MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 6 ppsx

178 Structural elements Figure 3.16. Local and global coordinate systems structure since the local reference frames differ necessarily along the horizontal beam and along the legs. It is precisely this transformation which produces the coupling terms between bending and axial terms which arise in the [K] matrix. Let us consider the beam element of Figure 3.16, which is assumed to lie in the Oxz plane. The local frame can be transformed into the global frame by the product of the translation −→ IO and the plane rotation α. Neither the forces nor the displacements are modified by a translation, which thus can be disregarded. Obviously, the same is not true as far as the rotation is concerned. The rotation is described by the matrix [R] which relates the displacements by [q]=[R][q ′ ]. The transformation rule of the node labelled J is:          X J Y J Z J ψ xJ ψ yJ ψ zJ          =        cos α 0 sin α 000 010 000 −sin α 0 cos α 000 000cosα 0 sin α 000 010 000−sin α 0 cos α                  X ′ J Y ′ J Z ′ J ψ ′ xJ ψ ′ yJ ψ ′ zJ           [3.110] More generally the rotation matrix takes the form: [R]=  [R][0] [0][R]  ; with [R]=   ℓ x ′ m x ′ n x ′ ℓ y ′ m y ′ n y ′ ℓ z ′ m z ′ n z ′   [3.111] Straight beam models: Hamilton’s principle 179 where ℓ i ′ , m i ′ , n i ′ are the director cosines of the local axes as expressed in the global system. Since energy is invariant through a transformation of axes, the element matrices and force vectors are found to comply with the following conditions: [ q I q J ][K (n) ]  q I q J  =[ q ′ I q ′ J ][K ′(n) ]  q ′ I q ′ J  ⇒[K (n) ]=[R] T [K ′(n) ][R] [ ˙q I ˙q J ][M (n) ]  ˙q I ˙q J  =[ ˙q ′ I ˙q ′ J ][M ′(n) ]  ˙q ′ I ˙q ′ J  ⇒[M (n) ]=[R] T [M ′(n) ][R] [ q I q J ]  Q I Q J  =[ q ′ I q ′ J ]  Q ′ I Q ′ J  ⇒  Q I Q J  =[R] T  Q ′ I Q ′ J  [3.112] Once the element matrices and vectors are suitably transformed into the global system of coordinates, the finite element model can be assembled according to the same rules as described in subsection 3.4.2.5. Colour Plate 3 shows the elastic response of a portal frame to a concentrated force as computed by using the finite element software CASTEM 2000 [CAS 92]. The finite element model uses two nodes beam elements. The material is steel and the beams have uniform rectangular cross-sections 20 cm × 10 cm. The height of the vertical legs is 8 m and the length of the horizontal beam (lintel) is 12 m. The legs are assumed to be clamped at the lower ends. A concentrated force is applied at some point of the lintel, load magnitude is 2.5 kN. which is assumed successively to be vertical (case a), along the horizontal beam (case b), and finally in the transverse Oy direction (case c). For each loading case, two figures are shown. The figures on the left-hand side show the global reference frame in green, the finite element mesh in black, the external force represented as a red arrow and finally the support reactions as blue arrows for the forces and magenta arrows for the moments. The length of the arrows is determined by a magnifying factor 1000. The figures on the right-hand side show the undeformed and the deformed frame (the black and red lines respectively). Real displacements are magnified by a factor 100 to make deformations clearly noticeable. In case (a), thelintel and the legs are deflected. As expected, maximumdeflection occurs along the lintel. The reaction forces comprise a vertical component equal to the shear forces at the ends of the lintel, transmitted to the legs as compressive forces, and horizontal components in the plane of the frame, which are equal to the shear forces at the clamped ends of the legs. The moments are perpendicular to the plane of the frame. They are clearly related to the bending of the legs. As the loaded point is nearer to one leg than the other, the two legs are not flexed by the same amount, but in the ratio of the lengths of the lever arms defined by the loaded point and the extremities of the lintel. 180 Structural elements In case (b), maximum deflection occurs in the legs, which are deflected by the same amount. The bending moments at the junction between the legs and the lintel induce in-plane bending of the latter. Deflection is found to be antisymmetric because the moments exerted at the ends of the lintel are the same. As a consequence, one leg is compressed and the other is stretched. In case (c), the legs and the lintel are deflected out of the frame plane. However, due to the asymmetrical position of the loaded point with respect to the two legs, the portal frame is also twisted about the vertical direction. Accordingly, though the major component of the moment reactions is horizontal and, in the plane of the frame, equal to half the load times the length of the legs, they also have a vertical component which differs from one leg to the other and a horizontal component in the out-of-plane direction. 3.4.3.4 Transverse load resisted by string and bending stresses in a roof truss The example of the portal frame was used in the last subsection to bring out that in an assembly of connected straight beams transverse shear stresses can be transmitted as longitudinal stresses to another part of the assembly due to geometrical effects. This leads also to a coupling of the basic modes of deformation of the individual beams. However, because of the large difference which usually exist between the coefficients of the longitudinal and flexure stiffness matrices [3.98] and [3.106], suitably scaled by the coefficients K ℓ = ES/ℓ and K b = EI/ℓ 3 respectively, coupling is barely detectable when looking at the deformed portal frame. Indeed, the axial displacements of the vertical legs are found to be quite negligible in comparison with the transverse deflection due to bending. To analyse this point further, it is instructive to study the assembly depicted in Figure 3.17 which stands Figure 3.17. Beam assembly used as roof truss Straight beam models: Hamilton’s principle 181 for a roof truss made of two identical straight beams (A 1 A 2 ) and (A 2 A 3 ) tilted by the angle α with respect to the horizontal Ox direction. The beams are rigidly connected to each other at (A 2 ) and the rooting points (A 1 ) and (A 3 ) are assumed to be either pinned or clamped. A vertical force  F 0 is applied to (A 2 ).Ifα = 0, the assembly responds to the load as a pinned-pinned, or clamped-clamped flexed beam and if α = 90 ◦ , it responds as a compressed beam. It is thus of interest to analyse the response of the roof truss when α is varied from 0 to 90 ◦ . An analytical solution of this problem can be obtained conveniently by using a finite element model reduced to two beam elements and to the nodal displacements of interest. Let us consider first the pinned-pinned configuration. In the local coordinate system of the first element (A 1 A 2 ), the displacement variables are denoted U 1 , U 2 (axial displacements), W 1 , W 2 (transverse displacements in the Oxz plane), ϕ 1 , ϕ 2 (small rotations about the Oy axis). These variables define the vector displacement [U 1 W 1 ϕ 1 U 2 W 2 ϕ 2 ] T . By using the matrix [3.109], the stiffness matrix of the first element is written as: K ℓ        10 0−10 0 012γ −6γℓ 0 −12γ −6γℓ 0 −6γℓ 4γℓ 2 06γℓ 2γℓ 2 −10 0 1 0 0 0 −12γ 6γℓ 012γ 6γℓ 0 −6γℓ 2γℓ 2 06γℓ 4γℓ 2        where γ = K b K ℓ = I Sℓ 2 ≃ 1 12η 2 ≪ 1 η is the slenderness ratio of the element, assumed to be much larger than one. In the same way, the vector displacement of the second element (A 2 A 3 ) is written as:  U 2 W 2 ϕ 2 U 3 W 3 ϕ 3  T The stiffness matrix is the same as that of the first element. However, due to the boundary conditions, U 1 = U 3 = W 1 = W 3 = 0 and the displacement vectors and stiffness matrices can be reduced as follows:  U 1 W 1 ϕ 1 U 2 W 2 ϕ 2  T ⇒  ϕ 1 U 2 W 2 ϕ 2  T 182 Structural elements K ℓ        10 0−10 0 012γ −6γℓ 0 −12γ −6γℓ 0 −6γℓ 4γℓ 2 06γℓ 2γℓ 2 −10 0 1 0 0 0 −12γ 6γℓ 012γ 6γℓ 0 −6γℓ 2γℓ 2 06γℓ 4γℓ 2        ⇒ K (1) = K ℓ     4γℓ 2 06γℓ 2γℓ 2 010 0 6γℓ 012γ 6γℓ 2γℓ 2 06γℓ 4γℓ 2      U 2 W 2 ϕ 2 U 3 W 3 ϕ 3  T ⇒  U 2 W 2 ϕ 2 ϕ 3  T K ℓ        10 0−10 0 012γ −6γℓ 0 −12γ −6γℓ 0 −6γℓ 4γℓ 2 06γℓ 2γℓ 2 −10 0 1 0 0 0 −12γ 6γℓ 012γ 6γℓ 0 −6γℓ 2γℓ 2 06γℓ 4γℓ 2        ⇒ K (2) = K ℓ     1000 012γ −6γℓ −6γℓ 0 −6γℓ 12γ 2γℓ 2 0 −6γℓ 2γℓ 2 4γℓ 2     Then, K (1) is rotated by the angle α by using the coordinate transformation rule:    ψ 1 X 2 Z 2 ψ 2    =    10 0 0 0 C −S 0 0 SC0 00 0 1       ϕ 1 U 2 W 2 ϕ 2    ; where C = cos α and S = sin α The rotated stiffness matrix is found to be: K (1) α = K ℓ     4γℓ 2 6γℓS 6γℓC 2γℓ 2 6γℓS C 2 + 12γS 2 CS(12γ −1) 6γℓS 6γℓC CS(12γ −1)S 2 + 12γC 2 6γℓC 2γℓ 2 6γℓS 6γℓC 4γℓ 2     Straight beam models: Hamilton’s principle 183 K (2) is rotated by the angle −α by using the coordinate transformation rule:    X 2 Z 2 ψ 2 ψ 3    =    C +S 00 −SC00 0010 0001       U 2 W 2 ϕ 2 ϕ 3    The rotated stiffness matrix is found to be: K (2) α = K ℓ     C 2 + 12γS 2 −CS(12γ −1) 6γℓS 6γℓS −CS(12γ −1)S 2 + 12γC 2 −6γℓC −6γℓC 6γℓS −6γℓC 4γℓ 2 2γℓ 2 6γℓS −6γℓC 2γℓ 2 4γℓ 2     The finite element model of the problem is assembled as:       4γℓ 2 6γℓS 6γℓC 2γℓ 2 0 6γℓS 2(C 2 + 12γS 2 ) 012γℓS 6γℓS 6γℓC 02(S 2 + 12γC 2 ) 0 −6γℓC 2γℓ 2 12γℓS 08γℓ 2 2γℓ 2 06γℓS −6γℓC 2γℓ 2 4γℓ 2            ψ 1 X 2 Z 2 ψ 2 ψ 3      =      0 0 −F 0 0      Finally, the solution can be conveniently obtained by noticing that because of the symmetry of the problem with respect to the Oz axis, it is necessary that X 2 = 0 and ψ 3 =−ψ 1 . As a consequence: Z 2 (α) = −F 2(S 2 + 3γC 2 )K ℓ = −F 2(K ℓ (sin α) 2 + 3K b (cos α) 2 ) [3.113] and ψ 3 =−ψ 1 = 3Z 2 cos α 2ℓ The solution [3.113] brings out that the structure resists the vertical force essen- tially through its longitudinal stiffness as soon as the tilt angle is larger than 1/η. Such a result is of paramount importance for designing structures resistant 184 Structural elements Figure 3.18. Deflection versus the tilt angle of the roof truss Figure 3.19. Roof truss in pinned-pinned configuration, from the above to the bottom plots: unloaded and loaded structure, plots of the vertical displacement in meters and flexure angle in radians Straight beam models: Hamilton’s principle 185 Figure 3.20. Roof truss in the clamped-clamped configuration: unloaded and loaded structure Figure 3.21. Roof truss α =45 ◦ , in pinned-pinned configuration: unloaded and loaded structure, plots of the vertical displacement in meters and flexure angle in radians to transverse loads, as will be further detailed in Chapters 7 and 8 devoted to curved structures. Figure 3.18 refers to a roof truss made of wood beams (E = 10 9 Pa, ℓ = 10 m, square cross-sections a = 20 cm) loaded by a vertical force F = 10 kN applied to (A 2 ). It plots the vertical displacement of (A 2 ) versus the tilt angle in degree. The dashed lines stand for the limit cases α = 0 and α = 90 ◦ . As soon as α>5 ◦ magnitude of the displacement is less than 1% than its value at α = 0. Figure 3.19 refers to the pinned-pinned configuration. It shows the deflection of the whole structure for α = 5 ◦ , as computed by using a finite element model comprising 40 two node beam elements. The load vector and the non-deformed structure are also represented. Vertical displacement and flexure angle are plotted along the beams. Figure 3.20 refers to the clamped-clamped configuration. The magnitude of the maximum deflection is less than in the pinned-pinned configuration (10.9 cm instead of 14.7 cm) as the result of a stiffening effect in bending. A simple analytical 186 Structural elements calculation carried out in the same manner as above gives for the clamped-clamped configuration: Z 2 (α) = −F 2(K ℓ (sin α) 2 + 12K b (cos α) 2 ) and ψ 1 = ψ 2 = ψ 3 = 0 Finally, Figure 3.21 refers to a roof truss α = 45 ◦ , in pinned-pinned con- figuration. The deflection is drastically reduced in comparison with the case α = 5 ◦ . 3.4.4 Saving DOF when modelling deformable solids The methodology followed to model deformable solids as equivalent beams is worth brief mention to emphasize the practical importance of minimizing judi- ciously the number of degrees of freedom when modelling continuous systems. This can be sketched graphically, as shown in Figure 3.22. Figure 3.22a sketches the ‘real’ structure as a 3D solid, supporting a load distributed as a 2D pressure field. In Figure 3.22b we take advantage of the slenderness of the body to reduce the size of the problem by shifting from a 3D medium to an equivalent 1D medium, via a beam model. Finally, in Figure 3.22c the beam is discretized in finite ele- ments. In this way, the final model to be solved has a finite number of DOF 1 Z 1 = 0 ′ ′ Z 1 = 0F n 2N Z N +1 = 0 ′ ′ Z N +1 = 0 N+1 (c) f(x) (b) P (a) Figure 3.22. From the deformable solid to the finite element model Straight beam models: Hamilton’s principle 187 and can be solved by using a computer. The whole process justifies the interest in discretizing a continuum; one other specificity of this modelling procedure is its ability to identify the most pertinent DOF in order to minimize their number. This step is partially fulfilled by a good appreciation of the structure behaviour. In the following chapter another efficient method for suppressing superfluous degrees of freedom is described, based on the natural modes of vibration of the structures. [...]... modes of vibration are mixing, bending, stretching and torsion of the beam components As shown in Figure 4 .20 , f1 = 0.957 Hz M1 = 27 10 kg f 2 = 1 .20 Hz M2 = 21 80 kg f3 = 2. 21 Hz M3 = 1370 kg f4 = 2. 45 Hz M4 = 933 kg f5 = 5.70 Hz M5 = 1540 kg f6 = 6. 09 Hz M6 = 21 20 kg f 8 = 9 .24 Hz M5 = 1590 kg f 9 = 12. 1 Hz M6 = 1 720 kg f 7 = 6. 88 Hz M7 = 1180 kg Figure 4 .20 First natural modes of vibration of a portal... 2 2 dξ 2 dξ where ̟ = ωL ; c0 2 = du dξ = 0; 1 2 νR L 0 = u(1) = 0 2 ; u= X ; L ξ= u = Aekξ + Be−kξ ⇒ k2 + ̟ 2 − k2 γ 2 ̟ 2 = 0 ⇒ k = ± where k = Lk is a dimensionless wave number x L i̟ 1 − γ 2 2 1 /2 200 Structural elements The condition ϕ(1) = 0 leads to: 2 kn = i(2n + 1)π /2 ⇒ ϕn (ξ ) = cos (|kn |ξ ) ; ̟n = |kn |2 1 + |kn |2 γ 2 In accordance with the variation of the phase speed displayed in... the beam This leads to the Love–Rayleigh equation [3.11], written here in the case of a cylindrical rod as: 2 c0 ¨ ∂ 2X ∂ 2X ¨ − X + β 2 2 = 0; 2 ∂x ∂x where β 2 = ν 2R2 2 The phase speed of the Love–Rayleigh model, as expressed in terms of the wave number k, is found to be: c = c0 1 1 + 1 (kνR )2 2 ; where k = 2 λ [4 .6] These waves √ thus found to be dispersive and c decreases as kR increases, are... functional, which can also be interpreted as a generalized mass is: 2 Eκ = Mg = (M + 0 .24 Mb )ZM ; L where 0 2 Z 2 (x) dx = 0 .24 LZM Finally, the first natural frequency of the system is given by the Rayleigh quotient: 2 ω1 = Kg F0 3EI = = Mg (M + 0 .24 Mb )ZM (M + 0 .24 Mb )L3 Modal analysis methods 20 9 Figure 4. 16 Beam on elastic foundation 4 .2. 3 .2 Beam on elastic foundation In some instances, one has to deal... = ; 2 2 Kn = 1 ES 2 ̟ 2 L n [4. 12] The modal coefficients ̟n , a, b, pertinent to standard boundary conditions, are reported in Table 4.1 It is worth emphasizing the remarkable result that the natural Modal analysis methods 197 Table 4.1 Longitudinal modes of vibration of a straight beam B.C ̟n free-free fixed-fixed free-fixed nπ nπ (1 + 2n) π 2 (1 + 2n) π 2 fixed-free a b n = 0, 1, 2, n = 1, 2, ... wavelength of the mode For this reason the calculation is detailed below From the equations [3.34] and [3.35] the following modal equations are derived: 2 ω ρ ρ ∂ 2 Zs = − 2 Z; where Z = Zs + Zb κG ∂x 2 ∂ 4Z ρI E ∂ 2Z + EI 4 = 0 2 −S Z+I 1+ κG κG ∂x 2 ∂x [4.19] The pinned ends imply that: Z(0) = Z(L) = 0; ∂ 2 Zb ∂x 2 0,L =0 [4 .20 ] However, as a consequence of the flexure equation [4.19], [4 .20 ] can be... ρSL /2 = M /2; if j = k 0; otherwise ρSL̟j2 /2; 0; if j = k otherwise Turning now to the Rayleigh–Love model, the mode shapes are found to be the same as those related to the basic model, but the modal frequencies are lower, as could be expected The modal system for a beam in the free-fixed support configuration is written in terms of dimensionless variables as follows: d 2u d 2u + 2 u − 2 2 dξ 2 dξ... is: ̟ 2 − 1 + (1 + γ )k2 ̟ + γ k4 = 0 n n [4 .22 ] Only one of the two solutions of [4 .22 ] is found to be positive, as appropriate to the present problem Finally, the relative displacement due to bending compared 20 6 Structural elements Figure 4.13 Natural frequencies versus modal index n Figure 4.14 Ratio of bending over shear modal displacement versus reduced wave number Modal analysis methods 20 7 to... + EI 4 = 0 κG ∂t 4 ∂x 2 ∂t 2 ∂x The related dispersion equation produces the two following branches for the phase velocity: 1+ 3 + 2 (kR )2 + 4 1+ 3 + 2 (kR )2 4 2 csb = cs 1+ 3 + 2 (kR )2 − 4 1+ 3 + 2 (kR )2 4 2 cbs = cs where cs = G ρ − (1 + ν) (kR)4 2 − (1 + ν)(kR)4 ; 2 [4.9] In Figure 4.3, they are plotted versus the reduced wavelength λ/R The lower branch identifies with the Bernoulli–Euler branch... 14.1; 17.3; (2n + 1)π /2 ̟n 2. 37; 5.50; 8 .64 11.8; 14.9; (4n − 1)π/4 ̟n 1.88; 4 .69 ; 7.85 11.0; 14.1; (2n − 1)π /2 ̟n (0); 3.93; 7.07; 10 .2 13.4; 16. 5; (4n + 1)π/4 The corresponding natural frequencies and related mode shapes are: ωn = 2 ̟n cb ; L2 where cb = EI and ξ = x/L ρS [4.17] ϕn (ξ ) = cos(̟n ξ ) + εn cosh(̟nξ ) − an {sin(̟n ξ ) + sinh(̟n ξ )} The modal stiffness and mass coefficients are of the type: . K ℓ     4γℓ 2 06 ℓ 2 ℓ 2 010 0 6 ℓ 0 12 6 ℓ 2 ℓ 2 06 ℓ 4γℓ 2      U 2 W 2 ϕ 2 U 3 W 3 ϕ 3  T ⇒  U 2 W 2 ϕ 2 ϕ 3  T K ℓ        10 0−10 0 0 12 6 ℓ 0 − 12 6 ℓ 0 6 ℓ 4γℓ 2 06 ℓ 2 ℓ 2 −10. follows:  U 1 W 1 ϕ 1 U 2 W 2 ϕ 2  T ⇒  ϕ 1 U 2 W 2 ϕ 2  T 1 82 Structural elements K ℓ        10 0−10 0 0 12 6 ℓ 0 − 12 6 ℓ 0 6 ℓ 4γℓ 2 06 ℓ 2 ℓ 2 −10 0 1 0 0 0 − 12 6 ℓ 0 12 6 ℓ 0 6 ℓ 2 ℓ 2 06 ℓ 4γℓ 2        ⇒. 6 ℓS −CS( 12 −1)S 2 + 12 C 2 6 ℓC 6 ℓC 6 ℓS 6 ℓC 4γℓ 2 2γℓ 2 6 ℓS 6 ℓC 2 ℓ 2 4γℓ 2     The finite element model of the problem is assembled as:       4γℓ 2 6 ℓS 6 ℓC 2 ℓ 2 0 6 ℓS 2( C 2 +