Vibration and Shock Handbook 14 Every so often, a reference book appears that stands apart from all others, destined to become the definitive work in its field. The Vibration and Shock Handbook is just such a reference. From its ambitious scope to its impressive list of contributors, this handbook delivers all of the techniques, tools, instrumentation, and data needed to model, analyze, monitor, modify, and control vibration, shock, noise, and acoustics. Providing convenient, thorough, up-to-date, and authoritative coverage, the editor summarizes important and complex concepts and results into “snapshot” windows to make quick access to this critical information even easier. The Handbook’s nine sections encompass: fundamentals and analytical techniques; computer techniques, tools, and signal analysis; shock and vibration methodologies; instrumentation and testing; vibration suppression, damping, and control; monitoring and diagnosis; seismic vibration and related regulatory issues; system design, application, and control implementation; and acoustics and noise suppression. The book also features an extensive glossary and convenient cross-referencing, plus references at the end of each chapter. Brimming with illustrations, equations, examples, and case studies, the Vibration and Shock Handbook is the most extensive, practical, and comprehensive reference in the field. It is a must-have for anyone, beginner or expert, who is serious about investigating and controlling vibration and acoustics.
14 Reinforced Concrete Structures Y.L Mo University of Houston 14.1 Introduction 14-1 14.2 Analytical Models 14-6 Basic Concepts † Natural Frequencies † Viscous Damping † Damped Harmonic Excitation Model-Based Simulation † Flexural Behavior Behavior † Time-History Analysis † Shear 14.3 Beams under Harmonic Excitations 14-18 14.4 Design for Explosions/Shocks 14-21 Mechanical Properties Column † † Design for Machine Vibration Shear Walls Summary This chapter concerns vibration and shock in reinforced concrete structures, addressing how to design reinforced concrete structures that will be subjected to vibration or shock, for example, due to a missile attack The chapter is intended to apply vibration theory to concrete structures Section 14.1 describes the fundamental properties of vibration theory Section 14.2 proposes analytical models for reinforced concrete structures that are subjected to flexure, shear, and dynamic loading Section 14.3 discusses reinforced concrete beams under harmonic excitations, including mechanical properties and designs for machine vibration Section 14.4 indicates the step-by-step procedure for designing reinforced concrete columns or walls that will be subjected to seismic loads or shocks, respectively 14.1 14.1.1 Introduction Basic Concepts Reinforced concrete structures are capable of free and forced vibration when disturbed from their equilibrium configuration (see Chapter and Chapter 2) Free vibration takes place when a structure vibrates under the action inherent to the structure itself without being impressed by external forces The structure under free vibration vibrates at its natural frequency, which is one of the dynamic properties of the structure Forced vibration takes place under the excitation of an external force and at the frequency of the exciting force, which is independent of the natural frequency of the structure When the frequency of the exciting force coincides with the natural frequency of a structure, resonance occurs and dangerously large amplitudes may result Therefore, the calculation of natural frequency and the examination of resonance are of practical importance Also, in reality, the influence of viscous dampening on a vibrating structure should be taken into account In this section, the fundamental dynamic properties are introduced 14-1 © 2005 by Taylor & Francis Group, LLC 14-2 14.1.2 Vibration and Shock Handbook Natural Frequencies An undamped single-degree-of-freedom (single-DoF) system (Figure 14.1) is used to explain natural frequency Figure 14.1(c) shows the free-body diagram with inclusion of the inertial force m€y: This force is equal to the mass, m; multiplied by the acceleration, y€ ; and should always be directed negatively with respect to the corresponding coordinate; note that k is the spring constant of the system The summation of forces in the y-direction directly gives the following equation of motion: my ỵ ky ẳ 14:1ị y ẳ A cos wt 14:2ị y ẳ B sin wt ð14:3Þ To solve Equation 14.1, we assume or where A and B are constants depending on the initiation of the motion and w is a quantity denoting a physical characteristic of the system The substitution of Equation 14.2 or Equation 14.3 into Equation 14.1 gives 2mw2 ỵ kịA cos wt ẳ 14:4ị If this equation is to be satisfied at any time, the factor in parentheses must be equal to zero: 2mw2 ỵ k ẳ The positive root of Equation 14.5 14:5ị s k rad=sec wẳ m ð14:6Þ is known as the circular or angular natural frequency of the system (Paz, 1997; Clough and Penzien, 1993) Note: The Greek symbol v is commonly used rather than w to denote this natural frequency in the literature, which is expressed in rad/sec and is called “angular” natural frequency The “cyclic” natural frequency f is usually expressed in hertz (Hz) or cycles per second (cps) The quantity w differs from the natural frequency f only by the constant factor, 2p: The period, T; is usually expressed in seconds per cycle (i.e., the value reciprocal to the natural frequency, f ) Therefore, it gives f ¼ w ¼ T 2p ð14:7Þ Example 14.1 Determine the natural frequency of the system shown in Figure 14.2, consisting of a weight of W ¼ 230 kN attached to a horizontal reinforced concrete cantilever beam through the coil spring k2 : y k m (a) mg ky ky (b) mg N (c) mÿ N FIGURE 14.1 Free-body diagrams: (a) showing single-DoF; (b) showing only external forces; (c) showing external and inertial forces © 2005 by Taylor & Francis Group, LLC Reinforced Concrete Structures 14-3 The cantilever beam has a cross section of 305 mm hị Ê 152 mm bị; modulus of elasticity E ẳ 30 kN/mm2, and a length L ¼ m The coil spring has stiffness k2 ¼ 200 N/mm 3000 mm A A k2 = 200 N/mm Solution The deflection, D, at the free end of a cantilever beam that is acted upon by a static force, P, at the free end is given by - #3 (fy = 413.8 MPa) PL3 3EI The corresponding spring constant, k1 , is then k1 ¼ W = 230 kN 305.0 mm 276.0 mm 23.8 mm D¼ (a) P 3EI ¼ D L where I ẳ 1=12ịbh3 (for rectangular section) when the contribution of reinforcing bars is neglected The cantilever and the coil spring of this system are connected as springs in series Consequently, the equivalent spring constant is f'c = 39 N/mm2 #2@89 mm A's = 142 mm2 As = 852 mm2 - #6 (fy = 475.85 MPa) 152.4 mm (b) FIGURE 14.2 System for Example 14.1 1 ẳ ỵ ke k1 k2 Substituting corresponding numerical values, we obtain Iẳ k1 ẳ Ê 152305ị3 ẳ 3:59 Ê 108 mm4 12 £ 30;000 £ 3:59 £ 108 ẳ 1197 N=mm 3000ị3 and 1 ỵ ẳ ke 1197 200 ke ¼ 171 N=mm The natural frequency for this system is then given by Equation 14.6 as pffiffiffiffiffiffi v ¼ ke =m ðm ¼ W=g and g ¼ 9:8 m=sec2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ¼ 171 £ 9:8=230 v ¼ 2:7 rad=sec or, using Equation 14.7 f ¼ 0:43 Hz 14.1.3 Viscous Damping Let us assume that we have modeled a structural system as a simple oscillator with viscous damping, as shown in Figure 14.3 In this figure, m and k are the mass and spring constant of the oscillator, respectively, and c is the viscous damping coefficient (or damping constant) The summation of forces in © 2005 by Taylor & Francis Group, LLC 14-4 Vibration and Shock Handbook the y-direction gives the differential equation of motion m€y ỵ c_y ỵ ky ẳ 14:8ị m c pt To solve Equation 14.8, we try y ¼ C e : Substituting this function into Equation 14.8 results in the equation mCp2 e pt ỵ cCp e pt ỵ kC e pt ẳ y k (a) 14:9ị ky cyã y mÿ (b) FIGURE 14.3 (a) Viscous damped oscillator; (b) freebody diagram After cancellation of the common factors, Equation 14.9 reduces to an equation called the characteristic equation for the system, given by mp2 ỵ cp ỵ k ẳ The roots of this quadratic equation are c ^ p1 ; p2 ẳ 2m 14:10ị s c k 2m m ð14:11Þ For a system oscillating with critical damping, the expression under the radical in Equation 14.11 is equal to zero, that is ccr 2m k ¼0 m p ccr ẳ km 14:12ị 14:13ị where ccr designates the critical damping value (Clough and Penzien, 1993; Paz, 1997) In general, the damping of the system is expressed as c ẳ j ccr 14:14ị where j is the damping ratio of the system Note: Often the symbol z is used instead of j to denote the damping ratio The dampening value depends on structural materials According to Dowrick (1987), the values for concrete structures are listed in Table 14.1 It can be seen from this table that for reinforced and prestressed concrete the values are and 2%, respectively Example 14.2 A reinforced concrete beam consists of a weight of 5.0 kN and a stiffness of k ¼ 4:0 kN/cm Find the undamped natural frequency the damping coefficient TABLE 14.1 Typical Damping Ratios for Concrete Structures Type of Construction Damping j; % of Critical Concrete frame, with all walls of flexible construction Concrete frame, with stiff cladding and all internal walls flexible Concrete frame, with concrete or masonry shear walls Concrete and/or masonry shear wall buildings Prestressed concrete 10 10 Notes: (1) The term frame indicates beam and column bending structures as distinct from shear structures (2) The term concrete includes both reinforced and prestressed concrete in buildings For isolated prestressed concrete members such as in bridge decks, damping values less than 5% may be appropriate, e.g., a value of to 2% may apply if the structure remains substantially uncracked © 2005 by Taylor & Francis Group, LLC Reinforced Concrete Structures 14-5 y k Fo sin ωt c (a) ky · cy mÿ Fo sin ωt (b) FIGURE 14.4 (a) Damped oscillator harmonically excited; (b) free-body diagram Solution pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi w ẳ k=m ẳ 4:0 kN=cm Ê 980 cm=sec2 ị=5:0 kN ¼ 28:0 rad=sec: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c ¼ jccr ¼ 0:05 Ê Ê 5:0 Ê 4:0ị=980 ẳ 14 N sec=cm: 14.1.4 Damped Harmonic Excitation Now consider the case of one-DoF system in Figure 14.4, vibration due to an external load of a sine function under the influence of viscous damping The differential equation of motion is obtained by equating to zero the sum of the forces in the free-body diagram of Figure 14.4(b) Hence, my ỵ c_y ỵ ky ¼ F0 sin wt ð14:15Þ The total response is then obtained by summing the complementary solution and the particular solution yst sin4t uị ytị ẳ e2jwt A cos wD t ỵ B sin wD tị ỵ p r2 ị2 ỵ 2rjị2 Y Dẳ ẳ p yst r2 ị2 ỵ 2r jị2 0.15 0.2 0.25 x = 0.4 x = 0.7 ð14:17Þ D varies with the frequency ratio, r; and the damping ratio, j: Equation 14.17 is plotted in Figure 14.5 (Paz, 1997; Clough and Penzien, 1993) It can be seen from this figure that 0:8 # r # 1:2 is in the resonance zone Therefore, for design, r 1:2 or r , 0:8 is required to avoid resonance © 2005 by Taylor & Francis Group, LLC 0.125 x=0 Dynamic magnification factor D where r is the frequency ratio that is equal to forced vibration frequency, 4; divided by natural frequency, w: Note: A further discussion of this topic is found in Chapter By examining the transient component of the response, it may be seen that the presence of the exponential factor, e2jwt , will cause this component to vanish, leaving only the steadystate motion, Y; which is given by the second term of Equation 14.16 The ratio of the steady-state amplitude of Y to the static deflection yst ; defined above, is known as the dynamic magnification factor, D: ð14:16Þ x=1 2 Frequency ratio γ = ω /ω FIGURE 14.5 Dynamic magnification factor as a function of the frequency ratio for various amounts of damping 14-6 14.2 14.2.1 Vibration and Shock Handbook Analytical Models Model-Based Simulation When a structure is analyzed for its vibrational characteristics, it first must be presented by a simple model that reflects its mechanical properties adequately In many analyses, mass is assumed to be concentrated at the nodes of the models By using this assumption, a single-story structure can be simplified as a single-DoF system subjected to a time-varying force, FðtÞ: In general, dynamic models of reinforced concrete structures depend on the structural systems Figure 14.6 indicates the structural systems and the corresponding dynamic models (Mo, 1994) 14.2.2 Flexural Behavior Using the trilinear theory (Mo, 1992), the primary curve (the load –deflection curve) of a reinforced/ prestressed concrete beam can be determined The trilinear theory is described as follows To find the load –deflection curve of a beam, first the moment –curvature relationship of each section needs to be determined The trilinear moment –curvature relationship is shown in Figure 14.7 The first branch of the trilinear curve represents the behavior of the reinforced concrete section until flexural cracking ðMc ; cc Þ: The second branch describes the behavior from the cracking until the yielding of the longitudinal steel ðMy ; cy Þ: The third branch gives the postyield behavior until flexural failure ðMu ; cu Þ: For a given cross section, the shape of the moment –curvature curve can be determined by using the following equations Basically, the parabola –rectangle stress –strain curve of concrete specified in the CEB code (1978) and the elastic –plastic stress –strain curve of steel are used in the computation (a) (b) Structure Dynamic model m m Framed shearwall Single-degree-of-freedom m m One-story frame Single-degree-of-freedom m m (c) Simply supported beam FIGURE 14.6 © 2005 by Taylor & Francis Group, LLC Single-degree-of-freedom Structures and corresponding dynamic models Reinforced Concrete Structures 14-7 Moment –curvature curve: Cracking state bh bh Mc ¼ f ¼ r where cc ¼ and qffiffiffi 7:5 f 0c Mc EI M ð14:18Þ (yy, My) (yc, Mc) 14:19ị y Mc ẳ cracking moment f 0c ¼ concrete compressive strength in psi fr ¼ concrete rupture strength in psi cc ¼ cracking curvature b ¼ beam width h ¼ beam depth E ¼ concrete Young’s modulus I ¼ moment of inertia (yu, Mu) FIGURE 14.7 Trilinear moment –curvature curve ec Xy e sc Yielding state (Figure 14.8) 1c ẳ 1y xy d xy 14:20ị 1sc ¼ 1y xy d d xy 14:21ị My ẳ k1 f 0c bxy d 20:375xy ị þ xy 2d0 1c ðEs 2Ec ÞA0s ðd 2d0 Þ xy and cy ¼ 1c xy e y = 0.00207 (for fy = 60 ksi) ð14:22Þ FIGURE 14.8 Strain diagram at yielding state 14:23ị where k1 ẳ 1c 11 12 c 10 10 14:24ị My ẳ yielding moment yy ¼ yielding curvature 1sc ¼ compression steel strain 1c ¼ concrete strain when tension steel yields 1y ¼ steel yielding strain d ¼ effective depth d0 ¼ distance between surface of concrete compression block and center of compression steel xy ¼ distance between surface of concrete compression lock and neutral axis when tension steel yields A0s ¼ area of compression steel Es ¼ steel Young’s modulus Ec ¼ concrete Young’s modulus © 2005 by Taylor & Francis Group, LLC 14-8 Vibration and Shock Handbook d′ Xu 0.41Xu 0.0035 d es f ′s A′s 0.81f ′c b Xu fs As b FIGURE 14.9 If Stress and strain diagrams at ultimate state Ultimate state (Figure 14.9) xu $ d then Mu ẳ 0:81f 0c bxu d 0:41xu ị ỵ xu d £ 0:0035ðEs Ec ÞA0s ðd d Þ xu ð14:25aÞ If xu , d then Mu ẳ fs As d 0:41xu ị þ d xu £ 0:0035Es A0s ðd 0:41xu ị xu cu ẳ 14:25bị 0:0035 xu 14:26ị where Mu ¼ ultimate moment cu ¼ curvature corresponding to ultimate moment xu ¼ distance between surface of concrete compression block and neutral axis at ultimate state fs ¼ steel stress at ultimate state As ¼ area of tension steel Load –deflection curve: Once the trilinear moment –curvature is found, we can convert it into the load –deflection curve (Figure 14.10) M¼ P L ; 2 [P ¼ where M ¼ moment P ¼ load L ¼ beam length © 2005 by Taylor & Francis Group, LLC 4M L P M ð14:27Þ M P/2 P/2 L FIGURE 14.10 Simple beam with a concentrated load Reinforced Concrete Structures 14-9 Curvature diagram: Cracking state (Figure 14.11) L Mc ML ¼ c 2 EI 4EI L L L dc ¼ uA u ¼ u A A M L2 ¼ c 12EI where uc ẳ L 14:29ị Yielding state (Figure 14.12) FIGURE 14.11 A Curvature diagram at cracking state L1 L2 C B yc L3 yy uy ẳ area of triangleị ỵ area of trapezoidị 1 ẳ L1 cc ỵ cc ỵ cy ịL2 2 B yc 14:28ị uc ¼ rotation at point A at cracking state dc ¼ midspan deflection at cracking state C A FIGURE 14.12 Curvature diagram at yielding state 14:30ị dy ẳ first moment of triangleị ỵ first moment of trapezoidị 1 1 ẳ L1 L1 cc ỵ L3 cc ỵ cy ịL2 ẳ cc L21 ỵ cc ỵ cy ịL2 L3 2 where L3 ¼ L1 þ L2 L 2L2 þ ðcy cc Þ cc L2 ỵ L2 cy cc Þ cc L2 ð14:31Þ ð14:32Þ uy ¼ rotation at point A at yielding state dy ¼ midspan deflection at yielding state Ultimate state (Figure 14.13) uu ¼ ðarea of triangleị ỵ area of first trapezoidị ỵ area of second trapezoidị 1 ẳ L1 cc ỵ cc þ cy ÞL2 þ ðcy þ cu ÞL4 2 du ẳ first moment of triangleị ỵ first moment of first trapezoidị ỵ first moment of second trapezoidị 1 ẳ L1 L1 cc ỵ L3 cc þ cy ÞL2 þ L5 ðcy þ cu ÞL4 2 1 14:34ị ẳ cc L1 þ ðcc þ cy ÞL2 L3 L1 L2 ỵ cy ỵ cu ịL4 L5 A where L L cc L2 ỵ cy cc Þ L2 2 ð14:35Þ L3 ¼ L1 þ cc L2 þ L2 ðcy cc Þ © 2005 by Taylor & Francis Group, LLC yc L3 L5 FIGURE 14.13 L4 C ð14:33Þ B yy yu Curvature diagram at ultimate state 14-10 Vibration and Shock Handbook L5 ẳ L1 ỵ L2 ỵ L4 L þ ðcu cy Þ L4 2 cy L4 ỵ L4 cu cy ị cy L4 14:36ị uu ẳ rotation at a point A at ultimate state du ¼ midspan deflection at ultimate state In this section, the maximum concrete strain at ultimate state ð1cu Þ is assumed to be 0.0035 according to the CEB code (1978) If the ACI code (2002) is employed, the value is 0.003 However, in seismic structures there will be more stirrups In these situations, reinforced concrete beams are confined Therefore, the maximum concrete strain at ultimate state for confined concrete can be used as follows (Dowrick, 1987): 1cu ẳ 0:003 ỵ 0:02 b lc ỵ rv fyv 138 14:37ị where b ẳ beam width lc ẳ distance from the critical section to the point of contraflexure rv ¼ ratio of volume of confining steel (including the compression steel) to volume of concrete confined fyv ¼ yielding stress of confining steel 14.2.3 Shear Behavior Structural walls in a frame building should be so proportioned that they possess the necessary stiffness needed to reduce the relative inter-story distortions caused by explosion-induced motions Such walls are termed structural (or shear) walls because their behavior is governed by shear if the ratio of height to length is less than unity Their additional function is to reduce the possibility of damage to nonstructural elements that most buildings contain Buildings stiffened by structural walls are considerably more effective than rigid frame buildings with regard to damage control, overall safety, and integrity of the structure This performance is due to the fact that structural walls are considerably stiffer than regular frame elements and thus can respond to or absorb the greater lateral forces induced by the earthquake motions, while controlling inter-story drift The past three decades saw a rapid development of knowledge regarding shear in reinforced concrete Various rational models that are based on the smeared-crack concept can satisfy Navior’s three principles of mechanics of materials (i.e., they satisfy stress equilibrium, strain compatibility, and constitutive laws of materials) These rational or mechanics-based models on the “smeared-crack level” (in contrast to the “discrete-crack level” or “local level”) include: the compression field theory (CFT) (Vecchio and Collins, 1981); the rotating-angle softened truss model (RA-STM) (Belarbi and Hsu, 1994, 1995; Pang and Hsu 1995); the fixed-angle softened truss model (FA-STM) (Pang and Hsu, 1996; Hsu and Zhang, 1997; Zhang and Hsu, 1998); the softened membrane model (SMM) (Hsu and Zhu, 1999; Zhu, 2000); and the cyclic SMM (Mansour, 2001) Vecchio and Collins (1981) proposed the earliest rational theory, CFT, to predict the nonlinear behavior of cracked reinforced concrete membrane elements However, the CFT is unable to take into account the tension stiffening of the concrete in the prediction of deformations because the tensile stress of concrete was assumed to be zero © 2005 by Taylor & Francis Group, LLC Reinforced Concrete Structures 14.2.4 14-17 Time-History Analysis To accurately determine the dynamic behavior of concrete structures, the time-history analysis (Clough and Penzien, 1993; Paz, 1997) is preferred This section will show an example for single-DoF systems, such as simple beams, torsional box tubes, spandrel beams, continuous beams, one-story frames and one-story framed shear walls All of these structures will be discussed later In time-history analysis, a framed shear wall can be modeled as a nonlinear single-DoF system (Figure 14.19) The dynamic incremental equilibrium is shown in Figure 14.19(c) The equation of the equilibrium is mDyi ỵ ci D_yi ỵ ki Dyi ẳ DFi 14:66ị where m is the mass at the top ci and ki are calculated for values of velocity and displacement corresponding to time t and assumed to remain constant during the increment of time Dt: Incremental acceleration, incremental velocity, and incremental displacement are D€yi ; D_yi ; and Dyi ; respectively To perform the step-by-step integration of Equation 14.66, the linear acceleration method is employed In this method, it is assumed that the acceleration may be expressed by a linear function of time during the time interval Dt: Let ti and tiỵ1 ẳ ti ỵ Dt be, respectively, the designation for the time at the beginning and at the end of the time interval Dt: When the acceleration is assumed to be a linear function of time for the interval of time ti to tiỵ1 ẳ ti ỵ Dt; as shown in Figure 14.20, the acceleration may be expressed as y tị ẳ y i þ D€yi ðt ti Þ Dt m F(t) (a) y k m c (b) m∆y··i ki∆yi c ∆y· i (c) F(t) ∆Fi i FIGURE 14.19 (a) Framed shear wall with a mass m at the top; (b) model for a nonlinear single-DoF system, and (c) free body diagram showing the incremental inertial force, the incremental damping force, the incremental spring force and the incremental external force ÿ ∆ÿi ÿi ÿi+1 ∆t ti FIGURE 14.20 time interval t i+1 t Linear variation of acceleration during ð14:67Þ Integrating Equation 14.67 twice with respect to time between the limits ti and t ẳ ti ỵ Dt and using the incremental displacement Dy as the basic variable gives D€yi ¼ 6 Dy y 3€yi Dt i Dt i ð14:68Þ D_yi ¼ Dt Dy 3_yi y€ Dt i i 14:69ị and â 2005 by Taylor & Francis Group, LLC 14-18 Vibration and Shock Handbook The substitution of Equation 14.68 and Equation 14.69 into Equation 14.66 leads to the following form of the equation of motion: m 6 Dt y_ i 3yi ỵ ci Dyi 3_yi y ỵ ki Dyi ẳ DFi Dyi 2 Dt Dt Dt i ð14:70Þ Transferring all the terms containing the unknown incremental displacement, Dyi ; to the left-hand side gives ki Dyi ẳ DFi 14:71ị where ki ẳ ki ỵ 6m 3c ỵ i Dt Dt 14:72ị and DFi ẳ DFi ỵ m Dt y_ i ỵ 3yi ỵ ci 3_yi ỵ y Dt i ð14:73Þ It should be noted that Equation 14.71 is equivalent to the static incremental-equilibrium equation and may be solved for the incremental displacement by simply dividing the equivalent incremental load, DFi ; by the equivalent spring constant ki : The displacement yiỵ1 and the velocity y_ iỵ1 at time tiỵ1 ẳ ti ỵ Dt are yiỵ1 ẳ yi ỵ Dyi 14:74ị y_ iỵ1 ẳ y_ i ỵ D_yi 14:75ị and The acceleration y iỵ1 at the end of the time step is obtained directly from the differential equation of motion to avoid the errors that generally might tend to accumulate from step to step It follows y€ iỵ1 ẳ ẵFtiỵ1 ị ciỵ1_yi ỵ1 kiỵyi þ1 m ð14:76Þ where the coefficients ciþ1 and kiþ1 are now evaluated at time tiỵ1 After the displacement, velocity, and acceleration have been determined at time tiỵ1 ẳ ti þ Dt; the procedure just outlined is repeated to calculate these quantities at the following time step, tiỵ2 ẳ tiỵ1 þ Dt: In general, sufficiently accurate results can be obtained if the time interval is taken to be no longer than one tenth of the natural period of the structure (Clough and Penzien, 1993; Paz, 1997) 14.3 14.3.1 Beams under Harmonic Excitations Mechanical Properties A 6-m-long beam is simply supported The cross section and material properties are shown in Figure 14.2(b) f 0c ¼ 39 N=mm2 A0s ¼ 142 mm2 d ẳ 305 12:7 6:35 19:05ị ¼ 276 mm As ¼ 852 mm2 Find the moment ðMÞ– curvature ðf Þ relationship of the cross section at the midspan and the force ðPÞ –displacement ðdÞ relationship of the beam © 2005 by Taylor & Francis Group, LLC Reinforced Concrete Structures 14-19 Solution Moment –curvature relationship q E ẳ 1500 f 0c N=mm2 ị ẳ 29614:9 N=mm2 (563, 86.1) (122, 82.5) M (kN m) Using the derived equations in Section 14.2.2 gives (8.66, 9.26) Mu ¼ 86;122;000 N mm; y (ì107 rad/mm) cu ẳ 5:63 Ê 1025 rad=mm FIGURE 14.21 example beam My ¼ 82;491;700 N mm; cy ¼ 1:22 £ 1025 rad=mm Mc ¼ 9;263;000 N mm Moment – curvature curve for the cc ¼ 8:66 £ 1027 rad=mm They are plotted in Figure 14.21, and the corresponding loads are determined below Force–displacement relationship PU ¼ £ 86;122;000 ¼ 57;415 N 6000 PY ¼ £ 82;491;700 ¼ 54;994 N 6000 PC ¼ £ 9;263;000 ¼ 6175 N 6000 The corresponding deflections can be determined as follows: A B C Cracking state (Figure 14.22) ψc dc ¼ ¼ Mc L2 12EI 9;263;000Ê6000ị2 12Ê29;614:9ịÊ Ê152:4ịÊ305ị3 12 L FIGURE 14.22 ẳ 2:6 mm L1 L2 A Yielding state (Figure 14.23) 9;263;000 ¼ 336:9 mm L1 ¼ 3000£ 82;491;700 © 2005 by Taylor & Francis Group, LLC RA Curvature diagram at cracking state Yc L3 FIGURE 14.23 C B Yy Curvature diagram at yielding state 14-20 Vibration and Shock Handbook L2 ẳ 30002336:9 ẳ 2663:1mm L3 ẳ 336:9 ỵ 2663:1 2663:1 ỵ 1:22 Ê 1025 8:66 Ê 1027 ị £ £ £ 2663:1 2 25 25 27 8:66 Ê 10 Ê 2663:1 ỵ Ê 2663:1 Ê ð1:22 £ 10 8:66 £ 10 Þ 8:66 £ 1027 £ 2663:1 £ ¼ 2053 mm dy ¼ 13 Ê 8:66 Ê 1027 Ê 342:2ị2 ỵ 12 Ê 8:66 Ê 1027 ỵ 1:22 Ê 1025 ị Ê 2663:1ị £ 2053 ¼ 35:75 mm L1 Ultimate state (Figure 14.24) L4 A 9;263;000 ¼ 322:7 mm L1 ¼ 3000 £ 86;122;000 Y L3 c 82;491;600 322:7 86;122;000 ¼ 2550:8 mm L2 ¼ 3000 £ L5 C B Yy Yu L4 ¼ 3000 322:7 2550:8 ¼ 126:5 mm L3 ẳ 322:7 ỵ L2 FIGURE 14.24 Curvature diagram at yielding state 2550:8 2550:8 ỵ 1:22 Ê 1025 8:66 £ 1027 Þ £ £ £ 2550:8 2 27 25 27 8:66 £ 10 £ 2550:8 ỵ Ê 2550:8 Ê 1:22 Ê 10 8:66 £ 10 Þ 8:66 £ 1027 £ 2550:8 £ ẳ 1967:3 mm L5 ẳ 322:7ỵ2550:8ỵ 126:5 126:5 ỵ5:63Ê1025 21:22£1025 Þ£ £ £126:5 2 25 25 25 1:22Ê10 Ê126:5ỵ Ê126:5Ê5:63Ê10 21:22Ê10 ị 1:22Ê1025 Ê126:5Ê ẳ 2950:4 mm du ẳ Ê8:66Ê1027 Ê322:7ị2 ỵ Ê8:66Ê1027 ỵ1:22Ê1025 ị Ê 2550:8Ê1967:3 ỵ Ê5:63Ê1025 ỵ1:22Ê1025 Þ £ 126:5£2950:4 ¼ 45:6 mm (45.6, 57415) (35.75, 54994) P (N) (2.6, 6175) d (mm) The load – deflection Figure 14.25 curve © 2005 by Taylor & Francis Group, LLC is shown in FIGURE 14.25 Load – deflection curve Reinforced Concrete Structures 14.3.2 14-21 Design for Machine Vibration The same beam as described in Section 14.3.1 is supporting a machine with a rotating frequency of Hz at the midspan of the beam The mass of the rotating machine and the beam is N sec2/mm Check if resonance will occur Solution Find the stiffness, k; the natural frequency, f ; and the frequency ratio, r; for each of the three states, namely the elastic, cracking, and yield states Elastic state 617:5 ¼ 2375 N=mm 2:6 rffiffiffiffiffiffiffiffi 2375 ¼ 7:76 Hz f ¼ 2p 7:76 r¼ ¼ 0:862 k¼ Since 0:8 , r , 1:2; it is in the resonance zone The beam needs to be redesigned to avoid resonance Cracking state k¼ 5499:4 617:5 ¼ 1472:7 N=mm 35:75 2:6 rffiffiffiffiffiffiffiffiffiffi 1472:7 ¼ 6:11 Hz f ¼ 2p 6:11 ¼ 0:68 , 0:8 r¼ OK Yield state k¼ 5741:5 5499:4 ¼ 245:8 N=mm 45:6 35:75 rffiffiffiffiffiffiffiffi 245:8 ¼ 2:50 Hz f ¼ 2p 2:50 ¼ 0:28 , 0:8 r¼ OK 14.4 14.4.1 Design for Explosions/Shocks Column A 3-m-tall viaduct column supporting a mass of 50 N sec2/cm is subjected to a shock acceleration that is the same as the 1940 El Centro seismogram 14.4.1.1 Load –Displacement Relationship Using the same procedures presented in Section 14.3, the moment –curvature relationship of the cross section and the force –displacement relationship of the column can be determined It is assumed that the force –displacement relationship of the column was found to be the same as that shown in Figure 14.25 © 2005 by Taylor & Francis Group, LLC 14-22 Vibration and Shock Handbook 14.4.1.2 Dynamic Response To perform the dynamic response analysis, the computer program in Mo (1994) can be used The required input data may be determined as follows: Stiffness ¼ 0:6175 ¼ 0:2375 T=mm 2:6 Stiffness ¼ 5:4994 0:6175 ¼ 0:1473 T=mm 35:75 2:6 Stiffness ¼ 5:7415 5:4994 ¼ 0:246 T=mm 45:6 35:75 dc ¼ 2:6 mm dy ¼ 35:75 mm du ¼ 45:6 mm Result: Maximum deflection ¼ 38.33 mm 14.4.2 Shear Walls 14.4.2.1 Design Approach 14.4.2.1.1 Design Considerations To insure ductile failure, the two design limitations for overreinforcement and minimum reinforcement can be derived in terms of the inclination angle, a; of the diagonal concrete struts Overreinforcement Mo (1987) shows the inclination angle of the diagonal compression struts in the “lower balanced reinforcement,” alb ; to be 0vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi139 u