THE UNIVERSITY OF MICHIGAN INDUSTRY PROGRAM OF THE COLLEGE OF ENGINEERING DYNAMIC BEHAVIOR OF GUY CABLES SUBJECTED TO A SMALL PERIODIC END DISTURBANCE Hassan Sarvi A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in the University of Michigan Department of Civil Engineering 1968 December, 1968 IP-827

ACKNOWLEDGEMENTS The author wishes to express his appreciation to Professor Lawrence C. Maugh, Chairman of his doctoral committee, for the guidance and encouragement in the course of research and preparation of this dissertation. He also wishes to thank all other members of his doctoral committee for their helpful suggestions. The author is grateful for the computer time made available through the Computing Center of the University of Michigan for the numerical computations. The reproduction of the thesis was undertaken by the College of Engineering, Industry Program at the University of Michigan. For this valuable assistance the author is sincerely grateful. iii

TABLE OF CONTENTS Page ACKNOWLEDGEMENTS..................................................... iii LIST OF TABLES....................................................... vi LIST OF FIGURES............................... vii NOMENCLATURE......................................................... ix I INTRODUCTION....................................... 1 1. Statement of the Problem.............................. 1 2. Scope of the Research................................... 2 3. Prior Studies.......................................... 3 II STATIC FORCE-DISPLACEMENT RELATION FOR CABLES.... 6 1. Fundamental Equations.......... 6 2. A Procedure for Evaluating Force-Displacement Relation.. 8 3. Force-displacement Relation for a System of Guys........ 13 4. A Linear Approximation for the Force-Displacement Relation............................................... 18 III DERIVATION OF THE DIFFERENTIAL EQUATIONS OF MOTION FOR A CABLE.................................................... 24 1. Introduction............................................ 24 2. The General Equations of Motion of a Cable.............. 25 3. Modified Equations of Motion for a Guy Cable....,.......28 a. In-plane Vibrations.............................. 29 b. Out-of-plane Vibrations...................... o o. 40 4. Motion of a Guy in the Free Vibration of a Guyed-tower.. 41 IV SOLUTION TO THE LONGITUDINAL EQUATION OF MOTION FOR A PERIODIC END DISTURBANCE................................ 44 1. Introduction............................................ 44 2. Solution for Harmonic Force at one End............... o 44 a. Steady State Solution,.......................... 48 3. Solution for an Imposed Harmonic End Motion............ 49 a. Steady State Solution.........o................. 53 4. Force-displacement Relation for a Periodic End-disturbance..........6............................ 54 5. Resonance Frequencies.....................doo e... o o 57 6. Some Further Remarks.. d................ e eo........O 58 iv

TABLE OF CONTENTS (CONT'D) Page V SOLUTION TO THE TRANSVERSE EQUATION OF MOTION,.... o..... 60 1, Statement of the Problem................................. 60 2. Variation of Te and R ooo............................. 61 3o An Approximation for the Variation of Te o o 62 4o Method of Solution..... o o... o o.. o o o o... 64 5~ Solution of Equation (5026) by Power Serieso...... oo... 69 ao Method of Frobeniusoo,......oo.......... oooo,,oo. 71 bo The Complete Solution, oooo,,...... oo0...00 oooooo 74 6, Natural Frequencies,,,,..,.....,0.... 0,,,Ooooooo0o o.... 76 70 Mode Shape s.,,,o,.oo o,,, oo ooooooo ooo ooooo o. o.o oo 78 ao Orthogonality of Normal Modeso o..o o o o.. o....... 79 b, Normal Modes in Terms of the Variable s o, o,,,oo 81 80 The Original Problem of Boundary Motiono.....o,,o.o.... 81 9, Normalized Modeso o.,..... o..o. o. o o... o o o.. o..o 83 10, An Approximate Solution for the Transverse Equation of Motion................................................... 97 llo Dynamic Force-displacement Relatio o o o o o o o o o o o 100 ao A More Exact Evaluation of the Dynamic Guy Moduluso 103 12, Comparison of the Results with Kolousek-Davenport 0Solution..... o..o..o..o.ooo.o..ooo.. oo e.... oo. o o...o o 105 VI REPRESENTATION OF CABLES BY A DISCRETE PARAMETER MODEL...o...o 112 1. Methods of Constructing the Discrete Parameter Modelo.... 112 a, Method of Funicular Polygon............. o....o. 114 bo Method of Hung Masses o....................... 117 c, A Numerical Example Using the Method of Funicular Po lygon........................................... 119 2. Equations of In-plane Motion of Discrete-parameter System 121 35 Method of Solution,,..,,.......... a o oa oo o o o o o o.o 124 4, Size of the Time Interval o,, o.,o o o,,oo,,o.. oo,,,,oo,oo,, 126 5, Response to a Harmonic End Disturbance, o.,,,,,,,,,,, 128 6, Inclusion of Damping,,,oooooooooo oo.......... ooooo00oo, 129 a, Effect of Relative and Absolute Damping,,,o,,,oo o 136 bo Response of the Damped System,o................ 139 VII SUMMARY AND CONCLUSIONS,. o o o o.o..o o o o o. o o o o o o o o o o a o o 1o46 ~REFE~~~RENCES0o S,,, 0,.0 0 o 000o o o 0, o o o o0 oo149

LIST OF TABLES Tables Page I Normalized Transverse Modes of Guys and Strings with Immovable Ends................................... 86 II Orthonormality Check.................................. 94 III Parameters of the Five-lumped-mass System.............. 120 vi

LIST OF FIGURES Figure Page 2.1 Hanging Cable...,..................... 6 2.2 Static Force-displacement Relationship for a Guy............ 11 2.3 Four-way Arrangement of Guys.............................. 15 2.4 Static Force-displacement Relationship for Four-way Arrangement of Guys................................. 16 2.5 Static Force-displacement Relationship for Four-way Arrangement of Guys................................ o.... 17 2.6 Three-way Arrangement of Guys............... o..... o.... 18 2.7 Static Force-displacement Relationship for Three-way Arrangement of Guys.. o.................................... 19 35 1 Parameters of a Cable in Motion,.,.......................0. 26 3.2 Element of a Cable in Static Equilibrium and in Motion.,... 30 353 Projection of Tension in Vibrating Cable on Static Equilibrium Configuration. d.,......0..... o..a e 30 3.4 Displacement of an Element of a Straight Bar......... 34 3.5 Stretched Position of an Element of a Cableo.... o... o. o, 536 5.6 Longitudinal Motion of a Cable Restrained in Transverse Direction..... 0 o.....d..... d...........o o o... o. 36 307 Cable Displaced in Longitudinal Direction,.OdO,.o,,,,,..,, 38 4.1 Dynamic Magnification Factor for Sinusoidal Forcing Function 56 4,2 Dynamic Magnification Factor for Sinusoidal End Motion.....o 56 5.1 Inclination of a Guy,,,,. a..................... 61 5.2 Variation of Static Tensile Force,,,,.,. o..... o. 65 5.3 Comparison of Normalized Transverse Modes of a Guy and a String................. o -.......... c............ o o o o o o o o 75 5.4 Comparison of Normalized Transverse Modes of a Guy and a String................... o. o o.. o aJ o o ao. o.ee o o o 96 5.5 Dynamic Guy Modulus Parameters.,,..... o a e o o o o o o o o 100 vii

LIST OF FIGURES (CONT'D) Figure Page 5.6 Variation of Dynamic Guy Modulus with Excitation Frequency... 102 5.7 Relationship Between Dynamic Modulus and Excitation Frequency: Kolousek-Davenport Method...................... 109 5.8 Relationship Between Dynamic Modulus and Excitation Frequency: Equation (5.122)............................. 109 6.1 Funicular Polygon.......................................... 112 6.2 Geometry of Discrete-parameter System........................ 116 6.3 Parameters of Lumped-mass System............................ 118 6.4 Representation of Five-mass System by Method of Funicular Polygon............................... 119 6.5 Variation of Horizontal Resisting Force with Time for a Horizontal Sinusoidal End Motion............................. 130 6.6 Motion of Middle Mass for a Horizontal Sinusoidal End Motion. 131 6.7 Single Mass Attached to a Highly Stretched Wire.............. 133 6.8 Arrangement of Dampers to Produce Relative Damping........... 135 6.9 Horizontal Motion of Top Point.......................... 137 6.10 Variation of Horizontal Resisting Force with Time for Horizontal Disturbing End Motion of Figure 6.9............ 137 6.11 Motion of Middle Mass for Horizontal Disturbing End Motion of Figure 6.9............,,,................. 138 6.12 Variation of Horizontal Resisting Force with Time for a Horizontal Sinusoidal End Motion............................. 140 6.13 Motion of Middle Mass for a Horizontal Sinusoidal End Motion. 14L viii

NOMENCLATURE Symbols are defined when they first appear in the text; those which appear frequently are listed below for reference: A Cross sectiona area of cable A, Ai Integration constants B, Bi Integration constants C Constant given by Equation (5.118) Ci Integration constants Di Integration constants E Modulus elasticity of cable F Constant defined by Equation (5.130) Fn Normal component of external force per unit length F Tangential component of external force per unit length; amplitude of applied sinusoidal force at end G Constant defined by Equation (5.131) HE Horizontal component of cable tension Hn Harmonic series J Bessel function of the first kind and of order m K Guy modulus (ratio of horizontal reaction to horizontal displacement) L Total length of cable Lu Unstressed length of cable L Length of cable chord Mi i-th mass of the discrete parameter system N. i-th normalization factor for transverse modes of a guy, 1 Equation (5.99) Ni i-th normalization factor for transverse modes of a string, Equation (5.110) ix

NOMENCLATURE (CONT'D) Nm Bessel function of the second kind and of order m Q Time function in separation of variables relations R Radius of curvature of stationary cable R Resultant horizontal force for a group of guys R Average radius of curvature of stationary cable T Cable tension during vibration in Section 5.2; elsewhere, cable tension resulting from in-plane vibration Te Cable tension in static equilibrium Teo, TeL Cable tension in static equilibrium at ends Te Average cable tension in static equilibrium Tei Cable tension in i-th segment of the discrete parameter system in static equilibrium Ti Cable tension in i-th segment of the discrete parameter system during vibration Ui, Vi Functionsdefined by Equations (6.28) Wi i-th weight in the discrete parameter system X,Y,Z Components of external forces per unit length relative to fixed coordinates X,Y Spacial functions in separation of variables reliaLic nrl; X. Longitudinal mode shapes of a guy Yi Transverse mode shapes of a guy Yi Transverse mode shapes of a string Xi, Yi Components of resultant applied force in the discrete parameter system Y First solution of Equation (5.26), given by Equation (5.36) YII Second solution of Equation (5.26), given by Equation (5.48) Z Function defined by Equation (5.40) Z(j) Variables defined by Equation (6.36) i x

NOMENCLATURE (CONT'D) a Constant defined by Equation (4.la) al, a2 Integration constants in Chapter II b Function defined in Equation (5.13) b Constant defined by Equation (5.117) c Constant equal to unstressed length of cable for no temperature change c Subscript denoting complimentary solution cn Coefficients of power series, Equation (5.29) e Subscript denoting equilibrium condition f Function defined by Equation (4.12) in Chapter IV; function defined by Equation (5.88) in Chapter V ft'f,fh Applied forces per unit length of cable g Acceleration of gravity g Function defined in Equation (4.35) h Vertical projection of cable length h Subscript denoting out-of-plane horizontal direction ij Integer numbers mostly used in indices k Constant defined by Equation (5.61) k Static taut wire modulus Q ~Horizontal projection of cable length li Length of i-th line-segment of the discrete parameter system m Mass of cable per unit length n Subscript denoting normal direction n Integer number in power series relation, Equation (5.29) n Number of masses in the discrete parameter system p Circular frequency of excitation p Subscript denoting particular solution xi

NOMENCLATURE (CONT'D).i Modal coordinates w~ r Dimensionless cable parameter = W2H r Constant defined by Equation (6.10) 1 s Length along the arc of stationary cable Si Length along the arc of stationary cable at segmentary dividing points on cable s Length along the chord of cable t Time t Subscript denoting tangential direction u Function defined in Equation (4.$5) v Function defined in Equation (5.13) Uiyvi Displacement coordinates of the discrete parameter system relative to static equilibrium w Weight of cable per unit length wtwn Tangential and normal components of gravity force per unit length x,yz Coordinates in a Cartesian coordinate system fixed in space Xi ~y i Coordinates of segmentary dividing points on stationary rab! in a Cartesian coordinate system fixed in space xi,Yi Coordinates of stationary lumped-mass points in a Cartesian coordinate system fixed in space z Constant defined by Equation (5.60a) zi Roots of frequency Equation (5.62) xii

NOMENCLATURE (CONT D) a Factor defined in the force-displacement relation, Equation (3.20).i Constant defined by Equation (5.63a) Pi ~Damping factor defined in Equation (6.45) Y Inclination of tangent to stationary cable 7 Inclination of cable chord 7 Inclination of i-th line-segment of the discrete parameter system i 5 Horizontal displacement of top point of a cable 5~ In-plane horizontal displacement of top point of a cable oA Incremental value of the argument A Amplitude of in-plane horizontal end motion of a guy A Amplitude of guy top motion in normal direction n A, Amplitude of guy top motion in tangential direction AH Change in horizontal component of cable tension Al Change in horizontal projection of cable length AL Change in length of cable relative to unstressed length AT Change in cable tension as a result of vibration At Time step CE Increment of strain in cable as a result of vibration be Increment of strain due to longitudinal vibration Out-of-plane displacement coordinate of guy cable r[ ~Displacement coordinate of in-plane motion of guy cable in a direction normal to stationary cable r( ~Function defined by Equation (5.114) Q Function defined by Equation (5.24) Q Angle between direction of guy top displacement and plane of guy xiii

NOMENCLATURE (CONT D) X\ Constant defined by Equation (5,27) \h Values of X corresponding to natural frequencies ti Mass density v Parameter defined in Equation (5.29) Displacement coordinate of guy cable along the tangent to stationary cable Function defined by Equations (4.31) and (4e54) ar Length along the arc of cable in motion T Increment of projection, on stationary cable, of guy tension during vibration relative to static tension Normalized transverse mode shapes of guy, Equation (5.100) Normalized transverse mode shapes of string, Equation (5.111) cp Function defined by Equation (5.52) ~cp Function defined by Equation (51553) cp Angle between tangents to stationary and vibrating cable *i ~ Normalized modal coordinate "Ai Natural frequency of guy _i ~Natural frequency of string Frequency ration) 1 xiv

CHAPTER I INTRODUCTION 1. Statement of the Problem The collapse of several guyed towers due to wind pressures well below the lateral design load has emphasized the need for a thorough investigation of the dynamic behavior of this type of structure. The current design procedure for lateral forces is based on the simple assumption that the effects of wind or earthquake on a structure can be represented by some static lateral force applied to the structure. It is apparent, however, that this assumption does not prevail in reality. When subjected to either type of disturbance the structure undergoes a vibrating motion and internal forces are developed which vary with time. Moreover, some of the important aspects of the excitation due to wind or earthquake cannot be expressed in terms of forces. In particular, during excitation a resonance phenomenon may develop, a condition whose detection is possible only by a study of the dynamic behavior of the structure. A dynamic study of guyed towers is, therefore, of considerable practical importance. For such an analysis it is essential to first investigate the dynamic properties of the guy cables particularly with respect to two factors of primary importance, i.e. (a) the natural frequencies of a guy, and (b) the relation between tower motion at a supporting point and the restraining forces which are developed by the guys attached to the tower at that point. Once this phase of the investigation is completed the tower may be studied as a continuous system on elastic supports with known dynamic properties. -1

-22. Scope of the Research A study of the dynamic properties of guy cables is presented here as the primary subject of this research. This problem presents considerable difficulty because of the following factors: (a) the geometry of the guys, (b) the variation of the tensile forces along a guy cable, and (c) the fact that during any motion of a guyed tower the guys experience a longitudinal vibration as well as transverse oscillations. The extent of the complexity of the problem is such that no comprehensive dynamic analysis of guy cables has been achieved as yet. Strictly speaking, the dynamics of a guy cable presents a non-linear problem whose exact solution is, as yet, not feasible. In this investigation two approximate solutions to the problem will be considered. In the first method a guy is studied as a distributed mass system and the non-linear differential equations governing its motion are reduced to linear equations for small oscillations. The linearized equations of motion are then made tractable by a suitable approximation as to the variation in tensile forces along the cable when in static equilibrium. As a second method, a lumped mass approximation to the original system is employed and a numerical procedure is used to integrate the equations of motion. As part of the research, the static force-displacement relation will also be evaluated and the results will be compared to the relation obtained dynamically. Moreover, attempt will be made to investigate the effects of concentrated hung masses (which are sometimes used in practice) on the overall behavior of the guy vibration. Finally, the dynamic behavior of a number of guys joined at a level of the tower will be determined,

-5In brief, it is the purpose of this research to investigate the primary factors involved in the dynamics of cables as commonly used in guyed towers. It is hoped this study will shed light on the subject and pave the way for a dynamic analysis of guyed towers. 3. Prior Studies The first analytical approach to the problem of vibration of cables was made by Rohr( ) in 1851. He examined a nearly flat cable suspended under the action of gravity from two points on the same horizontal level and evaluated the first two symmetrical natural modes for small oscillations. Rohr's treatment of the problem was based on the assumption of inextensible strings, hence his results correspond to the third and fifth modes of an extensible cable. The general equations governing the motion of a cable were set (2) forth by Routh in 1884. The equations were given both for inextensible and elastic strings. Although these equations are theoretically sufficient to determine the three components of motion and the tensile force acting on the string an analytical integration for the free vibration is possible only for small oscillations, which reduces the system to linear equations, and moreover, for some particular variation of mass along the cable. For an inelastic cable suspended under gravity from both ends on the same horizontal line, Routh obtained a solution for a case in which the cable hangs in equilibrium in the form of a cycloid. This solution was given for small free vibrations in the vertical plane of the cable by expanding the dependent variable into a harmonic series

-4with respect to time. He also deduced a transcendental equation for the natural frequencies of vibration for a flat cycloidal arc. In 1949. Pugsley( derived approximate formulae for the natural frequencies of an inextensible, nearly horizontal catenary based on the propagation of transverse waves in a vertical plane along the cable. He then modified the coefficients of the formulae for the first three frequencies to obtain better agreement with experimental results and the frequencies given by Routh. In 1953, Saxon and Cahn(4) analyzed the small free vibrations of an inelastic uniform cable by developing the solution into power series and obtained natural frequencies for a flat cable. In their paper Saxon and Cahn also demonstrated the accuracy of the analysis by comparison with experimental results. The natural frequencies obtained in References2 and 4 are in good agreement for a flat cable. This is expected because the variation in the mass per unit length along a flat cycloidal arc is small. In recent years, a few studies have been made on the vibration of guy cables. In 19479 Kolousek) analyzed the small vibrations of a taut guy cable caused by a harmonic horizontal motion at the top point and developed a formula for the horizontal force acting at the upper point of the cable. In KoloUsek's analysis the static tensile force is assumed constant throughout the guy. Moreover, the change in the cable tension during vibration is considered a function of time only. (6) (7) In his dissertation in 1955, and a later publication in 1961, Dean attempted a solution assuming the tensile force at any point to remain the same as in the static equilibrium.

-5Davenport() followed Kolousek's analysis and simplified the (9) final results in 1959. In a later paper in 1965, he extended the analysis by introducing viscous damping. An investigation of the forces on mooring cables due to the bobbing up and down of ships was reported in 1959 by Walton and (10) Polachek using a lamped mass approximation and assuming inextensible lines. All these works provide a good background for the study undertaken in the following chapters. Less directly related work has also been done on the vibration of cables suspended from one end and on the deposit of submarine cables.

CHAPTER II STATIC FORCE-DISPLACEMENT RELATION FOR CABLES 1. Fundamental Equations The equilibrium equation for a cable subjected to distributed vertical loads (Figure 2.1) is d2 ds H = w (2.1) dx2- dx where H represents the horizontal component of the cable tension, and w is the intensity of the vertical loads along the cable. When w is XX-r Figure 2.1. Hanging Cable. constant, i.e., when the vertical loads are uniformly distributed along the cable axis, Equation (2.1) is readily integrated to give(7) dy = sinh ( x + al) (2.2) dx H and H w y = - cosh (H x + a) + a2 (2.3) For the cable shown in Figure 2.1, the constants of integration al and a2 are a = sinh1 ( h ) - r (2.4) -6

-7a2 = - cosh a (25) where w~ r = (2.6) The length along the cable from the lower end is dy21/2 X s = J -dx=f [1 + () / dx f cosh ( x + al)dx o o H 1 or H w s = w [sinh (H x + al) - sinh al] (2.7) The length of the cable will then be H 2H L = [sinh (2r+al) -sinh al] = [cosh (r+al) sinh r](2.8a) using Equation (2.4), Equation (2.8a) gives 2 4H2 2 1/2 2 2 sinh2 rl/2 I = [h2 + - sink h r]= [h + I (2o8b) 22 r2 or 2H sinh r L2-h2= w- sinh r = (2.8c) The tensile force Te at any point x will be Te = ds = H cosh (W x + al) (2.9) The increase in the length from the unstressed length of the cable, defined as the length when the cable lies on a horizontal support under no load, is 1 L H ~ 2 w au AE J T ds = f cosh ( x + a1)dx (2.10a) where A is the area and E is te o where A is the area and E is the modulus of elasticity of the cable.

-8Carrying out the integration and substituting from Equation (2o4), one obtains, after simplification Hi i_ sinh (2r) h2 r (2lb) Lu = 2AE 1 + 2r +2 tanh r1b) The unstressed length Lu is obtained from definition as the difference between cable length under tensile forces and the change in length due to these forces; ie,, Lu = L -ALu (2.lla) or, when ALu is replaced from Equation (2,10b) Hi r EC sinh (2r) h2 r Lu = L AE[+ 2r + 2 2tanh r] (2 llb) 2AE 2 2. A Procedure for Evaluating Force-Displacement Relation In Reference 7 an approximate procedure is given for the determination of the stresses due to variations in loading and movement of the supports. A more accurate method is presented here based on the relations (2.8c) and (2,11b)o The procedure is outlined as follows: 1o Given the initial variables ~, h, w and H the length of the cable L is found by Equation (2,8c) and the unstressed length Lu is computed by Equation (2 llb). 2. The unstressed length thus obtained will remain a constant quantity for a change in any other variable except temperature. Thus, assuming no temperature change, Equation (2.11b) may be written as H sinh (2r) h2 r L - 2-E 1 + + 2 7 tanh = c (2.12) in which c is a constant equal to the unstresses length evaluated in Step 1o

-93. Equations (2.8c) and (2.12) represent two fundamental relations between five parameters I, h, w, L, and H. By specifying any three parameters these equations may be solved for the other two variables. For instance, the quantities L and H may be found due to given changes in ~, h, and w. The static analysis of a cable, therefore, represents no problem other than simultaneously solving Equations (2.8c) and (2.12). A case of interest is the variation AH of the horizontal cable force as a function of the change A2 in the horizontal projection of the cable. This relation is found here for a cable with the following dimensions: A = 2.42 in.2 E = 24 x 106 lbs/in2 w = 9 lbs/ft I = 950 ft. h = 1349 ft. H = 35300 lbs. The computation is carried out as follows: Equation (2.6): r =.121105 Equation (2.8a): L = 1651.2787 ft. Equation (2.10b): ALu = 1.7502 ft. Equation (2.11a): c = L = 1651.2787 - 1.7502 = 1649.5285 ft. Equations (2.8c) and (2.12) call for a trial and error method for the solution. These equations appear to be sensitive with respect to changes in H, therefore, the problem is attacked by assuming a value for H and solving the equations for ~ and L.

-10As sume AH = 1000 lbSo First trial: H = 36300 lbso, I = 950 fto r = o117769 Equation (2o10b): A Lu = 1o7993 ft. Equation (2,,11a): L = 1649.5285 + 1.7993 = 165103278 ft. Equation (2,8c): I = L 2/sinh r 950.2108 ft Second trial: H = 36300, = 950o2108 r -= 117795 ALu. 1.7792 L = 1649 5285 + lo7792 = 1651o3277 9 = 950.2096 A third trial gives the same result for.O Thus, AH = 1000 lbs. for LA = 22096 ft. The computations for other increments AH were carried out in a similar manner in the range - 17650 < AH < 35300 i:n which four trials proved sufficient to give an accuracy of at least four decimal points for A~ o The result of the solution is shown in Figure 2,2, For guy cables, the parameter r is usually small so that the hyperbolic terms may be replaced by a few terms of their power series without a significant loss of accuracy. If the series for the functions for r which appear in Equation (2o8c) and (2.12) are truncated after three terms, one obtains

AH IN KIPS -50.- 950 FT.. - 40 h = 1349 FT. W = 9 LB/FT. -30 H = 35.3 K A = 2.42 SQ, IN. -20_ E = 24 x 103 K/SQ. IN. a -7 -6 -5 -4 -3 -2 0 1 2 3 4 5 — 1 ~A__t0_A,IN FEET — 20 — -30 Figure 2.2. Static Force-displacement Relationship for a Guy.

-12-.kl-g (2.13) r 6 120 2 4 sinh(2r) ^ 1+ 21 + + (2.14) 2r 3 15 r ~....1 r2 2r4 r22r42 tanh r - (r- L - 1+ 5 15 = 1+ 3- 4 (2.15) Equations (2.8c) and (2.10b) will then become L2_-h2 = (1 - + - + ) (2.16) 6 120 2 2 r4 h2 ILu [(l + r)(l + 2+ (1 ) (2.17) AE 3 )2 15 312 A numerical example will be worked out below using Equations (2 16) and (2.17). A = 2.42 in.2 E = 24 x 106 lbs/in2 w = 9 lbs/ft = 950 ft. h = 1349 ft. H = 35,000 lbs. Find Al for AH = -17,000 lbs. Solution: r = 9 90.1211, r2.0147 2 x 35300 since r is very small terms involving r may also be neglected for this solution. Hence, from Equation (2.16) L = 1651.28 ft.

-13Equation (2.17) will then give ALu = 1.75 ft. Therefore, c = Lu = 1651.28 - 1.75 = 1649.53 ft. First trial: H = 35300 - 17000 = 18300 lbs. ~ = 950 ft. 2 r =.2336, r2 =.0546 ALu =.92 ft. L = 1649.53 +.92 = 1650.45 ft. I = 942.30 ft. Second trial: H = 18500 lbs. I = 942.30 ft. r =.217 r2 =.0557 ALu =.94 ft. L = 1649.53 +.94 = 1650.47 ft. i = 942.48 ft. A third trial will yield ~ = 942.44 ft. which compares with good accuracy with a value of 942.4280 ft. obtained by solving Equations (2.8c) and (2.12). 3. Force-Displacement Relation for a System of Guys To find the resultant horizontal force, R, equivalent to a group of guys attached at one level of the tower, the horizontal displacement of the attachment point, 6, is resolved into a change 8~ in the horizontal projection of any guy and a displacement perpendicular

-14to this directiono The effect of the latter displacement on the variation of the cable tension is small and may be neglectedo That is, the cable is assumed to offer no resistance to small displacements in a perpendicular direction. In any lateral deflection, the tower also undergoes an elastic axial deformation due to the change in the resultant vertical reaction of the guys. For actual guyed towers the elastic change in height, at any level of the guy attachment, is small compared to the horizontal displacement so that the effect of a change in h on the restraining force R may also be neglected. In a precise analysis in which this effect is included, it will be necessary to take into account the change in height at any level resulting from the action of the guys at all levelso With these simplifying assumptions the graph in Figure 2.2 can easily be used to construct the relation between the horizontal restraining force R and the horizontal displacement 6o Consider a four-way symmetrical arrangement of the guys as shown in Figure 2o3a in which the direction of the displacement is indicated by the angle ~o When 0 = 0 the resistance is caused by guys 1 and 3 only for which the variation in the horizontal forces Hi and H3 shown in Figure 2~4 are obtained from Figure 2.2. (note the positive directions of force and displacement on Figure 2.3b)o In Figure 2o4 is also shown the restraining force R = H + H (2o18) which, as seen from the graph, exhibits a nearly linear relation for this particular case.

-154 I f-V 2 ^^ ~(b) " Figure 2.5. Four-way Arrangement of Guys. If the displacement 5 is considered in a direction for which G = 450, then 6~ =51 = 6 cos 450 =.7076 (2.19) 6 = 6" = -6 cos 450 = -.7076 (2.20) The restraining force, which because of symmetry will be in the direction of the displacement, can be found as follows: R = LHi cos ~ + M2 sin ~ - AH cos ~ - AH4 sin ~ =.707 (,HM + M2 - LAH3 - H4) (2.21) since AlT AlT and AH -l H1 2 and H AH 4 (2,22) then R = 1.414 (AH1 - ALH) (2.25) where AH-1 and AH are found from Figure 2.2 at points 6,, and 6p3 give byEquaion (~.~) nd (.~0. Te reultis sownin Fg.1 6^ given by Equations (2.19) and (2.20). The result is shown in Figure 2.5,

70 - - 60___ _ 50 40 30 I - - FT - 0720 z " 40 O 0 _ I -10 ~N -20 FE T 0 I -30 -40 -50 -60 -701 -5 -4 -3 -2 -1 0 1 2 3 4 5 8 IN FEET Figure 2.4. Static Force-displacement Relationship for Four-way Arrangement of Guys, 9 = 0.

-1770 - - - -- T| — ----- - 60 50.... 40....................-... /.~ -,C 30- w 10 I... IN F 0 0 L. 0,. z 0 N -20 034 -30.. — -40 -60 -70............. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 8 IN FEET Figure 2.5. Static Force-displacement Relationship for Four-way Arrangement of Guys, 9 = 45~

-18In a three-way symmetrical guying for a displacement in a plane of symmetry (Figure 2.6). 3 // Figure 2.6. Three-way /U/ _ Arrangement of Guys. 51 = 5 (;2.24) 52 = 56 = - 5 cos 60 = - (2.25) 12 13 2 R = H1 - (AH2 + LAH) cos 600 = = H1 - AH2 (2.26) where AH1 is given in Figure 2.2 at 51l = 6, and AH2 is found from the same figure at &2 = -. The result is illustrated in Figure 2.7. ). A Linear Approximation for the Force-Displacement Relation From Figures 2.4, 2.5 and 2.7 it is seen that the relation R vs 5 may be approximated, for small displacements, by the tangent to the curve at the point R = 0, 5 = 0. The slope of the tangent can be found as follows: = H sinh r (2.8c) nLu = BH [1 + sinh (2r) + 2 r ta- s n (2.10b) Lu L 2AE- l (21 tanha) Lu = L - ALu (2.11a)

-19 - 70 -.. 40 30 - ^ 20 — o 0 -10 - 0 420 -50 -60 -70 -5 -4 -3 -2 -1 0 1 2 3 4 5 8 IN FEET Figure 2.7. Static Force-displacement Relationship for Three-way Arrangement of Guys.

-20when h and w are assumed constants, Equation(2.8c) is differentiated to give LdL 2 E = ( sinh r - H cosh r) dH + cosh r di or _____ 2 i L2h2 dL (- sinh r - - cosh r) dH + cosh r d L w H L (2.27) From Equations (2.10b) and (2.11a) one also obtains dLU = 0 = dL - d(ALu) or dL = d(Au) (2,28) but a(mLu) + (aLu) d(ALU) = dH + l d (2.29) where (ALu) Q [1 + inh (2r) + 2 r ] aH 2AE 2r E2 tanh r (2.30) H [4r cosh (2r)-2 sinh (2r) h2 tanh r -r/cosl rr 2E E (2r)2 + 2 2 t anh2r ] H Substituting 6r _ r (2.31) dH H and simplifying 6(ALLu) __sinh ( 2r ) h2 r2 ](2~32) ( =Lu) [1 + sinh (2r) - cosh (2r) + 2 h2 si r 6H 2AE r I sinh r Also, (AL_) H sinh(2r) h2 r a7 2AE + 2r + 2 tanh r 2 Hi [4r cosh (2r)-2 sinh (2r) h2 tanh r r/cosh r +2AE r[ + 2 tanh2 r 17 + LHL 4 h r ( +2AE [ - 3 tanh r.)

-21ozr r with - Equation (2.33) can be simplified to give H h2 r2 = HE[1 + cosh (2r) - 2 -i2 - (2.34) di 2AE 7 sinh2 r Substituting Equations (2.32) and (2.34) into Equation (2.29) dL = [ + sinh (2r) cosh (2r) + 2 - 2 dH HA_ _ r - h2 sinh r 2 r2 + H [1 + cosh (2r) - 2 ] di (2.35) 2AE ~2 sinh2 r Eliminating dL between Equations (2.27) and (2.35) L2t- (2 sinh r - cosh r) dH + L cosh r d L w H L -[l sinh r - cosh (2r) + 2 -2 i2 dH 2AE r 2 sinh2 r H h2 0 r2 + - [1 + cosh (2r) - 2 h sn2 r di (2.36) 2AE 2 sinh2 r or H (cosh h2 r2 -L2_ h2 dH AE 7' sinh2 r L 2 sinh (2r) h 2 L2-h2 sinh r AE (sinh2 r - 2r ~2sin (cosh r r ) AEL 2r 12 sinh2 r L H (2.37) Equation (2.37) will be evaluated when Q = 950 ft. h = 1349 ft. w = 9 lb/ft H = 35,300 lb. A = 2.42 in.2 E = 24 x 106 lbs/in2

-22for which r =.121105 sinh r =.121401, cosh r = 1.007343, tanh r =.120516 sinh r r sinh (2r) = 1 00980 r = 1.002444, sinh.997562 2r = 1.00980 AE = 58.08 x 103 kips., =.607782 x 103 L = 1651.2787 ft., h2 = 2.0164 ~2 _L__-_ = 71 - ih/LA =.5767 L dH.607782 x 10 3(1.01474-200658) -.580933 di 16.356749 x 10-3(.01474-1.00980-2.00658) - 15.52079(1.007343-1.002444) -.6028 x 10-3 -.580933 = -581536 4.647 kips/ft - 49.097 x 10-3 -.076036 -. 125133 This value is the slope of the tangent at the origin in Figure 2.2. It will be of interest to comparethis value with the so-called "taut wire" modulus k AE cos (28) k _- (2.38) in which y is the inclination and L is the length of the cable chord. For the guy under consideration Equation (2.38) gives k = 11.68 kips/ft The slope of the tangent to the R curve at the origin can be found thus: a) Figure 2.4 - Since R = H + H3 (2.18) then dR dH1 dH3 d6 d6 d5

-23but dH1 dH3 - = -- at 5 = 0 d5 d5 so that dR = 2 dH = 2(4.647) = 9.294 kips/ft d6 d8 b) Figure 2.5 - R = 1.414 (H1+H3) dH1 dH3 dR = 1.414 ( + - ) d6 d6 ds but, dH1 dH3 1 dH -- - at 6 =0 db db 1.414 di dR - 2 dH = 2 x 4.647 = 9.294 kips/ft d5 d5 c) Figure 2.7 - R = H1 + H2 dR dH1 dH2 d5 d5 d5 but, dH1 dH dH2 1 dH d5 dd' d5 2 di dR 3 dH= 3 (4.647) = 6.970 kips/ft d 2 d 2 The tangents are drawn by a dotted line in the figures.

CHAPTER III DERIVATION OF THE DIFFERENTIAL EQUATIONS OF MOTION FOR A CABLE 1. Introduction When aguyed-tower is vibrating, the upper end of a guy cable is subjected to a disturbance which induces a motion in the cableo The vibration of a guy is, therefore, a dynamic problem with time dependent boundary conditions. Cables used in such structures as suspension bridges and guyedtowers are commonly assumed flexible. With this assumption, the dynamic problem is reduced to that of a string. A flexible string is characterized by its inability to resist bending so that only tensile forces act on any section of the string at any time. The vibration of a guy cable is exceedingly complicated because there exist three components of motiono In a rigorous study of the dynamic problem of a guyed-tower, it will be necessary to take into account the effects of the inertia forces along a guy on the resistance which the guy offers to the tower motiono Since such a resistance is developed by all three components of guy vibration, it must be determined by a simultaneous solution of the differential equations of motion. In the following sections, the equations of motion for a cable will be given; first, for the most general case in which the cable, when in static equilibrium, may assume any shape in space and secondly, for a cable for which the curve of static equilibrium lies entirely in a vertical plane. Finally, the nature of boundary conditions in a guy cable will be stated. -24

-252. The General Equations of Motion of a Cable The equations of motion of a string under the action of any forces were first stated by Routh.(2) Although Routh's equations of motion have not, in general, been solved mathematically, they will be given here to demonstrate the complex nature of the motion of cables. Moreover, the presentation of these equations will show the need in selecting suitable variables in order to make the equations of motion somewhat tractable. Consider, in general, any two point P and Q on a cable (Figure 3.1). Let s and a(s,t) be, respectively, the distance along the arc of the cable from P to Q when the cable is in static equilibrium and in motion. Let x(s,t), y(s,t), and z(s,t) be the rectangular coordinates of point Q relative to some orthogonal axes fixed in space. Finally, let Te(s) denote the equilibrium tension in the cable and let it become T(s,t) = Te(s) + AT(s,t) at time t. Then, by D'Alembert's principle, the equations of motion are: s2 X(S, t) d2x(s,t) dsx ds 5s m(s) = +t X(s,t) yt(st) ds rs t2 (1 m(s) y(st) [T(s,t) d ] + Y(s,t) (5.1) os ~st)l 5s 6s -

-26T(s,t) / / z(s,t) /, /'x(ot) at P y(o,t) T(o,t) z (o,t) Figure 3.1. Parameters of a Cable in Motion.

-27where m(s) is the mass per unit length along the cable and X(s,t), Y(s,t), and Z(s,t) are the components of the external forces per unit length. Equations (3.1) involve five dependent variables: x, y, z T, and a in terms of the independent variables s and t. Two other equations relating these variables must, therefore, be given to complete the equations of motion. An equation which relates elastic forces to displacements is written by Hooke's Law. If A(s) be the area, and E the modulus of elasticity of the cable, then i(s, t) ds s 1 + AT(s,t) (.2a) ds AE or, ist) - 1+ T(st)-Te(S) (.2b) 5s AE Moreover, from the geometric constraints, one obtains ~x 2 Ly 2 )z 2 2 ( ads) + ( ds)2+ ( ds)2 ( ds)2 (.3a) or (x)2 + ( + ()2 () (.53b) Equations (3.1) in conjunction with Equations (3.2b) and (353b) will mathematically suffice for a determination of x, y, z, T, and a in terms of s and t. But, in general, no analytical solution is available because of coupling and non-linear nature of the equations. If the cable be assumed inextensible, i.e., if the elastic deformations be ignored, a cable retains its length at all times during vibration so that o(s,t) = s and da = ds. The equations governing

-28the motion of the system are then reduced to four equations as follows: m = a aT -) + X m a2y = a (T Y) + Y (4) M (T (3.4) ot2 os os m 2- = - (T -) + Z at2 as as (ax2 + y 2 z 2 (5) as + s = s Routh's equations of motion, as given previously, were written in the most general case. In practical applications a cable hangs in equilibrium under the action of gravity forces. Thus, all points of the cable lie in a vertical plane. Moreover, the physical parameters of the cable which appear in the equations of motion, ie., mass and area are uniform throughout. Even so, an analytical solution to Routh's equations is not feasible. The problem must, therefore, be formulated in such a manner as to result in a simplification in the equations of motion. 5. Modified Equations of Motion for a Guy Cable The motion of a guy cable may bedefined by its longitudinal and transverse vibrations. By longitudinal vibration is meant the motion in the direction of the curve of static equilibrium, while displacements perpendicular to this curve are ref:erred to as transverse vibration. In general, the latter consists of two components, namely, in-plane transverse vibrations and out-of-plane lateral oscillations,

-29Any vibration of the cable may be resolved into in-plane and out-of-plane oscillations. The former vibration which consists, in general, of two components -longitudinal and in-plane transverse oscillations - is considered more significant in the study of interaction of a guy and the tower; while, the latter vibration i.e., the out-of-plane vibration, is of secondary importance. For this reason, and also for simplicity, the two types of vibrations will be studied separately. a. In-plane Vibrations When the motion of a cable is restricted to its own plane, the problem may be considered as a generalization of the classical vibrating string. This generalized string problem is complicated because of the following factors: (a) Longitudinal as well as transverse motions of the cable must be taken into account. (b) The tensile forces along the cable must be considered a function of space coordinates as well as time. To simplify the problem, the following assumptions are made: (1) All physical properties of the cable are uniform throughout. (2) Displacements from the equilibrium configuration are small compared to the length of the cable. Let the displacement of a point of the cable originally at a distance s be designated by its components ~ and r measured respectively, in the tangential and normal directions relative to the static configuration of the cable (Figure 3.2a). The forces acting at the end points of an element of the cable are shown in Figure 3.2b,

-30~\.,~q(s,t) + ds B A \ A\7 t(s,t) (a) dTe T (s) + d ds Te(s) + -. ds + AT(s,t) + ds B/ ds s / / A/, T s.~J ~ (b) Te(s) + AT(s,t) Figure 3.2. Element of a Cable in Static Equilibrium and in Motion. T( St) + a ds (s) s [T (S)+T(s,t)] tan cp \ /\T, I( a- +L{[T,(s)+T(S,t)]tan (j Te(S) + d /T(s,t) + ds R(s) T(st) S Te( s)+T(s, t)]tan (p Figure 3.3. Projection of Tension in Vibrating Cable on Static Equilibrium Configuration.

-31Similar to the formulation of the classical vibrating string the tensile force at any point of the cable in vibration may be resolved into two components; one in the direction of static tensile force and the other in a direction perpendicular to it. If the motion of the cable was merely a transverse vibration, i.e., a motion in which S(s,t) = 0 for all points, the tangential component of the force T(s,t) would be equal to the static force. This is commonly assumed in the classical problem of vibrating strings. Such an assumption is not valid, however, for a guy cable the upper end of which does not vibrate in a perpendicular direction. The problem will, therefore, be studied when both components of the motion - - and T - are present. Let the change, from the static force, of the tangential component of the force T(.s,t) be denoted by T(s,t) as shown in Figure 3.3 on which cp(s,t) is the angle between the tangents to the stationary and vibrating cable or, in other words, p(s,t) is the rotation of the tensile force acting on a section of the cable during vibration relative to the direction of the static force at the same section. In Figure 3.3 is also shown the normal component of the force T(s,t) expressed in terms of the tangential component Te(s) + T(s,t), and the angle cp(s,t). If one assumes that the force T(s,t) is only due to longitudinal vibrations, the equation of motion in this direction will be 2 ae(T +T) m t = + Ft (5.6) 6t2 6s where Ft(s,t) is the tangential component of the external forces per unit length along the cable.

-32From geometry the angle cp is given by: as tan cp = (37a) 1 t- 21 as Assuming small vibrations, the quantity a6/6s is small compared to 1 so that one may write: tan cp L (3.7b) Os which asserts that the change in the inclination of the cable due to longitudinal vibrations is of secondary order compared to the effect of transverse oscillations. Referring to Figure 3.3 and using Equation (3.7b) the equation of motion in the transverse direction will be: 2 6 6 T ~ Te+T Mt2 a= (Te 6s) + - a (T s - + Fn (.8) in which R(s) is the radius of curvature of the cable in static equilibrium and Fn is the normal component of the external forces per unit length. Let wt and wn be, respectively, the tangential and normal components of the gravity forces. Then from statics dT + wt = (3.9a) ds - and e + wn = 0 (.9b) R Substituting Equations (3.9a) and (3.9b) into Equations (3.6) and (3.8), one obtains

-33m = -+ ft (3.10) at2 as and m t- e = - a (Te ) + ( ) fn (3.11) which are the equations of motion from the static equilibrium. In these equations ft and fn are, respectively, the components of the applied forces per unit length along the cable, Equations (3.10) and (3.11) involve three dependent variables:, rn, and T. An equation which relates elastic forces to displacements must, therefore, be found in terms of these variables. The length of an element of the cable in vibration is da = (ds + It ds)2 + (31 ds)2 (5.12) 6s 6s from which it may be deduced that as = V (l+ ^ at 22 (3.135a) By expanding the right hand side of Equation (3.13a) in a binomial series, one obtains as 2 as as as - (- (3.6_b) -1 2[(4 +2 ( )2] + (313b) Neglecting terms of order three and higher s 1+ + (3.13c) a, a, 2 6s

-34But, a- = 1+ E (3 14) as where c is the strain in the cable due to vibration. From Equations (3.13c) and (3.14) one gets JI1 6 2 ~s 2 )2 The term (~s ) in this equation is the well-known strain quantity 2 5s for a straight bar the ends of which are merely displaced in a perpendicular direction. The term X, also, is simply the strain due to a displacement ds in the direction of a straight bar. Thus, according to Equation (3.15), if a bar is stretched in the longitudinal direction and at the same time is given a rotation (Figure 3.4) the two displacements may be treated separately to give the strain in the bar. ads ~~~ds C ~ ~ W I Figure 3.4. Displacement of an Element of a Straight Bar. Actually, when both displacements are present the strain due to the displacement i is s = (3.16) cos cp In order to find an equation which relates deformations to the force T appearing in the equations of motion, it was found easier to use Equation (3.16) rather than Equation (3515). Assuming Hooke's Law applies during

-355vibration of the cable, the change in the tensile force due to the displacement ~ will be (AT) = A^E = AE (3.17) cos Cp Recalling that for solely transverse vibrations the projection, on the direction of the static forces, of the tensile force during vibration is equal to the static force, one may conclude that only the change (AT) is responsible for the force T. Thus, T = (AT) cos c = AE (3.18) Substituting for T from Equation (3.18) into Equation (3.10) one gets m a = AE + ft (3.19) at2 8s2 t which is identical to the equation of motion for longitudinal vibration of a straight bar. According to Equation (3.19) longitudinal vibration of a cable is independent of the transverse vibration. In reality, this may not be true as shown by the discussion which follows. If a small element S of a cable is stretched in the tangential direction by an amount AS (Figure 3.5) the strain in the AS element will surely be smaller than the quantity because of the change in curvature of the element. Equation (3.16), however, was obtained by assuming that an element of infinitesimal length may be represented by a straight line. Such an assumption will give correct results for force-strain relation only if the curvature remains nearly constant.

-36D, Stretched Position / C Figure 3.5. Stretched Position of an Element of a Cable. This is demonstrated by Figure 3.6 in which a cable is assumed embedded in a pipe whose diameter is slightly larger than the cable diameter such that a transverse motion is somewhat restricted. For such a system Equation (5316) is valid so that Equation (35.19) represents the vibration of the cable along the pipe. /r- Direction of motion Figure 3.6. Longitudinal Motion of a Cable Restrained in Transverse Direction. For a cable hung in space, since any longitudinal vibration causes a change in curvature, Equation (3516), therefore, does not represent the true force strain relation. The correct relation between variables, T, and T when a change in curvature is also considered is complicated. Attempts were made to arrive at a simple relation between the tensile force and two components of motion by re-defining the vibration of the cable in terms of its displacements along the chord and

-37perpendicular to it. Such fixed coordinates proved more suitable for the force displacement relation, but resulted in a complicated differential equation for the longitudinal vibration (defined as the component of motion on the chord). It became evident that when both components of motion are considered the complexity of the motion is present, one way or the other, in one or more of the equations governing the motion of the cable depending on the choice of coordinate system. This may explain the reason for approximation of the problem as essentially a transverse motion by earlier investigators.(5 8) For a guy cable the curvature along the stationary cable varies slightly. It may,therefore, be assumed that a force applied on an element of the cable anywhere along the length, would result in a nearly constant strain. The force - strain relation may then be approximated by the equation T = a AE (3.20o) Is in which a is a constant smaller than unity whose value depends on the curvature of the guy cable. The value of a may be determined from statics as follows. Referring to Figure 3.7 if a guy is displaced in the direction of the tangent at the upper end by an amount d and if the change (AT)sta in the tensile force which causes this displacement is found, then one may write d oAE - = (T)sta. (3.21a) L sta. which gives L. (T) sta. a (=T-ta(3.2lb) AE d

-38/ (AT sta. / Figure 307. Caple Displaced in Longitudinal Direction. (AT) sta The ratio --- may exhibit a non-linear behavior. For small displacements, however, it may be approximated with sufficient accuracy by a linear relation as follows: Suppose the upper end of a guy is displaced in a horizontal direction by an amount Al (Figure 3o8). This displacement can be thought of as a superposition of two displacements A1 and A2 as shown in Figure 3.8. The effect of the latter displacement on the change in the A I /Az // Figure 3.8. Components of Top-displacement of a Guy. tensile force is negligible compared to that of the displacement A1 so that one may write

-39AT -= A (3.22a) cos 7 and A1 = A - cos 7 (3.22b) from which AT _ A 1 (3.22c) A1 Al cos2 ) Substituting Equation (3.22c) into Equation (3.21b) gives AH A~ E = 2- (3.22d) I cos 7 L which is the ratio of the guy modulus to that of the taut cable. The non-linearity of a is, therefore, of the same nature as that of the AH quantity A. Referring to Chapter II a linearized approximation for will be 4.647 398 11.68 =.98 Having determined the value of a, the equation approximating the motion of the guy in the longitudinal direction will become 2 2 m 2 c=AE - + ft (3.23) ~at2 as Referring to the equation of motion for the transverse vibration, Equation (3.11), the quantity - (T s) represents a non-linear os ds term. On account of the factor AE in Equation (3.20), thisnon-linear term may or may not be small depending on the geometry of the guy and the nature of the boundary motions. To make the equations analytically tractable it is assumed that the motion is small such, that; << Te. The

-40term - (T a) may now be neglected compared to - (T ) which also ds a s. (Te ss appears in Equation (3511). With this linearization, the equation of transverse vibration after substituting for T from Equation (3.20) into Equation (3.11), will become: m 2 = a (Te L) AE + f (3.24) at2 as as R 5s n The solution to Equations (3.23) and (3524) will be given in the following chapters. b. Out-of-plane Vibrations The vibration of a guy cable may involve a motion perpendicular to the plane of static equilibrium of the guy. If the out-of-plane motion of a point of the guy be denoted by r(s,t) the equation of motion will be m = a (T ) + f (3.25) at2 6s 6s where fh is the out-of-plane component of the applied force per unit length of the cable and, T is the projection, on the plane of static equilibrium, of the tensile force during vibration of the guy, i.e., T is the tensile force resulting from the in-plane vibration of the guy. If a guy experiences merely an out-of-plane vibration, the projection T would be equal to Te, i.e., the static tensile force. In the general case in which the cable oscillates in the plane of equilibrium as well as in a direction perpendicular to this plane, referring to Figure 3.3 one has Te + T T = e (3.26) cos cp

-41Strictly speaking, since T depends on both components of in-plane-motion, the motion ~ will be affected by - and. For small vibrations, one may assume cos cp = 1. With this approximation Equation (3.25), when T is replaced from Equation (3.26), becomes m = — (Te+ -l-( at2 a(T, ( e h: ) + f (3.27) ^t2 ds ds ds ds h If Equation (3.27) is linearized by neglecting: compared to Te it will be reduced to mI = a (Te s) + fh (3.28) dt2 as e s which is the equation of motion for a mere out-of-plane oscillation. The coupling between the out-of-plane and in-plane vibration is, therefore, of a non-linear nature. The linearized equation of out-of-plane motion, Equation (3.28), is identical to the linearized equation of in-plane transverse vibration, Equation (3.24), except for a coupling term which appears in the latter equation. The solution to the out-of-plane vibration may, therefore, be easily deduced from the solution for the in-plane transverse vibration. For this reason, only the in-plane vibration will be studied in the following chapters. 4. Motion of a Guy in the Free Vibration of a Guyed-tower The first task in attacking the dynamic problem of a guyedtower is a study of the free vibration. In the absence of any applied force along the cable the equations of motion will become

-422 2 m t = oAE a (3.29) 2 aAE 6 at2 as2 m - - (Te ) - X (3.30) t2 as s R as m (T - a (3 31) One must now turn to boundary conditions in order to obtain information concerning the free vibration. When a guyed-tower is vibrating freely, the upper end of a supporting guy is subjected to a force whose change with time is responsible for the vibration of the guyo The motion of a guy cable, as part of the free vibration of the guyed-tower structure, may therefore be considered a forced vibration imposed by a disturbance at the point ot attachment to the tower. The tower, in general, experiences a horizontal and a vertical motion. However, the latter which is caused by the change in the vertical forces exerted by the guys and by the.bending deformations of the tower, is small compared to horizontal displacements. Thus, only the horizontal vibration of the tower is assumed of importance in the vibration of the guys. A natural mode of vibration for a guyed-tower may consist of horizontal displacements in two directions and a rotation around the centroidal axis. Therefore, the horizontal motion of the point of attachment of a guy to the tower will not, in general, take place along a straight line. The rotational component is absent only for the modes in a plane of dynamic symmetry. For guyed-towers, the rotational component, if present in a modal vibration, is small however and may be

-43ignored so that the motion of the upper point of a guy may be taken as the motion of the centroid of the tower at the level of the guy attachment. For vibration of a guyed-tower in a natural mode, the boundary condition at the upper end of a guy will be of a periodic nature. If the periodic boundary condition be represented by a harmonic series in a Fourier analysis the solution to the problem at hand may then be obtained by a superposition of the solutions to each term of the harmonic series. The problem will be reduced therefore to that of the vibration of a guy subjected at one end to a harmonic disturbance. There are two distinct problems to be considered; the first, when it is assumed that the end disturbance is a given harmonic force, the second, when some assigned harmonic motion of the end point is assumed. These two boundary conditions lead to different resonance frequencies as shown in the following chapters. The force-displacement relation, however, will remain the same in both cases as expected.

CHAPTER IV SOLUTION TO THE LONGITUDINAL EQJATION OF MOTION FOR A PERIODIC END DISTURBANCE 1. Introduction In this chapter the solution to the equation 2 2 2 a2~ a2 2 (4.1) at' as where, a2 a AE (4 la) m will be given for the following boundary conditions: At the immovable lower support, one has t(o,t) = 0 (4.2) At the upper point two conditions will be considered as follows: Case (a) - An applied harmonic force in the form Ft sin pt for which the boundary condition is c~AE ~(L~t).AE Ls = Ft sin pt (4.5) Case (b) - An imposed harmonic displacement S(L,t) = At sin pt (4.4) 2. Solution for Harmonic Force at one End To solve the differential Equation (4.1) subjected to boundary conditions (4.2) and (4.3) one may resort to the normal mode technique. The differential equation with homogeneous boundary conditions may be written as -44

-452 2 82c = a2 c (4.5) 6t2 a2 c(o, t) = 0 ( 0 (4.6) 6tc(Lt) 0 bs A separation of variables in the form Sc(s,t) = X(s) Q(t) (4.7) will lead to the well-known solutions =sin 2i-1 [C sin 2i-1 rat + 2 Sati] (4.8) c[T T Ci sin T + Di CS T in which 2i-l..s Xi = sin 2 L i = 1,2,... (4.9) are the normal modes of vibration, and i i = 1,2,... (4.10) Wi = 2 L are the corresponding natural frequencies. The solution to the original problem involving the end disturbance (4.3) may now be obtained by considering the solution to the equation 2 2 a2 + 1 f(s,t) (4.11) 6t2 6s2 m in which f(s,t) = 0 for 0 < s < L (4.12a) and f(L,t) = F sin pt (4.12b) t

-46The solution to Equation (4.11) is found by a superposition of displacements corresponding to normal modes of vibration in the form 00 I(s,t) = ~E Xi(s) qi(t) (4.13) where Xi is given by Equation (4.9). Taking derivative of Equation (4153) 2 Z xi % as2 i i2 11 E X qi(4.14) 2 - Z Xi. q at and substituting into Equation (4.11) one gets Xiqi - a Xiqi = f(s,t) (4.15) 1 1 or XiXqi - a2 ZXiXj =X mX f(s,t) (4.16) Integrating Equation (4.16) from 0 to L and interchanging the order of integration and summation I i 1 i o a Using the orthogonality relations L f XiXjds = 0 for i j (4.18) L f X Xods = 0, for i j 0 Equation (4,17) is reduced to L 6 2 Xf(st)ds q. + 1 qi 1 (4.19) m f X2ds o 1

-47in which cD. is given by Equation (4.10). The function f(s,t) expressed by Equations (4.12) may now be written as f(s,t) = (Ft sin pt) $ ((420) where b is the Kroneker Delta. Then L L 2i-1 cs s f Xi f(s,t) ds = Ft sin pt f sin 2 - b (L) ds o uo L L = - (-1)i Ft sin pt (4,21) Moreover, since L L f X. ds f sin2 2 - ds = (4.22) o 2 L 2 Equation (4.19) is reduced to ~0 2 2 i i + Oq = - (-1) Ft sin pt (4.23) for which the solution is q =A. cos aoit + Bi sin cit - - (- F sin pt (4.24) i mML m2 2 t sin pt Substituting Equation (4.24) into Equation (4.15) gives t(st) = S sin 2i-1 [Ai cos cit + B. sin cit - Ft sin Pt] i=l,2,... 2 L i 2 2 (4.25) The constants Ai and Bi are found from the initial conditions. For instance, when the vibration starts from rest, one has S(s,o) = 0 (4.26) M (SO)= O

-48which result in the values A. = 0 Bi =2 (-1)i Ft mL o-jj(o-p_)p2 Equation (4.25) will then become ( ) 2Ft sin 2i-1 rs _-i sin pt + p sin at( i (st) mL 1-_,2,.. 2 ( -p2) (4.28) a. Steady State Solution From Equation (4.28) the steady state solution is (s,t) = 2 2 Ft sin pt 2 sin ( i1 s - (4.29) PmL 1 -P A simpler expression for the steady state solution may be found as follows: Considering the Equations (4.1), (4.2), and (43,) one may solve the problem as the sum of two solutions in the form S(s,t) (t) = c((St) (4.30) where c is the solution to the set of homogeneous Equations (4.5) and Sp is a particular solution which satisfies the differential equation and actual boundary conditions. The solution Sc has already been given in Equation (4.8). The particular solution must obviously be sinusoidal with respect to time with a frequency p. One may therefore write Sp(s,t) = i(s) sin pt (4.51)

-49which upon substitution into Equations (4.1), (4.2), and (43.) gives 2- + a2T, = (0O) = o (4.32).- Ft T (L) = t aAE for which the solution is readily obtained as F sin - s t(s) c= c o —— a (4.33) CAE P cos p L a a Thus, Fjt sin p s g (st) - a sin pt (4.34) a p The solution ~c is acted upon by damping and will soon decay. The solution p, however, indicates motion of a continuing nature. Thus, P Equation (4.34) represents the steady state solution. The two solutions given for steady state vibration by Equations (4.29) and (4.34) are identical, except for the value As (L,t) as can be a Shown by a Fourier series expansion of the function sin j^ e The resonance frequencies and also the force-displacement relation at the upper end are discussed later after a solution is found for vibration in the second case of interest, i.e., the motion due to an imposed harmonic displacement at the end. 3. Solution for an Imposed Harmonic End Motion The equation of motion and boundary conditions which were given previously are repeated here for convenience

-502 a2 2 (4.1) at2 2s ~(ot) = 0 (4.2) t(L,t) = At sin pt (4.4) In this case, contrary to the problem for harmonic force considered in previous sections, the nature of the boundary condition (4.4) is such that a simple separation of variables will not allow a solution to be found. One can, however, reduce the problem to one with homogeneous (21) boundary conditions by the Mindlin-Goodman method and use the separation of variables technique as follows: Let |(s,t) = u(s,t) + g(s) At sin pt (4.35) where u(s,t) and g(s) are functions yet undetermined. Taking second derivatives of Equation (4.35) 2 2 2 - 2 + g" At sin pt (4.36) 2 2 a t e a - gp2 4 sin pt and substituting into Equation (4.1), one obtains 2 2 u 2 a u 2 it 2 a -_ a2 v7 = (a2g + p2g) At sin pt (4.37) The boundary conditions (4.2) and (4.4) will also become u(ot) = - g(o) A sin pt (4.38) u(L,t) = [1 - g(L)]At sin pt

-51One must choose the function g(s) such that the right hand sides of Equation (4.38) vanish. This requires that g(o) = 0 (4.59) g(L) = 1 The simplest function which satisfies (4.39) is g(s) = L (4.40) With this choice for g(s), Equation (4.37) becomes 2 ~2 2 u 22 2 u 2s t2 - a2 - = p2 t sin pt (4.4la) subjected to the boundary conditions u(Ot) = 0 (4.41b) u(L,t) = 0 (4.41c) The set of Equations (4.41a), (4.41b), and (4.41c) may now be solved by the separation of variables technique as follows: Considering the homogeneous equations u 2 a = o2uc 2 2 u c __- a c 0 at2 at2 u (o,t) = 0 (4.42) uc(L,t) = 0 let uc(st) = X(s) Q(t) (4.45)

-52which upon differentiating twice with respect to s and t and substituting in Equations (4.42) leads to the normal modes Xi = sin - i = 122,... (4.44) 1 L and the natural frequencies iiL X i= 0i = 1,2,... (4.45) The solution to Equation (4.41) may now be written in the form u(s,t) = ZXi(s) qi(t) (4.46) i which, following a derivation for Equations (4.14) through (4.19), gives rise to the equation L o of X. sds i + ai qi = P o sin pt (4.47) L 2 ds 0 But, L 2 L 2 ijrs L f xi ds = f sin2 ds = (4.48) and L s i= L ds (4.49) f X sds = s sin - ds = (4.49) o i o L ixt so that Equation (4.47) becomes 2 i 2p2 qf + c = (1 sin t (4 50) for which the solution is (-1) i P2 qi = A. cos a.t + B. sin C.t - 2 (-1)i pt i X 1 i i ip 2 1 sin pt(4.51) 1 (4.551)

-53The complete solution to the set of Equations (4.41a), (4.41b), and (4.41c) is therefore iTs u(s,t) = Z sin L [Ai cos wit + Bi sin oit i=l 2~.... 2 i=l,2,.. - 2 -I — - 2 sin pt] (4.52) iXr (i-P The solution to Equations (4.1), (4.2), (4.4) is now readily obtained by Equation (4.35) in which u(s,t) is given by Equation (4.52) and g(s) by Equation (4.40). a. Steady State Solution The steady state solution for an imposed sinusoidal motion at s = L is obtained by retaining terms involving the frequency p in the Equation (4.52) and substituting the result into Equation (4.35) in which the function g(s) is also substituted from Equation (4.40). The result is ijrs s 2p. sin (st) = 2 —- (-1) ] At sin pt (4.53) p L (2 il) 2 i i(-p ) The steady state solution for an imposed harmonic displacement at one end may also be found as the particular solution to the Equations (4.1), (4.2), and (4.4). One may assume a solution in the form ~p(s,t) = T(s) sin pt (4.54) which upon substitution into differential Equation (4.1) using boundary conditions (4.2), and (4.4) leads to sin P s 5(S) = Ats (4.55) sin ~. L a

-54thus, the steady state solution is sin P s Sp(s,t) = At - sin pt (4.56) J- tsin k L a The solutions (4.53) and (4.56) are identical in every respect as s sin PS/a can be shown by a Fourier series expansion of the function - sin PL/a 4. Force-displacement Relation for a Periodic End-disturbance Re-writing the two solutions for the steady state motion Ft sin P s (S (st) t a sin pt (4.34) p aAE _ cos P L a a and sin s S (s,t) = At sin pt (4.56) P' sin p L a one may find the end force-displacement relation as follows: (a) - From Equation (4.34) the end displacement due to the applied harmonic force Ft sin pt is (L,t) = A in t sin pt (4.57) aAE p a (b) - When the motion is described by Equation (4.56) in terms of the end disturbance At sin pt the force necessary to cause this motion is obtained from Equation (4.56) thus J^p p a Ft sin pt = AE s = oAE At L sin pt (4.58) t s tans=L a From Equations (4.57) and (4.58) it is seen that both cases lead to the same relation between force and displacement such that

-55A4t L _ t an (4.59) -L — tan L aAE a which in the limit, when p approaches zero, reduces to the static relation At Ft (A F (-(4.60) L sta. QAE sta. Equation (4.59) may also be obtained by equating Equations (4.34) and (4.56). Equation (4.59) relates the amplitudes Ft and At with the frequency p. If the system is subjected to a harmonic force with conFt stant amplitude Ft, the quantity -- in Equation (4.59) may be ret oAE placed by (t) according to Equation (4.60) so that Equation (4.59) L sta. gives the dynamic magnification factor (t)dyn. tan E L (tdy. (4.61) sta. a whose variation is illustrated in Figure 4.1. On the other hand, if the system is excited with different frequencies in such a manner that the amplitude of the end displacement At remains constant the end force will depend on the frequency p At F Replacing in Equation (4.59) the constant quantity - by ( ) L cAE sta. from Equation (4.60), one gets (Ft) L — ~~t dy~ = a L-~ ~(4.62) (Ft) tan P L sta. a This function is demonstrated in Figure 4.2.

-560t- ifK 5- PL o 2.0 o 2..0 2 — 1 -, —-5' Figure 4.1 Dynamic Magnification Factor for Sinusoidal Forcing Function. c i.O C 1.0 fir 1.0- ----- PL I-0 I 2 v I I' " \ -..,, -4.o -.I..... IlL I___ -2.0r Figure 4.2. Dynamic Magnification Factor for Sinusoidal End Motion.

-575. Resonance Frequencies From Figures 4.1 and 4.2 it is seen that there are two sets of frequencies which merit consideration. The first set consists of the frequencies p. 2i-1 a i = 1~2.. (4.63) i 2 L for which: (a) - If the end movement is not restricted the motion tends to infinity due to any finite force (Figure 4.1); (b) - if the amplitude of the end movement is restricted the force required would be extremely small (Figure 4.2). The frequencies given in Equation (4.63) are the same as those shown by Equation (4.10) for the natural frequencies of longitudinal vibration when one end of the cable is fixed and the other end is free as described by the system of Equations(4.5). The second set of frequencies in Figures 4.1 and 4,2 are. -Ta P. = 1 i = 1,2. (4.64) for which: (a) - The end motion tends to zero for any finite force (Figure 4.1); and (b) - The force required to cause a finite harmonic end motion would be exceedingly large (Figure 4.2). The frequencies given by Equation (4.64) are the natural frequencies of longitudinal vibration when both ends of the cable are fixed, as previously found in Equation (4.46). From the above discussion and also from the Equations (4.34) and (4.56) it is seen that the frequencies given by Equation (4.65) are

-58the resonance frequencies of longitudinal vibration only if the end displacement is unrestricted. For an actual problem in which the cable is attached to a tower the motion of the end point is somewhat restricted by the tower so that resonance does not actually occur at these frequencies. These frequencies are, however, of importance since the force required to impose a longitudinal motion in one of these frequencies would be nearly zero. That is, the cable if vibrated at these frequencies would exert no restraining force on the tower during vibration except the static force. The resonance frequencies of the longitudinal motion of cables in the guyed-tower structure are, in fact, the frequencies given by Equation (4,64). As seen from Equation (4.56) there exists a resonance condition in spite of the fact that the end motion is limited by the tower. As discussed previously, such a motion would require an extremely large force even for very small end displacement. This means, theoretically, that if the structure could vibrate with a frequency which coincides with one of the longitudinal resonance frequencies of a particular guy, the attachment point will nearly be a nodal point on the vibration of the tower. But in reality such a condition cannot be materialized because the guy will fail due to the large force which must be developed. 6. Some Further Remarks In this chapter the longitudinal vibration was investigated as if it were a separate problem by itself similar to longitudinal vibration of bars. Actually, any longitudinal vibration of a guy cable is accompanied by a transverse in-plane vibration because of the curvature

-59 - of the guy. The change in the tensile force due to longitudinal vibration will act as an exciting force on the transverse vibration as shown previously by Equation (3.24). Although the lateral resistance the guys offer to tower movement is essentially due to longitudinal vibration, the transverse vibration must also be studied for its effects on force-displacement relation and, in particular, for resonance conditions. It will be shown in Chapter V that the natural frequencies of transverse vibration are proportional to vTe where Te is some average value of static tensile force. For longitudinal vibrations however, the proportionality factor is zAE. The ratio -A ranges approximately e from 20 to40 for cables which are functioning as guys so that the resonance will occur first in the transverse vibration because in any actual problem, the tower movement would be slow compared to the fundamental frequency of longitudinal vibration. Based on the work in this chapter, the transverse vibration will be studied in Chapter V only for an imposed end motion since such a boundary condition will suffice to give the force-displacement relation as well as the resonance frequencies for guy cables,

CHAPTER V SOLUTION TO THE TRANSVERSE EQUATION OF MOTION 1. Statement of the Problem This chapter deals with a solution for Equation (3030) derived in Chapter III which is repeated here for convenience m = a- (T ) AE a~ (5.1) 6t2- as ae s R -s As for the boundary conditions consideration will be given only to the case of an imposed motion at the upper point of the guy for reasons discussed in Chapter IV. When the support point of the tower undergoes a horizontal motion A sin pt in the plane of a guy the tangential and normal components of this motion at the upper end of the guy will be, respectively |(L,t) = At sin pt = (A cos L ) sin pt (5.2a) and n(Lt) = An sin pt = (A sin yL) sin pt (5.2b) where 7L is the angle between the direction of the tangent to the guy at the upper point and a horizontal line (Figure 5.1). Moreover, only the steady state solution to the problem will be investigated in which case the coupling term in Equation (5.1) will be cos L s aAE E __ _S a R as R a in a p At sin pt (55) R 5s R sin p L as obtained from Equation (4.56). -60

61 / Figure 5.1. Inclination of a Guy. The steady state solution to the transverse ibration will therefore be the particular solution to the set of equations 2 P CosLs Tj = (T aAE L- At s in pt (5.4a) at2 as e as R sin t a r(o,t) = 0 (5.4b) n(l,t) = An sin pt (5o4c) in which Te and R are variable coefficients. 2. Variation of Te and R In Chapter II, Te was given in terms of the coordinate x (Figure 2.1). For a complete description of the problem the variation of the parameters Te and R apperaring in Equation (5.4a) must be determined in terms of the independent variable s. Replacing the hyperbolic term in Equation (2,9) from Equation (2.7), one can easily get Te = H \l + (H + sinh al) (505)

-62But from Equation (2.2) sinh a = tan o (5.6) dx=O where y is the angle of inclination at lower end of the guy (Figure 5.1). Thus one has T = H (H + (ws + tan ) (57) The curvature of the guy in static equilibrium R is found as follows: R1 = ^(5.8) From equilibrium Equation (2.1) one has o 1/2 Hy" = w(l+y' ) (5.9) Substituting for y" in Equation (5.8) from Equation (5.9) one obtains 1 w 1 = w 2 w H2 _ z _Wcos = - ( R l+yt H H T or 1 wH R -T 2 (5-10) e Using Equation (5.7) Equation (5.10) becomes 1 w 1 sa = ws 2 — (5.zl) 1 + (H + tan 7)) 3. An Approximation for the Variation of Te There exists no general method of solution of Equation (5.4a) (11) with variable coefficients. In the particular case of highly stressed cables, however, a simplification is possible which leads to the solution to the differential equation~

-63In practice a guy cable is always highly tensioned to provide effective support. For such cables the sag-span ratio is small as a result of which the change per unit length of the tensile force along the cable is approximately constant. Thus, if Teo and TeL denote, respectively, the static tensile force at s = o and s = L one may assume T = To + (TL-oT ) - (5.12) To show the degree of approximation of Equation (5.12) the exact relation between T and s, i.e., Equation (5.7) is evaluated for a guy with the following dimensions: w = 9 Ibs/ft R = 950 ft. h = 1549 ft. H = 55,500 Ibs. The computation is carried out as follows: Equation (2.6): r =.121105 Equation (2.4): a1 = 1.026450 Equation (2.2): tan 7 = sinh a =1.216452 Equation (2,8c): L = 1651.2787 ft. Now from Equation (5.7) one gets 2 Te = H 1+ tan = 55,587 lbs. and TeL = H V/l + (L + tan 7o)2 = 67,728 lbs. TL H7/ ~ H

-64Equation (5.7) was also evaluated for several values of s The result is illustrated on Figure 5.2 from which it is seen that the linear relation given by Equation (5.12) is a good approximation for the variation of the static tensile force. In particular if the end points of a highly stressed cable lie on the same horizontal line Equation (5.12) leads to a constant tension throughout the length. Therefore, in the general case of highly stressed cables the assumption of linearly varying tensile force is as good an approximation as the assumption of constant tension in horizontally stretched cables. From Equations (2.3) and (2.9) one can easily obtain the relation Te(x) = Teo + wy(x) from which TeL = Teo + wh Equation (5.12) may therefore be written as Te() = To + wh (5.12a) 4. Method of Solution Contrary to the problem of tangential vibration the particular solution to the set of Equations (5.4a), (5.4b), and (5.4c), even with the simplifying approximation given by Equation (5.12), does not appear to represent a well-known function. One must, therefore, resort to the normal mode technique as the only analytical method available.

7000 Equation (5.7) ------ Equation (5.12) -o 6000 5000 0 500 1000 1500 s in ft. Figure 5.2. Variation of Static Tensile Force.

-66For the set of Equations (5.4) to be tractable by the method of normal modes the problem must be converted such as to have homogeneous boundary conditions. Following the Mindlin-Goodman method (21) let r(s,t) = v(s,t) + b(s) An sin pt (5.13) then = v + b An sin pt 8s as 2 2 ) (5.14) l = av 2 - = -V bpAn sin pt nt2 t n which upon substitution into Equations (5.4) give _v _ v aAE a cos s d 2 s (Te s) [- A s +A (Teb') -t o s R sin P L nL a + An mbp ] sin pt (5.15) and v(Ot) = - b(o) A sin pt ( (5.16) v(L,t) = [1 - b(L)] An sin pt The simplest function by which the right hand sides of Equations (5.16) vanish is b(s) =(5.17) Substituting Equation (5.17) into Equation (5.15) one gets m - (T ) = [ A + Te + mp s] sin pt m-2 a s e s R t sin P L L e L (5.18) subjected to the boundary conditions v(o,t) = 0 (5.19) v(L,t) = 0

-67Equation (5.18) shows that the system is excited in the normal direction because of (a) the transverse boundary motion An, and (b) the tangential vibration which acts as a forcing function on the transverse vibration so that even in the particular case in which An =0 there still exists a transverse vibration. From Equations (5.18) and (5.19) it is seen that the problem has been reduced to that of the vibration of a guy fixed at both ends and subjected to an excitation as expressed by the forcing function on the right hand side of Equation (5.18). The solution to such a problem by the normal mode technique requires first a solution to the homogeneous part, i.e., the set of equations.2 c vc 8 dv m - - (T — ) = 0 t2 s e as vc(ot) = 0 (5.20) vc(Lt) = 0 for which a separation of variables in the form vc(s,t) = Y(s) Q(t) (5.21) leads to the following: d (Te d) + mT2Y = 0 (5.22a) ds e ds Y(o) = 0 (5.22b) Y(L) = 0 (5.22c) Using Equation (5.12a) for Te Equation (5.22a) becomes wh d2Y wh dY w 2 (T L s)- + —-+ w Y = 0 (5.23)

-68which can be simplified by introducing a new independent variable as follows: Let T Teo h T = -+ = - _ s (5.24) Lw Lw L then oY _ Y d _ h dY (5.25a) ds d ds L d25a) d2Y d h dY d9 h2 d2y ds2 dG = G (525b Substituting Equations (5.25a) and (5.25b) into Equation (5.23) and simplifying, one gets 2 d-Y + d+Y = 0 (5.26) dO2 dO where L3 5.) 2 g (5.27) h2 g The boundary conditions (5.22b) and (5.22c) will become =0Y(~~ 0E),~ ~(5.28) Y(G,) = O T T Teo Teo h in which 9 -- and = -- + - inwhich Lw L Lw L The solution to Equation (5.26) is found as described in the following sections.

-695. Solution of Equation(5.26) by Power Series Equation (5.26) can be solved by the method of power series (12,13) as follows: Assume that the solution to Equation (5.26) may be written in the form Y(G) = no cn Gn+v (5.29) where Cn, n = 0,1,2,... and v are as yet undetermined constants. Differentiating Equation (5.29) and substituting into Equation (5.26) gives 00 / w N n+v-2 n+v-1 n+v 9 (n+v)(n+v-1l)cn + Z (n+v)c(n + Z (n cn = 0 n=o no n=o (5.50) which reduces to 00 0n ( cn+vv- 1 n +v = n=o n n=o Rewriting the second summation as X Z cn-l, Equation n=l (5.31) may be written in the form v2CoGV + Z [(n+v)2 cn + \Cn1] nv- = 0 (53.2) 0 n=l For the power series expressed by Equation (5~32) to be identically zero for all values of 9 one must equate to zero ervery coefficient of like power of 9, giving v2 c2 = 0 (5533a) (n+v)2cn + cn_ = 0 n = 1,2,.o (5.55b)

-70Equation (5.33a) has two identical zero roots for v. Inserting v = 0 in Equation (5.33b) one obtains Cn = 2 n- n= 1, (5.54) By Equation (5.34) every coefficient cn, n = 1,2,... can be expressed in terms of c. For instance c1 = - o - 0 c - c =- ~ c 2 2 1 22 0 c C2= 3 C2'' =22 3 2 0 and so on such that, in general, cn = ()n - co n = 0,1,2,... (5.55) (nfl) Noting that O0 = 1, Equation (5.35) is also valid for n = 0 o Thus one solution to Equation (5.26) is co yI(e) ZC (1)n (%Gn) (5.56) I no(n:)2 This solution can easily be verified by substituting Equation (5.56) and its derivatives into Equation (5.26). Moreover, it can be shown by the ratio test that the series given by Equation (5.36) converges for all values of. Since differential Equation (5.26) is of second order there must exist another solution to it. If the roots to Equation (5.33a), which is called indicial equation, were two distinct values, v1 and v2,

-71each would lead to a recursion relation for determination of coefficients Cn resulting in two distinct solutions to the differential equation. The second solution to Equation (5.26), for which the indicial equation has identical roots, can be found by the method of Frobenius(3) as follows. a. Method of Frobenius Retaining the parameter v in Equation (5.33b) one gets Ic - \. c c 1 ( 0+v)2 Co 22 (2+v) (l+v) (2+v)2 and so on such that IC n (-1)n ---— [ —' - ) 2+' co n=l,2,.o. cn () [(l+v)(2+v).. (n+v)]2 O n=1 57) Introducing the following abbreviation symbols v(n) [(l+v)(2+v)...(n+v)]2 n > 1 (5.38a) and V(o)_ 1 (5.38b) Equation (5.37) may be written as c = (-1)n c0 for n= 0,1,2,,o v (5.59) Now by defining a power series in the form ) o o = on j_ Z(G,v) =Z C ( (5.40) f1= n ( n=o n no

-727 2 and evaluating the derivatives and it may be found, after aZ aZ simplifying, that 2 n v-l b + z + Z = C, (_)n (n+v)2 n n+v-1 5G2 -I G O n=o v(n)G 0 n=o v n+l n+v (5.4l) + c~ E (_l)n \ n Rewriting the second summation as _ (-1)n xn+l gn+v - n+v () =0> Z (-1> nT n-o e n=l ~-1and noting that by Equation (5o38a) n + 1 (n-)- ~ I7 it is seen that all terms cancel on the right hand side of Equation (5.41) except the term n = 0 in the first series of the equation. Thus -'2 = cov2 V- (5.42) by which it is seen that the series given by Equation (5.40) will satisfy the differential Equation (5.26) if and only if v = 0 o Therefore, the series z(ev) = (-1) GF (5.43) s a souton to the dfferental Equaton ( 6) Observng that by is a solution to the differential Equation (5,26)o Observing that by Equation (5o38a) O(n) = (n)2 it is seen that Equation (5,43) is identical to Equation (5.36) which was previously found by directly setting v = 0 in the recursion formula,

-73Equation (5o33b). It is shown in the following that 5Z(Gv) is 6v v=O the second solution to Equation (5.26). Differentiating both sides of Equation (5o42) with respect to the parameter v, one has aa l + I + %) = c~(-) By interchanging the order of derivatives on the left side and performing the derivation on the right side one gets o a2 (V) + G (v) + X v = co(2+v log G)vvS (5o45) which indicates that Z is also a solution to Equation (5o26). av v=O Equation (5.40) may be written in the form oo Z(Gv) = v + Z (-)n (5.46) in which the coefficient c which appears to be an arbitrary constant o has been taken as unityo Then, dZ = lo + nn+ log (-1) 1 + 1 + 1 -2l+ 2+v n+v n- 9n+v (5.47) By setting v = 0 in Equation (5o47)., one gets the second solution as YI(o) = [1 + (_1)n 2] log 9 II n=l (nV)2 1 1 - 2 (- )n (X1)n (5.48) The s in the bracket is Y ) as gve by Equaton ( ) so that The sum in the bracket is Yi(G) as given by Equation (5o36) so that

-7400 n Hn n Y (G) = Y (9) log 9 - 2 Z (-1) - (G. ) ( 5.49) i zn=l (nf,) where Hn = 1 + 1 + 1(550) 2 nf is the so-called harmonic serieso A verification of the solution given by Equation (5.49) as well as the convergence of its series can be easily established. bo The Complete Solution The two solutions YI(G) and YI1(G) given respectively by Equations (5.536) and (5o49) are linearly independent as seen from the presence of a constant term in the series for YI(O) o The complete solution to the differential Equation (5~26) will therefore be Y(G) = A YI(e) + B YII(e) (5o51) where A and B are constants which must be evaluated from the boundary conditions (5o28)o The two solutions YI and YII are shown now to lead, respectively, to the Bessel functions of the first and second kind of order zero. Let cp = 2 /G (5.52) then Equation (5~36) becomes Y 1 G + ( () + -(2 + /2) (/ -2 +.. 2 (2f-2 (5?) + 2 (2T)2 (3)a2

-75(14) which is the definition of Bessel function of order zero J(p) o Thus, Y1(~) = Jo(2 ") (5o53) Likewise, from Equation (5.49) II YI(G) log - 2 [- %G + 1+ 12 ( -IG)] T = log (42..() + 2[ (2~)2 j2() log V2 2[()2() + ( 4:] or y = 2 [J (C)(log 2 log ) + (P)2 (l) + LI ~~ 2 2 ~ (2.v)2 (5-54) Since J (c) is a solution to the differential equation the quantity - log \f may be replaced by any arbitrary constant c o In particular, if one chooses c as EulerVs constant(4) Ce =.5772156649 (5o55) and if one divides Equation (5054) by z the result represents the Bessel function of the second kind of order zero,(l4) or the so-called Neumann's function N (c). The second solution to differential Equation (5~26) is, therefore 2 2+C~)l~p~2 ___~i~ 2 L No(P) = i [Jo(P)(log 2 + C) + () () +... (556) The complete solution to Equation (5~26) may then be written as Y(e) = A J0 (2 Je9) + B No(2 I7) (5.57)

-766 Natural Frequencies Applying the boundary conditions (5~28) to Equation (5 27)) one obtains A Jo(2 _Go) + B No(2 J0o) = 0 (5..58) A Jo(2 diGL) + B No(2 \ TL) = 0 A non-trivial solution to Equations (5~58) exists if and only if the determinant of the coefficients is zero, ioe,, one must have Jo (2s J o) No(2 JGL) - No(2 J 0) Jo(2a dGL) = 0 (5o59) Equation (5,59) is the frequency equation of the system. Let z = 2 J/G (5o60a) and kz = 2 \N/TL (5060b) where k = GL/7Go = \TeL/Teo (5,61) is a constant greater than unityo The frequency Equation (5059) will then become JO(z) N(k) Nz) JN(kz) = 0 (5 62) (14) This equation has infinitely many real roots. For k-> 1 the asymp(14) totic expansion of i-th zero is 2 ki k2-k1 z = + + + o i = 1 2, OO (5w63) where (5.63a) 1k

-77k - k - 8k k2 25(k -1) (5.63b) 6(4k)3(k-1) k _lo70(k55-1) 53 -5(4k)5 (k-l) The natural frequencies are then obtained as follows: From Equation (5.60a) one has ( zi/2)( x, = ^ (5.64) Also from Equation (5.27) 2 X L- 1 (5.65) h2 g By eliminating %o between Equations (5064) and (5065) and replacing G0 from Equation (5024) one gets for the natural frequencies of the transverse vibration of a guy fixed at both ends h z! gw Wi h Zi (566) L 2 X Teo in which zo is given by Equation (5653)o The natural frequencies of the vibration of a string of length (23) L with constant tension Te Teo is e eo W = ^-T (5.67) 1 L w It is shown in the following that Equation (5067) may be derived from Equation (5.66): Replacing TeL in Equation (5o61) from Equation (5o12a), one has

-78k 1 +wh Teo When wh is very small compared to Teo one gets approximately k, 1 + wh 2eo eo a quantity which is slightly greater than 1 o The value Pi given by Equation (5.63a) will therefore be large so that Equation (5063) may be written approximately as Z = 1 = 1 wh 2 T eo which when substituted into Equation (5o66) results in the Equation (5.67)0 70 Mode Shapes The modes of vibration will be obtained from Equation (5~57) when X is replaced by values given by Equation (5664), Thus Yi(G) = AJ(2 o XiG) + BiN (2 1ij) (5.68) Equation (5.68) must satisfy the boundary conditions (5o28) so that one must have AiJo(2,iGio) + BiNo(2 1i9o) = 0 (5~69a) and Aio0(2 J/i ) + BNo(2 iL) = (5.69b) Since the determinant of the coefficients of Equations (5,69a) and (5o69b) is zero., ie., since one has Jo(2 fIo0) No(2N /JxL) - No(2 VG J).J(2 JxG) = o (5o70)

-79it is seen that the constants Ai and Bi are not independent of each other. Equations (5.69a) and (5.69b) will therefore give only the ratio of the coefficients Ao and Bi From Equations (5,69a) and (5.69b) one gets respectively Ai No(2 /o) BAi N(2 - (571a) and (5.7lb) Bi Jo(2 7ieL) These ratios are identical as seen by Equation (5o70). Moreover, the Ai ratio i is not ambiguous because the zeros of the functions J and ~Bi ((14) No are interlaced. Using Equation (5o71a) one may therefore take Ai = - No(2 Ji9o) (5.72a) Bi = Jo(2 i1o) (5j72b) so that one has for the mode shapes Yi(~) = -No(2 4 o9O ) Jo(2 \) + Jo(2 9I0) No(2 \i) (5753) which satisfies the boundary conditions: Yi(@o) = 0 ~~Yi~~~~(G0) =O~ ~(5074) Yi(L) = a. Orthogonality of Normal Modes Equation (5073) when substituted for Y in Equation (5.26) will give d2Yi dYi e -- + ^d- + iYi = 0~ (5075a)

-8 - which may be written as d dY. de (9 de) + %iYi = 0 (575b) Multiplying Equation (5.75b) by Y. and integrating from 9 to GL' one gets GL d dYi L Yj ( O )dO + \i f YiYjdG = 0 (5.76) Integrating by parts the first integral one obtains (OL d dYi yOL 0L dYi dYj f, ^^ (9 ^ d[GO —- -d- =f G- dG 0 (5.77) Y ( d) = Yj do dG d The first term on the right hand side of Equation (5,77) vanishes because of Equations (5~74) so that Equation (5.76) yields - f dYi d — dO + fi L Y Yo d~ = 0 (5.78) G d9 d9 o 0 Interchanging the subscripts i and j one gets 9L dY dY QL -f d d dG + % fj Y YidG = ~ 579) dO dO J 0 0 0 Subtracting Equation (5.79) from Equation (5.78) will result in QL (xi-xj) f Y~YjdO = 0 (5.80) 0 Since \i $ j. one therefore has f YiYj dG 0 for i / j (5,81) o Then from Equation (5.78) 9L dYi dYj _ - d- d dG = 0 for i j (5.82) Equations (5.81) and (5.o 82) represent the orthogonality of the transverse mode shapes

-81b. Normal Modes in Terms of the Variable s In previous sections the normal modes, ioeo the solutions t;o Equation (5.26) subjected to Equations (5.28), were found as functions of the variable 0. The solution to the original set of Equations (5.22a), (5.22b), and (5o22c) can easily be found by replacing 0 in Equation (5.73) by its expression from Equation (5.24)o The mode shapes in terms of the variable s will therefore be T T eoh Teo, *h Yi(S) = - No(22) Jo(2 X( -L s)) + J(2) No(2PV (| w s, Lw L ~ (5,83) for which the orthogonality relations (5o81) and (5.82) will become respectively L S Yi(s) Y(s) ds = 0 for i / j (5~84) o L dYi dYj Te ds ds ds = 0 for i / j (5.85) Equation (5.83) can be verified as the solution to Equation (5.22a) by a direct substitution. 8. The Original Problem of Boundary Motion The solution to the set of Equations (5.18) and (5o19) may now be obtained by a superposition of displacements corresponding t;o the normal modes, ie., by assuming a solution in the form v(s,t) _ Z Yi(s) qi(t) (5.86) Substituting Equation (5.86) into Equation (5o18) one gets g 12s oi - d (TeY )qi = f(s) sin pt (5.87) -i= 12)o -- 0 ~ 2 Lds e- 00

-82in which P cos a s L~ p~w f(s) = - R At + +- Te + s (5.88) R sin p L L L Multiplying both sides of Equation (5.87) by Yj and integrating from 0 to L L 0.1 jL d L w fLE YiYjids Z Yj Z s (TeYi)qi ds = sin pt f Yjf(s) ds (5.89) gO o jds eioii pt Interchanging the order of integration and summation one obtains L L L q'Zi f Y Y ds qi f Yj (TeYi)ds = sin pt f Yjf(s) ds (5.90) g i0 tjds ei If one integrates the second integral by parts it yields L d L L o Y i ds (TeY )ds = [TeYjYi] - o Te YY ds (5.91) The bracketed expression vanishes at s = o and s = L because of the boundary conditions, Equations(5.22b) and (5.22c). Substituting Equation (5.91) into Equation (5.90) it is readily seen that, by virtue of the orthogonality relations (5.84) and (5.85), Equation (5.90) becomes w qi of Y2 ds + qi L Te Y. ds = sin pt f Yif(s)ds (5.92) g 1 oi o e i which may be written as L,2 L J TeYi ds Yif(s)ds q. + L i = w L 2 sin pt (5.93) g of Y ds f Y ds g o i ogo' i By rewriting Equation (5.22a) for the i-th mode and multiplying it by Y. and integrating from 0 to L one gets L d 2w L 2 f Yi (TeYi)d +w - f Y. ds = 0 (5.94)

-835 Integrating the first term by parts Equation (5994) becomes L 2 2 w L2 Te Yi ds + ci g f Y. ds = 0 (5095) from which it is seen that L'2 L Te Yi ds 2 w 7L 2 i (5.96) g f Y. ds o 0 Equation (5,93) for the modal coefficients will therefore become 2 of Yif(s)ds q1 + wi qi W L 2 sin pt (5o97) w f Yi ds 0 in which w, is the natural frequency given by Equation (5,66). The steady state solution to Equation (5097) can readily be obtained as L - /o Yi(s)f(s)ds qi(t) = Of Ys)s)s sin pt (5o98) w2-2 w L 2 Wi -p g / Yji(s)ds 0 The steady state solution to the original problem - that is the particular solution for the coordinate ] governed by the set of Equations (5o4a), (5,4b) and (5.4c) - may now be found by Equation (5o13) in which v(st) and b(s) are given, respectively, by Equations (5,86) and (5ol7). 9. Normalized Modes Let L 2 ds Ni = fJ Yi() L (5.99) and Yi O~~~~~~~~~~~i = ~(5.100) 1i

-84The functions i. are then the normalized modes, with the orthonormality properties L 0s s 0 if i j f; siy L (5.101) 0 1 L 1 if i j With Equations (5,99) and (5.100), Equation (5.98) can be written as ofL Oi(s)f(s) is (t) i( —) sin pt (5.102) 22 w wi-p where ~i(t) = Niqi(t) (5.103) The modal series (5.86) can then be converted to v(s,t) 0i(s) *v(t) (5.104) i=l,2,...' The upper guys of the Oklahoma City TV tower will be used again to illustrate the formulas presented in this chapter for the natural frequencies and the modal functions. The physical constants for these guys are listed on page 9. The computation was carried out by an IBM 7094 electronic computer for the first eight modes in the following steps: The natural frequencies were computed by Equation (5.63) and the modal shapes were evaluated by Equation (5.83) at forty intervals equally s 1 2 spaced along the guy, ie., for the values - = 0 -, o and so L 40' 40' on, The normalization factors, Equation (5o99), were then determined by applying Simpson's rule. Finally, the normalized mode shapes were obtained by Equation (5,100).

-85In order to show how a linearly varying tensile force affects the natural frequencies, and how the normalized modes of a guy differ from those of a cable with constant tension computation was also made for a flat cable with a span L = ~2 + h2 (5~105) and a constant tension equal to the median tensile forces in the guy Teo+TeL T = +T (5.106) e 2 (23) For such a cable the natural frequencies and the mode shapes are respectively ijt = gTe w' (5.107) L w and Y = sin - (5.108) 1- L in which s is the coordinate along the chord of the cable. Since L 2 J~ Yo ds 2 (5.109) the normalization factor will be L 2d'T 1 N = \Jf Y L =- ( 5. 110) so that the normalized modes are in i = - = 2 sin ^ (5111) NIi L The result of the computation is shown in Table I which follows. In s this table the notation SIGMA denotes the parameter - for the guy and _-~, L s - for the flat cable (string). The normalized mode shapes are plotted on Figures 5.5 and 5.4.

-86TABLE I NORMALIZED TRANSVERSE MODES OF GUYS AND STRINGS WITH IMMOVABLE ENDS GUY PROPERTIES L= 950.000 FT H=1349.00C FT W=.0090 KIPS/FT CAPH= 35.300 KIPS LHAR=1649.94C FT CABL=1651.279 FT TO= 55.587 KIPS TL= 67.728 KIPS STRING PROPERTIES SPAN=LBAR W=.009C KIPS/FT TENSION=CONSTANT=(TO+TL)/2= 61.658 KIPS MODE NO. 1 GUY FREOUENCY=.892016E 00 GUY PERIOD=.704380E 01 STRING FREQUENCY=.893937E 00 STRING PERIOD=.702867E 01 SIGMA GUY MODE STRING MODE.000 -.0000OE 00.00COOG 00.025.119371E 00.110958E 00.050.237288E 00.221232E 00 -.75.352975E 00.330142E 00 - 10I.465681E 00.437016E 00.125.574686E 00.541196E 00,150.679310E 00.642040E 00.175.778907E 00.738925E 00,200.872880E 00.831254E 00.225.960674E GO.918458E 00,250.104178E 01 1.00000E 00.275.111576E 01.107538E 01 3CO.118219E 01.114412E 01.325.124073E 01.12'582E 01.350.129110E 01.126007E 01.375.1333C5E 01.130656E 01.400.136641E 01,134500E 01.425.139105E 01.137514E 01.450.140691E 01.139680E 01.475.141398E 01.14C985E 01.500.141229E 01.141421E 01.525.140195E 01.140985E 01.550.138311E 01.139680E 01.575.135595L 01.137514E 01.600.132073E 01.134500E 01.625.127774E 01.13C656E 01.650.122730E 01.126007E 01.675.116978E 01.12C582E 01.700.11C561E 01.114412E 01.725.103521E 01.107538E 01.750.959061E 00.100000E 01.775.877661E 00.918458E 00.q00.791532E 00.831254E OC.825.701216E 00.738925E 00.850.607274E 00.642040E 00.875.51C279E 00.541196E 00.900.410811 00.437016E 00.925.309459E 00.33C142E 00.950.206814E 00.221232E 00.975.103466E 00.110958E 00 1 GO0.111967E-05 421468E-07

-87TABLE I (CONT'D) GUY PROPERTIES L= 950.000 FT H=1349.00C FT Wu.OC90 KIPS/FT CAPHs 35.300 KIPS LBAR=1649.940 FT CABLs1651.279 FT TO 55.587 KIPS TL= 67.728 KIPS STRING PROPERTIES SPAN=LBAR W=.0090 KIPS/FT TENSIONsCONSTANT=(TO+TL)/2= 61.658 KIPS MODE NO. 2 GUY FREOUFNCY=.17842CE 01 GUY PERIOD=.352158E 01 STRING FREQUENCY=.178787E 01 STRING PERIOD=.351433E 01 SIGMA GUY MODE STRING MODE.00C -.000COOE.0OCCOOCE CO.025.237929E OC.221232E 00.050.468146E 00.437016E 00.C75.684548E 00.642040E 00,100.881500E 00.831254E OC,125.105397E 01 1.0OCO00E 00.150.119766E 01.114412E 01.175.130908E 01.126007E 01.2CC.138567E 01.134500E 01.225.142578E 01.139680E 01.250.142876E 01.141421E 01.275.139492E 01.139680E 01.300.132549E 01.134500E 01 -325.122261E 01.126007E 01.350.108924E 01.114412E 01.375.929C33E 00.100000E 01.400.746300E 00.831254E 00.425.545825E 00.642040E 00.450.332773E 00.437016E 00.475.112538E 00.221232E 00.500 -.109385E 00.421468E-07.525 -.327543E 00 -.221232E 00,550 -.536652E 00 -.437016E 00.575 -.731728E 00 -.642039E 00.6CC -.908193E 00 -.831254E 00.625 -.106199E 01 -1.0000E 00.650 -.118965E 01 -.114412E 01.675 -.128839E 01 -.126007E 01.700 -.135614E C1 -.134500E 01.725 -.139160E 01 -.139680E 01.750 -.139423E 01 -.141421E 01.775 -.136429E 01 -.139680E 01.800 -.130277E 01 -.134500E 01.825 -.121138E 01 -.126007E 01.850 -.109251E 01 -.114412E 01.875 -.949134E 00 -.100000E 01.900 -.784748E 00 -.831254E 00.925 -.603279E 00 -.642040E 00.950 -.408982E 00 -.437016E 00.975 -.206351E 00 -.221232E 00 1.000 -.919007E-06 -. 895621E-07

-88TABLE I (CONT'D) GUY PRUPERTIES L= 950.000 FT H=1349.000 FT W=.0090 KIPS/FT CAPH= 35.300 KIPS LBAR=1649.94;' FT CABL=1651.279 FT TO= 55.587 AIPS TL= 67.728 KIPS STRING PROPERTIES SPAN=LBAR W=.009C KIPS/FT TENSION=CONSTANT=(TO+TL)/2= 61.658 KI'S MODE NO. 3 GUY FREQUENCY=.267634E 01 GUY PERIOD=.234768E C1 STRING FREQUENCY=.268181E 01 STRING PERIOD=.234289E 01 SIGMA GUY MUDE STRING MODE.COO.OOCOCCE OC.COOOCOE 00.025.354868E 00.33;142E 00 050.686320E 00.642C40L OC.C75.974623E 00.918458E 00.1CC.120294E 01.114412E 01.125.135830E 01.130656E 01.15C.143223E 01.13968CE Cl.175.142123E 01.14C985E 01.?.C,. ~.132683E 01.134500E 01.225.115541E 01.120582E 01,250.917717E OC.lCOOOOE Cl.275.628184E 00.738925E 00.C 3C.3987E 00.437016E 00.325 -.359642F-01.110958t 00.350 -.372154E 00 -.221232E 00.375 -.685579E 00 -.541196E 00.400 -.958791E 00 -.831254E 00.425 -.117688E 01 -.107538E 01.45C -.13282CE Cl -.126007E 01.475 -.14C502E 01 -.137514E 01.500 -.140382E 01 -.141421E 01.525 -.132543E 01 -.137514E 01.550 -.117489E 01 -.126007E 01.575 -.961088E 00 -.107538E 01.600 -.696228E 00 -.831254E G0.625 -.395C76E 00 -.541196E 00.650 -.741469E-01 -.221232E 00.675.249224E OG.11C958E 00.700.557823E 00.437016E C0.725.835455E 00.738924E OC.750.106779E 01 1.00000E 00.775.124306E C1.12C582E 01.800.135265E 01.134500E 01.825.139148E 01.14C985E 01.850.135818E 01.13968CE 01.875.125513E 01.13C656E 01.900.108824E 01.114412F 01.925.866605C 00.918458E 00.950.601961E 00.642040E 00.975.308079E OC.330142E 00 1.000.158318E-05.131709E-06

-89TABLE I (CONT'D) GUY PROPERTIES L= 950.0(C0 FT H=1349.000 FT W=.0090 KIPS/FT CAPH= 35.30C KIPS LBAR=1649.94) FT CABL=1651.279 FT TO= 55.587 KIPS TL= 67.728 KIPS STRING PROPERTIES SPAN=LBAR W=.0090 KIPS/FT TENSION=CONSTANT=(TC+TL)/2= 61.658 KIPS MODE NO. 4 GUY FREQULNCY=.356848E 01 GUY PERIOD=.176075E 01 STRING FREQ(tJENCY=.357575E 01 STRING PERIOD=.175717E 01 SIGMA GUY MODE STRING MODE. 000 -.O GOOO 00.00CCOOE 00.025.469393E 00.437016E 00 ~.i5C.885902E 00.831254E 00.'75. 12C561E 01.114412E 01,100.139560E 01.134500E 01.125.143715E 01.141421E 01.150.132745E 01.134500E 01.175.1C7954E 01.114412t 01.200.72G641E 00.831254E 00.225.289C22E 00.437C16E 00.250 -.170147E 00.421468E-07.275 -.609559E 00 -.437016E 00.300 -.984652E 00 -.831254E 00.325 -.125805E 01 -.114412E 01.350 -.14C321E 01 -.134500E 01.375 -.140683E 01 -.141421E 01.400 -.126998E 01 -.134500E 01.425 -.100762E 01 -.114412E 01.450 -.646939E 00 -.831254E 00.475 -.224373E 00 -.437016E 00.500.218103E 00 -.895621E-07.525.637198E 00.437016E 00.55C.992505E 00.831254E 00.575.125036E 01.114412E 01.600.138694E 01.134500E 01.625.139035E 01.141421E 01.650.126152E 01.134500E 01.675.101390E 01.114412E 01.700.671917E 00.831254E 00.725.268529E 00.437016E 00.750 -.158054E 00.131709E-06.775 -.567986E 00 -.437016E 00.800 -.923527E 00 -.831254E 00.825 -.119244E 01 -.114412E 01.850 -.135090E 01 -.134500E 01.875 -.138544E 1 -.141421E Q1.900 -.129405E 01 -.134500E 01.925 -.108613E 01 -.114412E 01.950 -.781416E 00 -.831254E 00.975 -.408077E 00 -.437016E 00 1. 003 -. 14341E -05 - 179124E-0 6

-90TABLE I (CONT'D) GUY PROPERTIES L= 950.000 FT H=1349.000 FT W=.0090 KIPS/FT CAPH= 35.300 KIPS LhAR=1649.94C FT CAA3L=1651.279 FT TO= 55.587 KIPS TL= 67.728 KIPS STRIN'G PROPERTIES SPAN=LBAR W=.0090 KIPS/FT TENSION=CONSTANT=(TC+TL)/2= 61.658 KIPS MODE NO. 5 GUY FREQUENCY=.446061E 01 GUY PERIOD=.140859E 01 STRING FREQUENCY=.446968E 01 STRING PERIUD=.140573E 01 S I G,A GUY M)ODE STRI NG ODEE.3)00.00000J 00.00CO0GE 00.0"25.580723E 00.541196E 00.5O.106148E 01 1.00COOOE 00.': 75.136352E 01.13C656E 01 *.10'.143884E 01.141421E 01.125.127744E 01.13r656E 01 150.908138E 00.1CCOOOE 01.175.393128E 00.541196E 00.200 -.12833E 00,842937E-07 225 -.726454E OC -.541196E 00.250 -.115107E 01 -1.000000E 00.275 -.139026E 01 -.13C656E 01.330 -.140800E 01 -.141421E 01 325 -.12C372t 01 -.13C656E 01.350 -.811673E 03 -.10COOOE 01.375 -.294883E 00 -.541196E 00.400.265153E 00 -.173856E-06.425.781496E 00.541196E 00.450.117517E 01 1.00COOOE 00.475.138716E 01.130656E 01.500.138693E 01.141421E 01.525.117657E 01.13C656E 01.550.789797E 00.100000E 01.575.286350E 00.541196E 00. o00 -.257485E 00.300296E-06.625 -.760502E 00 -.541196E 00.650 -.114867E 01 -1.0000OCE OC.675 -.136591E 01 -.130656E 01.700 -.138199E 01 -.141421E 01.725 -.119643E 01 -.130656E 01.750 -.838019E 00 -.10000G E 01.775 -.36G149E 00 -.541196E OC.800.167357E 00 -.347711E-06.825.668510E 00.541196E 00.850.107212E 01 1.000000E 00.875.132181E 01.130656E 01.900.138370E 01.141421E 01.925.125075E 01.130656E 01.950.943316E 00.100000E C1.975.505785E 00.541197E 00 1.000.239784E-05.221271E-06

-91TABLE I (CONT'D) GUY PROPERTIES L= 950.000 FT H=1349.0CG FT W=.0090 KIPS/FT CAPH= 35.300 KIPS LBAR=1649.940 FT CARL=1651.279 FT TC= 55.587 KIPS TL= 67.728 KIPS STRING PROPERTIES SPAN=LBAR W=.0090 KIPS/FT TENSION=CONSTANT=(TG+TL)/2= 61.658 KIPS MODE NO. 6 GUY FREQUENCY=.5352'4E 01 GUY PERIOD=.117383E Cl STRING FREQUENCY=.536362E 31 STRING PERIOD=.117144E 01 SIMA GUY MODE STRING MODE.9003.00'0 O 30E 00.0OO')COOE O0.025.688104E 00.642040F 00.050.120930E 01.114412t 01 r.75.143876E 01.139680E 01.100.132803E 01.13450GE 01.125.905672E 00.10C'00E 01.150.273653E 00.437016E 00.175 -.418826E 00 -.221232E 00.200 -.101088E 01 -.831254E 00.225 -.136719E 01 -.126007E 01.250 -.140850E 01 -.141421E 01.275 -.112855E 01 -.126007E 01.300 -.594038E 0 -.831254E 00.325.718988E-01 -.221232E 00.350.718433E 00.437016E 00.375.120132E 01 1.00GC00E 00.403.141480E 01.134500E 01.425.131427E 01.139680E 01.453.924907E 00.114412E 01.475.334785E 00.642040E 00.503 -.325516E 00.131709E-06.525 -.912C75E 00 -.642C39E 00.550 -.129895E 01 -.114412E 01.575 -.140489E 01 -.139680E 01.600 -.120989E 01 -.134500E 01.625 -.758264E 00 -.10C00OE 01.650 -.148093E 00 -.437016E 00.675.490498E 00.221231E 00.703.132323E 01.831254E 00.725.133978E 01.126007E 01.750.137631E 01.141421E 01.775.112782E 01.126007E 01.800.648379E 00.831254E 00.825.375745E-01.221232E 00.850 -.577886E 00 -.437015E 00.875 -.107317E 01 -1.000000E 00.900 -.134914E 31 -.134500E 01,925 -.135215E 01 -139680E 01.950 -.108402E 01 -.114412E 01.975 -.600653E 00 -.642040E 00 1.000 -.270284E-05 -.2686866-06

-'2TABLE I (CONT'D) GUY PRUPERTIES L= 950.000 FT H=1349.000 FT W=.0090 KIPS/FT CAPH= 35.300 KIPS LtAR=1649.94D FT CABL=1651.279 FT TO= 55.587 KIPS TL= 67.728 KIPS STRING'RUPERTIES SPAN=LBAAR W=.0090 KIPS/FT TENSION=CONSTANT=(TO+TL)/2= 61.658 KIPS MODE NO. 7 GUY Fr-I:Ul (E?'CY=.624487E 01 GUY PERIUD=.100614E 31 STRING FT.E(W1PCY=.525756E 01 STRING PERIOD=.10041CE Cl SIGMA GUY MODE STRING MODE ~.^ S-.000000E O0.OOCOOOE 00.025.79G804E 00.738925t O0 ~J *O 5.132239E 01.126007E 01.C75.142678E 01.140985E 01.130.107504E 01.114412E 01.125.383566E 00.541196E 00.150 -.425673E 00 -.221232E 00,175 -.109704E 01 -.918458E 00,20, -.142192E 1l -.134500C 01.225 -.130267E 01 -.137514E 01.25r -.78G643E 00 -.COOcOOE 01.275 -.206680E-01 -.330142E 00 30CC.741955E 00.437016E 00.325.127457E 01.107538t 01.350.141778E 01.13968GE 01.375.113212E 01.130656E 01 4,)4C'.5C7585E OC.831254E 00.425 -.265792E 00.11C958E 00.450 -.956404E 00 -.642039E 00.475 -.136C51E 01 -.120582E 01.500 -.136172F 01 -.141421E 01.525 -.963478E 00 -.120582E 01.550 -.285808E 00 -.642C40E 00.575.471779E 00.110957E 00.630.108945E 01.831253E 00.625.139073E 01.13C656E 01.650.129222E 01.139680E 01.675.825687E O0.107538E 01.700.126906E 00.437017E 00.725 -.604583E 00 -.330141E 00.750 -.116278E 01 -1.0C0OOOE 00.775 -.139300E 01 -.137514E 01.800 -.123401F 01 -.134500E 01.825 -.733417E 00 -.918459E 00.850 -.324896E-01 -.221232E 00.875.674301E 00.541195E 00.900.119347E 01.114412E 01.925.138521E 01.14C985E 01.950.120037E 01.126007E 01.975.692151E 00.738925E 00 1.S000.317438E-05.479420E-06

-93TABLE I (CONT'D) GUY PROPERTIES L= 950.000 FT H=1349.000 FT W=.0090 KIPS/FT CAPH= 35.300 KIPS LBAR=1649.940 FT CAHL=1651.279 FT TO= 55.587 KIPS TL= 67.728 KIPS STRING PROPERTIES SPAN=LBAR W=.0090 KIPS/FT TENSIUN=CONSTANT=(TO+TL)/2= 61.658 KIPS MODE NO. 8 GUY FREQUENJCY=.713700E 01 GUY PERIOD=.880368E 00 STRING FREQULENCY=.715149E 01 STRING PERIOD=.878584E 00 SIGMA GUY MODE STRING MODE. O^.OOOOOOE 00.OOOOCCE 00.025.888125E 00.831254E 00.050.140064E 01.134500E 01.075.132831E 01.134500E 01.100.706955E 00.831254E 00.125 -.202212E 00.421468E-07.150 -.102414E 01 -.831254E 00.175 -.142493E 01 -.134500E 01.200 -.124637E 01 -.134500E 01.225 -.566170E 00 -.831254E 00,250.337878E 00 -.895621E-07.275.110271E 01.831254E 00.300.142593E 01.134500E 01.325.118407E 01.134500E 01 3350.477689E 00.831254E 00.375 -.412316E 00.131709E-06.400 -.113756E 01 -.831254E 00.425 -.141856E 01 -.134500E 01.450 -.115113E 01 -.134500E 01.475 -.443074E 00 -.831254E 00.500.430978E 00 -.179124E-06.525.113717E 01.831254E 00.55C.140984E 01.134500E 01.575.115030E 01.134500E 01.600.460742E 00.831254E 00.625 -.397627E 00.221271E-06.650 -.11C448E 01 -.831254E 00.675 -.139989E 01 -.134500E 01.700 -.117889E 01 -.134500E 01.725 -.527064E 00 -.831254E 00.750.314070E 00 -.268686E-06.775.103752E 01.831253E 00.800.138295E 01.134500E 01.825.122943E 01.134500E 01.850.636339E 00 831255E 00.875 -.181042E 00.479420E-06.900 -.930673E 00 -.831253E 00.925 -.134825E 01 -.134500E 01.950 -.128976E 01 -.134500E 01.975 -.779770E 00 -.831254E 00 1.000 -.571468E-05 -.358248E-06

TABLE II ORTHONORMALITY CHECK ORTHCNORMALITY CHECK NORM( 1,1...NORM(8,8) 1O.OOOO E-O1 -4.552616E-08 1.388292E-07 4.529797E-07 5.767836E-07 8.063701E-07 1.947706E-07 1.175298E-06 -4.552616E-08 10.000000E-01 4.099C61E-07 5.526468E-07 1.168313E-06 3.462657E-07 1.736109E-06 4.498909E-07 1.388292fE-7 4.099061E-07 1.OOOOOOE OC 1.338745E-06 6.382043E-07 2.265908E-06 8.758779E-07 3.275213E-06 4.52979TE-07 5.526468E-07 1.338745E-06 10.000000E-01 2.344511E-06 1.216121E-06 3.814821E-06 1.544505E-06 5.767836E-07 1.168313E-06 6.382043E-07 2.344511E-06 10.OOOOOOE-01 3.855427E-06 1.806517E-06 6.541361E-0& 8.063701E-07 3.462657E-07 2.265908E-06 1.216121E-06 3.855427E-06 1.OOOOOOE 00 6.108855E-06 1.028O05E-06 1.947706E-07 1.736109E-06 8.758779E-07 3.814821E-06 1.806517E-06 6.108855E-06 1.OOOOOOE 00 1.OZ2319E-05 1.175298E-06 4.498909E-07 3.275213E-06 1.5445C5E-06 6.541361E-06 1.028056E-06 1.022319E-05 I.OOOOOOE 00

-95 - Guy Cable -- --— String,,.1i 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1. 0 I _-1 Li__ll_1 slt Mode Ll l l^ l I I I 111 l l 3r2nd Mode. |_ ll —._ __ ___ 43 rd Mode o _ _ _, _ _ _ _ _ __________I.__4th Mode______ Figure 5.3, Comparison of Normalized Transverse Modes of a Guy and a String.

0 o o -- o. o -,- 0...... I. 00I F-b t ^^ ^ ~f -' )0 0 //VI D) ______ __ — -- — ____ I OD O 5.4 0o F-CL CL. CL (I CD CD rD CO ~^ (D0 0

-97As it is seen the corresponding natural frequencies of these two systems differ slightly - the difference being about.2 per cent in all modes. The modal shapes also show good agreement on the account that the discrepancies in the corresponding modal shapes are tolerable from an engineering point of view. In order to assure the accuracy of the computation and in particular to show the suitability of the interval L0 chosen for the Simpson's rule, the orthonormality relation Equation (5.101) was also evaluated. The result revealed a degree of accuracy of at least six decimal places as shown in Table II. 10. An Approximate Solution for the Transverse Equation of Motion In the preceding section it was shown that the transverse mode shapes of a guy are nearly the same as the corresponding mode shapes of a cable with comparable dimensions whose initial tension is taken a constant equal to the average tension in the actual guy. The natural frequencies of the two systems were also shown to be practically the same. The modal coordinates given by Equation (5.102) as well as the modal series, Equation (5.104), will, therefore, yield practically the same result for the two systems, This finding suggests that one may attack the problem of the transverse motion of a guy by treating the initail tension as a o~natant throughout. If one now assumes that the esffe~t on the respoanso~f the variation in the curvature along the guy is small, i.se, if the cu rvatur is also treated as a constant, it results in a set of equations of moti:on whose solution can readily be obtained as follows:

-98Equation (5.4) will become 2 2 t sin pt g Te a2 - = RAE a a (5.112a) g at2 e as2 R sin p L t a q(o,t) = 0 (5.112b) r(Lt) = An sin pt (5o112c) in which T and R are, respectively, some average value for the tensile force and the curvature of the stationary guy. Similar to the analysis for the longitudinal vibration one may assume a solution in the form T(s,t) = fc(St) + Qp(SSt) (5.113) in which qc is the solution to the homogeneous part and, Tp is the particular solution satisfying the actual differential equation and the boundary conditions. The solution Tc represents a free vibration and will soon decay. The solution Ti, however, will be a continuing sinusoidal motion with the frequency p and, therefore, it gives the steady state solution to the problem. Let np(st) = (s) sin pt (5.114) Substituting Equation (5 114) into Equations (5o112) yields _ w 2- aAE a p Ten + P EL -A cos s (5o115a) = sin At a T(0) = 0 (5.115b) -T(L) = An (5.115c) The solution to Equation (5oll15a) is ^(s) / = A sin s + B cos + C cos s (5 ll6) b

-99where, Te g b = (5.117) A g CAE t Cu" t C = g^ 2 t-2 — (5. 118) R (a2-b2) P sin PL a a Applying boundary conditions, Equations(5.115b) and (5.115c), one gets B C and An-C(cos P L - cos L) p sin P L Equation (5.116) will therefore become pL pL sin p p n(s) = [An-C(cos a - cos -)] p + C(cos s - cos s) sin -- ~~~b ~(5o120) in the limit when p approaches zero, ioeo, for a displacement of the top point of a guy over a long period of time, Equation (50120) yields (s) = nL oAE (s -Ls) (5.121) L 2%RT L which shows that the effect of a displacement of the top point in a normal direction is of a linear nature while a displacement in the tangential direction gives rise to a parabolic deflectiono Moreover, it is easily seen from Equation (5.121) that the deflections due to the components An and At of a horizontal displacement at the top are in opposite directions as expected,

-10011. Dynamic Force-Displacement Relation With the derivations of the previous sections one can easily find the dynamic guy modulus which is defined as the ratio of the change in the horizontal component of the restraining force at the upper point to the horizontal displacement of the same point. Referring to Figure 5o5 this ratio is T(L) cos 7L + [T (L) s sin K = (5.122) A sin pt Te(L) + T(L Te(L) / - /X/ [T (L) + T(L) )] J s=L / / Figure 5 D5. Dynamic Guy Modulus Parameters. in which the non-linear term, T(L), which is of second order if as s=L the vibration is small, has been neglected compared to Te(L) 1 6s s=L The term T(L) appearing in Equation (5.122) has been given before by Equation (4o58). The quantity a s=L can easily be obtained from the solution T for the transverse vibration previously given. If the approximate solution of the preceding section is used, one has from Equations (5.114) and (5.120)

-101= sin pt p p {[A4 - C(cos - - cos P)] b b C (P sin P s - sin ss)}sin pt; a b asin p a a b b sin t b (5:123) so that - =- {[4, - C(cos P L - cos b L)] E C( P sin P - s Ln )}si..n pt s=L tan P(5, 124) Equation (5o122) is illustrated by a numerical example for the top guys of the Oklahoma City Tower when the upper point undergoes a sinusoidal horizontal motion with an amplitude A l= LO fto and a frequency p =.1 rad/sec. For the static tensile force an average value of Te = 61.66 kips, as previously obtained. will. be usedo The computat:ion is carried out as follows: Equation (4.1a): a =\/ AEg =: 10,055 ft/sec Teg Equation (5 117): b = = 469.5 ft/sec w -2 T Equation (5.10): R = = 11,966 ft, Equation (5.2a): A = A cos 7L =.521 ft. Equation (5.2b): An = A sin YL = 853 ft. Equation (5.118): C = 266 ft. Equation (5,124): 0105 sin pt a s s=-L Equation (4,58): T(L) = 8.89 sin pt, kips Finally, Equation (5.122) gives K= (8.89)(o52l) + (67.7)5)(. 5)(.855) 1,0 4636 +.61:= 5.24 kips/ft,

----- BASED ON CONSTANT STATIC TENSILE FORCE | —-- BASED ON VARYING STATIC TENSILE FORCE 0 C 1.5|. C) t.0 0 0 R) 0 ____________________ < 0.5 F 0 1.0 2.0 3.0 4.0 ta::~ ~FREQUENCY RATIO P/~w Figure 5.6. Variation of Dynamic Guy Modulus with Excitation Frequency.

-103This value compares well with a value of AH = 5.22 kips for Al = 1,o ft. obtained in Chapter II (Figure 2.2). The dynamic guy modulus, Equation (5.122) was evaluated in a like manner by an IBM 7094 electronic computer for several values of the frequency p. The result is illustrated in Figure 5.6 from which it is seen that the dynamic guy modulus is nearly the same as the static guy modulus except for values of the frequency p in the vicinity of the natural frequencies of the guy. a. A More Exact Evaluation of the Dynamic Guy Modulus The quantity L appearing in the dynamic guy modulus s=L formula, Equation (5.122), may be determined more accurately from the modal analysis presented previously as follows: Referring to Equations (5.13), (5.17), and (5o86) one gets A in^ =,n sin pt + Z Y.i(s)qi(t) (5.125) 5s L i=12.o. L Using the well-known relations( ) Jo(x) = - J1(x) and No(x) = - N1(x) for the Bessel functions one obtains from Equation (5.83) Ys [( Teo Teo h Yi(s) = [No( i ) (2 X(L - (5.126) + Te~') TeN~~~o( w h 7(5~ 126) h -+ JN2eo)' A(2 eo + ))] Lw i Teo s)

-104where J1 and N1 are the Bessel functions of the first order of the first and second kind respectively. Substituting for qi in Equation (5.125) from Equation (5.103) and using Equation (5.100) one gets = n sin pt + Z i(5.127) L ids in which.i(t) is given by Equation (5.102) and the quantity d^i Y - (5- (5.128) ds Ni is determined from Equation (5.126) A numerical procedure was developed for carrying out the computation on a digital computer as described in the following: The forcing function f(s) appearing in Equation (5.102) is first evaluated by Equation (5.88) at forty intervals equally spaced along the guy. The radius of curvature of the guy R in Equation (5.102) is considered a variable as given by Equations (5o10) or (5 11). The modal coordinates?i(t) are then evaluated by Equation (5.102) for the first d44 eight modes of vibration using SimpsonVs ruleo The slope -- at s = L ds is determined from Equations (5.126) and (5.128) using the normalization factors Ni previously found in normalizing the mode shapeso Finally the slope at the upper point is computed by Equations (5,127) using the first eight modes of the vibration. The slope a s=L so determined is then used to evaluate the dynamic guy modulus, Equation (5.122), as shown previously. The computation by the normal mode technique for the guy modulus showedpractically the same results as the approximate method of the

-105preceding section over the range of frequencies p previously considered (see Figure 5.6). This indicates that the variation in the curvature is also of a negligible effect on the guy modulus. The approximate method presented in the preceding section gives, therefore, satisfactory results. It is particularly advantageous because of its simplicity while the normal mode technique requires a great deal of computation for each individual guy and for any particular frequency of the excitation. 12. Comparison of the Results with Kolousek-Davenport Solution It is of interest to compare the method presented in the preceding (5) sections with a solution originally developed in 1947 by Kolousek and (8) simplified later by Davenport. Kolousek's method of analysis is briefly outlined as follows: This approach deals with the transverse equation of motion for the guy cable assuming the static tensile force a constant throughout the length. The change in the tensile force during vibration is.further assumed independent of the space coordinate, i.e., it is assumed a function of time only. The differential equation of transverse motion is then reduced to a linear one by assuming small oscillations, The effect of the longitudinal motion of the guy is taken into account by equating the longitudinal displacement of the upper end with the change in length of the guy. The latter is considered as the sum of two terms: First, an elastic change in length due to the change in the tensile force using Hookers Law; and second, a geometric change in length resulting from the guy transverse displacement assuming the original guy curvature a constant, The longitudinal motion is, therefore, accounted for by its overall effect on the guy.

-106These approximations reduce the problem to an integral differential equation in terms of coordinate of the transverse motion for which a solution is obtained by expanding the dependent variable in a sine Fourier series and evaluating its coefficients. The change in the tensile force is then found by the integral equation relating the displacement at the upper point to the change in the guy length. Before presenting the final result of the analysis it is noted that the Fourier series obtained for the coordinate of motion consists of terms representing all modal shapes of a string fixed at both ends, The series for the change in the tensile force, however, consists only of odd modes because the contribution of even modes cancel when they are integrated over the length of the guy in the integral equation for the change in the tensile force. This leads, as demonstrated later, to a large magnification factor for the guy modulus at exciting frequencies only in the vicinity of odd natural frequencies of transverse vibrationo In a recent paper, Davenport(8) simplified Kolousek's approach and obtained the following expression for the guy modulus F 02 _ 1 K = k [1 F (2) 1 (5-129) in which 2G 2 -T F = it sin y __e (5130) 2 wL2k 2 Te G = - 2(5,o131) L w k an p d(5.152) and p( 2 [1 2 (5.133) 2

-107The quantity k is the taut wire modulus previously given by Equation (2.38) and, wl is the fundamental frequency for the guy cable as given by Equation (5,107). In the following the variation of the guy modulus, Equation (5.129), as a function of the frequency ratio Q will be determined by a numerical example and the results will be compared with those obtained (8) by Equation (50122). The comparison is made for a guy used by Davenport with the following dimensions: 2 A =.60 in. E = 20 x 106 lbs/in2 w = 1.9 lbs/ft L = 125 ft. 7 = 46~ Te = 12,000 lbso Substituting these into equations for the parameters E = 45.2 kips/ft F = 3580 G = 53.4 From Equation (5.129) K li^ *?.580 42 - I dyn. = 452 53 -.^ 5-4 - - In order to represent the result in a dimensionless form, the variation in the horizontal component of the static tensile force with a horizontal displacement at the upper end was evaluated by the method presented in Chapter II. This variation proved to be nearly a linear

-108one with a static modulus K 44.5 kips/ft sta. The variation of the relation Kdyn = 1.017 [1 -.380 1] Ksta. 53.4 cp()-l with the frequence ratio is given in Figure 5.7. The dynamic guy modulus is also computed by Equation (5.122) using Equation (5.124). The parameters required for such a computation are ~ =- 86.832 ft. h = 89.912 ft. H = 8336 lbs. r =.009896 cos 6L =.6897 sin YL =.7241 T L = 12,085 lbs. T =11,915 lbs. eo Te = 12,000 lbs. a =.985 a = 14,150 ft/sec b = 450.8 ft/sec R = 9093 ft. 1 1= 11.33 rad/sec The result of the computation is illustrated in Figure 5.8.

-1092.0 ~ 210 _ I..................I D3 1.5 C3 0 I) o >- 05 0 0.0 1.0 2.0 3.0 4.0 5.0 FREQUENCY RATIO P/w1 Figure 5.7. Relationship Between Dynamic Modulus and Excitation Frequency: Kolousek-Davenport Method. 2.0 0 1.5 C) o_ 1.0 0 >- 0.5 0 I r 0 Cr i1. 2.0 3.0 4.0 5.0 FREQUENCY RATIO P/w1 Figure 5.8. Relationship Between Dynamic Modulus and Excitation Frequency: Equation (5.122). Frequency. Equation (5. 122).

-110A comparison of Figures 5,7 and 5,8 show the following discrepancies: i. In Kolousek-Davenport method even modes do not affect the modulus. This is because, as mentioned earlier, the asymmetric deflections of even modes produce no net change in guy length based on the stress-strain relation used in their analysis, ii. The modulus becomes infinite in opposite directions. A study of the results of computation by the methods of present investigation indicates that the slope -(Lt) is positive at as t = ~ in the range 0 < K < 1 varying from a value of P 01 An aAE At approximately + _ for p = 0 to larger values for L 2Te R frequencies approaching wl from the lower side, This results in an increase in the modulus so that it becomes infinite in a positive sense when p approaches wl (Figures 5.7 and 5,8 ). iii, Except in the immediate vicinity of the critical transition frequencies the magnitude of the dynamic modulus is nearly a constant in Figure 5,7 while it gradually decreases in Figure 5.8. The latter variation is attributed to a decrease in the quantity T (equivalently, a decrease in the change in the tensile force) with increasing frequencies. In addition to the above mentioned descrepancies, it is noted that the transition frequencies of Kolousek-Davenport method are for values of Q such that [Gcp()-1] - 0. The first transition frequency is sensitive with respect to changes in magnitude of G such that it (9) shifts generally in the range 1.0 < Q1 < 200. The smaller the value

-111of G is the larger the shift is to the right. The higher transition frequencies, however, are always in the vicinity of the odd modes (Q v 3,5,7, etc.). No explanation is given by Davenport for the shift in the first critical transition frequency.

CHAPTER VI REPRESENTATION OF CABLES BY A DISCRETE PARAMETER MODEL In this chapter attempt is made for a solution to the problem of in-plane vibrations of cables by replacing the original system by a simplified model in which the distributed mass of cable is replaced by a series of lumped masses attached to a weightless, extensible line. The cable is, therefore, represented by a discrete mass-spring model. 1. Methods of Constructing the Discrete Parameter Model The discrete parameter system may be defined in a number of ways. One method is to divide the cable, in its static configuration, into a number of segments and draw an equilibrium funicular polygon for the gravity forces of segments and the end tensile forces of the cable (Figure 6.1). The funicular polygon thus obtained may be thought of as the equilibrium configuration of a weightless line acted upon by the weights Wi. e Tea. /@ 3 a WI 1 1/ ^ Teo Figure 6.1. Funicular Polygon. -112

-113From statics it is easy to see that the force in any line segment of this system is equal the static tensile force in the cable at the corresponding dividing point. The line segments of the system must therefore be tangent to the stationary cable configuration at dividing points. An analytical method may thus be used for constructing the discrete parameter system as a funicular polygon as described in the following: The coordinates of some selected points along the curve of the stationary cable are obtained and equations are written for the tangents to the curve at these points. This defines the shape of the funicular polygon. The mass of the cable between two successive points is then evaluated and is assigned to the intersection of corresponding tangents. Finally, the static tensile force in any line segment of the polygon is determined by evaluating the tensile force in the original system (cable) at the corresponding tangent point. As a second method one may define the discrete model by first considering a vertical weightless string of the same length as the original cable. Along the length of this string concentrated masses are attached such as to represent the mass distribution of the original system. When the extreme ends of such a string are displaced to coincide with the lower and upper ends of the cable, sufficient number of equations can be set up to solve for the coordinates of all mass points and the tensile forces in the system. Althoqugh, these equations appear to be non-linear, the problem is a statically determinate one even if the elastic deformations of the string in its final position are considered.

-114This method results in a mass-spring system in which the static tensile forces are somewhat larger than those in the original cable while the funicular polygon method yields to a system with a total length somewhat greater than the length of the cable. These discrepancies become practically negligible for a sufficiently large number of lumped masses. The formulation of the two methods described above will be given in the following sections. ao Method of Funicular Polygon This method is mathematically easy to use when the mass distribution along the cable is uniform with no concentrated masses hung on the cableo In such a case the cable along its length may be divided into n equal segments. Referring to Equations (2o7) and (208) one may find the abscissa xi of the dividing points as follows s = L =H [sinh (w xi + al) - sinh al], i = 012) oon n w H 1 (6.1) in which So is the length along the cable from the lower end to the i-th dividing point. Equation (6 1) yields sinh (w xi + a) sinh a1 + L (62) H - L n Hl Replacing the hyperbolic term on the left side by its exponential definition and simplifying, one gets w~a 1 i' L wH -t- -\I (si~a~h al 4 H iL w wT xa1iL w - / Lw\2 e = sin sinh al + ) + (1 (63) from which xa = H {-al + log [sinh al + L +A/(sinh al + iL ) + ]2 nHxila l nH)n + l](6 (6~4)

-115 Substituting Equation (6.4) into Equation (2~3) yields y. = -H - cosh al + cosh log [sinh al + + snh i, + ( )2 + 1w} i1 Vw J n H + sinh al + ) n +i (6.5) It can easily be shown that Equations (6.4) and (605). for i=o and i=n, result in the coordinates of the end points of the cableo The slope of the cable at any point can be found either by Equation (2.2) or, by the following statics relation iL w tan y = tan y + - 1 o n H in which(7) tan y = sinh a1 t n (h - L) (6y7) 0o- 2H tanh r From Equation (6,6), Equations (6o4) and (6.5) may be written as HK~~~~~~ 1 Xi = w {- al + log [tan 7y + T- } (6.8) Yi = H {- cosh al + cosh log [tan Yi + - } (6.9) w 1 cos y7 T'he quantity al appearing in the formulas given previously is defined by Equation (2.4). For its evaluation when a digital electronic computer is used Equation (204) may be converted as follows: Let sinh- (-) r (6.10) 2H sinh r 1 then sinh r = wh (6. 11) 1 2H sinh r which gives r = log wh + /( wh ) +1] (6.12) 2H sinh r V2H sinhr

-116thus a1 = rl - r (6.13) Equation (6.7) may also be used to evaluate a1 giving a1 = log [tan 7Y + cs 7 (6.14) Referring to Figure 6.2 the coordinates of the mass points of the polygon are given by Equations xi tan yi - xi-l tan 7i-l- Yi + Yi+l x. _ (6. 1) tan 7i - tan 7i-l (xi-xi-l)tan 7i tan 7i-l-yi tan 7i_+ Yi-_ tan 7i. y = (6.16) tan yi - tan 7il for i = 12,..,n. (Xi+l, Yi+i) x-i+li+l)' (Xi'Yi) - i-lX i, Y)l) Figure 6.2. Geometry of Discrete-parameter System.

-117The geometry of the discrete parameter system is thus completely defined. The static tensile force in any line of the system may be either found by Equation (2.9) using the abscissa xi or, more easily, by the relation Tei H (6.17) cos 7o Finally, the value of every mass in the system is given by M1 =.. = M = = (6,18) 1 d- n ng- b. Method of Hung Masses This method is particularly advantageous when there are concentrated weights hung on the cable. Choosing the line segments ~o, Q1''.. Rn and the lumped masses M1, M21,.0o, Mn so as to represent the mass distribution along the cable (including the uniform weight of the cable) then from statics (Figure 6.3) H tan 7i - H tan Yi = Mg i = 1, O oo n (6019) and, from geometry ~o cos y7 + ~1 cos 71 + 0o + in cos 7n = ~ (6.20) so sin 70 + ~1 sin 7 +,o + ~n sin 7n = h Equations (6,19) and (6.20) involve n + 2 equations governing H and n + 1 unknown angles, yl,.o., Yn Rewriting the equilibrium Equations (6.19) as ta i = tan Y Mi (62 tan 7i = tan 7 + E Z M. (6o2l) 1 0 H j=l1 J' /

-i8 n Figure 6.3. Parameters of Lumped-mass System. and using the relations cos 7 = ---- /1 + tan2 Yi - < - 2 (6 22) 2- i-2 tan yi sin yi l1 + tan2 Yi the geometric Equations (6.20) will become ~_ _ _o ~1 ~-.i __ _ -i. J1 + tan2 7 \/1 +(tan yo + M1) \/1 +(tan o + M )2 H n + - —? = + whch anvolve only +(tan yu + m)k 1 +tanan yo yoMj n wi+(tan Yo + n k.)

-119Since Equations (6023) are non-linear an iterative method is necessary to solve for the unknowns. Once H and tan y are found the other unknown angles in the system could be easily determined by Equations (6.21). The geometry of the system is thus completely defined. c. A Numerical Example Using the Method of Funicular Polygon The upper guys of the Oklahoma City tower will be used again to illustrate the numerical use of the formulas presented previously for constructing the discrete parameter system by the method of funicular polygon. The physical constants for each of these cables are listed on Page 9 Here the cable is divided into five equal segments giving five lumped masses as shown in Figure 6.4. MM (xi Yi) i ei fi( Figure 6.4, Representation of Five-mass System by Method of Funicular Polygon.

-120The results of the computations carried out by a digital electronic computer are listed below and in Table III. r =.121105 a = 1.026450 L = 1651.2787 ft. M =.092382 kips-sec2/ft. TABLE III PARAMETERS OF THE FIVE-LUMPED-MASS SYSTEM i 0 1 2 3 4 tan yi 1.216432 1.300633 1.3848355 1.469036 1.5535257 1.637438 xi, ft. 0 205.4725 402.7541 592.3075 774.5804 950.0 yip ft. 0O 258.5346 523.3754 793.8030 1069.1925 1349.0 xi, ft. 0 103.4411 304.7776 498.1570 689.0330 862.8478 yv r ft. 0 125.8291 387.6940 655.4925 928.5512 1206.2930 Tei kips 55 587 57.914 60.298 62.731 65.210 67.728 ir, ft. 162.8896 330.5174 330.3205 330.3195 330.3259 167.2148 The total length of the line segments of the discrete parameter system is 5 Z ~i = 1651.3877 ft. i=0 which differs by only.109 ft. from the length of the cable. Checking for consistancy, cos i = 950. 0003 i=o 1 ~? Q sin 7, = 1549.9997 = h i=o i i

-121The static equilibrium conditions in both horizontal and vertical directions also check with a high accuracy for every mass of the system. 2. Equations of In-plane Motion of Discrete-parameter System Let ui and vi denote, respectively, the horizontal and vertical motion of the i-th mass relative to its static equilibriumo The equations of motion will then be 00 Miui = Ti cos 7i - Ti1 cos 7i-l + Xi i=1,2, o,n(6.24) Miv. = T, sin 7i - T.l sin 7i-l + Y - Mig where Ti(t) = tension in segment of line between stations i and i + 1 when in motion 7 (t) = angle between horizontal and line segment i when in motion X.(t) = horizontal component of resultant applied force at station i Yi(t) = vertical component of resultant applied force at station i. Since the motion is considered relative to the stationary configuration of the system the applied forces exclude gravity. In Equations (6.24) angles 7i are functions of position only such that

-122cos Yi(t) = 7.(t) (6.25) L[i+l+ Vi+l(t)] - [Yi + vi(t)] sin yi(t) = i(t) in which 2 2 Qi(t) =- i+1+ui+l(t)-xi-ui(t)] +l [Y++vi+l(t)-y-vi(t)] (6.26) is the length of line segment i in motion. The tensile forces Ti are, however, functions of position and of the physical properties of the cable since the motion is assumed governed by the condition of extensibility of the cable. Accordingly, one may write o (t) - o To(t) =T + -i AE (6.27) 1 i A i in which the superscript o indicates initial condition, i.e., values relevant to static equilibrium. Up to this point an explicit formula has been given for the evaluation of every term in the equations of motion. Equation (6.26) is not~ however, suitable for a numerical evaluation because the components of motion Uo and Vo are very small compared to the coordinates xi and yi Consequently, any loss of accuracy in evaluating ~i by Equation (6.26) will result in a much greater inaccuracy in the value of To on account of the large quantity of the factor AE in Equation (6.27). For this reason a finite difference equation is used for evaluating Ii This equation may be found as follows:

-123Let o (6.28) 1 1 Vi = V. + 6V b where o Ui = Xi+l- Xi o (6.29) Vi = Yi+l- Yi and 5U = Ui+l- ui (6.30) 5Vi = vi- then ei= u 2- V (6031) Expanding Equation (6.31) in a Taylor series about the point (Ui, V), i.e., about the stationary position, one gets o o ~i 5i i(Ui,vi) = li(uivi) + -1 6Ui + a- 5V 1Ui=U i Ui=Ui Vi=VQ Vi=V~( (6.32) 2 ~(u2 2 )(vi ~ui — 5.) -- (Vi) + —- --- (6U.)(6V.) o1? 5 + 1 1 1 o. + higher order terms After performing the derivatives and simplifying 0 0f \02 \ ( 1u2 )___2_ i - i = cos yi(SUi) + sin yi(5Vi) + - sin2 Yi 2 7oU2 i1i 2 i 10 12 0 ('v)2 (6Ui) + Y in c ose s ( (v o) O O i 1

-124in which terms of order of three and higher have been neglected. 3. Method of Solution Equations of motion (6.24), after substituting for the variables on the right hand side by Equations (6.25), (6.27) and (6.33) are of the form - f(U, u v t) ~ ui = (ui+l, iuiu-lvi+lvivi-lt) (6.4) 00V = f(u. uu. v v nt) Mi-vi = f2(uilii+lV iV i_ v il, V i ) There are several methods for a numerical solution to such set of equations. One may, for instance, employ the finite difference method in which the following difference equivalents are used, (t) u,(t+At) - 2ui(t) + ~i(t-At) ui(t) = (At)2 (6 (6.55) iv(t) i(t+At) - 2vi(t) + vi(t-At) (At)2 The computational procedure for this method consists of first, determining the accelerations at time t from the equations of motion; and secondly, solving the difference equations for the coordinates at time t + At o This method projects the solution to time t + At using the values of the variables at two earlier times. Thus, special forms of difference equations are required for projecting the solution in the first time step. Because of this difficulty resort is made to other numerical methods. Most numerical methods for the solution to a set of equations of a type at hand are based on an approximation by several leading terms of the Taylor series expansion of the dependent variables. The Euler method, for instance, uses the following truncated forms,

-125ui(t+At) = ui(t) + At ui(t) ui(t+At) ui(t) + At ui(t) + (At) u(t) and similar expressions for the vo variables. This method, which is a second order approximation, evaluates the solution at time t + At from the values of the variables at time t. To assure accuracy for the solution to the problem of cable vibrations, which is very sensitive to small changes in the mass coordinates because of extensibility condition, a fourth order Runge-Kutta method is usedo This method considers a Taylor series expansion with leading terms up to, and including the term with the factor (At)4 The Runge-Kutta methods are suitable for a numerical solution to a set of simultaneous first order differential equations. The equations of motion must therefore be converted to first order equations. Let ( 1) (2) u = (3). = z(4) 1 1 m i Equations (6.34) may then be written as.(l) =1 (3) 2(3) z(5),(4) (4) (4) t.( ) 1 (3) (3) (3) (4) (4) (4) z() - f [z Z] = ZZ t] -- i+1 - i+V Zi i-1 (4) f [z(2)] -z() i 5 i i for i = 1,2,..., n

-126in which the motion of extreme points of the system, i.e,, values for i = 0, and i = n+l are considered known functions of time. Converting the double subscript notation Z( to a single subscript by the relations 1 Z^l) = Z(l) z(2) z(2) Z^ = Z(2) (6.58) Zj = Z[4(i-1) + j] i (4) Zn = Z(4n) Equations (6.37) may then be classified in the general form z(l) = cp[z(l),z(2),...,Z(N),t] Z(2) = c2[z(1),z(2),...,Z(N),t] (63.9) Z(N) = eN[Z(1),Z(2),.oo,Z(N),t] The number of such equations will be equal to N = 4n, where n is the number of lumped masses. These differential equations will be (15) solved by the Gill's version of the fourth-order Runge-Kutta procedure, available as a subroutine at the University of Michigan Computing Centero 4. Size of the Time Interval In order to compute the changes occurring in a time interval the conditions at the beginning of the time step are usedo The time interval

-127must therefore be sufficiently small for the solution to give reasonable accuracy. Moreover, it is necessary to insure the stability of the numerical solution. The line segments of the discrete parameter system were assumed extensible so as to give rise to changes in tensile forces due to elastic deformations, The system is therefore subject to wave transmission both in the longitudinal and the transverse directions. The velocity of propagation of stress waves along a bar is w \ / vw V (6.40) where is the mass density of the materialm For the discrete parameter system the travel time in the line segment I, is i I t. = - = / (6.41) 1 vW 1 E The smallestvalue of t. from Equation (6.41) is the critical time 1 interval. (16) The travel time given by Equation (6.41) is of the same order of magnitude as the lowest natural period of longitudinal vibration. Thus, a smaller time step would be required for a larger number of lumped masses in the system, Obviously, in order to account for the highest natural mode of vibration the time interval must be chosen smaller than the critical time interval given by Equation (6.41). A reduction factor of 25 was used to improve the accuracy as well as to assure stabilityo This gives a value of At =.0018 second for the lumped mass system examplified previously. From the foregoing discussion it is seen that if the number of masses is doubled the time interval is to be reduced by half while there

-128will be as many as twice equations. As a result the computation time would increase by a factor of four. 5. Response to a Harmonic End Disturbance A method of analysis was developed for use on a high-speed digital computer to carry out the step-by-step numerical integration of the set of simultaneous differential equations of in-plane motion when the system is subjected to a horizontal sinusoidal motion at the upper endo The lower end was assumed immovable and the motion was assumed to start from rest i,e., with zero initial conditions. The essential steps in the sequence of computation are as follows: 1. The incremental displacement of the upper point occurring in a time interval is computed. 2. The relative joint displacements are computed by Equations (6.30) and are used to evaluate the change in length of line segments by Equation (6.33). 35 The tensile forces are computed by Equation (6.27) and the geometry of the system is determined by Equations (6,25). 4. Values determined in Step 3 are used to evaluate the accelerating functions of Equations (6.24) or, in fact, the functions Pi of Equations (6.39). The Runge-Kutta subroutine is then used to yield the incremental displacements and velocities of all mass points. 5. The forces and displacements in the system at the end of the time interval are obtained by adding the changes during this time interval to the corresponding beginning values.

-129A repetition of this cycle of computations gives a progressive numerical solution to the dynamic problem. A number of solutions were carried out for varying excitation frequencies using the five-mass system illustrated previously. The amplitude of the disturbance at the upper point was taken as 1.0 ft. In Figure 6.5 a typical plot is given for the change in the horizontal component of the tensile force at the top line segment as a function of time for an excitation frequency of 3.14 cycles per second. The motion of the mid-point mass, M3, is also illustrated in Figure 6.6 for the same frequency. From Figures 6.5 and 6.6 it is seen that the motion obtained appears not to posses a periodic character. The variation of AH reflects the sensitivity of Equation (6.27) to small inaccuracies in evaluation of motion. Moreover, the magnitude of the change AH is greater than values obtained by the analytical method of Chapter V. Since the results presented in Chapter V were based on a steady-state analysis attempt was made to obtain a steady-state solution to the problem at hand by including damping in the discrete parameter system as described in the following sections. 6. Inclusion of Damping Frictional effects in guy cables present a complicated problem since resistance to the motion supplied by frictional forces are obscure, even for limited ranges of motion. The damping sources for stranded cables may be classified as: a. the resistance to motion by the surrounding air, b. the internal rubbing of the strands of the cable during motion,

20 I l l''l FIVE-MASS UNDAMPED SYSTEM 7o 1.7 2 37 (0 I: -10 -20......___________ 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Time in seconds Figure 6.5. Variation of Horizontal Resisting Force with Time for a Horizontal Sinusoidal End Motion.

3. 1. 2 3 FIVE-MASS UNDAMPED SYSTEM 2F 6.6. Motion ofMddlU3 u_ f 2. c. -L - ^^ -7 - ^- -\^/ —-^-^?1~V3.1c~~ -2 - -^ —-— ^-^ —-—.~~_-I 0., 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Time in seconds Figure 6.6. Motion of Middle Mass for a Horizontal Sinusoidal End Motion.

-132c. the internal frictional resistance associated with the molecular structure of the material of which the cable is made, d. damping in the end connections. In still air, the resistance to motion of the cable in air is negligible in comparison with damping forces supplied by other sources. The damping forces at end connections also amount to very little and, can (17) be neglected. Experiments conducted by A.T. Yu indicate that the source of internal damping of stranded cables consists essentially of the interstrand dry friction, Yu also concluded that the solid internal friction of the wire material is small so that for practical purposes, it may be assumed that only dry friction exists. Yu's experiments were limited in scope since he only considered friction of cables in bending under small axial tension. Not much is known about the nature and extent of the internal resistance caused by tensile forces acting on a straight cable, i.e., about damping in cables when vibrating in a longitudinal direction. This source of resisting force will be negligible if all strands undergo equal longitudinal strains such that there will be little rubbing of the strands if any at all. However, the tensile forces will certainly increase the dry frictional resisting forces which are induced by a change in curvature. The dry frictional resistance, or Columb damping, is a constant force whose direction must be taken so as to oppose the motion; in other words the sign depends on the direction of the relative velocity of the parts in contact. This fact complicates a treatment of the problem of damped vibration. Jacobsen(l8) has shown, however, that dry friction may be replaced, for practical purposes, by equivalent viscous

-133damping which causes the same amount of energy dissipation per cycle of vibration. The assumption is thus made here that damping is of a viscous type because mathematically it can be handled with relative ease. The inclusion of equivalent viscous damping in the discrete parameter system under investigation involves some rather troublesome problems. In a modal analysis for linear systems the usual practice is to assume a reasonable percentage of critical damping in each mode. For a forced vibration of the problem at hand, in which case a numerical solution is attempted here, one cannot be sure what arrangement of viscous dampers to assume and, what coefficients to assign in order to represent an equivalent viscous damping for the dry friction of the strands in the original system. The limited experimental information available makes the estimate of the viscous damper coefficients more difficult. Because of the shortcomings mentioned above, a simple procedure is used to represent damping in the discrete parameter system based on the following consideration: The free vibration along the x-axis of a mass M attached to a highly stretched wire (Figure 6.7) is, for very small displacements, a simple harmonic with a frequency(19) ir,\ \ >-A Figure 6.7. Single Mass Attached _^ -- -( -.-~' — Ito a Highly!t ^~~~~~ % | {Stretched Wire. J I

-134= 2 T (6.42) where T is the initial tension in the wire. Assuming viscous damping is present, the critical damping for the system of Figure 6.7 is given by cr = 2Mw = 4 \ M (6.43) The actual damping may then be represented as c = p Ccr = 4 TM (644) where p is the fraction of critical damping. If the system of Figure 6.7 is assumed to represent a simple version of the discrete parameter system between mid-points of any two successive line segments then damping coefficients may be taken as Te.W. i = 4 Xei (6045) 1 g in which the quantity Tei is the average value of initial tensile forces in line segments adjacent to the mass Mi and $ is a damping factor. If one now assumes an arrangement of dampers as shown in Figure 6.8 to account for resistance to changes in curvature then damping forces acting on the i-th mass will be ui+l1+ ___i i + i-o Fd + c[ 2 ui Ui+l Ui-1] F u1 Cull i,x 1 2 2 4 4 (6.46a) and vv + vi and;Vi+l Vi Vi i-l. d.. + cVi -o i- 1l F = cC[+ - 2~ = ci[ - 4 i 1 2 (6.46b) respectively, for the motion in horizontal and vertical directions.

-135/ /0r 00/00 W Figure 6.8. Arrangement of Dampers to Produce Relative Damping. The equations of motion for the discrete parameter system taking into account damping can be written d Miui = Ti cos Yi Til cos Yi-1 Fi x i X6 (6.47) Mivi = Ti sin i - Til sin 7i-l - Fd + Yi - Mg which are of the form Miui = fl(Ui+lUiUi-_lUi+lUiUil, Vi+lViVilvi+l9viVi-lyt) (6.48) Mivi = f2(ui+luiUi-l' Ui+l'Ui Ui-l'Vi+l' ViVi-l' i+l' i Vi-_lt) In terms of the variables defined by Equations (6i36), Equations (6.48) are written as M k i ( 3z) - zt2)

-136This set of equations can be integrated by Runge-Kutta subroutine with the same ease as the equations for undamped system. ao Effect of Relative and Absolute Damping In order to determine the reasonability of the damping effects as represented in the foregoing section and, to find a guideline for estimating the damping factor P in Equations (6.45), the response of the five-mass system previously considered was evaluated for F = 1.0 when the system is subjected to a horizontal displacement at the top point with the following time variation: a. A sinusoidal motion with an amplitude of 1.0 fto and a duration equal to one-quarter of the period of disturbance, b, a constant displacement of 1.0 ft. thereafter. The disturbing function is illustrated in Figure 6.9 for a frequency of 3.14 cycles per second for which Figure 6.10 shows the variation of the change AH in the top line segment. The motion of the lower and middle masses are also given in Figure 6.11. As a case of interest the response of the system was also evaluated for absolute damping, i.e., when Equations (6,46) are replaced by Fd = ciu( 1Ynx 1 1(6o50) Fd = ci iy the plots using the latter equations are given by dashed lines in Figures 6010 and 6.11. The results of computation show that in the case of absolute damping the motion may be characterized as a critically damped one,

I.c.0 *- t0 E 0 I - 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.90. 10.0 Time in seconds Figure 6.9. Horizontal Motion of Top Point. 10 2.'5 Tm in O,ecn —d- Relative Damping Figure 6~.10 ariation o —---- Absolute Damping Motion of Figure 6.9. z 5 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Time in seconds Figure 6.10. Variation of Horizontal Resisting Force with Time for Horizontal Disturbing End Motion of Figure 6.9.

3. l^ V3 Y Q =1A0 n --- Relative damping Absolute damping n o rol lnlol l lll -Om -2. Time in seconds Figure 6.11. Motion of Middle Mass for Horizontal Disturbing End Motion of Figure 6.9. Figure 6.1i. Motion of~ Middle Mass for Horizontal Disturbing Enid Motion of' Figure 6.9.

-139whereas for relative damping considerable fluctuations occur (see Figure 6.11). Both systems, however, converge to the same equilibrium position after a period of time. This new equilibrium position was also determined by statics as follows: In Chapter II a method was given to determine the static equilibrium of a cable for changes in Ai. Considering the static equilibrium of the cable at hand for Al = 1.0 ft. the static coordinates of lumped mass points were computed by the method of funicular polygon discussed previously in the present chapter. The dynamic solution for ui and vi, when the motion has nearly damped out, showed good agreement with the static displacements of lumped masses from the original position (Ae=O) to the new position (A~=l.0). Since damping has been introduced in the analysis only to obtain a steady state motion the absolute damped model will be used in later computations because this model appears to be more effective to bring about a steady state motion in a shorter time resulting in a saving in computer time. Moreover, the use of Equations (6.50), which are simpler than Equations (6..46), will reduce the amount of computation. From the foregoing test for D = 1.0 it is seen that the damping factor ( can be chosen of the same order of magnitude as a fraction of critical damping. b. Response of the Damped System The results of a number of tests on the five-mass system for absolute damping indicate that the response to a harmonic end motion is little influenced by changes in the excitation frequency or the value of damping factor. In Figures 6.12 and 6.13 typical plots are given for

20. FIVE-MASS DAMPED SYSTEM =0.2 i 0. C) -20. - 2 0. -------- ------------------------------------------------------------- 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Time in seconds Figure 6.12. Variation of Horizontal Resisting Force with Time for a Horizontal Sinusoidal End Motion.

2.iI FIVE-MASS DAMPED SYSTEM = 0.2 ___ - us V3 0 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 90 10.0 — 2. Time in seconds Figure 6.13. Motion of Middle Mass for a Horizontal Sinusoidal End Motion.

-142an excitation frequency of 3.14 cycles per second and a disturbance amplitude of 1.0 ft. and, for a damping factor f = 0.2. The results show that although the damped model responds in a uniform manner compared to the undamped system (see Figures 6.5 and 6.6) but the magnitude of the change in the horizontal component of the tensile force at the top line segment is still considerably greater than values obtained by the analytical method of Chapter V. To determine the cause of the discrepancy a ten-mass system was constructed by the method of funicular polygon previously outlined and tests were made on its dynamic response. The results showed little changes in tensile force from those for the five-mass system. Any refinement in modeling the cable by increasing the number of masses seems, therefore, to have little effect on the response. As a case of interest the response of a single mass model was also investigated for which the results proved of the same order of magnitude as the five and ten-mass systems. This indicates that the response of the lumped mass system is not appreciably controlled by the number of masses. The amplitude of the change AH in the three lumped mass systems considered for the cable under investigation is about twice as large as the amplitude obtained by a dynamic analysis in Chapter V and is comparable to the taut-wire modulus obtained in Chapter II. At first glance this seems to indicate that a lumped mass system for a cable yields to the properties of a taut-wire. This conclusion must be disregarded however because the curvature of the cable was taken into account in constructing the lumped mass models. The distribution of masses in the

-143lumped-mass system is shown to be responsible for the discrepancy mentioned above based on the following considerations: The previously given method of constructing a model leads in the case of the single-mass system, to a concentrated mass equal to the total distributed mass of the cable. However, in modeling other continuous media, for instance a uniform beam, the mass of the equivalent single-degree-of-freedom system is always smaller than the mass of the original system. The reduction factor is about half in the case of a (19) simply supported beam. If the single-mass model of a guy cable is constructed such as to yield the same natural frequency as the fundamental frequency of the guy then, from Equations (5,107) and (6.42), it is seen that only about 0.4 of the mass of the cable must be considered. Such a reduction in the mass of the single-mass system, which results in higher initial accelerations will reduce the level of tensile forces, hence yielding lower values for AH In the case of multi-mass models for the cable the discrepancy cannot be explained by the total mass of the model since this quantity must obviously approach the mass of the cable. However, a careful examination of previous computations revealed that the intensity of the computed dynamic tensile force is, in general, proportional to the quantity M*/~* where ~* is the length of the top line-segment and M* is the mass attached at the lower end of the same line-segmentO If a proportionate mass of the cable is assigned to the upper and lower ends of the cable, as in the case of the single-mass system, the result would be a system with almost equally spaced masseso (Notice the spacing, or.i, in Table III.) This means that for a five-mass system

-144the length of every line-segment would be about i 6518/ = 165128/6 = 275.21 ft. while each intermediate mass should be taken as M. = 5 x.092382/6 =.076986 The increase in length of the top line-segment and the decrease in the mass of lumped weights will have a two-fold effect in reducing the computed tensile forces. In general, for a system with equally spaced masses e* y L (6.51) n+l M* = X L g nl+ while for the systems considered previously in response computations 1 = - n*.v L (6.52) gM* The ratio - for the two systems indicates a discrepancy factor of 2, This factor is believed to be responsible for the discrepancies between the solution to the problem given in Chapter V and the results of computation presented in this chapter. The work in the present chapter was primarily intended for a study of the effects of concentrated hung masses on the dynamic properties of a guy cable, To insure the reasonability of the method of solution computation was made on models for a uniform guy with no concentrated masses hung on the cable for which the results of Chapter V provided a

-145 - means of comparison, The investigation revealed discrepancies which led to the conclusion that the distribution of the mass of the cable on the lumped-mass model is of a critical importance. Because of shortage in time the necessary modifications in modeling the system properly and repetition of the response computations were not completed. Although the basic objective of the study of a lumped-parameter system was not completely fulfilled it is believed that the work presented will be helpful in future investigations.

CHAPTER VII SUMMARY AND CONCLUSIONS This dissertation presents a study of the dynamic properties of guy cables directed toward the solution of oscillation problems in guyed towers. In Chapter II, formulas were developed for use in accurately calculating the static effects of a change in the cable variables. While the purpose of developing these equations was for use in determining the static force-displacement relation in guy cables, all the derivations were nevertheless given in a completely general manner so as to be applicable in the analysis of other cable structures. By making use of the results for a single guy, it was shown that a three-way or four-way guying system exhibits a rather linear behavior. In Chapter III, the equations of motion for a cable were presented and the difficulties in making suitable assumptions as to make the equations analytically tractable were discussed. A method of approximation was given which resulted in linear coupled differential equations with variable coefficients. Finally, the nature of the boundary conditions was stated, In Chapters IV and V, solutions were given for the equations of motion of a guy cable when subjected to a periodic end disturbance. The solutions involved a classical approach that is commonly used in the dynamics of continuous media. The motion of the guy was expanded in a series with time dependent coefficients with respect to the natural modes of longitudinal and transverse vibration. The solution to the spacial -146

-147equation of the transverse oscillation presented a complicated problem because of the variable coefficients involved, namely, the variation of the tensile force and the curvature along the guyo A basic assumption made in the analysis which led to a solution by the method of power series was that the static tensile force would vary linearlyo The assumption was shown to be reasonably good for cables functioning as guys. The mode shapes and corresponding natural frequencies were obtained from the solution of the free oscillation problem. Finally, the usual superposition-of-modes method was employed to determine the dynamic force-displacement relation for varying excitation frequencies. The analyses, as presented in Chapter IV and V, show the following important characteristics for guy cables, which however may not necessarily be generalized to other systems and forcing functions: 1. The lateral resistance the guys offer to tower motion is essentially due to longitudinal vibrationo 2. In any actual problem the tower movement would be slow enough, as compared with the fundamental frequency of tangential vibration, for a static analysis to be sufficient for forcedisplacement relation (see Figure 5o8). 3. The resonance frequencies will be governed by the natural frequencies of transverse vibration of guys when fixed at both ends. 4. The mode shapes and natural frequencies of transverse motion of a guy cable are practically the same as those for a comparable string with a constant tensile force so that the guys may be replaced by the simpler system of strings of the same dynamic properties,

-1485. The resistance-to-motion relation in the range of practical frequencies is nearly the same as the static relation except for values of the excitation frequency in the vicinity of the transverse natural frequencies of the guyo 60 Based on the results of static and dynamic analyses it is further concluded that a system of guys, in the free vibration of a guyed tower, exhibits a nearly linear behavior. The second phase of the work, presented in Chapter VI, consisted of a study of a lumped-mass representation for cables. Each lumped mass was assumed to have two degrees of freedom, i.e,, translations in the plane of the system. The tensile forces were assumed governed by condition of extensibility resulting in a discrete mass-spring model. A description of the method of analysis, which was developed for use on a high-speed digital computer, was given and the equations of motion were solved by a Runge-Kutta fourth order numerical procedure. The response of the model to a horizontal periodic end motion was computed by an incremental technique both for damped and undamped systemso The results of the computer analyses were presented along with the discussion and the observations derived therefrom. The investigation presented in the preceding work was made to cover the problem of guy motions in a free vibration of guyed towers, The results and the observations based upon them are strictly valid for the types of cables and the forcing functions considered in this study. The author feels that the work constitutes an addition to the published literature on cables and that some aspects of the results are very significant and contribute to a better understanding of the dynamic behavior of guyed towers, thus, fulfilling the basic objective of this dissertation,

REFERENCES lo Rohr, J, Ho,, "On the Oscillations of a Suspension Chain," Transactions, Cambridge Philosophical Society, Vol. 9, 1851. 2, Routh, E. J,, The Advanced Part of a Treatise on the Dynamics of a System of Rigid Bodies, Macmillan, 1884. 35 Pugsley, A. G,, "On the Natural Frequencies of Suspension Chains," Quarterly Journal of Mechanics and Applied Mathematics, Vol. II, 1949. 4, Saxon, D, S., and Cahn, Ao So, "Modes of Vibration of a Suspended Chain," Quarterly Journal of Mechanics and Applied Mathematics, Vol. VI, 19553 5. Kolousek, V., "Solution Statique et Dynamique des Pylones d'Antenne Haubans," Publication, International Association for Bridge and Structural Engineering, Vol. VIII, 1947. 6. Dean, D. L,, A Static and Dynamic Analysis of Tall Flexible Towers, Ph.D. Thesis, University of Michigan, Ann Arbor, Michigan, 1955. 7. Dean, Do L., "Static and Dynamic Analysis of Guy Cables," Proceedings, ASCE, No. ST 1i 1961, 8. Davenport, A.G., "The Wind-induced Vibration of Guyed and Selfsupporting Cylindrical Columns," Transactions, Engineering Institute of Canada, Vol. 3, No. 4, 1959. 9. Davenport, A. G,, "Dynamic Behavior of Massive Guy Cables," Proceedings, ASCE, No. ST2, 1965, 10. Walton, T. S., and Polachek, H., Calculation of Nonlinear Transient Motion of Cables, U.So Department of the Navy, David Taylor Model Basin, Report 1279, 1959. 11. Slater, J. Co, and Nathaniel, F. H,, Mechanics, McGraw-Hill Book Company, Inc,, New York, 1947. 12, Rainville, E. D,, Intermediate Differential Equations, Macmillan, 2nd ed., 1964o 15o Miller, K.., Engineering Mathematics, Dover Publications, Inc,, New York, 1956. 14. Abramowitz, M., Handbook of Mathematical Functions, UoSo Department of Commerce, NBS Applied Mathematics Series, No. 55, 1964. 15. Gill, S., "A Procedure for the Step-by-Step Integration of Differential Equations in an Automatic Computing Machine," Proceedings, Cambridge Philosophical Society, Vol.47, 1951, -149

itj3 @l0Cl@l9 -150160 Smith, E. A. L., "Impact and Longitudinal Wave Transmission," Transactions, ASME, 1955. 17. Yu, Ai-Tang,'Vibration Damping of Stranded Cables," Proceedings, Society of Experimental Stress Analysis, Vol. 9, No. 2, 1952. 18. Jacobsen, L. S., "Steady Forced Vibration as Influenced by Damping," Transactions, ASME, 1930. 19. Timoshenko, S., Vibration Problems in Engineering, 3rd ed,, in collaboration with D. H. Young, D. van Nostrand Company, Inc., Princeton, New Jersey, 1955. 20. Rayleigh, J. W. S., The Theory of Sound, Vol. I, Dover Publications, Inc., New York, 1945. 21, Mindlin, R. D., and Goodman, L. E., "Beam Vibrations with TimeDependent Boundary Conditions," Transactions, ASME, Vol. 72, 1950. 22. Tolstov, G. P., Fourier Series, Translated from the Russian by Richard A. Silverman, Prentice-Hall, Inc., Englewood Cliffs, New Jersey, 1962. 23. Churchill, R. V., Fourier Series and Boundary Value Problems, McGraw-Hill Book Company, Inc. New York, 1941.