MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 5 pptx

138 Structural elements The Lagrangian is: L =−  L 0 (E es + E eb )dx + F 0 Z(L) = L s + L b [3.23] where: L s =−  L 0 (E es )dx + F 0 Z s (L); L b =−  L 0 (E eb )dx + F 0 Z b (L) Since the Lagrangian can be split into the sum of a shear term L s and a bending term L b , it is anticipated that shear and bending deflections are uncoupled from each other. Furthermore, bending is still governed by the Bernoulli–Euler model. Hence, the variational calculation is detailed here for the shear term only: δL s =−  L 0 κGS  ∂Z s ∂x  ∂(δZ s ) ∂x  dx +F 0 δZ s (L) = 0 [3.24] Once more, expression [3.24] is integrated by parts and the cofactors of δZ s are set to zero. δL s =−  F 0 δZ s − κGS  ∂Z s ∂x  δZ s  L 0 +  L 0  κGS  ∂ 2 Z s ∂x 2  δZ s  dx = 0 [3.25] The deflection due to elastic shear is thus governed by the equation: κGS  ∂ 2 Z s ∂x 2  = 0 [3.26] with the boundary conditions: Z s (0) = 0; κGS  ∂Z s ∂x      L = F 0 [3.27] The system [3.26], [3.27] is consistent with equations [2.33] and [2.39], except the shear weighting factor κ. The total deflection is: Z(x) = Z s (x) + Z b (x) = F 0  x κGS + (3x 2 L − x 3 ) 6EI  [3.28] This result may be used to discuss the relative importance of shear and bending deflections in relation to the slenderness ratio of the beam. Let us consider for Straight beam models: Hamilton’s principle 139 Figure 3.3. Transverse deflection due to elastic shear and bending for two cantilever rods (upper plot L/R =10, lower L/R =2) instance a circular cylindrical rod of radius R and Poisson’s ratio ν = 0.3, κ = 0.95 ∼ = 1. We obtain: Z(x) = F 0 L πR 2 E  2(1 + ν)  x L  + 2 3  L 2 R 2  3  x L  2 −  x L  3  [3.29] The result [3.29] is evidence that the relative contribution of bending to the total deflection increases as (L/R) 2 , validating thus the Bernoulli–Euler model for slender beams, as illustrated in Figure 3.3. 3.2.2.3 The Rayleigh–Timoshenko dynamic model The same problem as in the last subsection is studied now in dynamics. Focus- ing on the case of beams of small slenderness ratios, for which shear effects are 140 Structural elements important, the kinetic energy due to the rotation of the cross-sections with respect to the bending axis can no longer be neglected. The kinetic energy of a cross-section is thus written as: E κ = 1 2  L 0 ρS( ˙ Z s + ˙ Z b ) 2 + 1 2 ρI  ∂ ˙ Z b ∂x  2 dx [3.30] In contrast with the elastic energy, shear and bending deflections are found to be coupled due to the cross term arising in the translational kinetic energy. The variation of [3.30] is:  t 2 t 1 δ[E κ ]dt = ρS  t 2 t 1  L 0 −( ¨ Z s + ¨ Z b )(δZ s + δZ b ) + ρI  ∂ 2 ¨ Z b ∂x 2  δZ b dt dx [3.31] Making use of the static results already derived in subsection 3.2.2.2, the following coupled dynamic equations are found: ρS ¨ Z −ρI ∂ 2 ¨ Z b ∂x 2 + EI ∂ 4 Z b ∂x 4 = 0 ρS ¨ Z −κGS ∂ 2 Z s ∂x 2 = 0 Z = Z s + Z b [3.32] Such equations can be further transformed to produce finally a single equation expressed in terms of the sole variable Z. At first, Z b is eliminated from the bending equation to produce the intermediate result: ρS ¨ Z −ρI ∂ 2 ¨ Z ∂x 2 + ρI ∂ 2 ¨ Z s ∂x 2 + EI ∂ 4 Z ∂x 4 − EI ∂ 4 Z s ∂x 4 = 0 [3.33] Then, from the second equation [3.32] we get: ∂ 2 Z s ∂x 2 = ρ κG ¨ Z ⇒ ∂ 4 Z s ∂x 4 = ρ κG ∂ 2 ¨ Z ∂x 2 [3.34] Relations [3.34] can be used to eliminate Z s from [3.33], providing thus the final result: ρS ∂ 2 Z ∂t 2 − ρI  1 + E κG  ∂ 4 Z ∂x 2 ∂t 2 + ρ 2 I κG ∂ 4 Z ∂t 4 + EI ∂ 4 Z ∂x 4 = 0 [3.35] Obviously, such an elimination of the variables Z s and Z b is possible only if the beam cross-sections are uniform. Straight beam models: Hamilton’s principle 141 3.2.3 Bending of a beam prestressed by an axial force As is well known from common experience, the bending response of a beam can be significantly modified when an axial force of sufficient magnitude is applied to it. This effect is used in particular to adjust accurately the pitch of musical string instruments. More generally, in many instances one is interested in investigating bending resistant structures when their initial state of static equilibrium is stressed, due to the presence of an initial loading. The variational method is particularly well suited to modelling this kind of problem, because it brings out in a logical manner the central importance of nonlinear strains, even if the interest is restricted to the formulation of linear equations. Indeed, starting from an elastic medium whose initial state of static equilibrium is characterized by the local stress tensor σ (0) , a further deformation ε referred to this initial state induces an additional stress field denoted σ (1) , and the actual stress tensor can be written as: σ = σ (0) + σ (1) [3.36] The central point of the problem is to realize that when such a prestressed system is set in motion, strain energy is varied due to the work done by the initial stress σ (0) as well as that done by σ (1) . Both of them are related to the Green–Lagrange strain tensor [1.17] which comprises a linear plus a quadratic form of the displacement gradient. On the other hand, if the aim is restricted to derive linear equations of motion, it is suitable to approximate strain energy as a quadratic form of the displacement derivatives, as is already the case in the absence of σ (0) . Because the elastic stress σ (1) is proportional to ε, whereas the initial stress σ (0) is independent of ε, it turns out that the elastic energy must be related to the linear component of the Green–Lagrange strain tensor, whereas the prestress energy must be related to the exact form of the Green–Lagrange strain tensor. This is further explained below taking the example of a cantilevered beam axially stressed by a static axial force  T 0 applied initially at its free end B, see Figure 3.4. Figure 3.4. Axially stressed cantilever 142 Structural elements 3.2.3.1 Strain energy and Lagrangian In so far as bending is described by the Bernoulli–Euler model, the stress tensor reduces to the axial component: σ xx = σ (0) xx + σ (1) xx = T 0 S + Eε xx [3.37] Making use of the relationship [1.47], the variation of the strain energy per unit volume is found to be: δ[e s ]=σ xx δε xx =  T 0 S + Eε xx  δε xx [3.38] which implies: e s = T 0 S ε xx + Eε 2 xx 2 [3.39] The longitudinal component of the Green–Lagrange tensor [1.17] is written as: ε xx = ∂ξ x ∂x + 1 2   ∂ξ x ∂x  2 +  ∂ξ z ∂x  2  [3.40] where ξ z is the local transverse displacement in the bending plane Oxz and ξ x is the local axial displacement of a current point of the beam at a distance z from the bending axis. These components are related to the global displacement field defined in the prestressed state of equilibrium by: ξ z (x, z) = Z(x); ξ x (x, z) =−z ∂Z ∂x [3.41] Substituting [3.41] into [3.40] we get: ε xx =−z ∂ 2 Z ∂x 2 + 1 2  ∂Z ∂x  2 + 1 2  z ∂ 2 Z ∂x 2  2 [3.42] Since the aim is to derive linear equilibrium equations, it follows that strain energy density [3.39] must be expressed as a quadratic form in terms of the Z derivatives. Therefore, it is appropriate to use the exact form of [3.42] to formu- late the energy involving the initial stress and to use its linear approximation to formulate the elastic energy, which is already quadratic in ε 2 xx . After making such manipulations and integrating e s over the cross-sectional area, the following strain Straight beam models: Hamilton’s principle 143 energy per unit beam length is obtained:  (S) e s dS = T 0  + 1 2  ∂Z ∂x  2 + I 2S  ∂ 2 Z ∂x 2  2  + EI 2  ∂ 2 Z ∂x 2  2 [3.43] Further, if the beam is sufficiently slender, only the first term proportional to the initial load may be retained, as the second one is less by a factor (D/L) 2 , where D and L are the characteristic length scales in the transverse and in the axial directions respectively. The Lagrangian of the flexed beam is thus written as: L = 1 2  L 0  ρS ˙ Z 2 − T 0  ∂Z ∂x  2 − EI  ∂ 2 Z ∂x 2  2  dx [3.44] 3.2.3.2 Vibration equation and boundary conditions Starting from the Lagrangian [3.44], Hamilton’s principle is written as: δ[A]=  t 2 t 1 dt  L 0 −  ρS ¨ ZδZ + T 0  ∂Z ∂x  ∂δZ ∂x  +EI  ∂ 2 Z ∂x 2  ∂ 2 δZ ∂x 2  dx = 0 [3.45] leading to the vibration equation: ρS ¨ Z −T 0 ∂ 2 Z ∂x 2 + EI ∂ 4 Z ∂x 4 = 0 [3.46] provided with the elastic boundary conditions:  T 0 ∂Z ∂x − EI ∂ 3 Z ∂x 3  δZ  L 0 = 0;  EI  ∂ 2 Z ∂x 2  δ  ∂Z ∂x  L 0 = 0 [3.47] In addition to the usual inertia and elastic stiffness terms, the vibration equation includes a prestress stiffness term proportional to the initial load T 0 , which thus may be positive or negative, depending on the sign of T 0 . On the other hand, the first boundary condition [3.47] related to the conjugate components of transverse shear stress and transverse displacement depends explicitly on the initial load T 0 . In particular, if the loaded end is left free, as it is the case of the cantilevered beam, the boundary condition at x = L is:  EI ∂ 3 Z ∂x 3 − T 0 ∂Z ∂x  L = 0 [3.48] 144 Structural elements The second boundary condition, which is related to the conjugate components of bending moment and rotation of the cross-section, is independent of T 0 . note. – Newtonian approach Equations [3.46] provided with the suitable boundary conditions can also be derived by using the Newtonian approach. However, the balance of forces and moments must be written by referring to the deflected configuration of the beam, contrasting with the usual case in which the non-deflected configuration of static equilibrium is used as a reference. The effect of the preload (initial load) is shown in Figure 3.5. The preload vector is assumed to be constant; in particular it remains parallel to the beam axisdirection of the non-deflected configuration, independently from the beam deflection. The forces −T 0 and T 0 act on the left- and right-hand side cross-sections bounding the infinitesimal beam element considered. As a first order approximation, used for projecting the forces, the small rotation of the cross- sections is assumed to be constant in the beam element and given by −∂Z/∂x. The projection of  T 0 onto a cross-sectional plane leads to the initial shear force: Q T ≃ T s ≃ T 0 ψ y =−T 0 ∂Z ∂x [3.49] The torque induced by the equilibrated axial forces ±  T 0 acting at the ends of the beam element is: Q T dx = T 0 ∂Z ∂x dx [3.50] Figure 3.5. Bent configuration of a beam stressed axially by an initial load Straight beam models: Hamilton’s principle 145 In agreement with the Bernoulli–Euler model, the moment balance is written as: M y (x + dx) − M y (x) + Q z dx +T 0 ∂Z ∂x dx = 0 [3.51] In so far as material behaves elastically, the resulting transverse shear force is: Q z =−EI y ∂ 3 Z ∂x 3 + T 0 ∂Z ∂x [3.52] Substituting the expression [3.52] into the transverse equation [2.18], equa- tion [3.46] is recovered. On the other hand, the boundary condition [3.48] is also recovered by cancelling the shear force Q z at the free end of the cantilevered beam. 3.2.3.3 Static response to a transverse force and buckling instability The physical effect of an axial preload on the bending response of the beam can be conveniently illustrated by solving the following static problem: EI d 4 Z dx 4 − T 0 d 2 Z dx 2 = 0 Z(0) = dZ dx     0 = 0; d 2 Z dx 2     L = 0 EI d 3 Z dx 3     L − T 0 dZ dx     L =−F 0 [3.53] where the inhomogeneous boundary condition stands for a transverse force F 0 applied to the free end of the cantilevered beam. The general solution of the homogeneous differential equation is found to be: Z(x) = ae k 0 x + be −k 0 x + cx +d [3.54] where k 2 0 = T 0 EI Depending on the sign of the initial axial load T 0 , k 0 is real or purely imaginary. Determination of the constants by using the boundary conditions of the problem 146 Structural elements presents no difficulty and the final result may be written as: Tensioned beam: T 0 ≥ 0 Z(x) = F 0 |T 0 |  x − sinh(k(x −L)) +sinh(kL) k cosh(kL)  Compressed beam: T 0 ≤ 0 Z(x) = F 0 |T 0 |  −x + sin(k(x −L)) +sin(kL) k cos(kL)  [3.55] where k =|k 0 |=  |T 0 | EI In Figure 3.6 the deflection of the beam is represented in a reduced form for a few values of the tensile axial force, characterized by the dimensionless parameter  = kL = L √ T 0 /EI , where L is the beam length and ξ = x/L is the dimension- less abscissa along the beam. Z m = Z(L;  = 0) is chosen as a relevant scaling factor to reduce the beam deflection. As is conspicuous in Figure 3.6, the mag- nitude of the deflection is significantly reduced as soon as >0.1, which means that initial tension enhances the effective stiffness of the deflected beam, as expec- ted. Furthermore, the shape of the deflection curves is also modified being less curved as  increases, which is a mere consequence of the increasing importance of the tensioning term in comparison with the flexure term in the local equilibrium equation [3.53]. Figure 3.6. Static response of a tensioned beam to a transverse force Straight beam models: Hamilton’s principle 147 On the other hand, an initial compression decreases the effective stiffness, in such a way that the analytical response tends to infinity for the following infinite sequence of k values: k n = (2n + 1)π 2L ; n = 0, 1, 2 [3.56] The physical interpretation of such a result is as follows. When the magnitude of the compressive load is progressively increased starting from zero, the effective stiffness of the beam is progressively reduced, leading to a larger lateral deflection for the same transverse load F 0 , see Figure 3.7. A critical compressive load T c exists for which the effective stiffnessis zero and the analyticaldeflection is infinite. Using [3.56], T c is given by: T c =−EI  π 2L  2 [3.57] If the magnitude of the compressive load is further increased, according to the mathematical model [3.53] the effective stiffness becomes negative, as can be checked by looking at the sign of the analytical deflection [3.55], see the plots in dashed lines of Figure 3.7. As already indicated in [AXI 04], based on a few discrete systems such as articulated rigid bars, the axial compression of a beam can lead to static instability, commonly called ‘buckling’ and validity of the linear model [3.53] to describe its physical response to F 0 is strictly restricted to the subcritical domain T 0 >T c . note. – Limit case of strings and cables Figure 3.7. Static response of a compressed beam to a transverse force [...]... K1 Z + d dx L2 [Z(L)] = K2 Z − d dx d2 dx 2 EI Z d2 dx 2 EI Z x=0 =0 x=L =0 L3 [Z(0)] = K3 d 2Z dZ + EI 2 dx dx x=0 =0 L4 [Z(L)] = K4 dZ d 2Z − EI 2 dx dx x=L =0 where K1 to K4 are the stiffness coefficients of the elastic supports The integration by parts of W , K[Z] W , K[Z] = W + (L) d dx dZ dx − Z, K[W ] EI EI d 2Z dx 2 d 2W dx 2 (L) gives: (L) −Z d dx EI d 2W dx 2 − dW dx EI d 2Z dx 2 L 0 Then,... is written in terms of distributions as: 2 ∂x 2 EIy ∂ 2Z ∂x 2 +ρS ∂ 2Z = Fz(e) (t)δ(x − x0 ) ∂t 2 [3.71] 154 Structural elements A similar calculation as that made just above shows immediately that [3.71] is equivalent to the system: 2 ∂x 2 EIy ∂ ∂x ∂ 2Z EIy 2 ∂x ∂ 2Z ∂x 2 +ρS ∂ 2Z = 0; ∂t 2 ∂ − ∂x x0 + ∀x ∈ [0, L] ∂ 2Z EIy 2 ∂x x0 − = [3. 72] Fz(e) (t) example 3 – Beam loaded by a local transverse... example 2 – Beam loaded by a concentrated transverse force Again, we start from the classical case of a beam bent by a transverse external force field distributed along the beam axis The local equation of the problem is (see equation [2. 71]): 2 ∂x 2 EIy ∂ 2Z ∂x 2 +ρS ∂ 2Z = Fz(e) (x; t) ∂t 2 If the external load is concentrated at x0 , the equation is written in terms of distributions as: 2 ∂x 2 EIy ∂ 2Z... static version of the problem Starting thus from the local equation: 2 ∂x 2 EIy ∂ 2Z ∂x 2 = ∂M(e) (x) y ∂x (e) A moment applied at x0 is written as the distribution My (t)δ(x − x0 ) When substituted into the right-hand side of the above equation we obtain: 2 ∂x 2 EIy ∂ 2Z ∂x 2 = M(e) y d(δ(x − x0 )) = M(e) δ ′ (x − x0 ) y dx [3.73] which is readily integrated to produce: ∂ ∂x EIy ∂ 2Z ∂x 2 = M(e) δ(x... equation in terms of action, in a consistent way with the relation of definition [3. 65] Therefore, we shift from the local formulation [3.66] to the integral formulation: x2 ρS x1 ∂ 2X ∂ − ∂x ∂t 2 ES ∂X ∂x x2 dx = x1 (e) Fx (t)δ(x − x0 ) dx [3.67] Here x1 and x2 may be chosen arbitrarily, except that they must comply with the condition of nonzero action of the loading, that is x0 ∈ [x1 , x2 ] Accordingly,... x2 x1 f ′ (x)δ(x − x0 ) dx = f ′ (x0 ) = [f (x)δ(x − x0 )]x2 − x1 x2 x1 f (x)δ ′ (x − x0 ) dx [3. 75] Straight beam models: Hamilton’s principle 155 As δ(x) is zero everywhere except at x = x0 , it follows that the dipole action is found to be: x2 x1 f (x)δ ′ (x − x0 ) dx = −f ′ (x0 ); 0; if x0 ∈ [x1 , x2 ] otherwise [3.76] note – Informal differentiation of δ(x) Let us assume that the derivative of. .. operator of a beam, provided with elastic boundary conditions: K[Z] = d2 dx 2 EI (x) d 2Z dx 2 + elastic supports conditions Elastic supports at the beam ends can be described either by linear relationships between the conjugate displacement and stress variables, as already seen in Chapter 2 subsection 2. 2 .5, or by concentrated stiffness forces as described in subsection 3.3.3.3 Adopting the first point of. .. to the free end of a cantilevered beam [3 .59 ] Straight beam models: Hamilton’s principle 149 Hence, the static problem now takes the form: d 4Z d 2Z − T0 2 = 0 4 dx dx d 2Z dZ =0 = 0; Z(0) = dx 0 dx 2 L EI EI d 3Z dx 3 L [3.60] = −F0 Though at first sight the change when shifting from the system [3 .53 ] to [3.60] may be thought as rather benign, the nature of the problem is actually profoundly modified... system [3 .53 ]: d 4Z d 2Z − T0 2 dx 4 dx d 3Z = 0; EI dx 3 K[Z] = EI Z(0) = dZ dx 0 d 2Z dx 2 = 0; L L − T0 dZ dx L =0 It is easily shown that the initial stress operator is also formally self-adjoint and so is K[Z] Furthermore, the concomitant [3.84] is found to be: EI dZ d 3Z − T0 3 dx dx + EI W − EI dW d 3W − T0 3 dx dx d 2W d 2Z dW dZ − EI dx 2 dx dx 2 dx L Z 0 L 0 =0 Accordingly, bending of an axially... vicinity of a stable position of static equilibrium must be constant because the mechanical energy is also constant Straight beam models: Hamilton’s principle 163 3 These properties are a direct extension of those established in [AXI 04], in the case of conservative discrete mechanical systems This indicates that it is possible to shift from the mathematical description of a discrete mechanical system . is equivalent to the system: ∂ 2 ∂x 2  EI y ∂ 2 Z ∂x 2  + ρS ∂ 2 Z ∂t 2 = 0; ∀x ∈[0, L] ∂ ∂x  EI y ∂ 2 Z ∂x 2      x 0 + − ∂ ∂x  EI y ∂ 2 Z ∂x 2      x 0 − = F (e) z (t) [3. 72] example 3.–Beam. 143 energy per unit beam length is obtained:  (S) e s dS = T 0  + 1 2  ∂Z ∂x  2 + I 2S  ∂ 2 Z ∂x 2  2  + EI 2  ∂ 2 Z ∂x 2  2 [3.43] Further, if the beam is sufficiently slender, only the first. =−z ∂Z ∂x [3.41] Substituting [3.41] into [3.40] we get: ε xx =−z ∂ 2 Z ∂x 2 + 1 2  ∂Z ∂x  2 + 1 2  z ∂ 2 Z ∂x 2  2 [3. 42] Since the aim is to derive linear equilibrium equations, it follows