The theory is based on orthogonal functions and the results
indicate that the governing differential equation can be transformed into
a series of coupled ordinary differential equations which is the solution
for the corresponding moving distributed force. The resulting governing
differential equation is solved by numerical approach (Finite central
difference method).
In conclusion, the deflection due to moving mass is greater than that
due to moving force.