Moving load

This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Moving load" – news · newspapers · books · scholar · JSTOR (June 2021) (Learn how and when to remove this message)

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

• 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]}

− 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]}

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]}

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]}

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]}

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]}

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 w_st 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]\ .}