Symmetry analysis of an elastic beam with axial load
aa r X i v : . [ m a t h . A P ] J a n Symmetry analysis of an elastic beam with axial load
Bidisha Kundu ∗ , Ranjan GanguliDepartment of Aerospace Engineering, Indian Institute of Science, Bangalore 560012, [email protected], [email protected] ∗ Corresponding author
We construct the closed form solution of an elastic beam with axial load using Lie symmetrymethod. A beam with spatially varying physical properties such as mass and second moment of in-ertia is considered. The governing fourth order partial differential equation with variable coefficientswhich is not amenable to simple methods of solution, is solved using Lie symmetry. We incorporateboundary conditions and then compare with the numerical solution.
Keywords : Lie symmetry, elastic beam, closed-form solution
I. INTRODUCTION
An elastic beam is a three dimensional structure whose axial extension is more than any other dimension orthogonalto it. It is a fundamental model which pervades every corner of physics and engineering. However, the integrability infinite numbers of terms or exact solution of the governing equation of this model is an open question. The deflectionis dependent not only on the mass of the beam or external applied force but also on the elastic properties of thematerial and geometry of the beam. The equation of motion of the beam is a fourth-order linear partial differentialequation with variable coefficients. The equation can be derived from generalised Hooke’s law and force balance orby minimizing the energy of the system. Though the equation is linear in nature, due to the presence of variablecoefficients, it is very arduous to get the solution analytically. In [1], the closed form solution of vibrating axiallyloaded beam was studied with different boundary conditions. However, to the best of our knowledge, in this work, forthe first time, Euler-Bernoulli beam with axial force is studied using the Lie-method. Here we follow the Lie-groupmethod to get the similarity solution.The main motivation to use the Lie method is to get the solution of the mathematical model in different formswhich are very easy to use. First, Sophus Lie applied this method to partial differential equations and later thismethod was further developed by Ovsjannikov [2] and Matschat and M¨uller[3].Lie method has been applied to various types of ordinary differential equations (ODE) and partial differentialequations(PDE). A systematic approach for applying Lie method to ODE and PDE is found in Refs. [[4],[5],[6], [7]].Bluman et al. used this method for different types of mathematical physics problems such as wave equation, diffusionequation etc. [8, 9]. Torrisi et al. studied diffusion equation by equivalence transformation [10]. Ibragimov appliedthis method to some real life problems [11, 12] related to tumour growth model and also to metallurgical industrymathematical models. In [13], the procedure to calculate all point symmetries of linear and linearizable differentialequations was studied. An algorithm for integrating systems of two second-order ordinary differential equations withfour symmetries was given in [14]. The Lie symmetry method is also employed in ordinary difference equation [15].The Lie-method is also available in recent research addressing equations in mathematical physics [16–18]. In Ref.[16], Kang and Qu have studied the relationship between Lie point symmetry and fundamental solution for systems ofparabolic equations. The invariant analysis of space-time fractional nonlinear systems of partial differential equationsis performed using Lie symmetry method in Ref. [17]. Hau, et al. presented a unifying solution framework for thelinearized compressible equations for two-dimensional linearly sheared unbounded flows [18].The symmetry analysis for a physical problem is very important. In Lorentz transformation used in special relativity,Yang-Mills theory [19, 20], and Schr¨odinger equation, the Lie symmetry analysis has been used extensively. In Ref.[21], Lie symmetries and canonical transformations are applied to construct the explicit solutions of Schr¨odingerequation with a spatially inhomogeneous nonlinearity from those of the homogeneous nonlinear Schr¨odinger equation.In Ref. [22], symmetry analysis is used to understand the fluid flow in a pipe or channel structure.The Lie-method has also been employed in the field of mechanics [23]. In Ref. [24], a beam moving with time-dependent axial velocity is studied using equivalence transformation. Wafo discussed the Euler-Bernoulli (EB) beamfrom a symmetry stand-point [25]. The general beam equation is studied by Bokhari et al. for symmetries andintegrability with Lie method [26]. Bokhari et al also found the complete Lie symmetry classification of the fourth-order dynamic EB beam equation with load dependent on normal displacement. Johnpillai et al studied the EB beamequation from the Noether symmetry viewpoint [27].We search for one-parameter group of transformation which leaves the governing PDE invariant and we get thecorresponding Lie-algebras. Using the infinitesimal generator of this transformation, we solve the newly reduceddifferential equation. In Section 2, we describe our problem. The mathematical theories for the Lie-symmetryapproach, the procedure to get the invariant solution and application of the Lie method to our problem are describedin Section 3. The closed form solutions for different cases and boundary condition imposition are given in Section 4and Section 5, respectively.
II. FORMULATION
We fix the coordinate axes X , Y, Z along the length, breadth and height of the beam, respectively. We considerthat the beam is slender and of length l . Here, u ( x, t ) and M ( x, t ) are the out-plane bending displacement along Zand bending moment, respectively, at the point x and the instant t where x ∈ [0 , l ] and t ∈ R + . We assume thestiffness functions EI : [0 , l ] → R + and m : [0 , l ] → R + are continuously differentiable functions. Here R + is the setof all positive real numbers.The equation of motion of the Euler-Bernoulli beam with axial force is given by ∂ ∂x (cid:18) EI ( x ) ∂ u∂x (cid:19) + m ( x ) ∂ u∂t − ∂∂x (cid:18) T ( x ) ∂u∂x (cid:19) = 0 . (1)For rotating beam T ( x ) = R lx m ( x )Ω xdx where Ω is the rotating speed. Another important practical case is thegravity loaded beam. For this problem T ( x ) = R lx m ( x ) gdx where g is the gravitational force acting on the system.In the case of a stiff-string or piano string T is constant [28]. This formulation is valid for all tensile loads and forcompressive loads less than the critical buckling load. In general, for this type of beam, cantilever boundary conditionat the left end and free at the right end is considered. Here, a schematic diagram of an axially loaded beam (undertension), is shown in Figure (1). , u(x,t) T(x)
FIG. 1: Schematic diagram of an axially loaded beam (under tension).
III. THE LIE SYMMETRY METHOD
Eq. (1) is a linear PDE of order four in x and order two in t . The Lie symmetry method will help us to extract theclosed form solution via the symmetry analysis. We define the original Eq. (1) in geometric form.Consider X = { ( x, t ) | x, t ∈ R } as the two-dimensional manifold of the independent variables and U ⊂ R as the onedimensional manifold of dependent variable. Equation (1) is defined on the manifold M ⊂ X × U of dimension three.In order to establish a concrete geometric structure of the differential equation, we have to construct a manifold J of dimension p which includes the independent variables, dependent variables, and all possible partial derivatives ofthe dependent variable with respect to independent variables up to order of the given differential equation. Here thedimension p depends on the dimension of X, U as well as on the order of the differential equation given. We “prolong”the original manifold M ⊂ X × U to J which also has manifold structure.The main motivation to study a differential equation with Lie symmetry method is to find a favourable coordinatetransformation which can produce another equivalent form of the given equation. In the new coordinate systems, thenew equation may have closed form solution. To get the favourable coordinate transformations, we should take careof the derivatives of the dependent variables which is related to the vector fields of M . We also want to measure thecontribution of the vector fields related to the given differential equation in the prolonged space J which is called theprolongation of the vector field.As this is a PDE of order four, we need the prolongation of order four, i.e. pr (4) v . For a PDE with one dependentvariable and two independent variables, the one-parameter Lie group of transformations is x ∗ = X ( x, t, u ; ǫ ) = x + ǫξ ( x, t, u ) + O ( ǫ ) t ∗ = T ( x, t, u ; ǫ ) = t + ǫτ ( x, t, u ) + O ( ǫ ) u ∗ = U ( x, t, u ; ǫ ) = u + ǫη ( x, t, u ) + O ( ǫ ) (2)The prolongation of order four pr (4) v is given by pr (4) v = ξ ∂∂x + τ ∂∂t + η ∂∂u + η x ∂∂u x + η t ∂∂u t + η xx ∂∂u xx + ...η xxt ∂∂u xxt + ... + η ttt ∂∂u ttt + η xxxx ∂∂u xxxx + η xxxt ∂∂u xxxt + ... + η tttt ∂∂u tttt (3)From the Fundamental theorem [Theorem 2 .
31, [4]] of transformations (2) admitted by the PDE (1), applying pr (4) v on Eq.(1) yields ξ ( x, t, u ( x, t )) (cid:16) EI (3) ( x ) u xx ( x, t ) + 2 EI ′′ ( x ) u xxx ( x, t )+ EI ′ ( x ) u xxxx ( x, t ) + m ′ ( x ) u tt ( x, t ) − T ′′ ( x ) u x ( x, t ) − T ′ ( x ) u xx ( x, t )) + η xx EI ′′ ( x ) + 2 η xxx EI ′ ( x )+ η xxxx EI ( x ) + η tt m ( x ) − η x T ′ ( x ) − η xx T ( x ) = 0 (4)Eq. (4) should be satisfied to yield the admissible transformations. The value of u xxxx ( x, t ) from Eq. (1) is substitutedinto the above Eq. (4). Now substituting the values of η x , η xx , η tt , η xxx , η xxxx from Theorem 32.3.5 [29], Eq. (4) issimplified and a polynomial equation is formed in all possible derivatives u i,j of u upto order four with respect to x , t where u i,j = ∂ i + j u∂x i ∂t j (5)This polynomial equation should be satisfied for arbitrary x , t and u i,j which requires the coefficients of all u i,j andall its product terms to be equal to zero. The determining equations for ξ, τ, η are ∂τ∂x = 0 (6) ∂τ∂u = 0 (7) ∂ξ∂t = 0 (8) ∂ξ∂u = 0 (9) m ( x ) ∂ η∂t − T ′ ( x ) ∂η∂x − T ( x ) ∂ η∂x + EI ′′ ( x ) ∂ η∂x + 2 EI ′ ( x ) ∂ η∂x + EI ( x ) ∂ η∂x = 0 (10) − ξ ( x, t, u )( EI ′ ( x )) EI ( x ) + 2 ξ ( x, t, u ) EI ′′ ( x ) + 2 EI ′ ( x ) ∂ξ∂x + 4 EI ( x ) ∂ η∂x∂u − EI ( x ) ∂ ξ∂x = 0 (11) T ( x ) ξ ( x, t, u ) EI ′ ( x ) EI ( x ) − ξ ( x, t, u ) T ′ ( x ) − ξ ( x, t, u ) EI ′ ( x ) EI ′′ ( x ) EI ( x ) + ξ ( x, t, u ) EI ′′′ ( x ) − T ( x ) ∂ξ∂x + 2 EI ′′ ( x ) ∂ξ∂x + 6 EI ′ ( x ) ∂ η∂x∂u + 6 EI ′ ( x ) ∂ ξ∂x + 6 EI ( x ) ∂ η∂x ∂u + 4 EI ( x ) ∂ ξ∂x = 0 (12) T ′ ( x ) ξ ( x, t, u ) EI ′ ( x ) EI ( x ) − T ′′ ( x ) ξ ( x, t, u ) − m ( x ) ∂ ξ∂t − T ′ ( x ) ∂ξ∂x − T ( x ) ∂ η∂x∂u + 2 EI ′′ ( x ) ∂ η∂x∂u + T ( x ) ∂ ξ∂x − EI ( x ) ∂ ξ∂x + 6 EI ′ ( x ) ∂ η∂x ∂u − EI ′ ( x ) ∂ ξ∂x + 4 EI ( x ) ∂ η∂x ∂u + EI ( x ) ∂ ξ∂x = 0 (13) − m ( x ) ξ ( x, t, u ) EI ′ ( x ) EI ( x ) + ξ ( x, t, u ) m ′ ( x ) − m ( x ) ∂τ∂t + 4 ∂ξ∂x = 0 (14)2 m ( x ) ∂ η∂t∂u − m ( x ) ∂ τ∂t + T ′ ( x ) ∂τ∂x + T ( x ) ∂ τ∂x − EI ′′ ( x ) ∂ τ∂x − EI ′ ( x ) ∂ τ∂x − EI ( x ) ∂ τ∂x = 0 (15) ∂ η∂u = 0 (16) From Eqs.(6, 7) τ ( x, t, u ) = τ ( t ), from Eqs. (8, 9) ξ ( x, t, u ) = ξ ( x ) and from Eq. (16) η ( x, t, u ) = A ( x, t ) u + B ( x, t )for some arbitrary functions A ( x, t ), B ( x, t ). We assume ∂B∂x = 0 and ∂ B∂t = 0, i.e., B ( x, t ) = d + d t where d , d are constants. From the Eq.(11), we observe that ∂ η∂x∂u = ∂A∂x should be free from time variable t . Hence, A ( x, t ) = f ( x ) + f ( t ).Now from Eq. (15), we see that 2 m ( x ) ∂ η∂t∂u − m ( x ) ∂ τ∂t = 0 (17)Again from Eq. (14), ∂τ∂t should be a constant. Assume, 2 ∂τ∂t = ω which implies τ ( t ) = ω t + t . As from Eq. (17),2 f ( t ) = ∂τ∂t = ω f ( t ) = ω and η ( x, t, u ) = ( f ( x ) + ω ) u + d + d t .Now we consider two cases; ( a ) when f ( x ) is constant, ( b ) when f ( x ) is not constant. For the first case, there aretwo subcases; ( a.
1) when d ξdx = k for some constant k i.e., d ξdx = 0 , d ξdx = 0,( a.
2) when d ξdx is not a constant. All thecombinations of the stiffness, mass, axial force, and transformations are given in Table I. Case ( a ) when f ( x ) is constant. − For the case ( a. ξ ( x ) = f ( EI ( x )) (19)and for the assumption d ξdx = k , EI ( x ) = k x f . For the case ( a. d ξdx is not a constant. Here, we assume thecoefficients of ξ ( x ) are zero in the Eqs. (11), (12), (13) and the form of EI ( x ) , m ( x ), and T ( x ) are evaluated. Case ( b ) when f ( x ) is not a constant. − In this case also, if we assume the coefficients of ξ ( x ) are zero in the Eqs.(11), (12), (13) and the form of EI ( x ) , m ( x ) and T ( x ) are evaluated.Based on the assumptions, there may be more than these combinations of physical properties and coordinatetransformations which can lead us to find a closed form solution. In Table I, the possible list of these combinationsare listed. It is observed that the transformation rule for the spatial coordinate is strongly dependent on the stiffnessof the beam, i.e., the geometry of beam of certain material. It can be shown that for a beam of polynomial varyingstiffness EI ( x ) = ( a + a x ) n where a = 0 any real number and n > G ( x ) is G ( x ) = ( a + a x ) − n − √ a A √ n − a + a x ) − √ a n − n − T a / √ n − + n +3 ! a / ( n − √ n − p a ( n − n − + 4 T − √ a A √ n − a + a x ) √ a n − n − T a / √ n − + n +3 ! a / ( n − √ n − − p a ( n − n − + 4 T − A ( a + a x ) a ( n − (20) IV. CLOSED FORM SOLUTION
For the case ( a. v = kx ∂∂x + ( ω t + t ) ∂∂t + (( α + ω u + d + d t ) ∂∂u (21)If we choose, ( α + ω ) = k , then for vector field X = x ∂∂x + u ∂∂u the characteristic equation is dx x = dt duu (22) Case Physical Properties Transformations u ( x, t )( a. EI ( x ) = k x f ξ ( x ) = kx e − /x (cid:0) A e λt + A e − λt (cid:1) m ( x ) = m e − c kx x τ ( t ) = ω t + t T ( x ) = T x − k x f η = ( α + ω ) u + d + d t ( a. EI ( x ) = a e a x ξ ( x ) = f e a x a e xr ( A + A t ) m ( x ) = m e a − c e − a x a f − x τ ( t ) = ω t + t T ( x ) = − a a e a x η = ( α + ω ) u + d + d t ( b ) EI ( x ) = a e − vx ξ ( x ) = e vx v e vx ( A + A t ) m ( x ) = m e v ( − c e − vx − x ) τ ( t ) = ω t + t T ( x ) = 2 a v e − vx η = (cid:16) e vx v + ω (cid:17) u + d + d t ( c ) EI ( x ) = ( a + a x ) n ξ ( x ) = ( a + a x ) a n t + G ( x ) m ( x ) = m ( a + a x ) f n − nωf τ ( t ) = ω t + t T ( x ) = T ( a + a x ) n − a ( n − η = ω u + d + d t TABLE I: The transformations and solutions for the corresponding physical properties. Here λ = − √ √ f T − k f / √ m and G ( x ) isgiven in (20). which implies u ( x, t ) = exp (cid:18) − x (cid:19) F ( t ) (23)for some arbitrary function F ( t ). Now substituting this u ( x, t ) in the original equation Eq. (1) for EI ( x ) = k x f , T ( x ) = T x − k x f and m ( x ) = m e − c kx x , a second order ordinary differential equation in F ( t ) is found. Solvingthis equation for F ( t ), the given u ( x, t ) in Table I is found. Following the similar way, we get the invariant solutionsfor other cases given in the Table I. V. BOUNDARY CONDITION IMPOSITION
The symmetry analysis of a boundary value problem (BVP) may not always be successful [30]. For a BVP, thedomain is fixed and the symmetry of a BVP requires not only the invariance of the given differential equation, butalso the invariance of the boundary data. There is no available systematic procedure for the symmetry analysis of aBVP [31]. However, in this paper, the invariant solution is found by weakening the conditions given by Bluman andKumei in section 4 . . u (0 , t ) = 0 , ∂u∂x (0 , t ) = 0 , ∂ u∂x (1 , t ) = 0 , ∂ u∂x (1 , t ) = 0 , and the initial conditions are u ( x,
0) = h ( x ), ∂u∂t ( x,
0) = 0. Satisfying all the boundary condition for the deflection, i.e. for u , of an elastic beam which is fixed atthe left end and free at the right end, the feasible stiffness, mass, and the axial load are given by; EI ( x ) = g x (6 m + x (3 m + 2 m x )) ( m + x ( m + m x )) (24) m ( x ) = m + m x + m x (25) T ( x ) = g x (6 m + x (3 m + 2 m x )) ( m + x ( m + m x )) (cid:0) − m + 8 m x (17 m x − m )+12 m x (cid:0) − m + 8 m m x + 2 m x (cid:1) + 2 m x (cid:0) − m + 9 m m x + 24 m m x + 12 m x (cid:1) − m x (cid:0) m + 14 m m x + 6 m x (cid:1)(cid:1) (26) ÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷ ì ì ì ì ì ì ì ì ì ì ì ì ì ì ì ì ì ì ì ì ì ÷ Numerical ì Analytical x D e f l ec t i o n Comparison of deflections at t = ÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷ ììììììììììììììììììììììììììììììììììììììììì ÷ Numerical ì Analytical - - t D e f l ec t i o n Comparison of deflections at tip
FIG. 2: Comparison of numerical and analytical solutions and the transformations are; ξ ( x ) = m x + m x + m x m + x ( m + m x ) (27) τ ( t ) = c (28) η = ωu + d + d t (29)where m , m are dependent on m and the general expressions are m = q √ m + 9 m / √ q m q √ m + 9 m − m m = m (5 √ m + 195 m + 8 √ (cid:0) (cid:0) √ (cid:1)(cid:1) / (cid:0) m (cid:1) / m + m √ m +9 m + √ / m ( √ ) / ( m ) / −
120 3 / q √ m √ m −
24 5 / q (cid:0) √ (cid:1) p m m )We choose ω = 2, g = 1, g = , m = 1. Then m = − . , m = 0 . ξ (0) = 0 but it is notpossible to impose ξ (1) = 0 which provides non-feasible physical properties. The corresponding analytical solution is; u ( x, t ) = (cid:18) m x + m x m x (cid:19) (cid:16) A cos (cid:16) √ √ g t (cid:17) + A sin (cid:16) √ √ g t (cid:17)(cid:17) (30)For zero initial velocity, A = 0. Considering A = 1, u ( x, t ) = (cid:18) m x + m x m x (cid:19) (cid:16) cos (cid:16) √ t (cid:17) + sin (cid:16) √ t (cid:17)(cid:17) (31)The comparison between the analytical form of the solution is matching well with the numerical solution given in Fig.2. For the numerical solution, NDSolve command in M athematica . h ( x ) = (cid:16) m x + m x + m x (cid:17) . VI. CONCLUSION
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