51 Part I: Passive Control Chapter 2 Optimal stiffness distribution 2.1 Introduction This chapter is concerned with the first step in passive motion control, establishing a distribution of structural stiffness which produces the desired displacement profile. When the design loading is quasi-static, the stiffness parameters are determined by solving the equilibrium equations in an inverse way. Dynamic loading is handled by selecting the stiffness parameters such that the fundamental mode shape has the desired displacement profile. The implicit assumptions here are that one can incorporate sufficient damping to minimize the contributions of the higher modes, and the fundamental mode shape is independent of damping. The latter assumption is reasonable for lightly damped structures. Discrete systems are governed by algebraic equations, and the problem reduces to finding the elements of the system stiffness matrix. The static case involves solving (2.1) for , where and are the prescribed displacement and loading vectors. Some novel numerical procedures for solving eqn (2.1) are presented in a later section. KU * P * = K U * P * 52 Chapter 2: Optimal Stiffness Distribution In the dynamic case, the equilibrium equation for undamped periodic excitation of the fundamental mode is used: (2.2) where is the discrete mass matrix, is a scaled version of the desired displacement profile, and is the fundamental frequency. Taking (2.3) (2.4) reduces eqn (2.2) to (2.5) The solution technique for eqn (2.1) also applies for eqn (2.5). Once is known, the stiffness can be scaled by specifying the frequency, . An appropriate value for is established by converting the system to an equivalent one degree-of-freedom system and using the SDOF design approaches discussed in the introduction. Continuous systems such as beams are governed by partial differential equations, and the degree of complexity that can be dealt with analytically is limited. The general strategy of working with equilibrium equations is the same, but now one has to determine analytic functions rather than discrete values for the stiffness. Analytical solutions are useful since they allow the key dimensionless parameters to be identified and contain generic information concerning the behavior. In what follows, the first topic discussed concerns establishing the stiffness distribution for static loading applied to a set of structures consisting of continuous cantilever beams, building type structures modeled as equivalent discontinuous beams with lumped masses, and truss-type structures. Closed form solutions are generated for the continuous cantilever beam example. The next topic concerns establishing the stiffness distribution for the case of dynamic loading applied to beam-type structures. The process of calibration of the KΦ * ω 1 2 MΦ * = M Φ * ω 1 K' 1 ω 1 2 K= P' MΦ * = K'Φ * P'= K' ω 1 ω 1 2.2 Governing Equations - Transverse Bending of Planar Beams 53 fundamental frequency is described and illustrated for both periodic and seismic excitation. The last section of the chapter deals with the situation where the higher modes cannot be ignored. An iterative numerical scheme is presented and applied to a representative range of beam-type structures. 2.2 Governing equations - transverse bending of planar beams In this section, the governing equations for a specialized form of a beam are developed. The beam is considered to have a straight centroidal axis and a cross- section that is symmetrical with respect to a plane containing the centroidal axis. Figure 2.1 shows the notation for the coordinate axes and the displacement measures (translations and rotations) that define the motion of the member. The beam cross-section is assumed to remain a plane under loading. This restriction is the basis for the technical theory of beams, and reduces the number of displacement variables down to three translations and three rotations which are functions of x and time. When the loading is constrained to act in the symmetry plane for the cross- section, the behavior involves only those motion measures associated with this plane. In this discussion, the plane is taken as the plane of symmetry, and , , and are the relevant displacement variables. If the loading is further restricted to act only in the y direction, the axial displacement measure, , is identically equal to zero. The behavior for this case is referred to as transverse bending. In what follows, the governing equations for transverse bending of a continuous planar beam are derived. The derivation is then extended to deal with discontinuous structures such as trusses and frames that are modelled as equivalent beams. x-yu x u y β z u x 54 Chapter 2: Optimal Stiffness Distribution Fig. 2.1: Notation - planar beam. Planar deformation-displacement relations Figure 2.2 shows the initial and deformed configurations of a differential beam element. The cross-sectional rotation, , is assumed to be sufficiently small such that . In this case, linear strain-displacement relations are acceptable. Letting denote the transverse shearing strain and the extensional strain at an arbitrary location from the reference axis, and taking and , the deformation relations take the form (2.6) (2.7) (2.8) where denotes the bending deformation parameter. x y z u x β x β y β z u y u z Note: x-y plane is a plane of symmetry for the cross-section z y β z β z 2 << 1 γε y ββ z ≡ uu y ≡ ε yχ–= γ u∂ x∂ β–= χ β∂ x∂ = χ 2.2 Governing Equations - Transverse Bending of Planar Beams 55 Fig. 2.2: Initial and deformed elements. Optimal deformation and displacement profiles Optimal design from a motion perspective corresponds to a state of uniform shear and bending deformation under the design loading. This goal is expressed as (2.9) (2.10) Uniform deformation states are possible only for statically determinate structures. Building type structures can be modeled as cantilever beams, and therefore the goal of uniform deformation can be achieved for these structures. Consider the vertical cantilever beam shown in Fig. 2.3. Integrating eqns (2.7) and (2.8) and enforcing the boundary conditions at leads to (2.11) (2.12) The deflection at the end of the beam is given by (2.13) β γ O A y O A y X Y' Y X' X B B γγ ∗ = χχ ∗ = x 0= βχ ∗ x= u γ ∗ x χ ∗ x 2 2 += uH() γ ∗ H χ ∗ H 2 2 += 56 Chapter 2: Optimal Stiffness Distribution where is the contribution from shear deformation and is the contribution from bending deformation. For actual buildings, the ratio of height to width (i.e. aspect ratio) provides an indication of the relative contribution of shear versus bending deformation. Buildings with aspect ratios on the order of unity tend to display shear beam behavior and . On the other hand, buildings with aspect ratios greater than about display bending beam behavior and . Fig. 2.3: Simple cantilever beam. One establishes the values of , based on the performance constraints imposed on the motion, and selects the stiffness such that these target deformations are reached. Introducing a dimensionless factor , which is equal to the ratio of the displacement due to bending and the displacement due to shear at x=H, (2.14) transforms eqn (2.13) to a form that is more convenient for low rise buildings. (2.15) A shear beam is defined by . Tall buildings tend to have . γ ∗ H χ ∗ H 2 2⁄ χ 0≈ 7 γ 0≈ b(x) x y , u H γ ∗ χ ∗ s s H 2 χ ∗ 2   γ*H()⁄ Hχ* 2γ* == uH() 1 s+()γ ∗ H= s 0= s 1≈ 2.2 Governing Equations - Transverse Bending of Planar Beams 57 Equilibrium equations Figure 2.4 shows a differential beam element subjected to an external transverse loading, , and restrained by the internal transverse shear, , and bending moment, . By definition, (2.16) (2.17) where and are the stresses acting on the cross-section. Summing forces and moments leads to (2.18) (2.19) where , J are the mass and rotatory inertia per unit length. When the member is supported only at (see Fig. 2.3), the equilibrium equations can be expressed in the following integral form (2.20) (2.21) bV M V τ Ad ∫ = Myσ Ad ∫ –= τσ V∂ x∂ b+ ρ m t 2 2 ∂ ∂ u = M∂ x∂ V+ J t 2 2 ∂ ∂β = ρ m x 0= Vx() b ρ m t 2 2 ∂ ∂ u –    xd x H ∫ = Mx() VJ t 2 2 ∂ ∂β –    xd x H ∫ = 58 Chapter 2: Optimal Stiffness Distribution Fig. 2.4: Forces acting on a differential element. In the case of static loading, the acceleration terms are equal to 0, and V and M can be determined by integrating eqns (2.20) and (2.21). Force-deformation relations The force-deformation relations, also referred to as the constitutive relations, depend on the characteristics of the materials that make up the beam. For the case of static loading and linear elastic behavior, the expressions relating the shear force and bending moment to the shear deformation and bending deformation respectively are expressed as (2.22) (2.23) where and are defined as the transverse shear and bending rigidities. These equations have to be modified when the deformation varies with time. This aspect is addressed in Section 2.4. Examples which illustrate how to determine the rigidity coefficients for a range of beam cross-sections are presented below. x y dx V M M ∂M ∂x dx+ V ∂V ∂x dx+ b σ ∂σ ∂x dx+ τ ∂τ ∂x dx+ σ τ Vx() D T x()γx()= Mx() D B x()χx()= D T D B 2.2 Governing Equations - Transverse Bending of Planar Beams 59 Example 2.1. Composite sandwich beam Figure 2.5 shows a sandwich beam composed of 2 face plates and a core. Fig. 2.5: Composite beam cross section. The face material is usually much stiffer than the core material, and therefore the core is assumed to carry only shear stress. Noting eqn (2.6), the strains in the face and core are (2.24) (2.25) The face thickness is also assumed to be small in comparison to the depth. Considering the material to be linear elastic, the expressions for shear and moment are: (2.26) (2.27) where is the shear modulus for the core and is the Young’s modulus for the face plate. The corresponding rigidity coefficients are: (2.28) b d t f ε f d 2 ±χ= γ c γ= Vbd()τ c bdG c ()γ== Mbt f d()σ f bt f d 2 2 E f    χ== G c E f D T bdG c = 60 Chapter 2: Optimal Stiffness Distribution (2.29) Example 2.2. Equivalent rigidities for a discrete Truss-beam The term, truss-beam, refers to a beam type structure composed of a pair of chord members and a diagonal bracing system. Figure 2.6 illustrates an x-bracing scheme. Truss beams are used as girders for long span horizontal systems. Truss beams are also deployed to form rectangular space structures which are the primary lateral load carrying mechanisms for very tall buildings. The typical “mega-truss” has large columns located at the 4 corners of a rectangular cross- section, and diagonal bracing systems placed on the outside force planes. These structures are usually symmetrical, and the behavior in one of the symmetry directions can be modelled using an equivalent truss beam. When the spacing, h, is small in comparison to the overall length, one can approximate the discrete structure as a continuous beam having equivalent properties. In this example, approximate expressions for these equivalent properties are derived for the case of x-bracing. Fig. 2.6: Parameters and internal forces - Truss-beam. The key assumption is that the members carry only axial force. This approximation is reasonable when the members are slender, and diagonal or chevron bracing is used. Noting Fig. 2.6, the cross section force quantities are related to the member forces by D B bt f d 2 2 E f = h B θ E d E c F c F c F d F d V M A d A c [...]... Fig 2. 10 ˙˙ p1 = V 1 – V 2 + m1 u1 ˙˙ p2 = V 2 + m2 u2 Substituting for V 1 and V 2 , eqn (2. 56) expands to (2. 56) 2. 2 Governing Equations - Transverse Bending of Planar Beams ˙˙ p1 = k1 u1 + k2 ( u1 – u2 ) + m1 u1 ˙˙ p2 = k2 ( u2 – u1 ) + m2 u2 67 (2. 57) It is convenient to express eqn (2. 57) in matrix form The various matrices are defined as: U = P = u1 (2. 58) u2 p1 (2. 59) p2 M = m1 0 K = k1 + k2 –k2... 2 = k2 ( u2 – u1 ) 65 (2. 48) Noting the definition of transverse shear strain, γ 1 = u1 ⁄ h1 u2 – u1 γ 2 = h2 (2. 49) one can relate the k ‘s to the equivalent transverse shear rigidity factors: V 1 = D T, 1 γ 1 , V 2 = D T, 2 γ 2 ⇒ D T, i = h i k i , i = 1, 2 Lb h2 m2 interior u2, p2 k2 exterior u1, p1 m1 h1 (2. 50) k1 Fig 2. 9: A discrete shear beam model m2 u2, p2 V2 V2 m1 u1, p1 V1 V1 Fig 2. 10:... -   L  member i (2. 104) Using a geometric analysis, the member extensions are expressed in terms of the nodal displacements: 2 e 1 = ( u 1 + u 2 ) 2 e2 = u2 (2. 105) 1 3 e 3 = – u 1 + u 2 2 2 The nodal and member forces are related by the nodal force equilibrium equations 2 1 P 1 = F 1 – F 3 2 2 (2. 106) 2 3 P 2 = F 1 + F 2 + F 3 2 2 Since there are 3 extensions and only 2. .. k2 –k2 –k2 k2 (2. 60) 0 m2 (2. 61) With these definitions, eqn (2. 57) has the form ˙˙ P = KU + M U (2. 62) Equation 2. 20 expresses the shear force in a continuous beam as an integral of the applied lateral loading The corresponding equations for this discrete system are obtained from eqn (2. 56) by combining the individual equations: ˙˙ V 2 = p2 – m2 u2 ˙˙ ˙˙ V 1 = p1 + p – m1 u1 –m2 u2 (2. 63) 2 In general,... related to the bending deformation χ , (2. 37) 62 Chapter 2: Optimal Stiffness Distribution ∆β χ = -h (2. 38) the strain can be expressed as c Bχ ε = -2 (2. 39) +u +β ∆h ∆β ∆v h θ θ x B y B Fig 2. 7: Deformed truss-beam section c d Substituting for ε and ε and combining eqns (2. 30 )-( 2. 33) results in c c 2 A E B M = - χ 2 d d V = [ A E sin 2 cos θ ]γ (2. 40) (2. 41) Comparing these expressions with... Case 1 P 1 = 100kN P 2 = 20 0kN Least square: V 1 = 31.37 ⋅ 10 7 V 2 = 16.86 ⋅ 10 7 V 3 = – 2. 29 ⋅ 10 7 Mean value least square: V 1 = 35.37 ⋅ 10 7 V 2 = 11.40 ⋅ 10 7 V 3 = 12. 26 ⋅ 10 7 Case 2 P 1 = 100kN Least square: V 1 = 30. 02 ⋅ 10 7 V 2 = 2. 71 ⋅ 10 7 V 3 = – 7 .25 ⋅ 10 7 P 2 = 100kN 86 Chapter 2: Optimal Stiffness Distribution Mean value least square: V 1 = 32. 23 ⋅ 10 7 V 2 = – 0. 32 ⋅ 10 7 V 3 = 0.86... optimization problem is 2 2 1 2 f = ( V 1 + V 2 + V 3 ) 2 (2. 115) Substituting for V 1 and V 2 transforms f to f = co + c1 V 3 + c2 V 3 2 (2. 116) Requiring f to be stationary with respect to a change in V 3 , ∂f = c1 + 2 c2 V 3 = 0 ∂V3 (2. 117) leads to c1 V 3 = – 2c 2 (2. 118) The problem with this approach is that it tends to eliminate redundant members, 2. 5 Stiffness Distribution - Truss under Static... denote the mean value, 1 V m = ( V 1 + V 2 + V 3 ) 3 (2. 119) and V i represent the deviation from the mean value for member i, Vi = Vi – Vm (2. 120 ) The optimization problem is stated as 2 2 1 2 minimize f = ( V 1 + V 2 + V 3 ) 2 (2. 121 ) subject to the constraints on V 1 and V 2 Substituting for V 1 and V 2 transforms f to 2 f = do + d1 V 3 + d2 V 3 (2. 122 ) The remaining steps are the same as for... (2. 110) as the member volume variable Expressing k i in terms of V i, AE 1 k i =  -  = - V i  L i L i2 (2. 111) and substituting for k i in the constraint equations transforms these equations to 80 Chapter 2: Optimal Stiffness Distribution 2 2 V 1 = L1 k1 + a1 ( L1 ⁄ L3 ) V 3 2 2 V 2 = L2 k2 + a2 ( L2 ⁄ L3 ) V 3 (2. 1 12) The optimization problem can be stated as: minimize f ( V ) = V 1 + V 2. .. using eqn (2. 111) transforms eqn (4) to V 1 = 0 .27 5V 3 + 32 ⋅ 10 7 V 2 = – 0.375V 3 + 32 ⋅ 10 7 (5) Least square procedure The objective function is 2 2 2 1 f = ( V 1 + V 2 + V 3 ) = f ( V 3 ) 2 (6) Differentiating with respect to V 3 , ∂V 2 ∂V 1 ∂f = V1 + V2 + V3 ∂V3 ∂V3 ∂V3 (7) substituting for V 1 and V 2 , and then setting the resulting expression equal to zero results in 7 1 .21 6V 3 – 3 .22 2 ⋅ 10 . as p 1 k 1 u 1 k 2 u 1 u 2 –()m 1 u ˙˙ 1 ++= p 2 k 2 u 2 u 1 –()m 2 u ˙˙ 2 += U u 1 u 2 = P p 1 p 2 = M m 1 0 0 m 2 = K k 1 k 2 + k 2 – k 2 – k 2 = PKU MU ˙˙ += V 2 p 2 m 2 u ˙˙ 2 –= V 1 p 1 p+ 2 m 1 u ˙˙ 1 m 2 u ˙˙ 2 ––= n 68. V 2 k 2 u 2 u 1 –()= γ 1 u 1 h 1 ⁄= γ 2 u 2 u 1 – h 2 = k V 1 D T 1, γ 1 = V 2 D T 2, γ 2 = D Ti, ⇒ h i k i = i 12, = h 1 h 2 k 1 k 2 m 1 m 2 u 1 , p 1 u 2 , p 2 exterior interior L b m 2 u 2 ,. beam properties: (2. 42) (2. 43) χ ∆β h = ε c Bχ 2 = B B h ∆ h θ θ ∆ v ∆β +u +β x y ε c ε d M A c E c B 2 2 χ= VA d E d 2 θcossin[]γ= D B A c E c B 2 2 = D T A d E d 2 θcossin= 2. 2 Governing Equations - Transverse