An Explicit Stiffness Matrix for Parabolic Beam Element 1

This study is devoted to strain-based formulation for a curved beam. Arches with parabolic geometry, which have a variety of applications, belong to this structural type. Dependency of the curvature radius to the arch length creates some complexities in the solution process. To analyze these complex structures, a twonode beam with six degrees of freedom is suggested by utilizing closed-form solution and the stiffness-based finite element method. Considering the effect of shear deformation, and incorporating equilibrium conditions into the finite element model, lead to the exact strains. Displacements and explicit stiffness matrix are found based on these exact strains. To validate the efficiency of the author's formulation, seven numerical tests are performed. The outcomes demonstrate that by employing only a single element, the locking-free answers can be found.

Latin American Journal of Solids and Structures 13 (2016) 1782-1801 Raveendranath and his colleagues (1999) assumed cubic polynomial for radial displacement.By taking advantage of the equilibrium equations, they suggested new displacement functions Raveendranath et al. (2001).Similarly, this procedure was extended to create a three-nodded element.Furthermore, it was stated that the consistent-field approach can find the sources of highstiffening errors.Afterward, many formulations were presented based on this effective technique Babu and Prathap (1995) Prathap and Naganarayana (1990).
All mentioned investigators studied about curved beams with the constant radius of curvature.In addition, few papers have focused on the variable-curvature structures.Marquis and Wang (1989) used the energy principles to solve parabolic arches.It is worth emphasizing that most of the proposed schemes have not offered a general finite element model.In fact, these solution techniques have only considered a few particular cases that were more reachable.These investigators calculated the stiffness matrix by considering the effect of special boundary conditions Gutierrez et al. (1989) Lin and Huang (2007) Lin and Hsieh (2007) Lee and Wilson (1989) Lee et al. (2008) Tarnopolskaya et al. (1996).Haung et al. (1998) utilized polynomial functions and power series to model the behavior of beams with variable curvatures and cross sections.In 1999, Oh et al. (1999) solved equilibrium equations numerically, and found the first four natural frequencies of sinusoidal, elliptical and parabolic beams for the special cases.This procedure was used by many researchers Huang et al. (1998) Oh et al. (2000) Gimena et al. (2010).Another way of finding the structural stiffness matrix is the flexibility-based method.Litewka and Rakowski (1998), Molari andUbertini (2006), andAttarnejad et al. (2013) utilized this approach in their study.Attarnejad and his coworkers (2013) defined Basic Displacement Function (BDF) as the nodal displacement by applying unit load technique.On the other hand, Molar & Ubertini (2006) employed a parametric cubic interpolation to model geometry of the structure.These investigators considered two parameters for the versatility of interpolation function.
In this research, a scheme for analyzing general curved beam with the variable curvature is suggested.Parabolic shapes are widely used in the group of arches, for their simple second-order polynomial form.By employing the finite element method and closed-form solution, a novel parabolic beam element is proposed.Choosing required interpolation function and satisfying the equilibrium conditions have the most significant role in this procedure.In fact, by employing just one element per member, precise responses are found.This is due to the exact strains that obtained by solving differential equations.Moreover, an explicit stiffness matrix is obtained, which can be used in the structural analysis.It should be noted that all shear, axial and bending effects are considered, simultaneously.The results of seven numerical tests show no locking error in the answers.All entries of the suggested stiffness matrix are explicitly given in the appendix.

EQUILIBRIUM EQUATIONS
According to the classical Timoshenko beam theory, normals to the axis of the beam remain straight after deformation.However, they are not required to remain perpendicular.This assumption is used in the present article to develop a curved beam element with variable curvature.As it is shown in Figure 1, the structure has two nodes and six degrees of freedom.Each node contains three displacements and three forces.Throughout this study, u, v and θ refer to axial, radial and rotational displacements, respectively.The width and thickness of the rectangular cross section are Latin American Journal of Solids and Structures 13 (2016) 1782-1801 assumed as w and t.It should be noted that the geometry of suggested element is based on the second-order polynomial function.Using the equations 2 . .
, the radius of curvature function, R, can be achieved.In these formulas, φ denotes the angle of tangential slope at a general point.
Strain functions for the neutral axis have the following form: (2) By integrating the stresses over the thickness of cross section, the compliance material matrix can be found.If the effect of thickness in strain equations is negligible, the following simple and approximate material matrix will be obtained: Approximate material matrix is based on the assumption of t/R<<1.Furthermore, the first three terms of Taylor's series are utilized for and the membrane-bending interaction is eliminated.With these assumptions and ε σ , the next equality is held: Latin American Journal of Solids and Structures 13 (2016) 1782-1801 The Young's modulus, shear modulus, area of the cross-section, bending moment of inertia about the neutral axis, and a shear correction factor are indicated as E, G, A, I, k, respectively.A set of equilibrium equations can be achieved by optimization of the following total potential energy: In these equations, subscript s demonstrates the differentiating with respect to the longitudinal axis s.The related answers for the set of equilibrium equations can be written in the below form: The unknowns 1 C ˆand 2 C ˆ are expressed as: 3 DISPLACEMENT FUNCTIONS By assuming t/R<<1, the membrane-bending interaction is omitted.Consequently, internal forces cause the next strains in the neutral axis. . .
Based on equation (2), the following equalities can be achieved: Latin American Journal of Solids and Structures 13 (2016) 1782-1801 Then, the coming displacement fields can be found by solving the equilibrium equations: Furthermore, the vector of nodal unknowns is obtained as bellow: Finally, the next strain and displacement interpolation functions can be derived: Latin American Journal of Solids and Structures 13 (2016) 1782-1801 Latin American Journal of Solids and Structures 13 (2016) 1782-1801

FINITE ELEMENT FORMULATION
If φ is inserted in the equation ( 16), the vector of the nodal displacement will be found.It should be noted, φ for the first node is zero while for the second one is unknown.The structural geometry is utilized to find the amount of φ .
At first, the displacement and strain interpolation functions are determined according to the vector of nodal displacements.Then, the following shape functions become available: Total potential energy can be written in terms of the strain interpolation function: By optimization Π, a general finite element formulation is obtained for each member: Calculating the exact integration over the arch length leads to the precise elemental stiffness matrix.It should be added that general form of the stiffness matrix entries, ij S , are based on the approximation of the compliance material matrix.All of these entries are explicitly given in the appendix.

NUMERICAL STUDIES
To verify the performance of the suggested element, some structures will be examined numerically.It should be mentioned that the effect of shear deformation is taken into account in the presented examples.Hence, in all tests, k is assumed 5/6 for rectangular cross section.In the following, the results of these tests are compared with the exact answers of the Castigliano's theory, which are characterized by the subscript C. Structures 13 (2016) 1782-1801 5.1 A Two-End Fixed Beam A beam with a radial load at its middle point and two fixed supports is shown in Figure 2. Elasticity modulus of Young, thickness and radius of curvature are 30,000( k/in 2 ), 6( in) and 1200 (in), respectively.Also, the tangent angle of the beam end is α =11.31 ° .Figure 2, illustrates the geometry of structure.Due to symmetry, this arch can be modeled with a single element.Marquis and Wang (1989) analyzed parabolic beams by taking advantage of potential energy principles.In this section, the answers of the proposed method can be compared with the responses of their study.It should be noted that they ignored the shear effect to solve this structure.The obtained results are available in Table 1.According to the results, there is no locking error in the answers of the suggested element.

Verifying Responses
In this section, the displacements of parabolic beams are calculated and compared with the available solution.Some cases of loading and slenderness ratios are considered in this study.Two boundary conditions are employed for this test.In addition, a nodal load is applied at the middle point.The structure has different slenderness ratios and is carrying radial, tangential and rotational loads, separately.The main aim of this study is to evaluate the new element efficiency, when it is used for thin and thick members.Constant radius of the curvature and variable thickness lead to the different slenderness ratios.Load's magnitude is assumed to be unit and the radius of curvature at φ=0 is presumed to be 25.Tangent angles of both beam ends, Young's modulus and shear correction factor in all cases are 38.66 °, 10.5E+06 and 4.0E+06, respectively.The results of this analysis and exact answers are inserted in Results of this study demonstrate the ability of the proposed element in modeling of the thin and thick structures.In fact, all tests with distinctive features give the precise answers.The outcomes indicate the extensive performance of the novel element.

A Two-End Hinged Beam
After examination of the displacement's quality, it is better to verify the internal forces, since the suggested element is formulated based on the equilibrium equations.For this purpose, the beam with mechanical and geometrical properties similar to the previous test is utilized.Figure 3    According to the Figures 4, 5 and 6, by employing the proposed method, all internal forces are precise.Based on this benchmark's outcomes, the accuracy of the suggested element in evaluating the structural internal forces is concluded.Structures 13 (2016) 1782-1801

Cantilever Parabolic Beam
Figure 7 shows a cantilever parabolic beam, with 25 R 0  , which is tested in this section.Mechani- cal properties of this structure are similar to the structure in section 5.2.Radial and tangential displacements of this beam are obtained for the radial concentrated unit load at the free-end.Tangent angle of the beam's end is equal to α =45 °.Table 3 illustrates all displacements of this structure versus the slenderness ratio.The answers are compared with the obtained results, for the composite curved beams Lin and Hsieh (2007).By assuming the equal transverse and longitudinal modulus, the anisotropic material reduces to an isotropic one, and in this case, the answers are comparable.For this cantilever parabolic beam, tangential and radial displacements are calculated from equation ( 27).

Parabolic Beam in Pure Bending
The geometry of a parabolic beam, with 0 R 25 = and a couple of moments at two ends, is shown in Figure 8.Both end moments are assumed to be unit.Mechanical properties in this benchmark are considered to be the same as one in section 5.2.It is clear that this load produces pure-bending behavior in the structure.By utilizing the symmetrical characteristics, the parabolic beam can be replaced by a cantilever beam having a moment at its free end.For a constant slenderness ra-Latin American Journal of Solids and Structures 13 (2016) 1782-1801 , the radial displacement is achieved for variable α.Table 4 demonstrates the results of proposed method, along with the one obtained by Lin and Hsieh (2007).Table 4: Radial displacement at the middle point of a parabolic beam in pure bending.

Sinusoidal Loading
The last example is a parabolic beam under a distributed load.Figure 13  case and the second one are pinned-pinned and pinned-roller, respectively.Two pinned supports make the beam statically indeterminate, but a pinned-roller beam is a determinate structure.A sinusoidal distributed load is applied to the beam with a maximum amount of a unit at the middle point.It should be noted, the geometry, cross section properties and mechanical characteristics are similar to the upper part of the structure shown in Figure 9.In this solution, only the bending effect is taken into account, due to negligibility of the axial and shear effects.Obtained results of the displacements with different meshes are shown in Table 6.By utilizing only one element, the answers for this complex loading are precise with zero errors.This is the result of using the exact solution and also satisfying the equilibrium equations. (1) (2) Displacements of node-3 in Fig. 13-2 Displacements of node-2 in Fig. 13

Figure 1 :
Figure 1: Geometry of a parabolic beam.

Figure 2 :
Figure 2: Geometry of a two-end fixed parabolic beam Marquis and Wang (1989).

Figure 3 :
Figure 3: A two-end hinged parabolic beam with a discontinuity in bending moment.

Figure 4 :Figure 5 :
Figure 4: Distribution of bending moment in the two-end hinged parabolic beam.

Figure 6 :
Figure 6: Distribution of shear force in the two-end hinged parabolic beam.

Figure 9 .
This arch is modeled with three elements.Element 1 both elements are the same.The parabolic vertex is carrying a moment of z m 2 = .All of mechanical properties are considered to be similar to the one in section 5.2.Table5gives the middle-point displacements.Figures 10, 11 and 12 illustrate internal forces versus horizontal distance from the center line.These distances are calculated by employing equation (29).

Figure 9 :
Figure 9: Geometry of an arch structure.
Middle-point displacements in arch structure.

Figure 10 :Figure 11 :
Figure 10: Distribution of the bending-moment in arch structure.

Figure 12 :
Figure 12: Distribution of the shear force in arch structure.
depicts two beams with different boundary conditions and identical loading and geometry.Boundary conditions of the first Latin American Journal of Solids and Structures 13 (2016) 1782-1801

Table 2 :
Load point displacements in parabolic beams with different conditions.

Table 3 :
Load point displacements in cantilever parabolic beam.

Table 6 :
Displacements of the beam with distributed load.