Theories of Straight Beams

Transcription

1 EVPM3ed /6/10 7:20 page 71 #25 This is a part of the revised chapter in the new edition of the tetbook Energy Principles and Variational Methods in pplied Mechanics, which will appear in These notes should help you see how the von Karmnan nonlinear equations of Euler-Bernoulli and Timoshenko beams are derived. JN Reddy 2.5. THEORIES OF STRIGHT BEMS 19 Oct Theories of Straight Beams Introction Most practical engineering structures, microscale or macroscale, consist of members that can be classified as beams, plates, and shells, called structural members. Beams are structural members that have a ratio of length-to-crosssectional dimensions very large, say 10 to 100 or more, and subjected to forces both along and transverse to the length and moments that tend to rotate them about an ais perpendicular to their length [see Fig a]. When all applied loads are along the length only, they are often called bars i.e., bars eperience only tensile or compressive strains and no bending deformation. Cables may be viewed as very fleible form of bars, and they can only take tension and not compression. Plates are two-dimensional versions of beams in the sense that one of the cross-sectional dimensions of a plate can be as large as the length. The smallest dimension of a plate is called its thickness. Thus, plates are thin bodies subjected to forces in the plane as well as in the direction normal to the plane and bending moments about either ais in the plane [see Fig b]. shell is a thin structure with curved geometry [see Fig c] and can be subjected to distributed as well as point forces and moments. Because of their geometries and loads applied, beams, plates, and shells are stretched and bent from their original shapes. The difference between structural elements Current Cantilever beam FM atomic micro scope tip a b c Fig : a n FM cantilever beam. b narrow plate strip subjected to a vertical force. c hyperboloidal shell. The deformations shown for b and c are eaggerated.

2 EVPM3ed /6/10 7:20 page 72 #26 72 REVIEW OF EQUTIONS OF SOLID MECHNICS and three-dimensional solid bodies, such as solid blocks and spheres, that have no restrictions on their geometric make up, is that the latter may change their geometries but they may not show significant bending deformation. lthough all solids can be analyed for stress and deformation using the elasticity equations reviewed in the preceding sections of this chapter, their geometries allow us to develop theories that are simple and yet yield results that are accurate enough for engineering analysis and design. Structural mechanics is a branch of solid mechanics that deals with the study of beams frames, plates, and shells using theories that are derived from three-dimensional elasticity theory by making certain simplifying assumptions concerning the deformation and stress states in these members. The present section is devoted to the development of structural theories of straight beams in bending for the static case. The equations governing beams will be referenced in the coming chapters and their development in this chapter will help the reader in understanding the material of the subsequent chapters. Plates will be studied in a chapter devoted entirely for them see Chapter 7. The geometric description of shells is more involved and they are not covered in this book to keep the book sie within reasonable limits see, for additional information, Timoshenko and Woinowsky-Krieger [124] and Reddy [50, 51]. In this section we consider two most commonly used theories of straight beams. They differ from each other in the representation of displacement and strain fields. The first one is the Bernoulli Euler beam theory 3, and it is the theory that is covered in all undergraate mechanics of materials books. In the Bernoulli Euler beam theory, the transverse shear strain is neglected, making the beam infinitely rigid in the transverse direction. The second one is a refinement to the Bernoulli Euler beam theory, known as the Timoshenko beam theory, which accounts for the transverse shear strain. These two beam theories will be developed assuming infinitesimal deformation. Therefore, we use σ ij to mean S ij, the second Piola Kirchhoff stress tensor, and i in place of X i. We consider straight beams of length L and symmetric about the 2 -ais cross section of area. The 1 -coordinate is taken along the centroidal ais of the beam with the 3 -coordinate along the thickness the height and the 2 -coordinate along the width of the beam into the plane of the page, as shown in Fig a. In general, the cross-sectional area can be a function of 1. Suppose that the beam is subjected to distributed aial force f 1 and transverse load q 1, and let u 1, u 2, u 3 denote the total displacements along the coordinates 1 =, 2 = y, 3 =. For simplicity of the developments only stretching along the length of the beam and bending about the 2 -ais are considered here. 3 Jacob Bernoulli was one of the many prominent Swiss mathematicians in the Bernoulli family. He, and along with his brother Johann Bernoulli, was one of the founders of the calculus of variations. Leonhard Euler was a pioneering Swiss mathematician and physicist.

3 EVPM3ed /6/10 7:20 page 73 # THEORIES OF STRIGHT BEMS The Bernoulli Euler Beam Theory The Bernoulli Euler beam theory is based on certain simplifying assumptions, known as the Bernoulli Euler hypothesis, concerning the kinematics of bending deformation. Figure The hypothesis states that straight lines perpendicular to the beam ais before deformation remain a straight, b inetensible, and c perpendicular to the tangent line to the beam ais after deformation. 3 = q q =- dw f a 1 = Undeformed edge b w u+ q u Deformed edge q q q M M +DM f N N +DN f V V +DV D c N M V d f f s s Fig : a typical beam with loads. b Kinematics of deformation of an Bernoulli Euler beam theory. c Equilibrium of a beam element. d Definitions or internal equilibrium of stress resultants. The Bernoulli Euler hypothesis leads to the following displacement field [see Fig b] u 1, y, = u dw, u 2 = 0, u 3, y, = w, where u, v, w are the displacements of a point on the -ais in the 1 =, 2 = y and 3 = coordinate directions, respectively. The infinitesimal strains are ε = u 1 1 = d2 w 2, 2ε = u u 3 1 = dw + dw = and all other strains are identically ero. free-body-diagram of an element of length is shown, with all its forces, in Fig c. Summing the forces in the - and -directions and summing

4 EVPM3ed /6/10 7:20 page 74 #28 74 REVIEW OF EQUTIONS OF SOLID MECHNICS the moments at the right end of the beam element, we obtain the following equilibrium equations: F = 0 : F = 0 : My = 0 : dn dv V dm = f, = q, = where N is the net aial force, M is the net bending moment about the y-ais, and V is the net transverse shear force [see Fig d] on the beam cross section: N = σ d, M = σ d, V = σ d where d denotes an area element of the cross section d = dyd. The set N, M, V are known as the stress resultants because they result from the stresses in the beam. The stresses in the beam can be computed using the linear elastic constitutive relation for an isotropic but possibly inhomogeneous material i.e., the material properties may be functions of and, we have σ, = E, ε, = E d2 w 2, σ, = 2G, ε, = 0. lthough the transverse shear stress computed using constitutive equation is ero hence, V = 0, it cannot be ero in a beam because the force equilibrium, Eq is violated. This is the inconsistency resulting from the Bernoulli Euler beam hypothesis. Therefore, to respect the physical requirement of V 0, we do not compute the shear stress from σ = 2Gε ; it is computed using the shear force V obtained from the equilibrium condition in Eq , namely, V = dm/. The stress resultants N, M now can be related to the displacements u and w and back to the stress σ, as discussed net. Using Eq , we obtain N = M = σ d = + B σ d = B + D d2 w 2 d2 w 2,, where is the aial stiffness, B is the bending-stretching stiffness, and D is the bending stiffness = E, d, B = E, d, D = 2 E, d 2.5.9

5 EVPM3ed /6/10 7:20 page 75 # THEORIES OF STRIGHT BEMS 75 When E is a function of only, Eq simplifies to = E d = E, B = E d = 0, D = E 2 d = EI, I = 2 d, where we have used the fact that the -ais coincides with the centroidal ais: d = Then N, M of Eq rece to N = E, M = EI d2 w Equations and can be inverted to epress / and d 2 w/ 2 in terms of N and M. We obtain = D N B M D B 2, d2 w 2 = M B N D B Substituting these relations for / and d 2 w/ 2 into the epression for σ in Eq , we obtain E σ, = D B 2 [D N B M + M B N] When E is not a function of, we obtain = E and D = EI = N E, d2 w 2 = M EI and σ = N + M I We note that, when E is not a function of, N is independent of w and M is independent of u. Therefore, Eq is independent of Eqs and In that case the aial deformation of members can be determined by solving Eq independent of the bending deformation and vice versa. lso, Eqs and can be combined to read d2 M = q, which can be epressed in terms of the transverse deflection w using the second equation in Eq as d 2 2 EI d2 w 2 = q

6 EVPM3ed /6/10 7:20 page 76 #30 76 REVIEW OF EQUTIONS OF SOLID MECHNICS We now return to the computation of shear stress σ in the Bernoulli Euler beam theory. The shear stress is computed from equilibrium considerations. For beams with E = E, the epression for σ is given by see pages of Fenner and Reddy [98] for a derivation, σ, = V Q, Q = Ib c1 bd, where Q denotes the first moment of the hatched area [see Fig ] about the ais y of the entire cross section, and b is the width of the cross section at where the longitudinal shear stress σ is computed. The hatched area is only a portion of the total area of cross section that lies below the surface on which the shear force acts; Q is the maimum at the centroid i.e., = 0, and σ is the maimum wherever Q/b is the maimum. For eample, for a rectangular cross section beam of height 2h and width b, Q takes the form Q = b and the shear stress becomes h σ, = V Q Ib d = b 2 h = 1 2I V h Thus, the shear stress varies quadratically through the beam height, vanishing at the top = h and bottom = h. The maimum shear stress is σfigure ma = σ 2.5.3, 0 = 3 4bhV. Thus, the actual maimum occurs wherever V is the maimum. σ, q D σ +D, s c 1 c 2 s y b Fig : Transverse shear stress σ e to bending The Timoshenko Beam Theory The Timoshenko beam theory is based on a relaation of the normality condition, namely, Part c of the Bernoulli Euler hypothesis. In other words, the transverse normal has rotation φ that is not the same as the slope dw/. The difference between these two quantities is the transverse shear strain. The relaed Bernoulli Euler hypothesis leads to the following displacement field for the Timoshenko beam theory see Fig

7 EVPM3ed /6/10 7:20 page 77 #31 Figure THEORIES OF STRIGHT BEMS 77 q =- dw f g w u q u+ f Undeformed edge Deformed edge Fig : Kinematics of deformation in the Timoshenko beam theory. u 1, y, = u + φ, u 2 = 0, u 3, y, = w, where φ is the rotation of a transverse normal line. The infinitesimal strains are ε = u 1 = 1 + dφ, 2ε = u 1 + u 3 = φ + dw 3 1 f dw.- dw and all other strains are identically ero. The free-body-diagram of a typical element of the beam remains the same as depicted in Fig c. Therefore, the equilibrium equations given in Eqs are also valid for the Timoshenko beam theory, ecept that the stress resultants N, M, V are related dw g to = fu, + w, φ in a different way, as discussed in the following paragraphs. The stresses in the beam are - σ, = E, ε, = E + dφ, u w σ, = 2G, ε, = G, φ + dw Note that the shear strain hence, shear force is not ero, removing the inconsistency of the Bernoulli Euler beam theory. The stress resultants N, M in the Timoshenko beam theory have the form N = M = V = σ d = + B σ d = B + D K s σ d = K s S φ + dw dφ, dφ, ,

8 EVPM3ed /6/10 7:20 page 78 #32 78 REVIEW OF EQUTIONS OF SOLID MECHNICS where K s is the shear correction coefficient introced to correct the energy loss e to the constant state 4 of transverse shear stress σ and S is the shear stiffness S = G, d Inverting Eqs , we obtain = D N B M D B 2, dφ = M B N D B 2, φ + dw = V K s S Substituting for /, dφ /, and φ + dw/ from Eq into the epressions for σ and σ in Eq , we obtain E σ, = D B 2 σ, = GV. K s S [D N B M + M B N], When E is not a function of, we obtain = E, B = 0, and D = EI σ, = N + M I, σ = V K s and the equations of equilibrium in Eqs in terms of the generalied displacements u, w, φ take the form d E = f, d [ ] dw K s G + φ = q, dw K s G + φ d EI dφ = s in the case of the Bernoulli Euler beam theory, the aial deformation of beams can be determined by solving Eq independent of the bending equations, Eqs and Eample illustrates several ideas from this chapter. Eample Consider the problem of an isotropic cantilever beam of rectangular cross section 2h b height 2h and width b, bent by an upward transverse load F 0 applied at the free end, as shown in 4 We note that the shear stress in the Timoshenko beam theory is not a function of, implying that it has a constant value at any section of the beam, violating the condition that it be ero at the top and bottom of the beam.

9 EVPM3ed /6/10 7:20 page 79 # THEORIES OF STRIGHT BEMS 79 Fig ssume that the beam is isotropic, linearly elastic, and homogeneous. a Compute the stresses using Eqs and , b compute the strains using the uniaial strain stress relations, c show that the strains computed satisfy the compatibility condition in the 1 3-plane, and d determine the two-dimensional displacement field u 1, u 2 that satisfies the geometric boundary conditions. Figure Solution: a The stresses in the beam are σ 11 = M3 I = F013, σ 13 = V Q I Ib = F0 2I h where I is the moment of inertia about the 2 ais, 2h is the height of the beam, and b is the width of the beam. 3 3 F0 F0 2 2h 1 M =-F 0 1 b L 1 V =-F 0 Fig : Cantilever beam bent by a point load, F 0. b The strains ε 11 and ε 13 are known from this section ecept for a change of coordinates from X i to i; ε 11 is the same from the Bernoulli Euler or the Timoshenko beam theory, and we use the shear strain obtained from the equilibrium conditions: ε 11 = σ11 E = M13 EI ε 13 = σ13 2G = V Q 2IbG = F013 EI ε 22 = ε 33 = νε 11 = νf013, EI, 1 + νf0 = h 2 2 3, 2 2EI where ν is the Poisson ratio, E is Young s molus, and G is the shear molus. c Net, we determine if the strains computed are compatible. Substituting ε ij into the third equation in Eq , we obtain = 0. Thus, the strains satisfy the compatibility equations in two dimensions i.e., in the 1 3-plane. lthough the two-dimensional strains are compatible, the three-dimensional strains are not compatible. For eample, using the additional strains, ε 33 = ε 22 = νε 11 and ε 12 = ε 23 = 0, one can show that all of the equations ecept the fifth equation in Eq are satisfied. We shall seek the two-dimensional displacement field u 1, u 3 associated with the 1 3-plane. d Integrating the strain-displacement equations, we obtain u 1 = ε 11 = F013 1 EI u 3 = ε 33 = νf013 3 EI or u 1 = F EI or u 3 = νf EI + f 3, 3 + g 1, 4 where f 3 and g 1 are functions of integration. Substituting u 1 and u 3 into the definition of 2ε 13, we obtain 2ε 13 = u1 + u3 = F EI + df + νf EI + dg. 5 1

10 EVPM3ed /6/10 7:20 page 80 #34 80 REVIEW OF EQUTIONS OF SOLID MECHNICS But this must be equal to the shear strain known from Eq. 2: F0 2EI df + νf0 3 2EI dg 1 + ν = F 0h EI Separating the terms that depend only on 1 and those depend only on 3 the constant term can go with either one, we obtain dg + F EI 2 νf0h2 1 = df EI νf EI Since the left side depends only on 1 and the right side depends only on 3, and yet the equality must hold, it follows that both sides should be equal to a constant, say c 0: df νf0 2 3 = c 0, 2EI Integrating the epressions for f and g, we obtain f 3 = g 1 = dg + F EI 2 νf0h2 1 = c 0. EI 2 + νf c c 1, 6EI F0 6EI νf0h2 1 c c 2, EI where c 1 and c 2 are constants of integration. Thus, the most general form of displacement field u 1, u 3 that corresponds to the strains in Eq. 2 is given by u 1 1, 3 = F0 2EI u 3 1, 3 = 2 + νf c c 1, 6EI 1 + νf0h2 1 + νf0 EI 2EI F0 6EI 3 1 c c 2. The constants c 0, c 1, and c 2 are determined using suitable boundary conditions. We impose the following boundary conditions that eliminate rigid-body displacements that is, rigid-body translation and rigid-body rotation: u 1L, 0 = 0, u 3L, 0 = 0, Ω 13 = 1 1 =L, 3 = u3 u1 = =L, 3 =0 Imposing the boundary conditions from Eq. 10 on the displacement field in Eq. 9, we obtain u 1L, 0 = 0 c 1 = 0, Thus, we have u 3L, 0 = 0 c 0L c 2 = 1 + νf0h2 L EI + F0L3 6EI, u3 u1 = 0, c 0 = 1 3 F0L2 1 + νf0h2 1 =L, 3 =0 2EI 2EI c 0 = F0L2 2EI νf0h2, c 1 = 0, c 2 = F0L3 2EI 3EI νf0h2 L. 12 2EI Then the final displacement field in Eq. 9 becomes [ ] u 1 1, 3 = F0L ν 2 3 h ν, 6EI L 2 L2 L 2 [ ] u 3 1, 3 = F0L ν h ν EI L L 2 L3 L 2 L 13

11 EVPM3ed /6/10 7:20 page 81 # THEORIES OF STRIGHT BEMS 81 In the Euler Bernoulli beam theory EBT, where one assumes that L >> 2h and ν = 0, we have u 1 = 0, and u 3 is given by u EBT 3 1, 3 = F0L EI L + 3 1, 14 L 3 while in the Timoshenko beam theory TBT we have u 1 = 0 [E = 21 + νg, I = h 2 /3, and = 2bh], and u 3 is given by u TBT 3 1, 3 = F0L EI L F0L L 3 K sg 1 1 L. 15 Here K s denotes the shear correction factor. Thus, the Timoshenko beam theory with shear correction factor of K s = 4/3 predicts the same maimum deflection, u 30, 0, as the twodimensional elasticity theory. Both beam theory solutions, in general, are in error compared to the plane elasticity solution primarily because of the Poisson effect The von Kármán Theory of Beams Preliminary discussion In the preceding sections on beam theories, we made an assumption of infinitesimal strains to write the linear strain displacement relations [see Eqs and ]. When one assumes the strain are small but the rotation of the transverse normal line is moderate, the strain fields of the Bernoulli Euler and Timoshenko beam theories will have additional term, which is nonlinear in the transverse deflection w, in the etensional part of the strain. The nonlinear strain is often referred to as the von Kármán strain. To see how this nonlinear term enters the calculation, we first make certain assumptions concerning the magnitude of the various terms in the Green strain tensor components and compute the nonero strains for beams. We begin with the assumption that the aial strain 1 / and the curvature d 2 w/ 2 are of order ɛ, and dw/ is of order ɛ, where ɛ << 1 is a small parameter. Then the nonero Green strain tensor components from Eq are with X i i ε = u 2 1 u3, ε = u1 + u Thus, the underlined term in Eq is new and it is nonlinear. The beam theory resulting from the inclusion of this nonlinear term is termed the von Kármán beam theory. Here we present the complete development for each of the beam theories considered before. Before we embark on the development of the beam theories, we must consider suitable equations of equilibrium for this nonlinear case. We adopt the

12 EVPM3ed /6/10 7:20 page 82 #36 82 REVIEW OF EQUTIONS OF SOLID MECHNICS equations of equilibrium in terms of the second Piola Kirchhoff stress tensor S. The vector equation of equilibrium in terms of S is [see Eqs , 2.2.9, and 2.3.7] 0 [S T I + 0 u ] + ρ 0 f = 0, which in component form is X J [ δ KI + u I X K S KJ ] + ρ 0 f I = 0, where f I are body forces measured per unit mass. In the present case, the equilibrium equations associated with balance of forces in the - and -directions are S + u 1 S + u 1 S + S + u 1 S + f = 0, S + u 3 S + u3 S + f = 0, where f = ρ 0 f 1 and f = ρ 0 f 3 are the body forces measured per unit volume. They are assumed, in the present case, to be only functions of The Bernoulli Euler beam theory The displacement field of the Bernoulli Euler beam theory is given in Eq Integration of Eqs and over the beam area of cross section yields d N + N d2 w 2 M dw V d d2 w V + dw N V + f = 0, q = 0, where f and q are the aially and transversely distributed forces per unit length. We note that S + u 1 S d = S + u 1 S d, where we have used the assumption that S = 0 on the surface of the beam. Net, multiply Eq with, integrate over the beam area of cross section, and use the identity in Eq to obtain d M + M d2 w 2 P dw R V V + d2 w 2 R = 0,

13 EVPM3ed /6/10 7:20 page 83 # THEORIES OF STRIGHT BEMS 83 where P, R are higher-order stress resultants P = σ 2 d, R = σ d Note that area integrals of f and f vanish because f and f are only functions of and the -ais coincides with the geometric centroidal ais. Using the order-of-magnitude assumption of various quantities, Eqs , , and can be simplified as dn = f, dv d dw N = q, dm + V = 0, where the underlined epression is the only new term e to the von Kármán nonlinearity when compared to Eqs Of course, the stress resultant N will also contain nonlinear term when it is epressed in terms of the displacements. Since in the Bernoulli Euler beam theory V cannot be determined independently, we use V = dm/ from Eq and substitute for V into Eq the following equations of equilibrium for the Bernoulli Euler beam theory: d2 M 2 dn = f d dw N = q In view of the displacement field in Eq , we obtain the following simplified Green strain tensor component E ε = d2 w dw 2 ε 0 + ε 1, γ = 0, where ε 0 = dw 2, ε 1 = d2 w 2, and the stress resultants N, M can be epressed in terms of the displacements u, w when linear stress strain relations are used as [see Eqs and2.5.9] dw N = M = σ d = [ σ d = B [ dw ] 2 + B d2 w 2 2 ] + D d2 w 2,

14 EVPM3ed /6/10 7:20 page 84 #38 84 REVIEW OF EQUTIONS OF SOLID MECHNICS Relations in Eq take the form + 1 dw 2 = D N B M 2 D B 2, d2 w 2 = M B N D B Consequently, Eq and remain unchanged The Timoshenko beam theory Using the displacement field in Eq and following the procere as in Eqs , we obtain d N + N + dφ M + φ V + dφ V + f = 0, d V + dw N + q = 0, d M + M + dφ P + φ R V V dφ R = Using the order-of-magnitude assumption of various quantities, Eqs are simplified to those listed in Eqs The simplified Green strain tensor components of the Timoshenko beam theory are E ε = + dφ where ε 0 = dw 2 ε 0 + ε 1, γ = φ + dw dw 2, ε 1 = dφ The stress resultants N, M, V are known in terms of the generalied displacements u, w, φ as [ N = σ d = + 1 ] dw 2 dφ + B, [ M = σ d = B + 1 ] dw 2 dφ + D, V = K s σ d = K s S φ + dw Relations in Eq take the form + 1 dw 2 = D N B M 2 D B 2, dφ = M B N D B 2, φ + dw = V. K s S

15 EVPM3ed /6/10 7:20 page 85 # SUMMRY 85 Therefore, Eqs and remain unchanged. This completes the development of the Bernoulli Euler and Timoshenko beam theories with the von Kármán nonlinear strain. The equations governing bars can be obtained as a special case from the equations of beams by setting bending related quantities to ero i.e., w = 0, φ = 0, M = 0, V = 0, and q = 0. Thus, the kinematic, constitutive, and equilibrium equations of bars written in terms of forces as well as displacements are ε =, σ = Eε, N = E dn [ + f = 0 d E ] + f = 0, 0 < < L Here u denotes the aial displacement and f is the distributed aial force. The governing equations for cables also termed rods, ropes, strings, and so on in a plane are the same as those for bars, with the eception that aial stiffness E is replaced with the tension T in the cable. Thus, we have [ d T + f = 0 ], 0 < < L, where u is the transverse displacement and f is the distributed force transverse to the cable. 2.6 Summary In this chapter a summary of the equations of elasticity are presented. These include the equations of motion, strain displacement equations, and constitutive or stress strain relations. The stress and strain transformations and compatibility conditions for strains are also discussed. lso, two beam theories, namely the Bernoulli Euler and Timoshenko beam theories, are also fully developed in this chapter. Finally, the von Kármán nonlinear beam theories using the Bernoulli Euler and Timoshenko kinematics are developed. The equations governing the two beam theories will be used etensively in the coming chapters. The main equations of this chapter for linearied elasticity and bending of beams are summaried here. Equations of motion [Eqs and 2.2.8] σ + ρf = ρ 2 u t 2 σji j + ρf i = ρ 2 u i t Strain displacement relations [Eq ] ε = 1 2 [ u + u T] [ ε ij = 1 ui 2 + u ] j j i

16 EVPM3ed /6/10 7:20 page 86 #40 86 REVIEW OF EQUTIONS OF SOLID MECHNICS Strain compatibility equations [Eqs and ] ε T 2 ε ij = ε kl = 2 ε lj + k l i j k i 2 ε ki l j Stress strain relations [modified Eqs and to account for thermal strains; see Problem 2.39] σ = σ ij = E 1+ν ε + νe Eα 1+ν1 2ν trε I 1 2ν T T 0 I E 1+ν ε νe ij + 1+ν1 2ν ε kk δ ij Eα 1 2ν T T 0δ ij Equilibrium equations of the Bernoulli Euler beam theory [Eqs and ; set the nonlinear terms underlined to obtain the linear equations] dn = f, M d2 2 d dw N = q Kinematic relations of the Bernoulli Euler beam theory [Eq ] ε = dw 2 d2 w 2, γ = Constitutive relations of the homogeneous Bernoulli Euler beam theory [Eqs and ] [ N = E + 1 ] dw 2, M = EI d2 w Equilibrium equations of the Timoshenko beam theory [Eqs ; set the nonlinear terms underlined to obtain the linear equations] dn = f, dv d dw N = q, dm + V = Kinematic relations of the Timoshenko beam theory [Eqs and ] ε = dw 2 + dφ, γ = φ + dw Constitutive relations of the homogeneous Timoshenko beam theory [Eqs ] [ N = E + 1 ] dw 2, M = EI dφ 2, V = GK s φ + dw

17 EVPM3ed /6/10 7:20 page 87 # SUMMRY 87 Table contains a summary of governing equations of homogeneous, isotropic, and linear elastic bodies in terms of displacements i.e., equilibrium equations, kinematics relations, and constitutive equations are combined to epress the governing equations in terms of displacements. The nonlinear equilibrium equations of beams with the von Kármán nonlinearity can be epressed in terms of displacements with the help of equations presented in Section 2.5. In the later chapters of this book we shall make reference to the equations developed in this chapter and summaried in this section. Some of the equations developed here may be re-derived using the energy approach in Chapters 3 through 5. Table 2.6.1: Summary of equations of elasticity, membranes, cables, bars, and beams. 1. Linearied Elasticity µ and λ are Lamé constants, u is the displacement vector, and f is the body force vector measured per unit volume µ 2 u + λ + µ u + f = 0 in Ω Membranes a 11 and a 22 are tensions in the - and y-coordinate directions, respectively, u is the transverse deflection, and f is the distributed transverse force measured per unit area u a 11 u a 22 = f in Ω y y 3. Cables T is the tension in the cable, u is the transverse deflection, and f is the distributed transverse load d T = f, 0 < < L Bars E is Young s molus, is the area of cross section, u is the aial displacement, and f is the distributed aial force d E = f, 0 < < L Bernoulli Euler beam theory E is Young s molus; I is the moment of inertia; w is the transverse displacement, and q is the distributed load d 2 EI d2 w = q, 0 < < L Timoshenko beam theory E is Young s molus, G is the shear molus, is the area of cross section, K s shear correction coefficients, I is the moment of inertia; w is the transverse displacement, φ rotation of a transverse normal line, and q is the distributed load d EI dφ d + GK s φ + dw [GK s φ + dw = 0 ] = q