MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 4 pptx

98 Structural elements Figure 2.21. Warping function of rectangular cross-sections that which governs the transverse displacement of a stretched membrane. More gen- erally, the problem can be solved numerically by using the finite element method. As a general result, the following inequality holds: J T ≤ J [2.66] which means that the stiffness due to the elastic deformation of the cross-sections is less with warping than without. The equality holds for a circular cross-section only. The values of the torsion constant for various cross-sectional shapes may be found in several books, see for instance [BLE 79]. A few results are quoted below: Ellipse: J T = πa 3 b 3 a 2 + b 2 Square: J T = 0,1406 a 4 Rectangle: J T = γa 3 b 3 a 2 + b 2 with a/b 248∞ γ 0.286 0.299 0.312 1/3 Straight beam models: Newtonian approach 99 Hollow rectangle (thickness: h a and h b ): J T = 2h a h b (a −h a ) 2 (b −h b ) 2 ah a + bh b − h 2 a − h 2 b Open thin walled cross-sections: J T = Ph 3 /3(P length, h thickness) Closed thin walled cross-sections: J T = 4Sh P (P perimeter, h thickness, S area of the closed part) A few numerical applications of theseformulaeareuseful to illustrate the import- ance of warping. Taking a square cross-section as a first example, the ratio J T /J is found to be 0.1406 × 6 ∼ = 0.84. For a rectangular section of aspect ratio a/b = 2, J = (5/6)b 4 and J T = 0.2860×(5/6)b 4 ∼ = 0.46b 4 or J T /J ∼ = 0.55. Finally, if the aspect ratio of the rectangle is very large (a/b ≫ 1) it is found that J T ≃ ab 3 /3, so J T /J ≃ 4(a/b) 2 ≪ 1. To conclude on the subject, it may be added that if warping is prevented by clamping both ends, local axial stresses are generated, which are especially important in the case of open thin walled cross-sections. The reader inter- ested in such aspects of the problem may be referred for instance to [TIM 51], [PIL 02]. 2.2.4 Pure bending mode of deformation As a preliminary we must emphasize that, in contrast to a frequently used definition according to which pure bending of a beam means that no shear forces arise, by pure bending mode of deformation we mean here that the beam is assumed to be flexed without any superposition of torsional or shear strains. 2.2.4.1 Simplifying hypotheses of the Bernoulli–Euler model As a first approximation, transverse shear deformation is neglected. Such a simplification gives rise to the Bernoulli–Euler model, which is based on the three following simplifying assumptions: 1. As the beam is flexed, the cross-sections remain rigid and perpendicular to the centroid line in the deformed configuration. 2. The transverse shear forces are derived from the equilibrium equation written in terms of moments, and do not arise from elastic shear stresses. 3. The rotational inertia of the beam cross-sections is negligible in comparison with the translation inertia. 100 Structural elements Figure 2.22. Beam element: bending without shear Such assumptions, represented schematically in Figure 2.22, are adopted for math- ematical convenience and their range of validity will be examined in Chapter 3 subsection 3.2.2. The first assumption implies that the magnitude of the small rotation angle due to bending is equal to the slope of the deformed centroid line. In agreement with the sign convention for rotations in a direct reference frame, the following is obtained: ψ y =− ∂Z ∂x ; ψ z =+ ∂Y ∂x ⇒ χ yy =− ∂ 2 Z ∂x 2 ; χ zz = ∂ 2 Z ∂y 2 [2.67] The global strains χ zz and χ yy may be interpreted as small (linear) curvatures of the deformed centroid line. According to the second assumption, the trans- verse shear forces are identified with the forces induced by the holonomic relations [2.67]. The third assumption implies the nullity of the inertia terms in the two last equations [2.21], which reduce thus to quasi-static equations. Validity of this sim- plification shall be discussed in Chapter 3, based on energy considerations. It will be shown that the kinetic energy due to the bending rotation of the cross-sections remains much smaller than the kinetic energy due to the transverse displacements provided the slenderness ratio of the beam is sufficiently large. 2.2.4.2 Local equilibrium As shear is neglected, the relations [2.12] cannot be used to calculate the trans- verse internal forces which are required to ensure the equilibrium under transverse loading. This is the reason why Q y and Q z are defined from the static form of the two last equations [2.21]. It becomes: Q y =− ∂M z ∂x − M (e) z (x; t) Q z =+ ∂M y ∂x + M (e) y (x; t) [2.68] Straight beam models: Newtonian approach 101 Substitution of [2.67] into [2.13] leads to: M y =−EI y ∂ 2 Z ∂x 2 ; M z =+EI z ∂ 2 Y ∂x 2 [2.69] and with [2.68] transverse shear forces are given as: Q y =− ∂ ∂x  EI z ∂ 2 Y ∂x 2  − M (e) z (x; t) Q z =− ∂ ∂x  EI y ∂ 2 Z ∂x 2  + M (e) y (x; t) [2.70] The vibration equations are then obtained by substituting [2.70] into the transverse equilibrium equations [2.18]: ∂ 2 ∂x 2  EI z ∂ 2 Y ∂x 2  + ρS ∂ 2 Y ∂t 2 = F (e) y (x; t) − ∂M (e) z (x; t) ∂x ∂ 2 ∂x 2  EI y ∂ 2 Z ∂x 2  + ρS ∂ 2 Z ∂t 2 = F (e) z (x; t) + ∂M (e) y (x; t) ∂x [2.71] Equations [2.71] were established first by Daniel Bernoulli 1735 and solutions found by Euler 1744. The external loading is expressed in terms of force and moment densities per unit length. The sketches of Figure 2.23 are helpful to understand the equivalence between a non uniform bending moment and a transverse force field. Indeed, if  M (e) is depicted as torques distributed along the beam, it becomes clear that the axial variation grad(  M (e) )dx is equivalent to a transverse force density, which arises as the resultant of the torque components acting at abscissa x. The force balance is: −ρS ¨ Zdx+(Q z (x + dx) −Q z (x)) +  M (e) y (x + dx) −M (e) y (x)  +F (e) z dx = 0 ⇒ ρS ¨ Z − ∂Q z ∂x = F (e) z + ∂M (e) y ∂x −ρS ¨ Ydx+(Q y (x + dx) −Q y (x)) −  M (e) z (x + dx) −M (e) z (x)  +F (e) y dx = 0 ⇒ ρS ¨ Y − ∂Q y ∂x = F (e) y − ∂M (e) z ∂x 102 Structural elements Figure 2.23. Representation of the term ∂  M (e) /∂x 2.2.4.3 Elastic boundary conditions Let a beam be bent in the plane Oxz. The end x = 0 is supported by a spring of stiffness coefficient K z acting in the Oz direction. The corresponding boundary condition is: Q z (0; t) −K z Z(0; t) = 0 ⇒ EI y ∂ 3 Z ∂x 3     x=0 + K z Z(0; t) = 0 [2.72] If the supporting spring acts on the rotation with a stiffness coefficient K ψ y , the condition becomes: M y (0; t) −K ψ y ψ y (0; t) = 0 ⇒−EI y ∂ 2 Z ∂x 2     x=0 + K ψ y ∂Z ∂x     x=0 = 0 [2.73] The two limit cases for which the spring stiffness tends to zero, or alternatively tends to infinity, provide us with the so called ‘standard’ boundary conditions which have specific names: free end: ∂ 2 Z/∂x 2 = ∂ 3 Z/∂x 3 = 0 pinned end: Z = ∂ 2 Z/∂x 2 = 0 sliding support: ∂Z/∂x = ∂ 3 Z/∂x 3 = 0 clamped end: Z = ∂Z/∂x = 0 [2.74] A symbolic representation of these conditions is shown in Figure 2.24. In the three-dimensional case (bending in the Oxy and the Oxz plane) for a clamped end, all the six displacement components are zero and for a pinned end the three translations are zero and the two transverse rotations are free. The pinned support is represented here as a pair of triangles, which stand for two knife edge supports, Straight beam models: Newtonian approach 103 Figure 2.24. Bending of a beam: symbolic representation of standard support conditions instead of only one as is most often accepted. This is to stress that the support condition is bilateral, i.e. transverse displacement is prohibited whatever its sign may be. Indeed, a unilateral support condition such as Z ≥ 0 is nonlinear in nature, as further discussed in Chapter 4, in relation to problems of impacted beams. 2.2.4.4 Intermediate supports Restricting the study to the case of elastic supports, the equations [2.73] and [2.74] become: EI y  ∂ 3 Z ∂x 3     x 0+ − ∂ 3 Z ∂x 3     x 0−  + K z Z(x 0 ; t) = 0 EI y  ∂ 2 Z ∂x 2     x 0+ − ∂ 2 Z ∂x 2     x 0−  − K ψ y ∂Z ∂x     x 0 = 0 [2.75] 2.2.4.5 Concentrated loads There are two types of concentrated loads, namely transverse forces and bending moments, which induce the following finite discontinuities: Q Z (x 0+ ) − Q Z (x 0− ) + F (e) z = 0 ⇒ EI y  ∂ 3 Z ∂x 3     x 0+ − ∂ 3 Z ∂x 3     x 0−  = F (e) z (t) M y (x 0+ ) − M y (x 0− ) + M (e) y = 0 ⇒ EI y  ∂ 2 Z ∂x 2     x 0+ − ∂ 2 Z ∂x 2     x 0−  = M (e) y (t) [2.76] 104 Structural elements 2.2.4.6 General solution of the static and homogeneous equation The beam is supposed to be homogeneous, of constant cross-section, and bent in the Oxz plane. In statics, if the right-hand side of the second equation [2.71] is zero, the solution reduces to: Z(x) = ax 3 + bx 2 + cx + d [2.77] where a, b, c, d are constants which are determined from the boundary conditions. 2.2.4.7 Application to some problems of practical interest example 1.–Cantilevered beam loaded by its own weight Let us consider a beam of constant cross-section set in a cantilevered configuration, i.e. clamped at one end and free at the other, see Figure 2.25. The beam lies horizontally in a uniform gravity field described by the acceleration vector g. It is thus loaded uniformly by its own weight  F 0 = ρSLg. The loading function is the force density per unit length of beam −F 0 /L = f 0 . The boundary value problem is written as: EI d 4 Z dx 4 = f 0 Z(0) = Z ′ (0) = 0 Z ′′ (L) = Z ′′′ (L) = 0 where the prime stands for a derivation with respect to x. Figure 2.25. Cantilevered beam loaded by its own weight Straight beam models: Newtonian approach 105 The solution is of the type: Z(x) = ax 4 + bx 3 + cx 2 + dx + e; the constants are determined by the loading and the boundary conditions as follows. Clamped end at x = 0 → d = e = 0 → Z(x) = ax 4 + bx 3 + cx 2 Loading: 24a = f 0 EI → a = f 0 24EI Free end at x = L Z ′′′ (L) = 24ax +6b = 0 ⇒ b =− f 0 L 6EI Z ′′ (L) = f 0 L 2 2EI − f 0 L 2 EI + 2c = 0 → c = f 0 L 2 4EI then: Z(ξ) = Z 0 (ξ 4 − 4ξ 3 + 6ξ 2 ) where ξ = x L and Z 0 = F 0 L 3 24EI The maximum deflection is Z max = Z(1) = 3Z 0 = F 0 L 3 /(8EI) The reactions at the clamped end balance exactly the global internal stresses exerted on the corresponding cross-section. This result agrees both with the action- reaction principle and with the sign convention about stresses, according to which the stresses are the efforts (forces and moments) exerted by the right-hand part of the beam on the left-hand part. Vice versa, the efforts on the supports are the efforts exerted by the left-hand part on the right-hand part. So a clamped support acts as a force and as a moment given by: R z =−Q z = EI d 3 Z dx 3     x=0 = 6bEI =−f 0 L =−F 0 M Ry =−M y = EI d 2 Z dx 2     x=0 =−2cEI =+ F 0 L 2 As expected, the reactions are found to be independent of the material law. Further, as the global equilibrium of the beam is concerned, the above results indicate that the external loading is equivalent to the weight  F 0 concentrated at mid-span of the beam. This resultant is necessarily balanced by the support reaction R z + F 0 = 0. Then the support reaction is R z =−Q z =−F 0 . In the same way, the external loading moment at x = 0isM (e) =−F 0 L/2 and is exactly balanced by the reaction moment M Ry = F 0 L/2. example 2.–Loads applied to the free end 106 Structural elements Deflection of a cantilevered beam loaded by a static force concentrated at x = L is found to be: Z(ξ) = Z 0 (3ξ 2 − ξ 3 ) where ξ = x L and Z 0 = F (e) z L 3 6EI If the end conditions and the load location are reversed the last result becomes: Z(ξ) = Z 0 (3(1 − ξ) 2 − (1 −ξ) 3 ) = Z 0 (ξ 3 − 3ξ + 2) If an external moment M (e) y is exerted at x = L, instead of an external force, the deflection is: Z(ξ) =− M (e) y L 2 2EI ξ 2 If the end conditions and the load location are reversed the last result becomes: Z(ξ) = M (e) y L 2 2EI (ξ − 1) 2 The deflection in each of these cases is sketched in Figure 2.26. example 3.–Transverse force applied to a sliding support Figure 2.27 shows a beam which is supported at x = 0 by a sliding support and at x = L by a pinned support. The external load is a transverse force applied at x = 0. It is recalled that, in the unloaded case, a sliding support is described by the conditions Z ′ (0) = 0 (no rotation) and Z ′′′ (0) = 0 (no shear force). In the presence of the external load, these conditions have to be modified. To find out Figure 2.26. Schematic view of deflected cantilevered beams subject to static end loads Straight beam models: Newtonian approach 107 Figure 2.27. Loading on a sliding support Figure 2.28. Equilibrium of the loaded section how, the simplest way is to adapt the Figure 2.14 to the case of a transverse loading, as shown in Figure 2.28. The balance of the transverse forces implies: F 0 + Q z (+ε) −Q z (−ε) = 0; but Q z (−ε) = 0 since −ε lies outside from the beam. On the other hand, lim ε→0 Q z (+ε) =−EI∂ 3 Z/∂x 3 , which leads to the load condition EI∂ 3 Z/∂x 3 = F 0 . Thus the boundary value problem to be solved is written as: EI d 4 Z dx 4 = 0 Z ′ (0) = Z ′′′ (0 − ) = 0; Z(L) = Z ′′ (L) = 0 EIZ ′′′ (0 + ) = F 0 [...]... results of Chapter 2 subsection 2. 1.5, the elastic energy Ee is expressed as: Ee = 1 2 L ES 0 1 + 2 2 ∂X ∂x L + GS ∂ψx ∂x GJT 0 ∂Ys ∂x 2 + EIz 2 + ∂ 2 Yb ∂x 2 ∂Zs ∂x 2 dx 2 + EIy ∂ 2 Zb ∂x 2 2 dx [3 .2] The kinetic energy Eκ is: Eκ = 1 2 L 0 1 + 2 ˙ ˙ ˙ ˙ ˙ ρS{X 2 + (Ys + Yb )2 + (Zs + Zb )2 } dx L 0 2 2 2 ρ(J ψx + Iy ψy + Iz ψz ) dx [3.3] 1 32 Structural elements In [3.3] the rotatory inertia of the... ∂ 2X − z 2 − α θ(z) ∂x ∂x σxx (z) = E [2. 85] ˜ ˜ The global stresses N (x) and My (x) are then given by: ˜ N (x) = (S) σxx (z)dS = ES ∂X ˜ − αEa N (x) = ES ∂x ˜ My (x) = (S) ∂X − ∂x a /2 Eaz −a /2 ∂ 2Z dz − αEa ∂x 2 a /2 θ (z)dz −a /2 [2. 86] a /2 θ (z) dz −a /2 σxx (z)z dS = −EI ∂ 2Z − αEa ∂x 2 a /2 θ (z)z dz −a /2 a4 12 [2. 87] where I = Figure 2. 37 Beam subjected to a transverse temperature gradient 122 Structural... elastic-plastic state of the beam, since according to the Bernoulli-Euler model it is independent of the material law The bending moment becomes: My = −Ea ∂ 2Z ∂x 2 z0 −z0 z2 dz − 2 0 a 3 2Eaz0 ∂ 2 Z My = − + σ0 a 3 ∂x 2 h 2 h /2 z0 2 z dz ⇒ [2. 99] 2 − z0 (e) This moment balances the torque of the applied force −Fz (L − x) Hence, the limit load for failure is given by: Ffail = σ0 ah h 4L ⇒ 3 Ffail = Fz0 2 [2. 100]... L3 192EI Turning to the non-symmetrical case ξ0 = 0.5, the coefficients are found to be: F0 L 3 F 0 L3 2 3 2 3 (3ξ0 − 2 0 − 1); A2 = (3ξ0 − 2 0 ) 6EI 6EI F 0 L3 2 F0 L 3 B1 = ξ0 (ξ0 − 1 )2 ; B2 = ξ (1 − ξ0 ) 2EI 2EI 0 A1 = Figure 2. 30 refers to a steel beam (E = 2. 11011 Pa) of length L = 1m and square cross-section (a = 2 cm) The transverse load of 1 kN is applied at ξ0 = 0 .25 The abscissa ξm of maximum... h χ0 2 χ [2. 1 02] then, My = − 3 Eah3 Eah3 χ0 χ0 − 8 24 χ 2 [2. 103] The elastic limit of the bending moment is introduced by the elastic condition: M0 = Eah3 χ0 12 [2. 1 04] The relation between the bending moment and the curvature is finally written as: sign(χ χ ) > 0 ⇒ ˙ My 1 3− = M0 2 χ0 χ My χ − χp sign(χ χ ) < 0 ⇒ ˙ = M0 χ0 2 [2. 105] 128 Structural elements [2. 105] is known as a global law of elastic-plasticity... parts of the crosssections which are plastic and that which is still in the elastic domain, as shown in Figure 2. 42 Hence, one starts from the definition of the bending moment in terms of local stresses: a /2 My (x) = h /2 −a /2 −h /2 σxx (x, y, z)z dz dy [2. 93] The local elastic stress is: σxx (x, z) = −zE ∂ 2Z ∂x 2 [2. 94] x Its maximum magnitude is reached at the lateral boundaries of the beam (z = ±h /2) ... Z2 (0.5) = Z1 (0.5) = 0 Calculation of the four constants is quite straightforward: ′ Z1 (0) = Z1 (0) = 0 ⇒ Z1 (ξ ) = A1 ξ 3 + B1 ξ 2 3A1 4 F 0 L3 ′′′ 3 EI Z1 (0.5) = −F0 L /2 ⇒ 6EI A1 = −F0 /2 ⇒ A1 = − 12EI ′ Z1 (0.5) = 0 ⇒ 3A1 (0.5 )2 + 2B1 (0.5) = 0 ⇒ B1 = − 110 Structural elements Thus, the beam deflection is given by: F0 L 3 2 3 ξ −ξ ξ ≤ 0.5 12EI 4 F0 L 3 1 Z2 (ξ ) = (1 − ξ )2 ξ − ξ ≥ 0.5 12EI 4. .. 2 Z 2 ∂x 2 [2. 95] x The curvature is given by: My = −Fz(e) (L − x) = −EIy (e) My ∂ 2Z 12Fz (L − x) ∂ 2Z = ⇒ = [2. 96] EIy ∂x 2 ∂x 2 Eah3 126 Structural elements then, (e) (|σxx (x)|)skin = 6Fz (L − x) ah2 [2. 97] As could be expected, the local longitudinal stress is maximum at the clamped end The load beyond which plasticity is initiated in the skin layer is given by: σ0 = 6F0 L σ0 ah2 ⇒ F0 = 6L ah2... failure of the structure 2. 4 .2 Elastic-plastic behaviour under bending As shown in Figure 2. 41 , we consider now a cantilevered beam loaded by a (e) transverse force Fz , applied at the free end Figure 2. 40 Static equilibrium of a uniformly tensioned beam element Figure 2. 41 Cantilevered beam loaded at the free end by a transverse force Straight beam models: Newtonian approach 125 Figure 2. 42 σxx stress... expression of the bending moment, we get: My = − =− 3 2Eaz0 χ + σ0 a 3 h 2 3 2Eaz0 χ + Eaz0 χ 3 2 2 − z0 h 2 2 2 − z0 3 Eaz0 Eah2 z0 − 4 3 = χ [2. 101] It is convenient to introduce, as a pertinent scaling factor, the curvature χ0 corresponding to the elastic limit beyond which plasticity is initiated somewhere in the beam χ0 is defined by the elastic condition σ0 = Eh /2 0 The boundary of the elastic . Z(x) = ax 4 + bx 3 + cx 2 Loading: 24 a = f 0 EI → a = f 0 24 EI Free end at x = L Z ′′′ (L) = 24 ax +6b = 0 ⇒ b =− f 0 L 6EI Z ′′ (L) = f 0 L 2 2EI − f 0 L 2 EI + 2c = 0 → c = f 0 L 2 4EI then: Z(ξ). be: A 1 = F 0 L 3 6EI (3ξ 2 0 − 2 3 0 − 1); A 2 = F 0 L 3 6EI (3ξ 2 0 − 2 3 0 ) B 1 = F 0 L 3 2EI ξ 0 (ξ 0 − 1) 2 ; B 2 = F 0 L 3 2EI ξ 2 0 (1 − ξ 0 ) Figure 2. 30 refers to a steel beam (E = 2. 110 11 Pa) of length. results are quoted below: Ellipse: J T = πa 3 b 3 a 2 + b 2 Square: J T = 0, 140 6 a 4 Rectangle: J T = γa 3 b 3 a 2 + b 2 with a/b 24 8∞ γ 0 .28 6 0 .29 9 0.3 12 1/3 Straight beam models: Newtonian approach