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Load that changes in time

Examples 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]


In computational models, load is usually applied as

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]

N 2 w ( x , t ) x 2 + ρ A 2 w ( x , t ) t 2 = δ ( x v t ) P   . {\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]

w ( x , t ) = 2 P ρ A l j = 1 1 ω ( j ) 2 ω 2 ( sin ( ω t ) ω ω ( j ) sin ( ω ( j ) t ) ) sin j π x l   , {\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}}\ ,}


ω = j π v l   , {\displaystyle \omega ={\frac {j\pi v}{l}}\ ,}

and the natural circular frequency of the string

ω ( j ) 2 = j 2 π 2 l 2 N ρ A   . {\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]

N 2 w ( x , t ) x 2 + ρ A 2 w ( x , t ) t 2 = δ ( x v t ) P δ ( x v t ) m d 2 w ( v t , t ) d t 2   . {\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}}}\ .}
Convergence of the solution for different number of terms.

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
Vibrations of a string under a moving massless force (v=0.1c); c is the wave speed.
Vibrations of a string under a moving massless force (v=0.5c); c is the wave speed.
Vibrations of a string under a moving inertial force (v=0.1c); c is the wave speed.
Vibrations of a string under a moving inertial force (v=0.5c); c is the wave speed.

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

Vibrations of the Timoshenko beam: red lines – beam axes in time, black line – mass trajectory (w0– static deflection).

The Renaudot approach vs. the Yakushev approach[edit]

Renaudot approach[edit]

δ ( x v t ) d d t [ m d w ( v t , t ) d t ] = δ ( x v t ) m d 2 w ( v t , t ) d t 2   . {\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[edit]

d d t [ δ ( x v t ) m d w ( v t , t ) d t ] = δ ( x v t ) m v d w ( v t , t ) d t + δ ( x v t ) m d 2 w ( v t , t ) d t 2   . {\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[edit]

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]

N 2 w ( x , t ) x 2 = δ ( x v t ) P δ ( x v t ) m d 2 w ( v t , t ) d t 2   . {\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]

Massless string and a moving mass – mass trajectory.
y ( τ ) = w ( v t , t ) w s t   ,         τ   =   v t l   , {\displaystyle y(\tau )={\frac {w(vt,t)}{w_{st}}}\ ,\ \ \ \ \tau \ =\ {\frac {vt}{l}}\ ,}

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

y ( τ ) = 4 α α 1 τ ( τ 1 ) k = 1 i = 1 k ( a + i 1 ) ( b + i 1 ) c + i 1 τ k k !   , {\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 :

α = N l 2 m v 2 > 0             α 1   . {\displaystyle \alpha ={\frac {Nl}{2mv2}}\,>\,0\ \ \ \wedge \ \ \ \alpha \,\neq \,1\ .}

Parameters a, b and c are given below

a 1 , 2 = 3 ± 1 + 8 α 2   ,           b 1 , 2 = 3 1 + 8 α 2   ,           c = 2   . {\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\ .}
Massless string and a moving mass – mass trajectory, α=1.

In the case of α=1, the considered problem has a closed solution:[citation needed] y ( τ ) = [ 4 3 τ ( 1 τ ) 4 3 τ ( 1 + 2 τ ln ( 1 τ ) + 2 ln ( 1 τ ) ) ]   . {\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]\ .}