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. Several publications deal with similar problems.

The fundamental monograph is devoted to massless loads. Inertial load in numerical models is described in

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. 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}}\ ,}

where

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}}}\ .}

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]

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

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

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

Consider a massless string, which is a particular case of moving inertial load problem. The first to solve the problem was Smith. The analysis will follow the solution of Fryba. 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]

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\ .}

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]\ .}