# Exact Static Analysis of In-Plane Curved Timoshenko Beams with Strong Nonlinear Boundary Conditions

Authors: Sen-Yung Lee; Qian-Zhi Yan
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

Analytical solutions have been developed for nonlinear boundary problems. In this paper, the shifting function method is applied to develop the static deflection of in-plane curved Timoshenko beams with nonlinear boundary conditions. Three coupled governing differential equations are derived via the Hamilton’s principle. The mathematical modeling of the curved beam system can be decomposed into a complete sixth-order ordinary differential characteristic equation and the associated boundary conditions. It is shown that the proposed method is valid and performs well for problems with strong nonlinearity.


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## Body

## 1. Introduction

Curved beam structures are widely used in engineering fields, such as mechanical, civil, and aerospace engineering. Reviews of research on such structures have been conducted by Henrych [1], Markus and Nanasi [2], Chidamparam and Leissa [3], and Auciello and De Rosa [4]. The in-plane and out-of-plane problems of plane curved beams have been studied. In most general problems, they are coupled. However, if the cross section of the curved beam is doubly symmetric and the plane is a principal plane of the cross section, then the in-plane and out-of-plane problems are uncoupled.Many investigators have studied linear problems about the static and free vibration behaviors of curved beams. Washizu [5] obtained the equations for the coupled motions of a curved and pretwisted bar. Rao [6] used Hamilton’s principle to derive the governing equations of motion in consideration of the effects of rotatory inertia and shearing deformation. Fettahlioglu and Mayers [7] studied the static deflection of a ring. Bickford and Maganty [8] used a formulation similar to that of Rao and developed equations of motion for out-of-plane vibrations of symmetrical cross-section thick rings, accounting for curvature variation through the thickness. Their frequency predictions were validated with the experimental data of Kuhl [9]. Based on the moment-displacement relationships presented by Rao, Silva and Urgueira [10] derived the dynamic stiffness matrices for the out-of-plane vibration of curved beams using dynamic equilibrium equations. Lee and Chao [11] developed the exact out-of-plane vibration solutions of curved nonuniform beams. In the book by Cook and Young [12], the exact static analysis of extensional circular curved Timoshenko beams with some special conditions was revealed. Based on the generalized Green function, Lin [13] developed the exact solution of extensible curved Timoshenko beams. Lee and Wu [14] studied the exact in-plane vibration solutions of extensible curved nonuniform Timoshenko beams.For the beams with time dependent boundary conditions, Lee and Lin [15] generalized the solution method of Mindlin and Goodman [16] and developed the shifting function method to study the problems with general time dependent elastic boundary conditions. Recently, Lee et al. [17, 18] extend the shifting function method to study the exact large deflection of a Bernoulli-Euler beam and Timoshenko beam with nonlinear boundary conditions. From the existing literatures, it shows that exact solutions for curved beam problems with nonlinear boundary conditions are not available.In the previous studies [17, 18], the governing differential equations are fourth-order differential equations. In the present study, one extends the previous studies and the shifting function method [15] to study the exact large static deflection of in-plane curved Timoshenko beams with nonlinear boundaries. The beam with doubly symmetric cross section is considered. The three coupled governing differential equations for the in-plane curved uniform beams of constant radius are derived via Hamilton’s principle. These three coupled governing differential equations are decoupled and reduced into a sixth-order differential equation with nonlinear boundary conditions. Consequently, the shifting function method is extended and applied to develop the exact solution of the system. It can be shown that the proposed method is valid for problems with strong nonlinearity.


## 2. Mathematical Modeling of the Curved Beam System

Consider the static response of an extensional curved uniform Timoshenko beam resting on an elastic foundation with nonlinear boundary conditions, subjected to loadsPv(s), Pu(s), and Mz(s), as shown in Figure 1. If the thickness of the curved beam is small in comparison with the radius of the curved beam without considering the warping effects, the displacement fields of the curved beam in cylindrical coordinates are(1)urr,s,z=vs,uθr,s,z=us-r-ψzs,uzr,s,z=0,where ur, uθ, and uz denote the displacement of the curved beam in the r, θ, and z directions, respectively, and s is the arc length along the neutral axis. R is the constant radius of the curved beam and s=Rθ, v and u are the neutral axis displacements of the curved beam in the r and θ directions, respectively, and ψz is the angle of rotation due to bending in the z direction. r- is measured inward from the neutral axis in the r direction.Figure 1
Geometry and coordinate system of uniform curved beam with nonlinear boundary conditions, subjected to loadsPv(s), Pu(s), and Mz(s) in r, θ, and z directions, respectively.Substituting (1) into the strain-displacement relations in cylindrical coordinates, only two nonzero strains, namely, εθθ and γrθ, are obtained:(2)εθθ=RR-r-∂u∂s-vR-r-∂ψz∂s,γrθ=-RR-r-∂v∂s+uR-ψz.Whenr- is small in comparison with R, the two strains reduce to(3)εθθ=∂u∂s-vR-r-∂ψz∂s,γrθ=-∂v∂s+uR-ψz.The strain energy of the curved beam is(4)V∗=12∫0L∫AE∂u∂s-vR2-2r-∂ψz∂s∂u∂s-vR+r2∂ψz∂s2dAds+12∫0L∫AkzG∂v∂s+uR-ψz2dAds,where L is the length of the neural axis. E, G, and A denote Young’s modulus, shear modulus, and cross-section area of the curved beam, respectively, and kz denotes the shear correction factors of the curved beam section about the z-axes. Since the cross section of the curved beam considered is doubly symmetric, the integral of the second term in the square brackets vanishes. Equation (4) becomes(5)V∗=12∫0LEA∂u∂s-vR2+EIz∂ψz∂s2ds+12∫0LkzGA∂v∂s+uR-ψz2ds,where Iz denotes the second area moments of inertia of the curved beam section about the z-axes.Considering the linear and nonlinear spring and moment on the boundary of the curved beam end and the force and moment loads on the curved beam, the potential energy and work are, respectively,(6)U∗=∫KVLv0+KNVLv30dv0+∫KVRvL+KNVRv3LdvL+∫KULu0+KNULu30du0+∫KURuL+KNURu3LduL+∫KθLψz0+KNθLψz30dψz0+∫KθRψzL+KNθRψz3LdψzL,(7)W∗=∫0LPvsvs+Pusus+Mzsψzsds+f1v0+f2u0+f3ψz0+f4vL+f5uL+f6ψzL,where KVL, KUL, KθL, KVR, KUR, and KθR are the linear spring and rotational stiffness constants in the r, θ, and z directions at the left and right ends of the curved beam, respectively. KNVL, KNUL, KNθL, KNVR, KNUR, and KNθR are the nonlinear spring and rotational stiffness constants in the r, θ, and z directions at the left and right ends of the curved beam, respectively. Pv(s), Pu(s), and Mz(s) are the force and moment loads in the r, θ, and z directions, respectively, and f1, f2, f3, f4, f5, and f6 are the force and moment loads in the r, θ, and z directions at the left and right ends of the curved beam, respectively.The general form of Hamilton’s principle is(8)δ∫t1t2-V∗-U∗+W∗dt=0.Based on Hamilton’s principle, the governing differential equations and the associated boundary conditions for the curved uniform Timoshenko beam can be derived. The governing differential equations for the in-plane are three coupled differential equations:(9)-kzGA∂2v∂s2+1R∂u∂s-∂ψz∂s-EAR∂u∂s-vR=Pv,-EA∂2u∂s2-1R∂v∂s+kzGAR∂v∂s+uR-ψz=Pu,-EIz∂2ψz∂s2-kzGA∂v∂s+uR-ψz=Mz.The associated boundary conditions are as follows:(i)
Ats=0(10)-KVLv-KNVLv3+kzGA∂v∂s+uR-ψz=-f1,-KULu-KNULu3+EA∂u∂s-vR=-f2,-KθLψz-KNθLψz3+EIz∂ψz∂s=-f3.(ii)
Ats=L(11)KVRv+KNVRv3+kzGA∂v∂s+uR-ψz=f4,KURu+KNURu3+EA∂u∂s-vR=f5,KθRψz+KNθRψz3+EIz∂ψz∂s=f6.In terms of the following nondimensional quantities,(12)Vξ=vL,Uξ=uL,Ψξ=ψz,P-vξ=PvL3EIz,P-uξ=PuL3EIz,M-zξ=MzL2EIz,f1∗=f1L2EIz,f2∗=f2L2EIz,f3∗=f3LEIz,f4∗=f4L2EIz,f5∗=f5L2EIz,f6∗=f6LEIz,βVL=KVLL3EIz,βNVL=KNVLL5EIz,βUL=KULL3EIz,βNUL=KNULL5EIz,βθL=KθLLEIz,βNθL=KNθLLEIz,βVR=KVRL3EIz,βNVR=KNVRL5EIz,βUR=KURL3EIz,βNUR=KNURL5EIz,βθR=KθRLEIz,βNθR=KNθRLEIz,θ0=LR,μ=EIzkzGAL2,ς=AL2Iz,ξ=sL,the nondimensional coupled governing characteristic differential equations are(13)-1μ∂2V∂ξ2+θ0∂U∂ξ-∂Ψ∂ξ-ςθ0∂U∂ξ-θ0V=P-v,(14)-ς∂2U∂ξ2-θ0∂V∂ξ+θ0μ∂V∂ξ+θ0U-Ψ=P-u,(15)-∂2Ψ∂ξ2-1μ∂V∂ξ+θ0U-Ψ=M-z.The associated boundary conditions are as follows:(i)
Atξ=0(16)-βVLV-βNVLV3+1μ∂V∂ξ+θ0U-Ψ=-f1∗,(17)-βULU-βNULU3+ς∂U∂ξ-θ0V=-f2∗,(18)-βθLΨ-βNθLΨ3+∂Ψ∂ξ=-f3∗.(ii)
Atξ=1(19)βVRV+βNVRV3+1μ∂V∂ξ+θ0U-Ψ=f4∗,(20)βURU+βNURU3+ς∂U∂ξ-θ0V=f5∗,(21)βθRΨ+βNθRΨ3+∂Ψ∂ξ=f6∗.The coupled differential equations (13)–(15) can be decoupled to get two equations, Ψ and V:(22)Ψ=1θ0δ1δ2U4+δ3U2+δ1θ02U+1ς-μθ021ς+μP-u+δ2δ2θ0P-v1+1ςP-u2+θ0M-z,(23)V=-1ςθ03δ1U5-1ςθ0δ1U3+1θ0U1+1ςθ02P-v-δ2ςθ02δ1P-v2+δ2ςθ0δ1P-u1-1ςθ03δ11ςP-u3+θ0M-z1,where the notation(n) denotes the nth-order differentiation with respect to ξ. Consider (24)δ1=1+θ021ς+μ,δ2=1ς+μ,δ3=1+2θ021ς+μ.Substituting (22)–(24) into (15) and (16)–(21) yields the complete sixth-order ordinary differential characteristic equation and the associated boundary conditions in the θ direction, respectively. Consider(25)U6+2θ02U4+θ04U2=θ0P-v1-δ2θ0P-v3-θ02P-u+μθ02P-u2-1ςP-u4-θ03M-z-θ0M-z2.(i)
Atξ=0(26)βVLςθ03δ1U5-1θ0δ1U4+βVLςθ0δ1U3-θ0δ1U2-βVLθ0·U1-βNVLV3=βVL1ςθ02P-v-δ2ςθ02δ1P-v2+δ2ςθ0δ1P-u1-1ςθ03δ11ςP-u3+θ0M-z1+1δ1δ2P-v1+1ςθ0P-u2-δ2θ0P-u-δ2θ02M-z+M-z-f1∗,1δ1θ02U5+1δ1U3-βULU-βNULU3=-1δ1θ0δ2P-v2+1ςθ0P-u3-δ2θ0P-u1+M-z1+1θ0P-v-f2∗,δ2θ0δ1U5-δ2βθLθ0δ1U4+δ3θ0δ1U3-δ3βθLθ0δ1U2+θ0U1-θ0βθLU-βNθLΨ3=βθLθ0δ11ς-μδ2θ02P-u+δ2δ2θ0P-v1+1ςP-u2+θ0M-z-1θ0δ11ς-μδ2θ02P-u1+δ2δ2θ0P-v2+1ςP-u3+θ0M-z1-f3∗.(ii)
Atξ=1(27)-βVRςθ03δ1U5-1δ1θ0U4-βVRςθ0δ1U3-θ0δ1U2+βVRθ0·U1+βNVRV3=-βVR1ςθ02P-v-δ2ςθ02δ1P-v2+δ2ςθ0δ1P-u1-1ςθ03δ11ςP-u3+θ0M-z1+1δ1δ2P-v1+1ςθ0P-u2-δ2θ0P-u-δ2θ02M-z+M-z+f4∗,1δ1θ02U5+1δ1U3+βURU+βNURU3=-1δ1θ0δ2P-v2+1ςθ0P-u3-δ2θ0P-u1+M-z1+1θ0P-v+f5∗,δ2θ0δ1U5+δ2βθRθ0δ1U4+δ3θ0δ1U3+δ3βθRθ0δ1U2+θ0U1+θ0βθRU+βNθRΨ3=-βθRθ0δ11ς-μδ2θ02P-u+δ2δ2θ0P-v1+1ςP-u2+θ0M-z-1θ0δ11ς-μδ2θ02P-u1+δ2δ2θ0P-v2+1ςP-u3+θ0M-z1+f6∗.


## 3. The Shifting Function Method

In this paper, the solution for the sixth-order differential equation (25) with nonlinear boundary conditions (26)-(27) is derived. The shifting function method developed by Lee and Lin [15], given below, is used:(28)Uξ=a-ξ+∑i=16f-ig-iξ,where(29)f-1=βNVLV30,f-2=βNULU30,f-3=βNθLΨ30,f-4=-βNVRV31,f-5=-βNURU31,f-6=-βNθRΨ31,and g-i(ξ), i=1,2,3,4,5,6, are the shifting functions to be specified, and a-(ξ) is the transformed function. Substituting (28)-(29) into (25)–(27) yields the differential equation for a-(ξ) and the associated boundary conditions:(30)a-6+2θ02a-4+θ04a-2=-∑i=16f-ig-i6+2θ02g-i4+θ04g-i2+θ0P-v1-δ2θ0P-v3-θ02P-u+μθ02P-u2-1ςP-u4-θ03M-z-θ0M-z2.(i)
Atξ=0(31)βVLςθ03δ1a-5-1θ0δ1a-4+βVLςθ0δ1a-3-θ0δ1a-2-βVLθ0a-1=f-1-∑i=16f-iβVLςθ03δ1g-i5-1θ0δ1g-i4+βVLςθ0δ1g-i3-θ0δ1g-i2-βVLθ0g-i1+βVL1ςθ02P-v-δ2ςθ02δ1P-v2+δ2ςθ0δ1P-u1-1ςθ03δ11ςP-u3+θ0M-z1+1δ1δ2P-v1+1ςθ0P-u2-δ2θ0P-u-δ2θ02M-z+M-z-f1∗,1δ1θ02a-5+1δ1a-3-βULa-=f-2-∑i=16f-i1δ1θ02g-i5+1δ1g-i3-βULg-i-1δ1θ0δ2P-v2+1ςθ0P-u3-δ2θ0P-u1+M-z1+1θ0P-v-f2∗,δ2θ0δ1a-5-δ2βθLθ0δ1a-4+δ3θ0δ1a-3-δ3βθLθ0δ1a-2+θ0a-1-θ0βθLa-=f-3-∑i=16f-iδ2θ0δ1g-i5-δ2βθLθ0δ1g-i4+δ3θ0δ1g-i3-δ3βθLθ0δ1g-i2+θ0g-i1-θ0βθLg-i+βθLθ0δ11ς-μδ2θ02P-u+δ2δ2θ0P-v1+1ςP-u2+θ0M-z-1θ0δ11ς-μδ2θ02P-u1+δ2δ2θ0P-v2+1ςP-u3+θ0M-z1-f3∗.(ii)
Atξ=1(32)-βVRςθ03δ1a-5-1δ1θ0a-4-βVRςθ0δ1a-3-θ0δ1a-2+βVRθ0a-1=f-4-∑i=16f-i-βVRςθ03δ1g-i5-1δ1θ0g-i4-βVRςθ0δ1g-i3-θ0δ1g-i2+βVRθ0g-i1-βVR1ςθ02P-v-δ2ςθ02δ1P-v2+δ2ςθ0δ1P-u1-1ςθ03δ11ςP-u3+θ0M-z1+1δ1δ2P-v1+1ςθ0P-u2-δ2θ0P-u-δ2θ02M-z+M-z+f4∗,1δ1θ02a-5+1δ1a-3+βURa-=f-5-∑i=16f-i1δ1θ02g-i5+1δ1g-i3+βURg-i-1δ1θ0δ2P-v2+1ςθ0P-u3-δ2θ0P-u1+M-z1+1θ0P-v+f5∗,δ2θ0δ1a-5+δ2βθRθ0δ1a-4+δ3θ0δ1a-3+δ3βθRθ0δ1a-2+θ0a-1+θ0βθRa-=f-6-∑i=16f-iδ2θ0δ1g-i5+δ2βθRθ0δ1g-i4+δ3θ0δ1g-i3+δ3βθRθ0δ1g-i2+θ0g-i1+θ0βθRg-i-βθRθ0δ11ς-μδ2θ02P-u+δ2δ2θ0P-v1+1ςP-u2+θ0M-z-1θ0δ11ς-μδ2θ02P-u1+δ2δ2θ0P-v2+1ςP-u3+θ0M-z1+f6∗.If the shifting functionsg-i(ξ), i=1,2,3,4,5,6, in (28) are chosen to satisfy the differential equation(33)g-i6+2θ02g-i4+θ04g-i2=0and the following boundary conditions,(i)
atξ=0(34)βVLςθ03δ1g-i5-1θ0δ1g-i4+βVLςθ0δ1g-i3-θ0δ1g-i2-βVLθ0g-i1=δij,j=1,1δ1θ02g-i5+1δ1g-i3-βULg-i=δij,j=2,δ2θ0δ1g-i5-δ2βθLθ0δ1g-i4+δ3θ0δ1g-i3-δ3βθLθ0δ1g-i2+θ0g-i1-θ0βθLg-i=δij,j=3,(ii)
atξ=1(35)-βVRςθ03δ1g-i5-1δ1θ0g-i4-βVRςθ0δ1g-i3-θ0δ1g-i2+βVRθ0g-i1=δij,j=4,1δ1θ02g-i5+1δ1g-i3+βURg-i=δij,j=5,δ2θ0δ1g-i5+δ2βθRθ0δ1g-i4+δ3θ0δ1g-i3+δ3βθRθ0δ1g-i2+θ0g-i1+θ0βθRg-i=δij,j=6,where δij is a Kronecker symbol, then the differential equation (30) and the associated boundary conditions (31)-(32) can be reduced to(36)a-6+2θ02a-4+θ04a-2=θ0P-v1-δ2θ0P-v3-θ02P-u+μθ02P-u2-1ςP-u4-θ03M-z-θ0M-z2.(i)
Atξ=0(37)βVLςθ03δ1a-5-1θ0δ1a-4+βVLςθ0δ1a-3-θ0δ1a-2-βVLθ0a-1=βVL1ςθ02P-v-δ2ςθ02δ1P-v2+δ2ςθ0δ1P-u1-1ςθ03δ11ςP-u3+θ0M-z1+1δ1δ2P-v1+1ςθ0P-u2-δ2θ0P-u-δ2θ02M-z+M-z-f1∗,1δ1θ02a-5+1δ1a-3-βULa-=-1δ1θ0δ2P-v2+1ςθ0P-u3-δ2θ0P-u1+M-z1+1θ0P-v-f2∗,δ2θ0δ1a-5-δ2βθLθ0δ1a-4+δ3θ0δ1a-3-δ3βθLθ0δ1a-2+θ0a-1-θ0βθLa-=βθLθ0δ11ς-μδ2θ02P-u+δ2δ2θ0P-v1+1ςP-u2+θ0M-z-1θ0δ11ς-μδ2θ02P-u1+δ2δ2θ0P-v2+1ςP-u3+θ0M-z1-f3∗.(ii)
Atξ=1(38)-βVRςθ03δ1a-5-1δ1θ0a-4-βVRςθ0δ1a-3-θ0δ1a-2+βVRθ0a-1=-βVR1ςθ02P-v-δ2ςθ02δ1P-v2+δ2ςθ0δ1P-u1-1ςθ03δ11ςP-u3+θ0M-z1+1δ1δ2P-v1+1ςθ0P-u2-δ2θ0P-u-δ2θ02M-z+M-z+f4∗,1δ1θ02a-5+1δ1a-3+βURa-=-1δ1θ0δ2P-v2+1ςθ0P-u3-δ2θ0P-u1+M-z1+1θ0P-v+f5∗,δ2θ0δ1a-5+δ2βθRθ0δ1a-4+δ3θ0δ1a-3+δ3βθRθ0δ1a-2+θ0a-1+θ0βθRa-=-βθRθ0δ11ς-μδ2θ02P-u+δ2δ2θ0P-v1+1ςP-u2+θ0M-z-1θ0δ11ς-μδ2θ02P-u1+δ2δ2θ0P-v2+1ςP-u3+θ0M-z1+f6∗.Once the transformed functiona-(ξ) and the shifting functions g-1(ξ), g-2(ξ), g-3(ξ), g-4(ξ), g-5(ξ), and g-6(ξ) are determined, they are substituted into (28), yielding(39)Uξ=a-ξ+βNVLV30g-1ξ+βNULU30g-2ξ+βNθLΨ30g-3ξ-βNVRV31g-4ξ-βNURU31g-5ξ-βNθRΨ31g-6ξ.Substituting (39) into (22) and (23) yields the solutions Ψ(ξ) and V(ξ). It can be observed that final solutions include the superposition of the linear and the nonlinear parts. The shifting function method can deal with the nonlinear parts of the boundary conditions very well.


## 4. Verification and Examples

The previous analysis is illustrated using the following example.Example. Consider the deflection of a beam subjected to uniform distributed load P. The curved beam is clamped at the left end and supported at the right end with linear and nonlinear springs in the r direction. The corresponding coefficients are(40)P-u=M-z=βNVL=βNUL=βNθL=βUR=βNUR=βθR=βNθR=f1∗=f2∗=f3∗=f5∗=f6∗=0,βUL,βVL,βθL⟶∞,P-v=P,where P is constant.Equations (22) and (23) become(41)Ψ=1θ0δ1δ2U4+δ3U2+δ1θ02U,V=-1ςθ03δ1U5-1ςθ0δ1U3+1θ0U1+Pςθ02.The complete sixth-order ordinary differential characteristic equation in (25) and the associated boundary conditions in (26)-(27) become(42)U6+2θ02U4+θ04U2=0.(i)
Atξ=0(43)U5+θ02U3-ςθ02δ1U1=θ0δ1P,U=0,U4+δ3δ2U2+θ02δ1δ2U=0.(ii)
Atξ=1(44)-βVRςθ03δ1U5-1δ1θ0U4-βVRςθ0δ1U3-θ0δ1U2+βVRθ0U1+βNVRV3=-βVRPςθ02+f4∗,1δ1θ02U5+1δ1U3=Pθ0,δ2θ0δ1U5+δ3θ0δ1U3+θ0U1=0.Let(45)Uξ=a-ξ+f-4g-4ξ,where f-4=-βNVRV3(1).Hereg-4(ξ) is the shifting function to be specified. a-(ξ) is the transformed function which satisfies the differential equation (42) and the associated boundary conditions (43)-(44). Consider(46)a-6+2θ02a-4+θ04a-2=0.The boundary conditions are as follows:(i)
Atξ=0(47)a-5+θ02a-3-ςθ02δ1a-1=θ0δ1P,a-=0,a-4+δ3δ2a-2+θ02δ1δ2a-=0.(ii)
Atξ=1(48)-βVRςθ03δ1a-5-1δ1θ0a-4-βVRςθ0δ1a-3-θ0δ1a-2+βVRθ0a-1=-βVRPςθ02+f4∗,1δ1θ02a-5+1δ1a-3=Pθ0,δ2θ0δ1a-5+δ3θ0δ1a-3+θ0a-1=0.It can be found that the functiona-(ξ) is(49)a-ξ=K-1+K-2ξ+K-3sin⁡θ0ξ+K-4cos⁡θ0ξ+K-5ξsin⁡θ0ξ+K-6ξcos⁡θ0ξ,where K-1, K-2, K-3, K-4, K-5, and K-6 are given in Appendix.The shifting functiong-4(ξ) satisfies the following differential equation and boundary conditions:(50)g-46+2θ02g-44+θ04g-42=0.(i)
Atξ=0(51)g-45+θ02g-43-ςθ02δ1g-41=0,g-4=0,g-44+δ3δ2g-42+θ02δ1δ2g-4=0.(ii)
Atξ=1(52)-βVRςθ03δ1g-45-1δ1θ0g-44-βVRςθ0δ1g-43-θ0δ1g-42+βVRθ0g-41=1,1δ1θ02g-45+1δ1g-43=0,δ2θ0δ1g-45+δ3θ0δ1g-43+θ0g-41=0.It can be found that the functiong-4(ξ) is(53)g-4ξ=J-1+J-2ξ+J-3sin⁡θ0ξ+J-4cos⁡θ0ξ+J-5ξsin⁡θ0ξ+J-6ξcos⁡θ0ξ,where J-1, J-2, J-3, J-4, J-5, and J-6 are given in Appendix.Substituting (45), (49), and (53) into (41) yields the exact solutions of U(ξ), V(ξ), and Ψ(ξ), respectively.Considering uniform straight Timoshenko beams asθ0=0 and f4∗=0 with (41), (45), (49), and (53), one has(54)Ψξ=a1ξ+a2ξ2+16Pξ3,Vξ=Pξ3-12Pξμ+12ξa1+8ξ2a2-48μa2ξ24,where (55)a1=--12P+24βNVRV31-PβVR83+βVR+3μβVR,a2=-24P-24βNVRV31+5PβVR+12μPβVR163+βVR+3μβVR.With both linear and nonlinear spring stiffness constants being zeros (i.e.,βNVR=βVR=0), (54) reduce to(56)Ψξ=ξ3-3ξ2+3ξ6P,Vξ=ξ4-4ξ3+61-2μξ2+24μξ24P.For a Bernoulli-Euler beam without shear deformation (i.e.,μ=0), (54) and (55) reduce to(57)Ψξ=a3ξ+a4ξ2+16Pξ3,Vξ=Pξ3+12ξa3+8ξ2a4ξ24,where(58)a3=--12P+24βNVRV31-PβVR83+βVR,a4=-24P-24βNVRV31+5PβVR163+βVR.It is the exact nondimensional deflection of cantilevered Timoshenko and Bernoulli-Euler curved beams subjected to uniform nondimensional distributed loadP. Equations (54)–(58) are exactly the same as those given by Lee et al. [18].Table1 shows that the exact solutions of the shifting function method are the same as Lin’s results [13]. With increasing distributed load, the amount of deflection in the r direction increases at a given location on the curved beam. Table 2 shows the neutral axis displacements in the r and θ directions and the angle of rotation due to bending in the z direction. The curved beam with linear and nonlinear boundaries has lower deflection in the r direction compared to that of a beam without such boundaries.Table 1
Neutral axis displacements inr direction of cantilever curved beam subjected to unit force in r direction at right end of cantilever curved beam [θ0 = π/2, ς = 100, μ = 0.01, βVR = 0, βNVR = 0, f4∗ = 1].
ξ
V
(
ξ
)
P = 100
P = 10
P = 1
P = 0
P = 0∗
0
0.00000
0.00000
0.00000
0.00000
0.00000
0.1
0.26284
0.02928
0.00592
0.00333
0.00333
0.2
0.87395
0.09922
0.02175
0.01314
0.01314
0.3
1.75342
0.20141
0.04621
0.02896
0.02896
0.4
2.81779
0.32677
0.07767
0.05000
0.05000
0.5
3.98364
0.46603
0.11427
0.07518
0.07518
0.6
5.17105
0.61000
0.15390
0.10322
0.10322
0.7
6.30683
0.75005
0.19437
0.13263
0.13263
0.8
7.32739
0.87835
0.23344
0.16179
0.16179
0.9
8.18120
0.98824
0.26894
0.18902
0.18902
1.0
8.83071
1.07445
0.29882
0.21264
0.21264
∗Lin’s result [13].Table 2
Neutral axis displacements inr and θ directions and angle of rotation due to bending in z direction for curved beam with linear and nonlinear boundaries subjected to uniform load in r direction [θ0 = π/2, ς = 100, μ = 0.01, βVR = 1, βNVR = 1, f4∗ = 0, P = 1].
ξ
U
(
ξ
)
Ψ
(
ξ
)
V
(
ξ
)
0
0.00000
0.00000
0.00000
0.1
−0.00037
0.03283
0.00236
0.2
−0.00004
0.05949
0.00767
0.3
0.00140
0.08033
0.15178
0.4
0.00423
0.09584
0.02411
0.5
0.00860
0.10662
0.03372
0.6
0.01455
0.11342
0.04331
0.7
0.02201
0.11705
0.05228
0.8
0.03083
0.11844
0.06011
0.9
0.04079
0.11854
0.06643
1.0
0.05161
0.11835
0.07101Of note, for the curved beam subjected to uniform load in ther direction, the deflections at the right end of the curved beam vary, as shown in Figures 2–4. According to (45), (49), and (53), the nonlinear spring stiffness constant is zero, and thus it is reasonable that the deflection at the right end of the curved beam is linearly related to uniform load in the r direction. When the nonlinear spring stiffness constant is nonzero value, the relationship can be curvilinear. The spring at the right end of the curved beam is strong enough to reduce the deflection variation, as shown in Figure 2. Figure 3 shows the influence of curvature on the static deflection of the beam in the r direction for a given set of boundary conditions. When the center angle is increased, the deflection will decrease. The curved beam itself can be regarded as a spring mechanism. Figure 4 shows that the center angle and boundary conditions significantly affect deflection in the r direction. If the load on the curved beam is insufficient and the value of the deflection is less than 1, a linear spring has a greater effect than that of a nonlinear spring. On the contrary, if the value of the deflection is over 1, a nonlinear spring has a greater effect than that of a linear spring.Figure 2
Deflections inr direction at right end of curved beam (θ0=π/2) varying with the uniform load for various boundary conditions and constants [ς=100, μ=0.01, f4∗=0].Figure 3
Deflections inr direction at right end of curved beam varying with the uniform load for various center angles and constants [ς=100, μ=0.01, βVR=1, βNVR=1, f4∗=0].Figure 4
Deflections inr direction at right end of curved beam varying with center angles for various boundary conditions and constants [ς=100, μ=0.01, f4∗=0, P=1].


## 5. Conclusion

In this paper, the shifting function method is applied to develop the static deflection of in-plane curved Timoshenko beams with nonlinear boundary conditions. Three coupled governing differential equations are derived via Hamilton’s principle and decomposed into a complete sixth-order ordinary differential characteristic equation that depends on the neutral axis displacement in theθ direction. The explicit relations among the neutral axis displacement in the r direction, the angle of rotation due to bending in the z direction, and the neutral axis displacement in the θ direction are revealed. An example was used to illustrate the analysis. It can be found that the center angle and the linear and nonlinear spring stiffness constants will have significant influence on the static deflection of the beam.



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*Source: 646391-2015-07-29.xml*