MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 13 ppsx

458 Structural elements  β 2 β 1  α 2 α 1  M αα g β ∂g α ∂β  δ  ∂Z ∂β  dα dβ =  α 2 α 1  M αα g β ∂g α ∂β δZ  β 2 β 1 dα −  β 2 β 1  α 2 α 1 ∂ ∂β  M αα g β ∂g α ∂β  δZ dα dβ −  β 2 β 1  α 2 α 1 M αα δχ αα g α g β dα dβ =  β 2 β 1 1 g α  g β M αα δ  ∂Z ∂α  − ∂(g β M αα ) ∂α ∂Z  α 2 α 1 dβ +  α 2 α 1  M αα g β ∂g α ∂β δZ  β 2 β 1 dα +  β 2 β 1  α 2 α 1  ∂ ∂α  ∂(g β M αα ) g α ∂α  − ∂ ∂β  M αα g β ∂g α ∂ β  δZ dα dβ [A.4.2] In the same way: −  β 2 β 1  α 2 α 1 M ββ δχ ββ g α g β dα dβ =  α 2 α 1 1 g β  g α M ββ δ  ∂Z ∂α  − ∂(g β M ββ ) ∂β ∂Z  β 2 β 1 dα +  β 2 β 1  M ββ g α ∂g β ∂α δZ  α 2 α 1 dβ +  β 2 β 1  α 2 α 1  ∂ ∂β  ∂(g α M ββ ) g β ∂β  − ∂ ∂α  M ββ g α ∂g β ∂ α  δZ dα dβ [A.4.3] 2  β 2 β 1  α 2 α 1 (M αβ δχ αβ )g α g β dα dβ =  β 2 β 1  α 2 α 1  M αβ g α g β ∂ ∂β  1 g 2 α δ  ∂Z ∂α  g α g β dα dβ +  β 2 β 1  α 2 α 1  M βα g β g α ∂ ∂α  1 g 2 β δ  ∂Z ∂β   g α g β dα dβ  β 2 β 1  α 2 α 1 M αβ  g α g β ∂ ∂β  1 g 2 α δ  ∂Z ∂α  g α g β dα dβ =  α 2 α 1  M αβ δ  ∂Z ∂α  β 2 β 1 dα −  β 2 β 1  α 2 α 1 1 g 2 α ∂ ∂β  g 2 α M αβ  δ  ∂Z ∂α  dα dβ Appendices 459 =  α 2 α 1  M αβ δ  ∂Z ∂α  β 2 β 1 dα −  β 2 β 1  1 g 2 α ∂ ∂β (g 2 α M αβ )δZ  α 2 α 1 dβ +  β 2 β 1  α 2 α 1 ∂ ∂α  1 g 2 α ∂ ∂β (g 2 α M αβ )  δZ dα dβ Finally:  β 2 β 1  α 2 α 1 M βα g β g α ∂ ∂α  1 g 2 β δ  ∂Z ∂β   g α g β dα dβ =  β 2 β 1  M βα δ  ∂Z ∂α  α 2 α 1 dβ −  α 2 α 1  1 g 2 β ∂  g 2 α M βα  ∂α δZ  β 2 β 1 dα +  β 2 β 1  α 2 α 1 ∂ ∂β   1 g 2 β ∂  g 2 β M βα  ∂α   δZ dα dβ [A.4.4] A.4.2 Equation of local equilibrium in terms of shear forces Gathering together the surface terms of [A.4.1] and changing their sign, we get: δA =  t 2 t 1 dt  β 2 β 1  α 2 α 1 δZ  ρh ¨ Zg α g β − ∂ ∂α  1 g α ∂(g β M αα ) ∂α + 1 g 2 α ∂(g 2 α M αβ ) ∂β − M ββ g α ∂g β ∂α  − ∂ ∂β  1 g β ∂  g α M ββ  ∂β + 1 g 2 β ∂(g 2 β M βα ) ∂α − M αα g β ∂g α ∂β  dα dβ = 0 [A.4.5] Transverse shear forces acting in the normal direction to the plate midplane are defined as: Q αz = 1 g α g β  ∂(g β M αα ) ∂α + 1 g α ∂  g 2 α M αβ  ∂β − M ββ ∂g β ∂α  Q βz = 1 g α g β   ∂(g α M ββ ) ∂β + 1 g β ∂  g 2 β M βα  ∂α − M αα ∂g α ∂β   [A.4.6] 460 Structural elements Hamilton’s principle is written in a more compact form than [4.1] as: δA =  t 2 t 1 dt  β 2 β 1  α 2 α 1 δZ  ρh ¨ Zg α g β − ∂(g β Q αz ) ∂α − ∂(g α Q βz ) ∂β  dα dβ = 0 The equation of local equilibrium follows as: ρh ¨ Z − 1 g α g β  ∂(g β Q αz ) ∂α + ∂(g α Q βz ) ∂β  = 0 [A.4.7] A.4.3 Boundary conditions: effective Kirchhoff’s shear forces and corner forces Gathering together the boundary terms of [A.4.1] associated with the moments and rotations, we obtain:  β 2 β 1  g β g α M αα δ  ∂Z ∂α  + M βα δ  ∂Z ∂β  α 2 α 1 dβ = 0  α 2 α 1  g α g β M ββ δ  ∂Z ∂β  + M αβ δ  ∂Z ∂α  α 2 α 1 dα = 0 [A.4.8] The important point is that, in such expressions, the rotations must be considered as dependent variables. Accordingly, the torsion term is integrated to express the variation in terms of δZ solely, so [A.4.11] becomes:  β 2 β 1  g β g α M αα δ  ∂Z ∂α  dβ  α 2 α 1 +  [M βα δZ] α 2 α 1  β 2 β 1 −  β 2 β 1  ∂M βα ∂β δZ  α 2 α 1 dβ = 0  α 2 α 1  g α g β M ββ δ  ∂Z ∂β  α 2 α 1 dα +  [M αβ δZ] α 2 α 1  β 2 β 1 −  α 2 α 1  ∂M αβ ∂α δZ  β 2 β 1 dα = 0 [A.4.9] Whence, the homogeneous boundary conditions: along the edges α 1 and α 2 : M αα = 0; or ∂Z/∂α = 0 along the edges β 1 and β 2 : M ββ = 0; or ∂Z/∂β = 0 at the corners: M αβ + M βα = 2M αβ = 0; or Z = 0 [A.4.10] Appendices 461 Gathering the boundary terms associated with the shear forces and displacements we get:  β 2 β 1  − 1 g α ∂(g β M αα ) ∂α − 1 g 2 α ∂ ∂β  g 2 α M αβ  + M ββ g α ∂g β ∂α − ∂M βα ∂β  α 2 α 1 δZ dβ =0  α 2 α 1  − 1 g β ∂(g α M ββ ) ∂β − 1 g 2 β ∂ ∂α  g 2 β M βα  + M αα g β ∂g α ∂β − ∂M αβ ∂α  β 2 β 1 δZ dα [A.4.11] Dividing the first row by g β and the second row by g α , we get:  β 2 β 1  −Q αz − ∂M βα ∂β  α 2 α 1 δZ dβ = 0  α 2 α 1  −Q βz − ∂M αβ ∂α  β 2 β 1 δZ dα = 0 [A.4.12] Then, to the shear forces already defined by [A.4.9], the term arising from the integration of [A.4.11] must be added, resulting in the effective Kirchhoff shear force per unit length, expressed as: V αz = Q αz + ∂M βα g β ∂β = 1 g α g β  ∂(g β M αα ) ∂α + 1 g α ∂ ∂β  g 2 α M αβ  − M ββ ∂g β ∂α  + ∂M βα g β ∂β V βz = Q βz + ∂M αβ g α ∂α = 1 g α g β  ∂(g α M ββ ) ∂β + 1 g β ∂ ∂α (g 2 β M βα ) − M αα ∂g α ∂β  + ∂M αβ g α ∂α [A.4.13] The free and fixed boundary conditions are: along the edges α 1 and α 2 : V αz = 0; or Z = 0 along the edges β 1 and β 2 : V βz = 0; or Z = 0 [A.4.14] A.5. Static equilibrium of a sagging cable loaded by its own weight A cable of length L 0 , is stretched between two points A and B, of coordinates x A =−L/2, z A = 0, x B =+L/2, z B = 0, where L<L 0 , see Figure A.5.1. The problem is to determine the static equilibrium configuration z(x) of the cable in 462 Structural elements A B –L/ 2 L/ 2 O z x → g → → → → T (x) T z (x) T x (x) T x (x+dx) T z (x+dx) T (x+dx) k –gSds i Figure A.5.1. Cable subjected to its own weight a uniform gravity field −g  k. As soon as the tensile stress in the cable is sufficiently large, the elastic deflection of the cable can be safely neglected. The equilibrium equation will be derived by using successively two distinct methods. The first follows the Newtonian approach, which consists in writing down directly the force balance for a cable element of infinitesimal length ds(x). The second method consists in deriving the Lagrange equation of the cable constrained by the condition of length invariance. It turns out that the last condition can be prescribed either in the global scale of the whole structure or in the local scale of a cable element. A.5.1 Newtonian approach The tensile force acting on the cable at the Cartesian abscissa x is denoted  T(x) = T x (x)  i + T z (x)  k. It is interpreted as the string stress related to the length invariance of the cable. The force balances are written as:          dT x dx = 0 ⇒ T x = T 0 dT z dx = ρgS ds dx = ρgS  1 +  dz dx  2 [A.5.1] On the other hand, the following relationship holds: T z T x = dz dx ⇒ T z = T 0 dz dx [A.5.2] whence the nonlinear differential equation: d 2 z dx 2 − γ  1 +  dz dx  2 = 0, where γ = ρgS T 0 [A.5.3] Appendices 463 The solution is found to be of the type: dz dx = sinh(γ x) ⇒ z(x) = cosh(γ x) γ + a [A.5.4] The boundary conditions imply the solution: z(x) L = T 0 ρgSL  cosh  ρgSx T 0  − cosh  ρgSL 2T 0  T z (x) = T 0 sinh  ρgSx T 0  [A.5.5] The horizontal component of the tensile force is obtained by stating that the resultant of the vertical components of the support reactions must balance the weight of the cable; whence the following transcendental equation: sinh  ρgSL 2T 0  = ρgSL 0 2T 0 [A.5.6] It can be easily checked that [A.5.6] has only one solution if L ≤ L 0 . It is also possible to express the Cartesian components of the deflected cable in terms of the arc length s: ds = cosh(γ x) ⇒ s = 1 γ sinh(γ x) ⇒ x = 1 γ (sinh(γ s)) −1 z = 1 γ  cosh  ( sinh ( γs )) −1  − cosh  γL 2  [A.5.7] A.5.2 Constrained Lagrange’s equations, invariance of the cable length The constraint condition is written as:  L/2 −L/2 ds =  L/2 −L/2  1 +  dz dx  2 dx = L 0 [A.5.8] The constrained Lagrangian is: L ′ =−  L/2 −L/2 {ρgSz(x) +λ}  1 +  dz dx  2 dx + λL 0 [A.5.9] 464 Structural elements The Euler–Lagrange equations are:            ρgS  (z ′ ) 2 + 1 − d dx  z ′ (ρgSz + λ)  (z ′ ) 2 + 1  = 0 − d dx  (ρgSz + λ)  (z ′ ) 2 + 1  = 0 ⇒ (ρgSz + λ)  (z ′ ) 2 + 1 = C where z ′ = dz dx [A.5.10] The Lagrange multiplier λ can be eliminated between the two equations [5.10] to obtain the nonlinear differential equation: ρgS  (z ′ ) 2 + 1 −Cz ′′ = 0 ⇐⇒  (z ′ ) 2 + 1 − z ′′ γ = 0 where γ = ρgS C and z ′′ = d 2 z dx 2 [A.5.11] The solution is found to be of the type: dz dx = sinh(γ x) ⇒ z(x) = cosh(γ x) γ + a [A.5.12] The boundary condition determines the constant a: a =− cosh(γ L/2) γ [A.5.13] On the other hand, λ is given by: (ρgSz + λ)  (z ′ ) 2 + 1 = C ⇒ C + λ − C cosh(γ L/2) cosh(γ x) = C [A.5.14] and finally: λ = C cosh(γ L/2) [A.5.15] The constant C is a force per unit length determined by using the condition [A.5.8], which implies that: L 0 =  L/2 −L/2 cosh(γ x) dx = 2 γ sinh  γL 2  [A.5.16] γ is a solution of the transcendental equation: γL 0 2 = sinh  γL 2  [A.5.17] Appendices 465 [A.5.17], which is of the same type as [5.6], gives the physical meaning of C. Indeed, the horizontal T x and the vertical T z components of the support reaction must verify the following relation: T z T x = ρgSL 0 2T x = dz dx     L/2 = sinh  γL 2  = γL 0 2 = ρgSL 0 2C ⇒ T x = C [A.5.18] This final result can be used to check that [A.5.17] is identical to [A.5.6]. A.5.3 Constrained Lagrange’s equations: length invariance of a cable element The constraint condition is now written as:  dx ds  2 +  dy ds  2 = 1 [A.5.19] The constrained Lagrangian is: L ′ =−  L/2 −L/2  ρgSz(s) +λ(s)   dx ds  2 +  dz ds  2 − 1  ds [A.5.20] Here, Lagrange’s multiplier depends on s since the constraint condition is written at the local scale of a cable element. The Euler–Lagrange equations are:          d ds  λ dx ds  = 0 d ds  λ dz ds  − ρgS = 0 [A.5.21] The first equation [A.5.21] gives: λ = α ds dx [A.5.22] Substituting [A.5.22] into the second equation [A.5.21], one gets: d ds  dz dx  = ρgS α [A.5.23] Equation [A.5.23] can be identified with equation [A.5.3] by expressing z as a function of the variable x. Starting from the relation, ds = dx   dz dx  2 + 1 [A.5.24] 466 Structural elements equation [A.5.23] is written as: d 2 z/dx 2  1 + (dz/dx) 2 = ρgS α [A.5.25] Accordingly, the solution can be obtained in the same way as in subsection A.5.2, α being identified with T 0 . A.6. Mechanical properties of some solids in common use Metals Stainless steel E ∼ = 1.9410 11 Pa, ν = 0.265, ρ = 7970 kgm −3 Aluminium E ∼ = 6.910 10 Pa, ν = 0.33, ρ = 2700 kgm −3 Brass E ∼ = 1.210 11 Pa, ρ = 8800 kgm −3 Chromium E ∼ = 4.810 9 Pa, ρ = 7200 kgm −3 Copper E ∼ = 1.110 11 Pa, ρ = 8970 kgm −3 Iron E ∼ = 2.110 11 Pa, ρ = 7860 kgm −3 Magnesium E ∼ = 4.510 10 Pa, ρ = 1740 kgm −3 Gold E ∼ = 7.410 10 Pa, ρ = 19 300 kgm −3 Nickel E ∼ = 2.110 11 Pa, ρ = 8910 kgm −3 Lead E ∼ = 1.810 10 Pa, ρ = 11300 kgm −3 Titane E ∼ = 1.210 11 Pa, ρ = 4540 kgm −3 Tungstene E ∼ = 3.410 11 Pa, ρ = 19 300 kgm −3 Zinc E ∼ = 8.310 10 Pa, ρ = 7140 kgm −3 Zirconium E ∼ = 7.610 10 Pa, ρ = 6370 kgm −3 Composites material using carbon fibers 1.510 11 ≤ E ≤ 810 11 Pa, 1500 ≤ ρ ≤ 2000 kgm −3 Appendices 467 Thermoplastics 410 8 ≤ E ≤ 10 9 Pa, 1000 ≤ ρ ≤ 2000 kgm −3 Glass E ∼ = 710 10 Pa, 2300 ≤ ρ ≤ 2600 kgm −3 Wood 3.510 9 (balsa) ≤ E ≤ 210 10 Pa(ebony), 100(balsa) ≤ ρ ≤ 1000 kgm −3 (ebony) [...]... causality 34 of superposition 29 of virtual work 151 projection 151, 171, 21 7 propagation delay 31, 35 speed 34 pseudo-mode 22 8 pure bending mode 99 pure shear model 86, 87 479 quadratic form 21 , 1 62, 171, 26 7 quasi-inertial range 22 5, 22 8 quasi-static correction 22 8, 23 5 mode 22 8 range 22 5, 22 8 Rayleigh minimum principle 29 4, 29 9 Rayleigh–Love model 199 Rayleigh–Ritz method 24 2, 29 3, 29 5, 335, 345, 436 Rayleigh–Timoshenko... Rayleigh–Timoshenko model 195, 20 5 Rayleigh’s quotient 20 7, 21 1, 29 5, 335, 434 reciprocity theorem 1 62 rectangular plate 26 2, 26 5, 27 5, 28 6 rectilinear mode 29 1, 300 reflected wave 43, 44, 46, 54 reflecting plane 43 surface 43 reflection coefficient 46 law 44, 46 wave 40 relation of constraint 2 resonant range 22 5, 22 6 response 22 6 response spectrum 22 7 resultant stress 72 rigid body 24 , 26 2 mode 23 6 motion 8 rigid... expansion method 23 0, 29 4 force 21 9, 22 0 frequency 199 mass 196, 21 0 model 24 8 oscillator 21 9 projection method 189, 21 7, 23 3, 23 8 series 190, 21 9 stiffness 196, 21 0 truncation 22 2, 23 3 vector 58 wavelength 1 32, 133 mode conversion 41, 53 of deformation 70 of propagation 50 shape 55, 57, 190, 197, 29 0 shape expansion method 197 shape localization 24 7 shape, natural, vibration 53 wavelength 21 0 model coupling... 19, 21 density 21 , 26 7 Kirchhoff, G.R 320 Kirchhoff–Love assumptions 27 8 Index hypotheses 3 12 model 26 2, 3 12, 313, 314, 315, 369 Kirchoff shear forces 460 Kirchoff’s theorem 30 knife edge support 1 02 Lagrange’s equations 19, 24 , 463 multiplier 19, 24 , 74, 80, 1 72, 173, 315 Lagrangian 12, 19, 20 , 184, 26 5 description 2 displacement 3 Lamé’s parameters 3, 16, 304, 307, 348, 350 Laplace operator 28 , 449... out -of- phase mode 24 4, 25 7, 29 2 out -of- plane load 26 1 motion 26 1 ovalization 61 P wave 42 reflection 43, 44 Parallelepiped 57 penalty coefficient 173 method 173 phase angle 34, 42 function 38 in-mode motion 25 9 et seq shift 34 velocity 38, 52 physical coordinate 21 7 pinned support 1 02 plane of constant phase 42, 50 harmonic wave 32 layer 52 mode 50, 64 strain 27 3 plane stress 27 2, 27 3 stress 92 wave... cross-section 26 2 manifold 29 5 masonry dome 383 mass matrix 27 4 operator 150, 157 material law 21 , 24 , 25 , 167 line 71 point 2, 3, 436 oscillations 31 wave 2, 32 MATLAB 39, 435 matrix notation 6 mean curvature 374 mean square value 22 7 mechanical energy 28 , 30 impedance 116 membrane 26 1 component 26 4 displacement 26 3 equation 414 equilibrium 26 5 mode 28 9, 29 0, 29 3, 300 strain tensor 305, 313 strains 26 3 stress... 65, 3 02 wave 28 9 meridian 371 line 374 plane 373 stress 374 mesh 163, 168 method of variables separation 32 metric tensor 25 7 midplane 26 1, 3 12 midsurface 26 0, 367 mixed formulation of equilibrium equations 16 modal analysis 100, 188 et seq, 189, 3 32 basis 190, 21 7 (truncation of ) 22 2 coordinate system 21 7 density 24 5, 29 2, 338, 428 displacement 190, 21 9 478 Index modal (continued) equation 196, 28 9... principle 130 et seq longitudinal motion 1 32 Newtonian approach 66 et seq variational formulation of equations 1 32 vibration modes 188, 190 strain density 26 8 energy 21 , 25 , 131 , 1 42, 305 hardening 123 , 128 isotherm tensors 28 4 rate 124 tensor 4, 24 vector 27 2, 305 strain/stress relationship 27 2 Index stress 11 coefficient 137 discontinuity 83, 84, 27 1 elastic 74 energy density 189 isotherm tensor 28 4 local... deformation 124 failure 125 , 126 flow 125 strain 124 zone 126 Index plate bending 457 contour 26 7, 322 geometry 26 0 loaded 29 9 out -of- plane motion 311 prestressed 23 6 Pochhammer 53, 191 Poisson effect 79, 1 32, 27 4, 28 8 Poisson ratio 16, 47 polar coordinates 95, 307 polarization 43, 481 polynomial interpolation 166, 171 portal frame 174, 179, 21 1 21 2 position vector 2, 5 post-buckling behaviour 21 4 potential... axisymmetry 364 balance of moments 50 bandwidth 22 4 Barré de Saint-Venant 60, 92 Index beam 66 clamped-clamped 1 12 elastic-plastic element 115, 117, 174 geometry 67 pinned-pinned 1 12 pinned supported 111 traction-compression 167 transverse loads 355 beats 39 bending 77 angle 69 axis 118 of beam 141, 20 7 element 174, 21 2 equations 316 mode 51, 21 2 mode of vibration 20 0 20 1, 21 2 moment 73, 125 , 22 2 operator 159 . 69 axis 118 of beam 141, 20 7 element 174, 21 2 equations 316 mode 51, 21 2 mode of vibration 20 0 20 1, 21 2 moment 73, 125 , 22 2 operator 159 plane 71 and shear modes 20 5 stiffness coefficient 327 strain. 21 density 21 , 26 7 Kirchhoff, G.R. 320 Kirchhoff–Love assumptions 27 8 Index 477 hypotheses 3 12 model 26 2, 3 12, 313, 314, 315, 369 Kirchoff shear forces 460 Kirchoff’s theorem 30 knife edge support 1 02 Lagrange’s equations. value 22 7 mechanical energy 28 , 30 impedance 116 membrane 26 1 component 26 4 displacement 26 3 equation 414 equilibrium 26 5 mode 28 9, 29 0, 29 3, 300 strain tensor 305, 313 strains 26 3 stress 65, 3 02 wave