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Mathematical Differential Equation in Calculating the Strength of Building Beam Structure Impact Resistance

Pubblicato online: 15 Jul 2022
Volume & Edizione: AHEAD OF PRINT
Pagine: -
Ricevuto: 17 Feb 2022
Accettato: 28 Apr 2022
Dettagli della rivista
License
Formato
Rivista
eISSN
2444-8656
Prima pubblicazione
01 Jan 2016
Frequenza di pubblicazione
2 volte all'anno
Lingue
Inglese
Introduction

The elastic impact of structures dates back to the early twentieth century. Some scholars have established a nonlinear integral equation for the simply supported beam problem with the impact of the center of the steel ball. They solved it using piecewise numerical quadrature. Later, many researchers did more in-depth research on this issue. But these studies are based on the Hertz contact theory [1]. It does not represent all real contact processes between objects, nor can it reflect the propagation process of waves when objects collide with each other.

Another direct mode superposition method (DMSM) has solved elastic collision problems in recent years. It regards the structural elastic impact problem as the initial value problem of the vibration of the impacting structural system. This can simplify complex collision problems [2]. This method can be used to analyze the collision problem of translational structures and analyze various elastic locking problems of the mechanism. Some scholars have obtained analytical solutions to the problem of mid-span lateral collision between particles and rods, respectively, and simply supported beams. They discussed the dynamic response characteristics of the collision problem employing the numerical analysis method and analyzed the convergence of the impact load. Some scholars have expanded the study's scope of application to discuss the impact force's change at any position on the beam with the impact point.

Since a particle impacts the beam for a very short time, it is important to understand its transient deformation and stress variation. The vibration solution is difficult to obtain satisfactory results due to its wide frequency band. So this paper adopts the integral transformation method to solve [3]. This paper expands the scope of application of the study. This paper studies the impact problem where the impact point is located anywhere on the beam. The study adds the effect of axial force to the governing differential equation. We use the Laplace transform, and the inverse Crump transform to obtain the analytical expressions for the linear elastic collision between beams and mass points.

Solve the basic equations in the frequency domain

The governing differential equation of the Euler-Bernoulli beam for the axial force is: 4yx4+N2yEIx2+ρA2yEIt2=0 {{{\partial^4}y} \over {\partial {x^4}}} + {{N{\partial^2}y} \over {EI\partial {x^2}}} + {{\rho A{\partial^2}y} \over {EI\partial {t^2}}} = 0

Where x is the axis coordinate of the beam. y is the deflection of the beam. t is time. E is the elastic modulus of the beam. I is the moment of inertia of the section. N is the axial force. ρ is the density. A is the cross-sectional area. Suppose α=ρAEI \alpha = {{\rho A} \over {EI}} , β=NEI \beta = {N \over {EI}} . Since the initial displacement and velocity of the beam is zero (y|t=0=0,yt|t=0=0) \left({y{{\left| {_{t = 0} = 0,\,{{\partial y} \over {\partial t}}} \right|}_{t = 0}} = 0} \right) , the Laplace transform of t is performed on the above equation. 4y˜x4+β2y˜x2+αp2y˜=0 {{{\partial^4}\tilde y} \over {\partial {x^4}}} + \beta {{{\partial^2}\tilde y} \over {\partial {x^2}}} + \alpha {p^2}\tilde y = 0

The characteristic equation of the above formula is: λ4+βλ2+αp2=0 {\lambda^4} + \beta {\lambda^2} + \alpha {p^2} = 0

Its four eigenvalues are n1, −n1, n2, −n2. Then the basic solution form of equation (2) is y%(x,p)=c1en1x+c2en1x+c3en2x+c4en2x(0xl2)y%(x,p)=a1en1x+a2en1x+a3en2x+a4en2x(l1x0) \matrix{{y\% \left({x,p} \right) = {c_1}{e^{n1x}} + {c_2}{e^{- n1x}} + {c_3}{e^{n2x}} + {c_4}{e^{- n2x}}\left({0 \le x \le {l_2}} \right)} \hfill \cr {y\% \left({x,p} \right) = {a_1}{e^{n1x}} + {a_2}{e^{- n1x}} + {a_3}{e^{n2x}} + {a_4}{e^{- n2x}}\left({- {l_1} \le x \le 0} \right)} \hfill \cr}

Where c1, c2, c3, c4, a1, a2, a3, a4 is the coefficient to be determined, and l1, l2 is the beam length on both sides of the collision. The relationship between the angle of rotation θ and the moment M of the beam at any point on the beam, the shear force Q, and the deflection y of the beam are θ=yx \theta = {{\partial y} \over {\partial x}} , M=EI2yx2 M = - EI{{{\partial^2}y} \over {\partial {x^2}}} , Q=EI3yx3 Q = - EI{{{\partial^3}y} \over {\partial {x^3}}} , respectively [4]. We perform Laplace transform on it. We can get the following relation by substituting it into equation (4) θ˜(x,p)=n1c1en1xn1c2en1x+n2c3en2xn2c4en2x \tilde \theta \left({x,p} \right) = {n_1}{c_1}{e^{n1x}} - {n_1}{c_2}{e^{- n1x}} + {n_2}{c_3}{e^{n2x}} - {n_2}{c_4}{e^{- n2x}} M˜(x,p)=k1c1en1x+k1c2en1x+k2c3en2xk2c4en2x \tilde M\left({x,p} \right) = {k_1}{c_1}{e^{n1x}} + {k_1}{c_2}{e^{- n1x}} + {k_2}{c_3}{e^{n2x}} - {k_2}{c_4}{e^{- n2x}} Q˜(x,p)=k3c1en1xk3c2en1x+k4c3en2xk4c4en2x \tilde Q\left({x,p} \right) = {k_3}{c_1}{e^{n1x}} - {k_3}{c_2}{e^{- n1x}} + {k_4}{c_3}{e^{n2x}} - {k_4}{c_4}{e^{- n2x}}

Where k1=EIn12 {k_1} = - EIn_1^2 , k2=EIn22 {k_2} = - EIn_2^2 , k3=EIn13 {k_3} = - EIn_1^3 , k4=EIn23 {k_4} = - EIn_2^3 . We change ci in formulas (5)(7) to ai. The expressions for rotation angle, bending moment, and shear force when a(−l1x ≤ 0) can also be obtained.

It is difficult to solve the beam collision problem accurately. Some scholars have proposed a simplified method called linearization. In this way, the problem is simplified to the initial value problem of free vibration of a linear elastic system. Some scholars have proved the simplicity and good accuracy of this calculation method [5]. This method provides a practical method for analyzing the impact response of flexible systems. This paper uses the linearization method. m body of mass A strikes a simply supported beam laterally with velocity V. The contact between the object and the impact beam is modeled with a spring with a compliance coefficient of f (Fig. 1). The contact equilibrium conditions are as follows m2ut2+1f(uy)=01f(uy)=Q1Q2 \matrix{{m{{{\partial^2}u} \over {\partial {t^2}}} + {1 \over f}\left({u - y} \right) = 0} \hfill \cr {{1 \over f}\left({u - y} \right) = {Q_1} - {Q_2}} \hfill \cr}

Figure 1

Schematic diagram of impact

Among them, u(t) is the displacement of the impactor. Q1, Q2 is approximately the shear force value of the positive and negative sides of the impact point. Q1Q|x=0, Q2Q|x=0+. We perform the Laplace transform [6]. Since the initial displacement of the impactor is zero and the initial velocity is V (u = 0, dudt=V {{du} \over {dt}} = V ), equation (8) can be simplified as shown in (9). mp2y˜|x=0+(Q˜1|x=0Q˜2|x=0+)×(mp2f+1)=mV m{p^2}\tilde y\left| {_{x = 0} + \left({{{\tilde Q}_1}\left| {_{x = {0^ -}} -} \right.{{\tilde Q}_2}\left| {_{x = {0^ +}}} \right.} \right) \times \left({m{p^2}\,f + 1} \right) = mV} \right.

The displacement, rotation angle, and bending moment on the positive and negative sides of the impact are continuous: y˜|x=0+=y˜|x=0,y˜x|x=0+=y˜x|x=0,EI2y˜x2|x=0+=EI2y˜x2|x=0 \tilde y\left| {_{x = {0^ +}} = \tilde y\left| {_{x = {0^ -}},{{\partial \tilde y} \over {\partial x}}} \right.\left| {_{x = {0^ +}} = {{\partial \tilde y} \over {\partial x}}\left| {_{x = {0^ -}},} \right. - EI{{{\partial^2}\tilde y} \over {\partial {x^2}}}} \right.\left| {_{x = {0^ +}}} \right. = - EI{{{\partial^2}\tilde y} \over {\partial {x^2}}}\left| {_{x = {0^ -}}} \right.} \right.

The boundary conditions of the beam have the following relation y˜|x=l1=0,EI2y˜x2|x=l1=0y˜|x=l2=0,EI2y˜x2|x=l2=0 \matrix{{\tilde y{{\left| {_{x = - {l_1}} = 0,\, - EI{{{\partial^2}\tilde y} \over {\partial {x^2}}}} \right|}_{x = - {l_1}}} = 0} \hfill \cr {\tilde y{{\left| {_{x = {l_2}} = 0,\, - EI{{{\partial^2}\tilde y} \over {\partial {x^2}}}} \right|}_{x = {l_2}}} = 0} \hfill \cr}

We can obtain the coefficient c1, c2, c3, c4, a1, a2, a3, a4 by substituting equations (4)(7) into (9)(11). This way, we can obtain expressions for the impact force and various dynamic responses.

Numerical inversion

One of the main difficulties in applying the Laplace transform to find the time domain response is obtaining its inversion quickly. This problem is often not easily solved for simple rational functions. Research shows that the inverse Laplace transform is ill-conditioned [7]. It is generally difficult for us to obtain analytical inverse Laplace transform results. We generally use the numerical inverse transformation method. The more common Laplace numerical inversion methods include the Duber method, the Abate method, the Durbin method, the Stefest method, the Crump method, the Schapery method, the Legendre polynomial method, and the Laguerre polynomial method. We choose the Crump method according to the specific characteristics of this problem and the advantages and disadvantages of various methods.

The Laplace transformation of the function f(t) is defined as: F(s)=0f(t)estdt F\left(s \right) = \int_0^\infty {f\left(t \right){e^{- st}}dt}

The formula for inverse Laplace transform to find f(t) is: f(t)=12πiβiβ+iF(t)estdt f\left(t \right) = {1 \over {2\pi i}}\int_{\beta - i\infty}^{\beta + i\infty} {F\left(t \right){e^{st}}dt}

We use the Crump inverse transform method for numerical inversion. Its inversion formula is as follows f(t)eatT{12Re[F(a)]+k=1NRe[F(a+ikπT)]×cos(kπT)tk=1NIm[F(a+ikπT)]sin(kπT)t} f\left(t \right) \approx {{{e^{at}}} \over T}\left\{{{1 \over 2}{\mathop{\rm Re}\nolimits} \left[ {F\left(a \right)} \right] + \sum\limits_{k = 1}^N {{\mathop{\rm Re}\nolimits} \left[ {F\left({a + i{{k\pi} \over T}} \right)} \right] \times \cos \left({{{k\pi} \over T}} \right)t - \sum\limits_{k = 1}^N {{\mathop{\rm Im}\nolimits} \left[ {F\left({a + i{{k\pi} \over T}} \right)} \right]\sin \left({{{k\pi} \over T}} \right)t}}} \right\}

i is an imaginary unit. Re(s) is the real part function. Im(s) is the function to find the imaginary part. The value a = α − ln(E′)/2T;α must satisfy |f(t)|≤Meat. M is an arbitrary constant. E′ is the allowable error, and we can take it as 10−5 : 10−8 ; T = 0.8tmax as needed. tmax is the desired maximum time, but since t is closer to tmax, the convergence is slower. Therefore, tmax can also be taken as twice the maximum time of desire. This ensures the accuracy of the calculation results within the required period.

When t = 0, the summation of infinite series in equation (14) can be accelerated by Euler transform method or FFT method to improve the convergence speed [8]. A concurrent calculation can determine the summation terms of an infinite series. If the accelerated algorithm is not used, the sum term generally needs more than 500 terms to converge.

Example analysis

The density of the simply supported beam is 7.83 × 103 (kg/m3). The modulus is 2.07 × 102 (Gpa), the length of the beam is 5 (m), and the section of the beam is 0.2 × 0.2 (m2). The mass of the impact body is 500 (kg), the impact velocity is 3 (m/s), and the compliance coefficient of the contact spring is 0.2 (m/MN).

Verification of the algorithm

The comparison between the impact force response calculated by the method in this paper and the impact force response obtained by ANSYS finite element is shown in Figure 2. It can be seen from the figure that the solution in this paper is in good agreement with the finite element solution. This shows that the method is correct and has better accuracy [9]. It can be seen from Figure 2 that the maximum impact force of the object hitting the beam is 119.71kN. This is less than the impact force of 150kN hitting a rigid wall. The impact time is 51ms, which is longer than 31.4ms hitting a rigid wall. The results show that the stiffness of the entire impact system is reduced due to beam flexibility. This reduces the impact force and increases the impact time.

Figure 2

Impact force curve

Figures 3 to 5 are the response diagrams of displacement, bending stress, and shear force, respectively. It can be seen from the figure that the displacement response and the bending stress response are positively symmetrical distributions. The maximum displacement and bending stress are both at mid-span [10]. The shear response is antisymmetrically distributed, and the position where the shear force varies the most is on the boundary at both ends.

Figure 3

Displacement Response Plot

Figure 4

Bending stress response graph

Figure 5

Shear response graph

Parameter influence analysis

Figure 6 shows the impact force at the impact positions at 1/2, 2/5, and 1/5 of the beam length, respectively [11]. The closer the impact position is to the boundary, the greater the maximum impact force and the shorter the impact time.

Figure 6

Impact force diagram at different impact locations

Figure 7 shows the impact force diagrams with axial pressures of 0kN, 2500kN, 5000kN, and 7500kN, respectively [12]. It can be seen from the figure that with the increase of the axial pressure, the maximum impact force decreases slightly, and the impact time also decreases.

Figure 7

Impact force diagram for different axial pressures

Figure 8 shows the impact force diagrams with impact masses of 250kg, 500kg, and 750kg, respectively. It can be seen from the figure that with the increase of the impact mass, the maximum impact force increases, and the impact time also increases.

Figure 8

Impact force diagram for different impact masses

Figure 9 shows the impact force diagrams with impact velocities of 3 m/s, 5 m/s, and 7 m/s, respectively. As the impact velocity increases, the maximum impact force increases linearly, and the impact velocity has nothing to do with the impact time.

Figure 9

Impact force diagram for different impact velocities

Figure 10 shows the impact force diagrams with contact compliance coefficients of 0.2 m/MN, 0.4 m/MN, and 0.8 m/MN, respectively. With the increase of the compliance coefficient, the large impact force decreases, and the impact time increases.

Figure 10

Impact diagram for different compliance coefficients

Conclusion

We perform Laplace transformation on the governing differential equations, boundary conditions, and continuity conditions of Euler-Bernoulli beams in the collision system. At this point, an analytical expression in the frequency domain is obtained. Then we use the Crump inverse transform method to perform numerical inversion to obtain various dynamic responses in the time domain. It is found that the results obtained by the method in this paper are in good agreement with the finite element solution. This proves that the method is correct and has high precision.

Figure 1

Schematic diagram of impact
Schematic diagram of impact

Figure 2

Impact force curve
Impact force curve

Figure 3

Displacement Response Plot
Displacement Response Plot

Figure 4

Bending stress response graph
Bending stress response graph

Figure 5

Shear response graph
Shear response graph

Figure 6

Impact force diagram at different impact locations
Impact force diagram at different impact locations

Figure 7

Impact force diagram for different axial pressures
Impact force diagram for different axial pressures

Figure 8

Impact force diagram for different impact masses
Impact force diagram for different impact masses

Figure 9

Impact force diagram for different impact velocities
Impact force diagram for different impact velocities

Figure 10

Impact diagram for different compliance coefficients
Impact diagram for different compliance coefficients

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