# Dynamic Response of Double Elastic Cantilever Beam Attributed to Variable Uniformly Distributed Load

Authors: Wei He; Yanjing Wei
Journal: Mathematical Problems in Engineering
 (2019)

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## Abstract

Based on the double-layer elastic foundation beam theory, the rails and the wall plates of the electromagnetic launching device are modeled as a double-layer elastic cantilever foundation beam. After establishing the kinetic differential equation and setting the boundary condition of the cantilever beam, the displacement solution of double-layer elastic cantilever foundation beam under no load condition is obtained. Applying the Heaviside Function, the deflection equation of the upper and lower beam, the expression of the bending moment, and the stress are obtained. In the case of given motion parameters and structural parameters, the analytical solutions of the rail and the wall plate are calculated. The ANSYS numerical analysis is carried out under the same condition and the results of both solutions are in good agreement. The results can provide theoretical basis for the design of strength and stiffness of the electromagnetic launch device.


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## Body

## 1. Introduction

Since the beginning of the last century, the electromagnetic railguns have seen considerable progress in development. As the electromagnetic railgun has unusual advantages, many countries attach great importance to it. Recently, many experts and researchers have used different theoretical models to study the stiffness, strength, and vibration problem of the electromagnetic launcher elements [1, 2]. Based on classic material mechanical theory in which the electromagnetic launching device is modeled as the elastic foundation beam [3–5], Tzeng investigated the strength and the deformation of launching device shell, deducing the control equation solution, and simulated the strain and stress field in the track under the magnetic field pressure through ANSYS software. Johnson et al. also simplified the electromagnetic launching device as elastic foundation beam [1], using material mechanical method for calculation, and preliminarily analyzed the passing characteristics of stress wave for electromagnetic track under electromagnetic pressure [6–8]; Jin et al. studied dynamic response of the rail for the electromagnetic railguns with the rail being modeled as an Euler–Bernoulli beam and used the Matlab software to carry out the numerical simulation [9], and Che et al. established a dynamic model of rail of single-layer beam based on the principle of Bernoulli–Euler, having discussed the influence of different constraints and pretightening force on vibration and stiffness in railgun [10]. Cao et al. used the Winkler beam to analyze the dynamic deformation of the rails [11], and Young-Hyun Lee et al. used the Timoshenko beam model to calculate the dynamic response of electromagnetic launcher with a C-shaped armature [12]. Tian et al. simplified the composite rail as an elastic foundation beam and analyzed its mechanical characteristics [13]. Hassanabadi, Attari, and Nikkhoo et al. made several investigations on the dynamics response of thin/thick subjected to moving mass by semi-analytical approaches [14–17].However, only simplifying the electromagnetic launcher rail as a single-layer elastic foundation to analyze bending moment, stress, and vibration is far from the actual condition. Our group discussed the double-layer elastic foundation beam model in the condition of electromagnetic track beam as simply supported [18–21] despite the factual model that is close to cantilever beam. Hence, simplifying electromagnetic launching device under force as a double-layer elastic cantilever beam will meet the requirement of actual working condition well.This paper takes rectangular-caliber electromagnetic launcher as research object, and the cover plates and insulation, which provide support to the rails, are modeled as elastics supports to the rails and wall plates (shown in Figures1 and 2). The insulation between the guide rails and the wall plates is equivalent to a layer of elastic foundation, and the wall plates are usually connected with the cover plates by bolts. Cover plates and bolts in work state of deformation are equivalent to another layer elastic foundation for the wall plate. Calculating the deformation and stress state of the rails and the wall plates should be carried out in launching process [18–21]. During launching process when electric current is guided into rails, electric armature is subjected to powerful electromagnetic force and, due to import of the current, two guide rails produce mutual repulsion q. The repulsion q is seen as uniformly distributed load, so due to the movement of armature the whole device can be simplified as a mechanics scenario in which the model of double-layer elastic cantilever beam is exposed to variable and uniformly distributed electromagnetic force q.Figure 1
Schematic of railgun.
(a)
Model of railgun (b)
Section of railgunFigure 2
Exterior of rectangular electromagnetic railgun.


## 2. Mechanical Analysis

### 2.1. Mechanics Model and Differential Equations

The launch rails and wall plates are simplified as a double-layer cantilever beam system on Winkler elastic foundation as shown in Figure3, where the rail is viewed as the upper beam and the wall plate is viewed as the lower beam. l represents the length of the beam and l1 marks the location of the moving armature. q is the electromagnetic force applied to the rails. v is the velocity of the armature. E1I1 and E2I2 are the bending stiffness of upper and lower beam, respectively. Elastic constant between the upper and lower beam is c1 and elastic constant between the lower beam and the foundation is c2. v is the velocity of armature.Figure 3
Mechanical model of double-layer cantilever beam.The dynamic equations of upper and lower beam are(1)E1I1∂4w1∂x4+m1∂2w1∂t2+c1w1-w2=fx,tE2I2∂4w2∂x4+m2∂2w2∂t2+c2w2=-c1w2-w1where w1 and w2 represent the deflection of the upper and lower beam, respectively. m1=ρ1S1 and m2=ρ2S2 are the mass per unit length of upper and lower beam, respectively. ρ1 and ρ2 are the mass density of upper and lower beam, respectively. S1 and S2 are the cross-sectional area of upper and lower beam, respectively. t is the time with armature’s move. f (x, t) is the uniformly distributed load electromagnetic force per unit length applied to the rail.


### 2.2. Homogeneous Solution

Based on interactions between upper and lower beam in the case of no load, the dynamic equations of upper and lower beam are(2)E1I1∂4w1∂x4+m1∂2w1∂t2+c1w1-w2=0(3)E2I2∂4w2∂x4+m2∂2w2∂t2+c1+c2w2-c1w1=0From (2), we obtain(4)w2=E1I1c1∂4w1∂x4+m1c1∂2w1∂t2+w1Applying (4) to (3) and simplifying the solution, we obtain(5)E1I1E2I2∂8w1∂x8+E2I2m1+E1I1m2∂6w1∂x4∂t2+c1E2I2+c1+c2E1I1∂4w1∂x4+m1m2∂4w1∂t4+c1m2+c1+c2m1∂2w1∂t2+c1c2w1=0According to the displacement forms of the beam, we assume the homogeneous solution of w1 and w2, respectively, as(6)w1=∑i=1∞XiAisin⁡Pit+Bicos⁡Pit=∑i=1∞XiYi(7)w2=∑i=1∞E1I1c1∂4Xi∂x+1-m1Pi2c1XiYi=∑i=1∞ZiYiwith Pi as circular frequency,Yi=AisinPit+BicosPit, and Zi=E1I1/c1∂4Xi/∂x+1-m1Pi2/c1Xi.Applying (6) to (5), we obtain(8)αi∂8Xi∂x8+βi∂4Xi∂x4+γiXi=0where αi=E1I1E2I2, βi=c1E2I2+(c1+c2)E1I1-Pi2(E2I2m1+E1I1m2), and γi=m1m2Pi4+c1c2-Pi2[c1m2+(c1+c2)m1].The solution of (8) can be expressed as(9)Xi=C1cos⁡Mix+C2sin⁡Mix+C3chMix+C4shMix+C5cos⁡Nix+C6sin⁡Nix+C7chNix+C8shNixwhere C1~C8 is undetermined coefficients and the coefficients(10)Mi=-βi+βi2-4αiγi1/22αi1/4andNi=-βi-βi2-4αiγi1/22αi1/4.Applying (9) to (7), we can obtain from w2(11)Zi=C1Jicos⁡Mix+C2Jisin⁡Mix+C3JichMix+C4JishMix+C5Kicos⁡Nix+C6Kisin⁡Nix+C7KichNix+C8KishNixwith Ji=E1I1Mi4+c1-m1Pi2/c1 and Ki=(E1I1Ni4+c1-m1Pi2)/c1.Accordingly, we obtain the bending moment of upper and lower beam as follows:(12)Mupper=E1I1∂2w1∂2x=E1I1∑i=1∞∂2Xi∂x2Yi=∑i=1∞E1I1-C1Mi2cos⁡Mix-C2Mi2sin⁡Mix+C3Mi2chMix+C4Mi2shMix-C5Ni2cos⁡Nix-C6Ni2sin⁡Nix+C7Ni2chNix+C8Ni2shNixAisin⁡Pit+Bicos⁡Pit(13)Mlower=E2I2∂2w2∂2x=E2I2∑i=1∞∂2Zi∂x2Yi=∑i=1∞E2I2-C1JiMi2cos⁡Mix-C2JiMi2sin⁡Mix+C3JiMi2chMix+C4JiMi2shMix-C5KiNi2cos⁡Nix-C6KiNi2sin⁡Nix+C7KiNi2chNix+C8KiNi2shNixAisin⁡Pit+Bicos⁡PitThe rotation angles of upper and lower beam are(14)θ1=E1I1∂w1∂x=E1I1∑i=1∞∂Xi∂xYi=∑i=1∞E1I1-C1Misin⁡Mix+C2Micos⁡Mix+C3MishMix+C4MichMix-C5Nisin⁡Nix+C6Nicos⁡Nix+C7NishNix+C8NichNix(15)θ2=E2I2∂w2∂x=E2I2∑i=1∞∂Zi∂xYi=∑i=1∞E2I2-C1JiMisin⁡Mix+C2JiMicos⁡Mix+C3JiMishMix+C4JiMichMix-C5KiNisin⁡Nix+C6KiNicos⁡Nix+C7KiNishNix+C8KiNichNixThe shear forces of upper and lower beam are(16)Q1=E1I1∂3w1∂x3=E1I1∑i=1∞∂3Xi∂x3Yi=∑i=1∞C1Mi3sin⁡Mix-C2Mi3cos⁡Mix+C3Mi3shMix+C4Mi3chMix+C5Ni3sin⁡Nix-C6Ni3cos⁡Nix+C7Ni3shNix+C8Ni3chNix(17)Q2=E2I2∂3w2∂x3=E2I2∑i=1∞∂3Zi∂x3Yi=∑i=1∞E2I2C1JiMi3sin⁡Mix-C2JiMi3cos⁡Mix+C3JiMi3shMix+C4JiMi3chMix+C5KiNi3sin⁡Nix-C6KiNi3cos⁡Nix+C7KiNi3shNix+C8KiNi3chNixwhere the constants C1~C8 can be determined from boundary conditions of the rail. According to the double-layer beam model where one end of the rail is constrained and the another is under condition of freedom, we can obtain the boundary conditions of both ends as follows:(18)w1x=0=0θ1x=0=0,w2x=0=0θ2x=0=0,M1x=l=0Q1x=l=0,M2x=l=0Q2x=l=0Substituting (6), (7), (14), and (15) into (18), from the left boundary conditions we can obtain(19)C1+C3+C5+C7=0C2Mi+C4Mi+C6Ni+C8Ni=0C1Ji+C3Ji+C5Ki+C7Ki=0C2JiMi+C4JiMi+C6KiNi+C8KiNi=0Substituting (12), (13), (16), and (17) into (18), from the right boundary conditions we can obtain(20)-C1Mi2cos⁡Mil-C2Mi2sin⁡Mil+C3Mi2chMil+C4Mi2shMil-C5Ni2cos⁡Nil-C6Ni2sin⁡Nil+C7Ni2chNil+C8Ni2shNil=0C1Mi3sin⁡Mil-C2Mi3cos⁡Mil+C3Mi3shMil-C4Mi3chMil+C5Ni3sin⁡Nil-C6Ni3cos⁡Nil+C7Ni3shNil+C8Ni3chNil=0-C1JiMi2cos⁡Mil-C2JiMi2sin⁡Mil+C3JiMi2chMil+C4JiMi2shMil-C5KiNi2cos⁡Nil-C6KiNi2sin⁡Nil+C7KiNi2chNil+C8KiNi2shNil=0C1JiMi3sin⁡Mil-C2JiMi3cos⁡Mil+C3JiMi3shMil-C4JiJiMi3chMil+C5KiNi3sin⁡Nil-C6KiNi3cos⁡Nil+C7KiNi3shNil+C8KiNi3chNil=0From (19), we obtain(21)C1=-C3,C2=-C4,C5=-C7,C6=-C8From (20), we obtain(22)-C5cos⁡Nil-C6sin⁡Nil+C7chNil+C8shNil=0-C1cos⁡Mil-C2sin⁡Mil+C3chMil+C4shMil=0C5sin⁡Nil-C6cos⁡Nil+C7shNil+C8chNil=0C1sin⁡Mil-C2cos⁡Mil+C3shMil+C4chMil=0Applying (21) into (22), we obtain(23)C7cos⁡Nil+chNil+C8sin⁡Nil+shNil=0C3cos⁡Mil+chMil+C4sin⁡Mil+shMil=0C7-sin⁡Nil+shNil+C8cos⁡Nil+chNil=0C3-sin⁡Mil+shMil+C4cos⁡Mil+chMil=0From (23), we obtain(24)C7chNilcos⁡Nil+1=0C8chNilcos⁡Nil+1=0C3chNilcos⁡Nil+1=0C4chNilcos⁡Nil+1=0From (24), we obtain(25)chNilcos⁡Nil+1=0,chMilcos⁡Mil+1=0AssumeC3=1,C7=1 with(26)C1=C5=-1,C4=-shMil-sin⁡Milch⁡Mil+cos⁡Mil,C2=shMil-sin⁡MilchMil+cos⁡Mil,C6=shNil-sin⁡NilchNil+cos⁡Nil,C8=-shNil-sin⁡NilchNil+cos⁡NilAccordingly, the mode function for upper and lower beam can be obtained as follows:(27)Xi=-cos⁡Mix+chMix-shMil-sin⁡MilchMil+cos⁡MilshMix-sin⁡Mix-cos⁡Nix+chNix-shNil-sin⁡NilchNil+cos⁡NilshNix-sin⁡Nix(28)Zi=-Jicos⁡Mix+JichMix-shMil-sin⁡MilchMil+cos⁡MilJishMix-sin⁡Mix-Kicos⁡Nix+KichNix-shNil-sinNilchNil+cosNilKishNix-sinNixThen w1 and w2 can be expressed as(29)w1=∑i=1∞-cos⁡Mix+chMix-ςshMix-sin⁡Mix-cos⁡Nix+chNix-ξshNix-sinNixAisin⁡Pit+Bisin⁡Pit(30)w2=∑i=1∞-Jicos⁡Mix+JichMix-ςJishMix-sin⁡Mix-Kicos⁡Nix+KichNix-ξKishNix-sin⁡NixAisin⁡Pit+Bisin⁡Pitwith ς=(shMil-sin⁡Mil)/(chMil+cos⁡Mil) and ξ=(shNil-sin⁡Nil)/(chNil+cos⁡Nil).


### 2.3. Particular Solution of Upper Beam

Assumef (x, t) as the force imposed on upper beam, then substituting it into (2), and we obtain(31)E1I1∂4w1∂x4+m1∂2w1∂t2+c1w1-w2=fx,tMake w1=∑i=1∞Xi(x)Yi(t), and then we obtain(32)∑i=1∞E1I1Xi4Yi+m1Yi′′Xi+c1XiYi-1JXiYi=fx,tMake (32) multiplied by Xj(x), and integrate it in the whole beam, considering (27) and the orthogonality of mode function of upper beam and then obtaining(33)Yi′′+Pi2Yi=Qitwhere Qi(t)=∫0lf(x,t)Xi(x)dx named i-order generalized loads. The general solution of (33) is(34)Yit=1Pi∫0tQiτsin⁡Pit-τdτSolving (34), then we can obtain the dynamic response of upper beam under load as follows:(35)w1x,t=∑i=1∞-cos⁡Mix+chMix-ςshMix-sin⁡MixPi-cos⁡Nix+chNix-ξshNix-sin⁡NixPi∫0l-cosMix+chMix-ςshMix-sin⁡Mix-cos⁡Nix+chNix-ξshNix-sin⁡Nix∫0tfx,tsin⁡Pit-τdτdx


### 2.4. Particular Solution of Lower Beam

For lower beam, the load imposed on it is the pressure from upper beam. According to (3), we obtain(36)E2I2∂4w2∂x4+m2∂2w2∂t2+c1+c2w2-c1w1=fx,twhere c1w1 determined by (35) is the load imposed on lower beam. Make w2=∑i=1∞Zi(x)Yi(t), and we obtain(37)∑i=1∞E2I2Zi4Yi+m2Yi′′Zi+c1+c2ZiYi=c1w1 Make (37) multiplied by Zj(x), and integrate it in the whole beam, considering (28) and the orthogonality of mode function of lower beam and then obtaining(38)Y′′+Pi2+c1Jim2Yi=Qitwhere Qi(t)=mt/Ji2m2∫0lc1w1Zidx is defined as generalized loads imposed on the lower beam. Then make ni=(Pi2+c1/Jim2)1/2, and the general solution of (38) is(39)Yit=1ni∫0tQiτsin⁡nit-τdτ Substituting Qi(t), (28), and (39) into w2=∑i=1∞Zi(x)Yi(t), we obtain the dynamic response of lower beam under load as follows:(40)w2=∑i=1∞mt-Jicos⁡Mix+JichMix-ςJishMix-sin⁡Mixnim2Ji2mt-Kicos⁡Nix+KichNix-ξKishNix-sin⁡Nixnim2Ji2∫0lc1-Jicos⁡Mix+JichMix-ςJishMix-sin⁡Mix-Kicos⁡Nix+KichNix-ξKishNix-sin⁡Nix∫0tw1x,τsin⁡nit-τdτdxwhere w1 is the dynamic response of upper beam under load obtained from (35).


### 2.5. Expression of Uniformly Distributed Load

According to the loading status of upper and lower beam, the function for variable uniformly distributed load can be expressed as follows:(41)pqx,t=-qHvt-x0≤t≤lvwhere q is the distributed electromagnetic force between the rails and H(x) is the Heaviside Function.


## 2.1. Mechanics Model and Differential Equations

The launch rails and wall plates are simplified as a double-layer cantilever beam system on Winkler elastic foundation as shown in Figure3, where the rail is viewed as the upper beam and the wall plate is viewed as the lower beam. l represents the length of the beam and l1 marks the location of the moving armature. q is the electromagnetic force applied to the rails. v is the velocity of the armature. E1I1 and E2I2 are the bending stiffness of upper and lower beam, respectively. Elastic constant between the upper and lower beam is c1 and elastic constant between the lower beam and the foundation is c2. v is the velocity of armature.Figure 3
Mechanical model of double-layer cantilever beam.The dynamic equations of upper and lower beam are(1)E1I1∂4w1∂x4+m1∂2w1∂t2+c1w1-w2=fx,tE2I2∂4w2∂x4+m2∂2w2∂t2+c2w2=-c1w2-w1where w1 and w2 represent the deflection of the upper and lower beam, respectively. m1=ρ1S1 and m2=ρ2S2 are the mass per unit length of upper and lower beam, respectively. ρ1 and ρ2 are the mass density of upper and lower beam, respectively. S1 and S2 are the cross-sectional area of upper and lower beam, respectively. t is the time with armature’s move. f (x, t) is the uniformly distributed load electromagnetic force per unit length applied to the rail.


## 2.2. Homogeneous Solution

Based on interactions between upper and lower beam in the case of no load, the dynamic equations of upper and lower beam are(2)E1I1∂4w1∂x4+m1∂2w1∂t2+c1w1-w2=0(3)E2I2∂4w2∂x4+m2∂2w2∂t2+c1+c2w2-c1w1=0From (2), we obtain(4)w2=E1I1c1∂4w1∂x4+m1c1∂2w1∂t2+w1Applying (4) to (3) and simplifying the solution, we obtain(5)E1I1E2I2∂8w1∂x8+E2I2m1+E1I1m2∂6w1∂x4∂t2+c1E2I2+c1+c2E1I1∂4w1∂x4+m1m2∂4w1∂t4+c1m2+c1+c2m1∂2w1∂t2+c1c2w1=0According to the displacement forms of the beam, we assume the homogeneous solution of w1 and w2, respectively, as(6)w1=∑i=1∞XiAisin⁡Pit+Bicos⁡Pit=∑i=1∞XiYi(7)w2=∑i=1∞E1I1c1∂4Xi∂x+1-m1Pi2c1XiYi=∑i=1∞ZiYiwith Pi as circular frequency,Yi=AisinPit+BicosPit, and Zi=E1I1/c1∂4Xi/∂x+1-m1Pi2/c1Xi.Applying (6) to (5), we obtain(8)αi∂8Xi∂x8+βi∂4Xi∂x4+γiXi=0where αi=E1I1E2I2, βi=c1E2I2+(c1+c2)E1I1-Pi2(E2I2m1+E1I1m2), and γi=m1m2Pi4+c1c2-Pi2[c1m2+(c1+c2)m1].The solution of (8) can be expressed as(9)Xi=C1cos⁡Mix+C2sin⁡Mix+C3chMix+C4shMix+C5cos⁡Nix+C6sin⁡Nix+C7chNix+C8shNixwhere C1~C8 is undetermined coefficients and the coefficients(10)Mi=-βi+βi2-4αiγi1/22αi1/4andNi=-βi-βi2-4αiγi1/22αi1/4.Applying (9) to (7), we can obtain from w2(11)Zi=C1Jicos⁡Mix+C2Jisin⁡Mix+C3JichMix+C4JishMix+C5Kicos⁡Nix+C6Kisin⁡Nix+C7KichNix+C8KishNixwith Ji=E1I1Mi4+c1-m1Pi2/c1 and Ki=(E1I1Ni4+c1-m1Pi2)/c1.Accordingly, we obtain the bending moment of upper and lower beam as follows:(12)Mupper=E1I1∂2w1∂2x=E1I1∑i=1∞∂2Xi∂x2Yi=∑i=1∞E1I1-C1Mi2cos⁡Mix-C2Mi2sin⁡Mix+C3Mi2chMix+C4Mi2shMix-C5Ni2cos⁡Nix-C6Ni2sin⁡Nix+C7Ni2chNix+C8Ni2shNixAisin⁡Pit+Bicos⁡Pit(13)Mlower=E2I2∂2w2∂2x=E2I2∑i=1∞∂2Zi∂x2Yi=∑i=1∞E2I2-C1JiMi2cos⁡Mix-C2JiMi2sin⁡Mix+C3JiMi2chMix+C4JiMi2shMix-C5KiNi2cos⁡Nix-C6KiNi2sin⁡Nix+C7KiNi2chNix+C8KiNi2shNixAisin⁡Pit+Bicos⁡PitThe rotation angles of upper and lower beam are(14)θ1=E1I1∂w1∂x=E1I1∑i=1∞∂Xi∂xYi=∑i=1∞E1I1-C1Misin⁡Mix+C2Micos⁡Mix+C3MishMix+C4MichMix-C5Nisin⁡Nix+C6Nicos⁡Nix+C7NishNix+C8NichNix(15)θ2=E2I2∂w2∂x=E2I2∑i=1∞∂Zi∂xYi=∑i=1∞E2I2-C1JiMisin⁡Mix+C2JiMicos⁡Mix+C3JiMishMix+C4JiMichMix-C5KiNisin⁡Nix+C6KiNicos⁡Nix+C7KiNishNix+C8KiNichNixThe shear forces of upper and lower beam are(16)Q1=E1I1∂3w1∂x3=E1I1∑i=1∞∂3Xi∂x3Yi=∑i=1∞C1Mi3sin⁡Mix-C2Mi3cos⁡Mix+C3Mi3shMix+C4Mi3chMix+C5Ni3sin⁡Nix-C6Ni3cos⁡Nix+C7Ni3shNix+C8Ni3chNix(17)Q2=E2I2∂3w2∂x3=E2I2∑i=1∞∂3Zi∂x3Yi=∑i=1∞E2I2C1JiMi3sin⁡Mix-C2JiMi3cos⁡Mix+C3JiMi3shMix+C4JiMi3chMix+C5KiNi3sin⁡Nix-C6KiNi3cos⁡Nix+C7KiNi3shNix+C8KiNi3chNixwhere the constants C1~C8 can be determined from boundary conditions of the rail. According to the double-layer beam model where one end of the rail is constrained and the another is under condition of freedom, we can obtain the boundary conditions of both ends as follows:(18)w1x=0=0θ1x=0=0,w2x=0=0θ2x=0=0,M1x=l=0Q1x=l=0,M2x=l=0Q2x=l=0Substituting (6), (7), (14), and (15) into (18), from the left boundary conditions we can obtain(19)C1+C3+C5+C7=0C2Mi+C4Mi+C6Ni+C8Ni=0C1Ji+C3Ji+C5Ki+C7Ki=0C2JiMi+C4JiMi+C6KiNi+C8KiNi=0Substituting (12), (13), (16), and (17) into (18), from the right boundary conditions we can obtain(20)-C1Mi2cos⁡Mil-C2Mi2sin⁡Mil+C3Mi2chMil+C4Mi2shMil-C5Ni2cos⁡Nil-C6Ni2sin⁡Nil+C7Ni2chNil+C8Ni2shNil=0C1Mi3sin⁡Mil-C2Mi3cos⁡Mil+C3Mi3shMil-C4Mi3chMil+C5Ni3sin⁡Nil-C6Ni3cos⁡Nil+C7Ni3shNil+C8Ni3chNil=0-C1JiMi2cos⁡Mil-C2JiMi2sin⁡Mil+C3JiMi2chMil+C4JiMi2shMil-C5KiNi2cos⁡Nil-C6KiNi2sin⁡Nil+C7KiNi2chNil+C8KiNi2shNil=0C1JiMi3sin⁡Mil-C2JiMi3cos⁡Mil+C3JiMi3shMil-C4JiJiMi3chMil+C5KiNi3sin⁡Nil-C6KiNi3cos⁡Nil+C7KiNi3shNil+C8KiNi3chNil=0From (19), we obtain(21)C1=-C3,C2=-C4,C5=-C7,C6=-C8From (20), we obtain(22)-C5cos⁡Nil-C6sin⁡Nil+C7chNil+C8shNil=0-C1cos⁡Mil-C2sin⁡Mil+C3chMil+C4shMil=0C5sin⁡Nil-C6cos⁡Nil+C7shNil+C8chNil=0C1sin⁡Mil-C2cos⁡Mil+C3shMil+C4chMil=0Applying (21) into (22), we obtain(23)C7cos⁡Nil+chNil+C8sin⁡Nil+shNil=0C3cos⁡Mil+chMil+C4sin⁡Mil+shMil=0C7-sin⁡Nil+shNil+C8cos⁡Nil+chNil=0C3-sin⁡Mil+shMil+C4cos⁡Mil+chMil=0From (23), we obtain(24)C7chNilcos⁡Nil+1=0C8chNilcos⁡Nil+1=0C3chNilcos⁡Nil+1=0C4chNilcos⁡Nil+1=0From (24), we obtain(25)chNilcos⁡Nil+1=0,chMilcos⁡Mil+1=0AssumeC3=1,C7=1 with(26)C1=C5=-1,C4=-shMil-sin⁡Milch⁡Mil+cos⁡Mil,C2=shMil-sin⁡MilchMil+cos⁡Mil,C6=shNil-sin⁡NilchNil+cos⁡Nil,C8=-shNil-sin⁡NilchNil+cos⁡NilAccordingly, the mode function for upper and lower beam can be obtained as follows:(27)Xi=-cos⁡Mix+chMix-shMil-sin⁡MilchMil+cos⁡MilshMix-sin⁡Mix-cos⁡Nix+chNix-shNil-sin⁡NilchNil+cos⁡NilshNix-sin⁡Nix(28)Zi=-Jicos⁡Mix+JichMix-shMil-sin⁡MilchMil+cos⁡MilJishMix-sin⁡Mix-Kicos⁡Nix+KichNix-shNil-sinNilchNil+cosNilKishNix-sinNixThen w1 and w2 can be expressed as(29)w1=∑i=1∞-cos⁡Mix+chMix-ςshMix-sin⁡Mix-cos⁡Nix+chNix-ξshNix-sinNixAisin⁡Pit+Bisin⁡Pit(30)w2=∑i=1∞-Jicos⁡Mix+JichMix-ςJishMix-sin⁡Mix-Kicos⁡Nix+KichNix-ξKishNix-sin⁡NixAisin⁡Pit+Bisin⁡Pitwith ς=(shMil-sin⁡Mil)/(chMil+cos⁡Mil) and ξ=(shNil-sin⁡Nil)/(chNil+cos⁡Nil).


## 2.3. Particular Solution of Upper Beam

Assumef (x, t) as the force imposed on upper beam, then substituting it into (2), and we obtain(31)E1I1∂4w1∂x4+m1∂2w1∂t2+c1w1-w2=fx,tMake w1=∑i=1∞Xi(x)Yi(t), and then we obtain(32)∑i=1∞E1I1Xi4Yi+m1Yi′′Xi+c1XiYi-1JXiYi=fx,tMake (32) multiplied by Xj(x), and integrate it in the whole beam, considering (27) and the orthogonality of mode function of upper beam and then obtaining(33)Yi′′+Pi2Yi=Qitwhere Qi(t)=∫0lf(x,t)Xi(x)dx named i-order generalized loads. The general solution of (33) is(34)Yit=1Pi∫0tQiτsin⁡Pit-τdτSolving (34), then we can obtain the dynamic response of upper beam under load as follows:(35)w1x,t=∑i=1∞-cos⁡Mix+chMix-ςshMix-sin⁡MixPi-cos⁡Nix+chNix-ξshNix-sin⁡NixPi∫0l-cosMix+chMix-ςshMix-sin⁡Mix-cos⁡Nix+chNix-ξshNix-sin⁡Nix∫0tfx,tsin⁡Pit-τdτdx


## 2.4. Particular Solution of Lower Beam

For lower beam, the load imposed on it is the pressure from upper beam. According to (3), we obtain(36)E2I2∂4w2∂x4+m2∂2w2∂t2+c1+c2w2-c1w1=fx,twhere c1w1 determined by (35) is the load imposed on lower beam. Make w2=∑i=1∞Zi(x)Yi(t), and we obtain(37)∑i=1∞E2I2Zi4Yi+m2Yi′′Zi+c1+c2ZiYi=c1w1 Make (37) multiplied by Zj(x), and integrate it in the whole beam, considering (28) and the orthogonality of mode function of lower beam and then obtaining(38)Y′′+Pi2+c1Jim2Yi=Qitwhere Qi(t)=mt/Ji2m2∫0lc1w1Zidx is defined as generalized loads imposed on the lower beam. Then make ni=(Pi2+c1/Jim2)1/2, and the general solution of (38) is(39)Yit=1ni∫0tQiτsin⁡nit-τdτ Substituting Qi(t), (28), and (39) into w2=∑i=1∞Zi(x)Yi(t), we obtain the dynamic response of lower beam under load as follows:(40)w2=∑i=1∞mt-Jicos⁡Mix+JichMix-ςJishMix-sin⁡Mixnim2Ji2mt-Kicos⁡Nix+KichNix-ξKishNix-sin⁡Nixnim2Ji2∫0lc1-Jicos⁡Mix+JichMix-ςJishMix-sin⁡Mix-Kicos⁡Nix+KichNix-ξKishNix-sin⁡Nix∫0tw1x,τsin⁡nit-τdτdxwhere w1 is the dynamic response of upper beam under load obtained from (35).


## 2.5. Expression of Uniformly Distributed Load

According to the loading status of upper and lower beam, the function for variable uniformly distributed load can be expressed as follows:(41)pqx,t=-qHvt-x0≤t≤lvwhere q is the distributed electromagnetic force between the rails and H(x) is the Heaviside Function.


## 3. Displacement of Double-Layer Beam under Variable Uniformly Distributed Load

### 3.1. Derivation of Dynamic Deflection Curve Equations of Upper Beam

Based on (31), we can obtain the dynamic equation of upper beam attributed to uniformly distributed load as follows:(42)E1I1∂4w1∂x4+m1∂2w1∂t2+c1w1-w2=pqx,tSubstituting pq(x,t) into (42), we obtain(43)w1qx,tx,t=∑i=1∞-cos⁡Mix+chMix-ςshMix-sin⁡MixPi-cos⁡Nix+chNix-ξshNix-sin⁡NixPi∫0l-cos⁡Mix+chMix-ςshMix-sin⁡Mix-cos⁡Nix+chNix-ξshNix-sin⁡Nix∫0tpqx,τsin⁡Pit-τdτdx Considering the property of Heaviside Function and substituting pq(x,t)=-qH(vt-x) into (43), we can obtained the dynamic response of upper beam under uniformly distributed loadq through the integral as follows:(44)w1qx,tx,t=∑i=1∞-q-cos⁡Mix+chMix-ςshMix-sin⁡MixPi-cos⁡Nix+chNix-ξshNix-sin⁡NixPi-Pisin⁡Mivt-Mivsin⁡PitMi-Mi2v2+Pi2+PishMivt-Mivsin⁡PitMiMi2v2+Pi2-ςMi2v2cos⁡Pt-1+Pi2chMivt-1MiPi2Mi2v2+Pi2-Mi2v2cos⁡Pt-1-Pi2cos⁡Mivt-1Mi-PiMi2v2+Pi3+-Pisin⁡Nivt-Nivsin⁡PitNi-Ni2v2+Pi2+PishNivt-Nivsin⁡PitNiNi2v2+Pi2-ξNi2v2cos⁡Pt-1+Pi2chNivt-1NiPi2Ni2v2+Pi2-Ni2v2cos⁡Pt-1-Pi2cos⁡Nivt-1Ni-PiNi2v2+Pi3


### 3.2. Bending Moment and Stress of Upper Beam

Having got the dynamic deflection curve equations and then considering the expression ofM1 and σ1, we can directly obtain from (44)(45)M1=E1I1d2w1qdx2=∑i=1∞-E1I1qMi2cos⁡Mix+Mi2chMix-ςMi2shMix+sin⁡MixPi+Ni2cos⁡Nix+Ni2chNix-ξNi2shNix+sin⁡NixPi-Pisin⁡Mivt-Mivsin⁡PitMi-Mi2v2+Pi2+PishMivt-Mivsin⁡PitMiMi2v2+Pi2-ςMi2v2cos⁡Pt-v2+Pi2chMivt-1MiPi2Mi2v2+Pi2-Mi2v2cos⁡Pt-v2-Pi2cos⁡Mivt-1Mi-PiMi2v2+Pi3+-Pisin⁡Nivt-Nivsin⁡PitNi-Ni2v2+Pi2+PishNivt-Nivsin⁡PitNiNi2v2+Pi2-ξNi2v2cos⁡Pt-v2+Pi2chNivt-1NiPi2Ni2v2+Pi2-Ni2v2cos⁡Pt-v2-Pi2cos⁡Nivt-1Ni-PiNi2v2+Pi3(46)σ1=E1h12d2w1qdx2=∑i=1∞-E1h1q2Mi2cos⁡Mix+Mi2chMix-ςMi2shMix+sin⁡MixPi+Ni2cos⁡Nix+Ni2chNix-ξNi2shNix+sin⁡NixPi-Pisin⁡Mivt-Mivsin⁡PitMi-Mi2v2+Pi2+PishMivt-Mivsin⁡PitMiMi2v2+Pi2-ςMi2v2cos⁡Pt-v2+Pi2chMivt-1MiPi2Mi2v2+Pi2-Mi2v2cos⁡Pt-v2-Pi2cos⁡Mivt-1Mi-PiMi2v2+Pi3+-Pisin⁡Nivt-Nivsin⁡PitNi-Ni2v2+Pi2+PishNivt-Nivsin⁡PitNiNi2v2+Pi2-ξNi2v2cos⁡Pt-v2+Pi2chNivt-1NiPi2Ni2v2+Pi2-Ni2v2cos⁡Pt-v2-Pi2cos⁡Nivt-1Ni-PiNi2v2+Pi3


### 3.3. Derivation of Dynamic Deflection Curve Equations of Lower Beam

According to the determinedw1q(x,t), similarly we obtain(47)w2qx,t=∑i=0∞mt-Jicos⁡Mix+JichMix-ςJishMix-sin⁡Mixnim2Ji2+mt-Kicos⁡Nix+KichNix-ξKishNix-sin⁡Nixnim2Ji2∫0lc1-Jicos⁡Mix+JichMix-ςJishMix-sin⁡Mix-Kicos⁡Nix+KichNix-ξKishNix-sin⁡Nix∫0tw1qx,τsin⁡nit-τdτdxSubstituting (44) into (47) and simplifying the solution, we can obtain(48)w2qx,t=-qc1mtnim2Ji2X1+X2H1+H2T1+T2where the expression of H1, H2, T1, and T2 can be reviewed in the appendix.


### 3.4. Bending Moment and Stress of Lower Beam

(49) M 2 = - q c 1 m t E 1 I 1 n i m 2 J i 2 X 1 ′ ′ + X 2 ′ ′ H 1 + H 2 T 1 + T 2 (50) σ 2 = - q c 1 m t E 1 h 2 2 n i m 2 J i 2 X 1 ′ ′ + X 2 ′ ′ H 1 + H 2 T 1 + T 2where X1′′=Mi2Ji(cosh⁡Mix+cos⁡Mix)-ςMi2Ji(sinh⁡Mix+sin⁡Mix) and X2′′=Ni2Ki(cosh⁡Nix+cos⁡Nix)-ξNi2Ki(sinh⁡Nix+sin⁡Nix).


## 3.1. Derivation of Dynamic Deflection Curve Equations of Upper Beam

Based on (31), we can obtain the dynamic equation of upper beam attributed to uniformly distributed load as follows:(42)E1I1∂4w1∂x4+m1∂2w1∂t2+c1w1-w2=pqx,tSubstituting pq(x,t) into (42), we obtain(43)w1qx,tx,t=∑i=1∞-cos⁡Mix+chMix-ςshMix-sin⁡MixPi-cos⁡Nix+chNix-ξshNix-sin⁡NixPi∫0l-cos⁡Mix+chMix-ςshMix-sin⁡Mix-cos⁡Nix+chNix-ξshNix-sin⁡Nix∫0tpqx,τsin⁡Pit-τdτdx Considering the property of Heaviside Function and substituting pq(x,t)=-qH(vt-x) into (43), we can obtained the dynamic response of upper beam under uniformly distributed loadq through the integral as follows:(44)w1qx,tx,t=∑i=1∞-q-cos⁡Mix+chMix-ςshMix-sin⁡MixPi-cos⁡Nix+chNix-ξshNix-sin⁡NixPi-Pisin⁡Mivt-Mivsin⁡PitMi-Mi2v2+Pi2+PishMivt-Mivsin⁡PitMiMi2v2+Pi2-ςMi2v2cos⁡Pt-1+Pi2chMivt-1MiPi2Mi2v2+Pi2-Mi2v2cos⁡Pt-1-Pi2cos⁡Mivt-1Mi-PiMi2v2+Pi3+-Pisin⁡Nivt-Nivsin⁡PitNi-Ni2v2+Pi2+PishNivt-Nivsin⁡PitNiNi2v2+Pi2-ξNi2v2cos⁡Pt-1+Pi2chNivt-1NiPi2Ni2v2+Pi2-Ni2v2cos⁡Pt-1-Pi2cos⁡Nivt-1Ni-PiNi2v2+Pi3


## 3.2. Bending Moment and Stress of Upper Beam

Having got the dynamic deflection curve equations and then considering the expression ofM1 and σ1, we can directly obtain from (44)(45)M1=E1I1d2w1qdx2=∑i=1∞-E1I1qMi2cos⁡Mix+Mi2chMix-ςMi2shMix+sin⁡MixPi+Ni2cos⁡Nix+Ni2chNix-ξNi2shNix+sin⁡NixPi-Pisin⁡Mivt-Mivsin⁡PitMi-Mi2v2+Pi2+PishMivt-Mivsin⁡PitMiMi2v2+Pi2-ςMi2v2cos⁡Pt-v2+Pi2chMivt-1MiPi2Mi2v2+Pi2-Mi2v2cos⁡Pt-v2-Pi2cos⁡Mivt-1Mi-PiMi2v2+Pi3+-Pisin⁡Nivt-Nivsin⁡PitNi-Ni2v2+Pi2+PishNivt-Nivsin⁡PitNiNi2v2+Pi2-ξNi2v2cos⁡Pt-v2+Pi2chNivt-1NiPi2Ni2v2+Pi2-Ni2v2cos⁡Pt-v2-Pi2cos⁡Nivt-1Ni-PiNi2v2+Pi3(46)σ1=E1h12d2w1qdx2=∑i=1∞-E1h1q2Mi2cos⁡Mix+Mi2chMix-ςMi2shMix+sin⁡MixPi+Ni2cos⁡Nix+Ni2chNix-ξNi2shNix+sin⁡NixPi-Pisin⁡Mivt-Mivsin⁡PitMi-Mi2v2+Pi2+PishMivt-Mivsin⁡PitMiMi2v2+Pi2-ςMi2v2cos⁡Pt-v2+Pi2chMivt-1MiPi2Mi2v2+Pi2-Mi2v2cos⁡Pt-v2-Pi2cos⁡Mivt-1Mi-PiMi2v2+Pi3+-Pisin⁡Nivt-Nivsin⁡PitNi-Ni2v2+Pi2+PishNivt-Nivsin⁡PitNiNi2v2+Pi2-ξNi2v2cos⁡Pt-v2+Pi2chNivt-1NiPi2Ni2v2+Pi2-Ni2v2cos⁡Pt-v2-Pi2cos⁡Nivt-1Ni-PiNi2v2+Pi3


## 3.3. Derivation of Dynamic Deflection Curve Equations of Lower Beam

According to the determinedw1q(x,t), similarly we obtain(47)w2qx,t=∑i=0∞mt-Jicos⁡Mix+JichMix-ςJishMix-sin⁡Mixnim2Ji2+mt-Kicos⁡Nix+KichNix-ξKishNix-sin⁡Nixnim2Ji2∫0lc1-Jicos⁡Mix+JichMix-ςJishMix-sin⁡Mix-Kicos⁡Nix+KichNix-ξKishNix-sin⁡Nix∫0tw1qx,τsin⁡nit-τdτdxSubstituting (44) into (47) and simplifying the solution, we can obtain(48)w2qx,t=-qc1mtnim2Ji2X1+X2H1+H2T1+T2where the expression of H1, H2, T1, and T2 can be reviewed in the appendix.


## 3.4. Bending Moment and Stress of Lower Beam

(49) M 2 = - q c 1 m t E 1 I 1 n i m 2 J i 2 X 1 ′ ′ + X 2 ′ ′ H 1 + H 2 T 1 + T 2 (50) σ 2 = - q c 1 m t E 1 h 2 2 n i m 2 J i 2 X 1 ′ ′ + X 2 ′ ′ H 1 + H 2 T 1 + T 2where X1′′=Mi2Ji(cosh⁡Mix+cos⁡Mix)-ςMi2Ji(sinh⁡Mix+sin⁡Mix) and X2′′=Ni2Ki(cosh⁡Nix+cos⁡Nix)-ξNi2Ki(sinh⁡Nix+sin⁡Nix).


## 4. Instance Analysis

### 4.1. Parameters

The dynamic model for electromagnetic launching rail due to uniformly distributed load is shown in Figure4. The length of the beam isl=3000mm. The section sizes of upper beam are H1=45mm and h1=15mm, respectively. The section sizes of lower beam are H2=75mm and h2=30mm, respectively. The material of upper beam is copper of which the elastic modulus is E1=110GPa and the mass density is ρ1=8290kg/m3; the material of lower beam is copper of which the elastic modulus is E2 =28.3GPa and the mass density is ρ2 =980kg/m3. The elastic constant between upper and lower beam is c1=3MPa; the elastic constant between lower beam and foundation is c2=6MPa.The uniformly distributed force is q=3000N/m. The velocity of the armature is assumed as1000m/s.Figure 4
The cross-section of electromagnetic rail launcher.


### 4.2. Consequence and Analysis

In the case of given motion parameters and structure parameters, we have calculated the displacement, bending moment, and stress of double-layer elastic cantilever foundation beam and obtained analytical solutions and numerical results got from ANSYS, which are shown in Figures5–12.Figure 5
Comparison of analytical solution with numerical solution for deflection of upper beam.Figure 6
Upper beam x = 1.5m, comparison of analytical solution with numerical solution for response.Figure 7
Comparison of analytical solution with numerical solution for bending moment distribution of upper beam.Figure 8
Comparison of analytical solution with numerical solution of stress distribution on upper beam.Figure 9
Comparison of analytical solution with numerical solution for deflection of lower beam.Figure 10
Lower beam x = 1.5m, Comparison of analytical solution with numerical solution for response.Figure 11
Comparison of analytical solution with numerical solution for bending moment distribution of lower beam.Figure 12
Comparison of analytical solution with numerical solution of stress distribution on lower beam.Figures5–8 show the dynamic response of the upper beam attributed to moving load. Analytical solution and numerical solution for deflection of upper beam are depicted in Figure 5 in which peak values of displacement of analytical and numerical solution are similar, and the values occur at close positions. However, the analytical solution begins to fluctuate near x=0.5m after which the numerical solution remains stable. Figure 6 shows the changing displacement of upper beam at x = 1.5m, in which the analytical solution and the numerical solution are in good agreement. Figures 7 and 8 depict the bending moment and stress of upper beam at different positions due to the motion of load. Due to the linear relationship between the bending moment and the stress, both of them are identical in trend. Similar to the change of displacement solution, the numerical solution and the analytical solution are close before x=0.5, and after that the analytical solution fluctuates.Meanwhile, dynamic response of the lower beam is investigated. The displacement results of analytical and numerical solution of lower beam differentiate obviously as shown in Figure9. Figure 10 shows the displacement of lower beam changing with time at x = 1.5m, in which, similar to upper beam, the analytical solution and the numerical solution are in good agreement in trend. In Figures 11 and 12, both analytical and numerical solution trend of bending moment and stress of lower beam follow the same patterns of upper beam in Figures 7 and 8.It is apparent that the peak values of dynamic response including displacement, bending moment, and stress of upper beam are bigger than that of lower beam because of upper beam dissipating majority of the vibration energy due to the moving load. There are some analytical solution fluctuations compared with numerical solution in Figures5, 7, 8, 9, 11, and 12, which may be aroused by neglecting the layer damping. In summary, the analytical and numerical solutions of the dynamic response of the upper and lower beams in the above mentioned figures are similar in the overall trend despite some acceptable deviation.In order to investigate the properties of analytical and numerical solutions more specifically, we compare the peak values of displacement, bending moment, and stress (Figures5–12) of the upper and lower beams of dynamic response, as shown in Table 1.Table 1
Analysis of calculation results.
Figure Maximum of numerical solution Maximum of analytical solution Errors% Minimum of numerical solution Minimum of analytical solution Errors% Relative numerical solution Relative analytical solution Relative numerical solution Relative analytical solution 5 0 2.536 —— —— -12.98 -13.62 4.9 4.7 6 3.431 3.593 4.7 4.5 -1.969 -2.078 5.5 5.2 7 244.7 257.8 5.4 5.1 -654.6 -624.2 4.6 4.9 8 145.3 152.7 5.1 4.8 -388.5 -369.9 4.8 5.0 9 0 1.884 —— —— 0 -2.009 —— —— 10 0 0 0 0 -2.939 -3.086 5.0 4.8 11 26.39 25.11 4.9 5.1 -23.19 -24.44 5.4 5.1 12 23.45 22.32 4.8 5.1 -20.59 -21.73 5.5 5.2It is clear that the relative error of the peak values between the analytical solution and the numerical solution obtained by ANSYS is within a reasonable range and fluctuates around 5%.In addition, to better illustrate the analytical dynamic response of double-layer elastic cantilever foundation beams under moving loads, the deflection, bending moment, and stress of upper and lower beams are given in Figures13, 14, and 15 and Figures 16, 17, and 18, respectively.Figure 13
Surface changes of upper beam deflection with time and position.Figure 14
Surface changes of upper beam bending moment with time and position.Figure 15
Surface changes of upper beam stress with time and position.Figure 16
Surface changes of lower beam deflection with time and positionFigure 17
Surface changes of lower beam bending moment with time and position.Figure 18
Surface changes of lower beam stress with time and position.


## 4.1. Parameters

The dynamic model for electromagnetic launching rail due to uniformly distributed load is shown in Figure4. The length of the beam isl=3000mm. The section sizes of upper beam are H1=45mm and h1=15mm, respectively. The section sizes of lower beam are H2=75mm and h2=30mm, respectively. The material of upper beam is copper of which the elastic modulus is E1=110GPa and the mass density is ρ1=8290kg/m3; the material of lower beam is copper of which the elastic modulus is E2 =28.3GPa and the mass density is ρ2 =980kg/m3. The elastic constant between upper and lower beam is c1=3MPa; the elastic constant between lower beam and foundation is c2=6MPa.The uniformly distributed force is q=3000N/m. The velocity of the armature is assumed as1000m/s.Figure 4
The cross-section of electromagnetic rail launcher.


## 4.2. Consequence and Analysis

In the case of given motion parameters and structure parameters, we have calculated the displacement, bending moment, and stress of double-layer elastic cantilever foundation beam and obtained analytical solutions and numerical results got from ANSYS, which are shown in Figures5–12.Figure 5
Comparison of analytical solution with numerical solution for deflection of upper beam.Figure 6
Upper beam x = 1.5m, comparison of analytical solution with numerical solution for response.Figure 7
Comparison of analytical solution with numerical solution for bending moment distribution of upper beam.Figure 8
Comparison of analytical solution with numerical solution of stress distribution on upper beam.Figure 9
Comparison of analytical solution with numerical solution for deflection of lower beam.Figure 10
Lower beam x = 1.5m, Comparison of analytical solution with numerical solution for response.Figure 11
Comparison of analytical solution with numerical solution for bending moment distribution of lower beam.Figure 12
Comparison of analytical solution with numerical solution of stress distribution on lower beam.Figures5–8 show the dynamic response of the upper beam attributed to moving load. Analytical solution and numerical solution for deflection of upper beam are depicted in Figure 5 in which peak values of displacement of analytical and numerical solution are similar, and the values occur at close positions. However, the analytical solution begins to fluctuate near x=0.5m after which the numerical solution remains stable. Figure 6 shows the changing displacement of upper beam at x = 1.5m, in which the analytical solution and the numerical solution are in good agreement. Figures 7 and 8 depict the bending moment and stress of upper beam at different positions due to the motion of load. Due to the linear relationship between the bending moment and the stress, both of them are identical in trend. Similar to the change of displacement solution, the numerical solution and the analytical solution are close before x=0.5, and after that the analytical solution fluctuates.Meanwhile, dynamic response of the lower beam is investigated. The displacement results of analytical and numerical solution of lower beam differentiate obviously as shown in Figure9. Figure 10 shows the displacement of lower beam changing with time at x = 1.5m, in which, similar to upper beam, the analytical solution and the numerical solution are in good agreement in trend. In Figures 11 and 12, both analytical and numerical solution trend of bending moment and stress of lower beam follow the same patterns of upper beam in Figures 7 and 8.It is apparent that the peak values of dynamic response including displacement, bending moment, and stress of upper beam are bigger than that of lower beam because of upper beam dissipating majority of the vibration energy due to the moving load. There are some analytical solution fluctuations compared with numerical solution in Figures5, 7, 8, 9, 11, and 12, which may be aroused by neglecting the layer damping. In summary, the analytical and numerical solutions of the dynamic response of the upper and lower beams in the above mentioned figures are similar in the overall trend despite some acceptable deviation.In order to investigate the properties of analytical and numerical solutions more specifically, we compare the peak values of displacement, bending moment, and stress (Figures5–12) of the upper and lower beams of dynamic response, as shown in Table 1.Table 1
Analysis of calculation results.
Figure Maximum of numerical solution Maximum of analytical solution Errors% Minimum of numerical solution Minimum of analytical solution Errors% Relative numerical solution Relative analytical solution Relative numerical solution Relative analytical solution 5 0 2.536 —— —— -12.98 -13.62 4.9 4.7 6 3.431 3.593 4.7 4.5 -1.969 -2.078 5.5 5.2 7 244.7 257.8 5.4 5.1 -654.6 -624.2 4.6 4.9 8 145.3 152.7 5.1 4.8 -388.5 -369.9 4.8 5.0 9 0 1.884 —— —— 0 -2.009 —— —— 10 0 0 0 0 -2.939 -3.086 5.0 4.8 11 26.39 25.11 4.9 5.1 -23.19 -24.44 5.4 5.1 12 23.45 22.32 4.8 5.1 -20.59 -21.73 5.5 5.2It is clear that the relative error of the peak values between the analytical solution and the numerical solution obtained by ANSYS is within a reasonable range and fluctuates around 5%.In addition, to better illustrate the analytical dynamic response of double-layer elastic cantilever foundation beams under moving loads, the deflection, bending moment, and stress of upper and lower beams are given in Figures13, 14, and 15 and Figures 16, 17, and 18, respectively.Figure 13
Surface changes of upper beam deflection with time and position.Figure 14
Surface changes of upper beam bending moment with time and position.Figure 15
Surface changes of upper beam stress with time and position.Figure 16
Surface changes of lower beam deflection with time and positionFigure 17
Surface changes of lower beam bending moment with time and position.Figure 18
Surface changes of lower beam stress with time and position.


## 5. Conclusion

Based on the double-layer elastic foundation beam theory, this paper establishes the dynamic equations considering the boundary conditions, where one end of beam is constrained and another end is free, and investigates the homogeneous solution for displacement of double-layer elastic cantilever foundation beam without loading. Introducing the Heaviside Function, we obtain the electromagnetic force function. Then the deflection curves equation and the expression of bending moment and stress for upper and lower beam under load are derived. We use the model to calculate a specific example.( 1 ) Through an instance, the analytical solution of deflection, stress, and bending moment of the upper and lower beams is obtained with given geometric parameters, motion parameters, and armature force. Using ANSYS, the paper obtains the numerical solution and simultaneously compares the analytic solution with the numerical solution of the dynamic response of the double-layer elastic foundation beams. The comparison shows that both solutions are not completely compatible, but the trends of analytical solution and numerical solution are basically in good agreement despite some acceptable numerical differences. The reliability of the analytical model is verified.( 2 ) The peak values of dynamic response including deflection, bending moment, and stress are very important in the design of electromagnetic railgun. In this paper, the relative error between the peak values of analytical solution and that of numerical solution is within a reasonable range. Therefore, the analytical model of double-layer elastic cantilever beam could be used to guide the actual design of electromagnetic gun.( 3 ) The oscillation of analytical solutions in this paper may be due to the lack of consideration of interlayer damping, which needs to be further improved in the future research.( 4 ) The derived analytical solution of the dynamic response of the double-layer elastic cantilever beam can improve the theoretical research on the electromagnetic launch device and can be used for reference of the engineering design in related fields.



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*Source: 2657271-2019-03-13.xml*