Pantograph

Train

Rifle

Examples of a moving load.

Force

Oscillator

Mass

Types of a moving load.

In structural dynamics, a moving load changes the point at which the load is applied over time.^{[citation needed]} Examples include a vehicle that travels across a bridge^{[citation needed]} and a train moving along a track.^{[citation needed]}

Properties

In computational models, load is usually applied as

• a simple massless force,^{[citation needed]}
• an oscillator,^{[citation needed]} or
• an inertial force (mass and a massless force).^{[citation needed]}

Numerous historical reviews of the moving load problem exist.^[1][2] Several publications deal with similar problems.^[3]

The fundamental monograph is devoted to massless loads.^[4] Inertial load in numerical models is described in ^[5]

Unexpected property of differential equations that govern the motion of the mass particle travelling on the string, Timoshenko beam, and Mindlin plate is described in.^[6] It is the discontinuity of the mass trajectory near the end of the span (well visible in string at the speed v=0.5c).^{[citation needed]} The moving load significantly increases displacements.^{[citation needed]} The critical velocity, at which the growth of displacements is the maximum, must be taken into account in engineering projects.^{[citation needed]}

Structures that carry moving loads can have finite dimensions or can be infinite and supported periodically or placed on the elastic foundation.^{[citation needed]}

Consider simply supported string of the length l, cross-sectional area A, mass density ρ, tensile force N, subjected to a constant force P moving with constant velocity v. The motion equation of the string under the moving force has a form^{[citation needed]}

${\displaystyle -N{\frac {\partial ^{2}w(x,t)}{\partial x^{2))}+\rho A{\frac {\partial ^{2}w(x,t)}{\partial t^{2))}=\delta (x-vt)P\ .}$

Displacements of any point of the simply supported string is given by the sinus series^{[citation needed]}

${\displaystyle w(x,t)={\frac {2P}{\rho Al))\sum _{j=1}^{\infty }{\frac {1}{\omega _{(j)}^{2}-\omega ^{2))}\left(\sin(\omega t)-{\frac {\omega }{\omega _{(j)))}\sin(\omega _{(j)}t)\right)\sin {\frac {j\pi x}{l))\ ,}$

where

${\displaystyle \omega ={\frac {j\pi v}{l))\ ,}$

and the natural circular frequency of the string

${\displaystyle \omega _{(j)}^{2}={\frac {j^{2}\pi ^{2)){l^{2))}{\frac {N}{\rho A))\ .}$

In the case of inertial moving load, the analytical solutions are unknown.^{[citation needed]} The equation of motion is increased by the term related to the inertia of the moving load. A concentrated mass m accompanied by a point force P:^{[citation needed]}

${\displaystyle -N{\frac {\partial ^{2}w(x,t)}{\partial x^{2))}+\rho A{\frac {\partial ^{2}w(x,t)}{\partial t^{2))}=\delta (x-vt)P-\delta (x-vt)m{\frac ((\mbox{d))^{2}w(vt,t)}((\mbox{d))t^{2))}\ .}$

The last term, because of complexity of computations, is often neglected by engineers.^{[citation needed]} The load influence is reduced to the massless load term.^{[citation needed]} Sometimes the oscillator is placed in the contact point.^{[citation needed]} Such approaches are acceptable only in low range of the travelling load velocity.^{[citation needed]} In higher ranges both the amplitude and the frequency of vibrations differ significantly in the case of both types of a load.^{[citation needed]}

The differential equation can be solved in a semi-analytical way only for simple problems.^{[citation needed]} The series determining the solution converges well and 2-3 terms are sufficient in practice.^{[citation needed]} More complex problems can be solved by the finite element method^{[citation needed]} or space-time finite element method.^{[citation needed]}

massless load	inertial load

The discontinuity of the mass trajectory is also well visible in the Timoshenko beam.^{[citation needed]} High shear stiffness emphasizes the phenomenon.^{[citation needed]}

The Renaudot approach vs. the Yakushev approach

Renaudot approach

${\displaystyle \delta (x-vt){\frac {\mbox{d))((\mbox{d))t))\left[m{\frac ((\mbox{d))w(vt,t)}((\mbox{d))t))\right]=\delta (x-vt)m{\frac ((\mbox{d))^{2}w(vt,t)}((\mbox{d))t^{2))}\ .}$^{[citation needed]}

Yakushev approach

${\displaystyle {\frac {\mbox{d))((\mbox{d))t))\left[\delta (x-vt)m{\frac ((\mbox{d))w(vt,t)}((\mbox{d))t))\right]=-\delta ^{\prime }(x-vt)mv{\frac ((\mbox{d))w(vt,t)}((\mbox{d))t))+\delta (x-vt)m{\frac ((\mbox{d))^{2}w(vt,t)}((\mbox{d))t^{2))}\ .}$^{[citation needed]}

Massless string under moving inertial load

Consider a massless string, which is a particular case of moving inertial load problem. The first to solve the problem was Smith.^[7] The analysis will follow the solution of Fryba.^[4] Assuming ρ=0, the equation of motion of a string under a moving mass can be put into the following form^{[citation needed]}

${\displaystyle -N{\frac {\partial ^{2}w(x,t)}{\partial x^{2))}=\delta (x-vt)P-\delta (x-vt)\,m{\frac ((\mbox{d))^{2}w(vt,t)}((\mbox{d))t^{2))}\ .}$

We impose simply-supported boundary conditions and zero initial conditions.^{[citation needed]} To solve this equation we use the convolution property.^{[citation needed]} We assume dimensionless displacements of the string y and dimensionless time τ:^{[citation needed]}

${\displaystyle y(\tau )={\frac {w(vt,t)}{w_{st))}\ ,\ \ \ \ \tau \ =\ {\frac {vt}{l))\ ,}$

where w_st is the static deflection in the middle of the string. The solution is given by a sum

${\displaystyle y(\tau )={\frac {4\,\alpha }{\alpha \,-\,1))\,\tau \,(\tau -1)\,\sum _{k=1}^{\infty }\,\prod _{i=1}^{k}{\frac {(a+i-1)(b+i-1)}{c+i-1))\;{\frac {\tau ^{k)){k!))\ ,}$

where α is the dimensionless parameters :

${\displaystyle \alpha ={\frac {Nl}{2mv2))\,>\,0\ \ \ \wedge \ \ \ \alpha \,\neq \,1\ .}$

Parameters a, b and c are given below

${\displaystyle a_{1,2}={\frac {3\,\pm \,{\sqrt {1+8\alpha ))}{2))\ ,\ \ \ \ \ b_{1,2}={\frac {3\,\mp \,{\sqrt {1+8\alpha ))}{2))\ ,\ \ \ \ \ c=2\ .}$

In the case of α=1, the considered problem has a closed solution:^{[citation needed]} ${\displaystyle y(\tau )=\left[{\frac {4}{3))\tau (1-\tau )-{\frac {4}{3))\tau \left(1+2\tau \ln(1-\tau )+2\ln(1-\tau )\right)\right]\ .}$